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Unformatted text preview: 11.2 ArcColoured Directed Multigraphs 623 some u ∈ ∪ k i =1 ( V ( P i ) ∪ V ( Q i )). Let ‘ be the literal of c j corresponding to u (that is, if u ∈ P i , then by (d) and the definition of D , ‘ = x i and if u ∈ Q i , then ‘ = x i ). If u ∈ V ( P i ), then C uses the path Q i and the truth assignment above will put ‘ = x i = 1. If u ∈ V ( Q i ), then C uses P j and x i is assigned the value 0, implying that ‘ = x i = 1 . This shows that the clause c j is satisfied. Since this argument is valid for all clauses we see that the truth assignment described above satisfies F . Suppose now that F has a satisfying truth assignment t = { t 1 ,t 2 ,...,t k } (see Section 1.10). Then we can fix, for each clause c j one literal l j which is true according to this assignment. Let ‘ 1 ,‘ 2 ,...,‘ m denote these fixed literals. Note that since t is a truth assignment, none of the chosen literals is the negation of another. By the construction of D there is a unique path C j = c j u j c j +1 which corresponds to the choice of ‘ j (that is, u j ∈ P i if ‘ j = x i and u j ∈ Q j if ‘ j = x i ). Furthermore if ‘ j 1 = ‘ j 2 , for some j 1 6 = j 2 , then u j 1 6 = u j 2 . For each i = 1 , 2 ,...,k fix one of the paths P i ,Q i as follows: if ‘ r = x i for some r ∈ { 1 , 2 ,...,m } , then T i = Q i , otherwise T i = P i . By the comment above this assignment of subpaths always chooses one subpath for each i = 1 , 2 ,...,k . Now it is easy to see that the following is an alternating directed cycle in D C 1 C 2 ...C m T 1 T 2 ...T k c 1 . This completes the proof of the lemma. ut To complete the proof of Theorem 11.2.2, it suffices to observe that the digraph D can be constructed in polynomial time for any given instance of the 3SAT problem. ut We do not know what the complexity of the ADC problem is when re stricted to tournaments. Problem 11.2.5 [371] Does there exist a polynomial algorithm to check whether a 2arccoloured tournament has an alternating cycle? Figure 11.8 illustrates the difficulty of this problem. In contrast to the ‘un coloured’ case, the 2arccoloured tournament T in Figure 11.8 has a unique alternating cycle, which is hamiltonian. Therefore, a reduction to ‘short’ al ternating cycles may well be impossible. Proposition 11.2.6 The cycle C in the tournament T of Figure 11.8 is hamiltonian and consists of the matching of colour 1 from R to B and the matching of colour 2 from B to R . If we reverse any arc of colour 1 in C , we obtain a tournament with no alternating cycle. Proof: Exercise 11.31. ut 624 11. Generalizations of Digraphs 1 1 1 1 2 2 2 2 2 2 2 B R 1 1 1 2 Figure 11.8 A 2arccoloured tournament with a unique alternating cycle C . All arcs within R ( B ) are of colour 1 (2). The cycle C is hamiltonian and consists of the matching of colour 1 from R to B and the matching of colour 2 from B to R ....
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This document was uploaded on 08/10/2011.
 Spring '11
 Algorithms

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