OFFSET
1,2
COMMENTS
The sum of symmetric elements a[i, j] and a[i, n+1-j] in a horizontally symmetric reduced Latin square of order n is constant and equal to n+1 for all pairs of elements (with rows and columns numbered from 1 to n and element values from 1 to n), based on the definition given in A287649 by Eduard I. Vatutin. Horizontal symmetry is of interest in the Latin stable marriage problem because the even-sized maximal examples given in A351413 have this property, as do the lower bounds preference profiles specified in A005154 and A065982.
LINKS
A. T. Benjamin, C. Converse, and H. A. Krieger, Note. How do I marry thee? Let me count the ways, Discrete Appl. Math. 59 (1995) 285-292 (Section 3 defines vertical reflection relation).
E. G. Thurber, Concerning the maximum number of stable matchings in the stable marriage problem, Discrete Math., 248 (2002), 195-219 (Section 10).
EXAMPLE
For order 4, the a(2)=2 solutions are:
1234...1234
2143...2413
3412...3142
4321...4321
For order 6, the example shown in A351413 is:
123456
214365
365214
456123
541632
632541
PROG
(MiniZinc)
int: n = 8; %reduced Latin squares of even-size order n with horizonal symmetry
% 1, 2, 80, 428032
set of int: Elements = 1..n;
array[Elements, Elements] of var Elements: L; % the Latin square
include "alldifferent.mzn";
constraint forall(i in Elements) ( % Latin square constraint
alldifferent([L[i, j] | j in Elements]) /\
alldifferent([L[j, i] | j in Elements]));
constraint forall(i in Elements) ( % reduced form
L[1, i] = i /\
L[i, 1] = i);
constraint forall(r in Elements, i in 1..n div 2) ( % horizontal symmetry
L[r, i] + L[r, n + 1 - i] = n + 1);
solve satisfy;
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Dan Eilers, Mar 22 2026
STATUS
approved
