Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 339 / 2 / g / Solution

The Krivine rounding constant has $t=c_K\pi /2=\log (1+\sqrt{2})<\pi /2$. Since $|X_{ij}|\le 1$, the principal real inverse sine satisfieson every cross-block entry. The matrix $A$ has zero diagonal blocks, so those cross blocks are the only contributors to its Frobenius inner product with $\arcsin [Y]$. Explicitly,No analogous identity is needed for the diagonal blocks involving $\sinh$.

Combining with the preceding expectation formula and optimality of $X$ gives the bipartite sign rounding boundAt least one feasible rounded outcome attains at least this expectation. This is an expectation-based approximation guarantee for the specific bipartite sign problem; no claim is made that $c_K$ is the largest possible constant. The inequalities remain valid when the optimal objective is zero.