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The question: enter image description here

My thoughts:

  • If $t=2$, then in each coloring we have either a red edge (satisfying the "red H" condition), or the entire graph is colored blue. So $R(K_s, P_2)=s$. Same goes for $s=2$.
  • By hand, I obtained $R(K_3, P_3)=5$.
  • If $t\geq3, s\geq2$, we can consider the following: Take a entirely-blue $K_{s-1}$ and $t-3$ additional vertices, that have all their edges colored red. Such graph has no blue $K_s$ and the maximal red path has $t-1$ vertices. So $R(K_s, P_t)>(s-1)+(t-3)$, i.e $R(K_s, P_t)\geq s+t-3$.

Any other ideas?

1 Answers 1

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The lower bound can be significantly improved: take $s-1$ vertex-disjoint red copies of $P_{t-1}$ and join them all by blue edges. This shows $R(K_s, P_t)\geq N:=(s-1)(t-1)+1$.

In fact, the inequality is exact and you can prove this e.g. by counting the edges of the red subgraph in a red-blue colouring of $K_N$ without blue copies of $K_s$.