Games of No Chance 3 2009
DOI: 10.1017/cbo9780511807251.024
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Unsolved problems in Combinatorial Games

Abstract: Abstract. This periodically updated reference resource is intended to put eager researchers on the path to fame and (perhaps) fortune.As in our earlier articles, WW stands for Winning Ways [Berlekamp et al. 1982]. We say that the nim-value of a position is n when its value is the nimber * n.

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Cited by 20 publications
(31 citation statements)
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“…By Proposition 4 we thus see that (14) holds also for G 6 . In conclusion, S = P for G 6 , where the first few entries of S are displayed in Table 6.…”
Section: Hammer Wins Even When the Theory Fails!mentioning
confidence: 60%
See 1 more Smart Citation
“…By Proposition 4 we thus see that (14) holds also for G 6 . In conclusion, S = P for G 6 , where the first few entries of S are displayed in Table 6.…”
Section: Hammer Wins Even When the Theory Fails!mentioning
confidence: 60%
“…for all 0 m < n. As we pointed out, 2 5 = 7 > 2 6 = 6, so without monotonicity, we do not seem to be able to get (14). We use an auxiliary result.…”
Section: Hammer Wins Even When the Theory Fails!mentioning
confidence: 99%
“…Consider the 5-pendant graceful (4, 5)-labelling (4,9,3,7,10,2,11,1,8,6,5) for P 11 . Using this along with the 5-pendant graceful labelling (5, 1, 4, 2, 3) for P 5 leads to the following terrace for Z 16 , which has match list (4, 5):…”
Section: Examplementioning
confidence: 99%
“…Reversing this and lifting to Z 32 can give the symmetrically sectionable directed terrace (0, 1,31,18,30,9,24,6,13,3,12,20,23,27,21,26,10,5,11,7,4,28,19,29,22,8,25,14,2,15,17,16) and hence a cyclic solution to OP (4,4,5,5,15) from the 2-factor [6,13,3,12,20], [23,27,21,26], [10,5,11,7], [4,28,19,29,22], [8, 25, 14, 2, 15, 17, 16, ∞, 0, 1, 31, 18, 30, 9, 24].…”
Section: Examplementioning
confidence: 99%
“…The well-known Oberwolfach problem [3] was first formulated by Ringel in 1967: "Is it possible to seat an odd number n of people at s round tables T 1 , T 2 ,... , T s (where each T i can accommodate t i ≥ 3 people and ∑ti = n) for k different meals so that each person has every other person for a neighbor exactly once?" In terms of graph theory, this problem is equivalent to asking for an odd integer n, is it possible for K n to have a 2-factorization in which each 2-factor consists of cycles of lengths t 1 , t 2 ,... , t s ?…”
Section: Introductionmentioning
confidence: 99%