4 ms·
This is actually fewer bits than they're using in the article, but that's not clear because they refer to their numbers by how many "[decimal] digits", not how
by aSanchezStern 5y ago
This is actually fewer bits than they're using in the article, but that's not clear because they refer to their numbers by how many "[decimal] digits", not how many bits. The hardware doubles they're starting with are 64-bits wide, representing 15.5 decimal digits, but then they move into 128- and 256-bit representations. 80-bit math, such as that available on the old x87 floating point co-processors, is I believe about 21 decimal digits.
- shmageggy 5y agoI found that very unintuitive. Is it typical in numerical analysis to talk about the number of decimal digits represented by a binary floating point number? I don't see what is gained by changing base.
- erosenbe0 5y agoIt's a bad idea, or okay as an informal thing! It is just using the result that Log10(2^53) is 15.95 to say that 53 bits has 15.95 digits of 'information.'