On the crossing number of K13

Closing in on Hill's Conjecture

“…An elementary counting using cr(K 11 ) = H(11) = 100 shows that cr(K 13 ) ≥ 217. McQuillan, Pan, and Richter [30] have ruled out the possibility that cr(K 13 ) = 217, and since cr(K 13 ) is an odd number [31], it follows that cr(K 13 ) ∈ {219, 221, 223, 225}. This was further narrowed in [1], finding that cr(K 13 ) ∈ {223, 225}.…”

A note on the cross-index of a complete graph based on a linear tree

“…Example 2.4. For n = 14, σ(a (1,7) ) is obtained as follows (see Figure 7): σ(a (1,7) ) = The components a (k,l) of M n satisfying 1 < k < 7 < l and a (k,l) = a (1,7) = 1 are a (2,8) , a (2,9) , a (2,10) , a (2,11) , a (2,12) , a (3,8) , a (3,9) , a (3,10) , a (3,11) , a (4,8) , a (4,9) , a (4,10) , a (5,8) , a (5,9) and a (6,8) .…”

On the Crossing Number of Kn without Computer Assistance

“…Each of the other two vK 6 -edges has at most two additional crossings, so the resulting drawing has no more crossings than the original. Thus, we may assume (v, x, y) is a face boundary of K (3) 5 .…”

“…We still have not been able to determine the number of optimal K 11 s. The proof by contradiction in [6] did not generate these drawings. They were not explicitly recorded in [5]; there were too many of them to check for isomorphism, and in any case they were only generated up to some natural equivalence. This complicates trying the same avenue for K 13 .…”

“…Then cr(D) ≥ 36.Proof. For each vertex v of K 9 not in the K5 , cr(D[K 5 + v]) = 5 = cr(D[K 5 ]). Therefore, Theorem 3.4 shows D[K 5 ] is natural and D[K 5 + v] is the natural extension of D[K 5 ].Thus, the other four vertices of K 9 are all in the same face F of D[K 5 ].…”