Home Answer below questions in homework 1 document with 100+ words each in Apa format with no plagiarism and 1+ reference

Answer below questions in homework 1 document with 100+ words each in Apa format with no plagiarism and 1+ reference

Introduction to Data Mining

Instructor’s Solution Manual

Pang-Ning Tan
Michael Steinbach
Vipin Kumar

Contents

1 Introduction

2 Data

3 Exploring Data 1

4 Classification: Basic Concepts, Decision Trees, and Model
Evaluation 25

5 Classification: Alternative Techniques 45

6 Association Analysis: Basic Concepts and Algorithms 71

7 Association Analysis: Advanced Concepts 95

8 Cluster Analysis: Basic Concepts and Algorithms 125

9 Cluster Analysis: Additional Issues and Algorithms 14

10 Anomaly Detection 1

iii

Introduction

1. Discuss whether or not each of the following activities is a data mining
task

(a) Dividing the customers of a company according to their gender.
No. This is a simple database query.

(b) Dividing the customers of a company according to their prof

–

itability.
No. This is an accounting calculation, followed by the applica-
tion of a threshold. However, predicting the profitability of a new
customer would be data mining.

(d) Sorting a student database based on student identification num-
bers.
No. Again, this is a simple database query.

(e) Predicting the outcomes of tossing a (fair) pair of dice.
No. Since the die is fair, this is a probability calculation. If th

die were not fair, and we needed to estimate the probabilities of
each outcome from the data, then this is more like the problems
considered by data mining. However, in this specific case, solu-
tions to this problem were developed by mathematicians a long
time ago, and thus, we wouldn’t consider it to be data mining.

(f) Predicting the future stock price of a company using historical
records.
Yes. We would attempt to create a model that can predict the
continuous value of the stock price. This is an example of the

2 Chapter 1 Introduction

area of data mining known as predictive modelling. We could use
regression for this modelling, although researchers in many fields
have developed a wide variety of techniques for predicting time
series.

(g) Monitoring the heart rate of a patient for abnormalities.
Yes. We would build a model of the normal behavior of heart
rate and raise an alarm when an unusual heart behavior occurred.
This would involve the area of data mining known as anomaly de-
tection. This could also be considered as a classification problem
if we had examples of both normal and abnormal heart behavior.

(h) Monitoring seismic waves for earthquake activities.
Yes. In this case, we would build a model of different types of
seismic wave behavior associated with earthquake activities an

raise an alarm when one of these different types of seismic activit

was observed. This is an example of the area of data mining
known as classification.

(i) Extracting the frequencies of a sound wave.
No. This is signal processing.

2. Suppose that you are employed as a data mining consultant for an In-
ternet search engine company. Describe how data mining can help the
company by giving specific examples of how techniques, such as clus-
tering, classification, association rule mining, and anomaly detection
can be applied.

The following are examples of possible answers.

• Clustering can group results with a similar theme and present
them to the user in a more concise form, e.g., by reporting the
10 most frequent words in the cluster.

• Classification can assign results to pre-defined categories such as
“Sports,” “Politics,” etc.

• Sequential association analysis can detect that that certain queries
follow certain other queries with a high probability, allowing for
more efficient caching.

• Anomaly detection techniques can discover unusual patterns of
user traffic, e.g., that one subject has suddenly become much
more popular. Advertising strategies could be adjusted to take
advantage of such developments.

3. For each of the following data sets, explain whether or not data privacy
is an important issue.

(a) Census data collected from 1900–1950. No

(b) IP addresses and visit times of Web users who visit your Website.
Yes

(d) Names and addresses of people from the telephone book. No

(e) Names and email addresses collected from the Web. No

Dat

1. In the initial example of Chapter 2, the statistician says, “Yes, fields 2 and
3 are basically the same.” Can you tell from the three lines of sample data
that are shown why she says that?

Field 2
Field 3 ≈ 7 for the values displayed. While it can be dangerous to draw con-
clusions from such a small sample, the two fields seem to contain essentially
the same information.

2. Classify the following attributes as binary, discrete, or continuous. Also
classify them as qualitative (nominal or ordinal) or quantitative (interval or
ratio). Some cases may have more than one interpretation, so briefly indicate
your reasoning if you think there may be some ambiguity.

Example: Age in years.

Answer:

Discrete, quantitative, ratio

(a) Time in terms of AM or PM. Binary, qualitative, ordinal

(b) Brightness as measured by a light meter. Continuous, quantitative,
ratio

(d) Angles as measured in degrees between 0◦ and 360◦. Continuous, quan-
titative, ratio

(e) Bronze, Silver, and Gold medals as awarded at the Olympics. Discrete,
qualitative, ordinal

(f) Height above sea level. Continuous, quantitative, interval/ratio (de-
pends on whether sea level is regarded as an arbitrary origin

)

(g) Number of patients in a hospital. Discrete, quantitative, ratio

(h) ISBN numbers for books. (Look up the format on the Web.) Discrete,
qualitative, nominal (ISBN numbers do have order information, though)

6 Chapter 2 Data

(i) Ability to pass light in terms of the following values: opaque, translu-
cent, transparent. Discrete, qualitative, ordinal

(j) Military rank. Discrete, qualitative, ordinal

(k) Distance from the center of campus. Continuous, quantitative, inter-
val/ratio (depends)

(l) Density of a substance in grams per cubic centimeter. Discrete, quan-
titative, ratio

(m) Coat check number. (When you attend an event, you can often give
your coat to someone who, in turn, gives you a number that you can
use to claim your coat when you leave.) Discrete, qualitative, nominal

3. You are approached by the marketing director of a local company, who be-
lieves that he has devised a foolproof way to measure customer satisfaction.
He explains his scheme as follows: “It’s so simple that I can’t believe that
no one has thought of it before. I just keep track of the number of customer
complaints for each product. I read in a data mining book that counts a

ratio attributes, and so, my measure of product satisfaction must be a ratio
attribute. But when I rated the products based on my new customer satisfac-
tion measure and showed them to my boss, he told me that I had overlooked
the obvious, and that my measure was worthless. I think that he was just
mad because our best-selling product had the worst satisfaction since it had
the most complaints. Could you help me set him straight?”

(a) Who is right, the marketing director or his boss? If you answered, his
boss, what would you do to fix the measure of satisfaction?

The boss is right. A better measure is given by

Satisfaction(product)

number of complaints for the product
total number of sales for the product

(b) What can you say about the attribute type of the original product
satisfaction attribute?

Nothing can be said about the attribute type of the original measure.
For example, two products that have the same level of customer satis-
faction may have different numbers of complaints and vice-versa.

4. A few months later, you are again approached by the same marketing director
as in Exercise 3. This time, he has devised a better approach to measure the
extent to which a customer prefers one product over other, similar products.
He explains, “When we develop new products, we typically create several
variations and evaluate which one customers prefer. Our standard procedure
is to give our test subjects all of the product variations at one time and then

ask them to rank the product variations in order of preference. However, our
test subjects are very indecisive, especially when there are more than two
products. As a result, testing takes forever. I suggested that we perform
the comparisons in pairs and then use these comparisons to get the rankings.
Thus, if we have three product variations, we have the customers compare
variations 1 and 2, then 2 and 3, and finally 3 and 1. Our testing time with
my new procedure is a third of what it was for the old procedure, but the
employees conducting the tests complain that they cannot come up with a
consistent ranking from the results. And my boss wants the latest product
evaluations, yesterday. I should also mention that he was the person who
came up with the old product evaluation approach. Can you help me?”

(a) Is the marketing director in trouble? Will his approach work for gener-
ating an ordinal ranking of the product variations in terms of customer
preference? Explain.

Yes, the marketing director is in trouble. A customer may give incon-
sistent rankings. For example, a customer may prefer 1 to 2, 2 to 3,
but 3 to 1.

(b) Is there a way to fix the marketing director’s approach? More generally,
what can you say about trying to create an ordinal measurement scale
based on pairwise comparisons?

One solution: For three items, do only the first two comparisons.

more general solution: Put the choice to the customer as one of order-
ing the product, but still only allow pairwise comparisons. In general,
creating an ordinal measurement scale based on pairwise comparison is
difficult because of possible inconsistencies.

(c) For the original product evaluation scheme, the overall rankings of each
product variation are found by computing its average over all test sub-
jects. Comment on whether you think that this is a reasonable ap-
proach. What other approaches might you take?

First, there is the issue that the scale is likely not an interval or ratio
scale. Nonetheless, for practical purposes, an average may be good
enough. A more important concern is that a few extreme ratings might
result in an overall rating that is misleading. Thus, the median or a
trimmed mean (see Chapter 3) might be a better choice.

5. Can you think of a situation in which identification numbers would be useful
for prediction?

One example: Student IDs are a good predictor of graduation date.

6. An educational psychologist wants to use association analysis to analyze test
results. The test consists of 100 questions with four possible answers each.

8 Chapter 2 Data

(a) How would you convert this data into a form suitable for association
analysis?

Association rule analysis works with binary attributes, so you have to
convert original data into binary form as follows:

Q1 = A Q1 = B Q1 = C Q1 = D … Q100 = A Q100 = B Q100 = C Q100 = D
1 0 0 0 … 1 0 0

0 0 1 0 … 0 1 0 0

(b) In particular, what type of attributes would you have and how
many of them are there?

400 asymmetric binary attributes.

7. Which of the following quantities is likely to show more temporal autocorre-
lation: daily rainfall or daily temperature? Why?

A feature shows spatial auto-correlation if locations that are closer to each
other are more similar with respect to the values of that feature than loca-
tions that are farther away. It is more common for physically close locations
to have similar temperatures than similar amounts of rainfall since rainfall
can be very localized;, i.e., the amount of rainfall can change abruptly from
one location to another. Therefore, daily temperature shows more spatial
autocorrelation then daily rainfall.

8. Discuss why a document-term matrix is an example of a data set that has
asymmetric discrete or asymmetric continuous features.

The ijth entry of a document-term matrix is the number of times that term
j occurs in document i. Most documents contain only a small fraction of
all the possible terms, and thus, zero entries are not very meaningful, either
in describing or comparing documents. Thus, a document-term matrix has
asymmetric discrete features. If we apply a TFIDF normalization to terms
and normalize the documents to have an L2 norm of 1, then this creates a
term-document matrix with continuous features. However, the features are
still asymmetric because these transformations do not create non-zero entries
for any entries that were previously 0, and thus, zero entries are still not very
meaningful.

9. Many sciences rely on observation instead of (or in addition to) designed ex-
periments. Compare the data quality issues involved in observational science
with those of experimental science and data mining.

Observational sciences have the issue of not being able to completely control
the quality of the data that they obtain. For example, until Earth orbit-

ing satellites became available, measurements of sea surface temperature re-
lied on measurements from ships. Likewise, weather measurements are often
taken from stations located in towns or cities. Thus, it is necessary to wor

with the data available, rather than data from a carefully designed experi-
ment. In that sense, data analysis for observational science resembles data
mining.

10. Discuss the difference between the precision of a measurement and the terms
single and double precision, as they are used in computer science, typically
to represent floating-point numbers that require 32 and 64 bits, respectively.

The precision of floating point numbers is a maximum precision. More ex-
plicity, precision is often expressed in terms of the number of significant digits
used to represent a value. Thus, a single precision number can only represent
values with up to 32 bits, ≈ 9 decimal digits of precision. However, often the
precision of a value represented using 32 bits (64 bits) is far less than 32 bits
(64 bits).

11. Give at least two advantages to working with data stored in text files instead
of in a binary format.

(1) Text files can be easily inspected by typing the file or viewing it with a
text editor.
(2) Text files are more portable than binary files, both across systems and
programs.
(3) Text files can be more easily modified, for example, using a text editor
or perl.

12. Distinguish between noise and outliers. Be sure to consider the following
questions.

(a) Is noise ever interesting or desirable? Outliers?
No, by definition. Yes. (See Chapter 10.)

(b) Can noise objects be outliers?
Yes. Random distortion of the data is often responsible for outliers.

(d) Are outliers always noise objects?
No. Often outliers merely represent a class of objects that are different
from normal objects.

(e) Can noise make a typical value into an unusual one, or vice versa?
Yes.

10 Chapter 2 Data

13. Consider the problem of finding the K nearest neighbors of a data object. A
programmer designs Algorithm 2.1 for this task.

Algorithm 2.1 Algorithm for finding K nearest neighbors.
1: for i = 1 to number of data objects do
2: Find the distances of the ith object to all other objects.
3: Sort these distances in decreasing order.

(Keep track of which object is associated with each distance.)
4: return the objects associated with the first K distances of the sorted list
5: end for

(a) Describe the potential problems with this algorithm if there are dupli-
cate objects in the data set. Assume the distance function will only
return a distance of 0 for objects that are the same.

There are several problems. First, the order of duplicate objects on a
nearest neighbor list will depend on details of the algorithm and the
order of objects in the data set. Second, if there are enough duplicates,
the nearest neighbor list may consist only of duplicates. Third, an
object may not be its own nearest neighbor.

(b) How would you fix this problem?

There are various approaches depending on the situation. One approach
is to to keep only one object for each group of duplicate objects. In
this case, each neighbor can represent either a single object or a group
of duplicate objects.

14. The following attributes are measured for members of a herd of Asian ele-
phants: weight, height, tusk length, trunk length, and ear area. Based on
these measurements, what sort of similarity measure from Section 2.4 would
you use to compare or group these elephants? Justify your answer and ex-
plain any special circumstances.

These attributes are all numerical, but can have widely varying ranges of
values, depending on the scale used to measure them. Furthermore, the
attributes are not asymmetric and the magnitude of an attribute matters.
These latter two facts eliminate the cosine and correlation measure. Eu-
clidean distance, applied after standardizing the attributes to have a mean
of 0 and a standard deviation of 1, would be appropriate.

15. You are given a set of m objects that is divided into K groups, where the ith

group is of size mi. If the goal is to obtain a sample of size n < m, what is the difference between the following two sampling schemes? (Assume sampling with replacement.)

(a) We randomly select n ∗ mi/m elements from each group.
(b) We randomly select n elements from the data set, without regard for

the group to which an object belongs.

The first scheme is guaranteed to get the same number of objects from each
group, while for the second scheme, the number of objects from each group
will vary. More specifically, the second scheme only guarantes that, on aver-
age, the number of objects from each group will be n ∗ mi/m.

16. Consider a document-term matrix, where tfij is the frequency of the ith word
(term) in the jth document and m is the number of documents. Consider
the variable transformation that is defined by

tf ′ij = tfij ∗ log
m

, (2.1)

where dfi is the number of documents in which the ith term appears and is
known as the document frequency of the term. This transformation is
known as the inverse document frequency transformation.

(a) What is the effect of this transformation if a term occurs in one docu-
ment? In every document?

Terms that occur in every document have 0 weight, while those that
occur in one document have maximum weight, i.e., log m.

(b) What might be the purpose of this transformation?

This normalization reflects the observation that terms that occur in
every document do not have any power to distinguish one document
from another, while those that are relatively rare do.

17. Assume that we apply a square root transformation to a ratio attribute

to obtain the new attribute x∗. As part of your analysis, you identify an
interval (a, b) in which x∗ has a linear relationship to another attribute y.

(a) What is the corresponding interval (a, b) in terms of x? (a2, b2)

(b) Give an equation that relates y to x. In this interval, y = x2.

18. This exercise compares and contrasts some similarity and distance measures.

(a) For binary data, the L1 distance corresponds to the Hamming distance;
that is, the number of bits that are different between two binary vec-
tors. The Jaccard similarity is a measure of the similarity between two
binary vectors. Compute the Hamming distance and the Jaccard simi-
larity between the following two binary vectors.

12 Chapter 2 Data

x = 0101010001
y = 0100011000

Hamming distance = number of different bits = 3
Jaccard Similarity = number of 1-1 matches /( number of bits – number
0-0 matches) = 2 / 5 =

(b) Which approach, Jaccard or Hamming distance, is more similar to the
Simple Matching Coefficient, and which approach is more similar to the
cosine measure? Explain. (Note: The Hamming measure is a distance,
while the other three measures are similarities, but don’t let this confuse
you.)

The Hamming distance is similar to the SMC. In fact, SMC = Hamming
distance / number of bits.
The Jaccard measure is similar to the cosine measure because both
ignore 0-0 matches.

(c) Suppose that you are comparing how similar two organisms of different
species are in terms of the number of genes they share. Describe which
measure, Hamming or Jaccard, you think would be more appropriate
for comparing the genetic makeup of two organisms. Explain. (Assume
that each animal is represented as a binary vector, where each attribute
is 1 if a particular gene is present in the organism and 0 otherwise.)

Jaccard is more appropriate for comparing the genetic makeup of two
organisms; since we want to see how many genes these two organisms
share.

(d) If you wanted to compare the genetic makeup of two organisms of the
same species, e.g., two human beings, would you use the Hamming
distance, the Jaccard coefficient, or a different measure of similarity or
distance? Explain. (Note that two human beings share > 99.9% of the
same genes.)

Two human beings share >99.9% of the same genes. If we want to
compare the genetic makeup of two human beings, we should focus on
their differences. Thus, the Hamming distance is more appropriate in
this situation.

19. For the following vectors, x and y, calculate the indicated similarity or dis-
tance measures.

(a) x = (1, 1, 1, 1), y = (2, 2, 2, 2) cosine, correlation, Euclidean

cos(x,y) = 1, corr(x,y) = 0/0 ( ), Euclidean(x,y) = 2

(b) x = (0, 1, 0, 1), y = (1, 0, 1, 0) cosine, correlation, Euclidean, Jaccard

cos(x,y) = 0, corr(x,y) = −1, Euclidean(x,y) = 2, Jaccard(x,y) = 0

(d) x = (1, 1, 0, 1, 0, 1), y = (1, 1, 1, 0, 0, 1) cosine, correlation, Jaccard

cos(x,y) = 0.75, corr(x,y) = 0.25, Jaccard(x,y) =

(e) x = (2,−1, 0, 2, 0,−3), y = (−1, 1,−1, 0, 0,−1) cosine, correlation
cos(x,y) = 0, corr(x,y)

= 0

20. Here, we further explore the cosine and correlation measures.

(a) What is the range of values that are possible for the cosine measure?

[−1, 1]. Many times the data has only positive entries and in that case
the range is [0, 1].

(b) If two objects have a cosine measure of 1, are they identical? Explain.

Not necessarily. All we know is that the values of their attributes differ
by a constant factor.

(c) What is the relationship of the cosine measure to correlation, if any?
(Hint: Look at statistical measures such as mean and standard devia-
tion in cases where cosine and correlation are the same and different.)

For two vectors, x and y that have a mean of 0, corr(x, y) = cos(x, y).

(d) Figure 2.1(a) shows the relationship of the cosine measure to Euclidean
distance for 100,000 randomly generated points that have been normal-
ized to have an L2 length of 1. What general observation can you make
about the relationship between Euclidean distance and cosine similarity
when vectors have an L2 norm of 1?

Since all the 100,000 points fall on the curve, there is a functional rela-
tionship between Euclidean distance and cosine similarity for normal-
ized data. More specifically, there is an inverse relationship between
cosine similarity and Euclidean distance. For example, if two data
points are identical, their cosine similarity is one and their Euclidean
distance is zero, but if two data points have a high Euclidean distance,
their cosine value is close to zero. Note that all the sample data points
were from the positive quadrant, i.e., had only positive values. This
means that all cosine (and correlation) values will be positive.

(e) Figure 2.1(b) shows the relationship of correlation to Euclidean distance
for 100,000 randomly generated points that have been standardized
to have a mean of 0 and a standard deviation of 1. What general
observation can you make about the relationship between Euclidean
distance and correlation when the vectors have been standardized to
have a mean of 0 and a standard deviation of 1?

14 Chapter 2 Data

Same as previous answer, but with correlation substituted for cosine.

(f) Derive the mathematical relationship between cosine similarity and Eu-
clidean distance when each data object has an L2 length of 1.

Let x and y be two vectors where each vector has an L2 length of 1.
For such vectors, the variance is just n times the sum of its squared
attribute values and the correlation between the two vectors is their
dot product divided by n.

d(x,y) =

√√√√ n∑

k=1

(xk − yk)2

=
√√√√ n∑
k=1

x2k − 2xkyk + y2k

=
√

1 − 2cos(x,y) + 1
=

√
2(1 − cos(x,y))

(g) Derive the mathematical relationship between correlation and Euclidean
distance when each data point has been been standardized by subtract-
ing its mean and dividing by its standard deviation.

Let x and y be two vectors where each vector has an a mean of 0
and a standard deviation of 1. For such vectors, the variance (standard
deviation squared) is just n times the sum of its squared attribute values
and the correlation between the two vectors is their dot product divided
by n.

d(x,y) =
√√√√ n∑
k=1
(xk − yk)2
=
√√√√ n∑
k=1
x2k − 2xkyk + y2k
=
√

n − 2ncorr(x,y) + n
=

√
2n(1 − corr(x,y))

21. Show that the set difference metric given by

d(A,B) = size(A − B) + size(B − A)

satisfies the metric axioms given on page 70. A and B are sets and A−B is
the set difference.

0 0.2 0.4 0.6 0.8 1
Cosine Similarity

1.4

1.2

0.6

0.4

0.2

E
u

lid
ea

is
ta

nc
e

(a) Relationship between Euclidean
distance and the cosine measure.

0 0.2 0.4 0.6 0.8 1
Correlation

1.4
1.2
1

0.8

0.6
0.4
0.2
0
E
uc
lid
ea
n
D
is
ta
nc
e

(b) Relationship between Euclidean
distance and correlation.

Figure 2.1. Figures for exercise 20.

1(a). Because the size of a set is greater than or equal to 0, d(x,y) ≥ 0.
1(b). if A = B, then A − B = B − A = empty set and thus d(x,y) = 0
2. d(A,B) = size(A−B)+size(B−A) = size(B−A)+size(A−B) = d(B,A)
3. First, note that d(A,B) = size(A) + size(B) − 2size(A ∩ B).
∴ d(A,B)+d(B,C) = size(A)+size(C)+2size(B)−2size(A∩B)−2size(B∩
C)
Since size(A ∩ B) ≤ size(B) and size(B ∩ C) ≤ size(B),
d(A,B) + d(B,C) ≥ size(A) + size(C) + 2size(B) − 2size(B) = size(A)

size(C) ≥ size(A) + size(C) − 2size(A ∩ C) = d(A,C)
∴ d(A,C) ≤ d(A,B) + d(B,C)

22. Discuss how you might map correlation values from the interval [−1,1] to the
interval [0,1]. Note that the type of transformation that you use might depend
on the application that you have in mind. Thus, consider two applications:
clustering time series and predicting the behavior of one time series given
another.

For time series clustering, time series with relatively high positive correlation
should be put together. For this purpose, the following transformation would
be appropriate:

sim =
{

corr if corr ≥ 0
0 if corr < 0

For predicting the behavior of one time series from another, it is necessary to
consider strong negative, as well as strong positive, correlation. In this case,
the following transformation, sim = |corr| might be appropriate. Note that
this assumes that you only want to predict magnitude, not direction.

16 Chapter 2 Data

23. Given a similarity measure with values in the interval [0,1] describe two ways
to transform this similarity value into a dissimilarity value in the interval
[0,∞].

d = 1−ss and d = − log s.
24. Proximity is typically defined between a pair of objects.

(a) Define two ways in which you might define the proximity among a group
of objects.

Two examples are the following: (i) based on pairwise proximity, i.e.,
minimum pairwise similarity or maximum pairwise dissimilarity, or (ii)
for points in Euclidean space compute a centroid (the mean of all the
points—see Section 8.2) and then compute the sum or average of the
distances of the points to the centroid.

(b) How might you define the distance between two sets of points in Eu-
clidean space?

One approach is to compute the distance between the centroids of the
two sets of points.

(c) How might you define the proximity between two sets of data objects?
(Make no assumption about the data objects, except that a proximity
measure is defined between any pair of objects.)

One approach is to compute the average pairwise proximity of objects
in one group of objects with those objects in the other group. Other
approaches are to take the minimum or maximum proximity.

Note that the cohesion of a cluster is related to the notion of the proximity
of a group of objects among themselves and that the separation of clusters
is related to concept of the proximity of two groups of objects. (See Section
8.4.) Furthermore, the proximity of two clusters is an important concept in
agglomerative hierarchical clustering. (See Section 8.2.)

25. You are given a set of points S in Euclidean space, as well as the distance of
each point in S to a point x. (It does not matter if x ∈ S.)
(a) If the goal is to find all points within a specified distance ε of point y,

y = x, explain how you could use the triangle inequality and the al-
ready calculated distances to x to potentially reduce the number of dis-
tance calculations necessary? Hint: The triangle inequality, d(x, z) ≤
d(x,y) + d(y,x), can be rewritten as d(x,y) ≥ d(x, z) − d(y, z).
Unfortunately, there is a typo and a lack of clarity in the hint. The
hint should be phrased as follows:

Hint: If z is an arbitrary point of S, then the triangle inequality,
d(x,y) ≤ d(x, z)+d(y, z), can be rewritten as d(y, z) ≥ d(x,y)−d(x, z).

Another application of the triangle inequality starting with d(x, z) ≤
d(x,y) + d(y, z), shows that d(y, z) ≥ d(x, z) − d(x,y). If the lower
bound of d(y, z) obtained from either of these inequalities is greater
than �, then d(y, z) does not need to be calculated. Also, if the upper
bound of d(y, z) obtained from the inequality d(y, z) ≤ d(y,x)+d(x, z)
is less than or equal to �, then d(x, z) does not need to be calculated.

(b) In general, how would the distance between x and y affect the number
of distance calculations?

If x = y then no calculations are necessary. As x becomes farther away,
typically more distance calculations are needed.

(c) Suppose that you can find a small subset of points S′, from the original
data set, such that every point in the data set is within a specified
distance ε of at least one of the points in S′, and that you also have
the pairwise distance matrix for S′. Describe a technique that uses this
information to compute, with a minimum of distance calculations, the
set of all points within a distance of β of a specified point from the data
set.

Let x and y be the two points and let x∗ and y∗ be the points in S′

that are closest to the two points, respectively. If d(x∗,y∗) + 2� ≤ β,
then we can safely conclude d(x,y) ≤ β. Likewise, if d(x∗,y∗)−2� ≥ β,
then we can safely conclude d(x,y) ≥ β. These formulas are derived
by considering the cases where x and y are as far from x∗ and y∗ as
possible and as far or close to each other as possible.

26. Show that 1 minus the Jaccard similarity is a distance measure between two
data objects, x and y, that satisfies the metric axioms given on page 70.
Specifically, d(x,y) = 1 − J(x,y).

1(a). Because J(x,y) ≤ 1, d(x,y) ≥ 0.
1(b). Because J(x,x) = 1, d(x,x) = 0
2. Because J(x,y) = J(y,x), d(x,y) = d(y,x)
3. (Proof due to Jeffrey Ullman)
minhash(x) is the index of first nonzero entry of x
prob(minhash(x) = k) is the probability tha minhash(x) = k when x is ran-
domly permuted.
Note that prob(minhash(x) = minhash(y)) = J(x,y) (minhash lemma)
Therefore, d(x,y) = 1−prob(minhash(x) = minhash(y)) = prob(minhash(x) =
minhash(y))
We have to show that,
prob(minhash(x) = minhash(z)) ≤ prob(minhash(x) = minhash(y)) +
prob(minhash(y) = minhash(z)

18 Chapter 2 Data

However, note that whenever minhash(x) = minhash(z), then at least one of
minhash(x) = minhash(y) and minhash(y) = minhash(z) must be true.

27. Show that the distance measure defined as the angle between two data vec-
tors, x and y, satisfies the metric axioms given on page 70. Specifically,
d(x,y) = arccos(cos(x,y)).

Note that angles are in the range 0 to 180◦.
1(a). Because 0 ≤ cos(x,y) ≤ 1, d(x,y) ≥ 0.
1(b). Because cos(x,x) = 1, d(x,x) = arccos(1) = 0
2. Because cos(x,y) = cos(y,x), d(x,y) = d(y,x)
3. If the three vectors lie in a plane then it is obvious that the angle between
x and z must be less than or equal to the sum of the angles between x and
y and y and z. If y′ is the projection of y into the plane defined by x and
z, then note that the angles between x and y and y and z are greater than
those between x and y′ and y′ and z.

28. Explain why computing the proximity between two attributes is often simpler
than computing the similarity between two objects.

In general, an object can be a record whose fields (attributes) are of different
types. To compute the overall similarity of two objects in this case, we need
to decide how to compute the similarity for each attribute and then combine
these similarities. This can be done straightforwardly by using Equations 2.15
or 2.16, but is still somewhat ad hoc, at least compared to proximity measures
such as the Euclidean distance or correlation, which are mathematically well-
founded. In contrast, the values of an attribute are all of the same type,
and thus, if another attribute is of the same type, then the computation of
similarity is conceptually and computationally straightforward.

Exploring Data

1. Obtain one of the data sets available at the UCI Machine Learning Repository
and apply as many of the different visualization techniques described in the
chapter as possible. The bibliographic notes and book Web site provide
pointers to visualization software.

MATLAB and R have excellent facilities for visualization. Most of the fig-
ures in this chapter were created using MATLAB. R is freely available from
http://www.r-project.org/.

2. Identify at least two advantages and two disadvantages of using color to
visually represent information.

Advantages: Color makes it much easier to visually distinguish visual el-
ements from one another. For example, three clusters of two-dimensional
points are more readily distinguished if the markers representing the points
have different colors, rather than only different shapes. Also, figures with
color are more interesting to look at.

Disadvantages: Some people are color blind and may not be able to properly
interpret a color figure. Grayscale figures can show more detail in some cases.
Color can be hard to use properly. For example, a poor color scheme can be
garish or can focus attention on unimportant elements.

3. What are the arrangement issues that arise with respect to three-dimensional
plots?

It would have been better to state this more generally as “What are the issues
. . . ,” since selection, as well as arrangement plays a key issue in displaying a
three-dimensional plot.

The key issue for three dimensional plots is how to display information so
that as little information is obscured as possible. If the plot is of a two-
dimensional surface, then the choice of a viewpoint is critical. However, if the
plot is in electronic form, then it is sometimes possible to interactively change

20 Chapter 3 Exploring Data

the viewpoint to get a complete view of the surface. For three dimensional
solids, the situation is even more challenging. Typically, portions of the
information must be omitted in order to provide the necessary information.
For example, a slice or cross-section of a three dimensional object is often
shown. In some cases, transparency can be used. Again, the ability to change
the arrangement of the visual elements interactively can be helpful.

4. Discuss the advantages and disadvantages of using sampling to reduce the
number of data objects that need to be displayed. Would simple random
sampling (without replacement) be a good approach to sampling? Why or
why not?

Simple random sampling is not the best approach since it will eliminate most
of the points in sparse regions. It is better to undersample the regions where
data objects are too dense while keeping most or all of the data objects from
sparse regions.

5. Describe how you would create visualizations to display information that
describes the following types of systems.

Be sure to address the following issues:

• Representation. How will you map objects, attributes, and relation-
ships to visual elements?

• Arrangement. Are there any special considerations that need to be
taken into account with respect to how visual elements are displayed?
Specific examples might be the choice of viewpoint, the use of trans-
parency, or the separation of certain groups of objects.

• Selection. How will you handle a large number of attributes and data
objects?

The following solutions are intended for illustration.

(a) Computer networks. Be sure to include both the static aspects of the
network, such as connectivity, and the dynamic aspects, such as traffic.

The connectivity of the network would best be represented as a graph,
with the nodes being routers, gateways, or other communications de-
vices and the links representing the connections. The bandwidth of the
connection could be represented by the width of the links. Color could
be used to show the percent usage of the links and nodes.

(b) The distribution of specific plant and animal species around the world
for a specific moment in time.

The simplest approach is to display each species on a separate map
of the world and to shade the regions of the world where the species
occurs. If several species are to be shown at once, then icons for each
species can be placed on a map of the world.

(c) The use of computer resources, such as processor time, main memory,
and disk, for a set of benchmark database programs.

The resource usage of each program could be displayed as a bar plot
of the three quantities. Since the three quantities would have different
scales, a proper scaling of the resources would be necessary for this
to work well. For example, resource usage could be displayed as a
percentage of the total. Alternatively, we could use three bar plots, one
for type of resource usage. On each of these plots there would be a bar
whose height represents the usage of the corresponding program. This
approach would not require any scaling. Yet another option would be to
display a line plot of each program’s resource usage. For each program,
a line would be constructed by (1) considering processor time, main
memory, and disk as different x locations, (2) letting the percentage
resource usage of a particular program for the three quantities be the
y values associated with the x values, and then (3) drawing a line to
connect these three points. Note that an ordering of the three quantities
needs to be specified, but is arbitrary. For this approach, the resource
usage of all programs could be displayed on the same plot.

(d) The change in occupation of workers in a particular country over the
last thirty years. Assume that you have yearly information about each
person that also includes gender and level of education.

For each gender, the occupation breakdown could be displayed as an
array of pie charts, where each row of pie charts indicates a particu-
lar level of education and each column indicates a particular year. For
convenience, the time gap between each column could be 5 or ten years.

Alternatively, we could order the occupations and then, for each gen-
der, compute the cumulative percent employment for each occupation.
If this quantity is plotted for each gender, then the area between two
successive lines shows the percentage of employment for this occupa-
tion. If a color is associated with each occupation, then the area between
each set of lines can also be colored with the color associated with each
occupation. A similar way to show the same information would be to
use a sequence of stacked bar graphs.

6. Describe one advantage and one disadvantage of a stem and leaf plot with
respect to a standard histogram.

A stem and leaf plot shows you the actual distribution of values. On the
other hand, a stem and leaf plot becomes rather unwieldy for a large number
of values.

7. How might you address the problem that a histogram depends on the number
and location of the bins?

22 Chapter 3 Exploring Data

The best approach is to estimate what the actual distribution function of the
data looks like using kernel density estimation. This branch of data analysis
is relatively well-developed and is more appropriate if the widely available,
but simplistic approach of a histogram is not sufficient.

8. Describe how a box plot can give information about whether the value of an
attribute is symmetrically distributed. What can you say about the symme-
try of the distributions of the attributes shown in Figure 3.11?

(a) If the line representing the median of the data is in the middle of the
box, then the data is symmetrically distributed, at least in terms of the
75% of the data between the first and third quartiles. For the remain-
ing data, the length of the whiskers and outliers is also an indication,
although, since these features do not involve as many points, they may
be misleading.

(b) Sepal width and length seem to be relatively symmetrically distributed,
petal length seems to be rather skewed, and petal width is somewhat
skewed.

9. Compare sepal length, sepal width, petal length, and petal width, using
Figure 3.12.

For Setosa, sepal length > sepal width > petal length > petal width. For
Versicolour and Virginiica, sepal length > sepal width and petal length >
petal width, but although sepal length > petal length, petal length > sepal
width.

10. Comment on the use of a box plot to explore a data set with four attributes:
age, weight, height, and income.

A great deal of information can be obtained by looking at (1) the box plots
for each attribute, and (2) the box plots for a particular attribute across
various categories of a second attribute. For example, if we compare the box
plots of age for different categories of ages, we would see that weight increases
with age.

11. Give a possible explanation as to why most of the values of petal length and
width fall in the buckets along the diagonal in Figure 3.9.

We would expect such a distribution if the three species of Iris can be ordered
according to their size, and if petal length and width are both correlated to
the size of the plant and each other.

12. Use Figures 3.14 and 3.15 to identify a characteristic shared by the petal
width and petal length attributes.

There is a relatively flat area in the curves of the Empirical CDF’s and the
percentile plots for both petal length and petal width. This indicates a set
of flowers for which these attributes have a relatively uniform value.

13. Simple line plots, such as that displayed in Figure 2.12 on page 56, which
shows two time series, can be used to effectively display high-dimensional
data. For example, in Figure 56 it is easy to tell that the frequencies of the
two time series are different. What characteristic of time series allows the
effective visualization of high-dimensional data?

The fact that the attribute values are ordered.

14. Describe the types of situations that produce sparse or dense data cubes.
Illustrate with examples other than those used in the book.

Any set of data for which all combinations of values are unlikely to occur
would produce sparse data cubes. This would include sets of continuous
attributes where the set of objects described by the attributes doesn’t occupy
the entire data space, but only a fraction of it, as well as discrete attributes,
where many combinations of values don’t occur.

A dense data cube would tend to arise, when either almost all combinations of
the categories of the underlying attributes occur, or the level of aggregation is
high enough so that all combinations are likely to have values. For example,
consider a data set that contains the type of traffic accident, as well as its
location and date. The original data cube would be very sparse, but if it is
aggregated to have categories consisting single or multiple car accident, the
state of the accident, and the month in which it occurred, then we would
obtain a dense data cube.

15. How might you extend the notion of multidimensional data analysis so that
the target variable is a qualitative variable? In other words, what sorts of
summary statistics or data visualizations would be of interest?

A summary statistics that would be of interest would be the frequencies with
which values or combinations of values, target and otherwise, occur. From
this we could derive conditional relationships among various values. In turn,
these relationships could be displayed using a graph similar to that used to
display Bayesian networks.

24 Chapter 3 Exploring Data

16. Construct a data cube from Table 3.1. Is this a dense or sparse data cube?
If it is sparse, identify the cells that are empty.

The data cube is shown in Table 3.2. It is a dense cube; only two cells are
empty.

Table 3.1. Fact table for Exercise 16.
Product ID Location ID Number Sold

1 1

1 3 6
2 1 5
2 2 22

Table 3.2. Data cube for Exercise 16.
Location

1 2 3 Total
1 10 0 6 16
2 5 22 0

Total 15 22 6 43P
r
o
d
u
c
t

17. Discuss the differences between dimensionality reduction based on aggrega-
tion and dimensionality reduction based on techniques such as PCA and
SVD.

The dimensionality of PCA or SVD can be viewed as a projection of the
data onto a reduced set of dimensions. In aggregation, groups of dimensions
are combined. In some cases, as when days are aggregated into months or
the sales of a product are aggregated by store location, the aggregation can
be viewed as a change of scale. In contrast, the dimensionality reduction
provided by PCA and SVD do not have such an interpretation.

Classification: Basic
Concepts, Decision
Trees, and Model
Evaluation

1. Draw the full decision tree for the parity function of four Boolean attributes,
A, B, C, and D. Is it possible to simplify the tree?

C C C

D D D D D D D D

F F T F T T F F T TT F FFT T

A B C D Class
T T T T T
T T T F

T T F

T F

T T F F T
T F T T F
T F T F T
T F F T T
T F F F F
F T T T F
F T T F T
F T F T T
F T F F F
F F T T T
F F T F F
F F F T F
F F F F T

T F

T F T F

T F T F T F T F

T F T F T F T F T F T F T F T F

Figure 4.1. Decision tree for parity function of four Boolean attributes.

26 Chapter 4 Classification

The preceding tree cannot be simplified.

2. Consider the training examples shown in Table 4.1 for a binary classification
problem.

Table 4.1. Data set for Exercise 2.

Customer ID Gender Car Type Shirt Size Class
1 M Family Small C0
2 M Sports Medium C0
3 M Sports Medium C0
4 M Sports Large C0
5 M Sports Extra Large C0
6 M Sports Extra Large C0
7 F Sports Small C0
8 F Sports Small C0
9 F Sports Medium C0
10 F Luxury Large C0
11 M Family Large

12 M Family Extra Large C1
13 M Family Medium C1
14 M Luxury Extra Large C1
15 F Luxury Small C1
16 F Luxury Small C1
17 F Luxury Medium C1
18 F Luxury Medium C1
19 F Luxury Medium C1
20 F Luxury Large C1

(a) Compute the Gini index for the overall collection of training examples.

Answer:

Gini = 1 − 2 × 0.52 = 0.5.

(b) Compute the Gini index for the Customer ID attribute

.
Answer:

The gini for each Customer ID value is 0. Therefore, the overall gini
for Customer ID is 0.

(c) Compute the Gini index for the Gender attribute.
Answer:
The gini for Male is 1− 2× 0.52 = 0.5. The gini for Female is also 0.5.
Therefore, the overall gini for Gender is 0.5 × 0.5 + 0.5 × 0.5 = 0.5.

Table 4.2. Data set for Exercise 3.

Instance a1 a2 a3 Target Class
1 T T 1.0 +
2 T T 6.0 +
3 T F 5.0 −
4 F F 4.0 +
5 F T 7.0 −
6 F T 3.0 −
7 F F 8.0 −
8 T F 7.0 +
9 F T 5.0 −

(d) Compute the Gini index for the Car Type attribute using multiway
split.
Answer:
The gini for Family car is 0.375, Sports car is 0, and Luxury car is
0.2188. The overall gini is 0.1625.

(e) Compute the Gini index for the Shirt Size attribute using multiway
split.
Answer:
The gini for Small shirt size is 0.48, Medium shirt size is 0.4898, Large
shirt size is 0.5, and Extra Large shirt size is 0.5. The overall gini for
Shirt Size attribute is 0.4914.

(f) Which attribute is better, Gender, Car Type, or Shirt Size?
Answer:
Car Type because it has the lowest gini among the three attributes.

(g) Explain why Customer ID should not be used as the attribute test
condition even though it has the lowest Gini.
Answer:
The attribute has no predictive power since new customers are assigned
to new Customer IDs.

3. Consider the training examples shown in Table 4.2 for a binary classification
problem.

(a) What is the entropy of this collection of training examples with respect
to the positive class?
Answer:
There are four positive examples and five negative examples. Thus,
P (+) = 4/9 and P (−) = 5/9. The entropy of the training examples is
−4/9 log2(4/9) − 5/9 log2(5/9) = 0.9911.

28 Chapter 4 Classification

(b) What are the information gains of a1 and a2 relative to these training
examples?
Answer:
For attribute a1, the corresponding counts and probabilities are:

+ –

T 3 1
F 1 4

The entropy for a1 is

4
9

[
− (3/4) log2(3/4) − (1/4) log2(1/4)

]

5
9

[
− (1/5) log2(1/5) − (4/5) log2(4/5)

]
= 0.7616.

Therefore, the information gain for a1 is 0.9911 − 0.7616 = 0.2294.

For attribute a2, the corresponding counts and probabilities are:

a2 + –

T 2 3
F 2 2

The entropy for a2 is

5
9

[
− (2/5) log2(2/5) − (3/5) log2(3/5)

]

+
4
9

[
− (2/4) log2(2/4) − (2/4) log2(2/4)

]
= 0.9839.

Therefore, the information gain for a2 is 0.9911 − 0.9839 = 0.0072.
(c) For a3, which is a continuous attribute, compute the information gain

for every possible split.
Answer:

a3 Class label Split point Entropy Info Gain

1.0 + 2.0 0.8484 0.1427

3.0 – 3.5 0.9885 0.0026

4.0 + 4.5 0.9183 0.0728

5.0 –
5.0 – 5.5 0.9839 0.0072

6.0 + 6.5 0.9728 0.01

7.0 +
7.0 – 7.5 0.8889 0.1022

The best split for a3 occurs at split point equals to 2.

(d) What is the best split (among a1, a2, and a3) according to the infor-
mation gain?
Answer:
According to information gain, a1 produces the best split.

(e) What is the best split (between a1 and a2) according to the classification
error rate?
Answer:
For attribute a1: error rate = 2/9.
For attribute a2: error rate = 4/9.
Therefore, according to error rate, a1 produces the best split.

(f) What is the best split (between a1 and a2) according to the Gini index?
Answer:
For attribute a1, the gini index is

4
9

[
1 − (3/4)2 − (1/4)2

]
+

5
9

[
1 − (1/5)2 − (4/5)2

]
= 0.3444.

For attribute a2, the gini index is

5
9

[
1 − (2/5)2 − (3/5)2

]
+
4
9

[
1 − (2/4)2 − (2/4)2

]
= 0.4889.

Since the gini index for a1 is smaller, it produces the better split.

4. Show that the entropy of a node never increases after splitting it into smaller
successor nodes.

Answer:

Let Y = {y1, y2, · · · , yc} denote the c classes and X = {x1, x2, · · · , xk} denote
the k attribute values of an attribute X. Before a node is split on X, the
entropy is:

E(Y ) = −
c∑

j=1

P (yj) log2 P (yj) =
c∑

j=1

k∑

i=1

P (xi, yj) log2 P (yj), (4.1)

where we have used the fact that P (yj) =
∑k

i=1 P (xi, yj) from the law of
total probability.

After splitting on X, the entropy for each child node X = xi is:

E(Y |xi) = −
c∑

j=1

P (yj |xi) log2 P (yj |xi) (4.2)

30 Chapter 4 Classification

where P (yj |xi) is the fraction of examples with X = xi that belong to class
yj . The entropy after splitting on X is given by the weighted entropy of the
children nodes:

E(Y |X)

=
k∑

i=1

P (xi)E(Y |xi)

= −
k∑

i=1

c∑
j=1

P (xi)P (yj |xi) log2

P (yj |xi)

= −
k∑
i=1
c∑
j=1

P (xi, yj) log2 P (yj |xi), (4.3)

where we have used a known fact from probability theory that P (xi, yj) =
P (yj |xi)×P (xi). Note that E(Y |X) is also known as the conditional entropy
of Y given X.

To answer this question, we need to show that E(Y |X) ≤ E(Y ). Let us com-
pute the difference between the entropies after splitting and before splitting,
i.e., E(Y |X) − E(Y ), using Equations 4.1 and 4.3:

E(Y |X) − E(Y )

= −
k∑
i=1
c∑
j=1

P (xi, yj) log2 P (yj |xi) +
k∑

i=1
c∑
j=1

P (xi, yj) log2 P (yj)

=
k∑
i=1
c∑
j=1

P (xi, yj) log2
P (yj)

P (yj |xi)
=
k∑
i=1
c∑
j=1

P (xi, yj) log2
P (xi)P (yj)
P (xi, yj)

(4.4)

To prove that Equation 4.4 is non-positive, we use the following property of
a logarithmic function:

d∑
k=1

ak log(zk) ≤ log
( d∑

k=1

akzk
)
, (4.5)

subject to the condition that
∑d

k=1 ak = 1. This property is a special case
of a more general theorem involving convex functions (which include the
logarithmic function) known as Jensen’s inequality.

By applying Jensen’s inequality, Equation 4.4 can be bounded as follows:

E(Y |X) − E(Y ) ≤ log2
[ k∑

i=1
c∑
j=1

P (xi, yj)
P (xi)P (yj)
P (xi, yj)

]

= log2

[ k∑
i=1

P (xi)
c∑

j=1

P (yj)
]

= log2(1)
= 0

Because E(Y |X) − E(Y ) ≤ 0, it follows that entropy never increases after
splitting on an attribute.

5. Consider the following data set for a binary class problem.

A B Class Label
T F +
T T +
T T +
T F −
T T +
F F −
F F −
F F −
T T −
T F −

(a) Calculate the information gain when splitting on A and B. Which
attribute would the decision tree induction algorithm choose?
Answer:
The contingency tables after splitting on attributes A and B are:

A = T A = F
+ 4 0
− 3 3

B = T B = F
+ 3 1
− 1 5

The overall entropy before splitting is:

Eorig = −0.4 log 0.4 − 0.6 log 0.6 = 0.9710
The information gain after splitting on A is:

EA=T = −47 log

4
7

− 3

7
log

3
7

= 0.9852

EA=F = −33 log

3
3

− 0

3
log

0
3

= 0

∆ = Eorig − 7/10EA=T − 3/10EA=F = 0.2813

32 Chapter 4 Classification

The information gain after splitting on B is:

EB=T = −34 log
3
4
− 1

4
log

1
4

= 0.8

113

EB=F = −16 log

1
6

− 5

6
log

5
6

= 0.6500

∆ = Eorig − 4/10EB=T − 6/10EB=F = 0.25

Therefore, attribute A will be chosen to split the node.

(b) Calculate the gain in the Gini index when splitting on A and B. Which
attribute would the decision tree induction algorithm choose?
Answer:
The overall gini before splitting is:

Gorig = 1 − 0.42 − 0.62 = 0.48

The gain in gini after splitting on A is:

GA=T = 1 −
(

4
7

)

2
−

(
3
7

)2
= 0.4898

GA=F = 1 =
(

3
3
)2
−

(
0
3

)2
= 0

∆ = Gorig − 7/10GA=T − 3/10GA=F = 0.1371

The gain in gini after splitting on B is:

GB=T = 1 −
(

1
4
)2
−

(
3
4

)2
= 0.3750

GB=F = 1 =
(

1
6
)2
−

(
5
6

)2
= 0.2778

∆ = Gorig − 4/10GB=T − 6/10GB=F = 0.16

Therefore, attribute B will be chosen to split the node.

(c) Figure 4.13 shows that entropy and the Gini index are both monotonously
increasing on the range [0, 0.5] and they are both monotonously decreas-
ing on the range [0.5, 1]. Is it possible that information gain and the
gain in the Gini index favor different attributes? Explain.
Answer:
Yes, even though these measures have similar range and monotonous
behavior, their respective gains, ∆, which are scaled differences of the
measures, do not necessarily behave in the same way, as illustrated by
the results in parts (a) and (b).

6. Consider the following set of training examples.

X Y Z No. of Class C1 Examples No. of Class C2 Examples
0 0 0 5 40
0 0 1 0 15
0 1 0 10 5
0 1 1 45 0
1 0 0 10 5
1 0 1 25 0
1 1 0 5 20
1 1 1 0 15

(a) Compute a two-level decision tree using the greedy approach described
in this chapter. Use the classification error rate as the criterion for
splitting. What is the overall error rate of the induced tree?
Answer:
Splitting Attribute at Level 1.
To determine the test condition at the root node, we need to com-
pute the error rates for attributes X, Y , and Z. For attribute X, the
corresponding counts are:

X C1

0 60 60
1 40 40

Therefore, the error rate using attribute X is (60 + 40)/200 = 0.5.
For attribute Y , the corresponding counts are:

C1 C2

0 40 60
1 60 40

Therefore, the error rate using attribute Y is (40 + 40)/200 = 0.4.
For attribute Z, the corresponding counts are:

Z C1 C2

0 30 70
1 70

Therefore, the error rate using attribute Y is (30 + 30)/200 = 0.3.
Since Z gives the lowest error rate, it is chosen as the splitting attribute
at level 1.

Splitting Attribute at Level 2.
After splitting on attribute Z, the subsequent test condition may in-
volve either attribute X or Y . This depends on the training examples
distributed to the Z = 0 and Z = 1 child nodes.
For Z = 0, the corresponding counts for attributes X and Y are the
same, as shown in the table below.

34 Chapter 4 Classification

X C1 C2 Y C1 C2
0 15 45 0 15 45
1 15 25 1 15 25

The error rate in both cases (X and Y ) are (15 + 15)/100 = 0.3.
For Z = 1, the corresponding counts for attributes X and Y are shown
in the tables below.

X C1 C2 Y C1 C2
0 45 15 0 25 15
1 25 15 1 45 15

Although the counts are somewhat different, their error rates remain
the same, (15 + 15)/100 = 0.3.
The corresponding two-level decision tree is shown below.

or
Y

0 1

0 0

1 1

C2 C1 C1

X
or
Y

The overall error rate of the induced tree is (15+15+15+15)/200 = 0.3.

(b) Repeat part (a) using X as the first splitting attribute and then choose
the best remaining attribute for splitting at each of the two successor
nodes. What is the error rate of the induced tree?
Answer:
After choosing attribute X to be the first splitting attribute, the sub-
sequent test condition may involve either attribute Y or attribute Z.
For X = 0, the corresponding counts for attributes Y and Z are shown
in the table below.

Y C1 C2 Z C1 C2
0 5 55 0 15 45
1 55 5 1 45 15

The error rate using attributes Y and Z are 10/120 and 30/120, re-
spectively. Since attribute Y leads to a smaller error rate, it provides a
better split.
For X = 1, the corresponding counts for attributes Y and Z are shown
in the tables below.

Y C1 C2 Z C1 C2
0 35 5 0 15 25
1 5 35 1 25 15

The error rate using attributes Y and Z are 10/80 and 30/80, respec-
tively. Since attribute Y leads to a smaller error rate, it provides a
better split.
The corresponding two-level decision tree is shown below.

X
C2
0 1
0 01 1

C1 C1 C2

Y Y

The overall error rate of the induced tree is (10 + 10)/200 = 0.1.

(c) Compare the results of parts (a) and (b). Comment on the suitability
of the greedy heuristic used for splitting attribute selection.
Answer:
From the preceding results, the error rate for part (a) is significantly
larger than that for part (b). This examples shows that a greedy heuris-
tic does not always produce an optimal solution.

7. The following table summarizes a data set with three attributes A, B, C and
two class labels +, −. Build a two-level decision tree.

B C

Number of
Instances
+ −

T T T 5 0
F T T 0 20
T F T 20 0
F F T 0 5
T T F 0 0
F T F 25 0
T F F 0 0
F F F 0 25

(a) According to the classification error rate, which attribute would be
chosen as the first splitting attribute? For each attribute, show the
contingency table and the gains in classification error rate.

36 Chapter 4 Classification

Answer:
The error rate for the data without partitioning on any attribute is

Eorig = 1 − max( 50100 ,
50
100

) =
50
100

After splitting on attribute A, the gain in error rate is:

A = T A = F
+ 25 25
− 0 50

EA=T = 1 − max(2525 ,
0
25

) =
0
25

= 0

EA=F = 1 − max(2575 ,
50

) =
25
75

∆A = Eorig − 25100EA=T −
75
100

EA=F =
25
100

After splitting on attribute B, the gain in error rate is:

B = T B = F
+ 30 20
− 20 30

EB=T =
20
50

EB=F =
20
50

∆B = Eorig − 50100EB=T −
50
100

EB=F =
10
100

After splitting on attribute C, the gain in error rate is:

C = T C = F
+ 25 25
− 25 25

EC=T =
25
50

EC=F =
25
50

∆C = Eorig − 50100EC=T −
50
100

EC=F =
0

100
= 0

The algorithm chooses attribute A because it has the highest gain.

(b) Repeat for the two children of the root node.
Answer:
Because the A = T child node is pure, no further splitting is needed.
For the A = F child node, the distribution of training instances is:

B C Class label+ −
T T 0 20
F T 0 5
T F 25 0
F F 0 25

The classification error of the A = F child node is:

Eorig =
25
75

After splitting on attribute B, the gain in error rate is:

B = T B = F
+ 25 0
− 20 30

EB=T =
20
45

EB=F = 0

∆B = Eorig − 4575EB=T −
20
75

EB=F =
5
75

After splitting on attribute C, the gain in error rate is:

C = T C = F
+ 0 25
− 25 25

EC=T =
0
25

EC=F =
25
50

∆C = Eorig − 2575EC=T −
50
75

EC=F = 0

The split will be made on attribute B.

(c) How many instances are misclassified by the resulting decision tree?
Answer:
20 instances are misclassified. (The error rate is 20100 .)

(d) Repeat parts (a), (b), and (c) using C as the splitting attribute.
Answer:
For the C = T child node, the error rate before splitting is:
Eorig = 2550 .
After splitting on attribute A, the gain in error rate is:

A = T A = F
+ 25 0
− 0 25

EA=T = 0

EA=F = 0

∆A =
25
50

After splitting on attribute B, the gain in error rate is:

B = T B = F
+ 5 20
− 20 5

EB=T =
5
25

EB=F =
5
25

∆B =
15
50

Therefore, A is chosen as the splitting attribute.

38 Chapter 4 Classification

+ _ + _

B C
A

Instance
1
2
3
4
5
6
7
8
9
10

0
0

0
0
1
1
1

1
1
1

0
1
1
0
0

1
0
1
1

0
1
0
1
0

0
0
1
0
0

A B C

+
+

–
+

+
–
+
–
–

Class
Training:

Instance
11
12
13
14
15

0
0
1
1
1
0
1
1
0
0
0
1
0
1
0
A B C
+
+
+
–
+

Class
Validation:

0 1 0 1

Figure 4.2. Decision tree and data sets for Exercise 8.

For the C = F child, the error rate before splitting is: Eorig = 2550 .
After splitting on attribute A, the error rate is:

A = T A = F
+ 0 25
− 0 25

EA=T = 0

EA=F =
25
50

∆A = 0

After splitting on attribute B, the error rate is:

B = T B = F
+ 25 0
− 0 25

EB=T = 0
EB=F = 0

∆B =
25
50

Therefore, B is used as the splitting attribute.
The overall error rate of the induced tree is 0.

(e) Use the results in parts (c) and (d) to conclude about the greedy nature
of the decision tree induction algorithm.
The greedy heuristic does not necessarily lead to the best tree.

8. Consider the decision tree shown in Figure 4.2.

(a) Compute the generalization error rate of the tree using the optimistic
approach.
Answer:
According to the optimistic approach, the generalization error rate is
3/10 = 0.3.

(b) Compute the generalization error rate of the tree using the pessimistic
approach. (For simplicity, use the strategy of adding a factor of 0.5 to
each leaf node.)
Answer:
According to the pessimistic approach, the generalization error rate is
(3 + 4 × 0.5)/10 = 0.5.

(c) Compute the generalization error rate of the tree using the validation
set shown above. This approach is known as reduced error pruning.
Answer:
According to the reduced error pruning approach, the generalization
error rate is 4/5 = 0.8.

9. Consider the decision trees shown in Figure 4.3. Assume they are generated
from a data set that contains 16 binary attributes and 3 classes, C1, C2, and
C3. Compute the total description length of each decision tree according to
the minimum description length principle.

(a) Decision tree with 7 errors (b) Decision tree with 4 errors

C2 C3

C1
C2 C3
C1 C2

Figure 4.3. Decision trees for Exercise 9.

• The total description length of a tree is given by:
Cost(tree, data) = Cost(tree) + Cost(data|tree).

40 Chapter 4 Classification

• Each internal node of the tree is encoded by the ID of the splitting
attribute. If there are m attributes, the cost of encoding each attribute
is log2 m bits.

• Each leaf is encoded using the ID of the class it is associated with. If
there are k classes, the cost of encoding a class is log2 k bits.

• Cost(tree) is the cost of encoding all the nodes in the tree. To simplify
the computation, you can assume that the total cost of the tree is
obtained by adding up the costs of encoding each internal node and
each leaf node.

• Cost(data|tree) is encoded using the classification errors the tree com-
mits on the training set. Each error is encoded by log2 n bits, where n
is the total number of training instances.

Which decision tree is better, according to the MDL principle?

Answer:

Because there are 16 attributes, the cost for each internal node in the decision
tree is:

log2(m) = log2(16) = 4

Furthermore, because there are 3 classes, the cost for each leaf node is:

log2(k)� =
log2(3)� = 2

The cost for each misclassification error is log2(n).

The overall cost for the decision tree (a) is 2×4+3×2+7×log2 n = 14+7 log2 n
and the overall cost for the decision tree (b) is 4×4+5×2+4×5 = 26+4 log2 n.
According to the MDL principle, tree (a) is better than (b) if n < 16 and is worse than (b) if n > 16.

10. While the .632 bootstrap approach is useful for obtaining a reliable estimate
of model accuracy, it has a known limitation. Consider a two-class problem,
where there are equal number of positive and negative examples in the data.
Suppose the class labels for the examples are generated randomly. The clas-
sifier used is an unpruned decision tree (i.e., a perfect memorizer). Determine
the accuracy of the classifier using each of the following methods.

(a) The holdout method, where two-thirds of the data are used for training
and the remaining one-third are used for testing.
Answer:
Assuming that the training and test samples are equally representative,
the test error rate will be close to 50%.

(b) Ten-fold cross-validation.
Answer:
Assuming that the training and test samples for each fold are equally
representative, the test error rate will be close to 50%.

(c) The .632 bootstrap method.
Answer:
The training error for a perfect memorizer is 100% while the error rate
for each bootstrap sample is close to 50%. Substituting this information
into the formula for .632 bootstrap method, the error estimate is:

b∑
i=1

[
0.632 × 0.5 + 0.368 × 1

]
= 0.684.

(d) From the results in parts (a), (b), and (c), which method provides a
more reliable evaluation of the classifier’s accuracy?
Answer:
The ten-fold cross-validation and holdout method provides a better
error estimate than the .632 boostrap method.

11. Consider the following approach for testing whether a classifier A beats an-
other classifier B. Let N be the size of a given data set, pA be the accuracy
of classifier A, pB be the accuracy of classifier B, and p = (pA + pB)/2
be the average accuracy for both classifiers. To test whether classifier A is
significantly better than B, the following Z-statistic is used:

Z =
pA − pB√

2p(1−p)

Classifier A is assumed to be better than classifier B if Z > 1.96.

Table 4.3 compares the accuracies of three different classifiers, decision tree
classifiers, näıve Bayes classifiers, and support vector machines, on various
data sets. (The latter two classifiers are described in Chapter 5.)

42 Chapter 4 Classification

Table 4.3. Comparing the accuracy of various classification methods.

Data Set Size Decision näıve Support vector
(N) Tree (%) Bayes (%) machine (%)

Anneal 898 92.09 79.62 87.19
Australia 690 85.51 76.81 84.78
Auto 205 81.95 58.05 70.

Breast 699 95.14 95.99 96.42
Cleve 303 76.24 83.50 84.

Credit 690 85.80 77.54 85.07
Diabetes 768 72.40 75.91 76.82
German 1000 70.90 74.70 74.40
Glass 214 67.29 48.59 59.

Heart 270 80.00 84.07 83.70
Hepatitis 155 81.94 83.23 87.10
Horse 368 85.33 78.80 82.

Ionosphere 351 89.17 82.34 88.

Iris 150 94.67 95.33 96.00
Labor 57 78.95 94.74 92.98
Led7 3200 73.34 73.16 73.56
Lymphography 148 77.03 83.11 86.49
Pima 768 74.35 76.04 76.95
Sonar 208 78.85 69.71 76.92
Tic-tac-toe 958 83.72 70.04 98.33
Vehicle 846 71.04 45.04 74.94
Wine 178 94.38 96.63 98.88
Zoo 101 93.07 93.07 96.04

Answer:

A summary of the relative performance of the classifiers is given below:

win-loss-draw Decision tree Näıve Bayes Support vector
machine

Decision tree 0 – 0 – 23 9 – 3 – 11 2 – 7- 14
Näıve Bayes 3 – 9 – 11 0 – 0 – 23 0 – 8 – 15
Support vector machine 7 – 2 – 14 8 – 0 – 15 0 – 0 – 23

12. Let X be a binomial random variable with mean Np and variance Np(1−p).
Show that the ratio X/N also has a binomial distribution with mean p and
variance p(1 − p)/N .
Answer: Let r = X/N . Since X has a binomial distribution, r also has the
same distribution. The mean and variance for r can be computed as follows:

Mean, E[r] = E[X/N ] = E[X]/N = (Np)/N = p;

Variance, E[(r − E[r])2] = E[(X/N − E[X/N ])2]
= E[(X − E[X])2]/N2
= Np(1 − p)/N2
= p(1 − p)/N

Classification:
Alternative Techniques

1. Consider a binary classification problem with the following set of attributes
and attribute values:

• Air Conditioner = {Working, Broken}
• Engine = {Good, Bad}
• Mileage = {High, Medium, Low}
• Rust = {Yes, No}

Suppose a rule-based classifier produces the following rule set:

Mileage = High −→ Value = Low
Mileage = Low −→ Value = High
Air Conditioner = Working, Engine = Good −→ Value = High
Air Conditioner = Working, Engine = Bad −→ Value = Low
Air Conditioner = Broken −→ Value = Low

(a) Are the rules mutually exclustive?
Answer: No

(b) Is the rule set exhaustive?
Answer: Yes

(d) Do you need a default class for the rule set?
Answer: No because every instance is guaranteed to trigger at least
one rule.

46 Chapter 5 Classification: Alternative Techniques

2. The RIPPER algorithm (by Cohen [1]) is an extension of an earlier algorithm
called IREP (by Fürnkranz and Widmer [3]). Both algorithms apply the
reduced-error pruning method to determine whether a rule needs to be
pruned. The reduced error pruning method uses a validation set to estimate
the generalization error of a classifier. Consider the following pair of rules:

R1: A −→ C
R2: A ∧ B −→ C

R2 is obtained by adding a new conjunct, B, to the left-hand side of R1. For
this question, you will be asked to determine whether R2 is preferred over
R1 from the perspectives of rule-growing and rule-pruning. To determine
whether a rule should be pruned, IREP computes the following measure:

vIREP =
p + (N − n)

P + N
,

where P is the total number of positive examples in the validation set, N is
the total number of negative examples in the validation set, p is the number
of positive examples in the validation set covered by the rule, and n is the
number of negative examples in the validation set covered by the rule. vIREP
is actually similar to classification accuracy for the validation set. IREP
favors rules that have higher values of vIREP . On the other hand, RIPPER
applies the following measure to determine whether a rule should be pruned:

vRIPPER =
p − n
p + n

(a) Suppose R1 is covered by 350 positive examples and 150 negative ex-
amples, while R2 is covered by 300 positive examples and 50 negative
examples. Compute the FOIL’s information gain for the rule R2 with
respect to R1.
Answer:
For this problem, p0 = 350, n0 = 150, p1 = 300, and n1 = 50. There-
fore, the FOIL’s information gain for R2 with respect to R1 is:

Gain = 300 ×
[

log2
300
350

− log2
350
500

]
= 87.65

(b) Consider a validation set that contains 500 positive examples and 500
negative examples. For R1, suppose the number of positive examples
covered by the rule is 200, and the number of negative examples covered
by the rule is 50. For R2, suppose the number of positive examples
covered by the rule is 100 and the number of negative examples is 5.
Compute vIREP for both rules. Which rule does IREP prefer?

Answer:
For this problem, P = 500, and N = 500.
For rule R1, p = 200 and n = 50. Therefore,

VIREP (R1) =
p + (N − n)

P + N
=

200 + (500 − 50)
1000

0.65

For rule R2, p = 100 and n = 5.

VIREP (R2) =
p + (N − n)

P + N
=

100 + (500 − 5)
1000

= 0.595

Thus, IREP prefers rule R1.

VRIPPER(R1) =
p − n
p + n

=
150
250

= 0.6

VRIPPER(R2) =
p − n
p + n

=
95

105

0.9

Thus, RIPPER prefers the rule R2.

3. C4.5rules is an implementation of an indirect method for generating rules
from a decision tree. RIPPER is an implementation of a direct method for
generating rules directly from data.

(a) Discuss the strengths and weaknesses of both methods.
Answer:
The C4.5 rules algorithm generates classification rules from a global
perspective. This is because the rules are derived from decision trees,
which are induced with the objective of partitioning the feature space
into homogeneous regions, without focusing on any classes. In contrast,
RIPPER generates rules one-class-at-a-time. Thus, it is more biased
towards the classes that are generated first.

(b) Consider a data set that has a large difference in the class size (i.e.,
some classes are much bigger than others). Which method (between
C4.5rules and RIPPER) is better in terms of finding high accuracy
rules for the small classes?
Answer:
The class-ordering scheme used by C4.5rules has an easier interpretation
than the scheme used by RIPPER.

48 Chapter 5 Classification: Alternative Techniques

4. Consider a training set that contains 100 positive examples and 400 negative
examples. For each of the following candidate rules,

R1: A −→ + (covers 4 positive and 1 negative examples),
R2: B −→ + (covers 30 positive and 10 negative examples),
R3: C −→ + (covers 100 positive and 90 negative examples),

determine which is the best and worst candidate rule according to:

(a) Rule accuracy.
Answer:
The accuracies of the rules are 80% (for R1), 75% (for R2), and 52.6%
(for R3), respectively. Therefore R1 is the best candidate and R3 is the
worst candidate according to rule accuracy.

(b) FOIL’s information gain.
Answer:
Assume the initial rule is ∅ −→ +. This rule covers p0 = 100 positive
examples and n0 = 400 negative examples.
The rule R1 covers p1 = 4 positive examples and n1 = 1 negative
example. Therefore, the FOIL’s information gain for this rule is

4 ×
(

log2
4
5
− log2

100
500

)
= 8.

The rule R2 covers p1 = 30 positive examples and n1 = 10 negative
example. Therefore, the FOIL’s information gain for this rule is

30 ×
(

log2
30
40

− log2
100
500

)
= 57.2.

The rule R3 covers p1 = 100 positive examples and n1 = 90 negative
example. Therefore, the FOIL’s information gain for this rule is

100 ×
(

log2
100
190

− log2
100
500

)
= 139.6.

Therefore, R3 is the best candidate and R1 is the worst candidate ac-
cording to FOIL’s information gain.

= 1

and the expected frequency for the negative class is 5 × 400/500 = 4.
Therefore, the likelihood ratio for R1 is

2 ×
[
4 × log2(4/1) + 1 × log2(1/4)

]
= 12.

For R2, the expected frequency for the positive class is 40×100/500 = 8
and the expected frequency for the negative class is 40× 400/500 = 32.
Therefore, the likelihood ratio for R2 is

2 ×
[
30 × log2(30/8) + 10 × log2(10/32)

]
= 80.

For R3, the expected frequency for the positive class is 190×100/500 =
38 and the expected frequency for the negative class is 190×400/500 =
152. Therefore, the likelihood ratio for R3 is

2 ×
[
100 × log2(100/38) + 90 × log2(90/152)

]
= 143.09

Therefore, R3 is the best candidate and R1 is the worst candidate ac-
cording to the likelihood ratio statistic.

(d) The Laplace measure.
Answer:
The Laplace measure of the rules are 71.43% (for R1), 73.81% (for R2),
and 52.6% (for R3), respectively. Therefore R2 is the best candidate
and R3 is the worst candidate according to the Laplace measure.

(e) The m-estimate measure (with k = 2 and p+ = 0.2).
Answer:
The m-estimate measure of the rules are 62.86% (for R1), 73.38% (for
R2), and 52.3% (for R3), respectively. Therefore R2 is the best candi-
date and R3 is the worst candidate according to the m-estimate mea-
sure.

5. Figure 5.1 illustrates the coverage of the classification rules R1, R2, and R3.
Determine which is the best and worst rule according to:

(a) The likelihood ratio statistic.
Answer:
There are 29 positive examples and 21 negative examples in the data
set. R1 covers 12 positive examples and 3 negative examples. The
expected frequency for the positive class is 15 × 29/50 = 8.7 and the
expected frequency for the negative class is 15×21/50 = 6.3. Therefore,
the likelihood ratio for R1 is

2 ×
[
12 × log2(12/8.7) + 3 × log2(3/6.3)

]
= 4.71.

R2 covers 7 positive examples and 3 negative examples. The expected
frequency for the positive class is 10 × 29/50 = 5.8 and the expected

50 Chapter 5 Classification: Alternative Techniques

class = +

class = –

+
+

+
+
+
+

+ +

+
+
+
+
+
+

+ + + + +
+

+
+
+
+
+
+
+
+ +

–
–

–
–
–
–
–
–

– –
–

–
–

– –
–
–
–

– –
–

R3 R2

Figure 5.1. Elimination of training records by the sequential covering algorithm. R1, R2, and R3
represent regions covered by three different rules.

frequency for the negative class is 10 × 21/50 = 4.2. Therefore, the
likelihood ratio for R2 is

2 ×
[
7 × log2(7/5.8) + 3 × log2(3/4.2)

]
= 0.89.

R3 covers 8 positive examples and 4 negative examples. The expected
frequency for the positive class is 12 × 29/50 = 6.96 and the expected
frequency for the negative class is 12 × 21/50 = 5.04. Therefore, the
likelihood ratio for R3 is

2 ×
[
8 × log2(8/6.96) + 4 × log2(4/5.04)

]
= 0.5472.

R1 is the best rule and R3 is the worst rule according to the likelihood
ratio statistic.

(b) The Laplace measure.
Answer:
The Laplace measure for the rules are 76.47% (for R1), 66.67% (for
R2), and 64.29% (for R3), respectively. Therefore R1 is the best rule
and R3 is the worst rule according to the Laplace measure.

(c) The m-estimate measure (with k = 2 and p+ = 0.58).
Answer:
The m-estimate measure for the rules are 77.41% (for R1), 68.0% (for
R2), and 65.43% (for R3), respectively. Therefore R1 is the best rule
and R3 is the worst rule according to the m-estimate measure.

(d) The rule accuracy after R1 has been discovered, where none of the
examples covered by R1 are discarded).

Answer:
If the examples for R1 are not discarded, then R2 will be chosen because
it has a higher accuracy (70%) than R3 (66.7%).

(e) The rule accuracy after R1 has been discovered, where only the positive
examples covered by R1 are discarded).
Answer:
If the positive examples covered by R1 are discarded, the new accuracies
for R2 and R3 are 70% and 60%, respectively. Therefore R2 is preferred
over R3.

(f) The rule accuracy after R1 has been discovered, where both positive
and negative examples covered by R1 are discarded.
Answer:
If the positive and negative examples covered by R1 are discarded, the
new accuracies for R2 and R3 are 70% and 75%, respectively. In this
case, R3 is preferred over R2.

6. (a) Suppose the fraction of undergraduate students who smoke is 15% and
the fraction of graduate students who smoke is 23%. If one-fifth of the
college students are graduate students and the rest are undergraduates,
what is the probability that a student who smokes is a graduate student?
Answer:
Given P (S|UG) = 0.15, P (S|G) = 0.23, P (G) = 0.2, P (UG) = 0.8.
We want to compute P (G|S).
According to Bayesian Theorem,

P (G|S) = 0.23 × 0.2
0.15 × 0.8 + 0.23 × 0.2 = 0.277. (5.1)

(b) Given the information in part (a), is a randomly chosen college student
more likely to be a graduate or undergraduate student?
Answer:
An undergraduate student, because P (UG) > P (G).

(d) Suppose 30% of the graduate students live in a dorm but only 10% of
the undergraduate students live in a dorm. If a student smokes and lives
in the dorm, is he or she more likely to be a graduate or undergraduate
student? You can assume independence between students who live in
a dorm and those who smoke.
Answer:
First, we need to estimate all the probabilities.

52 Chapter 5 Classification: Alternative Techniques

P (G|DS) = 0.069 × 0.2

P (DS)

=
0.0138
P (DS)

P (UG|DS) = 0.015 × 0.8
P (DS)

=
0.012

P (DS)

Since P (G|DS) > P (UG|DS), he/she is more likely to be a graduate
student.

7. Consider the data set shown in Table 5.1

Table 5.1. Data set for Exercise 7.
Record A B C Class

1 0 0 0 +
2 0 0 1 −
3 0 1 1 −
4 0 1 1 −
5 0 0 1 +
6 1 0 1 +
7 1 0 1 −
8 1 0 1 −
9 1 1 1 +
10 1 0 1 +

(a) Estimate the conditional probabilities for P (A|+), P (B|+), P (C|+),
P (A|−), P (B|−), and P (C|−).
Answer:
P (A = 1|−) = 2/5 = 0.4, P (B = 1|−) = 2/5 = 0.4,
P (C = 1|−) = 1, P (A = 0|−) = 3/5 = 0.6,
P (B = 0|−) = 3/5 = 0.6, P (C = 0|−) = 0; P (A = 1|+) = 3/5 = 0.6,
P (B = 1|+) = 1/5 = 0.2, P (C = 1|+) = 2/5 = 0.4,
P (A = 0|+) = 2/5 = 0.4, P (B = 0|+) = 4/5 = 0.8,
P (C = 0|+) = 3/5 = 0.6.

(b) Use the estimate of conditional probabilities given in the previous ques-
tion to predict the class label for a test sample (A = 0, B = 1, C = 0)
using the näıve Bayes approach.
Answer:
Let P (A = 0, B = 1, C = 0) = K.

P (+|A = 0, B = 1, C = 0

)
=

P (A = 0, B = 1, C = 0|+) × P (+)
P (A = 0, B = 1, C = 0)

=
P (A = 0|+)P (B = 1|+)P (C = 0|+) × P (+)

= 0.4 × 0.2 × 0.6 × 0.5/K
= 0.024/K.

P (−|A = 0, B = 1, C = 0)
=

P (A = 0, B = 1, C = 0|−) × P (−)
P (A = 0, B = 1, C = 0)

=
P (A = 0|−) × P (B = 1|−) × P (C = 0|−) × P (−)

K
= 0/K

The class label should be ’+’.

(c) Estimate the conditional probabilities using the m-estimate approach,
with p = 1/2 and m = 4.
Answer:
P (A = 0|+) = (2 + 2)/(5 + 4) = 4/9,
P (A = 0|−) = (3 + 2)/(5 + 4) = 5/9,
P (B = 1|+) = (1 + 2)/(5 + 4) = 3/9,
P (B = 1|−) = (2 + 2)/(5 + 4) = 4/9,
P (C = 0|+) = (3 + 2)/(5 + 4) = 5/9,
P (C = 0|−) = (0 + 2)/(5 + 4) = 2/9.

(d) Repeat part (b) using the conditional probabilities given in part (c).
Answer:
Let P (A = 0, B = 1, C = 0) = K

54 Chapter 5 Classification: Alternative Techniques

P (+|A = 0, B = 1, C = 0)
=
P (A = 0, B = 1, C = 0|+) × P (+)
P (A = 0, B = 1, C = 0)
=
P (A = 0|+)P (B = 1|+)P (C = 0|+) × P (+)
K

=
(4/9) × (3/9) × (5/9) ×

0.5

K
= 0.0412/K

P (−|A = 0, B = 1, C = 0)
=
P (A = 0, B = 1, C = 0|−) × P (−)
P (A = 0, B = 1, C = 0)
=
P (A = 0|−) × P (B = 1|−) × P (C = 0|−) × P (−)
K

=
(5/9) × (4/9) × (2/9) × 0.5

K
= 0.0274/K

The class label should be ’+’.

(e) Compare the two methods for estimating probabilities. Which method
is better and why?
Answer:
When one of the conditional probability is zero, the estimate for condi-
tional probabilities using the m-estimate probability approach is better,
since we don’t want the entire expression becomes zero.

8. Consider the data set shown in Table 5.2.

(a) Estimate the conditional probabilities for P (A = 1|+), P (B = 1|+),
P (C = 1|+), P (A = 1|−), P (B = 1|−), and P (C = 1|−) using the
same approach as in the previous problem.
Answer:
P (A = 1|+) = 0.6, P (B = 1|+) = 0.4, P (C = 1|+) = 0.8, P (A =
1|−) = 0.4, P (B = 1|−) = 0.4, and P (C = 1|−)

= 0.2

(b) Use the conditional probabilities in part (a) to predict the class label for
a test sample (A = 1, B = 1, C = 1) using the näıve Bayes approach.
Answer:
Let R : (A = 1, B = 1, C = 1) be the test record. To determine its
class, we need to compute P (+|R) and P (−|R). Using Bayes theorem,

Table 5.2. Data set for Exercise 8.
Instance A B C Class

1 0 0 1 −
2 1 0 1 +
3 0 1 0 −
4 1 0 0 −
5 1 0 1 +
6 0 0 1 +
7 1 1 0 −
8 0 0 0 −
9 0 1 0 +
10 1 1 1 +

P (+|R) = P (R|+)P (+)/P (R) and P (−|R) = P (R|−)P (−)/P (R).
Since P (+) = P (−) = 0.5 and P (R) is constant, R can be classified by
comparing P (+|R) and P (−|R).
For this question,

P (R|+) = P (A = 1|+) × P (B = 1|+) × P (C = 1|+) = 0.192
P (R|−) = P (A = 1|−) × P (B = 1|−) × P (C = 1|−) = 0.032

Since P (R|+) is larger, the record is assigned to (+) class.
(c) Compare P (A = 1), P (B = 1), and P (A = 1, B = 1). State the

relationships between A and B.
Answer:
P (A = 1) = 0.5, P (B = 1) = 0.4 and P (A = 1, B = 1) = P (A) ×
P (B) = 0.2. Therefore, A and B are independent.

(d) Repeat the analysis in part (c) using P (A = 1), P (B = 0), and P (A =
1, B = 0).
Answer:
P (A = 1) = 0.5, P (B = 0) = 0.6, and P (A = 1, B = 0) = P (A =
1) × P (B = 0) = 0.3. A and B are still independent.

56 Chapter 5 Classification: Alternative Techniques

Distinguishing Attributes Noise

Attributes

Class A

Class B

Records

Attributes

Figure 5.2. Data set for Exercise 9.

9. (a) Explain how näıve Bayes performs on the data set shown in Figure 5.2.

Answer:
NB will not do well on this data set because the conditional probabilities
for each distinguishing attribute given the class are the same for both
class A and class B.

(b) If each class is further divided such that there are four classes (A1, A2,
B1, and B2), will näıve Bayes perform better?
Answer:
The performance of NB will improve on the subclasses because the
product of conditional probabilities among the distinguishing attributes
will be different for each subclass.

(c) How will a decision tree perform on this data set (for the two-class
problem)? What if there are four classes?
Answer:
For the two-class problem, decision tree will not perform well because
the entropy will not improve after splitting the data using the distin-
guishing attributes. If there are four classes, then decision tree will
improve considerably.

10. Repeat the analysis shown in Example 5.3 for finding the location of a decision
boundary using the following information:

(a) The prior probabilities are P (Crocodile) = 2 × P (Alligator).
Answer: x̂ = 13.0379.

(b) The prior probabilities are P (Alligator) = 2 × P (Crocodile).
Answer: x̂ = 13.9621.

(c) The prior probabilities are the same, but their standard deviations are
different; i.e., σ(Crocodile) = 4 and σ(Alligator) = 2.
Answer: x̂ = 22.1668.

11. Figure 5.3 illustrates the Bayesian belief network for the data set shown in
Table 5.3. (Assume that all the attributes are binary).

Mileage

Engine

Car
Value

Air
Conditioner

Figure 5.3. Bayesian belief network.

Table 5.3. Data set for Exercise 11.
Mileage Engine Air Conditioner Number of Records Number of Records

with Car Value=Hi with Car Value=Lo

Hi Good Working 3 4
Hi Good Broken 1 2
Hi Bad Working 1 5
Hi Bad Broken 0 4
Lo Good Working 9 0
Lo Good Broken 5 1
Lo Bad Working 1 2
Lo Bad Broken 0 2

(a) Draw the probability table for each node in the network.
P(Mileage=Hi)

= 0.5

P(Air Cond=Working) = 0.625
P(Engine=Good|Mileage=Hi) = 0.5
P(Engine=Good|Mileage=Lo) = 0.75

58 Chapter 5 Classification: Alternative Techniques

Battery

Gauge

Start

Fuel

P(B = bad) = 0.1 P(F = empty) = 0.2

P(G = empty | B = good, F = not empty) =

0.1

P(G = empty | B = good, F = empty)

= 0.8

P(G = empty | B = bad, F = not empty) = 0.2
P(G = empty | B = bad, F = empty) = 0.9

P(S = no | B = good, F = not empty) = 0.1
P(S = no | B = good, F = empty) = 0.8
P(S = no | B = bad, F = not empty) = 0.9
P(S = no | B = bad, F = empty)

= 1.0

Figure 5.4. Bayesian belief network for Exercise 12.

P(Value=High|Engine=Good, Air Cond=Working) = 0.750
P(Value=High|Engine=Good, Air Cond=Broken) = 0.6

P(Value=High|Engine=Bad, Air Cond=Working) = 0.222
P(Value=High|Engine=Bad, Air Cond=Broken) = 0

(b) Use the Bayesian network to compute P(Engine = Bad, Air Conditioner
= Broken).

P (Engine = Bad,Air Cond = Broken)

=
∑
αβ

P (Engine = Bad,Air Cond = Broken,Mileage = α, V alue = β)

=
∑
αβ

P (V alue = β|Engine = Bad,Air Cond = Broken)

× P (Engine = Bad|Mileage = α)P (Mileage = α)P (Air Cond = Broken)
= 0.1453.

12. Given the Bayesian network shown in Figure 5.4, compute the following prob-
abilities:

(a) P (B = good, F = empty, G = empty, S = yes).

Answer:

P (B = good, F = empty,G = empty, S = yes)
= P (B = good) × P (F = empty) × P (G = empty|B = good, F = empty)

×P (S = yes|B = good, F = empty)
= 0.9 × 0.2 × 0.8 × 0.2 = 0.0288.

(b) P (B = bad, F = empty, G = not empty, S = no).
Answer:

P (B = bad, F = empty,G = not empty, S = no)
= P (B = bad) × P (F = empty) × P (G = not empty|B = bad, F = empty)

×P (S = no|B = bad, F = empty)
= 0.1 × 0.2 × 0.1 × 1.0 = 0.002.

P (S = yes|B = bad)
=

∑
α

P (S = yes|B = bad, F = α)P (B = bad)P (F = α)

= 0.1 × 0.1 × 0.8
= 0.008

13. Consider the one-dimensional data set shown in Table 5.4.

Table 5.4. Data set for Exercise 13.

x 0.5 3.0 4.5 4.6 4.9 5.2 5.3 5.5 7.0 9.5
y − − + + + − − + − −

(a) Classify the data point x = 5.0 according to its 1-, 3-, 5-, and 9-nearest
neighbors (using majority vote).
Answer:
1-nearest neighbor: +,
3-nearest neighbor: −,
5-nearest neighbor: +,
9-nearest neighbor: −.

(b) Repeat the previous analysis using the distance-weighted voting ap-
proach described in Section 5.2.1.

60 Chapter 5 Classification: Alternative Techniques

Answer:
1-nearest neighbor: +,
3-nearest neighbor: +,
5-nearest neighbor: +,
9-nearest neighbor: +.

14. The nearest-neighbor algorithm described in Section 5.2 can be extended to
handle nominal attributes. A variant of the algorithm called PEBLS (Parallel
Examplar-Based Learning System) by Cost and Salzberg [2] measures the
distance between two values of a nominal attribute using the modified value
difference metric (MVDM). Given a pair of nominal attribute values, V1 and
V2, the distance between them is defined as follows:

d(V1, V2) =
k∑

i=1

∣∣∣∣ni1n1 −
ni2
n2

∣∣∣∣, (5.2)
where nij is the number of examples from class i with attribute value Vj and
nj is the number of examples with attribute value Vj .

Consider the training set for the loan classification problem shown in Figure
5.9. Use the MVDM measure to compute the distance between every pair of
attribute values for the Home Owner and Marital Status attributes.

Answer:

The training set shown in Figure 5.9 can be summarized for the Home Owner
and Marital Status attributes as follows.

Marital Status
Class Single Married Divorced
Yes 2 0 1
No 2 4 1

Home Owner
Class Yes No
Yes 0 3
No 3 4

d(Single, Married) = 1

d(Single, Divorced) = 0

d(Married, Divorced) = 1

d(Refund=Yes, Refund=No) = 6/7

15. For each of the Boolean functions given below, state whether the problem is
linearly separable.

(a) A AND B AND C
Answer: Yes

(b) NOT A AND B
Answer: Yes

(d) (A XOR B) AND (A OR B)
Answer: No

16. (a) Demonstrate how the perceptron model can be used to represent the
AND and OR functions between a pair of Boolean variables.
Answer:
Let x1 and x2 be a pair of Boolean variables and y be the output. For
AND function, a possible perceptron model is:

y = sgn
[
x1 + x2 − 1.5

]
.

For OR function, a possible perceptron model is:

y = sgn
[
x1 + x2 − 0.5

]
.

(b) Comment on the disadvantage of using linear functions as activation
functions for multilayer neural networks.
Answer:
Multilayer neural networks is useful for modeling nonlinear relation-
ships between the input and output attributes. However, if linear func-
tions are used as activation functions (instead of sigmoid or hyperbolic
tangent function), the output is still a linear combination of its input
attributes. Such a network is just as expressive as a perceptron.

17. You are asked to evaluate the performance of two classification models, M1
and M2. The test set you have chosen contains 26 binary attributes, labeled
as A through Z.

Table 5.5 shows the posterior probabilities obtained by applying the models to
the test set. (Only the posterior probabilities for the positive class are shown).
As this is a two-class problem, P (−) = 1 − P (+) and P (−|A, . . . , Z) = 1 −
P (+|A, . . . , Z). Assume that we are mostly interested in detecting instances
from the positive class.

62 Chapter 5 Classification: Alternative Techniques

Table 5.5. Posterior probabilities for Exercise 17.

Instance True Class P (+|A, . . . , Z, M1) P (+|A, . . . , Z, M2)
1 + 0.73 0.61
2 + 0.69 0.03
3 − 0.44 0.68
4 − 0.55 0.31
5 + 0.67

0.45

6 + 0.47 0.09
7 − 0.08 0.38
8 − 0.15

0.05

9 + 0.45 0.01
10 − 0.35 0.04

(a) Plot the ROC curve for both M1 and M2. (You should plot them on
the same graph.) Which model do you think is better? Explain your
reasons.
Answer:
The ROC curve for M1 and M2 are shown in the Figure 5.5.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0

0.1
0.2

0.3

0.4
0.5
0.6

0.7

0.8
0.9
1

M1 M2

FPR

TPR

Figure 5.5. ROC curve.

M1 is better, since its area under the ROC curve is larger than the area
under ROC curve for M2.

(b) For model M1, suppose you choose the cutoff threshold to be t = 0.5.
In other words, any test instances whose posterior probability is greater
than t will be classified as a positive example. Compute the precision,
recall, and F-measure for the model at this threshold value.

When t = 0.5, the confusion matrix for M1 is shown below.

+ –

Actual + 3 2
– 1 4

Precision = 3/4 = 75%.
Recall = 3/5 = 60%.
F-measure = (2 × .75 × .6)/(.75 + .6) = 0.667.

(c) Repeat the analysis for part (c) using the same cutoff threshold on
model M2. Compare the F -measure results for both models. Which
model is better? Are the results consistent with what you expect from
the ROC curve?
Answer:
When t = 0.5, the confusion matrix for M2 is shown below.

+ –

Actual + 1 4
– 1 4

Precision = 1/2 = 50%.
Recall = 1/5 = 20%.
F-measure = (2 × .5 × .2)/(.5 + .2) = 0.2857.
Based on F-measure, M1 is still better than M2. This result is consis-
tent with the ROC plot.

(d) Repeat part (c) for model M1 using the threshold t = 0.1. Which
threshold do you prefer, t = 0.5 or t = 0.1? Are the results consistent
with what you expect from the ROC curve?
Answer:
When t = 0.1, the confusion matrix for M1 is shown below.

+ –

Actual + 5 0
– 4 1

Precision = 5/9 = 55.6%.
Recall = 5/5 = 100%.
F-measure = (2 × .556 × 1)/(.556 + 1) = 0.715.
According to F-measure, t = 0.1 is better than t = 0.5.
When t = 0.1, FPR = 0.8 and TPR = 1. On the other hand, when
t = 0.5, FPR = 0.2 and TRP = 0.6. Since (0.2, 0.6) is closer to the
point (0, 1), we favor t = 0.5. This result is inconsistent with the results
using F-measure. We can also show this by computing the area under
the ROC curve

64 Chapter 5 Classification: Alternative Techniques

For t = 0.5, area = 0.6 × (1 − 0.2) = 0.6 × 0.8 = 0.48.
For t = 0.1, area = 1 × (1 − 0.8) = 1 × 0.2 = 0.2.
Since the area for t = 0.5 is larger than the area for t = 0.1, we prefer
t = 0.5.

18. Following is a data set that contains two attributes, X and Y , and two class
labels, “+” and “−”. Each attribute can take three different values: 0, 1, or
2.

X Y
Number of
Instances
+ −

0 0 0 100
1 0 0 0
2 0 0 100
0 1 10 100
1 1 10 0
2 1 10 100
0 2 0 100
1 2 0 0
2 2 0 100

The concept for the “+” class is Y = 1 and the concept for the “−” class is
X = 0 ∨ X = 2.

(a) Build a decision tree on the data set. Does the tree capture the “+”
and “−” concepts?
Answer:
There are 30 positive and 600 negative examples in the data. Therefore,
at the root node, the error rate is

Eorig = 1 − max(30/630, 600/630) = 30/630.
If we split on X, the gain in error rate is:

X = 0 X = 1 X = 2
+ 10 10 10
− 300 0 300

EX=0 = 10/310
EX=1 = 0
EX=2 = 10/310

∆X = Eorig − 310630

10
310

− 10
630

0 − 310
630

10
310

= 10/630.

If we split on Y , the gain in error rate is:

Y = 0 Y = 1 Y = 2
+ 0 30 0
− 200 200 200

EY =0 = 0
EY =1 = 30/230
EY =2 = 0

∆Y = Eorig − 230630
30
230

= 0.

Therefore, X is chosen to be the first splitting attribute. Since the
X = 1 child node is pure, it does not require further splitting. We may
use attribute Y to split the impure nodes, X = 0 and X = 2, as follows:

• The Y = 0 and Y = 2 nodes contain 100 − instances.
• The Y = 1 node contains 100 − and 10 + instances.

In all three cases for Y , the child nodes are labeled as −. The resulting
concept is

class =
{

+, X = 1;
−, otherwise.

(b) What are the accuracy, precision, recall, and F1-measure of the decision
tree? (Note that precision, recall, and F1-measure are defined with
respect to the “+” class.)
Answer: The confusion matrix on the training data:

Predicted
+ −

Actual + 10 20− 0 600

accuracy :
610
630

= 0.9683

precision :
10
10

= 1.0

recall :
10
30

= 0.3333

F − measure : 2 ∗ 0.3333 ∗ 1.0
1.0 + 0.3333

= 0.5

C(i, j) =

⎧⎨
⎩

0, if i = j;
1, if i = +, j = −;
Number of − instances
Number of + instances , if i = −, j = +.

(Hint: only the leaves of the old decision tree need to be changed.)
Does the decision tree capture the “+” concept?
Answer:
The cost matrix can be summarized as follows:

66 Chapter 5 Classification: Alternative Techniques

Predicted
+ −

Actual + 0 600/30=20− 1 0
The decision tree in part (a) has 7 leaf nodes, X = 1, X = 0 ∧ Y = 0,
X = 0 ∧ Y = 1, X = 0 ∧ Y = 2, X = 2 ∧ Y = 0, X = 2 ∧ Y = 1, and
X = 2 ∧ Y = 2. Only X = 0 ∧ Y = 1 and X = 2 ∧ Y = 1 are impure
nodes. The cost of misclassifying these impure nodes as positive class
is:

10 ∗ 0 + 1 ∗ 100 = 100
while the cost of misclassifying them as negative class is:

10 ∗ 20 + 0 ∗ 100 = 200.

These nodes are therefore labeled as +.
The resulting concept is

class =
{

+, X = 1 ∨ (X = 0 ∧ Y = 1) ∨ (X = 2 ∧ Y = 2);
−, otherwise.

(d) What are the accuracy, precision, recall, and F1-measure of the new
decision tree?
Answer:
The confusion matrix of the new tree

Predicted
+ −

Actual + 30 0− 200 400

accuracy :
430
630

= 0.6825

precision :
30
230

= 0.1304

recall :
30
30

= 1.0

F − measure : 2 ∗ 0.1304 ∗ 1.0
1.0 + 0.1304

= 0.2307

19. (a) Consider the cost matrix for a two-class problem. Let C(+,+) =
C(−,−) = p, C(+,−) = C(−,+) = q, and q > p. Show that min-
imizing the cost function is equivalent to maximizing the classifier’s
accuracy.
Answer:

Confusion Matrix + −
+ a b
− c d

Cost Matrix + –
+ p q
− q p

The total cost is F = p(a + d) + q(b + c).
Since acc = a+dN , where N = a + b + c + d, we may write

F = N
[
acc(p − q) + q].

Because p − q is negative, minimizing the total cost is equivalent to
maximizing accuracy.

(b) Show that a cost matrix is scale-invariant. For example, if the cost
matrix is rescaled from C(i, j) −→ βC(i, j), where β is the scaling
factor, the decision threshold (Equation 5.82) will remain unchanged.
Answer:
The cost matrix is:

Cost Matrix + −
+ c(+,+) c(+,−)
− c(−,+) c(−,−)

A node t is classified as positive if:

c(+,−)p(+|t) + c(−,−)p(−|t) > c(−,+)p(−|t) + c(+,+)p(+|t)
=⇒ c(+,−)p(+|t) + c(−,−)[1 − p(+|t)] > c(−,+)[1 − p(+|t)] + c(+,+)p(+|t)
=⇒ p(+|t) > c(−,+) − c(−,−)

[c(−,+) − c(−,−)] + [c(+,−) − c(+,+)]

The transformed cost matrix is:

Cost Matrix + −
+ βc(+,+) βc(+,−)
− βc(−,+) βc(−,−)

Therefore, the decision rule is:

p(+|t) > βc(−,+) − βc(−,−)
[βc(−,+) − βc(−,−)] + [βc(+,−) − βc(+,+)]

=
c(−,+) − c(−,−)

[c(−,+) − c(−,−)] + [c(+,−) − c(+,+)]
which is the same as the original decision rule.

(c) Show that a cost matrix is translation-invariant. In other words, adding
a constant factor to all entries in the cost matrix will not affect the
decision threshold (Equation 5.82).
Answer:
The transformed cost matrix is:

68 Chapter 5 Classification: Alternative Techniques

Cost Matrix + −
+ c(+,+) + β c(+,−) + β
− c(−,+) + β c(−,−) + β

Therefore, the decision rule is:

p(+|t) > β + c(−,+) − β − c(−,−)
[β + c(−,+) − β − c(−,−)] + [β + c(+,−) − β − c(+,+)]

=
c(−,+) − c(−,−)
[c(−,+) − c(−,−)] + [c(+,−) − c(+,+)]
which is the same as the original decision rule.

20. Consider the task of building a classifier from random data, where the at-
tribute values are generated randomly irrespective of the class labels. Assume
the data set contains records from two classes, “+” and “−.” Half of the data
set is used for training while the remaining half is used for testing.

(a) Suppose there are an equal number of positive and negative records in
the data and the decision tree classifier predicts every test record to be
positive. What is the expected error rate of the classifier on the test
data?
Answer: 50%.

(b) Repeat the previous analysis assuming that the classifier predicts each
test record to be positive class with probability 0.8 and negative class
with probability 0.2.
Answer: 50%.

(c) Suppose two-thirds of the data belong to the positive class and the
remaining one-third belong to the negative class. What is the expected
error of a classifier that predicts every test record to be positive?
Answer: 33%.

(d) Repeat the previous analysis assuming that the classifier predicts each
test record to be positive class with probability 2/3 and negative class
with probability 1/3.
Answer: 44.4%.

21. Derive the dual Lagrangian for the linear SVM with nonseparable data where
the objective function is

f(w) =
‖w‖2

2
+ C

( N∑
i=1

ξi

.
Answer:

LD =
N∑

i=1

λi − 12
∑
i,j

λiλjyiyjxi · xj − C
(∑

i
ξi

)2
.

Notice that the dual Lagrangian depends on the slack variables ξi’s.

22. Consider the XOR problem where there are four training points:

(1, 1,−), (1, 0,+), (0, 1,+), (0, 0,−).

Transform the data into the following feature space:

Φ = (1,
√

2×1,
√

2×2,
√

2x1x2, x21, x
2
2).

Find the maximum margin linear decision boundary in the transformed space.

Answer:

The decision boundary is f(x1, x2) = x1x2.

23. Given the data sets shown in Figures 5.6, explain how the decision tree, näıve
Bayes, and k-nearest neighbor classifiers would perform on these data sets.

Answer:

(a) Both decision tree and NB will do well on this data set because the
distinguishing attributes have better discriminating power than noise
attributes in terms of entropy gain and conditional probability. k-NN
will not do as well due to relatively large number of noise attributes.

(b) NB will not work at all with this data set due to attribute dependency.
Other schemes will do better than NB.

(c) NB will do very well in this data set, because each discriminating at-
tribute has higher conditional probability in one class over the other
and the overall classification is done by multiplying these individual
conditional probabilities. Decision tree will not do as well, due to the
relatively large number of distinguishing attributes. It will have an
overfitting problem. k-NN will do reasonably well.

(d) k-NN will do well on this data set. Decision trees will also work, but
will result in a fairly large decision tree. The first few splits will be quite
random, because it may not find a good initial split at the beginning.
NB will not perform quite as well due to the attribute dependency.

(e) k-NN will do well on this data set. Decision trees will also work, but
will result in a large decision tree. If decision tree uses an oblique split
instead of just vertical and horizontal splits, then the resulting decision
tree will be more compact and highly accurate. NB will not perform
quite as well due to attribute dependency.

(f) kNN works the best. NB does not work well for this data set due to
attribute dependency. Decision tree will have a large tree in order to
capture the circular decision boundaries.

70 Chapter 5 Classification: Alternative Techniques

Distinguishing
Attributes Noise Attributes

Class A
Class B
Records
Attributes

(a) Synthetic data set 1.

Distinguishing Attributes Noise Attributes

Class A
Class B
Records
Attributes

(b) Synthetic data set 2.

Distinguishing
Attribute set 1 Noise Attributes

Class A
Class B
Records
Attributes

Distinguishing
Attribute set 2

60% filled
with 1

40% filled
with 1

Class A Class B Class A Class B Class A

Class A Class B Class A Class BClass B

Class A Class B Class A Class B Class A
Class A Class B Class A Class BClass B

Attribute X

A
tt

ri
bu

te
Y

(d) Synthetic data set 4

Attribute X
A
tt
ri
bu
te
Y
Class A
Class B

(e) Synthetic data set 5.

Attribute X
A
tt
ri
bu
te
Y
Class A
Class B
Class B

(f) Synthetic data set 6.

Figure 5.6. Data set for Exercise 23.

Association Analysis:
Basic Concepts and
Algorithms

1. For each of the following questions, provide an example of an association rule
from the market basket domain that satisfies the following conditions. Also,
describe whether such rules are subjectively interesting.

(a) A rule that has high support and high confidence.
Answer: Milk −→ Bread. Such obvious rule tends to be uninteresting.

(b) A rule that has reasonably high support but low confidence.
Answer: Milk −→ Tuna. While the sale of tuna and milk may be
higher than the support threshold, not all transactions that contain milk
also contain tuna. Such low-confidence rule tends to be uninteresting.

(c) A rule that has low support and low confidence.
Answer: Cooking oil −→ Laundry detergent. Such low confidence rule
tends to be uninteresting.

(d) A rule that has low support and high confidence.
Answer: Vodka −→ Caviar. Such rule tends to be interesting.

2. Consider the data set shown in Table 6.1.

(a) Compute the support for itemsets {e}, {b, d}, and {b, d, e} by treating
each transaction ID as a market basket.
Answer:

72 Chapter 6 Association Analysis

Table 6.1. Example of market basket transactions.

Customer ID Transaction ID Items Bought
1 0001 {a, d, e}
1 0024 {a, b, c, e}
2 0012 {a, b, d, e}
2 0031 {a, c, d, e}
3 0015 {b, c, e}
3 0022 {b, d, e}
4 0029 {c, d}
4 0040 {a, b, c}
5 0033 {a, d, e}
5 0038 {a, b, e}

s({e}) = 8
10

= 0.8

s({b, d}) = 2
10

= 0.2

s({b, d, e}) = 2
10

= 0.2

(6.1)

(b) Use the results in part (a) to compute the confidence for the association
rules {b, d} −→ {e} and {e} −→ {b, d}. Is confidence a symmetric
measure?
Answer:

c(bd −→ e) = 0.2
0.2

= 100%

c(e −→ bd) = 0.2
0.8

= 25%

No, confidence is not a symmetric measure.
(c) Repeat part (a) by treating each customer ID as a market basket. Each

item should be treated as a binary variable (1 if an item appears in at
least one transaction bought by the customer, and 0 otherwise.)
Answer:

s({e}) = 4
5

= 0.8

s({b, d}) = 5
5

= 1

s({b, d, e}) = 4
5

= 0.8

(d) Use the results in part (c) to compute the confidence for the association
rules {b, d} −→ {e} and {e} −→ {b, d}.
Answer:

c(bd −→ e) = 0.8
1

= 80%

c(e −→ bd) = 0.8
0.8

= 100%

(e) Suppose s1 and c1 are the support and confidence values of an associa-
tion rule r when treating each transaction ID as a market basket. Also,
let s2 and c2 be the support and confidence values of r when treating
each customer ID as a market basket. Discuss whether there are any
relationships between s1 and s2 or c1 and c2.
Answer:
There are no apparent relationships between s1, s2, c1, and c2.

3. (a) What is the confidence for the rules ∅ −→ A and A −→ ∅?
Answer:
c(∅ −→ A) = s(∅ −→ A).
c(A −→ ∅) = 100%.

(b) Let c1, c2, and c3 be the confidence values of the rules {p} −→ {q},
{p} −→ {q, r}, and {p, r} −→ {q}, respectively. If we assume that c1,
c2, and c3 have different values, what are the possible relationships that
may exist among c1, c2, and c3? Which rule has the lowest confidence?
Answer:
c1 =

s(p∪q)

s(p)

c2 =
s(p∪q∪r)

s(p)

c3 =
s(p∪q∪r)

s(p∪r)
Considering s(p) ≥ s(p ∪ q) ≥ s(p ∪ q ∪ r)
Thus: c1 ≥ c2 & c3 ≥ c2.
Therefore c2 has the lowest confidence.

(c) Repeat the analysis in part (b) assuming that the rules have identical
support. Which rule has the highest confidence?
Answer:
Considering s(p ∪ q) = s(p ∪ q ∪ r)
but s(p) ≥ s(p ∪ r)
Thus: c3 ≥ (c1 = c2)
Either all rules have the same confidence or c3 has the highest confi-
dence.

74 Chapter 6 Association Analysis

(d) Transitivity: Suppose the confidence of the rules A −→ B and B −→ C
are larger than some threshold, minconf . Is it possible that A −→ C
has a confidence less than minconf?
Answer:
Yes, It depends on the support of items A, B, and C.
For example:
s(A,B) = 60% s(A) = 90%
s(A,C) = 20% s(B) = 70%
s(B,C) = 50% s(C) = 60%

Let minconf = 50% Therefore:
c(A → B) = 66% > minconf
c(B → C) = 71% > minconf
But c(A → C) = 22% < minconf

4. For each of the following measures, determine whether it is monotone, anti-
monotone, or non-monotone (i.e., neither monotone nor anti-monotone).

Example: Support, s = σ(X)|T | is anti-monotone because s(X) ≥
s(Y ) whenever X ⊂ Y .

(a) A characteristic rule is a rule of the form {p} −→ {q1, q2, . . . , qn}, where
the rule antecedent contains only a single item. An itemset of size k can
produce up to k characteristic rules. Let ζ be the minimum confidence
of all characteristic rules generated from a given itemset:

ζ({p1, p2, . . . , pk}) = min
[

c
({p1} −→ {p2, p3, . . . , pk}), . . .

c
({pk} −→ {p1, p3 . . . , pk−1}) ]

Is ζ monotone, anti-monotone, or non-monotone?
Answer:
ζ is an anti-monotone measure because

ζ({A1, A2, · · · , Ak}) ≥ ζ({A1, A2, · · · , Ak, Ak+1}) (6.2)

For example, we can compare the values of ζ for {A,B} and {A,B,C}.

ζ({A,B}) = min (c(A −→ B), c(B −→ A))
= min

(s(A,B)
s(A)

,
s(A,B)
s(B)

)
=

s(A,B)
max(s(A), s(B))

(6.3)

ζ({A,B,C}) = min (c(A −→ BC), c(B −→ AC), c(C −→ AB))
= min

(s(A,B,C)
s(A)

,
s(A,B,C)

s(B)
,
s(A,B,C)

s(C)
)

=
s(A,B,C)

max(s(A), s(B), s(C))
(6.4)

Since s(A,B,C) ≤ s(A,B) and max(s(A), s(B), s(C)) ≥ max(s(A), s(B)),
therefore ζ({A,B}) ≥ ζ({A,B,C}).

(b) A discriminant rule is a rule of the form {p1, p2, . . . , pn} −→ {q}, where
the rule consequent contains only a single item. An itemset of size k can
produce up to k discriminant rules. Let η be the minimum confidence
of all discriminant rules generated from a given itemset:

η({p1, p2, . . . , pk}) = min
[

c
({p2, p3, . . . , pk} −→ {p1}), . . .

c
({p1, p2, . . . pk−1} −→ {pk}) ]

Is η monotone, anti-monotone, or non-monotone?
Answer:
η is non-monotone. We can show this by comparing η({A,B}) against
η({A,B,C}).

η({A,B}) = min (c(A −→ B), c(B −→ A))
= min

(s(A,B)
s(A)
,
s(A,B)
s(B)
)
=
s(A,B)
max(s(A), s(B))

(6.5)

η({A,B,C}) = min (c(AB −→ C), c(AC −→ B), c(BC −→ A))
= min

(s(A,B,C)
s(A,B)

,
s(A,B,C)
s(A,C)

,
s(A,B,C)
s(B,C)

)
=

s(A,B,C)
max(s(A,B), s(A,C), s(B,C))

(6.6)

Since s(A,B,C) ≤ s(A,B) and max(s(A,B), s(A,C), s(B,C)) ≤ max(s(A), s(B)),
therefore η({A,B,C}) can be greater than or less than η({A,B}).
Hence, the measure is non-monotone.

76 Chapter 6 Association Analysis

Answer:
Let

ζ ′({A1, A2, · · · , Ak}) = max( c(A1 −→ A2, A3, · · · , Ak), · · ·
c(Ak −→ A1, A3 · · · , Ak−1))

ζ ′({A,B}) = max (c(A −→ B), c(B −→ A))
= max

(s(A,B)
s(A)
,
s(A,B)
s(B)
)
=

s(A,B)
min(s(A), s(B))

(6.7)

ζ ′({A,B,C}) = max (c(A −→ BC), c(B −→ AC), c(C −→ AB))
= max

(s(A,B,C)
s(A)
,
s(A,B,C)
s(B)
,
s(A,B,C)
s(C)
)
=
s(A,B,C)

min(s(A), s(B), s(C))
(6.8)

Since s(A,B,C) ≤ s(A,B) and min(s(A), s(B), s(C)) ≤ min(s(A), s(B)),
ζ ′({A,B,C}) can be greater than or less than ζ ′({A,B}). Therefore,
the measure is non-monotone.
Let

η′({A1, A2, · · · , Ak}) = max( c(A2, A3, · · · , Ak −→ A1), · · ·
c(A1, A2, · · ·Ak−1 −→ Ak))

η′({A,B}) = max (c(A −→ B), c(B −→ A))
= max

(s(A,B)
s(A)
,
s(A,B)
s(B)
)
=
s(A,B)
min(s(A), s(B))

(6.9)

η({A,B,C}) = max (c(AB −→ C), c(AC −→ B), c(BC −→ A))
= max

(s(A,B,C)
s(A,B)
,
s(A,B,C)
s(A,C)
,
s(A,B,C)
s(B,C)
)
=

s(A,B,C)
min(s(A,B), s(A,C), s(B,C))

(6.10)

Since s(A,B,C) ≤ s(A,B) and min(s(A,B), s(A,C), s(B,C)) ≤ min(s(A), s(B), s(C))
≤ min(s(A), s(B)), η′({A,B,C}) can be greater than or less than η′({A,B}).
Hence, the measure is non-monotone.

5. Prove Equation 6.3. (Hint: First, count the number of ways to create an
itemset that forms the left hand side of the rule. Next, for each size k
itemset selected for the left-hand side, count the number of ways to choose
the remaining d − k items to form the right-hand side of the rule.)
Answer:

Suppose there are d items. We first choose k of the items to form the left-
hand side of the rule. There are

(
d

)
ways for doing this. After selecting the

items for the left-hand side, there are
(
d−k

)
ways to choose the remaining

items to form the right hand side of the rule, where 1 ≤ i ≤ d− k. Therefore
the total number of rules (R) is:

=
d∑

k=1
(
d
k

) d−k∑
i=1

(
d − k

i
)
=
d∑
k=1
(
d
k

)(
2d−k − 1

)
=
d∑
k=1
(
d
k

)
2d−k −

d∑
k=1
(
d
k
)
=
d∑
k=1
(
d
k
)
2d−k −

[
2d + 1

]
,

where
n∑

i=1

(
n

)
= 2n − 1.

Since

(1 + x)d =
d∑

i=1
(
d
i

)
xd−i + xd,

substituting x = 2 leads to:

3d =
d∑

i=1
(
d
i

)
2d−i + 2d.

Therefore, the total number of rules is:

R = 3d − 2d −
[
2d + 1

]
= 3d − 2d+1 + 1.

78 Chapter 6 Association Analysis

Table 6.2. Market basket transactions.

Transaction ID Items Bought
1 {Milk, Beer, Diapers}
2 {Bread, Butter, Milk}
3 {Milk, Diapers, Cookies}
4 {Bread, Butter, Cookies}
5 {Beer, Cookies, Diapers}
6 {Milk, Diapers, Bread, Butter}
7 {Bread, Butter, Diapers}
8 {Beer, Diapers}
9 {Milk, Diapers, Bread, Butter}
10 {Beer, Cookies}

6. Consider the market basket transactions shown in Table 6.2.

(a) What is the maximum number of association rules that can be extracted
from this data (including rules that have zero support)?
Answer: There are six items in the data set. Therefore the total
number of rules is 602.

(b) What is the maximum size of frequent itemsets that can be extracted
(assuming minsup > 0)?
Answer: Because the longest transaction contains 4 items, the maxi-
mum size of frequent itemset is 4.

(
6
3

)
= 20.

(d) Find an itemset (of size 2 or larger) that has the largest support.
Answer: {Bread, Butter}.

(e) Find a pair of items, a and b, such that the rules {a} −→ {b} and
{b} −→ {a} have the same confidence.
Answer: (Beer, Cookies) or (Bread, Butter).

7. Consider the following set of frequent 3-itemsets:

{1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {2, 3, 4}, {2, 3, 5}, {3, 4, 5}.
Assume that there are only five items in the data set.

(a) List all candidate 4-itemsets obtained by a candidate generation proce-
dure using the Fk−1 × F1 merging strategy.
Answer:
{1, 2, 3, 4},{1, 2, 3, 5},{1, 2, 3, 6}.
{1, 2, 4, 5},{1, 2, 4, 6},{1, 2, 5, 6}.

{1, 3, 4, 5},{1, 3, 4, 6},{2, 3, 4, 5}.
{2, 3, 4, 6},{2, 3, 5, 6}.

(b) List all candidate 4-itemsets obtained by the candidate generation pro-
cedure in Apriori.
Answer:
{1, 2, 3, 4}, {1, 2, 3, 5}, {1, 2, 4, 5}, {2, 3, 4, 5}, {2, 3, 4, 6}.

8. The Apriori algorithm uses a generate-and-count strategy for deriving fre-
quent itemsets. Candidate itemsets of size k +1 are created by joining a pair
of frequent itemsets of size k (this is known as the candidate generation step).
A candidate is discarded if any one of its subsets is found to be infrequent
during the candidate pruning step. Suppose the Apriori algorithm is applied
to the data set shown in Table 6.3 with minsup = 30%, i.e., any itemset
occurring in less than 3 transactions is considered to be infrequent.

Table 6.3. Example of market basket transactions.

Transaction ID Items Bought
1 {a, b, d, e}
2 {b, c, d}
3 {a, b, d, e}
4 {a, c, d, e}
5 {b, c, d, e}
6 {b, d, e}
7 {c, d}
8 {a, b, c}
9 {a, d, e}
10 {b, d}

(a) Draw an itemset lattice representing the data set given in Table 6.3.
Label each node in the lattice with the following letter(s):

• N: If the itemset is not considered to be a candidate itemset by
the Apriori algorithm. There are two reasons for an itemset not to
be considered as a candidate itemset: (1) it is not generated at all
during the candidate generation step, or (2) it is generated during

80 Chapter 6 Association Analysis

the candidate generation step but is subsequently removed during
the candidate pruning step because one of its subsets is found to
be infrequent.

• F: If the candidate itemset is found to be frequent by the Apriori
algorithm.

• I: If the candidate itemset is found to be infrequent after support
counting.

Answer:

The lattice structure is shown below.

null

AB AC AD AE BC BD BE CD CE DE

A B C D E

ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE

ABCD ABCE ABDE ACDE BCDE

ABCDE

F F F F F

F I F F F F F F

N I I N N F I N F N

N N N N N

N
F

Figure 6.1. Solution.

(b) What is the percentage of frequent itemsets (with respect to all itemsets
in the lattice)?
Answer:
Percentage of frequent itemsets = 16/32 = 50.0% (including the null
set).

(c) What is the pruning ratio of the Apriori algorithm on this data set?
(Pruning ratio is defined as the percentage of itemsets not considered
to be a candidate because (1) they are not generated during candidate
generation or (2) they are pruned during the candidate pruning step.)
Answer:

{258}
{289}

{356}
{689}

{568}{168} {367}{346}
{379}
{678}

{459}
{456}
{789}

{125}
{158}
{458}

2,5,8

1,4,7

1,4,7
1,4,7

1,4,7

3,6,9

3,6,9
3,6,9

2,5,8
2,5,8

2,5,8 1,4,7

3,6,9
2,5,8

L1 L5 L6 L7 L8 L9 L11 L12

L2 L3

{246}
{278}

{145}
{178}

{127}
{457}

Figure 6.2. An example of a hash tree structure.

Pruning ratio is the ratio of N to the total number of itemsets. Since
the count of N = 11, therefore pruning ratio is 11/32 = 34.4%.

(d) What is the false alarm rate (i.e, percentage of candidate itemsets that
are found to be infrequent after performing support counting)?
Answer:
False alarm rate is the ratio of I to the total number of itemsets. Since
the count of I = 5, therefore the false alarm rate is 5/32 = 15.6%.

9. The Apriori algorithm uses a hash tree data structure to efficiently count
the support of candidate itemsets. Consider the hash tree for candidate 3-
itemsets shown in Figure 6.2.

(a) Given a transaction that contains items {1, 3, 4, 5, 8}, which of the hash
tree leaf nodes will be visited when finding the candidates of the trans-
action?
Answer:
The leaf nodes visited are L1, L3, L5, L9, and L11.

(b) Use the visited leaf nodes in part (b) to determine the candidate item-
sets that are contained in the transaction {1, 3, 4, 5, 8}.
Answer:
The candidates contained in the transaction are {1, 4, 5}, {1, 5, 8}, and
{4, 5, 8}.

10. Consider the following set of candidate 3-itemsets:

{1, 2, 3}, {1, 2, 6}, {1, 3, 4}, {2, 3, 4}, {2, 4, 5}, {3, 4, 6}, {4, 5, 6}

82 Chapter 6 Association Analysis

(a) Construct a hash tree for the above candidate 3-itemsets. Assume the
tree uses a hash function where all odd-numbered items are hashed
to the left child of a node, while the even-numbered items are hashed
to the right child. A candidate k-itemset is inserted into the tree by
hashing on each successive item in the candidate and then following the
appropriate branch of the tree according to the hash value. Once a leaf
node is reached, the candidate is inserted based on one of the following
conditions:

Condition 1: If the depth of the leaf node is equal to k (the root is
assumed to be at depth 0), then the candidate is inserted regardless
of the number of itemsets already stored at the node.

Condition 2: If the depth of the leaf node is less than k, then the
candidate can be inserted as long as the number of itemsets stored
at the node is less than maxsize. Assume maxsize = 2 for this
question.

Condition 3: If the depth of the leaf node is less than k and the
number of itemsets stored at the node is equal to maxsize, then
the leaf node is converted into an internal node. New leaf nodes
are created as children of the old leaf node. Candidate itemsets
previously stored in the old leaf node are distributed to the children
based on their hash values. The new candidate is also hashed to
its appropriate leaf node.

Answer:

1 3 4
2 3 4 2 4 5

4 5 6

1 2 3 1 2 6

3 4 6

Figure 6.3. Hash tree for Exercise 10.

83
null

ba c d e

decebeaeadacab

abc abd abe

abcd

acd

abcde

abce abde acde bcde

ace ade bcd bce bde cde

bdbc cd

Figure 6.4. An itemset lattice

(b) How many leaf nodes are there in the candidate hash tree? How many
internal nodes are there?
Answer: There are 5 leaf nodes and 4 internal nodes.

(c) Consider a transaction that contains the following items: {1, 2, 3, 5, 6}.
Using the hash tree constructed in part (a), which leaf nodes will be
checked against the transaction? What are the candidate 3-itemsets
contained in the transaction?
Answer: The leaf nodes L1, L2, L3, and L4 will be checked against
the transaction. The candidate itemsets contained in the transaction
include {1,2,3} and {1,2,6}.

11. Given the lattice structure shown in Figure 6.4 and the transactions given in
Table 6.3, label each node with the following letter(s):

• M if the node is a maximal frequent itemset,
• C if it is a closed frequent itemset,
• N if it is frequent but neither maximal nor closed, and
• I if it is infrequent.

Assume that the support threshold is equal to 30%.

84 Chapter 6 Association Analysis

Answer:
The lattice structure is shown below.
null
AB AC AD AE BC BD BE CD CE DE
A B C D E
ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE
ABCD ABCE ABDE ACDE BCDE
ABCDE

C C C C F

MC
I F F M

C
C F M

C
I

I I I

I I M
C

I I M
C
I

I I I I I

I
C

Figure 6.5. Solution for Exercise 11.

12. The original association rule mining formulation uses the support and confi-
dence measures to prune uninteresting rules.

(a) Draw a contingency table for each of the following rules using the trans-
actions shown in Table 6.4.

Rules: {b} −→ {c}, {a} −→ {d}, {b} −→ {d}, {e} −→ {c},
{c} −→ {a}.

Answer:
c c

b 3 4
b 2 1

d d
a 4 1
a 5 0

d d
b 6 1
b 3 0

c c
e 2 4
e 3 1

a a

c 2 3
c 3 2

(b) Use the contingency tables in part (a) to compute and rank the rules
in decreasing order according to the following measures.

Table 6.4. Example of market basket transactions.

Transaction ID Items Bought
1 {a, b, d, e}
2 {b, c, d}
3 {a, b, d, e}
4 {a, c, d, e}
5 {b, c, d, e}
6 {b, d, e}
7 {c, d}
8 {a, b, c}
9 {a, d, e}
10 {b, d}

i. Support.
Answer:

Rules Support Rank
b −→ c 0.3 3
a −→ d 0.4 2
b −→ d 0.6 1
e −→ c 0.2 4
c −→ a 0.2 4

ii. Confidence.
Answer:

Rules Confidence Rank
b −→ c 3/7 3
a −→ d 4/5 2
b −→ d 6/7 1
e −→ c 2/6 5
c −→ a 2/5 4

iii. Interest(X −→ Y ) = P (X,Y )P (X) P (Y ).
Answer:

Rules Interest Rank
b −→ c 0.214 3
a −→ d 0.72 2
b −→ d 0.771 1
e −→ c 0.167 5
c −→ a 0.2 4

iv. IS(X −→ Y ) = P (X,Y )√
P (X)P (Y )

.
Answer:

86 Chapter 6 Association Analysis

Rules IS Rank
b −→ c 0.507 3
a −→ d 0.596 2
b −→ d 0.756 1
e −→ c 0.365 5
c −→ a 0.4 4

v. Klosgen(X −→ Y ) = √P (X,Y )×(P (Y |X)−P (Y )), where P (Y |X) =
P (X,Y )
P (X) .

Answer:

Rules Klosgen Rank
b −→ c -0.039 2
a −→ d -0.063 4
b −→ d -0.033 1
e −→ c -0.075 5
c −→ a -0.045 3

vi. Odds ratio(X −→ Y ) = P (X,Y )P (X,Y )
P (X,Y )P (X,Y )

.
Answer:

Rules Odds Ratio Rank
b −→ c 0.375 2
a −→ d 0 4
b −→ d 0 4
e −→ c 0.167 3
c −→ a 0.444 1

13. Given the rankings you had obtained in Exercise 12, compute the correla-
tion between the rankings of confidence and the other five measures. Which
measure is most highly correlated with confidence? Which measure is least
correlated with confidence?

Answer:

Correlation(Confidence, Support) = 0.97.

Correlation(Confidence, Interest) = 1.

Correlation(Confidence, IS) = 1.

Correlation(Confidence, Klosgen) = 0.7.

Correlation(Confidence, Odds Ratio) = -0.606.

Interest and IS are the most highly correlated with confidence, while odds
ratio is the least correlated.

14. Answer the following questions using the data sets shown in Figure 6.6.
Note that each data set contains 1000 items and 10,000 transactions. Dark
cells indicate the presence of items and white cells indicate the absence of

items. We will apply the Apriori algorithm to extract frequent itemsets with
minsup = 10% (i.e., itemsets must be contained in at least 1000 transac-
tions)?

(a) Which data set(s) will produce the most number of frequent itemsets?
Answer: Data set (e) because it has to generate the longest frequent
itemset along with its subsets.

(b) Which data set(s) will produce the fewest number of frequent itemsets?
Answer: Data set (d) which does not produce any frequent itemsets
at 10% support threshold.

(d) Which data set(s) will produce frequent itemsets with highest maximum
support?
Answer: Data set (b).

(e) Which data set(s) will produce frequent itemsets containing items with
wide-varying support levels (i.e., items with mixed support, ranging
from less than 20% to more than 70%).
Answer: Data set (e).

15. (a) Prove that the φ coefficient is equal to 1 if and only if f11 = f1+ = f+1.
Answer:
Instead of proving f11 = f1+ = f+1, we will show that P (A,B) =
P (A) = P (B), where P (A,B) = f11/N , P (A) = f1+/N , and P (B) =
f+1/N . When the φ-coefficient equals to 1:

φ =
P (A,B) − P (A)P (B)√

P (A)P (B)
[
1 − P (A)][1 − P (B)] = 1

The preceding equation can be simplified as follows:[
P (A,B) − P (A)P (B)

]2
= P (A)P (B)

[
1 − P (A)][1 − P (B)]

P (A,B)2 − 2P (A,B)P (A)P (B) = P (A)P (B)[1 − P (A) − P (B)]
P (A,B)2 = P (A)P (B)

[
1 − P (A) − P (B) + 2P (A,B)]

We may rewrite the equation in terms of P (B) as follows:

P (A)P (B)2 − P (A)[1 − P (A) + 2P (A,B)]P (B) + P (A,B)2 = 0
The solution to the quadratic equation in P (B) is:

P (B) =
P (A)β − √P (A)2β2 − 4P (A)P (A,B)2

2P (A)
,

88 Chapter 6 Association Analysis

Tr
an

sa
ct

io
ns

2000

4000

6000

600 800400200

8000

Items

2000
4000
6000
600 800400200
8000
Items

(a)

(b)

Tr
an
sa
ct
io
ns
2000
4000
6000
600 800400200
8000
Items

(c)

2000
4000
6000
600 800400200
8000
Items

(d)

Tr
an
sa
ct
io
ns
Tr
an
sa
ct
io
ns
Tr
an
sa
ct
io
ns
Tr
an
sa
ct
io
ns
2000
4000
6000
600 800400200
8000
Items

(e)

2000
4000
6000
600 800400200
8000
Items

(f)

10% are 1s
90% are 0s

(uniformly distributed)

Figure 6.6. Figures for Exercise 14.

where β = 1 − P (A) + 2P (A,B). Note that the second solution, in
which the second term on the left hand side is positive, is not a feasible
solution because it corresponds to φ = −1. Furthermore, the solution
for P (B) must satisfy the following constraint: P (B) ≥ P (A,B). It
can be shown that:

P (B) − P (A,B)

=
1 − P (A)

2
−

√
(1 − P (A))2 + 4P (A,B)(1 − P (A))(1 − P (A,B)/P (A))

2
≤ 0

Because of the constraint, P (B) = P (A,B), which can be achieved by
setting P (A,B) = P (A).

(b) Show that if A and B are independent, then P (A,B) × P (A,B) =
P (A,B) × P (A,B).
Answer:
When A and B are independent, P (A,B) = P (A) × P (B) or equiva-
lently:

P (A,B) − P (A)P (B) = 0
P (A,B) − [P (A,B) + P (A,B)][P (A,B) + P (A,B)] = 0
P (A,B)[1 − P (A,B) − P (A,B) − P (A,B)] − P (A,B)P (A,B) = 0
P (A,B)P (A,B) − P (A,B)P (A,B) = 0.

Q =
[
f11f00 − f10f01
f11f00 + f10f01

]

Y =
[√

f11f00 −
√

f10f01√
f11f00 +

√
f10f01

]

are normalized versions of the odds ratio.
Answer:
Odds ratio can be written as:

α =
f11f00
f10f01

We can express Q and Y in terms of α as follows:

Q =
α − 1
α + 1

Y =
√

α − 1√
α + 1

90 Chapter 6 Association Analysis

In both cases, Q and Y increase monotonically with α. Furthermore,
when α = 0, Q = Y = −1 to represent perfect negative correlation.
When α = 1, which is the condition for attribute independence, Q =
Y = 1. Finally, when α = ∞, Q = Y = +1. This suggests that Q and
Y are normalized versions of α.

(d) Write a simplified expression for the value of each measure shown in
Tables 6.11 and 6.12 when the variables are statistically independent.
Answer:

Measure Value under independence
φ-coefficient 0
Odds ratio 1
Kappa κ 0
Interest 1

Cosine, IS
√

P (A,B)
Piatetsky-Shapiro’s 0
Collective strength 1

Jaccard 0 · · · 1
Conviction 1

Certainty factor 0
Added value 0

16. Consider the interestingness measure, M = P (B|A)−P (B)1−P (B) , for an association
rule A −→ B.
(a) What is the range of this measure? When does the measure attain its

maximum and minimum values?
Answer:
The range of the measure is from 0 to 1. The measure attains its max-
imum value when P (B|A) = 1 and its minimum value when P (B|A) =
P (B).

(b) How does M behave when P (A,B) is increased while P (A) and P (B)
remain unchanged?
Answer:
The measure can be rewritten as follows:

P (A,B) − P (A)P (B)
P (A)(1 − P (B)) .

It increases when P (A,B) is increased.
(c) How does M behave when P (A) is increased while P (A,B) and P (B)

remain unchanged?
Answer:
The measure decreases with increasing P (A).

(d) How does M behave when P (B) is increased while P (A,B) and P (A)
remain unchanged?
Answer:
The measure decreases with increasing P (B).

(e) Is the measure symmetric under variable permutation?

Answer: No.

(f) What is the value of the measure when A and B are statistically inde-
pendent?
Answer: 0.

(g) Is the measure null-invariant?
Answer: No.

(h) Does the measure remain invariant under row or column scaling oper-
ations?
Answer: No.

(i) How does the measure behave under the inversion operation?
Answer: Asymmetric.

17. Suppose we have market basket data consisting of 100 transactions and 20
items. If the support for item a is 25%, the support for item b is 90% and the
support for itemset {a, b} is 20%. Let the support and confidence thresholds
be 10% and 60%, respectively.

(a) Compute the confidence of the association rule {a} → {b}. Is the rule
interesting according to the confidence measure?
Answer:
Confidence is 0.2/0.25 = 80%. The rule is interesting because it exceeds
the confidence threshold.

(b) Compute the interest measure for the association pattern {a, b}. De-
scribe the nature of the relationship between item a and item b in terms
of the interest measure.
Answer:
The interest measure is 0.2/(0.25 × 0.9) = 0.889. The items are nega-
tively correlated according to interest measure.

(c) What conclusions can you draw from the results of parts (a) and (b)?
Answer:
High confidence rules may not be interesting.

(d) Prove that if the confidence of the rule {a} −→ {b} is less than the
support of {b}, then:

i. c({a} −→ {b}) > c({a} −→ {b}),
ii. c({a} −→ {b}) > s({b}),

92 Chapter 6 Association Analysis

where c(·) denote the rule confidence and s(·) denote the support of an
itemset.
Answer:
Let

c({a} −→ {b}) = P ({a, b})
P ({a}) < P ({b}),

which implies that

P ({a})P ({b}) > P ({a, b}).
Furthermore,

c({a} −→ {b}) = P ({a, b})
P ({a}) =

P ({b}) − P ({a, b})
1 − P ({a})

i. Therefore, we may write

c({a} −→ {b}) − c({a} −→ {b}) = P ({b}) − P ({a, b})
1 − P ({a}) −

P ({a, b})
P ({a})

=
P ({a})P ({b}) − P ({a, b})

P ({a})(1 − P ({a}))
which is positive because P ({a})P ({b}) > P ({a, b}).

ii. We can also show that

c({a} −→ {b}) − s({b}) = P ({b}) − P ({a, b})
1 − P ({a}) − P ({b})

=
P ({a})P ({b}) − P ({a, b})

1 − P ({a})
is always positive because P ({a})P ({b}) > P ({a, b}).

18. Table 6.5 shows a 2× 2× 2 contingency table for the binary variables A and
B at different values of the control variable C.

(a) Compute the φ coefficient for A and B when C = 0, C = 1, and

C = 0

or 1. Note that φ({A,B}) = P (A,B)−P (A)P (B)√

P (A)P (B)(1−P (A))(1−P (B)) .

Answer:

i. When C = 0, φ(A,B) = −1/3.
ii. When C = 1, φ(A,B) = 1.
iii. When C = 0 or C = 1, φ = 0.

(b) What conclusions can you draw from the above result?
Answer:
The result shows that some interesting relationships may disappear if
the confounding factors are not taken into account.

Table 6.5. A Contingency Table.

A
C = 0

C = 1

B
B
1
1
0
0
0
5
1
15
0
15
0
0
30
15

Table 6.6. Contingency tables for Exercise 19.

B B B B

A 9 1 A 89 1

A 1 89 A 1 9

(a) Table I. (b) Table II.

19. Consider the contingency tables shown in Table 6.6.

(a) For table I, compute support, the interest measure, and the φ correla-
tion coefficient for the association pattern {A, B}. Also, compute the
confidence of rules A → B and B → A.
Answer:
s(A) = 0.1, s(B) = 0.9, s(A,B) = 0.09.
I(A,B) = 9, φ(A,B) = 0.89.
c(A −→ B) = 0.9, c(B −→ A) = 0.9.

(b) For table II, compute support, the interest measure, and the φ correla-
tion coefficient for the association pattern {A, B}. Also, compute the
confidence of rules A → B and B → A.
Answer:
s(A) = 0.9, s(B) = 0.9, s(A,B) = 0.89.
I(A,B) = 1.09, φ(A,B) = 0.89.
c(A −→ B) = 0.98, c(B −→ A) = 0.98.

(c) What conclusions can you draw from the results of (a) and (b)?
Answer:
Interest, support, and confidence are non-invariant while the φ-coefficient
is invariant under the inversion operation. This is because φ-coefficient

94 Chapter 6 Association Analysis

takes into account the absence as well as the presence of an item in a
transaction.

20. Consider the relationship between customers who buy high-definition televi-
sions and exercise machines as shown in Tables 6.19 and 6.20.

(a) Compute the odds ratios for both tables.
Answer:
For Table 6.19, odds ratio = 1.4938.
For Table 6.20, the odds ratios are 0.8333 and 0.98.

(b) Compute the φ-coefficient for both tables.
Answer:
For table 6.19, φ = 0.098.
For Table 6.20, the φ-coefficients are -0.0233 and -0.0047.

(c) Compute the interest factor for both tables.
Answer:
For Table 6.19, I = 1.0784.
For Table 6.20, the interest factors are 0.88 and 0.9971.

For each of the measures given above, describe how the direction of associa-
tion changes when data is pooled together instead of being stratified.

Answer:

The direction of association changes sign (from negative to positive corre-
lated) when the data is pooled together.

Association Analysis:
Advanced Concepts

1. Consider the traffic accident data set shown in Table 7.1.

Table 7.1. Traffic accident data set.

Weather Driver’s Traffic Seat Belt Crash
Condition Condition Violation Severity

Good Alcohol-impaired Exceed speed limit No Major
Bad Sober None Yes Minor
Good Sober Disobey stop sign Yes Minor
Good Sober Exceed speed limit Yes Major
Bad Sober Disobey traffic signal No Major
Good Alcohol-impaired Disobey stop sign Yes Minor
Bad Alcohol-impaired None Yes Major
Good Sober Disobey traffic signal Yes Major
Good Alcohol-impaired None No Major
Bad Sober Disobey traffic signal No Major
Good Alcohol-impaired Exceed speed limit Yes Major
Bad Sober Disobey stop sign Yes Minor

(a) Show a binarized version of the data set.
Answer: See Table 7.2.

(b) What is the maximum width of each transaction in the binarized data?
Answer: 5

96 Chapter 7 Association Analysis: Advanced Concepts

Table 7.2. Traffic accident data set.

Good Bad Alcohol Sober Exceed None Disobey Disobey Belt Belt Major Minor
speed stop traffic = No = Yes

1 0 1 0 1 0 0 0 1 0 1 0
0 1 0 1 0 1 0 0 0 1 0 1
1 0 0 1 0 0 1 0 0 1 0 1
1 0 0 1 1 0 0 0 0 1 1 0
0 1 0 1 0 0 0 1 1 0 1 0
1 0 1 0 0 0 1 0 0 1 0 1
0 1 1 0 0 1 0 0 0 1 1 0
1 0 0 1 0 0 0 1 0 1 1 0
1 0 1 0 0 1 0 0 1 0 1 0
0 1 0 1 0 0 0 1 1 0 1 0
1 0 1 0 1 0 0 0 0 1 1 0
0 1 0 1 0 0 1 0 0 1 0 1

Answer: 5
The number of candidate itemsets from size 1 to size 3 is 10+28+3 = 41.
The number of frequent itemsets from size 1 to size 3 is 8+10+0 = 18.

(d) Create a data set that contains only the following asymmetric binary at-
tributes: (Weather = Bad, Driver’s condition = Alcohol-impaired,
Traffic violation = Yes, Seat Belt = No, Crash Severity = Major).
For Traffic violation, only None has a value of 0. The rest of the
attribute values are assigned to 1. Assuming that support threshold is
30%, how many candidate and frequent itemsets will be generated?
Answer:
The binarized data is shown in Table 7.3.

Table 7.3. Traffic accident data set.
Bad Alcohol Traffic Belt Major

Impaired violation = No
0 1 1 1 1
1 0 0 0 0
0 0 1 0 0
0 0 1 0 1
1 0 1 1 1
0 1 1 0 0
1 1 0 0 1
0 0 1 0 1
0 1 0 1 1
1 0 1 1 1
0 1 1 0 1
1 0 1 0 0

The number of candidate itemsets from size 1 to size 3 is 5+10+0 = 15.

The number of frequent itemsets from size 1 to size 3 is 5 + 3 + 0 = 8.

(e) Compare the number of candidate and frequent itemsets generated in
parts (c) and (d).

Answer:

The second method produces less number of candidate and frequent itemsets.

2. (a) Consider the data set shown in Table 7.4. Suppose we apply the fol-
lowing discretization strategies to the continuous attributes of the data
set.

D1: Partition the range of each continuous attribute into 3 equal-sized
bins.

D2: Partition the range of each continuous attribute into 3 bins; where
each bin contains an equal number of transactions

For each strategy, answer the following questions:

i. Construct a binarized version of the data set.
ii. Derive all the frequent itemsets having support ≥ 30%.

Table 7.4. Data set for Exercise 2.

TID Temperature Pressure Alarm 1 Alarm 2 Alarm 3
1 95 1105 0 0 1
2 85 1040 1 1 0
3 103 1090 1 1 1
4 97 1084 1 0 0
5 80 1038 0 1 1
6 100 1080 1 1 0
7 83 1025 1 0 1
8 86 1030 1 0 0
9 101 1100 1 1 1

Answer:
Table 7.5 shows the discretized data using D1, where the discretized
intervals are:

• X1: Temperature between 80 and 87,
• X2: Temperature between 88 and 95,
• X3: Temperature between 96 and 103,
• Y1: Pressure between 1025 and 1051,
• Y2: Pressure between 1052 and 1078,
• Y3: Pressure between 1079 and 1105.

98 Chapter 7 Association Analysis: Advanced Concepts

Table 7.5. Discretized data using D1.

TID X1 X2 X3 Y1 Y2 Y3 Alarm1 Alarm2 Alarm3
1 0 1 0 0 0 1 0 0 1
2 1 0 0 1 0 0 1 1 0
3 0 0 1 0 0 1 1 1 1
4 0 0 1 0 0 1 1 0 0
5 1 0 0 1 0 0 0 1 1
6 0 0 1 0 0 1 1 1 0
7 1 0 0 1 0 0 1 0 1
8 1 0 0 1 0 0 1 0 0
9 0 0 1 0 0 1 1 1 1

Table 7.6. Discretized data using D2.

TID X1 X2 X3 Y1 Y2 Y3 Alarm1 Alarm2 Alarm3
1 0 1 0 0 0 1 0 0 1
2 1 0 0 0 1 0 1 1 0
3 0 0 1 0 0 1 1 1 1
4 0 1 0 0 1 0 1 0 0
5 1 0 0 1 0 0 0 1 1
6 0 0 1 0 1 0 1 1 0
7 1 0 0 1 0 0 1 0 1
8 0 1 0 1 0 0 1 0 0
9 0 0 1 0 0 1 1 1 1

Table 7.6 shows the discretized data using D1, where the discretized
intervals are:

• X1: Temperature between 80 and 85,
• X2: Temperature between 86 and 97,
• X3: Temperature between 100 and 103,
• Y1: Pressure between 1025 and 1038,
• Y2: Pressure between 1039 and 1084,
• Y3: Pressure between 1085 and 1105.

For D1, there are 7 frequent 1-itemset, 12 frequent 2-itemset, and 5
frequent 3-itemset.
For D2, there are 9 frequent 1-itemset, 7 frequent 2-itemset, and 1
frequent 3-itemset.

(b) The continuous attribute can also be discretized using a clustering ap-
proach.

i. Plot a graph of temperature versus pressure for the data points
shown in Table 7.4.

Answer:
The graph of Temperature and Pressure is shown below.

Pressure vs

Temperature

1020

1030

1040

1050

1060

1070

1080

1090

1100

1110

75 80 85 90 95 100 105

Temperature

P
re

ss
u

re
C1
C2

Figure 7.1. Temperature versus Pressure.

ii. How many natural clusters do you observe from the graph? Assign
a label (C1, C2, etc.) to each cluster in the graph.
Answer: There are two natural clusters in the data.

iii. What type of clustering algorithm do you think can be used to
identify the clusters? State your reasons clearly.
Answer: K-means algorithm.

iv. Replace the temperature and pressure attributes in Table 7.4 with
asymmetric binary attributes C1, C2, etc. Construct a transaction
matrix using the new attributes (along with attributes Alarm1,
Alarm2, and Alarm3).
Answer:

Table 7.7. Example of numeric data set.

TID C1 C2 Alarm1 Alarm2 Alarm3
1 0 1 0 0 1
2 1 0 1 1 0
3 0 1 1 1 1
4 0 1 1 0 0
5 1 0 0 1 1
6 0 1 1 1 0
7 1 0 1 0 1
8 1 0 1 0 0
9 0 1 1 1 1

100 Chapter 7 Association Analysis: Advanced Concepts

v. Derive all the frequent itemsets having support ≥ 30% from the
binarized data.
Answer:
There are 5 frequent 1-itemset, 7 frequent 2-itemset, and 1 frequent
3-itemset.

3. Consider the data set shown in Table 7.8. The first attribute is continuous,
while the remaining two attributes are asymmetric binary. A rule is consid-
ered to be strong if its support exceeds 15% and its confidence exceeds 60%.
The data given in Table 7.8 supports the following two strong rules:

(i) {(1 ≤ A ≤ 2), B = 1} → {C = 1}
(ii) {(5 ≤ A ≤ 8), B = 1} → {C = 1}

Table 7.8. Data set for Exercise 3.

A B C
1 1 1
2 1 1
3 1 0
4 1 0
5 1 1
6 0 1
7 0 0
8 1 1
9 0 0
10 0 0
11 0 0
12 0 1

(a) Compute the support and confidence for both rules.
Answer:
s({(1 ≤ A ≤ 2), B = 1} → {C = 1}) = 1/6
c({(1 ≤ A ≤ 2), B = 1} → {C = 1}) = 1
s({(5 ≤ A ≤ 8), B = 1} → {C = 1}) = 1/6
c({(5 ≤ A ≤ 8), B = 1} → {C = 1}) = 1

(b) To find the rules using the traditional Apriori algorithm, we need to
discretize the continuous attribute A. Suppose we apply the equal width
binning approach to discretize the data, with bin-width = 2, 3, 4. For
each bin-width, state whether the above two rules are discovered by
the Apriori algorithm. (Note that the rules may not be in the same
exact form as before because it may contain wider or narrower intervals

101

for A.) For each rule that corresponds to one of the above two rules,
compute its support and confidence.
Answer:
When bin − width = 2:

Table 7.9. A Synthetic Data set

A1 A2 A3 A4 A5 A6 B C
1 0 0 0 0 0 1 1
1 0 0 0 0 0 1 1
0 1 0 0 0 0 1 0
0 1 0 0 0 0 1 0
0 0 1 0 0 0 1 1
0 0 1 0 0 0 0 1
0 0 0 1 0 0 0 0
0 0 0 1 0 0 1 1
0 0 0 0 1 0 0 0
0 0 0 0 1 0 0 0
0 0 0 0 0 1 0 0
0 0 0 0 0 1 0 1

Where

A1 = 1 ≤ A ≤ 2; A2 = 3 ≤ A ≤ 4;
A3 = 5 ≤ A ≤ 6; A4 = 7 ≤ A ≤ 8;

A5 = 9 ≤ A ≤ 10; A6 = 11 ≤ A ≤ 12;

For the first rule, there is one corresponding rule:

{A1 = 1, B = 1} → {C = 1}
s(A1 = 1, B = 1} → {C = 1}) = 1/6
c(A1 = 1, B = 1} → {C = 1}) = 1

Since the support and confidence are greater than the thresholds, the
rule can be discovered.
For the second rule, there are two corresponding rules:

{A3 = 1, B = 1} → {C = 1}
{A4 = 1, B = 1} → {C = 1}

For both rules, the support is 1/12 and the confidence is 1. Since
the support is less than the threshold (15%), these rules cannot be
generated.

102 Chapter 7 Association Analysis: Advanced Concepts

When bin − width = 3:

Table 7.10. A Synthetic Data set

A1 A2 A3 A4 B C
1 0 0 0 1 1
1 0 0 0 1 1
1 0 0 0 1 0
0 1 0 0 1 0
0 1 0 0 1 1
0 1 0 0 0 1
0 0 1 0 0 0
0 0 1 0 1 1
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 1 0 0
0 0 0 1 0 1

Where

A1 = 1 ≤ A ≤ 3; A2 = 4 ≤ A ≤ 6;
A3 = 7 ≤ A ≤ 9; A4 = 10 ≤ A ≤ 12;

For the first rule, there is one corresponding rule:

{A1 = 1, B = 1} → {C = 1}
s(A1 = 1, B = 1} → {C = 1}) = 1/6
c(A1 = 1, B = 1} → {C = 1}) = 2/3

Since the support and confidence are greater than the thresholds, the
rule can be discovered. The discovered rule is in general form than the
original rule.
For the second rule, there are two corresponding rules:

{A2 = 1, B = 1} → {C = 1}
{A3 = 1, B = 1} → {C = 1}

For both rules, the support is 1/12 and the confidence is 1. Since
the support is less than the threshold (15%), these rules cannot be
generated.

103

When bin − width = 4:

Table 7.11. A Synthetic Data set

A1 A2 A3 B C
1 0 0 1 1
1 0 0 1 1
1 0 0 1 0
1 0 0 1 0
0 1 0 1 1
0 1 0 0 1
0 1 0 0 0
0 1 0 1 1
0 0 1 0 0
0 0 1 0 0
0 0 1 0 0
0 0 1 0 1

Where

A1 = 1 ≤ A ≤ 4; A2 = 5 ≤ A ≤ 8;
A3 = 9 ≤ A ≤ 12;

For the first rule, there is one correspomding rule:

{A1 = 1, B = 1} → {C = 1}
s(A1 = 1, B = 1} → {C = 1}) = 1/6
c(A1 = 1, B = 1} → {C = 1}) = 1/2

Since the confidence is less than the threshold (60%), then the rule
cannot be generated.
For the second rule, there is one corresponding rule:

{A2 = 1, B = 1} → {C = 1}
s(A2 = 1, B = 1} → {C = 1}) = 1/6
c(A2 = 1, B = 1} → {C = 1}) = 1

Since the support and threshold are greater than thresholds, the the
rule can be discovered.

(c) Comment on the effectiveness of using the equal width approach for
classifying the above data set. Is there a bin-width that allows you to

104 Chapter 7 Association Analysis: Advanced Concepts

find both rules satisfactorily? If not, what alternative approach can you
take to ensure that you will find both rules?
Answer:

None of the discretization methods can effectively find both rules. One
approach to ensure that you can find both rules is to start with bin
width equals to 2 and consider all possible mergings of the adjacent
intervals. For example, the discrete intervals are:
1 <= A <= 2, 3 <= A <= 4, 5 <= A <= 6, ·, 11 <= A <= 12 1 <= A <= 4, 5 <= A <= 8 , 9 <= A <= 12

4. Consider the data set shown in Table 7.12.

Table 7.12. Data set for Exercise 4.

Age Number of Hours Online per Week (B)
(A) 0 – 5 5 – 10 10 – 20 20 – 30 30 – 40

10 – 15 2 3 5 3 2
15 – 25 2 5 10 10 3
25 – 35 10 15 5 3 2
35 – 50 4 6 5 3 2

(a) For each combination of rules given below, specify the rule that has the
highest confidence.

i. 15 < A < 25 −→ 10 < B < 20, 10 < A < 25 −→ 10 < B < 20, and 15 < A < 35 −→ 10 < B < 20. Answer: Both 15 < A < 25 −→ 10 < B < 20 and 10 < A < 25 −→ 10 < B < 20 have confidence 33.3%.

ii. 15 < A < 25 −→ 10 < B < 20, 15 < A < 25 −→ 5 < B < 20, and 15 < A < 25 −→ 5 < B < 30. Answer: The rule 15 < A < 25 −→ 5 < B < 30 has the highest confidence (83.3%).

iii. 15 < A < 25 −→ 10 < B < 20 and 10 < A < 35 −→ 5 < B < 30. Answer: The rule 10 < A < 35 −→ 5 < B < 30 has the highest confidence (73.8%).

(b) Suppose we are interested in finding the average number of hours spent
online per week by Internet users between the age of 15 and 35. Write
the corresponding statistics-based association rule to characterize the
segment of users. To compute the average number of hours spent online,

105

approximate each interval by its midpoint value (e.g., use B = 7.5 to
represent the interval 5 < B < 10). Answer: There are 65 people whose average age is between 15 and 35. The average number of hours spent online is

2.5×12/65+7.5×20/65+15×15/65+25×13/65+35×5/65 = 13.92.

Therefore the statistics-based association rule is:

15 ≤ A < 35 −→ B : µ = 13.82.

(c) Test whether the quantitative association rule given in part (b) is sta-
tistically significant by comparing its mean against the average number
of hours spent online by other users who do not belong to the age group.
For other users, the average number of hours spent online is:

2.5 × 6/35 + 7.5 × 9/35 + 15 × 10/35 + 25 × 6/65 + 35 × 4/65 = 14.93.

The standard deviations for the two groups are 9.786 (15 ≤ Age < 35) and 10.203 (Age < 15 or Age ≥ 35), respectively.

Z =
14.93 − 13.82√
9.7862

65 +
10.2032

= 0.476 < 1.64

The difference is not significant at 95% confidence level.

5. For the data set with the attributes given below, describe how you would con-
vert it into a binary transaction data set appropriate for association analysis.
Specifically, indicate for each attribute in the original data set

(a) How many binary attributes it would correspond to in the transaction
data set,

(b) How the values of the original attribute would be mapped to values of
the binary attributes, and

(c) If there is any hierarchical structure in the data values of an attribute
that could be useful for grouping the data into fewer binary attributes.

The following is a list of attributes for the data set along with their possible
values. Assume that all attributes are collected on a per-student basis:

• Year : Freshman, Sophomore, Junior, Senior, Graduate:Masters, Grad-
uate:PhD, Professional
Answer:

106 Chapter 7 Association Analysis: Advanced Concepts

(a) Each attribute value can be represented using an asymmetric bi-
nary attribute. Therefore, there are altogether 7 binary attributes.

(b) There is a one-to-one mapping between the original attribute values
and the asymmetric binary attributes.

• Zip code : zip code for the home address of a U.S. student, zip code
for the local address of a non-U.S. student
Answer:

(a) Each attribute value is represented by an asymmetric binary at-
tribute. Therefore, we have as many asymmetric binary attributes
as the number of distinct zipcodes.

(b) There is a one-to-one mapping between the original attribute values
and the asymmetric binary attributes.

(c) We can have a hierarchical structure based on geographical regions
(e.g., zipcodes can be grouped according to their corresponding
states).

• College : Agriculture, Architecture, Continuing Education, Education,
Liberal Arts, Engineering, Natural Sciences, Business, Law, Medical,
Dentistry, Pharmacy, Nursing, Veterinary Medicine
Answer:

(a) Each attribute value is represented by an asymmetric binary at-
tribute. Therefore, we have as many asymmetric binary attributes
as the number of distinct colleges.

(b) There is a one-to-one mapping between the original attribute values
and the asymmetric binary attributes.

(c) We can have a hierarchical structure based on the type of school.
For example, colleges of Medical and Medical might be grouped
together as Medical school while Engineering and Natural Sciences
might be grouped together into the same school.

• On Campus : 1 if the student lives on campus, 0 otherwise
Answer:

(a) This attribute can be mapped to one binary attribute.
(b) There is no hierarchical structure.

• Each of the following is a separate attribute that has a value of 1 if the
person speaks the language and a value of 0, otherwise.

– Arabic
– Bengali
– Chinese Mandarin
– English

107

– Portuguese
– Russian
– Spanish

Answer:

(a) Each attribute value can be represented by an asymmetric bi-
nary attribute. Therefore, we have as many asymmetric binary
attributes as the number of distinct dialects.

(b) There is a one-to-one mapping between the original attribute values
and the asymmetric binary attributes.

(c) We can have a hierarchical structure based on the region in which
the languages are spoken (e.g., Asian, European, etc.)

6. Consider the data set shown in Table 7.13. Suppose we are interested in
extracting the following association rule:

{α1 ≤ Age ≤ α2,Play Piano = Yes} −→ {Enjoy Classical Music = Yes}

Table 7.13. Data set for Exercise 6.

Age Play Piano Enjoy Classical Music
9 Yes Yes
11 Yes Yes
14 Yes No
17 Yes No
19 Yes Yes
21 No No
25 No No
29 Yes Yes
33 No No
39 No Yes
41 No No
47 No Yes

To handle the continuous attribute, we apply the equal-frequency approach
with 3, 4, and 6 intervals. Categorical attributes are handled by introducing
as many new asymmetric binary attributes as the number of categorical val-
ues. Assume that the support threshold is 10% and the confidence threshold
is 70%.

(a) Suppose we discretize the Age attribute into 3 equal-frequency intervals.
Find a pair of values for α1 and α2 that satisfy the minimum support
and minimum confidence requirements.

108 Chapter 7 Association Analysis: Advanced Concepts

Answer:
(α1 = 19, α2 = 29): s = 16.7%, c = 100%.

(b) Repeat part (a) by discretizing the Age attribute into 4 equal-frequency
intervals. Compare the extracted rules against the ones you had ob-
tained in part (a).
Answer:
No rule satisfies the support and confidence thresholds.

(c) Repeat part (a) by discretizing the Age attribute into 6 equal-frequency
intervals. Compare the extracted rules against the ones you had ob-
tained in part (a).
Answer:
(α1 = 9, α2 = 11): s = 16.7%, c = 100%.

(d) From the results in part (a), (b), and (c), discuss how the choice of
discretization intervals will affect the rules extracted by association rule
mining algorithms.
If the discretization interval is too wide, some rules may not have enough
confidence to be detected by the algorithm. If the discretization interval
is too narrow, the rule in part (a) will be lost.

7. Consider the transactions shown in Table 7.14, with an item taxonomy given
in Figure 7.25.

Table 7.14. Example of market basket transactions.

Transaction ID Items Bought
1 Chips, Cookies, Regular Soda, Ham
2 Chips, Ham, Boneless Chicken, Diet Soda
3 Ham, Bacon, Whole Chicken, Regular Soda
4 Chips, Ham, Boneless Chicken, Diet Soda
5 Chips, Bacon, Boneless Chicken
6 Chips, Ham, Bacon, Whole Chicken, Regular Soda
7 Chips, Cookies, Boneless Chicken, Diet Soda

(a) What are the main challenges of mining association rules with item
taxonomy?
Answer:
Difficulty of deciding the right support and confidence thresholds. Items
residing at higher levels of the taxonomy have higher support than those
residing at lower levels of the taxonomy. Many of the rules may also be
redundant.

109

(b) Consider the approach where each transaction t is replaced by an ex-
tended transaction t′ that contains all the items in t as well as their re-
spective ancestors. For example, the transaction t = { Chips, Cookies}
will be replaced by t′ = {Chips, Cookies, Snack Food, Food}. Use this
approach to derive all frequent itemsets (up to size 4) with support ≥
70%.
Answer:
There are 8 frequent 1-itemsets, 25 frequent 2-itemsets, 34 frequent
3-itemsets and 20 frequent 4-itemsets. The frequent 4-itemsets are:

{Food, Snack Food, Meat, Soda} {Food, Snack Food, Meat, Chips}
{Food, Snack Food, Meat, Pork} {Food, Snack Food, Meat, Chicken}
{Food, Snack Food, Soda, Chips} {Food, Snack Food, Chips, Pork}
{Food, Snack Food, Chips, Chicken} {Food, Meat, Soda, Chips}
{Food, Meat, Soda, Pork} {Food, Meat, Soda, Chicken}
{Food, Meat, Soda, Ham} {Food, Meat, Chips, Pork}
{Food, Meat, Chips, Chicken} {Food, Meat, Pork, Chicken}
{Food, Meat, Pork, Ham} {Food, Soda, Pork, Ham}
{Snack Food, Meat, Soda, Chips} {Snack Food, Meat, Chips, Pork}
{Snack Food, Meat, Chips, Chicken} {Meat, Soda, Pork, Ham}

(c) Consider an alternative approach where the frequent itemsets are gen-
erated one level at a time. Initially, all the frequent itemsets involving
items at the highest level of the hierarchy are generated. Next, we use
the frequent itemsets discovered at the higher level of the hierarchy to
generate candidate itemsets involving items at the lower levels of the hi-
erarchy. For example, we generate the candidate itemset {Chips, Diet
Soda} only if {Snack Food, Soda} is frequent. Use this approach to
derive all frequent itemsets (up to size 4) with support ≥ 70%.
Answer:
There are 8 frequent 1-itemsets, 6 frequent 2-itemsets, and 1 frequent
3-itemset. The frequent 2-itemsets and 3-itemsets are:

{Snack Food, Meat} {Snack Food, Soda}
{Meat, Soda} {Chips, Pork}
{Chips, Chicken} {Pork, Chicken}
{Snack Food, Meat, Soda}

(d) Compare the frequent itemsets found in parts (b) and (c). Comment
on the efficiency and completeness of the algorithms.
Answer:
The method in part (b) is more complete but less efficient compared to
the method in part (c). The method in part (c) is more efficient but
may lose some frequent itemsets.

8. The following questions examine how the support and confidence of an asso-
ciation rule may vary in the presence of a concept hierarchy.

110 Chapter 7 Association Analysis: Advanced Concepts

(a) Consider an item x in a given concept hierarchy. Let x1, x2, . . ., xk
denote the k children of x in the concept hierarchy. Show that s(x) ≤∑k

i=1 s(xi), where s(·) is the support of an item. Under what conditions
will the inequality become an equality?
Answer:
If no transaction contains more than one child of x, then s(x) =

∑k
i=1 s(xi).

(b) Let p and q denote a pair of items, while p̂ and q̂ are their corresponding
parents in the concept hierarchy. If s({p, q}) > minsup, which of the
following itemsets are guaranteed to be frequent? (i) s({p̂, q}), (ii)
s({p, q̂}), and (iii) s({p̂, q̂}).
Answer:
All three itemsets are guaranteed to be frequent.

(c) Consider the association rule {p} −→ {q}. Suppose the confidence of
the rule exceeds minconf . Which of the following rules are guaranteed
to have confidence higher than minconf? (i) {p} −→ {q̂}, (ii) {p̂} −→
{q}, and (iii) {p̂} −→ {q̂}.
Answer:
Only {p} −→ {q̂} is guaranteed to have confidence higher than minconf .

9. (a) List all the 4-subsequences contained in the following data sequence:

< {1, 3} {2} {2, 3} {4} >,
assuming no timing constraints.
Answer:

< {1, 3} {2} {2} > < {1, 3} {2} {3} >
< {1, 3} {2} {4} > < {1, 3}{2, 3} >
< {1, 3} {3} {4} > < {1} {2} {2, 3} >
< {1} {2} {2} {4} > < {1} {2} {3} {4} >
< {1} {2, 3} {4} > < {3} {2} {2, 3} >
< {3} {2} {2} {4} > < {3} {2} {3} {4} >
< {3} {2, 3} {4} > < {2} {2, 3} {4} >

(b) List all the 3-element subsequences contained in the data sequence for
part (a) assuming that no timing constraints are imposed.
Answer:

< {1, 3} {2} {2, 3} > < {1, 3} {2} {4} >
< {1, 3} {3} {4} > < {1, 3} {2} {2} >
< {1, 3} {2} {3} > < {1, 3} {2, 3} {4} >
< {1} {2} {2, 3} > < {1} {2} {4} >
< {1} {3} {4} > < {1} {2} {2} >
< {1} {2} {3} > < {1} {2, 3} {4} >
< {3} {2} {2, 3} > < {3} {2} {4} >
< {3} {3} {4} > < {3} {2} {2} >
< {3} {2} {3} > < {3} {2, 3} {4} >

111

(a)

(assuming the timing constraints are flexible).
Answer:
This will include all the subsequences in part (a) as well as the following:

< {1, 2, 3, 4} > < {1, 2, 3} {2} >
< {1, 2, 3} {3} > < {1, 2, 3} {4} >
< {1, 3} {2, 4} > < {1, 3} {3, 4} >
< {1} {2} {2, 4} > < {1} {2} {3, 4} >
< {3} {2} {2, 4} > < {3} {2} {3, 4} >
< {1, 2} {2, 3} > < {1, 2} {2, 4} >
< {1, 2} {3, 4} > < {1, 2} {2}{4} >
< {1, 2} {3} {4} > < {2, 3} {2, 3} >
< {2, 3} {2, 4} > < {2, 3} {3, 4} >
< {2, 3} {2}{4} > < {2, 3} {3} {4} >
< {1} {2, 3, 4} > < {1} {2} {2, 4} >
< {1} {2} {3, 4} > < {3} {2, 3, 4} >
< {3} {2} {2, 4} > < {3} {2} {3, 4} >
< {2} {2, 3, 4} >

(d) List all the 3-element subsequences contained in the data sequence for
part (a) (assuming the timing constraints are flexible).
Answer:
This will include all the subsequences in part (b) as well as the following:

< {1, 2, 3} {2} {4} > < {1, 2, 3} {3} {4} >
< {1, 2, 3} {2, 3} {4} > < {1, 2} {2} {4} >
< {1, 2} {3} {4} > < {1, 2} {2, 3} {4} >
< {2, 3} {2} {4} > < {2, 3} {3} {4} >
< {2, 3} {2, 3} {4} > < {1} {2} {2, 4} >
< {1} {2} {3, 4} > < {1} {2} {2, 3, 4} >
< {3} {2} {2, 4} > < {3} {2} {3, 4} >
< {3} {2} {2, 3, 4} > < {1, 3} {2} {2, 4} >
< {1, 3} {2} {3, 4} > < {1, 3} {2} {2, 3, 4} >

10. Find all the frequent subsequences with support ≥ 50% given the sequence
database shown in Table 7.15. Assume that there are no timing constraints
imposed on the sequences.

Answer:

< {A} >, < {B} >, < {C} >, < {D} >, < {E} >
< {A} {C} >, < {A} {D} >, < {A} {E} >, < {B} {C} >,
< {B} {D} >, < {B} {E} >, < {C} {D} >, < {C} {E} >, < {D, E} >

11. (a) For each of the sequences w =< e1e2 . . . ei . . . ei+1 . . . elast > given be-
low, determine whether they are subsequences of the sequence

< {1, 2, 3}{2, 4}{2, 4, 5}{3, 5}{6} >

112 Chapter 7 Association Analysis: Advanced Concepts

Table 7.15. Example of event sequences generated by various sensors.

Sensor Timestamp Events
S1 1 A, B

2 C
3 D, E
4 C

S2 1 A, B
2 C, D
3 E

S3 1 B
2 A
3 B
4 D, E

S4 1 C
2 D, E
3 C
4 E

S5 1 B
2 A
3 B, C
4 A, D

subjected to the following timing constraints:

mingap = 0 (interval between last event in ei and first event

in ei+1 is > 0)
maxgap = 3 (interval between first event in ei and last event

in ei+1 is ≤ 3)
maxspan = 5 (interval between first event in e1 and last event

in elast is ≤ 5)

ws = 1 (time between first and last events in ei is ≤ 1)

• w =< {1}{2}{3} >

Answer: Yes.
• w =< {1, 2, 3, 4}{5, 6} >

Answer: No.
• w =< {2, 4}{2, 4}{6} >

Answer: Yes.
• w =< {1}{2, 4}{6} >

Answer: Yes.
• w =< {1, 2}{3, 4}{5, 6} >

Answer: No.

(b) Determine whether each of the subsequences w given in the previous
question are contiguous subsequences of the following sequences s.

113

• s =< {1, 2, 3, 4, 5, 6}{1, 2, 3, 4, 5, 6}{1, 2, 3, 4, 5, 6} >
– w =< {1}{2}{3} >

Answer: Yes.
– w =< {1, 2, 3, 4}{5, 6} >

Answer: Yes.
– w =< {2, 4}{2, 4}{6} >

Answer: Yes.
– w =< {1}{2, 4}{6} >

Answer: Yes.
– w =< {1, 2}{3, 4}{5, 6} >

Answer: Yes.
• s =< {1, 2, 3, 4}{1, 2, 3, 4, 5, 6}{3, 4, 5, 6} >

– w =< {1}{2}{3} >
Answer: Yes.

– w =< {1, 2, 3, 4}{5, 6} >
Answer: Yes.

– w =< {2, 4}{2, 4}{6} >
Answer: Yes.

– w =< {1}{2, 4}{6} >
Answer: Yes.

– w =< {1, 2}{3, 4}{5, 6} >
Answer: Yes.

• s =< {1, 2}{1, 2, 3, 4}{3, 4, 5, 6}{5, 6} >
– w =< {1}{2}{3} >

Answer: Yes.
– w =< {1, 2, 3, 4}{5, 6} >
Answer: Yes.
– w =< {2, 4}{2, 4}{6} >

Answer: No.
– w =< {1}{2, 4}{6} >

Answer: Yes.
– w =< {1, 2}{3, 4}{5, 6} >

Answer: Yes.
• s =< {1, 2, 3}{2, 3, 4, 5}{4, 5, 6} >

– w =< {1}{2}{3} >
Answer: No.

– w =< {1, 2, 3, 4}{5, 6} >
Answer: No.

– w =< {2, 4}{2, 4}{6} >
Answer: No.

– w =< {1}{2, 4}{6} >
Answer: Yes.

114 Chapter 7 Association Analysis: Advanced Concepts

– w =< {1, 2}{3, 4}{5, 6} >
Answer: Yes.

12. For each of the sequence w = 〈e1, . . . , elast〉 below, determine whether they
are subsequences of the following data sequence:

〈{A,B}{C,D}{A,B}{C,D}{A,B}{C,D}〉

subjected to the following timing constraints:

mingap = 0 (interval between last event in ei and first event
in ei+1 is > 0)

maxgap = 2 (interval between first event in ei and last event
in ei+1 is ≤ 2)

maxspan = 6 (interval between first event in e1 and last event
in elast is ≤ 6)

ws = 1 (time between first and last events in ei is ≤ 1)

(a) w = 〈{A}{B}{C}{D}〉
Answer: Yes.

(b) w = 〈{A}{B,C,D}{A}〉
Answer: No.

(d) w = 〈{B,C}{A,D}{B,C}〉
Answer: No.

(e) w = 〈{A,B,C,D}{A,B,C,D}〉
Answer: No.

13. Consider the following frequent 3-sequences:

< {1, 2, 3} >, < {1, 2}{3} >, < {1}{2, 3} >, < {1, 2}{4} >,
< {1, 3}{4} >, < {1, 2, 4} >, < {2, 3}{3} >, < {2, 3}{4} >,
< {2}{3}{3} >, and < {2}{3}{4} >.

(a) List all the candidate 4-sequences produced by the candidate generation
step of the GSP algorithm.
Answer:

< {1, 2, 3} {3} >, < {1, 2, 3} {4} >, < {1, 2} {3} {3} >, < {1, 2} {3} {4} >,
< {1} {2, 3} {3} >, < {1} {2, 3} {4} >.

(b) List all the candidate 4-sequences pruned during the candidate pruning
step of the GSP algorithm (assuming no timing constraints).
Answer:

115

When there is no timing constraints, all subsequences of a candidate
must be frequent. Therefore, the pruned candidates are:

< {1, 2, 3} {3} >, < {1, 2} {3} {3} >, < {1, 2} {3} {4} >,
< {1} {2, 3} {3} >, < {1} {2, 3} {4} >.

(c) List all the candidate 4-sequences pruned during the candidate pruning
step of the GSP algorithm (assuming maxgap = 1).
Answer:
With timing constraint, only contiguous subsequences of a candidate
must be frequent. Therefore, the pruned candidates are:

< {1, 2, 3} {3} >, < {1, 2} {3} {3} >, < {1, 2} {3} {4} >,
< {1} {2, 3} {3} >, < {1} {2, 3} {4} >.

14. Consider the data sequence shown in Table 7.16 for a given object. Count
the number of occurrences for the sequence 〈{p}{q}{r}〉 according to the
following counting methods:

Assume that ws = 0, mingap = 0, maxgap = 3, maxspan = 5).

Table 7.16. Example of event sequence data for Exercise 14.

Timestamp Events
1 p, q
2 r
3 s
4 p, q
5 r, s
6 p
7 q, r
8 q, s
9 p
10 q, r, s

(a) COBJ (one occurrence per object).
Answer: 1.

(b) CWIN (one occurrence per sliding window).
Answer: 2.

(d) CDIST O (distinct occurrences with possibility of event-timestamp over-
lap).
Answer: 3.

116 Chapter 7 Association Analysis: Advanced Concepts

(e) CDIST (distinct occurrences with no event timestamp overlap allowed).
Answer: 2.

15. Describe the types of modifications necessary to adapt the frequent subgraph
mining algorithm to handle:

(a) Directed graphs

(b) Unlabeled graphs

(d) Disconnected graphs

For each type of graph given above, describe which step of the algorithm will
be affected (candidate generation, candidate pruning, and support counting),
and any further optimization that can help improve the efficiency of the
algorithm.

Answer:

(a) Adjacency matrix may not be symmetric, which affects candidate gen-
eration using vertex growing approach.

(b) An unlabeled graph is equivalent to a labeled graph where all the ver-
tices have identical labels.

(c) No effect on algorithm. If the graph is a rooted labeled tree, more
efficient techniques can be developed to encode the tree (see: M.J. Zaki,
Efficiently Mining Frequent Trees in a Forest, In Proc. of the Eighth
ACM SIGKDD Int’l Conf. on Knowledge Discovery and Data Mining,
2002).

16. Draw all candidate subgraphs obtained from joining the pair of graphs shown
in Figure 7.2. Assume the edge-growing method is used to expand the sub-
graphs.

Answer: See Figure 7.3.

17. Draw all the candidate subgraphs obtained by joining the pair of graphs
shown in Figure 7.4. Assume the edge-growing method is used to expand the
subgraphs.

Answer: See Figure 7.5.

18. (a) If support is defined in terms of induced subgraph relationship, show
that the confidence of the rule g1 −→ g2 can be greater than 1 if g1 and
g2 are allowed to have overlapping vertex sets.
Answer:
We illustrate this with an example. Consider the five graphs, G1, G2,
· · · , G5, shown in Figure 7.6. The graph g1 shown on the top-right hand

117

a
b
a

ab a

ab a
a
a
a
b
a
a
a
ab a

ac a

(a)
(b)

Figure 7.2. Graphs for Exercise 16.

a a

b
a a
aa
b
a a
aa

b
a a
aa
b
b
c
(a)
(b)

Figure 7.3. Solution for Exercise 16.

diagram is a subgraph of G1, G3, G4, and G5. Therefore, s(g1) = 4/5 =
80%. Similarly, we can show that s(g2) = 60% because g2 is a subgraph
of G1, G2, and G3, while s(g3) = 40% because g3 is a subgraph of

and G3.
Consider the association rule, g2 −→ g1. Using the standard definition
of confidence as the ratio between the support of g2 ∪ g1 ≡ g3 to the

118 Chapter 7 Association Analysis: Advanced Concepts

b b

(a)
(b)
b ba
b b
b ba
ab a

b aa

ac a

Figure 7.4. Graphs for Exercise 17.

b
bb

ba b

bb
ba
b
bb
ba
b
b
bb
ba
b
b
aa

aa b

aa
aa
b
aa
aa
b
C

b c

b
bb

ba b b

bb
ba

Figure 7.5. Solution for Exercise 17.

support of g2, we obtain a confidence value greater than 1 because
s(g3) > s(g2).

(b) What is the time complexity needed to determine the canonical label
of a graph that contains |V | vertices?
Answer:

119

a
c
b

Subgraph support = 80%
Induced subgraph support = 80%

Subgraph support = 60%
Induced subgraph support = 20%

Subgraph support = 40%
Induced subgraph support = 40%

e
d
1 1

1
a e1

1
1
a
c
b
e

d
1

1
1
1
1
G1

G3 G4

Subgraph g1

Subgraph g2

Subgraph g3

a
a
c
c
b
b
e
e
d
d
1 1
1
1
1

a e

d
1
1
1

11 1

Graph Data Set

a
e

1
a
e
d1
11

Figure 7.6. Computing the support of a subgraph from a set of graphs.

A näıve approach requires |V |! computations to examine all possible
permutations of the canonical label.

(c) The core of a subgraph can have multiple automorphisms. This will
increase the number of candidate subgraphs obtained after merging two
frequent subgraphs that share the same core. Determine the maximum
number of candidate subgraphs obtained due to automorphism of a core
of size k.
Answer: k.

(d) Two frequent subgraphs of size k may share multiple cores. Determine
the maximum number of cores that can be shared by the two frequent
subgraphs.
Answer: k − 1.

19. (a) Consider a graph mining algorithm that uses the edge-growing method
to join the two undirected and unweighted subgraphs shown in Figure
19a.

i. Draw all the distinct cores obtained when merging the two sub-
graphs.
Answer: See Figure 7.7.

120 Chapter 7 Association Analysis: Advanced Concepts

A A

A A
B
A A
A A
B
A A
B
A A
A A
B
A A
A A
B
A

Figure 7.7. Solution to Exercise 19.

ii. How many candidates are generated using the following core?

A A
A A
B

Answer: No candidate k + 1-subgraph can be generated from the
core.

20. The original association rule mining framework considers only presence of
items together in the same transaction. There are situations in which itemsets
that are infrequent may also be informative. For instance, the itemset TV,
DVD, ¬ VCR suggests that many customers who buy TVs and DVDs do not
buy VCRs.

In this problem, you are asked to extend the association rule framework to
negative itemsets (i.e., itemsets that contain both presence and absence of
items). We will use the negation symbol (¬) to refer to absence of items.
(a) A näıve way for deriving negative itemsets is to extend each transaction

to include absence of items as shown in Table 7.17.

i. Suppose the transaction database contains 1000 distinct items.
What is the total number of positive itemsets that can be gener-

121

Table 7.17. Example of numeric data set.

TID TV ¬TV DVD ¬DVD VCR ¬VCR . . .
1 1 0 0 1 0 1 . . .
2 1 0 0 1 0 1 . . .

ated from these items? (Note: A positive itemset does not contain
any negated items).
Answer: 21000 − 1.

ii. What is the maximum number of frequent itemsets that can be
generated from these transactions? (Assume that a frequent item-
set may contain positive, negative, or both types of items)
Answer: 22000 − 1.

iii. Explain why such a näıve method of extending each transaction
with negative items is not practical for deriving negative itemsets.
Answer: The number of candidate itemsets is too large, many
of the them are also redundant and useless (e.g., an itemset that
contains both items x and x).

(b) Consider the database shown in Table 7.14. What are the support and
confidence values for the following negative association rules involving
regular and diet soda?

i. ¬Regular −→ Diet.
Answer: s = 42.9%, c = 100%.

ii. Regular −→ ¬Diet.
Answer: s = 42.9%, c = 100%.

iii. ¬Diet −→ Regular.
Answer: s = 42.9%, c = 100%.

iv. Diet −→ ¬Regular.
Answer: s = 42.9%, c = 100%.

21. Suppose we would like to extract positive and negative itemsets from a data
set that contains d items.

(a) Consider an approach where we introduce a new variable to represent
each negative item. With this approach, the number of items grows
from d to 2d. What is the total size of the itemset lattice, assuming
that an itemset may contain both positive and negative items of the
same variable?
Answer: 22d.

(b) Assume that an itemset must contain positive or negative items of dif-
ferent variables. For example, the itemset {a, a, b, c} is invalid because
it contains both positive and negative items for variable a. What is the
total size of the itemset lattice?
Answer:

∑d
k=1

(
d
k

) ∑k
i=0

(
k
i

)
=
∑d
k=1
(
d
k

)
2k = 3d − 1.

122 Chapter 7 Association Analysis: Advanced Concepts

22. For each type of pattern defined below, determine whether the support mea-
sure is monotone, anti-monotone, or non-monotone (i.e., neither monotone
nor anti-monotone) with respect to increasing itemset size.

(a) Itemsets that contain both positive and negative items such as {a, b, c, d}.
Is the support measure monotone, anti-monotone, or non-monotone
when applied to such patterns?
Answer: Anti-monotone.

(b) Boolean logical patterns such as {(a ∨ b ∨ c), d, e}, which may contain
both disjunctions and conjunctions of items. Is the support measure
monotone, anti-monotone, or non-monotone when applied to such pat-
terns?
Answer: Non-monotone.

23. Many association analysis algorithms rely on an Apriori -like approach for
finding frequent patterns. The overall structure of the algorithm is given
below.

Algorithm 7.1 Apriori -like algorithm.
1: k = 1.
2: Fk = { i | i ∈ I ∧ σ({i})N ≥ minsup}. {Find frequent 1-patterns.}
3: repeat
4: k = k + 1.
5: Ck = genCandidate(Fk−1). {Candidate Generation}
6: Ck = pruneCandidate(Ck, Fk−1). {Candidate Pruning}
7: Ck = count(Ck, D). {Support Counting}
8: Fk = { c | c ∈ Ck ∧ σ(c)N ≥ minsup}. {Extract frequent patterns}
9: until Fk = ∅

10: Answer =
⋃

Fk.

Suppose we are interested in finding boolean logical rules such as

{a ∨ b} −→ {c, d},

which may contain both disjunctions and conjunctions of items. The corre-
sponding itemset can be written as {(a ∨ b), c, d}.

(a) Does the Apriori principle still hold for such itemsets?

(b) How should the candidate generation step be modified to find such
patterns?

123

(d) How should the support counting step be modified to find such pat-
terns?
Answer:
Refer to R. Srikant, Q. Vu, R. Agrawal: Mining Association Rules
with Item Constraints. In Proc of the Third Int’l Conf on Knowledge
Discovery and Data Mining, 1997.

Cluster Analysis:
Basic Concepts and
Algorithms

1. Consider a data set consisting of 220 data vectors, where each vector has
32 components and each component is a 4-byte value. Suppose that vec-
tor quantization is used for compression and that 216 prototype vectors are
used. How many bytes of storage does that data set take before and after
compression and what is the compression ratio?
Before compression, the data set requires 4 × 32 × 220 = 134, 217, 728 bytes.
After compression, the data set requires 4×32×216 = 8, 388, 608 bytes for the
prototype vectors and 2×220 = 2, 097, 152 bytes for vectors, since identifying
the prototype vector associated with each data vector requires only two bytes.
Thus, after compression, 10,485,760 bytes are needed to represent the data.
The compression ratio is 12.8.

2. Find all well-separated clusters in the set of points shown in Figure 8.1.
The solutions are also indicated in Figure 8.1.

Figure 8.1. Points for Exercise 2.

126 Chapter 8 Cluster Analysis: Basic Concepts and Algorithms

3. Many partitional clustering algorithms that automatically determine the
number of clusters claim that this is an advantage. List two situations in
which this is not the case.

(a) When there is hierarchical structure in the data. Most algorithms that
automatically determine the number of clusters are partitional, and
thus, ignore the possibility of subclusters.

(b) When clustering for utility. If a certain reduction in data size is needed,
then it is necessary to specify how many clusters (cluster centroids) are
produced.

4. Given K equally sized clusters, the probability that a randomly chosen initial
centroid will come from any given cluster is 1/K, but the probability that
each cluster will have exactly one initial centroid is much lower. (It should
be clear that having one initial centroid in each cluster is a good starting
situation for K-means.) In general, if there are K clusters and each cluster
has n points, then the probability, p, of selecting in a sample of size K one
initial centroid from each cluster is given by Equation 8.1. (This assumes
sampling with replacement.) From this formula we can calculate, for example,
that the chance of having one initial centroid from each of four clusters is
4!/44 = 0.0938.

p =
number of ways to select one centroid from each cluster

number of ways to select K centroids
=

K!nK

(Kn)K
=

KK
(8.1)

(a) Plot the probability of obtaining one point from each cluster in a sample
of size K for values of K between 2 and 100.

The solution is shown in Figure 4. Note that the probability is essen-
tially 0 by the time K = 10.

(b) For K clusters, K = 10, 100, and 1000, find the probability that a
sample of size 2K contains at least one point from each cluster. You can
use either mathematical methods or statistical simulation to determine
the answer.
We used simulation to compute the answer. Respectively, the proba-
bilities are 0.21, < 10−6, and < 10−6. Proceeding analytically, the probability that a point doesn’t come from a particular cluster is, 1 − 1K , and thus, the probability that all

points don’t come from a particular cluster is (1 − 1K )2K . Hence, the
probability that at least one of the 200 points comes from a particular
cluster is 1 − (1 − 1K )2K . If we assume independence (which is too
optimistic, but becomes approximately true for larger values of K), then
an upper bound for the probability that all clusters are represented in
the final sample is given by (1− (1− 1K )2K)K . The values given by this
bound are 0.27, 5.7e-07, and 8.2e-64, respectively.

127

0 10 20 30 40 50 60 70 80 90 100
0

0.05
0.1

0.15

0.2

0.25

0.3

0.35

0.4
0.45
0.5
K

P
ro

ba
bi

lit
y

Figure 8.2. Probability of at least one point from each cluster. Exercise 4.

5. Identify the clusters in Figure 8.3 using the center-, contiguity-, and density-
based definitions. Also indicate the number of clusters for each case and
give a brief indication of your reasoning. Note that darkness or the number
of dots indicates density. If it helps, assume center-based means K-means,
contiguity-based means single link, and density-based means DBSCAN.

(a) (b)

Figure 8.3. Clusters for Exercise 5.

(a) center-based 2 clusters. The rectangular region will be split in half.
Note that the noise is included in the two clusters.
contiguity-based 1 cluster because the two circular regions will be
joined by noise.
density-based 2 clusters, one for each circular region. Noise will be
eliminated.

(b) center-based 1 cluster that includes both rings.
contiguity-based 2 clusters, one for each rings.
density-based 2 clusters, one for each ring.

128 Chapter 8 Cluster Analysis: Basic Concepts and Algorithms

(c) center-based 3 clusters, one for each triangular region. One cluster is
also an acceptable answer.
contiguity-based 1 cluster. The three triangular regions will be joined
together because they touch.
density-based 3 clusters, one for each triangular region. Even though
the three triangles touch, the density in the region where they touch is
lower than throughout the interior of the triangles.

(d) center-based 2 clusters. The two groups of lines will be split in two.
contiguity-based 5 clusters. Each set of lines that intertwines be-
comes a cluster.
density-based 2 clusters. The two groups of lines define two regions
of high density separated by a region of low density.

6. For the following sets of two-dimensional points, (1) provide a sketch of how
they would be split into clusters by K-means for the given number of clusters
and (2) indicate approximately where the resulting centroids would be. As-
sume that we are using the squared error objective function. If you think that
there is more than one possible solution, then please indicate whether each
solution is a global or local minimum. Note that the label of each diagram
in Figure 8.4 matches the corresponding part of this question, e.g., Figure
8.4(a) goes with part (a).

(a)

(b)

(c)

Local minimum

Global minimum

(d)

Global minimum

Local minimum

(e)

Figure 8.4. Diagrams for Exercise 6.

(a) K = 2. Assuming that the points are uniformly distributed in the circle,
how many possible ways are there (in theory) to partition the points
into two clusters? What can you say about the positions of the two
centroids? (Again, you don’t need to provide exact centroid locations,
just a qualitative description.)

In theory, there are an infinite number of ways to split the circle into
two clusters – just take any line that bisects the circle. This line can

129

make any angle 0◦ ≤ θ ≤ 180◦ with the x axis. The centroids will lie
on the perpendicular bisector of the line that splits the circle into two
clusters and will be symmetrically positioned. All these solutions will
have the same, globally minimal, error.

(b) K = 3. The distance between the edges of the circles is slightly greater
than the radii of the circles.

If you start with initial centroids that are real points, you will necessar-
ily get this solution because of the restriction that the circles are more
than one radius apart. Of course, the bisector could have any angle, as
above, and it could be the other circle that is split. All these solutions
have the same globally minimal error.

The three boxes show the three clusters that will result in the realistic
case that the initial centroids are actual data points.

(d) K = 2.

In both case, the rectangles show the clusters. In the first case, the two
clusters are only a local minimum while in the second case the clusters
represent a globally minimal solution.

(e) K = 3. Hint: Use the symmetry of the situation and remember that
we are looking for a rough sketch of what the result would be.

For the solution shown in the top figure, the two top clusters are en-
closed in two boxes, while the third cluster is enclosed by the regions
defined by a triangle and a rectangle. (The two smaller clusters in the
drawing are supposed to be symmetrical.) I believe that the second
solution—suggested by a student—is also possible, although it is a lo-
cal minimum and might rarely be seen in practice for this configuration
of points. Note that while the two pie shaped cuts out of the larger cir-
cle are shown as meeting at a point, this is not necessarily the case—it
depends on the exact positions and sizes of the circles. There could be a
gap between the two pie shaped cuts which is filled by the third (larger)
cluster. (Imagine the small circles on opposite sides.) Or the boundary
between the two pie shaped cuts could actually be a line segment.

7. Suppose that for a data set

• there are m points and K clusters,
• half the points and clusters are in “more dense” regions,
• half the points and clusters are in “less dense” regions, and
• the two regions are well-separated from each other.

130 Chapter 8 Cluster Analysis: Basic Concepts and Algorithms

For the given data set, which of the following should occur in order to mini-
mize the squared error when finding K clusters:

(a) Centroids should be equally distributed between more dense and less
dense regions.

(b) More centroids should be allocated to the less dense region.

Note: Do not get distracted by special cases or bring in factors other than
density. However, if you feel the true answer is different from any given
above, justify your response.
The correct answer is (c). Less dense regions require more centroids if the
squared error is to be minimized.

8. Consider the mean of a cluster of objects from a binary transaction data
set. What are the minimum and maximum values of the components of the
mean? What is the interpretation of components of the cluster mean? Which
components most accurately characterize the objects in the cluster?

(a) The components of the mean range between 0 and 1.

(b) For any specific component, its value is the fraction of the objects in
the cluster that have a 1 for that component. If we have asymmetric
binary data, such as market basket data, then this can be viewed as
the probability that, for example, a customer in group represented by
the the cluster buys that particular item.

(c) This depends on the type of data. For binary asymmetric data, the
components with higher values characterize the data, since, for most
clusters, the vast majority of components will have values of zero. For
regular binary data, such as the results of a true-false test, the signifi-
cant components are those that are unusually high or low with respect
to the entire set of data.

9. Give an example of a data set consisting of three natural clusters, for which
(almost always) K-means would likely find the correct clusters, but bisecting
K-means would not.
Consider a data set that consists of three circular clusters, that are identical
in terms of the number and distribution of points, and whose centers lie on
a line and are located such that the center of the middle cluster is equally
distant from the other two. Bisecting K-means would always split the middle
cluster during its first iteration, and thus, could never produce the correct
set of clusters. (Postprocessing could be applied to address this.)

10. Would the cosine measure be the appropriate similarity measure to use with
K-means clustering for time series data? Why or why not? If not, what
similarity measure would be more appropriate?

131

Time series data is dense high-dimensional data, and thus, the cosine measure
would not be appropriate since the cosine measure is appropriate for sparse
data. If the magnitude of a time series is important, then Euclidean distance
would be appropriate. If only the shapes of the time series are important,
then correlation would be appropriate. Note that if the comparison of the
time series needs to take in account that one time series might lead or lag
another or only be related to another during specific time periods, then more
sophisticated approaches to modeling time series similarity must be used.

11. Total SSE is the sum of the SSE for each separate attribute. What does it
mean if the SSE for one variable is low for all clusters? Low for just one
cluster? High for all clusters? High for just one cluster? How could you use
the per variable SSE information to improve your clustering?

(a) If the SSE of one attribute is low for all clusters, then the variable is
essentially a constant and of little use in dividing the data into groups.

(b) if the SSE of one attribute is relatively low for just one cluster, then
this attribute helps define the cluster.

(d) If the SSE of an attribute is relatively high for one cluster, then it is
at odds with the information provided by the attributes with low SSE
that define the cluster. It could merely be the case that the clusters
defined by this attribute are different from those defined by the other
attributes, but in any case, it means that this attribute does not help
define the cluster.

(e) The idea is to eliminate attributes that have poor distinguishing power
between clusters, i.e., low or high SSE for all clusters, since they are
useless for clustering. Note that attributes with high SSE for all clusters
are particularly troublesome if they have a relatively high SSE with
respect to other attributes (perhaps because of their scale) since they
introduce a lot of noise into the computation of the overall SSE.

12. The leader algorithm (Hartigan [4]) represents each cluster using a point,
known as a leader, and assigns each point to the cluster corresponding to
the closest leader, unless this distance is above a user-specified threshold. In
that case, the point becomes the leader of a new cluster.
Note that the algorithm described here is not quite the leader algorithm
described in Hartigan, which assigns a point to the first leader that is within
the threshold distance. The answers apply to the algorithm as stated in the
problem.

(a) What are the advantages and disadvantages of the leader algorithm as
compared to K-means?

132 Chapter 8 Cluster Analysis: Basic Concepts and Algorithms

The leader algorithm requires only a single scan of the data and is thus
more computationally efficient since each object is compared to the
final set of centroids at most once. Although the leader algorithm is
order dependent, for a fixed ordering of the objects, it always produces
the same set of clusters. However, unlike K-means, it is not possible
to set the number of resulting clusters for the leader algorithm, except
indirectly. Also, the K-means algorithm almost always produces better
quality clusters as measured by SSE.

(b) Suggest ways in which the leader algorithm might be improved.

Use a sample to determine the distribution of distances between the
points. The knowledge gained from this process can be used to more
intelligently set the value of the threshold.

The leader algorithm could be modified to cluster for several thresholds
during a single pass.

13. The Voronoi diagram for a set of K points in the plane is a partition of all
the points of the plane into K regions, such that every point (of the plane) is
assigned to the closest point among the K specified points. (See Figure 8.5.)
What is the relationship between Voronoi diagrams and K-means clusters?
What do Voronoi diagrams tell us about the possible shapes of K-means
clusters?

(a) If we have K K-means clusters, then the plane is divided into K Voronoi
regions that represent the points closest to each centroid.

(b) The boundaries between clusters are piecewise linear. It is possible to
see this by drawing a line connecting two centroids and then drawing
a perpendicular to the line halfway between the centroids. This per-
pendicular line splits the plane into two regions, each containing points
that are closest to the centroid the region contains.

Figure 8.5. Voronoi diagram for Exercise 13.

133

14. You are given a data set with 100 records and are asked to cluster the data.
You use K-means to cluster the data, but for all values of K, 1 ≤ K ≤ 100,
the K-means algorithm returns only one non-empty cluster. You then apply
an incremental version of K-means, but obtain exactly the same result. How
is this possible? How would single link or DBSCAN handle such data?

(a) The data consists completely of duplicates of one object.

(b) Single link (and many of the other agglomerative hierarchical schemes)
would produce a hierarchical clustering, but which points appear in
which cluster would depend on the ordering of the points and the exact
algorithm. However, if the dendrogram were plotted showing the prox-
imity at which each object is merged, then it would be obvious that the
data consisted of duplicates. DBSCAN would find that all points were
core points connected to one another and produce a single cluster.

15. Traditional agglomerative hierarchical clustering routines merge two clusters
at each step. Does it seem likely that such an approach accurately captures
the (nested) cluster structure of a set of data points? If not, explain how
you might postprocess the data to obtain a more accurate view of the cluster
structure.

(a) Such an approach does not accurately capture the nested cluster struc-
ture of the data. For example, consider a set of three clusters, each of
which has two, three, and four subclusters, respectively. An ideal hi-
erarchical clustering would have three branches from the root—one to
each of the three main clusters—and then two, three, and four branches
from each of these clusters, respectively. A traditional agglomerative
approach cannot produce such a structure.

(b) The simplest type of postprocessing would attempt to flatten the hier-
archical clustering by moving clusters up the tree.

16. Use the similarity matrix in Table 8.1 to perform single and complete link
hierarchical clustering. Show your results by drawing a dendrogram. The
dendrogram should clearly show the order in which the points are merged.
The solutions are shown in Figures 8.6(a) and 8.6(b).

17. Hierarchical clustering is sometimes used to generate K clusters, K > 1 by
taking the clusters at the Kth level of the dendrogram. (Root is at level
1.) By looking at the clusters produced in this way, we can evaluate the
behavior of hierarchical clustering on different types of data and clusters,
and also compare hierarchical approaches to K-means.
The following is a set of one-dimensional points: {6, 12, 18, 24, 30, 42, 48}.

(a) For each of the following sets of initial centroids, create two clusters
by assigning each point to the nearest centroid, and then calculate the

134 Chapter 8 Cluster Analysis: Basic Concepts and Algorithms

Table 8.1. Similarity matrix for Exercise 16.

p1 p2 p3 p4 p5
p1 1.00 0.10 0.41 0.55 0.35
p2 0.10 1.00 0.64 0.47 0.98
p3 0.41 0.64 1.00 0.44

0.85

p4 0.55 0.47 0.44 1.00 0.76
p5 0.35 0.98 0.85 0.76 1.00

2 5 3 4 1
1

0.95

0.9
0.85
0.8

0.75

0.7
0.65
0.6

0.55

S
im

ila
rit

(a) Single link.

2 5 3 1 4
1

0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
S
im
ila
rit
y

(b) Complete link.

Figure 8.6. Dendrograms for Exercise 16.

total squared error for each set of two clusters. Show both the clusters
and the total squared error for each set of centroids.

i. {18, 45}
First cluster is 6, 12, 18, 24, 30.
Error = 360.
Second cluster is 42, 48.
Error = 18.
Total Error = 378

ii. {15, 40} First cluster is 6, 12, 18, 24 .
Error = 180.
Second cluster is 30, 42, 48.
Error = 168.
Total Error = 348.

(b) Do both sets of centroids represent stable solutions; i.e., if the K-means
algorithm was run on this set of points using the given centroids as the
starting centroids, would there be any change in the clusters generated?

135

Yes, both centroids are stable solutions.

The two clusters are {6, 12, 18, 24, 30} and {42, 48}.
(d) Which technique, K-means or single link, seems to produce the “most

natural” clustering in this situation? (For K-means, take the clustering
with the lowest squared error.)

MIN produces the most natural clustering.

(e) What definition(s) of clustering does this natural clustering correspond
to? (Well-separated, center-based, contiguous, or density.)

MIN produces contiguous clusters. However, density is also an accept-
able answer. Even center-based is acceptable, since one set of centers
gives the desired clusters.

(f) What well-known characteristic of the K-means algorithm explains the
previous behavior?

K-means is not good at finding clusters of different sizes, at least when
they are not well separated. The reason for this is that the objective of
minimizing squared error causes it to “break” the larger cluster. Thus,
in this problem, the low error clustering solution is the “unnatural”
one.

18. Suppose we find K clusters using Ward’s method, bisecting K-means, and
ordinary K-means. Which of these solutions represents a local or global
minimum? Explain.

Although Ward’s method picks a pair of clusters to merge based on mini-
mizing SSE, there is no refinement step as in regular K-means. Likewise,
bisecting K-means has no overall refinement step. Thus, unless such a refine-
ment step is added, neither Ward’s method nor bisecting K-means produces
a local minimum. Ordinary K-means produces a local minimum, but like the
other two algorithms, it is not guaranteed to produce a global minimum.

19. Hierarchical clustering algorithms require O(m2 log(m)) time, and conse-
quently, are impractical to use directly on larger data sets. One possible
technique for reducing the time required is to sample the data set. For ex-
ample, if K clusters are desired and

√
m points are sampled from the m

points, then a hierarchical clustering algorithm will produce a hierarchical
clustering in roughly O(m) time. K clusters can be extracted from this hier-
archical clustering by taking the clusters on the Kth level of the dendrogram.
The remaining points can then be assigned to a cluster in linear time, by
using various strategies. To give a specific example, the centroids of the K
clusters can be computed, and then each of the m − √m remaining points
can be assigned to the cluster associated with the closest centroid.

136 Chapter 8 Cluster Analysis: Basic Concepts and Algorithms

For each of the following types of data or clusters, discuss briefly if (1) sam-
pling will cause problems for this approach and (2) what those problems are.
Assume that the sampling technique randomly chooses points from the to-
tal set of m points and that any unmentioned characteristics of the data or
clusters are as optimal as possible. In other words, focus only on problems
caused by the particular characteristic mentioned. Finally, assume that K is
very much less than m.

(a) Data with very different sized clusters.

This can be a problem, particularly if the number of points in a cluster
is small. For example, if we have a thousand points, with two clusters,
one of size 900 and one of size 100, and take a 5% sample, then we
will, on average, end up with 45 points from the first cluster and 5
points from the second cluster. Five points is much easier to miss or
cluster improperly than 50. Also, the second cluster will sometimes
be represented by fewer than 5 points, just by the nature of random
samples.

(b) High-dimensional data.

This can be a problem because data in high-dimensional space is typi-
cally sparse and more points may be needed to define the structure of
a cluster in high-dimensional space.

By definition, outliers are not very frequent and most of them will be
omitted when sampling. Thus, if finding the correct clustering depends
on having the outliers present, the clustering produced by sampling will
likely be misleading. Otherwise, it is beneficial.

(d) Data with highly irregular regions.

This can be a problem because the structure of the border can be lost
when sampling unless a large number of points are sampled.

(e) Data with globular clusters.

This is typically not a problem since not as many points need to be
sampled to retain the structure of a globular cluster as an irregular
one.

(f) Data with widely different densities.

In this case the data will tend to come from the denser region. Note
that the effect of sampling is to reduce the density of all clusters by
the sampling factor, e.g., if we take a 10% sample, then the density of
the clusters is decreased by a factor of 10. For clusters that aren’t very
dense to begin with, this may means that they are now treated as noise
or outliers.

137

(g) Data with a small percentage of noise points.

Sampling will not cause a problem. Actually, since we would like to
exclude noise, and since the amount of noise is small, this may be
beneficial.

(h) Non-Euclidean data.

This has no particular impact.

(i) Euclidean data.

This has no particular impact.

(j) Data with many and mixed attribute types.

Many attributes was discussed under high-dimensionality. Mixed at-
tributes have no particular impact.

20. Consider the following four faces shown in Figure 8.7. Again, darkness or
number of dots represents density. Lines are used only to distinguish regions
and do not represent points.

(a) (b)

Figure 8.7. Figure for Exercise 20.

(a) For each figure, could you use single link to find the patterns represented
by the nose, eyes, and mouth? Explain.

Only for (b) and (d). For (b), the points in the nose, eyes, and mouth
are much closer together than the points between these areas. For (d)
there is only space between these regions.

(b) For each figure, could you use K-means to find the patterns represented
by the nose, eyes, and mouth? Explain.

Only for (b) and (d). For (b), K-means would find the nose, eyes, and
mouth, but the lower density points would also be included. For (d), K-

138 Chapter 8 Cluster Analysis: Basic Concepts and Algorithms

means would find the nose, eyes, and mouth straightforwardly as long
as the number of clusters was set to 4.

Clustering techniques can only find patterns of points, not of empty
spaces.

21. Compute the entropy and purity for the confusion matrix in Table 8.2.

Table 8.2. Confusion matrix for Exercise 21.
Cluster Entertainment Financial Foreign Metro National Sports Total Entropy Purity

#1 1 1 0 11 4 676 693 0.20 0.98
#2 27 89 333 827 253 33 1562 1.84 0.53
#3 326 465 8 105 16 29 949 1.70 0.49

Total 354 555 341 943 273 738 3204 1.44 0.61

22. You are given two sets of 100 points that fall within the unit square. One set
of points is arranged so that the points are uniformly spaced. The other set
of points is generated from a uniform distribution over the unit square.

(a) Is there a difference between the two sets of points?

Yes. The random points will have regions of lesser or greater density,
while the uniformly distributed points will, of course, have uniform
density throughout the unit square.

(b) If so, which set of points will typically have a smaller SSE for K=10
clusters?

The random set of points will have a lower SSE.

DBSCAN will merge all points in the uniform data set into one cluster
or classify them all as noise, depending on the threshold. There might
be some boundary issues for points at the edge of the region. However,
DBSCAN can often find clusters in the random data, since it does have
some variation in density.

23. Using the data in Exercise 24, compute the silhouette coefficient for each
point, each of the two clusters, and the overall clustering.
Cluster 1 contains {P1, P2}, Cluster 2 contains {P3, P4}. The dissimilarity
matrix that we obtain from the similarity matrix is the following:

139

Table 8.3. Table of distances for Exercise 23

P1 P2 P3 P4
P1 0 0.10 0.65 0.55
P2 0.10 0 0.70 0.60
P3 0.65 0.70 0 0.30
P4 0.55 0.60 0.30 0

Let a indicate the average distance of a point to other points in its cluster.
Let b indicate the minimum of the average distance of a point to points in
another cluster.
Point P1: SC = 1- a/b = 1 – 0.1/((0.65+0.55)/2)= 5/6 = 0.833
Point P2: SC = 1- a/b = 1 – 0.1/((0.7+0.6)/2) = 0.846
Point P2: SC = 1- a/b = 1 – 0.3/((0.65+0.7)/2) = 0.556
Point P2: SC = 1- a/b = 1 – 0.3/((0.55+0.6)/2) = 0.478

Cluster 1 Average SC = (0.833+0.846)/2 = 0.84
Cluster 2 Average SC = (0.556+0.478)/2 = 0.52
Overall Average SC = (0.840+0.517)/2 = 0.68

24. Given the set of cluster labels and similarity matrix shown in Tables 8.4 and
8.5, respectively, compute the correlation between the similarity matrix and
the ideal similarity matrix, i.e., the matrix whose ijth entry is 1 if two objects
belong to the same cluster, and 0 otherwise.

Table 8.4. Table of cluster labels for Exercise 24.

Point Cluster Label
P1 1
P2 1
P3 2
P4 2

Table 8.5. Similarity matrix for Exercise 24.

Point P1 P2 P3 P4
P1 1 0.8 0.65 0.55
P2 0.8 1 0.7 0.6
P3 0.65 0.7 1 0.9
P4 0.55 0.6 0.9 1

We need to compute the correlation between the vector x =< 1, 0, 0, 0, 0, 1 >
and the vector y =< 0.8, 0.65, 0.55, 0.7, 0.6, 0.3 >, which is the correlation
between the off-diagonal elements of the distance matrix and the ideal simi-
larity matrix.
We get:
Standard deviation of the vector x : σx = 0.5164
Standard deviation of the vector y : σy = 0.1703
Covariance of x and y: cov(x,y) = −0.200

140 Chapter 8 Cluster Analysis: Basic Concepts and Algorithms

Therefore, corr(x,y) = cov(x,y)/σxσy = −0.227

25. Compute the hierarchical F-measure for the eight objects {p1, p2, p3, p4, p5,
p6, p7, p8} and hierarchical clustering shown in Figure 8.8. Class A contains
points p1, p2, and p3, while p4, p5, p6, p7, and p8 belong to class B.

{p1, p2, p3, p4, p5, p6, p7, p8}

{p3, p6, p7, p8}

{p1, p2} {p4, p5} {p3, p6} {p7, p8}

{p1, p2, p4, p5,}

Figure 8.8. Hierarchical clustering for Exercise 25.

Let R(i, j) = nij/ni indicate the recall of class i with respect to cluster j.
Let P (i, j) = nij/nj indicate the precision of class i with respect to cluster
j.
F (i, j) = 2R(i, j)×P (i, j)/(R(i, j)+P (i, j)) is the F-measure for class i and
cluster j.
For cluster #1 = {p1, p2, p3, p4, p5, p6, p7, p8}:
Class = A:
R(A, 1) = 3/3 = 1, P (A, 1) = 3/8 = 0.375
F (A, 1) = 2 × 1 × 0.375/(1 + 0.375) = 0.55
Class = B:
R(B, 1) = 5/5 = 1, P (A, 1) = 5/8 = 0.625, F (A, 1) = 0.77

For cluster #2= {p1,p2,p4,p5}
Class = A:
R(A, 2) = 2/3, P (A, 2) = 2/4, F (A, 2) = 0.57
Class = B:
R(B, 2) = 2/5, P (B, 2) = 2/4, F (B, 2) = 0.44

For cluster #3= {p3, p6, p7, p8}
Class = A:
R(A, 3) = 1/3, P (A, 3) = 1/4, F (A, 3) = 0.29
Class =B:
R(B, 3) = 3/5, P (B, 3) = 3/4, F (B, 3) = 0.67

For cluster #4={p1, p2}
Class = A:

141

R(A, 4) = 2/3, P (A, 4) = 2/2, F (A, 4) = 0.8
Class =B:
R(B, 4) = 0/5, P (B, 4) = 0/2, F (B, 4) = 0

For cluster #5 = {p4, p5}
Class = A:
R(A, 5) = 0, P (A, 5) = 0, F (A, 5) = 0
Class =B:
R(B, 5) = 2/5, P (B, 5) = 2/2, F (B, 5) = 0.57

For cluster #6 = {p3, p6}
Class = A:
R(A, 6) = 1/3, P (A, 6) = 1/2, F (A, 6) = 0.4
Class =B:
R(B, 6) = 1/5, P (B, 6) = 1/2, F (B, 6) = 0.29

For cluster #7 = {p7, p8}
Class = A:
R(A, 7) = 0, P (A, 7) = 1, F (A, 7) = 0
Class = B:
R(B, 7) = 2/5, P (B, 7) = 2/2, F (B, 7) = 0.57

Class A: F (A) = max{F (A, j)} = max{0.55, 0.57, 0.29, 0.8, 0, 0.4, 0} = 0.8
Class B: F (B) = max{F (B, j)} = max{0.77, 0.44, 0.67, 0, 0.57, 0.29, 0.57} =
0.77
Overall Clustering: F =

∑2
1

ni
n maxi F (i, j) = 3/8∗F (A)+5/8∗F (B) = 0.78

26. Compute the cophenetic correlation coefficient for the hierarchical clusterings
in Exercise 16. (You will need to convert the similarities into dissimilarities.)
This can be easily computed using a package, e.g., MATLAB. The answers
are single link, 0.8116, and complete link, 0.7480.

27. Prove Equation 8.14.

142 Chapter 8 Cluster Analysis: Basic Concepts and Algorithms

|Ci|
∑

x∈Ci

∑
y∈Ci

(x − y)2 = 1
2|Ci|

∑
x∈Ci

∑
y∈Ci

((x − ci) − (y − ci))2

=
1

2|Ci|
( ∑

x∈Ci
∑
y∈Ci

(x − ci)2 − 2
∑

x∈Ci
∑
y∈Ci

(x − ci)(y − ci)

+
∑

x∈Ci
∑
y∈Ci

(y − ci)2
)

=
1
2|Ci|
( ∑
x∈Ci
∑
y∈Ci

(x − ci)2 +
∑

x∈Ci
∑
y∈Ci
(y − ci)2
)
=
1
|Ci|
∑

x∈Ci
|Ci|(x − ci)2

= SSE

The cross term
∑

x∈Ci
∑

y∈Ci(x − ci)(y − ci) is 0.
28. Prove Equation 8.15.

1
K

K∑
i=1

K∑
j=1

|Ci|(cj − ci)2 = 1
2K

K∑
i=1
K∑
j=1

|Ci|((m − ci) − (m − cj))2

=
1
2K

( K∑
i=1

K∑
j=1

|Ci|(m − ci)2 − 2
K∑

i=1
K∑
j=1

|Ci|(m − ci)(m − cj)

+
K∑
i=1
K∑
j=1

|Ci|(m − cj)2
)

=
1
2K
( K∑
i=1
K∑
j=1

|Ci|(m − ci)2 +
K∑

i=1
K∑
j=1
|Ci|(m − cj)2
)
=
1
K
K∑
i=1

K|Ci|(m − ci)2

= SSB

Again, the cross term cancels.

29. Prove that
∑K

i=1

∑
x∈Ci(x − mi)(m − mi) = 0. This fact was used in the

proof that TSS = SSE + SSB on page 557.

143

K∑
i=1
∑
x∈Ci

(x − ci)(c − ci)

=
K∑

i=1
∑
x∈Ci

(xc − cic − xci + c2i )

=
K∑
i=1
∑
x∈Ci

xc −
K∑

i=1
∑
x∈Ci

cic −
K∑

i=1
∑
x∈Ci

xci +

K∑
i=1
∑
x∈Ci

c2i

= micic − micic − mic2i + mic2i
= 0

30. Clusters of documents can be summarized by finding the top terms (words)
for the documents in the cluster, e.g., by taking the most frequent k terms,
where k is a constant, say 10, or by taking all terms that occur more fre-
quently than a specified threshold. Suppose that K-means is used to find
clusters of both documents and words for a document data set.

(a) How might a set of term clusters defined by the top terms in a document
cluster differ from the word clusters found by clustering the terms with
K-means?

First, the top words clusters could, and likely would, overlap somewhat.
Second, it is likely that many terms would not appear in any of the
clusters formed by the top terms. In contrast, a K-means clustering of
the terms would cover all the terms and would not be overlapping.

(b) How could term clustering be used to define clusters of documents?

An obvious approach would be to take the top documents for a term
cluster; i.e., those documents that most frequently contain the terms in
the cluster.

31. We can represent a data set as a collection of object nodes and a collection
of attribute nodes, where there is a link between each object and each at-
tribute, and where the weight of that link is the value of the object for that
attribute. For sparse data, if the value is 0, the link is omitted. Bipartite
clustering attempts to partition this graph into disjoint clusters, where each
cluster consists of a set of object nodes and a set of attribute nodes. The
objective is to maximize the weight of links between the object and attribute
nodes of a cluster, while minimizing the weight of links between object and
attribute links in different clusters. This type of clustering is also known
as co-clustering since the objects and attributes are clustered at the same
time.

(a) How is bipartite clustering (co-clustering) different from clustering the
sets of objects and attributes separately?

In regular clustering, only one set of constraints, related either to ob-
jects or attributes, is applied. In co-clustering both sets of constraints

144 Chapter 8 Cluster Analysis: Basic Concepts and Algorithms

are applied simultaneously. Thus, partitioning the objects and at-
tributes independently of one another typically does not produce the
same results.

(b) Are there any cases in which these approaches yield the same clusters?

Yes. For example, if a set of attributes is associated only with the
objects in one particular cluster, i.e., has 0 weight for objects in all other
clusters, and conversely, the set of objects in a cluster has 0 weight for
all other attributes, then the clusters found by co-clustering will match
those found by clustering the objects and attributes separately. To use
documents as an example, this would correspond to a document data
set that consists of groups of documents that only contain certain words
and groups of words that only appear in certain documents.

Co-clustering automatically provides a description of a cluster of objects
in terms of attributes, which can be more useful than a description
of clusters as a partitioning of objects. However, the attributes that
distinguish different clusters of objects, may overlap significantly, and
in such cases, co-clustering will not work well.

32. In Figure 8.9, match the similarity matrices, which are sorted according to
cluster labels, with the sets of points. Differences in shading and marker
shape distinguish between clusters, and each set of points contains 100 points
and three clusters. In the set of points labeled 2, there are three very tight,
equal-sized clusters.

Answers: 1 – D, 2 – C, 3 – A, 4 – B

145

1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1

0
0 0.2 0.4 0.6 0.8 1

x
y
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 0.2 0.4 0.6 0.8 1
x
y
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 0.2 0.4 0.6 0.8 1
x
y
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 0.2 0.4 0.6 0.8 1
x
y
1
3
2
4

Figure 8.9. Points and similarity matrices for Exercise 32.

Cluster Analysis:
Additional Issues and
Algorithms

1. For sparse data, discuss why considering only the presence of non-zero values
might give a more accurate view of the objects than considering the actual
magnitudes of values. When would such an approach not be desirable?

Consider document data. Intuitively, two documents are similar if they con-
tain many of the same words. Although we can also include the frequency
with which those words occur in the similarity computation, this can some-
times give a less reliable assessment of similarity. In particular, if one of
the words in a document occurs rather frequently compared to other words,
then this word can dominate the similarity comparison when magnitudes are
taken into account. In that case, the document will only be highly similar
to other documents that also contain the same word with a high frequency.
While this may be appropriate in many or even most cases, it may lead to the
wrong conclusion if the word can appear in different contexts, that can only
be distinguished by other words. For instance, the word, ‘game,’ appears
frequently in discussions of sports and video games.

2. Describe the change in the time complexity of K-means as the number of
clusters to be found increases.

As the number of clusters increases, the complexity of K-means approaches
O(m2).

3. Consider a set of documents. Assume that all documents have been normal-
ized to have unit length of 1. What is the “shape” of a cluster that consists
of all documents whose cosine similarity to a centroid is greater than some
specified constant? In other words, cos(d, c) ≥ δ, where 0 < δ ≤ 1.

148 Chapter 9 Cluster Analysis: Additional Issues and Algorithms

Once document vectors have been normalized, they lie on am n-dimensional
hypershpere. The constraint that all documents have a minimum cosine
similarity with respect to a centroids is a constraint that the document vectors
lie within a cone, whose intersection with the sphere is a circle on the surface
of the sphere.

4. Discuss the advantages and disadvantages of treating clustering as an opti-
mization problem. Among other factors, consider efficiency, non-determinism,
and whether an optimization-based approach captures all types of clusterings
that are of interest.

Two key advantage to treating clustering as an optimization problem are
that (1) it provides a clear definition of what the clustering process is do-
ing, and (2) it allows the use of powerful optimization techniques that have
been developed in a wide variety of fields. Unfortunately, most of these op-
timization techniques have a high time complexity. Furthermore, it can be
shown that many optimization problems are NP hard, and therefore, it is
necessary to use heuristic optimization approaches that can only guarantee
a locally optimal solution. Often such techniques work best when used with
random initialization, and thus, the solution found can vary from one run to
another. Another problem with optimization approaches is that the objective
functions they use tend to favor large clusters at the expense of smaller ones.

5. What is the time and space complexity of fuzzy c-means? Of SOM? How do
these complexities compare to those of K-means?

The time complexity of K-means O(I ∗ K ∗ m ∗ n), where I is the number
of iterations required for convergence, K is the number of clusters, m is the
number of points, and n is the number of attributes. The time required by
fuzzy c-means is basically the same as K-means, although the constant is
much higher. The time complexity of SOM is also basically the same as K-
means because it consists of multiple passes in which objects are assigned to
centroids and the centroids are updated. However, because the surrounding
centroids are also updated and the number of passes can be large, SOM will
typically be slower than K-means.

6. Traditional K-means has a number of limitations, such as sensitivity to out-
liers and difficulty in handling clusters of different sizes and densities, or with
non-globular shapes. Comment on the ability of fuzzy c-means to handle
these situations.

Fuzzy c-means has all the limitations of traditional K-means, except that it
does not make a hard assignment of an object to a cluster.

7. For the fuzzy c-means algorithm described in this book, the sum of the mem-
bership degree of any point over all clusters is 1. Instead, we could only
require that the membership degree of a point in a cluster be between 0 and
1. What are the advantages and disadvantages of such an approach?

149

The main advantage of this approach occurs when a point is an outlier and
does not really belong very strongly to any cluster, since in that situation,
the point can have low membership in all clusters. However, this approach is
often harder to initialize properly and can perform poorly when the clusters
are not are not distinct. In that case, several cluster centers may merge
together, or a cluster center may vary significantly from one iteration to
another, instead of changing only slightly, as in ordinary K-means or fuzzy
c-means.

8. Explain the difference between likelihood and probability.

Probability is, according to one common statistical definition, the frequency
with which an event occurs as the number of experiments tends to infinity.
Probability is defined by a probability density function which is a function
of the attribute values of an object. Typically, a probability density function
depends on some parameters. Considering probability density function to be
a function of the parameters yields the likelihood function.

9. Equation 9.12 gives the likelihood for a set of points from a Gaussian dis-
tribution as a function of the mean µ and the standard deviation σ. Show
mathematically that the maximum likelihood estimate of µ and σ are the
sample mean and the sample standard deviation, respectively.
First, we solve for µ.

∂�((µ, σ)|X )

∂µ

=
∂

∂µ

(
−

m∑
i=1

(xi − µ)2
2σ2

− 0.5m log 2π − m log σ
)

= −
m∑

i=1

2(xi − µ)
2σ2

Setting this equal to 0 and solving, we get µ = 1m
∑m

i=1 xi.
Likewise, we can solve for σ.

∂�((µ, σ)|X )

∂σ

=
∂
∂σ
(
−
m∑
i=1
(xi − µ)2
2σ2
− 0.5m log 2π − m log σ
)

=
m∑

i=1

2(xi − µ)2
2σ3

− m
σ

Setting this equal to 0 and solving, we get σ2 = 1m
∑m

i=1(xi − µ)2.
10. We take a sample of adults and measure their heights. If we record the gender

of each person, we can calculate the average height and the variance of the
height, separately, for men and women. Suppose, however, that this informa-
tion was not recorded. Would it be possible to still obtain this information?
Explain.

150 Chapter 9 Cluster Analysis: Additional Issues and Algorithms

The height of men and women will have separate Gaussian distributions with
different means and perhaps different variances. By using a mixture model
approach, we can obtain estimates of the mean and variance of the two height
distributions. Given a large enough sample size, the estimated parameters
should be close to those that could be computed if we knew the gender of
each person.

11. Compare the membership weights and probabilities of Figures 9.1 and 9.4,
which come, respectively, from applying fuzzy and EM clustering to the same
set of data points. What differences do you detect, and how might you explain
these differences?

The fuzzy clustering approach only assigns very high weights to those points
in the center of the clusters. Those points that are close to two or three
clusters have relatively low weights. The points that are on the far edges of
the clusters, away from other clusters also have lower weights than the center
points, but not as low as points that are near two or three clusters.

The EM clustering approach assigns high weights both to points in the center
of the clusters and those on the far edges. The points that are near two or
three clusters have lower weights, but not as much so as with the fuzzy
clustering procedure.

The main difference between the approaches is that as a point on the far
edge of a cluster gets further away from the center of the cluster, the weight
with which is belongs to a cluster becomes more equal among clusters for
the fuzzy clustering approach, but for the EM approach the point tends to
belong more strongly to the cluster to which it is closest.

12. Figure 9.1 shows a clustering of a two-dimensional point data set with two
clusters: The leftmost cluster, whose points are marked by asterisks, is some-
what diffuse, while the rightmost cluster, whose points are marked by circles,
is compact. To the right of the compact cluster, there is a single point
(marked by an arrow) that belongs to the diffuse cluster, whose center is
farther away than that of the compact cluster. Explain why this is possible
with EM clustering, but not K-means clustering.

In EM clustering, we compute the probability that a point belongs to a
cluster. In turn, this probability depends on both the distance from the
cluster center and the spread (variance) of the cluster. Hence, a point that
is closer to the centroid of one cluster than another can still have a higher
probability with respect to the more distant cluster if that cluster has a higher
spread than the closer cluster. K-means only takes into account the distance
to the closest cluster when assigning points to clusters. This is equivalent to
an EM approach where all clusters are assumed to have the same variance.

151

–10 –8 –6 –4 –2 0 2 4
–8

–6

–4

–2

0
2
4
6
x
y

Figure 9.1. Data set for Exercise 12. EM clustering of a two-dimensional point set with two clusters
of differing density.

13. Show that the MST clustering technique of Section 9.4.2 produces the same
clusters as single link. To avoid complications and special cases, assume that
all the pairwise similarities are distinct.

In single link, we start with with clusters of individual points and then succes-
sively join two clusters that have the pair of points that are closest together.
Conceptually, we can view the merging of the clusters as putting an edge
between the two closest points of the two clusters. Note that if both clusters
are currently connected, then the resulting cluster will also be connected.
However, since the clusters are formed from disjoint sets of points, and edges
are only placed between points in different clusters, no cycle can be formed.
From these observations and by noting that we start with clusters (graphs)
of size one that are vacuously connected, we can deduce by induction that
at any stage in single link clustering process, each cluster consists of a con-
nected set of points without any cycles. Thus, when the last two clusters are
merged to form a cluster containing all the points, we also have a connected
graph of all the points that is a spanning tree of the graph. Furthermore,
since each point in the graph is connected to its nearest point, the spanning
tree must be a minimum spanning tree. All that remains to establish the
equivalence of MST and single link is to note that MST essentially reverses
the process by which single link built the minimum spanning tree; i.e., by

152 Chapter 9 Cluster Analysis: Additional Issues and Algorithms

breaking edges beginning with the longest and proceeding until the smallest.
Thus, it generates the same clusters as single link, but in reverse order.

14. One way to sparsify a proximity matrix is the following: For each object
(row in the matrix), set all entries to 0 except for those corresponding to
the objects k-nearest neighbors. However, the sparsified proximity matrix is
typically not symmetric.

(a) If object a is among the k-nearest neighbors of object b, why is b not
guaranteed to be among the k-nearest neighbors of a?

Consider a dense set of k+1 objects and another object, an outlier, that
is farther from any of the objects than they are from each other. None
of the objects in the dense set will have the outlier on their k-nearest
neighbor list, but the outlier will have k of the objects from the dense
set on its k-nearest neighbor list.

(b) Suggest at least two approaches that could be used to make the sparsi-
fied proximity matrix symmetric.

One approach is to set the ijth entry to 0 if the jith entry is 0, or vice
versa. Another approach is to set the ijth entry to 1 if the jith entry is
1, or vice versa.

15. Give an example of a set of clusters in which merging based on the closeness
of clusters leads to a more natural set of clusters than merging based on the
strength of connection (interconnectedness) of clusters.
An example of this is given in the Chameleon paper that can be found at
http://www.cs.umn.edu/ karypis/publications/Papers/PDF/chameleon .
The example consists of two narrow rectangles of points that are side by side.
The top rectangle is split into two clusters, one much smaller than the other.
Even though the two rectangles on the top are close, they are not strongly
connected since the strong links between them are across a small area. On the
other hand, the largest rectangle on the top and the rectangle on the bottom
are strongly connected. Each individual connection is not as strong, because
these two rectangles are not as close, but there are more of them because
the area of connection is large. Thus, an approach based on connectivity will
merge the largest rectangle on top with the bottom rectangle.

16. Table 9.1 lists the two nearest neighbors of four points.
Calculate the SNN similarity between each pair of points using the definition
of SNN similarity defined in Algorithm 9.10.

The following is the SNN similarity matrix.

17. For the definition of SNN similarity provided by Algorithm 9.10, the cal-
culation of SNN distance does not take into account the position of shared

153

Table 9.1. Two nearest neighbors of four points.

Point First Neighbor Second Neighbor
1 4 3
2 3 4
3 4 2
4 3 1

Table 9.2. Two nearest neighbors of four points.

Point 1 2 3 4
1 2 0 0 1
2 0 2 1 0
3 0 1 2 0
4 1 0 0 2

neighbors in the two nearest neighbor lists. In other words, it might be de-
sirable to give higher similarity to two points that share the same nearest
neighbors in the same or roughly the same order.

(a) Describe how you might modify the definition of SNN similarity to give
higher similarity to points whose shared neighbors are in roughly the
same order.

This can be done by assigning weights to the points based on their
position in the nearest neighbor list. For example, we can weight the
ith point in the nearest neighbor list by n − i + 1. For each point, we
then take the sum or product of its rank on both lists. These values are
then summed to compute the similarity between the two objects. This
approach was suggested by Jarvis and Patrick [5].

(b) Discuss the advantages and disadvantages of such a modification.

Such an approach is more complex. However, it is advantageous if it is
the case that two objects are more similar if the shared neighbors are
roughly of the same rank. Furthermore, it may also help to compensate
for arbitrariness in the choice of k.

18. Name at least one situation in which you would not want to use clustering
based on SNN similarity or density.

When you wish to cluster based on absolute density or distance.

19. Grid-clustering techniques are different from other clustering techniques in
that they partition space instead of sets of points.

(a) How does this affect such techniques in terms of the description of the
resulting clusters and the types of clusters that can be found?

154 Chapter 9 Cluster Analysis: Additional Issues and Algorithms

In grid-based clustering, the clusters are described in terms of collec-
tions of adjacent cells. In some cases, as in CLIQUE, a more compact
description is generated. In any case, the description of the clusters is
given in terms of a region of space, not a set of objects. (However, such
a description can easily be generated.) Because it is necessary to work
in terms of rectangular regions, the shapes of non-rectangular clusters
can only be approximated. However, the groups of adjacent cells can
have holes.

(b) What kind of cluster can be found with grid-based clusters that cannot
be found by other types of clustering approaches? (Hint: See Exercise
20 in Chapter 8, page 564.)

Typically, grid-based clustering techniques only pay attention to dense
regions. However, such techniques could also be used to identify sparse
or empty regions and thus find patterns of the absence of points. Note,
however, that this would not be appropriate for a sparse data space.

20. In CLIQUE, the threshold used to find cluster density remains constant,
even as the number of dimensions increases. This is a potential problem
since density drops as dimensionality increases; i.e., to find clusters in higher
dimensions the threshold has to be set at a level that may well result in the
merging of low-dimensional clusters. Comment on whether you feel this is
truly a problem and, if so, how you might modify CLIQUE to address this
problem.

This is a real problem. A similar problem exists in association analysis. In
particular, the support of association patterns with a large number of items
is often low. To find such patterns using an algorithm such as Apriori is diffi-
cult because the low support threshold required results in a large number of
association patterns with few items that are of little interest. In other words,
association patterns with many items (patterns in higher-dimensional space)
are interesting at support levels (densities) that do not make for interesting
patterns when the size of the association pattern (number of dimensions) is
low. One approach is to decrease the support threshold (density threshold)
as the number of items (number of dimensions) is increased.

21. Given a set of points in Euclidean space, which are being clustered using
the K-means algorithm with Euclidean distance, the triangle inequality can
be used in the assignment step to avoid calculating all the distances of each
point to each cluster centroid. Provide a general discussion of how this might
work.

Charles Elkan presented the following theorem in his keynote speech at the
Workshop on Clustering High-Dimensional Data at SIAM 2004.
Lemma 1:Let x be a point, and let b and c be centers.
If d(b, c) ≥ 2d(x, b) then d(x, c) ≥ d(x, b).

155

Proof:
We know d(b, c) ≤ d(b, x) + d(x, c).
So d(b, c) − d(x, b) ≤ d(x, c).
Now d(b, c) − d(x, b) ≥ 2d(x, b) − d(x, b) = d(x, b).
So d(x, b) ≤ d(x, c).
This theorem can be used to eliminate a large number of unnecessary distance
calculations.

22. Instead of using the formula derived in CURE—see Equation 9.19—we could
run a Monte Carlo simulation to directly estimate the probability that a
sample of size s would contain at least a certain fraction of the points from
a cluster. Using a Monte Carlo simulation compute the probability that a
sample of size s contains 50% of the elements of a cluster of size 100, where
the total number of points is 1000, and where s can take the values 100, 200,
or 500.

This question should have said “contains at least 50%.”
The results of our simulation consisting of 100,000 trials was 0, 0, and 0.54
for the sample sizes 100, 200, and 500 respectively.

Anomaly Detection

1. Compare and contrast the different techniques for anomaly detection that
were presented in Section 10.1.2. In particular, try to identify circumstances
in which the definitions of anomalies used in the different techniques might
be equivalent or situations in which one might make sense, but another would
not. Be sure to consider different types of data.

First, note that the proximity- and density-based anomaly detection tech-
niques are related. Specifically, high density in the neighborhood of a point
implies that many points are close to it, and vice-versa. In practice, density
is often defined in terms of distance, although it can also be defined using a
grid-based approach.

The model-based approach can be used with virtually any underlying tech-
nique that fits a model to the data. However, note that a particular model,
statistical or otherwise, must be assumed. Consequently, model-based ap-
proaches are restricted in terms of the data to which they can be applied.
For example, if the model assumes a Gaussian distribution, then it cannot
be applied to data with a non-Gaussian distribution.

On the other hand, the proximity- and density-based approaches do not
make any particular assumption about the data, although the definition of
an anomaly does vary from one proximity- or density-based technique to
another. Proximity-based approaches can be used for virtually any type
of data, although the proximity metric used must be chosen appropriately.
For example, Euclidean distance is typically used for dense, low-dimensional
data, while the cosine similarity measure is used for sparse, high-dimensional
data. Since density is typically defined in terms of proximity, density-based
approaches can also be used for virtually any type of data. However, the
meaning of density is less clear in a non-Euclidean data space.

Proximity- and density-based anomaly detection techniques can often pro-
duce similar results, although there are significant differences between tech-

158 Chapter 10 Anomaly Detection

niques that do not account for the variations in density throughout a data set
or that use different proximity measures for the same data set. Model-based
methods will often differ significantly from one another and from proximity-
and density-based approaches.

2. Consider the following definition of an anomaly: An anomaly is an object
that is unusually influential in the creation of a data model.

(a) Compare this definition to that of the standard model-based definition
of an anomaly.

The standard model-based definition labels objects that don’t fit the
model very well as anomalies. Although these object often are unusu-
ally influential in the model, it can also be the case that an unusually
influential object can fit the model very well.

(b) For what sizes of data sets (small, medium, or large) is this definition
appropriate?

This definition is typically more appropriate for smaller data sets, at
least if we are talking about one very influential object. However, a
relatively small group highly influential objects can have a significant
impact on a model, but still fit it well, even for medium or large data
sets.

3. In one approach to anomaly detection, objects are represented as points in
a multidimensional space, and the points are grouped into successive shells,
where each shell represents a layer around a grouping of points, such as a
convex hull. An object is an anomaly if it lies in one of the outer shells.

(a) To which of the definitions of an anomaly in Section 10.1.2 is this defi-
nition most closely related?

This definition is most closely related to the distance-based approach.

(b) Name two problems with this definition of an anomaly.

i. For the convex hull approach, the distance of the points in a con-
vex hull from the center of the points can vary significantly if the
distribution of points in not symmetrical.

ii. This approach does not lend itself to assigning meaningful numer-
ical anomaly scores.

4. Association analysis can be used to find anomalies as follows. Find strong as-
sociation patterns, which involve some minimum number of objects. Anoma-
lies are those objects that do not belong to any such patterns. To make this
more concrete, we note that the hyperclique association pattern discussed in
Section 6.8 is particularly suitable for such an approach. Specifically, given a
user-selected h-confidence level, maximal hyperclique patterns of objects are

159

found. All objects that do not appear in a maximal hyperclique pattern of
at least size three are classified as outliers.

(a) Does this technique fall into any of the categories discussed in this
chapter? If so, which one?

In a hyperclique, all pairs of objects have a guaranteed cosine similar-
ity of the h-confidence or higher. Thus, this approach can be viewed
as a proximity-based approach. However, rather than a condition on
the proximity of objects with respect to a particular object, there is a
requirement on the pairwise proximities of all objects in a group.

(b) Name one potential strength and one potential weakness of this ap-
proach.

Strengths of this approach are that (1) the objects that do not belong to
any size 3 hyperclique are not strongly connected to other objects and
are likely anomalous and (2) it is computationally efficient. Potential
weaknesses are (1) this approach does not assign a numerical anomaly
score, but simply classifies an object as normal or an anomaly, (2)
it is not possible to directly control the number of objects classified as
anomalies because the only parameters are the h-confidence and support
threshold, and (3) the data needs to be discretized.

5. Discuss techniques for combining multiple anomaly detection techniques to
improve the identification of anomalous objects. Consider both supervised
and unsupervised cases.

In the supervised case, we could use ensemble classification techniques. In
these approaches, the classification of an object is determined by combining
the classifications of a number of classifiers, e.g., by voting. In the unsuper-
vised approach, a voting approach could also be used. Note that this assumes
that we have a binary assignment of an object as an anomaly. If we have
anomaly scores, then the scores could be combined in some manner, e.g., an
average or minimum, to yield an overall score.

6. Describe the potential time complexity of anomaly detection approaches
based on the following approaches: model-based using clustering, proximity-
based, and density. No knowledge of specific techniques is required. Rather,
focus on the basic computational requirements of each approach, such as the
time required to compute the density of each object.

If K-means clustering is used, then the complexity is dominated by finding
the clusters. This requires time proportional to the number of objects, i.e.,
O(m). The distance and density based approaches, usually require computing
all the pairwise proximities, and thus the complexity is often O(m2). In
some cases, such as low dimensional data, special techniques, such as the R∗

160 Chapter 10 Anomaly Detection

tree or k-d trees can be used to compute the nearest neighbors of an object
more efficiently, i.e., O(m log m), and this can reduce the overall complexity
when the technique is based only on nearest neighbors. Also, a grid-based
approach to computing density can reduce the complexity of density-based
anomaly detection to O(m), but such techniques are not as accurate and are
only effective for low dimensions.

7. The Grubbs’ test, which is described by Algorithm 10.1, is a more statistically
sophisticated procedure for detecting outliers than that of Definition 10.3. It
is iterative and also takes into account the fact that the z-score does not
have a normal distribution. This algorithm computes the z-score of each
value based on the sample mean and standard deviation of the current set of
values. The value with the largest magnitude z-score is discarded if its z-score
is larger than gc, the critical value of the test for an outlier at significance
level α. This process is repeated until no objects are eliminated. Note that
the sample mean, standard deviation, and gc are updated at each iteration.

Algorithm 10.1 Grubbs’ approach for outlier elimination.
1: Input the values and α

{m is the number of values, α is a parameter, and tc is a value chosen so that
α = prob(x ≥ tc) for a t distribution with m − 2 degrees of freedom.}

2: repeat
3: Compute the sample mean (x) and standard deviation (sx).
4: Compute a value gc so that prob(|z| ≥ gc) = α.

(In terms of tc and m, gc = m−1√m

√
t2c

m−2+t2c .)
5: Compute the z-score of each value, i.e., z = (x − x)/sx.
6: Let g = max |z|, i.e., find the z-score of largest magnitude and call it g.
7: if g > gc then
8: Eliminate the value corresponding to g.
9: m ← m − 1

10: end if
11: until No objects are eliminated.

(a) What is the limit of the value m−1√
m

√
t2c

m−2+t2c used for Grubbs’ test as
m approaches infinity? Use a significance level of 0.05.

Note that this could have been phrased better. The value of this ex-
pression approaches tc, but strictly speaking this is not a limit as tc
depends on m.

lim
m→∞

m − 1√
m

√
t2c

m − 2 + t2c
= lim

m→∞
m − 1√

m(m − 2 + t2c)
× tc = 1 × tc = tc.

161

Also, the value of tc will continue to increase with m, although slowly.
For m = 1020, tc = 93 for a significance value of 0.05.

(b) Describe, in words, the meaning of the previous result.

The distribution of g is becomes a t distribution as m increases.

8. Many statistical tests for outliers were developed in an environment in which
a few hundred observations was a large data set. We explore the limitations
of such approaches.

(a) For a set of 1,000,000 values, how likely are we to have outliers according
to the test that says a value is an outlier if it is more than three standard
deviations from the average? (Assume a normal distribution.)

This question should have asked how many outliers we would have since
the object of this question is to show that even a small probability of
an outlier yields a large number of outliers for a large data set. The
probability is unaffected by the number of objects.

The probability is either 0.00135 for a single sided deviation of 3 stan-
dard deviations or 0.0027 for a double-sided deviation. Thus, the num-
ber of anomalous objects will be either 1,350 or 2,700.

(b) Does the approach that states an outlier is an object of unusually low
probability need to be adjusted when dealing with large data sets? If
so, how?

There are thousands of outliers (under the specified definition) in a
million objects. We may choose to accept these objects as outliers or
prefer to increase the threshold so that fewer outliers result.

9. The probability density of a point x with respect to a multivariate normal
distribution having a mean µ and covariance matrix Σ is given by the equa-
tion

prob(x) =
1

(
√

2π)m|Σ|1/2 e
− (x−µ)Σ−1(x−µ)T2 . (10.1)

Using the sample mean x and covariance matrix S as estimates of the mean
µ and covariance matrix Σ, respectively, show that the log(prob(x)) is equal
to the Mahalanobis distance between a data point x and the sample mean x
plus a constant that does not depend on x.

log prob(x) = − log ((
√

2π)m|Σ|1/2) − 1
2
(x − µ)Σ−1(x − µ)T .

If we use the sample mean and covariance as estimates of µ and Σ, respec-
tively, then

log prob(x) = − log ((
√

2π)m|S|1/2) − 1
2
(x − x)S−1(x − x)T

162 Chapter 10 Anomaly Detection

The constant and the constant factor do not affect the ordering of this quan-
tity, only their magnitude. Thus, if we want to base a distance on this
quantity, we can keep only the variable part, which is the Mahalanobis dis-
tance.

10. Compare the following two measures of the extent to which an object belongs
to a cluster: (1) distance of an object from the centroid of its closest cluster
and (2) the silhouette coefficient described in Section 8.5.2.

The first measure is somewhat limited since it disregards that fact that the
object may also be close to another cluster. The silhouette coefficient takes
into account both the distance of an object to its cluster and its distance
to other clusters. Thus, it can be more informative about how strongly an
object belongs to the cluster to which it was assigned.

11. Consider the (relative distance) K-means scheme for outlier detection de-
scribed in Section 10.5 and the accompanying figure, Figure 10.10.

(a) The points at the bottom of the compact cluster shown in Figure 10.10
have a somewhat higher outlier score than those points at the top of
the compact cluster. Why?

The mean of the points is pulled somewhat upward from the center of
the compact cluster by point D.

(b) Suppose that we choose the number of clusters to be much larger, e.g.,
10. Would the proposed technique still be effective in finding the most
extreme outlier at the top of the figure? Why or why not?

No. This point would become a cluster by itself.
(c) The use of relative distance adjusts for differences in density. Give an

example of where such an approach might lead to the wrong conclusion.

If absolute distances are important. For example, consider heart rate
monitors for patients. If the heart rate goes above or below a specified
range of values, then this has an physical meaning. It would be incorrect
not to identify any patient outside that range as abnormal, even though
there may be a group of patients that are relatively similar to each other
and all have abnormal heart rates.

12. If the probability that a normal object is classified as an anomaly is 0.01 and
the probability that an anomalous object is classified as anomalous is 0.99,
then what is the false alarm rate and detection rate if 99% of the objects are
normal? (Use the definitions given below.)

detection rate =
number of anomalies detected

total number of anomalies

false alarm rate =
number of false anomalies

number of objects classified as anomalies

163

The detection rate is simply 99%.
The false alarm rate = 0.99m× 0.01/(0.99m× 0.01+0.01m× 0.99) = 0.50 =
50%.

13. When a comprehensive training set is available, a supervised anomaly detec-
tion technique can typically outperform an unsupervised anomaly technique
when performance is evaluated using measures such as the detection and false
alarm rate. However, in some cases, such as fraud detection, new types of
anomalies are always developing. Performance can be evaluated according
to the detection and false alarm rates, because it is usually possible to de-
termine, upon investigation, whether an object (transaction) is anomalous.
Discuss the relative merits of supervised and unsupervised anomaly detection
under such conditions.

When new anomalies are to be detected, an unsupervised anomaly detection
scheme must be used. However, supervised anomaly detection techniques are
still important for detecting known types of anomalies. Thus, both super-
vised and unsupervised anomaly detection methods should be used. A good
example of such a situation is network intrusion detection. Profiles or sig-
natures can be created for well-known types of intrusions, but cannot detect
new types of intrusions.

14. Consider a group of documents that has been selected from a much larger
set of diverse documents so that the selected documents are as dissimilar
from one another as possible. If we consider documents that are not highly
related (connected, similar) to one another as being anomalous, then all of
the documents that we have selected might be classified as anomalies. Is it
possible for a data set to consist only of anomalous objects or is this an abuse
of the terminology?

The connotation of anomalous is that of rarity and many of the definitions
of an anomaly incorporate this notion to some extent. However, there are
situations in which an anomaly typically does not occur very often, e.g., a
network failure, but has a very concrete definition. This makes it possible to
distinguish an anomaly in an absolute sense and for a situation to arise where
the majority of objects are anomalous. Also, in providing mathematical or
algorithmic definitions of an anomaly, it can happen that these definitions
produce situations in which many or all objects of a data set can be classified
as anomalies. Another viewpoint might say that if it is impossible to define
a meaningful normal situation, then all objects are anomalous. (“Unique” is
the term typically used in this context.) In summary, this can be regarded
as a philosophical or semantic question. A good argument (although likely
not an uncontested one) can be made that it is possible to have a collection
of objects that are mostly or all anomalies.

164 Chapter 10 Anomaly Detection

15. Consider a set of points, where most points are in regions of low density, but
a few points are in regions of high density. If we define an anomaly as a point
in a region of low density, then most points will be classified as anomalies.
Is this an appropriate use of the density-based definition of an anomaly or
should the definition be modified in some way?

If the density has an absolute meaning, such as assigned by the domain, then
it may be perfectly reasonable to consider most of the points as anomalous.
(See the answer to the previous exercise.) However, in many circumstances,
the appropriate approach would be to use an anomaly detection technique
that takes the relative density into account.

16. Consider a set of points that are uniformly distributed on the interval [0,1].
Is the statistical notion of an outlier as an infrequently observed value mean-
ingful for this data?

Not really. The traditional statistical notion of an outlier relies on the idea
that an object with a relatively low probability is suspect. With a uniform
distribution, no such distinction can be made.

17. An analyst applies an anomaly detection algorithm to a data set and finds
a set of anomalies. Being curious, the analyst then applies the anomaly
detection algorithm to the set of anomalies.

(a) Discuss the behavior of each of the anomaly detection techniques de-
scribed in this chapter. (If possible, try this for real data sets and
algorithms.)

(b) What do you think the behavior of an anomaly detection algorithm
should be when applied to a set of anomalous objects?

In some cases, such as the statistically-based anomaly detection techniques,
it would not be valid to apply the technique a second time, since the assump-
tions would no longer hold. This could also be true for other model-based
approaches. The behavior of the proximity- and density-based approaches
would depend on the particular techniques. Interestingly, the approaches
that use absolute thresholds of distance or density would likely classify the
set of anomalies as anomalies, at least if the original parameters were kept.
The relative approaches would likely classify most of the anomalies as normal
and some as anomalies.

Whether an object is anomalous depends on the the entire group of objects,
and thus, it is probably unreasonable to expect that an anomaly detection
technique will identify a set of anomalies as such in the absence of the original
data set.

BIBLIOGRAPHY 165

Bibliography
[1] W. W. Cohen. Fast effective rule induction. In Proc. of the 12th Intl. Conf. on Machine

Learning, pages 115–123, Tahoe City, CA, July 1995.

[2] S. Cost and S. Salzberg. A weighted nearest neighbor algorithm for learning with
symbolic features. Machine Learning, 10:57–78, 1993.

[3] J. Fürnkranz and G. Widmer. Incremental reduced error pruning. In Proc. of the 11th
Intl. Conf. on Machine Learning, pages 70–77, New Brunswick, NJ, July 1994.

[4] J. Hartigan. Clustering Algorithms. Wiley, New York, 1975.

[5] R. A. Jarvis and E. A. Patrick. Clustering using a similarity measure based on shared
nearest neighbors. IEEE Transactions on Computers, C-22(11):1025–1034, 1973.

<< /ASCII85EncodePages false /AllowTransparency false /AutoPositionEPSFiles true /AutoRotatePages /None /Binding /Left /CalGrayProfile (Dot Gain 20%) /CalRGBProfile (sRGB IEC61966-2.1) /CalCMYKProfile (U.S. Web Coated \050SWOP\051 v2) /sRGBProfile (sRGB IEC61966-2.1) /CannotEmbedFontPolicy /Error /CompatibilityLevel 1.4 /CompressObjects /Tags /CompressPages true /ConvertImagesToIndexed true /PassThroughJPEGImages true /CreateJDFFile false /CreateJobTicket false /DefaultRenderingIntent /Default /DetectBlends true /ColorConversionStrategy /LeaveColorUnchanged /DoThumbnails false /EmbedAllFonts true /EmbedJobOptions true /DSCReportingLevel 0 /SyntheticBoldness 1.00 /EmitDSCWarnings false /EndPage -1 /ImageMemory 1048576 /LockDistillerParams false /MaxSubsetPct 100 /Optimize true /OPM 1 /ParseDSCComments true /ParseDSCCommentsForDocInfo true /PreserveCopyPage true /PreserveEPSInfo true /PreserveHalftoneInfo false /PreserveOPIComments false /PreserveOverprintSettings true /StartPage 1 /SubsetFonts true /TransferFunctionInfo /Apply /UCRandBGInfo /Preserve /UsePrologue false /ColorSettingsFile () /AlwaysEmbed [ true ] /NeverEmbed [ true ] /AntiAliasColorImages false /DownsampleColorImages true /ColorImageDownsampleType /Bicubic /ColorImageResolution 300 /ColorImageDepth -1 /ColorImageDownsampleThreshold 1.50000 /EncodeColorImages true /ColorImageFilter /DCTEncode /AutoFilterColorImages true /ColorImageAutoFilterStrategy /JPEG /ColorACSImageDict << /QFactor 0.15 /HSamples [1 1 1 1] /VSamples [1 1 1 1] >>
/ColorImageDict << /QFactor 0.15 /HSamples [1 1 1 1] /VSamples [1 1 1 1] >>
/JPEG2000ColorACSImageDict << /TileWidth 256 /TileHeight 256 /Quality 30 >>
/JPEG2000ColorImageDict << /TileWidth 256 /TileHeight 256 /Quality 30 >>
/AntiAliasGrayImages false
/DownsampleGrayImages true
/GrayImageDownsampleType /Bicubic
/GrayImageResolution 300
/GrayImageDepth -1
/GrayImageDownsampleThreshold 1.50000
/EncodeGrayImages true
/GrayImageFilter /DCTEncode
/AutoFilterGrayImages true
/GrayImageAutoFilterStrategy /JPEG
/GrayACSImageDict << /QFactor 0.15 /HSamples [1 1 1 1] /VSamples [1 1 1 1] >>
/GrayImageDict << /QFactor 0.15 /HSamples [1 1 1 1] /VSamples [1 1 1 1] >>
/JPEG2000GrayACSImageDict << /TileWidth 256 /TileHeight 256 /Quality 30 >>
/JPEG2000GrayImageDict << /TileWidth 256 /TileHeight 256 /Quality 30 >>
/AntiAliasMonoImages false
/DownsampleMonoImages true
/MonoImageDownsampleType /Bicubic
/MonoImageResolution 1200
/MonoImageDepth -1
/MonoImageDownsampleThreshold 1.50000
/EncodeMonoImages true
/MonoImageFilter /CCITTFaxEncode
/MonoImageDict << /K -1 >>
/AllowPSXObjects false
/PDFX1aCheck false
/PDFX3Check false
/PDFXCompliantPDFOnly false
/PDFXNoTrimBoxError true
/PDFXTrimBoxToMediaBoxOffset [
0.00000
0.00000
0.00000
0.00000
]
/PDFXSetBleedBoxToMediaBox true
/PDFXBleedBoxToTrimBoxOffset [
0.00000
0.00000
0.00000
0.00000
]
/PDFXOutputIntentProfile ()
/PDFXOutputCondition ()
/PDFXRegistryName (http://www.color.org)
/PDFXTrapped /Unknown
/Description << /ENU (Use these settings to create PDF documents with higher image resolution for high quality pre-press printing. The PDF documents can be opened with Acrobat and Reader 5.0 and later. These settings require font embedding.) /JPN
/FRA
/DEU
/PTB
/DAN
/NLD
/ESP
/SUO
/ITA
/NOR
/SVE
>>
>> setdistillerparams
<< /HWResolution [2400 2400] /PageSize [612.000 792.000] >> setpagedevice

Homework

1. (5 pts) List two reasons why Traditional data analysis tools and techniques are not enough for Extracting useful information for today’s Data.

___________________________________________________________

b. ___________________________________________________________

2. (5 pts) Give one example from Chapter 1 showing how the field of Molecular biology understood the structure and function of genes

a. Before data mining

_____________________________________________________________________

b. And using Data mining

_____________________________________________________________________

3. (5 pts) Give two examples to show how Information retrieval techniques are different from Data

mining:

a. ___________________________________________________________
b. ___________________________________________________________

4. (5 pts) KDD is the Process of converting raw data into useful information using transformation steps. List those 4 steps below:

a. ___________________________________________________________
b. ___________________________________________________________

c. ___________________________________________________________

d. ___________________________________________________________

5. (5 pts) List the four Core data mining tasks that is covered in Chapter 1:

a. ___________________________________________________________
b. ___________________________________________________________
c. ___________________________________________________________
d. ___________________________________________________________

(5 pts) How Regression is different than Classification in Predictive modeling?

a. Regression:

___________________________________________________________
___________________________________________________________

b. Classification:

___________________________________________________________
___________________________________________________________

Turn in your highest-quality paper
Get a qualified writer to help you with

“ Answer below questions in homework 1 document with 100+ words each in Apa format with no plagiarism and 1+ reference ”

Get high-quality paper

NEW! AI matching with writer

Hire a Writer

Client Reviews

4.9

Sitejabber

4.6

Trustpilot

4.8

Our Guarantees

100% Confidentiality

Information about customers is confidential and never disclosed to third parties.

Original Writing

We complete all papers from scratch. You can get a plagiarism report.

Timely Delivery

No missed deadlines – 97% of assignments are completed in time.

Money Back

If you're confident that a writer didn't follow your order details, ask for a refund.

New to Your Trusted Assignment Help Service? Sign up & Save

Calculate the price of your order

Type of paper needed:

Pages:

You will get a personal manager and a discount.

Academic level:

We'll send you the first draft for approval by at

Total price:

$0.00

Power up Your Academic Success with the
Team of Professionals. We’ve Got Your Back.

Power up Your Study Success with Experts We’ve Got Your Back.

Order Now Order Now