Digraphs+Theory,+Algorithms+and+Applications_Part33

Digraphs+Theory,+Algorithms+and+Applications_Part33 - 11.2...

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Unformatted text preview: 11.2 Arc-Coloured 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 3-SAT 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 2-arc-coloured tournament has an alternating cycle? Figure 11.8 illustrates the difficulty of this problem. In contrast to the ‘un- coloured’ case, the 2-arc-coloured 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 2-arc-coloured 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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Digraphs+Theory,+Algorithms+and+Applications_Part33 - 11.2...

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