Missing a 'not'. You would use `bool( nn and not (nn&(nn-1)) )` Let's try for 15 start 0b1111 sans1 0b1110 anded 0b1110 not 0b0000 No

Missing a 'not'. You would use `bool( nn and not (nn&(nn-1)) )` Let's try for 15 start 0b1111 sans1 0b1110 anded 0b1110 not 0b0000 Now 8 start 0b1000 sans1 0b0111 anded 0b0000 not 0b0001 It's basically a claim that, in binary, only powers of two won't have any overlapping bits between the initial number and that number minus one. Doesn't work for 0 so you have to special case it.