On 6/4/2018 at 6:01 PM, GautamGreat said:
I already read that, that is not a proper solution.
for (int b = 0; b < 49; b++) for (int c = b + 1; c < 50; c++) if (huge[y++]) if (calc[b] & calc[c]) bit = !bit;
the expression will be negated only if calc=1 and calc[c]=1 and huge[y++]
this is easier in reverse order from the end to the begin:
last time checks last two bits: calc&calc
Still trying to find solution.