Fixed error in new comment.
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@ -1786,12 +1786,11 @@ least 2*shift+2 digits, and there's not obviously enough room for the
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extra two bits. We need a sharper analysis in this case. The major
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laziness was in the "the same is true of ah+al" clause: ah+al can't actually
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have shift+1 digits + 1 bit unless bsize is odd and asize == bsize. In that
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case, we actually have (2*shift+1)*2 - shift = 3*shift + 2 allocated digits
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remaining, and that's obviously plenty to hold 2*shift + 2 digits + 2 bits.
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case, we actually have (2*shift+1)*2 - shift = 3*shift+2 allocated digits
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remaining, and that's obviously plenty to hold 2*shift+2 digits + 2 bits.
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Else (bsize is odd and asize < bsize) ah and al each have at most shift digits,
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so ah+al has at most shift digits + 1 bit, and (ah+al)*(bh+bl) has at most
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2*shift+1 digits + 2 bits, and again 2*shift+2 digits + 2 bits is
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enough to hold it.
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2*shift+1 digits + 2 bits, and again 2*shift+2 digits is enough to hold it.
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*/
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/* b has at least twice the digits of a, and a is big enough that Karatsuba
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