search for: unroll_factor

...op peel and gvn). For a same debugloc, a unique number is assigned to each copy and encoded in the discriminator. Assume that the discriminator is uint32. The traditional discriminator is less than 256, let's take 8 bit for it. For duplication factor (type 1 duplication), we assume the maximum unroll_factor * vectorize_factor is less than 256, thus 8 bit for it. For unique number(type 2 duplication), we assume code is at most duplicated 32 times, thus 5 bit for it. Overall, we still have 11 free bits left in the discriminator encoding. Let's take the original source as an example, after loop unro...

...que number is assigned to each >> copy and encoded in the discriminator. >> >> Assume that the discriminator is uint32. The traditional discriminator is >> less than 256, let's take 8 bit for it. For duplication factor (type 1 >> duplication), we assume the maximum unroll_factor * vectorize_factor is >> less than 256, thus 8 bit for it. For unique number(type 2 duplication), we >> assume code is at most duplicated 32 times, thus 5 bit for it. Overall, we >> still have 11 free bits left in the discriminator encoding. >> >> Let's take the or...

...same debugloc, a unique number is assigned to each copy and > encoded in the discriminator. > > Assume that the discriminator is uint32. The traditional discriminator is > less than 256, let's take 8 bit for it. For duplication factor (type 1 > duplication), we assume the maximum unroll_factor * vectorize_factor is > less than 256, thus 8 bit for it. For unique number(type 2 duplication), we > assume code is at most duplicated 32 times, thus 5 bit for it. Overall, we > still have 11 free bits left in the discriminator encoding. > > Let's take the original source as an...

...same debugloc, a unique number is assigned to each copy and > encoded in the discriminator. > > Assume that the discriminator is uint32. The traditional discriminator is > less than 256, let's take 8 bit for it. For duplication factor (type 1 > duplication), we assume the maximum unroll_factor * vectorize_factor is > less than 256, thus 8 bit for it. For unique number(type 2 duplication), we > assume code is at most duplicated 32 times, thus 5 bit for it. Overall, we > still have 11 free bits left in the discriminator encoding. > > Let's take the original source as an...

...ed to each >>> copy and encoded in the discriminator. >>> >>> Assume that the discriminator is uint32. The traditional discriminator >>> is less than 256, let's take 8 bit for it. For duplication factor (type 1 >>> duplication), we assume the maximum unroll_factor * vectorize_factor is >>> less than 256, thus 8 bit for it. For unique number(type 2 duplication), we >>> assume code is at most duplicated 32 times, thus 5 bit for it. Overall, we >>> still have 11 free bits left in the discriminator encoding. >>> >>> L...

It looks like the example doesn't use the encoding described in the text? Assume that the discriminator is uint32. The traditional discriminator is less than 256, let's take 8 bit for it. For duplication factor (type 1 duplication), we assume the maximum unroll_factor * vectorize_factor is less than 256, thus 8 bit for it. For unique number(type 2 duplication), we assume code is at most duplicated 32 times, thus 5 bit for it. Overall, we still have 11 free bits left in the discriminator encoding. Let's take the original source as an example, after loop unro...

...c, a unique number is assigned to > each copy and encoded in the discriminator. > > > Assume that the discriminator is uint32. The traditional > discriminator is less than 256, let's take 8 bit for it. For > duplication factor (type 1 duplication), we assume the maximum > unroll_factor * vectorize_factor is less than 256, thus 8 bit for > it. For unique number(type 2 duplication), we assume code is at most > duplicated 32 times, thus 5 bit for it. Overall, we still have 11 > free bits left in the discriminator encoding. > > > Let's take the original source...

...> > > > > > > > > > Assume that the discriminator is uint32. The traditional > > > > discriminator is less than 256, let's take 8 bit for it. For > > > > duplication factor (type 1 duplication), we assume the maximum > > > > unroll_factor * vectorize_factor is less than 256, thus 8 bit for > > > > it. For unique number(type 2 duplication), we assume code is at > > > most > > > > duplicated 32 times, thus 5 bit for it. Overall, we still have 11 > > > > free bits left in the discriminato...

...> > > > > > discriminator is less than 256, let's take 8 bit for it. For > > > > > > > > > > > duplication factor (type 1 duplication), we assume the > > > > > maximum > > > > > > > > > > > unroll_factor * vectorize_factor is less than 256, thus 8 bit > > > > > for > > > > > > > > > > > it. For unique number(type 2 duplication), we assume code is > > > > > at > > > > > most > > > > > > > > &g...