Talk:Random Latin squares
Python Algorithm
I got the mention of Latin squares from a stack-overflow question and a link to some solution methods that I did not read (and so did not add to the task as a reference).
I worked out that if you have a smaller solution of:
0 1 1 0
then the next larger solution (ignoring randomisation), could be got by: 1. Insert a copy of the first row at the end:
0 1 1 0 0 1
2. Insert the new symbol along the diagonal
2 0 1 1 2 0 0 1 2
I randomise the selection of symbol to insert at each recursive stage, and at the end swap rows randomly and swap columns randomly as these transformations preserve "Latin-ness".
If you read the reference mentioned in the first sentence and think it is of use, then please add it to the task.
- --Paddy3118 (talk) 11:44, 9 June 2019 (UTC)
"Randomness"
I don't actually use the algorithm in earnest, but several of the references mention the randomness of the generation.
I have no measure of the randomness, but thought that any number should be as likely to appear in any position in the square so I created a counter of how many times each symbol appeared in each cell of the square grid and present the counts for a million runs:
Additional Python: <lang python>def distributions(n, count):
counters = [[defaultdict(int) for _ in range(n)] for __ in range(n)] range_n = range(n) for _ in range(count): square = rls(n) for r in range_n: for c in range_n: counters[r][c][ square[r][c] ] += 1 for symbol in range_n: symcounts = [[counters[r][c][symbol] for c in range_n] for r in range_n] mx = max(cnt for row in symcounts for cnt in row) mn = min(cnt for row in symcounts for cnt in row) print(f'\n{symbol} distribution: from {mn} to {mx}\n==') print(_to_text(symcounts))
distributions(5, 1000_000)</lang>
- Output:
0 distribution: from 199123 to 200657 == 199758 200268 199123 200194 200657 199878 200137 200165 199932 199888 200447 199564 200602 200024 199363 200094 200412 200266 199659 199569 199823 199619 199844 200191 200523 1 distribution: from 198960 to 200500 == 200057 200135 200085 199825 199898 200144 200249 199308 200118 200181 199297 200500 200268 200380 199555 200441 198960 200000 200317 200282 200061 200156 200339 199360 200084 2 distribution: from 198983 to 200849 == 200849 200156 200034 199978 198983 199747 199808 200144 200054 200247 199796 200405 199650 199895 200254 199820 199935 200178 199872 200195 199788 199696 199994 200201 200321 3 distribution: from 199433 to 200820 == 199816 199452 200527 200133 200072 200820 200003 199749 199598 199830 199769 199985 200143 199664 200439 199739 200558 199545 199932 200226 199856 200002 200036 200673 199433 4 distribution: from 199337 to 200691 == 199520 199989 200231 199870 200390 199411 199803 200634 200298 199854 200691 199546 199337 200037 200389 199906 200135 200011 200220 199728 200472 200527 199787 199575 199639
--Paddy3118 (talk) 14:10, 9 June 2019 (UTC)
Uniformity over all possible
I suspect this algorithm does not generate random latin squares uniformly. What matters is the uniformity of latin squares taken from the list of all latin squares. See https://math.stackexchange.com/questions/63131/generate-random-latin-squares . --Chunes (talk) 14:19, 9 June 2019 (UTC)
Hi Chunes, you're right!
I went back to the wikipedia article which has a table stating that the numbero f distinct n=4 LS is 576. This additional code: <lang python>In [97]: def distinct(n, count):
...: distinct = Counter(tuple(rls(n)) for _ in range(count)) ...: print(f"Found {len(distinct)} different {n}-by-{n} Latin squares in {count} trials") ...: ...:
In [98]: distinct(4, 576_00) Found 432 different 4-by-4 Latin squares in 57600 trials
In [99]: </lang>
Shows that my algorithm cannot generate 25% of the possible outputs. It would have been nice if it could, but I won't disallow the algorithm as the task description isn't that precise.
Ideally I would find examples of what is missing and try and work out an additional algorithm that generates from the whole set. .... If time allowed.
--Paddy3118 (talk) 14:55, 9 June 2019 (UTC)