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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
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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
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if visited[v] = f alse then
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v.color ← x.color × e.label
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result ← BinColor(v)
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if result! = T rue then
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return F alse
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end if
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else
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if v.color! = x.color × e.label then
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return F alse
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end if
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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.
 Spring '09
 Nabil

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