All-edges Exact Triangle in time $O(\kappa\, n^{3-143/68400}\log^2 n)$ from the pruned Lop detection

Assume Lop-AE-SparseTri detection has the running time of the thin regime, that is, $\mathrm{LopDetectP}$ holds: for some $C\ge0$ the problem is solved in a time $T_d$ with

Then All-Edges Exact Triangle is solved, uniformly in the size and the weights, in time

for some constant $K\ge1$, where $s$ is the number of vertices per part and $u$ an upper bound on the absolute values of the weights.

Here All-Edges Exact Triangle asks, given integer weight matrices $w_{AB},w_{BC},w_{AC}\in\mathbb Z^{s\times s}$, for the flag of every pair $(a,b)$ telling whether some $c$ has $w_{AB}(a,b)+w_{BC}(b,c)+w_{AC}(a,c)=0$; Lop-AE-SparseTri is Definition 13 of Alman and Vassilevska Williams. "Solved in time $T$" refers to programs of the library's light language, which compiles to the word RAM: one program solves every instance within the bound, with a polynomially bounded need of memory and word size.

The exponent comes from Theorem 17 for the all-edges host at the middle dimension with $D=\lfloor n^{13/228}\rfloor$ and $g=\lceil D^{11/300}\rceil$: the oracle is called at inner dimension about $D^{289/300}\approx n^{0.0549}$, inside the regime $D^{91}\le n^5$, and $\tfrac{143}{68400}=\tfrac{11}{300}\cdot\tfrac{13}{228}$. This bound is what the reduction from APSP to all-edges Exact Triangle consumes.

Formalization Note The statement is Claim.LopDetectP aeModel → Claim.ExactTriangleUniform aeModel (143 / 68400) 2, where ExactTriangleUniform M δ e says M.exactTriangle (uniformTime K δ e) for some K ≥ 1, and aeModel reads "Exact Triangle" as the all-edges task.