22



Modularity is useful for selecting the number of clusters:

Q Next time: Why not optimize Modularity directly?

J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

23



Undirected graph

𝑮(𝑽, 𝑬):

1 2



Bi-partitioning task:



3

Divide vertices into two disjoint groups 𝑨, 𝑩

4 5

A

1 5

B

2 3 4 6



Questions:

 How can we define a “good” partition of 𝑮 ?

 How can we efficiently identify such a partition?

J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

25

6



What makes a good partition?

 Maximize the number of within-group connections  Minimize the number of between-group connections

5 1 2 A 3 4 B 6

J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

26



Express partitioning objectives as a function of the “edge cut” of the partition



Cut:

Set of edges with only one vertex in a group: A

2 1 3 4 5 6

B

cut(A,B) = 2

J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

27

 

Criterion:

Minimum-cut

 Minimize weight of connections between groups

arg min A,B Degenerate case:

cut(A,B)

“Optimal cut” Minimum cut



Problem:

 Only considers external cluster connections  Does not consider internal cluster connectivity J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

28

[Shi-Malik] 

Criterion:

Normalized-cut

[Shi-Malik, ’97]  Connectivity between groups relative to the density of each group  𝒗𝒐𝒍(𝑨) : total weight of the edges with at least one endpoint in 𝑨 : 𝒗𝒐𝒍 𝑨 = 𝒊∈𝑨

Why use this criterion?

𝒌 𝒊   Produces more balanced partitions

How do we efficiently find a good partition?

 Problem: Computing optimal cut is NP-hard J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

29

  

A

: adjacency matrix of undirected G 

x A

ij

=1 if (𝒊, 𝒋) is an edge, else 0 is a vector in 

n

with components (𝒙 𝟏 , … , 𝒙 𝒏 )  Think of it as a label/value of each node of 𝑮

What is the meaning of

A



x

?



Entry

y i

is a sum of labels

x j

of neighbors of

i

J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

30

 

j th



coordinate of

A



x

:

 Sum of the

x

-values of neighbors of

j

 Make this a new value at node

j

Spectral Graph Theory:

 𝑨 ⋅ 𝒙 = 𝝀 ⋅ 𝒙 Analyze the “spectrum” of matrix representing 𝑮

Spectrum:

Eigenvectors 𝒙 𝒊 of a graph, ordered by the magnitude (strength) of their corresponding eigenvalues 𝝀 𝒊 : J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

31

  Suppose all nodes in 𝑮 and 𝑮 is connected have degree 𝒅

What are some eigenvalues/vectors of

𝑮

?

𝑨  𝒙 = 𝝀 ⋅ 𝒙 What is  ? What

x

?



Let’s try:

𝒙 = (𝟏, 𝟏, … , 𝟏) 

Then:

𝑨 ⋅ 𝒙 = 𝒅, 𝒅, … , 𝒅 = 𝝀 ⋅ 𝒙 . So: 𝝀 = 𝒅 

We found eigenpair of

𝑮

:

𝒙 = (𝟏, 𝟏, … , 𝟏)

,

𝝀 = 𝒅

Remember the meaning of

𝒚 = 𝑨  𝒙

:

J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

32

Details!

      G is d-regular connected, A is its adjacency matrix

Claim:

 d is largest eigenvalue of A,  d has multiplicity of 1 (there is only 1 eigenvector associated with eigenvalue d)

Proof: Why no eigenvalue

𝒅 ′ > 𝒅

?

 To obtain d we needed 𝒙 𝒊 = 𝒙 𝒋 for every 𝑖, 𝑗  This means 𝒙 = 𝑐 ⋅ (1,1, … , 1) for some const. 𝑐

Define:

𝑺 = nodes 𝒊 with maximum possible value of 𝒙 𝒊 Then consider some vector 𝒚 which is not a multiple of vector (𝟏, … , 𝟏) . So not all nodes 𝒊 (with labels 𝒚 𝒊 ) are in 𝑺 Consider some node node 𝒋 𝒋 ∈ 𝑺 and a neighbor gets a value strictly less than 𝒅 𝒊 ∉ 𝑺 then  So 𝑦 is not eigenvector! And so 𝒅 is the largest eigenvalue!

J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

33

  

What if

𝑮

is not connected?

 𝑮 has 2 components, each 𝒅 -regular

What are some eigenvectors?

A B

 𝒙 = Put all 𝟏 s on 𝑨 and   𝟎 s on 𝑩 or vice versa 𝒙′ = (𝟏, … , 𝟏, 𝟎, … , 𝟎) |A| |B| 𝒙′′ = (𝟎, … , 𝟎, 𝟏, … , 𝟏)

then then

𝐀 ⋅ 𝒙′ = 𝒅, … , 𝒅, 𝟎, … , 𝟎 𝑨 ⋅ 𝒙′′ = (𝟎, … , 𝟎, 𝒅, … , 𝒅)  And so in both cases the corresponding 𝝀 = 𝒅

A bit of intuition: A B A B

2 nd largest eigval. 𝜆 𝑛−1 now has value very close to 𝜆 𝑛 𝝀 𝒏 = 𝝀 𝒏−𝟏 𝝀 𝒏 − 𝝀 𝒏−𝟏 ≈ 𝟎 J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

34



More intuition:



A B A B

2 nd largest eigval. 𝜆 𝑛−1 now has value very close to 𝜆 𝑛 𝝀 𝒏 = 𝝀 𝒏−𝟏 𝝀 𝒏 − 𝝀 𝒏−𝟏 ≈ 𝟎 If the graph is connected (right example) then we already know that 𝒙 𝒏 = (𝟏, … 𝟏) is an eigenvector   Since eigenvectors are orthogonal then the components of 𝒙 𝒏−𝟏 sum to 0.

 Why? Because 𝒙 𝒏 ⋅ 𝒙 𝒏−𝟏 = 𝒊 𝒙 𝒏 𝒊 ⋅ 𝒙 𝒏−𝟏 [𝒊] So we can look at the eigenvector of the 2 eigenvalue and declare nodes with positive label in A and negative label in B. nd largest 

But there is still lots to sort out.

J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

35



Adjacency matrix (

A

):



n



n

matrix 

A=[a ij ], a ij =1

if edge between node

i

and

j

 2 1 3 4 5 6 1 1 2 3 4 0 5 6 1 0 0 1 1 2 1 0 1 0 0 0 3 1 1 0 1 0 0 4 0 0 1 0 1 1 5 1 0 0 1 0 1 6 0 0 0 1 1 0 J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

36



Degree matrix (D):



n



n

diagonal matrix 

D=[d ii ], d ii =

degree of node

i

2 1 3 4 5 6 5 6 1 2 3 4 1 3 0 0 0 0 0 2 0 2 0 0 0 0 3 0 0 3 0 0 0 4 0 0 0 3 0 0 5 0 0 0 0 3 0 6 0 0 0 0 0 2 J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

37



Laplacian matrix (L):



n



n

symmetric matrix 5 1 2 3 4 6 1 2 3 4 5 6 1 3 -1 -1 0 -1 0 2 -1 2 -1 0 0 0 3 -1 -1 3 -1 0 0 4 0 0 -1 3 -1 -1  

What is trivial eigenpair?

𝑳 = 𝑫 − 𝑨  𝒙 = (𝟏, … , 𝟏) then 𝑳 ⋅ 𝒙 = 𝟎 and so 𝝀 = 𝝀 𝟏

Important properties:

 Eigenvalues are non-negative real numbers  Eigenvectors are real and orthogonal = 𝟎 5 -1 0 0 -1 3 -1 6 0 0 0 -1 -1 2 J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

38

Details!

(a) All eigenvalues are

≥ 0

(b) (c)

𝑥 𝑇 𝐿𝑥 = 𝑖𝑗 𝐿 = 𝑁 𝑇 ⋅ 𝑁 𝐿 𝑖𝑗 𝑥 𝑖 𝑥 𝑗 ≥ 0  for every 𝑥  That is, 𝐿 is positive semi-definite

Proof:



(c)



(b):

𝑥 𝑇 𝐿𝑥 = 𝑥 𝑇 𝑁 𝑇 𝑁𝑥 = 𝑥𝑁   𝑇 𝑁𝑥 ≥ 0  As it is just the square of length of 𝑁𝑥

(b)

𝑥 𝑇  (a): Let 𝐿𝑥 ≥ 0 so 𝝀 𝑥 be an eigenvalue of 𝑳 . Then by (b) 𝑇 𝐿𝑥 = 𝑥 𝑇 𝜆𝑥 = 𝜆𝑥 𝑇 𝑥  𝝀 ≥ 𝟎

(a)

 (c): is also easy! Do it yourself.

J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

39



Fact: For symmetric matrix

M

:

 2  min

x x T M x x T x



What is the meaning of min

x

T

L x

 x T L x = 𝑛 𝑖,𝑗=1 𝐿 𝑖𝑗 𝑥 𝑖 𝑥 𝑗 = 𝑛 𝑖,𝑗=1  = 𝑖 𝐷 𝑖𝑖 𝑥 𝑖 2 − 𝑖,𝑗 ∈𝐸 2𝑥 𝑖 𝑥 𝑗

on G?

𝐷 𝑖𝑗 − 𝐴 𝑖𝑗 𝑥 𝑖 𝑥 𝑗  = 𝑖,𝑗 ∈𝐸 (𝑥 𝑖 2 + 𝑥 𝑗 2 − 2𝑥 𝑖 𝑥 𝑗 ) = 𝒊,𝒋 ∈𝑬 𝒙 𝒊 Node 𝒊 has degree 𝒅 𝒊 . So, value 𝒙 𝒊 𝟐 But each edge (𝒊, 𝒋) needs to be summed up 𝒅 𝒊 has two endpoints so we need 𝒙 𝒊 𝟐 +𝒙 𝟐 𝒋 times.

J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

− 𝒙 𝒋 𝟐 40

 2  min

x x T M x x T x

Details!

   Write 𝑥 in axes of eigenvecotrs 𝑤 1 , 𝑤 2 , … , 𝑤 𝑛 𝑴 . So, 𝑥 = 𝑖 𝑛 𝛼 𝑖 𝑤 𝑖 Then we get: 𝑀𝑥 = 𝑖 So, what is 𝒙 𝑻 𝑴𝒙 ?

𝛼 𝑖 𝑀𝑤 𝑖 𝝀 𝒊 𝒘 𝒊 = 𝑖 𝛼 𝑖 𝜆 𝑖 𝑤 𝑖 = 𝟎 of

if

𝒊 ≠ 𝒋

1 otherwise

 𝑥 𝑇 𝑀𝑥 = 𝑖 𝛼 𝑖 𝑤 𝑖 𝑖 𝛼 𝑖 𝜆 𝑖 𝑤 𝑖 = 𝑖𝑗 𝛼 𝑖 𝜆 𝑗 𝛼 𝑗 𝑤 𝑖 𝑤 𝑗   = 𝑖 𝛼 𝑖 𝜆 𝑖 𝑤 𝑖 𝑤 𝑖 = 𝒊 𝝀 𝒊 𝜶 𝒊 𝟐 To minimize this over all unit vectors x orthogonal to: w = min over choices of (𝛼 1 , … 𝛼 𝑛 ) 𝛼 𝑖 2 = 1 (unit length) 𝛼 𝑖 = 0 so that: (orthogonal to 𝑤 1 ) To minimize this, set 𝜶 𝟐 = 𝟏 and so 𝒊 𝝀 𝒊 𝜶 𝒊 𝟐 = 𝝀 𝟐 J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

41



What else do we know about

x

?

  𝒙 is unit vector: 𝒊 𝒙 𝒊 𝟐 = 𝟏 𝒙 𝒊 is orthogonal to 1

st

𝒙 𝒊 ⋅ 𝟏 = 𝒊 𝒙 𝒊 = 𝟎 eigenvector (𝟏, … , 𝟏) thus:  

Remember:

2  min All labelings  of nodes 𝑖 so (

i

, that 𝑥 𝑖 = 0 ) 

E j

 (

x i i x i

2 

x j

) 2

We want to assign values

𝒙 𝒊

to nodes that few edges cross 0.

i

(we want x i and x j such to subtract each other)

𝑥 𝑖 J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

0

Balance to minimize

𝑥 𝑗 42 x

  

Back to finding the optimal cut Express partition (A,B) as a vector

𝒚 𝒊 = +𝟏 −𝟏 𝒊𝒇 𝒊 ∈ 𝑨 𝒊𝒇 𝒊 ∈ 𝑩 We can minimize the cut of the partition by finding a non-trivial vector

x

that

minimizes

:

Can’t solve exactly. Let’s relax

𝒚

allow it to take any real value.

and

𝑦 𝑖 = −1 0 J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

𝑦 𝑗 = +1 43

𝑥 𝑖 0  𝝀 𝟐 = 𝐦𝐢𝐧 𝒚 𝒇 𝒚

:

The minimum value of given by the 2 nd smallest eigenvalue

λ 2

Laplacian matrix

L

𝒇(𝒚) of the is 𝑥 𝑗  𝐱 = 𝐚𝐫𝐠 𝐦𝐢𝐧 𝐲 𝒇 𝒚

:

The optimal solution for is given by the corresponding eigenvector 𝒙 ,

y

referred as the

Fiedler vector

J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

x 44

Details!

  Suppose there is a partition of G into A and B (# 𝑒𝑑𝑔𝑒𝑠 𝑓𝑟𝑜𝑚 𝐴 𝑡𝑜 𝐵) where 𝐴 ≤ |𝐵| , s.t.

𝜶 = 𝐴 then 2𝜶 ≥ 𝝀 𝟐  This is the approximation guarantee of the spectral clustering. It says the cut spectral finds is at most 2 away from the optimal one of score 𝜶 .

Proof:

 Let: a=|A|, b=|B| and e= # edges from A to B  Enough to choose some 𝒙 𝒊 2 that: 𝜆 2 ≤ 𝑥 𝑖 −𝑥 𝑗 𝑖 𝑥 𝑖 2 ≤ 2𝛼 based on A and B such (while also 𝑖 𝑥 𝑖 = 0 ) 𝝀 𝟐

is only smaller

J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

45

Details!



Proof (continued):



1) Let’s set:

𝒙 𝒊 = − + 𝟏 𝒂 𝟏 𝒃 𝒊𝒇 𝒊 ∈ 𝑨 𝒊𝒇 𝒊 ∈ 𝑩  Let’s quickly verify that 𝑖 𝑥 𝑖 = 0: 𝑎 − 1 𝑎 + 𝑏 1 𝑏 = 𝟎 

2) Then:

𝑥 𝑖 −𝑥 𝑗 𝑖 𝑥 𝑖 2 2 = 𝑖∈𝐴,𝑗∈𝐵 𝑎 − 1 𝑎 2 1 1 1 1 𝟐 𝑒 + ≤ 𝑒 + ≤ 𝒆 𝑎 𝑏 𝑎 𝑎 𝒂 e … number of edges between A and B 1 𝑏 +𝑏 + 1 𝑎 1 𝑏 2 2 = 𝟐𝜶 = 𝑒⋅ 1 𝑎 1 𝑎 + 1 𝑏 + 1 𝑏 2 = Which proves that the cost achieved by spectral is better than twice the OPT cost J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

46

Details!



Putting it all together:

 𝜶 𝟐 𝟐𝜶 ≥ 𝝀 𝟐 ≥ 𝟐𝒌 𝒎𝒂𝒙 where 𝑘 𝑚𝑎𝑥 in the graph  is the maximum node degree Note we only provide the 1 st part: 𝟐𝜶 ≥ 𝝀 𝟐   We did not prove 𝝀 𝟐 𝜶 𝟐 ≥ 𝟐𝒌 𝒎𝒂𝒙 Overall this always certifies that 𝝀 𝟐 useful bound always gives a J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

47

 

How to define a “good” partition of a graph?

Minimize a given graph cut criterion  

How to efficiently identify such a partition?

Approximate using information provided by the eigenvalues and eigenvectors of a graph 

Spectral Clustering

J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

48



Three basic stages:



1) Pre-processing

 Construct a matrix representation of the graph 

2) Decomposition

 Compute eigenvalues and eigenvectors of the matrix  Map each point to a lower-dimensional representation based on one or more eigenvectors 

3) Grouping

 Assign points to two or more clusters, based on the new representation J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

49



1)



Pre-processing:

Build Laplacian matrix

L

of the graph 1 5 6 2 3 4 1 3 -1 -1 0 -1 0 0 0 2 -1 2 -1 0 3 -1 -1 3 -1 0 0 0 -1 3 -1 -1 4 0 5 -1 0 0 -1 3 -1 0 0 -1 -1 2 6 0 

2) Decomposition:

 Find eigenvalues and eigenvectors

x

of the matrix

L

 

=

0.0

1.0

3.0

3.0

4.0

5.0

X =

0.4

0.4

0.4

0.4

0.4

0.4

0.3

0.6

0.3

-0.3

-0.3

-0.6

-0.5

0.4

0.1

0.1

-0.5

0.4

-0.2

-0.4

0.6

-0.4

0.4

-0.4

0.6

-0.2

0.4

0.4

-0.4

-0.4

-0.5

0.0

0.5

-0.5

0.5

0.0

 Map vertices to corresponding components of 

2

1 2 3 4 5 6

0.3

0.6

0.3

-0.3

-0.3

-0.6

J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

How do we now find the clusters?

50

 

3)



Grouping:

Sort components of reduced 1-dimensional vector  Identify clusters by splitting the sorted vector in two

How to choose a splitting point?

 Naïve approaches:  Split at 0 or median value  More expensive approaches:  Attempt to minimize normalized cut in 1-dimension (sweep over ordering of nodes induced by the eigenvector) 1 2 3 4 5 6 0.3

0.6

0.3

-0.3

-0.3

-0.6

Split at 0:

Cluster A: Positive points Cluster B: Negative points 1 0.3

4 -0.3

2 0.6

5 -0.3

3 0.3

6 -0.6

J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

A B 51

Rank in

x 2

J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

52

Components of x

2

Rank in

x 2

J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

53

Components of

x 1

Components of

x 3

J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

54



How do we partition a graph into

k

clusters?



Two basic approaches:



Recursive bi-partitioning

[Hagen et al., ’92]  Recursively apply bi-partitioning algorithm in a hierarchical divisive manner  Disadvantages: Inefficient, unstable 

Cluster multiple eigenvectors

[Shi-Malik, ’00]  Build a reduced space from multiple eigenvectors  Commonly used in recent papers  A preferable approach… J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

55

   

Approximates the optimal cut

[Shi-Malik, ’00]  Can be used to approximate optimal

k

-way normalized cut

Emphasizes cohesive clusters

 Increases the unevenness in the distribution of the data  Associations between similar points are amplified, associations between dissimilar points are attenuated  The data begins to “approximate a clustering”

Well-separated space

 Transforms data to a new “embedded space”, consisting of

k

orthogonal basis vectors Multiple eigenvectors prevent instability due to information loss J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

56

[Kumar et al. ‘99]  

Searching for small communities in the Web graph

What is the signature of a community / discussion in a Web graph?

Use this to define “topics”:

What the same people on the left talk about on the right

Remember HITS!

Dense 2-layer graph

Intuition:

Many people all talking about the same things J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

58



A more well-defined problem:

Enumerate complete bipartite subgraphs K

s,t

 Where K

s,t

:

to the same

t s

nodes on the “left” where each links other nodes on the “right” |X| = s = 3 |Y| = t = 4 X Y

K 3,4

Fully connected

J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

59

[Agrawal-Srikant ‘99]   

Market basket analysis.

Setting:   Market: Universe

U

of

n

items Baskets: m subsets of

U

:

S 1 , S 2 , …, S m

(

S i

is a set of items one person bought)   Support: Frequency threshold

f

Goal:



U

Find all subsets

(items in

T T

s.t.

T



S i

of at least

were bought together at least

f f

sets

S i

times)

What’s the connection between the itemsets and complete bipartite graphs?

J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

60

[Kumar et al. ‘99]

Frequent itemsets

=

complete bipartite graphs!



How?

 View each node

i

set

S i

of nodes

i

as a points to 

K s,t

= a set

Y

of size that occurs in

s

sets

t S i

 Looking for K

s,t

 set of frequency threshold to

s

and look at layer

t

– all frequent sets of size

t

i

X k j i

a b c d S i ={a,b,c,d}

a b c d Y

t s

… minimum support (|X|=s) … itemset size (|Y|=t) J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

61

[Kumar et al. ‘99] View each node i as a set S

i

of nodes i points to

a i b c d S i ={a,b,c,d}

Find frequent itemsets:

s

… minimum support

t

… itemset size

x

Say we find a

frequent itemset

Y={a,b,c} of supp s So, there are s nodes that link to all of {a,b,c}:

a a b c y b c z We found K

s,t

!

K s,t

= a set Y of size t that occurs in

s

sets S

i

X x y z J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

a b c Y 62

a b c

a c d e f Itemsets:

a = {b,c,d} b = {d} c = {b,d,e,f} d = {e,f} e = {b,d} f = {}

b

 

Support threshold s=2

 {b,d}: support 3  {e,f}: support 2

And we just found 2 bipartite subgraphs:

a c d b c e

J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

e f d

63



Example of a community from a web graph Nodes on the right Nodes on the left

[Kumar, Raghavan, Rajagopalan, Tomkins: Trawling the Web for emerging cyber-communities 1999] J. Leskovec, A. Rajaraman, J. Ullman: Mining of Massive Datasets, http://www.mmds.org

64