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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Statement

Definition (p. 185). f(k,ℓ1,ℓ2)f(k,\ell_1,\ell_2) is the smallest integer nn for which there is a graph G(n)G(n) on nn vertices containing no K(ℓ2)K(\ell_2) such that every coloring of the edges of G(n)G(n) by kk colors has a monochromatic K(ℓ1)K(\ell_1).

Values reported (p. 185). Graham (the paper's reference [12]) proved f(2,3,6)=8f(2,3,6)=8, and Irving (reference [13], whose name prints as "Inving" [sic] in the text) proved f(2,3,5)≤18f(2,3,5)\le18.

Problem (p. 185). Folkman's upper bound for f(2,3,4)f(2,3,4) is enormous, much bigger than the tower of seven tens 1010101010101010^{10^{10^{10^{10^{10^{10}}}}}}, and the same holds for the bound of Nešetřil and Rödl. Erdős offers, quoted, "max (100 dollars, 300 Swiss francs) for a proof or disproof of f(2,3,4)<1010f(2,3,4)<10^{10}."

Source. P. Erdős, Problems and results on finite and infinite graphs, Recent advances in graph theory (Proc. Second Czechoslovak Sympos., Prague, 1974), Academia, Prague, 1975, pp. 183--192; Section III, p. 185. The edition read is identified on the source card. Reference [12] is R. L. Graham, On edgewise 2-colored graphs with monochromatic triangles and containing no complete hexagon, J. Combinatorial Theory 4 (1968), 300; reference [13] is R. W. Irving, On a bound of Graham and Spencer for a graph colouring constant, J. Comb. Theory (Ser. B) 15 (1973), 200--203.

Read depth. Claims checked: the two paragraphs were read clause by clause on the printed page; the height of the tower was counted on an enlarged image of the page.

Proof pointer

None in this paper; the values are reported with references [12] and [13].

Dependencies

The conjecture of p. 184, whose case k=2k=2, Folkman's theorem, makes f(2,ℓ,ℓ+1)f(2,\ell,\ell+1) finite.

Bears on

  • Problem 582: f(2,3,4)f(2,3,4) is the least order of a graph of the kind the problem asks for; the prize offer concerns that least order, not the existence the problem asks about.