2012
DOI: 10.1002/jcd.21296
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2‐Starters, Graceful Labelings, and a Doubling Construction for the Oberwolfach Problem

Abstract: Every 1‐rotational solution of a classic or twofold Oberwolfach problem (OP) of order n is generated by a suitable 2‐factor (starter) of Kn or 2Kn, respectively. It is shown that any starter of a twofold OP of order n gives rise to a starter of a classic OP of order 2n−1 (doubling construction). It is also shown that by suitably modifying the starter of a classic OP, one may obtain starters of some other OPs of the same order but having different parameters. The combination of these two constructions leads to … Show more

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Cited by 22 publications
(51 citation statements)
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“…In the nonuniform case, it is known that OP( F ) has no solution if F{false[32false],false[34false],false[4,5false],false[32,5false]}, but that there is a solution for every other instance in which v40 . The problem OP( F ) has been solved if F has exactly two components (see also for 2‐factorizations of the complete multipartite graph), and if F is bipartite . Although complete solutions have been given for infinite families of orders , the general problem is still open.…”
Section: Introductionmentioning
confidence: 99%
“…In the nonuniform case, it is known that OP( F ) has no solution if F{false[32false],false[34false],false[4,5false],false[32,5false]}, but that there is a solution for every other instance in which v40 . The problem OP( F ) has been solved if F has exactly two components (see also for 2‐factorizations of the complete multipartite graph), and if F is bipartite . Although complete solutions have been given for infinite families of orders , the general problem is still open.…”
Section: Introductionmentioning
confidence: 99%
“…Application In , the authors found a method to construct 1‐rotational solutions to many different Oberwolfach problems. In particular, it is shown that for all α1 and k3 there exists a 1‐rotational solution to OP(2α+1, 2-0.15emk, 2α1-0.15em(2k)) when k is odd and OP(2α+1, 2α+1-0.15emk) when k is even (Proposition 3.9 in ).…”
Section: Using a 2‐starter Of Kbold-italicgtrue¯ To Obtain A 2n‐startmentioning
confidence: 99%
“…Application In , the authors found a method to construct 1‐rotational solutions to many different Oberwolfach problems. In particular, it is shown that for all α1 and k3 there exists a 1‐rotational solution to OP(2α+1, 2-0.15emk, 2α1-0.15em(2k)) when k is odd and OP(2α+1, 2α+1-0.15emk) when k is even (Proposition 3.9 in ). Therefore we obtain solutions to OP(p2α+1, 2p-0.15emk, p2α1-0.10em(2k)) for each odd prime p and OP(p2α+1, p(2α+1)-0.15emk) for all primes p after applying Theorem with the same conditions on α and k.…”
Section: Using a 2‐starter Of Kbold-italicgtrue¯ To Obtain A 2n‐startmentioning
confidence: 99%
“…First note that if T=[g1,g2,,g2n]scriptTbs(G), then T¯=[g¯1,g¯2,,g¯n]scriptTb(G¯). We need the following result due to Anderson whose proof will be sketched below following the doubling construction for Oberwolfach problems described in (recall that a HCS(2n+1) is nothing but a solution of the Oberwolfach problem OP(2n+1)). Theorem The counterimage of any basic terrace of G¯ under the map π:[g1,g2,,g2n]scriptTbs(G)[g¯1,g¯2,,g¯n]Tb(G¯),has size 2(n+a(G¯)1)/2, where a(G¯) is the number of involutions of G¯. Proof Let U=[g¯1,g¯2,,g¯n] be any given terrace of scriptTb(G¯).…”
Section: Enumerating 1‐rotational Hamiltonian Cycle Systems Up To Isomentioning
confidence: 99%
“…We need the following result due to Anderson [4] whose proof will be sketched below following the doubling construction for Oberwolfach problems described in [21] (recall that a HCS(2n + 1) is nothing but a solution of the Oberwolfach problem OP(2n + 1)).…”
Section: Enumerating 1-rotational Hamiltonian Cycle Systems Up To Isomentioning
confidence: 99%