Talk:Random Latin squares
Latin Squares in reduced form
Latin Squares in reduced form makes trivial uniformly generating random Latin Squares up to order 6. 3 random numbers are required. The first in the range 1 to number of reduced Latin Squares of order n is used to select a member of the set of reduced Latin Squares of order n. The second in the range 1 to n! is used to select a permutation of the columns of the selected reduced Latin Square. The third in the range 1 to (n-1)! is used to select a permutation of the rows 2 to n.--Nigel Galloway (talk) 09:51, 12 July 2019 (UTC)
Revert to draft
I think the task description needs to require and demonstrate that the algorithm must be capable of producing all valid latin squares of size n. As it stands starting from a valid latin square, which is easy to generate say:
0 1 2 3 4 1 2 3 4 0 2 3 4 0 1 3 4 0 1 2 4 0 1 2 3
there are 5! ways to arrange the columns and 5! ways to arrange the rows each of which is a valid random latin square. From the task description I see no reason why this would not be a valid solution as the task stands. I have added a reference to A002860 which is the number of latin squares of size n. For n>4 this simple solution produces random valid latin squares but can not produce all solutions. Nor does the Python algorithm which is just obfuscation which reduces to the above.--Nigel Galloway (talk) 18:38, 10 June 2019 (UTC)
Non python algorithm
Apart from selecting from almost none of the possible latin squares the 'python algorithm' has second disadvantage. Considering the above latin square if I move the rightmost column to in front of the leftmost, or the bottom row to above the top I produce the same result. That is there are varying number of ways of producing each of the small number of latin squares that this algorithm can produce. An alternative is to just permute the meaning of 0..4. This would produce a smaller subset of all possible latin squares, but almost none is almost none either way, and they would be uniform. This task was promoted from draft very quickly, I think it needs to go back to draft until this issue is resolved--Nigel Galloway (talk) 09:30, 12 June 2019 (UTC)
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.
- It needs to be precise. The task asks for random latin squares of size 5. At 5 this algorithm produces less than 9% of all possible. At 6 it produces about 0.05%.--Nigel Galloway (talk) 21:00, 10 June 2019 (UTC)
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)
- There is not just a little bit missing, above n=4 it produces almost none of the possible solutions.--Nigel Galloway (talk) 21:00, 10 June 2019 (UTC)
- It might be worth mentioning that non-uniform solutions are acceptable in the task description. From my cursory research, it appears that generating a uniformly-random latin square is a difficult problem which either requires the exponential-time brute force approach of generating random permutations for rows and restarting the row if a clash occurs, or else requires implementing the involved Jacobson and Matthews algorithm. This also absolves us of some mathematical pedantry. :) --Chunes (talk) 15:14, 9 June 2019 (UTC)
Task description second sentence clear ?
Leaving aside the missing full stop, I wonder if that sentence is clear and well-formed ? The kernel statement seems to be that 'A Latin square generates a Latin square'. Might that not risk furrowing a brow or two ? Perhaps an edit for clarity and transitive plausibility ?Hout (talk) 02:48, 12 June 2019 (UTC)
- Thanks. Changed. --Paddy3118 (talk) 14:33, 12 June 2019 (UTC)
- Well ... That still makes no sense at all ... How can a Latin square "generate" the values which constitute it ? Perhaps what the incoherence (and attempted circularity) of that sentence expresses is really a need for a slightly more solid concept of what is actually being asked for here ? Hout (talk) 17:34, 12 June 2019 (UTC)