search for: vectorize_factor

.... 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 unrolling and peeling,...

...signed 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...

...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...

...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...

...gt;> 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 o...

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 unrolling and peeling,...

...er 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, aft...

...> > > > > > > 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. > &...

...> > > 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 > > > > > > > > > > > duplic...