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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

"A graph is called even if every circuit of it has an even number of edges" (p. 122), that is, bipartite. With f(n−1)=⌊(n−1)2/4⌋f(n-1)=\lfloor(n-1)^2/4\rfloor:

Lemma 1 (p. 123). "Every Gf(n−1)+2(n)G^{(n)}_{f(n-1)+2} which is not even contains a triangle."

"Lemma 1 was found jointly by Gallai and myself. (The lemma was also found by Mr. Andrásfai independently.)"

The proof (pp. 123--124) shows more: if GG has nn vertices, is not even and contains no triangle, and α1,…,α2k+1\alpha_1,\dots,\alpha_{2k+1} is a shortest odd circuit (3<2k+1≤n3<2k+1\le n), then the α\alpha's span no other edge, every other vertex is joined to at most two of the α\alpha's, and the other n−2k−1n-2k-1 vertices span at most f(n−2k−1)f(n-2k-1) edges, so GG has at most 2k+1+2(n−2k−1)+f(n−2k−1)≤f(n−1)+12k+1+2(n-2k-1)+f(n-2k-1)\le f(n-1)+1 edges, "by a simple calculation (equality only for 2k+1=52k+1=5)". Hence a non-even triangle-free graph on nn vertices has at most f(n−1)+1f(n-1)+1 edges, and the lemma holds for every edge count at least f(n−1)+2f(n-1)+2.

The remark after the proof (p. 124): "Our proof in fact gives that a graph GG of nn vertices whose smallest odd circuit has 2k+12k+1 vertices, k>1k>1, has at most 2n−2k−1+f(n−2k−1)2n-2k-1+f(n-2k-1) edges, and the following simple example shows that this result is best possible": vertices α1,…,αv\alpha_1,\dots,\alpha_v, β1,…,βu\beta_1,\dots,\beta_u, γ1,…,γ2k+1\gamma_1,\dots,\gamma_{2k+1} with v=[(n−2k−1)/2]v=[(n-2k-1)/2], u=n−2k−1−vu=n-2k-1-v (the print has u=n−[(n−2k−1)/2]u=n-[(n-2k-1)/2], a misprint by a count made here: with β1,…,βu\beta_1,\dots,\beta_u that value gives v+u+2k+1=n+2k+1v+u+2k+1=n+2k+1 vertices, while u=n−2k−1−vu=n-2k-1-v gives nn vertices and exactly vu+2(v+u)+2k+1=2n−2k−1+f(n−2k−1)vu+2(v+u)+2k+1=2n-2k-1+f(n-2k-1) edges, the stated bound), and the edges (αi,βj)(\alpha_i,\beta_j), (γ1,αi)(\gamma_1,\alpha_i), (γ3,αi)(\gamma_3,\alpha_i) for 1≤i≤v1\le i\le v, (γ2,βi)(\gamma_2,\beta_i), (γ4,βi)(\gamma_4,\beta_i) for 1≤i≤u1\le i\le u, and the circuit edges (γi,γi+1)(\gamma_i,\gamma_{i+1}), 1≤i≤2k1\le i\le2k, (γ1,γ2k+1)(\gamma_1,\gamma_{2k+1}). At k=2k=2 the bound is 2n−5+f(n−5)=f(n−1)+12n-5+f(n-5)=f(n-1)+1 (a check made here: f(m)−f(m−1)=[m/2]f(m)-f(m-1)=[m/2], so f(n−1)−f(n−5)=2n−6f(n-1)-f(n-5)=2n-6), so the example is a triangle-free non-even graph with f(n−1)+1f(n-1)+1 edges for every n≥5n\ge5.

Source. P. Erdős, On a theorem of Rademacher-Turán, Illinois J. Math. 6 (1962), no. 1, 122--127; Lemma 1 and its proof on printed pp. 123--124 = PDF pp. 2--3 of the Rényi scan (1962-09.pdf), the remark and the example on p. 124 = PDF p. 3, read on the page images. The edition read is identified in the source digest.

Read depth. Claims checked: the lemma, the attribution, the remark and the example were read clause by clause on the page images; the proof was read for structure and its edge count followed as printed; the k=2k=2 arithmetic above is an authored check.

Proof pointer

The shortest-odd-circuit argument above, with Turán's theorem for the vertices off the circuit (p. 123--124). Ren, Wang, Wang and Yang restate the bound as their Theorem 1.2, "Let GG be a non-bipartite triangle-free graph on nn vertices. Then e(G)≤⌊(n−1)24⌋+1e(G)\le\lfloor\frac{(n-1)^2}4\rfloor+1", with the graph H0H_0 (a complete bipartite Turán graph on n−1n-1 vertices with an edge replaced by a path of length two) showing sharpness (theorem_1_2).

Dependencies

Turán's theorem.

Bears on

  • Problem 1011: for triangle-free graphs "not even" is chromatic number at least 33, so the lemma and the example give f3(n)=⌊(n−1)2/4⌋+2f_3(n)=\lfloor(n-1)^2/4\rfloor+2 for n≥5n\ge5, the value the site attributes to Erdős and Gallai.
  • Problem 1010: the first of the three lemmas behind the Theorem.