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H.A. Kierstead and W.T. Trotter, Explicit matchings in the middle levels of the Boolean lattice, Order 5 (1988), 163-171.

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The Antipodal Layers Problem - Hurlbert   (8 citations)  (Correct)

....conjecture that M n;k is Hamiltonian is what we call the Antipodal Layers Problem, and we will prove the conjecture is true for n ck 2 k, k large enough. 2 Middle Layers The Middle Layers Problem rst appeared in [H] and again as problem 2.5 in [K] It has been veri ed [MR] for k 11. In [KT] the problem is attacked with the understanding that any Hamiltonian cycle in M k is the union of two perfect matchings. There, Kierstead and Trotter generalize the notion of lexicographic matchings to lexical matchings, in hopes that some pair will work. In [DSW] it is shown that this cannot be ....

H.A. Kierstead and W.T. Trotter, Explicit matchings in the middle levels of the Boolean lattice, Order 5 (1988), 163-171.


A Hamilton Path Heuristic with Applications to the Middle Two .. - Shields, Savage   (Correct)

.... middle two levels problem is to determine if there is a Hamilton path or cycle in the subgraph M 2k 1 of H(B 2k 1 ) induced by the middle two levels B 2k 1 (k) and B 2k 1 (k 1) The problem has been variously attributed to Dejter, Erdos, Trotter, Havel, and Kelley (see [4] 5] 7] [8], 13] The Boolean lattice is isomorphic to the set of n bit binary numbers with two numbers being adjacent in the Hasse diagram i# they di#er in exactly one bit position. For n = 2k 1, the middle two levels are the 2k 1bit binary numbers whose representations are permutations of 0 k 1 ....

H. A. Kierstead and W. T. Trotter. Explicit matchings in the middle levels of the Boolean lattice. Order, 5(2):163--171, 1988.


A Hamilton Path Heuristic with Applications to the Middle Two .. - Shields, Savage   (Correct)

.... notorious middle two levels problem is to determine if there is a Hamilton path or cycle in the subgraph M 2k 1 of H(B 2k 1 ) induced by the middle two levels B 2k 1 (k) and B 2k 1 (k 1) The problem has been variously attributed to Dejter, Erdos, Trotter, Havel, and Kelley (see [4] 5] 7] [8], 13] The Boolean lattice is isomorphic to the set of n bit binary numbers with two numbers being adjacent in the Hasse diagram iff they differ in exactly one bit position. For n = 2k 1, the middle two levels are the 2k 1bit binary numbers whose representations are permutations of 0 k ....

H. A. Kierstead and W. T. Trotter. Explicit matchings in the middle levels of the Boolean lattice. Order, 5(2):163--171, 1988.


Long Cycles in the Middle Two Levels of the Boolean Lattice - Savage (1997)   (1 citation)  (Correct)

....problems in the area of Gray codes for sets, as well as Hamilton cycles in vertex transitive graphs, involves the problems about paths among certain levels of B n , the Boolean lattice of subsets of a n element set. Perhaps the best known is the middle two levels problem which is attributed in [KT] to Dejter, Erdos, and Trotter and by others to H avel and Kelley. This problem has been attacked by several researchers with no success. The question is whether it is possible to list all of the k element and k 1 element subsets of the set f1; 2k 1g in such a way that (i) each subset ....

....to try to form a Hamilton cycle as the union of two edge disjoint matchings. In [DSW] it was shown that a Hamilton cycle in the middle two levels cannot be the union of two lexicographic matchings. However, other matchings may work. Motivated by this approach to the problem, Kierstead and Trotter [KT] define a large class of matchings, called lexical matchings in the middle two levels. Is there at least a good lower bound on the length of the longest cycle in the bipartite graph formed by the middle two levels of the Boolean lattice No lower bound of the form c N is known. Since this graph ....

H. A. Kierstead and W. T. Trotter, "Explicit matchings in the middle levels of the Boolean lattice," Order 5 (1988) 163-171.


A Survey of Combinatorial Gray Codes - Savage (1996)   (23 citations)  (Correct)

....by an adjacent transposition or by the transposition of two bits that have a single 0 bit between them. There are several open problems about paths between levels of the Hasse diagram of 14 the Boolean lattice, B n . The most notorious is the middle two levels problem which is attributed in [KT88] to Dejter, Erdos, and Trotter and by others to H avel and Kelley. The middle two levels of B 2k 1 have the same number of elements and induce a bipartite, vertex transitive graph on the k and k 1 element subsets of [2k 1] The question is whether there is a Hamilton cycle in the middle two ....

....is to try to form a Hamilton cycle as the union of two edge disjoint matchings. In [DSW88] it was shown that a Hamilton cycle in the middle two levels cannot be the union of two lexicographic matchings. However, other matchings may work and new matchings in the middle two levels have been defined [KT88, DKS94]. The largest value of k for which a Hamilton cycle is known to exist is k = 11 (n = 23. See Figure 7 for an example when k = 3. This unpublished work was done by Moews and Reid using a computer search [MR] To speed up the search, they used a necklace based approach, gambling that there would ....

H. A. Kierstead and W. T. Trotter. Explicit matchings in the middle levels of the Boolean lattice. Order, 5:163--171, 1988.


Monotone Gray Codes and the Middle Levels Problem - Savage, Winkler (1995)   (6 citations)  (Correct)

.... n ) induced by V n (i) V n (i 1) It is an open problem to determine whether G n (i) has a path which includes every vertex of the smaller of the two sets V n (i) V n (i 1) In particular, the notorious middle levels problem is to determine whether G 2k 1 (k) has a hamilton path or cycle [3, 5, 11, 12, 16]. Recently, Felsner and Trotter have shown that G 2k 1 (k) has a cycle of length at least one fourth the length of a hamilton cycle, giving the first constant factor approximation [6] we show in Section 3 that monotone Gray codes do even better, yielding paths of more than half the ; 1 2 3 4 5 ....

....p i = y i ; x i 1 if i is even p i = x i 1 ; y i if i is odd 2 3. Paths Between Adjacent Levels One of the best known of combinatorial problems is the middle levels problem which has been variously attributed to Dejter, Erdos, Trotter, H avel, and Kelley (see [3, 5, 11, 12, 16]) The problem is to determine whether it is possible to list all of the k element and k 1 element subsets of the set f1; 2k 1g in such a way that (i) each subset occurs exactly once, ii) the k and k 1 element subsets occur alternately, and (iii) consecutive sets on the list ....

H. A. Kierstead and W. T. Trotter, Explicit matchings in the middle levels of the Boolean lattice, Order 5 (1988), 163-171.


The Bricklayer Problem And The Strong Cycle Lemma - Snevily, West   (Correct)

....Since k and k 1 are relatively prime, the number of (k; k 1) arrangements (and ballot sequences of length 2k) is exactly the Catalan number 1 2k 1 2k 1 k 1 = 1 k 1 2k k = C k : The Cycle Lemma thus explains one 0 in a (k; k 1) arrangement. Kierstead and Trotter [9] generalized this to give combinatorial meaning to each 0. We present their result in Section 4 and extend it slightly. Our task at present is to prove the Cycle Lemma and use it to count the q satisfying sequences of length m. Figure 2 illustrates the Cycle Lemma (and its proof) when (k; q; p) ....

....0 interval. We use w 0 (I) and w 1 (I) to denote the number of 0 s and 1 s in I, respectively. A 0 interval I is q good if w 0 (I) qw 1 (I) In every 0 linearization of a (k; qk p) arrangement with p 0, the trivial 0 interval is q good. For q = p = 1 and 1 i k 1, Kierstead and Trotter [9] proved that every (k; k 1) arrangement a has a unique 0 linearization such that exactly i of the 0 intervals are 1 good. They noted that this result is implicit in the work of Feller [6] and Narayana [11] and they used it to construct new explicit perfect matchings in the bipartite graph of ....

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H.A. Kierstead and W.T. Trotter, Explicit matchings in the middle levels of the Boolean lattice, Order 5 (1988), 163--171.


Gray Codes from Antimatroids - Pruesse, Ruskey (1993)   (3 citations)  (Correct)

....G If jAj = jBj 1 then is there a Hamilton path in G We conjecture that the answer is yes in both cases. Note that this conjecture contains the famous middle two levels of the boolean algebra lattice problem as a special case and so is most likely very difficult (e.g. Kierstead and Trotter [8] or Hurlbert [6] if indeed it is true. Conjecture 3 For every antimatroid (E; F) and every k, the graph J(F ; k) has a Hamilton path. This cannot be strengthened to Hamilton cycle since J(F ; k) is a path if F is the poset antimatroid of a poset consisting of two disjoint chains. ....

H.A. Kierstead and W.T. Trotter, "Explicit matchings in the middle levels of the Boolean lattice", Order 5 (1988) 163-171.

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