Decomposing Kn∪P into triangles

Abstract

In this paper, we extend the work on minimum coverings of Kn with triangles. We prove that when P is any forest on n vertices the multigraph G = Kn ∪ P can be decomposed into triangles if and only if three trivial necessary conditions are satisÿed: (i) each vertex in G has even degree, (ii) each vertex in P has odd degree, and (iii) the number of edges in G is a multiple of 3. This result is of particular interest because the corresponding packing problem where the leave is any forest is yet to be solved. We also consider some other families of packings, and provide a variation on a proof by Colbourn and Rosa which settled the packing problem when P is any 2-regular graph on at most n vertices.

Directed 3-cycle decompositions of complete directed graphs with quadratic leaves

“…, 10). Then D 11 − C 11 = {(0, 2, 7), (1,3,8), (3,5,10), (4,6,0), (5,7,1), (9, 4, 1), (0, 5, 2), (6,8,2), (7,9,3), (8, 10, 4), (10, 1, 6), (8, 3,2), (4,0,8), (2,10,7), (4,3,6), (3,9,5), (0, 10, 5), (0, 7, 3), (6,5,8), (0, 3, 1), (4,7,5), (7, 10, 8), (8, 5, 1), (9, 2, 5), (4, 2, 1), (8, 0, 9), (9, 1, 10), (4, 10, 2), (10, 6,3), (2,9,6), (6,1,7), (6,9,0), (4,9,7)}.…”

“…Let D 8 be defined on Z 8 and take C 8 = (0, 1,2,3,4,5,6,7). Then D 8 − C 8 = { (5,4,3), (4,7,2), (5,0,2), (6,0,3), (1,0,5), (3,0,7), (3,7,1), (1,5,3), (6,3,2), (7, 6, 5), (7, 4, 1), (5,2,7), (6,1,4), (4,0,6), (2,0,4), (1,6,2)}.…”

“…Let D 7 be defined on Z 7 and take C 6 = (0, 1,2,3,4,5). Then D 7 − C 6 = {(4, 0, 2), (1,3,2), (3,5,4), (3,0,6), (6,4,2), (5,2,0), (5,6,2), (6,1,4), (0, 4, 1), (6,0,3), (3,1,5), (1,6,5)}.…”

Achromatic numbers of Kneser graphs

“…Further, if K n has an even number of vertices, then there is a set F ⊆ E(K n ) such that E(K n ) \ F can be partitioned into triangles. More precisely, if n ≡ 0, 2 mod 6,then F is a perfect matching in K n , and if n ≡ 4 mod 6, then F induces a spanning forest of n/2 + 1 edges in K n with all vertices having an odd degree, see [12,15].…”

Achromatic numbers of Kneser graphs

“…Moreover, if K n has an even number of vertices, then K n − F contains a partition into triangles with F , a matching, for n ≡ 0, 2 mod 6; and with F , a spanning forest of n/2 + 1 edges such that each each vertex has odd degree, for n ≡ 4 mod 6, see [11,14].…”