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k_mul() comments: Simplified the simplified explanation of why ah*bh and 

al*bl "always fit":  it's actually trivial given what came before.
This commit is contained in:
Tim Peters 2002-08-14 17:07:32 +00:00
parent b8c20a723f
commit 48d52c0fcc
1 changed files with 3 additions and 6 deletions
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9
Objects/longobject.c
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@ -1792,12 +1792,9 @@ Else (bsize is odd and asize < bsize) ah and al each have at most shift digits,
so ah+al has at most shift digits + 1 bit, and (ah+al)*(bh+bl) has at most
2*shift+1 digits + 2 bits, and again 2*shift+2 digits is enough to hold it.
Note that the "lazy" analysis is enough to show that there's always enough
room to subtract al*bl and ah*bh. al and bl each have no more than shift
digits, so al*bl has no more than 2*shift, so there's at least one digit
to spare in the remaining allocated digits. The same is true for ah*bh when
bsize is even. When bsize is odd, ah*bh has at most 2*shift+2 digits, and
there are at least that many remaining allocated digits when bsize is odd.
Note that since there's always enough room for (ah+al)*(bh+bl), and that's
clearly >= each of ah*bh and al*bl, there's always enough room to subtract
ah*bh and al*bl too.
*/
/* b has at least twice the digits of a, and a is big enough that Karatsuba