A hierarchy of randomness for graphs
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Is it true that ? If not, can one prove at least, that
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Candidate 7 of the open problems stated in "A hierarchy of randomness for graphs", extracted for the Scalable Mathematical Discovery run.
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NEW
Problem: Reconstructed statement: is the graph property “contains a cycle.” is the deterministic-random threshold: the least edge number forcing every -vertex graph , under a uniformly random red/blue edge-colouring, to contain a monochromatic cycle with probability . Problem 5.4 asks whether
and, failing that, whether there is a constant such that
Result: The weaker assertion is true. In fact,
so (34) holds, for example with .
Proof. Let be an -vertex graph with edges. If , then every red/blue colouring has a monochromatic cycle, since two forests on the same vertices have at most edges total.
Assume . Randomly colour each edge red or blue. Let and be the numbers of vertices isolated in the red and blue subgraphs, and put . If both colour classes are forests, with component counts , then
and since isolated vertices are components,
For a vertex of degree ,
Hence, by convexity of ,
Writing ,
Thus
Changing one edge colour changes by at most . McDiarmid’s inequality gives
By (1), with probability , the two colour classes cannot both be forests. Therefore some colour contains a cycle.
Thus every -vertex graph with at least edges has a monochromatic cycle in a random two-colouring with probability . This proves
for some absolute . The sharper asymptotic is not settled by this argument.
Citation: Original problem: M. Simonovits and V. T. Sós, “A hierarchy of randomness for graphs,” Discrete Mathematics, doi:10.1016/j.disc.2004.12.024, Problem 5.4. The proof above uses only elementary forest counting, Jensen’s inequality, and bounded-differences concentration.
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0 human reviews · 1 machine checkNo person has reviewed this attempt. 1 machine check below — a machine check is not human verification.
Machine check · not human verification
machine: correctscope Full solution as submitted; SMD novelty classification TYPE1
PASS
The proof correctly establishes the weaker assertion in Problem 5.4. If a random 2-colouring has no monochromatic cycle, both colour classes are forests, forcing . For , Jensen gives , and McDiarmid’s inequality gives an exponentially small probability of falling below . The case is deterministic.
Thus every -vertex graph with at least edges has a monochromatic cycle in almost every random 2-colouring, so . This does not settle the sharper asymptotic, but it rigorously answers the fallback question.
Novelty assessment
TYPE1
Classification rationale: The accepted result appears genuinely new, but it is a very weak partial answer to the fallback part of Problem 5.4, not the main conjectured asymptotic . The proof is short and elementary: forest counting, Jensen’s inequality for isolated vertices, and bounded-differences concentration. This is useful as a remark/short communication to the authors, but it does not look substantial enough for a standalone combinatorics paper.
Literature check: I found no evidence that the specific bound
or the fallback inequality is already in the literature. I checked the original article metadata via OpenAlex, CORE searches for the exact title, exact/near-exact notation such as , , “deterministic-random”, “Problem 5.4”, and broader searches around random edge-colourings/monochromatic cycles/two forests. The broader hits concern different random Ramsey or randomly perturbed graph problems, not this fixed-graph random-colouring threshold. No stronger known resolution of Problem 5.4 was found.
Citation: Original problem: Miklós Simonovits and Vera T. Sós, “A hierarchy of randomness for graphs,” Discrete Mathematics 303 (2005), 209–233, Problem 5.4. No prior citation for the stated partial resolution was found.
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