Following is an interesting problem that I bumped into when I was doing my bachelor’s degree.

Given the top two three by three Mutually Orthogonal Latin Squares one needs to complete the two four by four Mutually Orthogonal Latin Squares below. A Latin Square has no duplicate across any row and across any column. Two n by n Latin Squares are called Mutually Orthogonal if their superimposition contains all the n square ordered pairs. Under these rules one can solve the above two four by four Mutually Orthogonal Latin Squares while trying to place the 16 pairs. It’s a good brain teaser.

See my one past blog here to find a Statistical approach to Latin Square

Leonhard Euler was the first person to make significant contribution to this problem. He could construct Mutually Orthogonal Latin Squares of all orders except 2, 6, 10 etc. That caused him to conjecture that MOLS of all orders exist except when the order is of the form n = 4k + 2. The conjecture stood unsettled for around one and half a century before it got disproved by Bose, Srikhande and Parker in 1969 as they proved that MOLS of all orders exist except n = 2 and 6.

There is the Galois Field – Kronecker Product construction of MOLS that works fine except when n = 4k + 2 form. I was thinking if Permutation Group approach could be helpful. Each row of a Latin Square is a permutation of numbers 1 to n and corresponds uniquely to a member in the Permutation Group. See Wikipedia for a ready reference of Permutation Group please. Now we have a possible theorem that if n permutations of numbers {1, …,n} form a subgroup of order n then they form the rows of a Latin Square. Clearly it is not necessary to form a subgroup of order n for the rows of an arbitrary Latin Square of order n.

If one starts with an acyclic (having no cycle) permutation a then a⁰, a¹, a², …, a^n-1 form a subgroup and form rows of a Latin Square. This provides a construction of non-trivial Latin Squares of order n for any general n. However, for a general construction of a subgroup of order n = r x s one needs to start with r cycles each of length s or the other way. This will provide all the Latin Squares of order n the rows of which form subgroups of order n.

Hopefully there will be ways to extend such a Latin Square (except 2 and 6) to its Orthogonal Mate. Possibly we will be able to deduce the first column of the Orthogonal Mate and construct the rows of the Orthogonal Mate by just copying the rows of the Latin Square. We will wait.