For any even n , say n = 2 m , a row complete Latin square of order n can be formed

Andy Erickson

Andy Erickson

Answered question

2022-05-23

For any even n, say n = 2 m, a row complete Latin square of order n can be formed by writing down
0 , 1 , 2 m 1 , 2 , 2 m 2 , 3 , , m + 1 , m
as the first row and then developing subsequent rows by adding 1 modulo n.
I'm not quite clear on how that goes, such as what the difference is between 2 m and m. 2 m stands for n, does m stand for modulo? I'd like to see and example with an even number or two, such as 4, and 6.
I tried, 2 m = 4
0 , 1 , 3 , 2 , 2
wait, that can't be right.

Answer & Explanation

Bruce Bridges

Bruce Bridges

Beginner2022-05-24Added 13 answers

If I understand your description correctly, for 2 m = 4:
[ 0 1 3 2 1 2 0 3 2 3 1 0 3 0 2 1 ]
For 2 m = 6:
[ 0 1 5 2 4 3 1 2 0 3 5 4 2 3 1 4 0 5 3 4 2 5 1 0 4 5 3 0 2 1 5 0 4 1 3 2 ]
In each case, the first row follows the pattern you gave; in subsequent rows, each number is one more than the number above it, but wrapping around at the size of the square (e.g. in the case where n = 2 m = 6, after 5 comes 0).
Let's look at n = 2 m = 12 ( m = 6 ) a bit more closely. The pattern that you gave said:
0 , 1 , 2 m 1 , 2 , 2 m 2 , 3 , , m + 1 , m
So, the first row is:
0 , 1 , 12 1 , 2 , 12 2 , 3 , , 6 + 1 , 6
which is
0 , 2 + 1 = 3 , 9 + 1 = 10 , 4 + 1 = 5 , 8 + 1 = 9 , 5 + 1 = 6 , 7 + 1 = 8 , 6 + 1 = 7
Continuing gives:
[ 0 1 11 2 10 3 9 4 8 5 7 6 1 2 0 3 11 4 10 5 9 6 8 7 2 3 1 4 0 5 11 6 10 7 9 8 3 4 2 5 1 6 0 7 11 8 10 9 ]

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