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Data Mining
Association Rules: Advanced Concepts and Algorithms

Lecture Notes for Chapter 7

Introduction to Data Mining

by

Tan, Steinbach, Kumar

Continuous and Categorical Attributes

Example of Association Rule:

{Number of Pages [5,10)  (Browser=Mozilla)}  {Buy = No}

How to apply association analysis formulation to non-asymmetric binary variables?

Session Id

Country

Session Length (sec)

Number of Web Pages viewed

Gender

Browser Type

Buy

1

USA

982

8

Male

IE

No

2

China

811

10

Female

Netscape

No

3

USA

2125

45

Female

Mozilla

Yes

4

Germany

596

4

Male

IE

Yes

5

Australia

123

9

Male

Mozilla

No

…

…

…

…

…

…

…

10

Handling Categorical Attributes

  • Transform categorical attribute into asymmetric binary variables

  • Introduce a new “item” for each distinct attribute-value pair
  • Example: replace Browser Type attribute with
  • Browser Type = Internet Explorer
  • Browser Type = Mozilla
  • Browser Type = Mozilla

Handling Categorical Attributes

  • Potential Issues
  • What if attribute has many possible values
  • Example: attribute country has more than 200 possible values
  • Many of the attribute values may have very low support

Potential solution: Aggregate the low-support attribute values

  • What if distribution of attribute values is highly skewed
  • Example: 95% of the visitors have Buy = No
  • Most of the items will be associated with (Buy=No) item

Potential solution: drop the highly frequent items

Handling Continuous Attributes

  • Different kinds of rules:
  • Age[21,35)  Salary[70k,120k)  Buy
  • Salary[70k,120k)  Buy  Age: =28, =4

  • Different methods:
  • Discretization-based
  • Statistics-based
  • Non-discretization based
  • minApriori

Handling Continuous Attributes

  • Use discretization
  • Unsupervised:
  • Equal-width binning
  • Equal-depth binning
  • Clustering

  • Supervised:

bin1

bin3

bin2

Attribute values, v

Class v1 v2 v3 v4 v5 v6 v7 v8 v9
Anomalous 0 0 20 10 20 0 0 0 0
Normal 150 100 0 0 0 100 100 150 100

Discretization Issues

  • Size of the discretized intervals affect support & confidence

  • If intervals too small
  • may not have enough support
  • If intervals too large
  • may not have enough confidence
  • Potential solution: use all possible intervals

{Refund = No, (Income = $51,250)}  {Cheat = No}

{Refund = No, (60K  Income  80K)}  {Cheat = No}

{Refund = No, (0K  Income  1B)}  {Cheat = No}

Discretization Issues

  • Execution time
  • If intervals contain n values, there are on average O(n2) possible ranges

  • Too many rules

{Refund = No, (Income = $51,250)}  {Cheat = No}

{Refund = No, (51K  Income  52K)}  {Cheat = No}

{Refund = No, (50K  Income  60K)}  {Cheat = No}

Approach by Srikant & Agrawal

  • Preprocess the data
  • Discretize attribute using equi-depth partitioning
  • Use partial completeness measure to determine number of partitions
  • Merge adjacent intervals as long as support is less than max-support

  • Apply existing association rule mining algorithms

  • Determine interesting rules in the output

Approach by Srikant & Agrawal

  • Discretization will lose information

  • Use partial completeness measure to determine how much information is lost

C: frequent itemsets obtained by considering all ranges of attribute values
P: frequent itemsets obtained by considering all ranges over the partitions

P is K-complete w.r.t C if P  C,and X  C,  X’  P such that:

1. X’ is a generalization of X and support (X’)  K  support(X) (K  1)
2. Y  X,  Y’  X’ such that support (Y’)  K  support(Y)

Given K (partial completeness level), can determine number of intervals (N)

X

Approximated X

Interestingness Measure

  • Given an itemset: Z = {z1, z2, …, zk} and its generalization Z’ = {z1’, z2’, …, zk’}
    P(Z): support of Z
    EZ’(Z): expected support of Z based on Z’

  • Z is R-interesting w.r.t. Z’ if P(Z)  R  EZ’(Z)

{Refund = No, (Income = $51,250)}  {Cheat = No}

{Refund = No, (51K  Income  52K)}  {Cheat = No}

{Refund = No, (50K  Income  60K)}  {Cheat = No}

Interestingness Measure

  • For S: X  Y, and its generalization S’: X’  Y’

P(Y|X): confidence of X  Y
P(Y’|X’): confidence of X’  Y’
ES’(Y|X): expected support of Z based on Z’

  • Rule S is R-interesting w.r.t its ancestor rule S’ if
  • Support, P(S)  R  ES’(S) or
  • Confidence, P(Y|X)  R  ES’(Y|X)

Statistics-based Methods

  • Example:

Browser=Mozilla  Buy=Yes  Age: =23

  • Rule consequent consists of a continuous variable, characterized by their statistics
  • mean, median, standard deviation, etc.
  • Approach:
  • Withhold the target variable from the rest of the data
  • Apply existing frequent itemset generation on the rest of the data
  • For each frequent itemset, compute the descriptive statistics for the corresponding target variable
  • Frequent itemset becomes a rule by introducing the target variable as rule consequent
  • Apply statistical test to determine interestingness of the rule

Statistics-based Methods

  • How to determine whether an association rule interesting?
  • Compare the statistics for segment of population covered by the rule vs segment of population not covered by the rule:

A  B:  versus A  B: ’

  • Statistical hypothesis testing:
  • Null hypothesis: H0: ’ =  + 
  • Alternative hypothesis: H1: ’ >  + 
  • Z has zero mean and variance 1 under null hypothesis

Statistics-based Methods

  • Example:

r: Browser=Mozilla  Buy=Yes  Age: =23

  • Rule is interesting if difference between  and ’ is greater than 5 years (i.e.,  = 5)
  • For r, suppose n1 = 50, s1 = 3.5
  • For r’ (complement): n2 = 250, s2 = 6.5

  • For 1-sided test at 95% confidence level, critical Z-value for rejecting null hypothesis is 1.64.
  • Since Z is greater than 1.64, r is an interesting rule

Min-Apriori (Han et al)

Example:

W1 and W2 tends to appear together in the same document

Document-term matrix:

Sheet1

TID W1 W2 W3 W4 W5 TID W1 W2 W3 W4 W5
D1 2 2 0 0 1 D1 0.4 0.4 0.0 0.0 0.2
D2 0 0 1 2 2 D2 0.0 0.0 0.2 0.4 0.4
D3 2 3 0 0 0 D3 0.4 0.6 0.0 0.0 0.0
D4 0 0 1 0 1 D4 0.0 0.0 0.5 0.0 0.5
D5 1 1 1 0 2 D5 0.2 0.2 0.2 0.0 0.4

Sheet2

Sheet3

Min-Apriori

  • Data contains only continuous attributes of the same “type”
  • e.g., frequency of words in a document

  • Potential solution:
  • Convert into 0/1 matrix and then apply existing algorithms
  • lose word frequency information
  • Discretization does not apply as users want association among words not ranges of words

Sheet1

TID W1 W2 W3 W4 W5 TID W1 W2 W3 W4 W5
D1 2 2 0 0 1 D1 0.4 0.4 0.0 0.0 0.2
D2 0 0 1 2 2 D2 0.0 0.0 0.2 0.4 0.4
D3 2 3 0 0 0 D3 0.4 0.6 0.0 0.0 0.0
D4 0 0 1 0 1 D4 0.0 0.0 0.5 0.0 0.5
D5 1 1 1 0 2 D5 0.2 0.2 0.2 0.0 0.4

Sheet2

Sheet3

Min-Apriori

  • How to determine the support of a word?
  • If we simply sum up its frequency, support count will be greater than total number of documents!
  • Normalize the word vectors – e.g., using L1 norm
  • Each word has a support equals to 1.0

Normalize

Sheet1

TID W1 W2 W3 W4 W5 TID W1 W2 W3 W4 W5
D1 2 2 0 0 1 D1 0.4 0.4 0.0 0.0 0.2
D2 0 0 1 2 2 D2 0.0 0.0 0.2 0.4 0.4
D3 2 3 0 0 0 D3 0.4 0.6 0.0 0.0 0.0
D4 0 0 1 0 1 D4 0.0 0.0 0.5 0.0 0.5
D5 1 1 1 0 2 D5 0.2 0.2 0.2 0.0 0.4

Sheet2

Sheet3

Sheet1

TID W1 W2 W3 W4 W5 TID W1 W2 W3 W4 W5
D1 2 2 0 0 1 D1 0.40 0.33 0.00 0.00 0.17
D2 0 0 1 2 2 D2 0.00 0.00 0.33 1.00 0.33
D3 2 3 0 0 0 D3 0.40 0.50 0.00 0.00 0.00
D4 0 0 1 0 1 D4 0.00 0.00 0.33 0.00 0.17
D5 1 1 1 0 2 D5 0.20 0.17 0.33 0.00 0.33

Sheet2

Sheet3

Min-Apriori

  • New definition of support:

Example:

Sup(W1,W2,W3)

= 0 + 0 + 0 + 0 + 0.17

= 0.17

Sheet1

TID W1 W2 W3 W4 W5 TID W1 W2 W3 W4 W5
D1 2 2 0 0 1 D1 0.40 0.33 0.00 0.00 0.17
D2 0 0 1 2 2 D2 0.00 0.00 0.33 1.00 0.33
D3 2 3 0 0 0 D3 0.40 0.50 0.00 0.00 0.00
D4 0 0 1 0 1 D4 0.00 0.00 0.33 0.00 0.17
D5 1 1 1 0 2 D5 0.20 0.17 0.33 0.00 0.33

Sheet2

Sheet3

Anti-monotone property of Support

Example:

Sup(W1) = 0.4 + 0 + 0.4 + 0 + 0.2 = 1

Sup(W1, W2) = 0.33 + 0 + 0.4 + 0 + 0.17 = 0.9

Sup(W1, W2, W3) = 0 + 0 + 0 + 0 + 0.17 = 0.17

Sheet1

TID W1 W2 W3 W4 W5 TID W1 W2 W3 W4 W5
D1 2 2 0 0 1 D1 0.40 0.33 0.00 0.00 0.17
D2 0 0 1 2 2 D2 0.00 0.00 0.33 1.00 0.33
D3 2 3 0 0 0 D3 0.40 0.50 0.00 0.00 0.00
D4 0 0 1 0 1 D4 0.00 0.00 0.33 0.00 0.17
D5 1 1 1 0 2 D5 0.20 0.17 0.33 0.00 0.33

Sheet2

Sheet3

Multi-level Association Rules

Multi-level Association Rules

  • Why should we incorporate concept hierarchy?
  • Rules at lower levels may not have enough support to appear in any frequent itemsets

  • Rules at lower levels of the hierarchy are overly specific
  • e.g., skim milk  white bread, 2% milk  wheat bread,
    skim milk  wheat bread, etc.
    are indicative of association between milk and bread

Multi-level Association Rules

  • How do support and confidence vary as we traverse the concept hierarchy?
  • If X is the parent item for both X1 and X2, then
    (X) ≤ (X1) + (X2)

  • If (X1  Y1) ≥ minsup,
    and X is parent of X1, Y is parent of Y1
    then (X  Y1) ≥ minsup, (X1  Y) ≥ minsup
    (X  Y) ≥ minsup

  • If conf(X1  Y1) ≥ minconf,
    then conf(X1  Y) ≥ minconf

Multi-level Association Rules

  • Approach 1:
  • Extend current association rule formulation by augmenting each transaction with higher level items

Original Transaction: {skim milk, wheat bread}

Augmented Transaction:
{skim milk, wheat bread, milk, bread, food}

  • Issues:
  • Items that reside at higher levels have much higher support counts
  • if support threshold is low, too many frequent patterns involving items from the higher levels
  • Increased dimensionality of the data

Multi-level Association Rules

  • Approach 2:
  • Generate frequent patterns at highest level first

  • Then, generate frequent patterns at the next highest level, and so on

  • Issues:
  • I/O requirements will increase dramatically because we need to perform more passes over the data
  • May miss some potentially interesting cross-level association patterns

Sequence Data

Sequence Database:

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Examples of Sequence Data

Sequence

E1
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E4

E2

Element (Transaction)

Event
(Item)

Sequence Database Sequence Element (Transaction) Event (Item)
Customer Purchase history of a given customer A set of items bought by a customer at time t Books, diary products, CDs, etc
Web Data Browsing activity of a particular Web visitor A collection of files viewed by a Web visitor after a single mouse click Home page, index page, contact info, etc
Event data History of events generated by a given sensor Events triggered by a sensor at time t Types of alarms generated by sensors
Genome sequences DNA sequence of a particular species An element of the DNA sequence Bases A,T,G,C

Formal Definition of a Sequence

  • A sequence is an ordered list of elements (transactions)

s = < e1 e2 e3 … >

  • Each element contains a collection of events (items)

ei = {i1, i2, …, ik}

  • Each element is attributed to a specific time or location

  • Length of a sequence, |s|, is given by the number of elements of the sequence

  • A k-sequence is a sequence that contains k events (items)

Examples of Sequence

  • Web sequence:

< {Homepage} {Electronics} {Digital Cameras} {Canon Digital Camera} {Shopping Cart} {Order Confirmation} {Return to Shopping} >

  • Sequence of initiating events causing the nuclear accident at 3-mile Island:
    (http://stellar-one.com/nuclear/staff_reports/summary_SOE_the_initiating_event.htm)

< {clogged resin} {outlet valve closure} {loss of feedwater}
{condenser polisher outlet valve shut} {booster pumps trip}
{main waterpump trips} {main turbine trips} {reactor pressure increases}>

  • Sequence of books checked out at a library:

<{Fellowship of the Ring} {The Two Towers} {Return of the King}>

Formal Definition of a Subsequence

  • A sequence <a1 a2 … an> is contained in another sequence <b1 b2 … bm> (m ≥ n) if there exist integers
    i1 < i2 < … < in such that a1  bi1 , a2  bi1, …, an  bin

  • The support of a subsequence w is defined as the fraction of data sequences that contain w
  • A sequential pattern is a frequent subsequence (i.e., a subsequence whose support is ≥ minsup)
Data sequence Subsequence Contain?
< {2,4} {3,5,6} {8} > < {2} {3,5} > Yes
< {1,2} {3,4} > < {1} {2} > No
< {2,4} {2,4} {2,5} > < {2} {4} > Yes

Sequential Pattern Mining: Definition

  • Given:
  • a database of sequences
  • a user-specified minimum support threshold, minsup

  • Task:
  • Find all subsequences with support ≥ minsup

Sequential Pattern Mining: Challenge

  • Given a sequence: <{a b} {c d e} {f} {g h i}>
  • Examples of subsequences:

<{a} {c d} {f} {g} >, < {c d e} >, < {b} {g} >, etc.

  • How many k-subsequences can be extracted from a given n-sequence?

<{a b} {c d e} {f} {g h i}> n = 9

k=4: Y _ _ Y Y _ _ _ Y

<{a} {d e} {i}>

Sequential Pattern Mining: Example

Minsup = 50%

Examples of Frequent Subsequences:

< {1,2} > s=60%

< {2,3} > s=60%

< {2,4}> s=80%

< {3} {5}> s=80%

< {1} {2} > s=80%

< {2} {2} > s=60%

< {1} {2,3} > s=60%

< {2} {2,3} > s=60%

< {1,2} {2,3} > s=60%

Extracting Sequential Patterns

  • Given n events: i1, i2, i3, …, in

  • Candidate 1-subsequences:

<{i1}>, <{i2}>, <{i3}>, …, <{in}>

  • Candidate 2-subsequences:

<{i1, i2}>, <{i1, i3}>, …, <{i1} {i1}>, <{i1} {i2}>, …, <{in-1} {in}>

  • Candidate 3-subsequences:

<{i1, i2 , i3}>, <{i1, i2 , i4}>, …, <{i1, i2} {i1}>, <{i1, i2} {i2}>, …,

<{i1} {i1 , i2}>, <{i1} {i1 , i3}>, …, <{i1} {i1} {i1}>, <{i1} {i1} {i2}>, …

Generalized Sequential Pattern (GSP)

  • Step 1:
  • Make the first pass over the sequence database D to yield all the 1-element frequent sequences

  • Step 2:

Repeat until no new frequent sequences are found

  • Candidate Generation:
  • Merge pairs of frequent subsequences found in the (k-1)th pass to generate candidate sequences that contain k items

  • Candidate Pruning:
  • Prune candidate k-sequences that contain infrequent (k-1)-subsequences

  • Support Counting:
  • Make a new pass over the sequence database D to find the support for these candidate sequences

  • Candidate Elimination:
  • Eliminate candidate k-sequences whose actual support is less than minsup

Candidate Generation

  • Base case (k=2):
  • Merging two frequent 1-sequences <{i1}> and <{i2}> will produce two candidate 2-sequences: <{i1} {i2}> and <{i1 i2}>

  • General case (k>2):
  • A frequent (k-1)-sequence w1 is merged with another frequent
    (k-1)-sequence w2 to produce a candidate k-sequence if the subsequence obtained by removing the first event in w1 is the same as the subsequence obtained by removing the last event in w2
  • The resulting candidate after merging is given by the sequence w1 extended with the last event of w2.

If the last two events in w2 belong to the same element, then the last event in w2 becomes part of the last element in w1

Otherwise, the last event in w2 becomes a separate element appended to the end of w1

Candidate Generation Examples

  • Merging the sequences
    w1=<{1} {2 3} {4}> and w2 =<{2 3} {4 5}>
    will produce the candidate sequence < {1} {2 3} {4 5}> because the last two events in w2 (4 and 5) belong to the same element

  • Merging the sequences
    w1=<{1} {2 3} {4}> and w2 =<{2 3} {4} {5}>
    will produce the candidate sequence < {1} {2 3} {4} {5}> because the last two events in w2 (4 and 5) do not belong to the same element

  • We do not have to merge the sequences
    w1 =<{1} {2 6} {4}> and w2 =<{1} {2} {4 5}>
    to produce the candidate < {1} {2 6} {4 5}> because if the latter is a viable candidate, then it can be obtained by merging w1 with
    < {1} {2 6} {5}>

GSP Example

< {1} {2} {3} > < {1} {2 5} >�< {1} {5} {3} > < {2} {3} {4} > < {2 5} {3} > < {3} {4} {5} > < {5} {3 4} >�

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< {1} {2 5} {3} >�

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Timing Constraints (I)

{A B} {C} {D E}

<= ms

<= xg

>ng

xg: max-gap

ng: min-gap

ms: maximum span

xg = 2, ng = 0, ms= 4

Data sequence Subsequence Contain?
< {2,4} {3,5,6} {4,7} {4,5} {8} > < {6} {5} > Yes
< {1} {2} {3} {4} {5}> < {1} {4} > No
< {1} {2,3} {3,4} {4,5}> < {2} {3} {5} > Yes
< {1,2} {3} {2,3} {3,4} {2,4} {4,5}> < {1,2} {5} > No

Mining Sequential Patterns with Timing Constraints

  • Approach 1:
  • Mine sequential patterns without timing constraints
  • Postprocess the discovered patterns

  • Approach 2:
  • Modify GSP to directly prune candidates that violate timing constraints
  • Question:
  • Does Apriori principle still hold?

Apriori Principle for Sequence Data

Suppose:

xg = 1 (max-gap)

ng = 0 (min-gap)

ms = 5 (maximum span)

minsup = 60%

<{2} {5}> support = 40%

but

<{2} {3} {5}> support = 60%

Problem exists because of max-gap constraint

No such problem if max-gap is infinite

Contiguous Subsequences

  • s is a contiguous subsequence of
    w = <e1>< e2>…< ek>
    if any of the following conditions hold:

s is obtained from w by deleting an item from either e1 or ek

s is obtained from w by deleting an item from any element ei that contains more than 2 items

s is a contiguous subsequence of s’ and s’ is a contiguous subsequence of w (recursive definition)

  • Examples: s = < {1} {2} >
  • is a contiguous subsequence of
    < {1} {2 3}>, < {1 2} {2} {3}>, and < {3 4} {1 2} {2 3} {4} >
  • is not a contiguous subsequence of
    < {1} {3} {2}> and < {2} {1} {3} {2}>

Modified Candidate Pruning Step

  • Without maxgap constraint:
  • A candidate k-sequence is pruned if at least one of its (k-1)-subsequences is infrequent

  • With maxgap constraint:
  • A candidate k-sequence is pruned if at least one of its contiguous (k-1)-subsequences is infrequent

Timing Constraints (II)

xg: max-gap

ng: min-gap

ws: window size

ms: maximum span

xg = 2, ng = 0, ws = 1, ms= 5

Data sequence Subsequence Contain?
< {2,4} {3,5,6} {4,7} {4,6} {8} > < {3} {5} > No
< {1} {2} {3} {4} {5}> < {1,2} {3} > Yes
< {1,2} {2,3} {3,4} {4,5}> < {1,2} {3,4} > Yes

Modified Support Counting Step

  • Given a candidate pattern: <{a, c}>
  • Any data sequences that contain

<… {a c} … >,
<… {a} … {c}…> ( where time({c}) – time({a}) ≤ ws)
<…{c} … {a} …> (where time({a}) – time({c}) ≤ ws)

will contribute to the support count of candidate pattern

Other Formulation

  • In some domains, we may have only one very long time series
  • Example:
  • monitoring network traffic events for attacks
  • monitoring telecommunication alarm signals
  • Goal is to find frequent sequences of events in the time series
  • This problem is also known as frequent episode mining

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E3 E5

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E1

E2

E3 E1

Pattern: <E1> <E3>

General Support Counting Schemes

Assume:

xg = 2 (max-gap)

ng = 0 (min-gap)

ws = 0 (window size)

ms = 2 (maximum span)

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Frequent Subgraph Mining

  • Extend association rule mining to finding frequent subgraphs
  • Useful for Web Mining, computational chemistry, bioinformatics, spatial data sets, etc

Graph Definitions

Representing Transactions as Graphs

  • Each transaction is a clique of items

Sheet1

Transaction Id Items
1 {A,B,C,D}
2 {A,B,E}
3 {B,C}
4 {A,B,D,E}
5 {B,C,D}

Sheet2

Sheet3

Representing Graphs as Transactions

Challenges

  • Node may contain duplicate labels
  • Support and confidence
  • How to define them?
  • Additional constraints imposed by pattern structure
  • Support and confidence are not the only constraints
  • Assumption: frequent subgraphs must be connected
  • Apriori-like approach:
  • Use frequent k-subgraphs to generate frequent (k+1) subgraphs
  • What is k?

Challenges…

  • Support:
  • number of graphs that contain a particular subgraph

  • Apriori principle still holds

  • Level-wise (Apriori-like) approach:
  • Vertex growing:
  • k is the number of vertices
  • Edge growing:
  • k is the number of edges

Vertex Growing

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  • Find frequent 1-subgraphs
  • Repeat
  • Candidate generation
  • Use frequent (k-1)-subgraphs to generate candidate k-subgraph
  • Candidate pruning
  • Prune candidate subgraphs that contain infrequent
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  • Support counting
  • Count the support of each remaining candidate
  • Eliminate candidate k-subgraphs that are infrequent

In practice, it is not as easy. There are many other issues

Example: Dataset

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Candidate Generation

  • In Apriori:
  • Merging two frequent k-itemsets will produce a candidate (k+1)-itemset

  • In frequent subgraph mining (vertex/edge growing)
  • Merging two frequent k-subgraphs may produce more than one candidate (k+1)-subgraph

Multiplicity of Candidates (Vertex Growing)

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Multiplicity of Candidates (Edge growing)

  • Case 1: identical vertex labels

Multiplicity of Candidates (Edge growing)

  • Case 2: Core contains identical labels

Core: The (k-1) subgraph that is common
between the joint graphs

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Multiplicity of Candidates (Edge growing)

  • Case 3: Core multiplicity

Adjacency Matrix Representation

  • The same graph can be represented in many ways

Sheet1

A(1) A(2) A(3) A(4) B(5) B(6) B(7) B(8)
A(1) 1 1 1 0 1 0 0 0
A(2) 1 1 0 1 0 1 0 0
A(3) 1 0 1 1 0 0 1 0
A(4) 0 1 1 1 0 0 0 1
B(5) 1 0 0 0 1 1 1 0
B(6) 0 1 0 0 1 1 0 1
B(7) 0 0 1 0 1 0 1 1
B(8) 0 0 0 1 0 1 1 1
A(1) A(2) A(3) A(4) B(5) B(6) B(7) B(8)
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B(7) 0 1 0 0 1 1 1 0
B(8) 0 0 0 1 1 1 0 1

Sheet2

Sheet3

Sheet1

A(1) A(2) A(3) A(4) B(5) B(6) B(7) B(8)
A(1) 1 1 1 0 1 0 0 0
A(2) 1 1 0 1 0 1 0 0
A(3) 1 0 1 1 0 0 1 0
A(4) 0 1 1 1 0 0 0 1
B(5) 1 0 0 0 1 1 1 0
B(6) 0 1 0 0 1 1 0 1
B(7) 0 0 1 0 1 0 1 1
B(8) 0 0 0 1 0 1 1 1
A(1) A(2) A(3) A(4) B(5) B(6) B(7) B(8)
A(1) 1 1 0 1 0 1 0 0
A(2) 1 1 1 0 0 0 1 0
A(3) 0 1 1 1 1 0 0 0
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B(5) 0 0 1 0 1 0 1 1
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Graph Isomorphism

  • A graph is isomorphic if it is topologically equivalent to another graph

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  • During candidate counting, to check whether a candidate is contained within another graph

Graph Isomorphism

  • Use canonical labeling to handle isomorphism
  • Map each graph into an ordered string representation (known as its code) such that two isomorphic graphs will be mapped to the same canonical encoding
  • Example:
  • Lexicographically largest adjacency matrix

String: 0010001111010110

Canonical: 0111101011001000

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Length

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Type

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2 China 811 10 Female Netscape No

3 USA 2125 45 Female Mozilla Yes

4 Germany 596 4 Male IE Yes

5 Australia 123 9 Male Mozilla No

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Frequent

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A(2)

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A(3)

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A(4)

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B(5)

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B(7)

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B(8)

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A(2)

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A(3)

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B(5)

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B(7)

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