Talk:Random Latin squares: Difference between revisions

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(→‎"restarting row" method: Nigel's conjecture is correct!)
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::: If the factor code is used to produce a large number of random Latin Squares of order 4, say 1 million, my conjecture is that c and d will be produced twice as often as a and b. Your conjecture is they will be produced equally. Do you fancy giving it a go?--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 01:29, 17 July 2019 (UTC)
::: If the factor code is used to produce a large number of random Latin Squares of order 4, say 1 million, my conjecture is that c and d will be produced twice as often as a and b. Your conjecture is they will be produced equally. Do you fancy giving it a go?--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 01:29, 17 July 2019 (UTC)

:::: My Go solution also uses the "restarting row" method which I certainly expected to be uniformly random but, when I ran Nigel's test, sure enough c and d occurred about twice as often as a and b. So it does demonstrate how easily one can be led astray by intuition in this sort of work. Anyway I'm going to rewrite the Go solution on the lines Nigel suggested earlier and will re-post later today. --[[User:PureFox|PureFox]] ([[User talk:PureFox|talk]]) 13:15, 17 July 2019 (UTC)


:: in particular, you may find this interesting: "The probability of finishing the entire LS is a combination of the previous series of probabilities, but not their product, as the rows are not independent to each other (i.e. row i depends of values on row i-1)." --[[User:Chunes|Chunes]] ([[User talk:Chunes|talk]]) 19:37, 16 July 2019 (UTC)
:: in particular, you may find this interesting: "The probability of finishing the entire LS is a combination of the previous series of probabilities, but not their product, as the rows are not independent to each other (i.e. row i depends of values on row i-1)." --[[User:Chunes|Chunes]] ([[User talk:Chunes|talk]]) 19:37, 16 July 2019 (UTC)