In fact, you can use the second algorithm mentioned there, with a slight modification:
In the special case of b
being of the form 0n1m
(a > b) == !!(a & ~b)
holds.
In fact, you can use the second algorithm mentioned there, with a slight modification:
In the special case of b
being of the form 0n1m
(a > b) == !!(a & ~b)
holds.