Squashing maximum packings of Kn with 8-cycles into maximum packings of Kn with 4-cycles

Abstract

An 8-cycle is said to be squashed if we identify a pair of opposite vertices and name one of them with the other (and thereby turning the 8-cycle into a pair of 4-cycles with exactly one vertex in common). The resulting pair of 4-cycles is called a bowtie. We say that we have squashed the 8-cycle into a bowtie. Evidently an 8-cycle can be squashed into a bowtie in eight different ways. The object of this paper is the construction, for every n > 8, of a maximum packing of Kn with 8-cycles which can be squashed in a maximum packing of Kn with 4-cycles.

Squashing maximum packings of 6-cycles into maximum packings of triples

“…An open problem is to solve the general case, i.e. squashing a maximum packing of K n with 2m-cycles into a maximum packing of K n with m-cycles; the case m = 4 is completely solved in [10] .…”

Squashing minimum coverings of 6-cycles into minimum coverings of triples

“…(2, 3, 4, 5, 0, 1), (1, 3, 6, 4, 7, 0), (5, 3, 0, 2, 4, 1), (4, 5, 7, 2, 6, 0), (2,3,7,1,6,5). Example 3.9.…”

“…On the set Y we place a copy of a maximum packing of K 8 with 6-cycles which squashes into a maximum packing with triples, with leave the set of edges {6, 9}, {7, 8}, {10, 13}, {11, 12}; see Example 2.2 in [4] for this. Then on X \ Y , with padding K 4 on {0, 1, 2, 3} together with the edges {4, 5}, {6, 7}, {8, 9}, {10, 11}, {12, 13}, we place the following 6-cycles which squash into triples: (3, 2, 1, 0, 10, 5), (6, 5, 0, 1, 4, 3), (1, 2, 0, 3, 10, 11), (0, 2, 10, 1, 13, 3), (0, 4, 12, 1, 3, 11), (1,8,4,6,0,9), (7, 1, 6, 2, 8, 0), (1, 3, 12, 0, 13, 5), (7,2,3,9,5,4), (4,10,13,12,11,2), (2,5,8,7,6,9), (12,5,11,4,13,2), (9,8,3,7,5,4).…”

“…Recently three papers by Lindner et al have dealt with so-called "squashing" of 6-cycles into triples ( [6], [4]) and 8-cycles into 4-cycles ( [5]). This paper completes the problem regarding the existence for all orders of squashable 6-cycle systems, by considering coverings.…”

“…Here the padding contains no triples and B = {(2, 3, 4), (2, 0, 1), (1,2,3), (1, 6, 0), (5, 3, 0), (5, 4, 1), (7, 1, 8), (7, 0, 4), (9, 2, 6), (9, 0, 4), (8, 2, 5), (8, 10, 0), (4,8,6), (4, 3, 10), (9, 3, 8), (9, 10, 1), (7,3,6), (7, 10, 2), (5, 6, 10), (5,7,9)}.…”