3 ms·
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
by guard-of-terra 15y ago
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?
- philh 15y agoOh, I read AND but was doing OR. My mistake. > 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.
- guard-of-terra 15y agoI have a same gut feeling, but one of my professors told us how the discoveries are being made: Everybody know that some thing is impossible to do. But one person who didn't know that accidentally does it. Right now I have one type-1 set having 36 numbers in it. The first numbers are 7, 25 and 42. I like it.