Digraphs+Theory,+Algorithms+and+Applications_Part18

Digraphs+Theory,+Algorithms+and+Applications_Part18 - 6.9...

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Unformatted text preview: 6.9 Oriented Hamiltonian Paths and Cycles 323 1 2 3 4 5 6 7 8 9 10 11 12 Figure 6.6 An oriented path with intervals [1 , 3] , [3 , 6] , [6 , 7] , [7 , 8] , [8 , 10] , [10 , 11] , [11 , 12]. For every set X ⊆ V in a tournament T = ( V,A ), we define the sets R + ( X ) ( R- ( X )) to be those vertices that can be reached from (can reach) the set X by a directed path. By definition X ⊆ R + ( X ) ∩ R- ( X ). A vertex u is an out-generator ( in-generator ) of T if R + ( u ) = V ( R- ( u ) = V ). Recall that by Theorem 1.4.5, every tournament T has at least one out-generator and at least one in-generator. In fact, by Proposition 4.10.2, a vertex is an out-generator (in-generator) if and only if it is the initial (terminal) vertex of at least one hamiltonian path in T . The next result, due to Havet and Thomass´ e, deals with oriented paths covering all but one vertex in a tournament. It plays an important role in the proof of Theorem 6.9.3 in [408]. Theorem 6.9.4 [408] Let T = ( V,A ) be a tournament on n + 1 vertices. Then (1) For every out-path P on n vertices and every choice of distinct vertices x,y such that | R + ( { x,y } ) | ≥ ‘ 1 ( P ) + 1 , either x or y is an origin of (a copy of) P in T . (2) For every in-path P on n vertices and every choice of distinct vertices x,y such that | R- ( { x,y } ) | ≥ ‘ 1 ( P ) + 1 , either x or y is an origin of (a copy of) P in T . The following is an easy corollary of Theorem 6.9.4. We state it now since we shall use it in the inductive proof below. Corollary 6.9.5 [694] Every tournament T on n vertices contains every oriented path P on n- 1 vertices. Moreover, every subset of ‘ 1 ( P )+1 vertices contains an origin of P . In particular, there are at least two distinct origins of P in T . ut Proof of Theorem 6.9.4: (We follow the proof in [408]). The proof is by induction on n and clearly holds for n = 1. Now suppose that the theorem holds for all tournaments on at most n vertices. It suffices to prove (1) since (2) can be reduced to (1) by considering the converses of T and P . Let P = u 1 u 2 ...u n be given and let x,y be distinct vertices such that | R + ( { x,y } ) | ≥ ‘ 1 ( P ) + 1. We may assume that x → y and hence R + ( x ) = R + ( { x,y } ). We consider two cases. 324 6. Hamiltonian Refinements Case 1 ‘ 1 ( P ) ≥ 2: If | N + ( x ) | ≥ 2, let z ∈ N + ( x ) be an out-generator of T h R + ( x )- x i and let t ∈ N + ( x ) be distinct from z . By the definition of z we have that | R + T- x ( { t,z } ) | = | R + ( x ) | - 1 > ‘ 1 ( * P ). Note that * P is an out-path, since ‘ 1 ( P ) > 1. By the induction hypothesis, either z or t is the origin of * P in T- x , implying that x is an origin of P in T ....
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Digraphs+Theory,+Algorithms+and+Applications_Part18 - 6.9...

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