COT 6410 Homework #2. Due next Tuesday in class (June 5, 2007)
Consider the following decision problem:
Disjoint Clique Cover (DCC)
Given: a graph G = (V, E) and an integer K.
Question: Can V be partitioned into k ≤ K sets, V1, V2, …, Vk where each set forms
a complete subgraph in G?
Find (and demonstrate the validity of) a decision problem for which there is a
polynomial transformation to, and from, DCC. {Hint: we have seen this problem.}
Solution: Use the Graph k-Colorability Problem.
Graph K-Color Problem (COL)
Given: a graph G and in integer k.
Question: Can G be properly colored using at most k colors?
"properly" means the end points of each edge in G must have a different color.
COL DCC
1) Construct the instance of DCC from an arbitrary instance of COL.
Let (G, k) be an arbitrary instance ICOL of COL. Construct the complement graph
G' of G and let the k of DCC be the k of COL. Thus, the constructed instance
f(ICOL) = IDCC of DCC is (G', k). Since this takes order the number of edges of G
time, and the number of edges is O(n2), the time is polynomial in n, the number of
vertices of G.
2) Show ICOL is true for COL if and only if f(ICOL) is true for DCC
Suppose ICOL is true. Then, there are k' ≤ k color classes C1, C2, …, Ck' of G
which partitions the set of vertices (i.e., every vertex is in one and only one of the
Ci's.). The vertices in each Ci are all mutually independent (no edges among
them). Therefore, in the constructed instance (G', k) the vertices in each Ci will
have all edges present (a complete subgraph, or "clique"). Since all vertices are
present, the Ci's form a disjoint clique cover of G'. Therefore, f(ICOL) is a true
instance of DCC.
Similarly, suppose the constructed instance f(ICOL) is true. Then, the set of
vertices of the k' ≤ k cliques identified, C1, C2, …, Ck' partition the set of vertices
in both G' and G. In G, each Ci is an independent set (since it is a clique in G').
Thus, the vertices can be "colored" i. Thus, k' ≤ k colors are used, and the original
instance ICOL is true for COL.
DCC COL
The transformation above is 1-1 onto, so has an inverse and the above proof suffices.
Otherwise, one can give the very similar argument as above.
Therefore, COL and DCC are polynomially equivalent problems.
