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Illustrating by example one good reason not to trust a proof <wink>.

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Tim Peters 2002-08-15 20:10:45 +00:00
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Objects/longobject.c
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@ -1791,9 +1791,9 @@ then we're asking whether asize digits >= f(bsize/2) digits + 2 bits. By #4,
asize is at least f(bsize/2)+1 digits, so this in turn reduces to whether 1 asize is at least f(bsize/2)+1 digits, so this in turn reduces to whether 1
digit is enough to hold 2 bits. This is so since SHIFT=15 >= 2. If digit is enough to hold 2 bits. This is so since SHIFT=15 >= 2. If
asize == bsize, then we're asking whether bsize digits is enough to hold asize == bsize, then we're asking whether bsize digits is enough to hold
f(bsize/2) digits + 2 bits, or equivalently (by #1) whether c(bsize/2) digits c(bsize/2) digits + 2 bits, or equivalently (by #1) whether f(bsize/2) digits
is enough to hold 2 bits. This is so if bsize >= 1, which holds because is enough to hold 2 bits. This is so if bsize >= 2, which holds because
bsize >= KARATSUBA_CUTOFF >= 1. bsize >= KARATSUBA_CUTOFF >= 2.
Note that since there's always enough room for (ah+al)*(bh+bl), and that's 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 clearly >= each of ah*bh and al*bl, there's always enough room to subtract