, Volume 37, Issue 2, pp 137–142 | Cite as

Square-root cancellation for the signs of Latin squares

Original Paper


Let L(n) be the number of Latin squares of order n, and let L even(n) and L odd(n) be the number of even and odd such squares, so that L(n)=L even(n)+L odd(n). The Alon-Tarsi conjecture states that L even(n) ≠ L odd(n) when n is even (when n is odd the two are equal for very simple reasons). In this short note we prove that
$$\left| {{L^{even}}\left( n \right) - {L^{odd}}\left( n \right)} \right| \leqslant L{\left( n \right)^{\frac{1}{2} + o\left( 1 \right)}}$$
, thus establishing the conjecture that the number of even and odd Latin squares, while conjecturally not equal in even dimensions, are equal to leading order asymptotically. Two proofs are given: both proceed by applying a differential operator to an exponential integral over SU(n). The method is inspired by a recent result of Kumar-Landsberg.

Mathematics Subject Classification (2000)

05B15 05A16 


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Copyright information

© János Bolyai Mathematical Society and Springer-Verlag Berlin Heidelberg 2017

Authors and Affiliations

  1. 1.Churchill CollegeUniversity of CambridgeCambridgeUK

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