Exact Second-Order Zarankiewicz Numbers for Complete-Graph Incidence Families

Abstract

For $n\ge6$, $m=\binom n2$, let the complete-graph incidence family on $K_n$ have the vertices of $K_n$ as columns, its edges as rows, and the incidence graph as one-edge graph. The universal cell bound of L\"ofberg and Qi gives $z_2(m,n)\le Z(n):=\lfloor n(n-1)(n+2)/4\rfloor$. We implement the nested one-factorization construction of that family and determine exactly what it certifies. For $n=2q$ with $q$ an odd prime, $q\ge5$, the construction has no hole and the cross-factor transfer equations apply verbatim, giving $z_2=z_{SL}=z_{RL}=Z(n)$; $n=6$ is settled by a separate cyclic witness. For odd $n=2p+1$ the near-perfect one-factorization yields Hamilton paths closed into odd cycles, and the transfer argument breaks: that scheme certifies only $R(G_p)\le Z(n)$. The orders $n=7,8,9,12,13,16,17,18,20,21$ are settled by explicit configurations of a different shape, each attaining the cell bound and satisfying $(\mathrm{RW}3^+)$, obtained from a larger configuration by deleting vertex stars and repairing the restricted grid. At each of these orders $z_2=z_{SL}=z_{RL}=Z(n)$; all remaining orders are conjectural. The first settled order $n=7$ is the only one whose grid has a hole, grounded by a zero-companion rule. Machine-readable configurations and a certificate checker accompany the paper.