4 ms·
Wouldn’t that produce 2^n possible results to choose from, where n is the number of chunks? That seems like a lot of additional (he-he) instructions executed.
by volemo 1y ago
Wouldn’t that produce 2^n possible results to choose from, where n is the number of chunks? That seems like a lot of additional (he-he) instructions executed.
- repelsteeltje 1y agoNope. Just 2n: each chunk pair is added once without carry, and once won't carry=1. For as long as radix=2, you either have a carry or you don't.
- mananaysiempre 1y agoFor a single addition, the radix is irrelevant, the carry is always zero or one: (r-1) + (r-1) = 2r - 2 < 2r.
- account42 1y agoBut for a single addition you will also not save any time because you still need to do the select in series. The whole point of the radix 2^51 trick is that you can defer the normalization over several additions, but in order to do that your carry needs to be more than a single bit.