A309345 a(n) is the minimum number of transversals in Latin squares of order n that have at least 1 transversal.

3, 8, 3, 8, 3, 8

Offset: 3

Author

Alvaro R. Belmonte, Eugene Fiorini, Peterson Lenard, Froylan Maldonado, Sabrina Traver, Wing Hong Tony Wong, Jul 24 2019

Keywords

• nonn
• hard
• more

Comments

We find all the transversals of the main class representatives of order n Latin squares then find the one with the fewest transversals.

Examples

			For n = 5,they are 2 main classes of Latin squares. One of them has representative M = [[1,2,3,4,5],[2,4,1,5,3],[3,5,4,2,1],[4,1,5,3,2],[5,3,2,1,4]], and it has 3 transversals; {M(1,1), M(5,2), M(3,3), M(2,4), M(4,5)}, {M(2,1), M(5,2), M(4,3), M(1,4), M(3,5)}, and {M(4,1), M(5,2), M(2,3), M(3,4), M(1,5)}. The other main class representative [[1,2,3,4,5],[2,3,4,5,1],[3,4,5,1,2],[4,5,1,2,3],[5,1,2,3,4]] has 15 transversals. Therefore, the minimum number of transversals in order 5 Latin squares is 3, i.e., a(5) = 3.
		

Links

• Brendan McKay, Latin squares

Programs

MATLAB

%This extracts entries from each column.  For an example, if
%A=[1 2 3 4; 5 6 7 8; 9 10 11 12; 13 14 15 16], and if list = (2, 1, 4),
%this code extracts the second element in the first column, the first
%element in the second column, and the fourth element in the third column.
function [output] = extract(matrix,list)
for i=1:length(list)
    output(i) = matrix(list(i),i);
end
end
%Searches matrix to find transversal and outputs the transversal.
function [output] = findtransversal(matrix)
n=length(matrix);
for i=1:n
    partialtransversal(i,1)=i;
end
for i=2:n
    newpartialtransversal=[];
    for j=1:length(partialtransversal)
        for k=1:n
            if (~ismember(k,partialtransversal(j,:)))&(~ismember(matrix(k,i),extract(matrix,partialtransversal(j,:))))
                newpartialtransversal=[newpartialtransversal;[partialtransversal(j,:),k]];
            end
        end
    end
    partialtransversal=newpartialtransversal;
end
output=partialtransversal;
end
%Takes input of n^2 numbers with no spaces between them and converts it
%into an n by n matrix.
function [A] = tomatrix(input)
n=sqrt(floor(log10(input))+2);
for i=1:n^2
    temp(i)=mod(floor(input/(10^(i-1))),10);
end
for i=1:n
    for j=1:n
        A(i,j)=temp(n^2+1-(n*(i-1)+j));
    end
end
A=A+ones(n);
end