Question

Let A and B be orthogonal Latin squares of order n, with symbols 0, 1 …, n – 1. Let B’ be obtained from B by permuting the symbols in B. Show that A and B’ are still orthogonal.

Answer 1

To answer this question, we will take help of group theory.

We are given 2 Latn Squares of order 'n', where the elements inside it are made from digits 0,1,2,..., (n-1).

Shown below is a Latin Square of order 4 for example.

Now, to design a Latin Square, we use a very simple method, as shown below:

A and B are 2 orthogonal Latin Squares. (Given) -------------------------------------------------------------------------- (1)

Hence when both of them are matched we know that they don't have the same elements. (Definition of Orthogonality of Latin Square)

if B' is obtained by permutating the B Latin Square. We know that B' is a Latin Square which is Orthogonal to B.

This can be seen as B' has been obtained by permutating the elements of B, hence they can't have the same elements making them Orthogonal.

To prove whether B' is a Latin Square or not, we check that claim first that this is a Latin square with symbols 0,1,...,…,n−1. Consider two entries in a row, say i+j and i+k. If i+j=i+j(mod n), then j = k. Thus, all entries of the row are distinct. --------------------------------------------------------------------------------------------------------------------------------- (2)

Hence B' is a Latin Square. ------------------------------------------------------------------------------------------------- (from 2)

Now let us assume that A and B' are non-orthogonal Latin Squares, which means every element in A is equal to that corresponding element in B'.

which means that A = B'.and further B' = A.

Making A = B, which is not possible as A and B are orthogonal.---------------------------------------------------------- (from 1)

This is a contradicting statement!

Hence our hypothesis is completely wrong. and A and B' are Orthogonal Latin Squares.