I want to generate $n$ by $n$ binary arrays, and then reduce the output from those arrays by performing a number of permutations and eliminating duplicates. The problem is that the number of arrays increases as $2^{n^2}$, so storing every tuple in memory and comparing each permutation is impractical for $n>4$.

A working solution when $n<4$ is given here: Reduce the output from tuples by including symmetry?.

My questions is: how can I generate $n$ by $n$ binary arrays and eliminate duplicates (as defined by certain permutations) in a memory efficient manner?

For instance, is it possible to generate a few tuples at a time, run the permutations, check for duplicates, and then continue generating the remaining tuples?

P.S. I asked a different version of this question here: Constructing and Reducing Tuples in a Memory Efficient Manner, and have been unable to implement the solution given here: Memory efficient generation and selection of tuples.


I think you can do something with these identities:

Table[Partition[IntegerDigits[i, 2, 4], 2, 2], {i, 0, 2^4 - 1}] === Tuples[{0, 1}, {2, 2}]

Table[Partition[IntegerDigits[i, 2, 9], 3, 3], {i, 0, 2^9 - 1}] === Tuples[{0, 1}, {3, 3}]

So in general, you can say that this is True:

Table[Partition[IntegerDigits[i, 2, (n*n)], n, n], {i, 0, 2^(n*n) - 1}] === Tuples[{0, 1}, {n, n}]

(I checked this up to 5).

But this then gives you a way to just get any individual one:

f[n_, i_] := Partition[IntegerDigits[i, 2, (n*n)], n, n]


f[10, 4938742893] //MatrixForm


enter image description here

Hope this helps.

| improve this answer | |
  • $\begingroup$ f[start_Integer, howMany_Integer, n_Integer] := Partition[#, n] & /@ Table[IntegerDigits[i, 2, n^2], {i, start, start + howMany - 1}] /; start >= 0 && howMany >= 1 && n > 0 && 0 < start + howMany - 1 <= 2^n^2 - 1; MatrixForm /@ f[0, 13, 2] $\endgroup$ – Dr. belisarius Mar 3 '16 at 2:47

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