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k_mul() comments: Explained why there's always enough room to subtract 

ah*bh and al*bl.  This is much easier than explaining why that's true
for (ah+al)*(bh+bl), and follows directly from the simple part of the
(ah+al)*(bh+bl) explanation.
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Tim Peters 2002-08-14 16:36:23 +00:00
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Objects/longobject.c
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@ -1791,6 +1791,13 @@ remaining, and that's obviously plenty to hold 2*shift+2 digits + 2 bits.
Else (bsize is odd and asize < bsize) ah and al each have at most shift digits, 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 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. 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.
*/ */
/* b has at least twice the digits of a, and a is big enough that Karatsuba /* b has at least twice the digits of a, and a is big enough that Karatsuba