6 ms·
Not quite: It's taking advantage of (1+a)(1+b) = 1 + a + b + ab. And where a and b are both small-ish, ab is really small and can just be ignored. So it turns
by mota7 2y ago
Not quite: It's taking advantage of (1+a)(1+b) = 1 + a + b + ab.
And where a and b are both small-ish, ab is really small and can just be ignored.
So it turns the (1+a)(1+b) into 1+a+b. Which is definitely not the same! But it turns out, machine guessing apparently doesn't care much about the difference.
- tommiegannert 2y agoPlus the 2^-l(m) correction term. Feels like multiplication shouldn't be needed for convergence, just monotonicity? I wonder how well it would perform if the model was actually trained the same way.
- amelius 2y agoYou might then as well replace the multiplication by the addition in the original network. In that case you're not even approximating anything. Am I missing something?
- dotnet00 2y agoThey're applying that simplification to the exponent bits of an 8 bit float. The range is so small that the approximation to multiplication is going to be pretty close.
- dsv3099i 2y agoThis trick is used a ton when doing hand calculation in engineering as well. It can save a lot of work. You're going to have tolerance on the result anyway, so what's a little more error. :)