If by summetry you mean the fact that all but one of the rows have the same number of 0s as 1s, you can replace 10011011 with 00011011 and have them all have that property. (The cofactor of the changed bit is 0.) You'll note that the columns are not particularly symmetric in this sense. You haven't described your hillclimbing algorithm, but to maximize the determinant while changing a single row, you clearly set each element in the row to 0 if its cofactor is negative and 1 if the cofactor is positive. (If the cofactor is 0, the element doesn't matter.) If you assume that a cofactor is as likely to be positive as negative, then you'd expect roughly the same number of 0s as 1s. --ms Gareth McCaughan wrote:
Generating a bunch of random boolean 8x8's, I've gotten every determinant from 0 to 32, and 34, 35, 36, 42. 73 different values, when you count the negative ones.
33: 00001111 01011100 11000100 01100001 10010101 10111110 01010110 11011011 40: 11101000 00110110 10001111 01011101 10010001 00011010 01100011 00101001 44: 00101010 10000011 11001000 10011110 11110010 01011011 01100101 10111001 45: 11001011 01110011 10010101 10100110 01010110 00101111 11111100 10011010 48: 11100100 01100011 10000111 01001110 10101010 10111101 11010010 00110110 56: 11001100 01101001 10011011 01010101 01100110 00111100 11110000 10100101
These were found by naive hillclimbing starting from random matrices, which makes it interesting to me how much symmetry there seems to be in the last one's rows. But maybe I'm being taken in by the Strong Law of Small Numbers myself now.
Anyone got a proof that bigger than 56 is impossible? Or a way to rule out (e.g.) 38 or 39?