# sol13 - CS577 Spring09: Solution to Homework 13∗ 1...

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Unformatted text preview: CS577 Spring09: Solution to Homework 13∗ 1 Butterﬂy Species Consistency This is an extension to the bipartiteness testing problem. Consider each butterﬂy specimen as a node in an undirected graph. The edges are deﬁned by the given set of judgements, labeled “+1” (same) or “-1” (different). Now the “consistency” question becomes: is there a partition of V into A and B, s.t. all “+1” edges connect nodes from the same group while all “-1” edges connect one node from A and the other from B. We can test this extended “bipartiteness” property using binary coloring. Algorithm 1 Extended Bipartiteness Testing 1: INPUT: An undirected graph G = (V, E) where V = {1, 2, · · · , n} and E is represented by adjacency lists N (v) = {x|(v, x) ∈ E or (x, v) ∈ E} for v ∈ V . Each node v ∈ V has a label v.color ∈ {0, 9, −9} initialized to be 0. Each edge e ∈ E has a label e.label ∈ {+1, −1} as deﬁned by the set of judgements. 2: OUTPUT: True iff G passes the test 3: DATA STRUCTURES: • Array visited[1 . . . n], all elements initially f alse 4: 5: 6: 7: 8: 9: 10: 11: while there exists some node u such that visited[u] = f alse do u.color ← 9 result ← BinColor(u) if result! = T rue then return F alse end if end while return T rue ∗ Please send any questions to Jason fniu@wisc.edu 1 Algorithm 2 BinColor(x) 1: visited[x] ← true 2: for all v ∈ N (x) and the corresponding edge e do 3: if visited[v] = f alse then 4: v.color ← x.color × e.label 5: result ← BinColor(v) 6: if result! = T rue then 7: return F alse 8: end if 9: else 10: if v.color! = x.color × e.label then 11: return F alse 12: end if 13: end if 14: end for 15: return T rue 2 ...
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## This note was uploaded on 10/02/2009 for the course CS 577 taught by Professor Nabil during the Spring '09 term at Lahore University of Management Sciences.

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sol13 - CS577 Spring09: Solution to Homework 13∗ 1...

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