3 ms·
And a 64b sweep is ~21 days. As a HW guy who tests fp units... just saying. The real issue is that the verification code is much, much, slower.
by thechao 4y ago
And a 64b sweep is ~21 days. As a HW guy who tests fp units... just saying.
The real issue is that the verification code is much, much, slower.
- dragontamer 4y ago> And a 64b sweep is ~21 days Per GPU. A consumer GPU at that (A100 is faster). 4x GPUs (the AMD 6900 XT) per node and maybe 10x nodes in a rack (4U per node, 42U cabinet) can do 64-bit sweep in half a day. > The real issue is that the verification code is much, much, slower. Isn't verification of 64b multipliers actually a 128-bit problem? (x * y == z, so you have 2x 64-bit numbers to work with). Also, I think they use BDDs, which although multiplication is exponential, its still small enough exponent that 128-bit BDDs fit on today's computers. So its not a brute force thing like discussed in this topic, but actually a very elegant data-structure (though with tons of complications, as is the nature of NP-complete and expspace problems).
- thechao 4y agoIt’s a lot easier to do formal verification… which is a fancy BDD, in many ways.
- insanitybit 4y agoAt that point I might just take the first 2^32 numbers, the last 2^32 numbers, and some range in between, and feel good about it.
- eddsh1994 4y agohttps://en.wikipedia.org/wiki/Bounds_checking https://en.wikipedia.org/wiki/Bounds_checking still seems sensible, if 2^32-1 works why wouldn't 2^32-2?