4 ms·
I don't think that follows? Just because a better method exists (the inverse) doesn't make the original perform better. Imagine the following method: take what
by mecsred 4y ago
I don't think that follows? Just because a better method exists (the inverse) doesn't make the original perform better.
Imagine the following method: take whatever the bit was, and set it to the opposite of that. 0% < 50%. Of course you can invert and get 100%, but that's a different method. It doesn't make the original any better.