> Do I understand correctly that you can shuffle the image's rows (or columns) without losing that effect?
Yes, but a dramatic reduction in complexity still leaves a lot of complexity.
Your conditions seem to be a lot simpler than they originally appear. If I interpret you correctly, we're drawing seventeen 4-tuples of seventeen-bit numbers such that
> every tuple is a subset of one of the sets in the second set
"the elements of each tuple have no bits in common", and
> every drawn number is contained in one of sets in the first set
"each number has at most one bit set".
But there are only eighteen seventeen-bit numbers with at most one bit set, or seventeen if you exclude 0 (which seems reasonable). So isomorphically, we're trying to find seventeen 4-tuples of distinct numbers less than seventeen.
This is indeed easy (there are (17 choose 4) = 2380 of them), but I don't understand what you hope to do with these. At any rate, I don't think they particularly help solve the problem at hand.
Nope, the second condition would be "each i-number is from a set where no two numbers share more than one common bit"
You can pre-compute that set of sets (of course throwing away numbers with just 0, 1 or 2 bits set because we suspect they would slow us down without delivering) fit them in memory and then crunch the problem like a bug.
I've thrown together two simple scripts (in perl) which show that the known solution fits both of my rules. Hopefully I can write a program that would find solutions for any given NxM (or tell that no solution exists) tomorrow or something like that.
Still I suspect the solution would hit the wall because e.g. sets would be too big therefore the program would take actual ages to run, but one can surely hope?
> Still I suspect the solution would hit the wall because e.g. sets would be too big therefore the program would take actual ages to run, but one can surely hope?
I suspect something will go wrong, and this seems to me the obvious place for it to fail. But I expect it to fail mostly based on "if it was that easy, someone would probably have noticed", not because I have any particular insight into the problem.
And the first rule would be "the elements have no bits in common, but together (ORed) they have all 17 bits"
That's almost XOR(i in S) = -1 but not really
Yes, but a dramatic reduction in complexity still leaves a lot of complexity.
Your conditions seem to be a lot simpler than they originally appear. If I interpret you correctly, we're drawing seventeen 4-tuples of seventeen-bit numbers such that
> every tuple is a subset of one of the sets in the second set
"the elements of each tuple have no bits in common", and
> every drawn number is contained in one of sets in the first set
"each number has at most one bit set".
But there are only eighteen seventeen-bit numbers with at most one bit set, or seventeen if you exclude 0 (which seems reasonable). So isomorphically, we're trying to find seventeen 4-tuples of distinct numbers less than seventeen.
This is indeed easy (there are (17 choose 4) = 2380 of them), but I don't understand what you hope to do with these. At any rate, I don't think they particularly help solve the problem at hand.