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Replacing k of the sets E_i by their complements yields a subsystem T_k, and the 2^d subsystems so obtained partition the chess-board. We prove that df(T_k) = (-1)^k df(T_0), and give a multilinear proof showing that no assumption on the sets E_i is needed and that the result holds verbatim for d-fold stochastic matrices. Consequently the partition has a single degree of freedom: the rook counts of all its 2^d members follow from their volumes and the single number df(T_0). For d = k = 3 this specializes to the identity of Cruse relating a brick and its remote mate, a necessary condition for a partial Latin square to be completable. We also prove that for d = 3 the chess-board represents precisely one main class.
From: Béla Jónás [view email]
[v1]
Mon, 8 Aug 2022 13:18:17 UTC (484 KB)
[v2]
Mon, 20 Jul 2026 15:56:29 UTC (484 KB)
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