Keep the remainder of % smaller than the divisor [patch] - #111
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Mod rounded the remainder to the operands' fewest significant digits, which could carry it up to the divisor itself (19.9 % 2 == 2, 9.96 % 5 == 5). A remainder is a subtraction, so round it to the fewest decimal places as Subtract does, and return zero when that rounding still reaches the divisor or beyond, since the remainder is then indistinguishable from a whole multiple. Update the README and class remarks, which said modulus followed the significant-digits rule. Fixes #109 Co-Authored-By: Claude Opus 5.5 <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_01QBrEwaLGeo2SvjVK3AnLrp
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Fixes #109
What was wrong
ModroundedPreciseNumber.Mod(left, right)to the operands' fewest significant digits. A remainder is usually much smaller than the operands, so that rounding could carry it up to the divisor itself:19.9 % 2 == 2,9.96 % 5 == 5,-19.9 % 2 == -2. That breaks the|x % n| < |n|invariant.Fix
left - right × quotient), so round it to the fewest decimal places, the waySubtractdoes, using the existingLowestDecimalDigitshelper.1.9rounded to 0 places is still2. And when the divisor has more decimals than the dividend, rounding can land past it:6.8 % 2.27is exactly 2.26, which rounds to 2.3. So when the rounded magnitude reaches|right|, the result is zero. At that precision, the remainder is indistinguishable from a whole multiple of the divisor.This touches only
Modand the remarks. It does not conflict with #110 (theLowestSignificantDigitsfix for #108). I checked that the two branches merge cleanly.Tests
The new
SignificantNumberModTestscovers:6.8 % 2.27|result| < |divisor|across a spread of operands, including a negative divisorWith the fix reverted, 10 cases fail. With it, the whole suite passes.
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https://claude.ai/code/session_01QBrEwaLGeo2SvjVK3AnLrp
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