Frequent Subgraph Mining Frequent Subgraph Mining (FSM) Outline FSM - PowerPoint PPT Presentation

Frequent Subgraph Mining

Frequent Subgraph Mining (FSM) Outline • FSM Preliminaries • FSM Algorithms – gSpan – complete FSM on labeled graphs – SUBDUE – approximate FSM on labeled graphs – SLEUTH – FSM on trees • Review

FSM In a Nutshell • Discovery of graph structures that occur a significant number of times across a set of graphs • Ex.: Common occurrences of hydroxide-ion • Other instances: – Finding common biological pathways among species. – Recurring patterns of humans interaction during an epidemic. Carbonic Acid – Highlighting similar data to reveal data set as a whole. H O H Sulfuric Acid Acetic Acid O O O C O H O S H C C Ammonia O O H H O H H H H N H

FSM Preliminaries • Support is some integer or frequency • Frequent graphs occur more than support number of times. O-H present in ¾ inputs � frequent if support <= 3 Carbonic Acid H O H Sulfuric Acid Acetic Acid O O O C O H O S H C C Ammonia O O H H O H H H H N H

What Makes FSM So Hard? • Isomorphic graphs have same structural properties even though they may look different. • Subgraph isomorphism problem : Does a graph contain a subgraph isomorphic to another graph? • FSM algorithms encounter this problem while buildings graphs. • This problem is known to be NP-complete ! A B A B Isomorphic under A,B,C,D labeling C D D C

Pattern Growth Approach • Underlying strategy of both traditional frequent pattern mining and frequent subgraph mining • General Process: – candidate generation : which patterns will be considered? For FSM, – candidate pruning : if a candidate is not a viable frequent pattern, can we exploit the pattern to prevent unnecessary work? • subgraphs and subsets exponentiate as size increases! – support counting : how many of a given pattern exist? • These algorithms work in a breadth-first or depth-first way. – Joins smaller frequent sets into larger ones. – Checks the frequency of larger sets.

Pattern Growth Approach – Apriori • Apriori principle: if an itemset is frequent, then all of its subsets are also frequent. – Ex. if itemset {A, B, C, D} is frequent, then {A, B} is frequent. – Simple proof: With respect to frequency, all sets trivially contain their subsets, thus frequency of subset >= frequency of set. – Same property applies to (sub)graphs! • Apriori algorithm exploits this to prune huge sections of the search space! ∅ If A is infrequent, no A B C supersets with A can be frequent! AB AC BC ABC

FSM Algorithms Discussed • gSpan – complete frequent subgraph mining – improves performance over straightforward apriori extensions to graphs through DFS Code representation and aggressive candidate pruning • SUBDUE – approximate frequent subgraph mining – uses graph compression as metric for determining a “frequently occuring” subgraph • SLEUTH – complete frequent subgraph mining – built specifically for trees

FSM – R package • R package for FSM is called subgraphMining • To import: install.packages(“subgraphMining”) • Package contains: gSpan, SUBDUE, SLUETH. • Also contains the following data sets: – cslogs – metabolicInteractions. • To load the data, use the following code: # The cslogs data set data(cslogs) # The matabolicInteractions data data(metabolicInteractions)

FSM Outline • FSM Preliminaries • FSM Algorithms – gSpan – SUBDUE – SLEUTH • Review

gSpan: Graph-Based Substructure Pattern Mining • Written by Xifeng Yan & Jiawei Han in 2002. • Form of pattern-growth mining algorithm. – Adds edges to candidate subgraph – Also known as, edge extension • Avoid cost intensive problems like – Redundant candidate generation – Isomorphism testing • Uses two main concepts to find frequent subgraphs – DFS lexicographic order – minimum DFS code

gSpan Inputs • Set of graphs, support • Graph of form � = ( � , � , � � , � � ) – � , � – vertex and edge sets – � � – vertex labels – � � – edge labels – label sets need not be one-to-one H O H � � = { � , � , � } � � = { single−bond, double−bond } O C O

gSpan Components Strategy: • build frequent subgraphs bottom-up , using DFS code as regularized representation • eliminate redundancies via minimal DFS codes based on code lexicographic ordering Depth-first Search (DFS) Code structured graph representation for building, comparing DFS Lexicographic minimal DFS code Order selection, pruning of canonical comparison subgraphs of graphs

Depth First Search Primer Todo…?

gSpan: DFS codes Code Edge # DFS Code: sequence of edges traversed during DFS 0 (0,1,X,a,Y) 1 (1,2,Y,b,X) Vertex discovery 2 (2,0,X,a,X) times 3 (2,3,X,c,Z) 0 X 4 (3,1,Z,b,Y) 5 (1,4,Y,d,Z) a a 1 Y d Format: ( � , � , � � , � ( � , � ) , � � ) 4 b Z � , � – vertices by time of discovery b X 2 � � , � � - vertex labels of � � , � � c � ( � , � ) – edge label between � � , � � 3 Z � < � : forward edge � > � : back edge

DFS Code: Edge Ordering • Edges in code ordered in very specific manner, corresponding to DFS process • � � = ( � � , � � ), � � = ( � � , � � ) • � � ≺ � � � � � appears before � � in code • Ordering rules: 1. if � � = � � and � � < � � � � � ≺ � � • from same source vertex, � � traversed before � � in DFS 2. if � � < � � and � � = � � � � � ≺ � � • � � is a forward edge and � � traversed as result of � � traversal if � � ≺ � � and � � ≺ � � , � � � ≺ � � 3. • ordering is transitive

DFS Code: Edge Ordering Example 0 Code Edge # X • Rule applications by edge # a a 0 (0,1,X,a,Y) • 0 ≺ 1 (Rule 2) 1 Y 1 (1,2,Y,b,X) d • 1 ≺ 2 (Rule 2) 2 (2,0,X,a,X) 4 b Z • 0 ≺ 2 (Rule 3) 3 (2,3,X,c,Z) b 2 X 4 (3,1,Z,b,Y) • 2 ≺ 3 (Rule 1) 5 (1,4,Y,d,Z) c • Exercise: what 3 Z others? Edge ordering can be recorded easily during the DFS!

Graphs have multiple DFS Codes! Exercise: Write the 2 rightmost graphs using DFS code 0 0 0 X Y X X d 4 b a a a a a a Z 1 1 X 1 X Y Y b d d 4 a a b b Z Z c b b b Y 2 X 2 X 2 X d b c c c Z 3 3 3 Z Z Z Z 4 solution to redundant DFS codes: lexical ordering, minimal code!

DFS Lexicographic Ordering vs. DFS Code • DFS code: Ordering of edge sequence of a particular DFS – E.g. DFS’s that start at different vertices may have different DFS codes • Lexicographic ordering: ordering between different DFS codes

DFS Lexicographic Ordering • Given lexicographic ordering of label set � , ≺ � • Given graphs � � , � � (equivalent label sets). • Given DFS codes – � = code � � , � � = � � , � � , … , � � – � = code � � , � � = � � , � � , … , � � – (assume � ≥ � ) • � ≤ � iff either of the following are true: – ∃ � , 0 ≤ � ≤ min � , � such that • � � = � � for � < � and • � � ≺ � � � – � � = � � ��� 0 ≤ � ≤ �

DFS Lex. Ordering: Edge Comparison • Given DFS codes – � = code � � , � � = � � , � � , … , � � – � = code � � , � � = � � , � � , … , � � – (assume � ≥ � ) • Given � such that � � = � � for � < � • Given � � = � � , � � , � � � , � � � , � � , � � � , � � = � � , � � , � � � , � � � , � � , � � � , • � � ≺ � � � if one of the following cases Case 1 : Both forward edges, AND… Case 3 : � � back , � � forward � � � ≺ � � � Case 2 : Both back edges, AND…

Edge Comparison: Case 1 (both forward) • Both forward edges, AND one of the following: – � � < � � (edge starts from a later visited vertex) • Why is this (think about DFS process)? – � � = � � AND labels of � lexicographically less than labels of � , in order of tuple. • Ex: Labels are strings, � � = __, __, m, e, x , � � = (__, __, m, u, x) – m = m, e < u � � � ≺ � � � • Note: if both forward edges, then � � = � � – Reasoning: all previous edges equal, target vertex discovery times are the same

Edge Comparison: Case 2 (both back) • Both back edges, AND one of the following: – � � < � � (edge refers to earlier vertex) – � � = � � AND edge label of � lexicographically less than � • Note: given that all previous edges equal, vertex labels must also be equal • Note: if both back edges, then � � = � � – Reasoning: all previous edges equal, source vertex discovery times are the same.

Code (A) Code (B) Code (C) Edge # 0 (0,1,X,a,Y) (0,1,Y,a,X) (0,1,X,a,X) 1 (1,2,Y,b,X) (1,2,X,a,X) (1,2,X,a,Y) 2 (2,0,X,a,X) (2,0,X,b,Y) (2,0,Y,b,X) 3 (2,3,X,c,Z) (2,3,X,c,Z) (2,3,Y,b,Z) 4 (3,1,Z,b,Y) (3,1,Z,b,X) (3,0,Z,c,X) 5 (1,4,Y,d,Z) (0,4,Y,d,Z) (2,4,Y,d,Z) 0 0 0 X X Y d a 4 a b a a Z 1 1 Y 1 X X d b 4 b a a Z b b c X 2 X 2 Y 2 d c c b Z 3 3 3 Z Z Z 4