4 ms·

> It does not preserve associativity of addition. It does, if you use the mathematically natural addition operation (see my reply to enriquto). In the example

> It does not preserve associativity of addition. It does, if you use the mathematically natural addition operation (see my reply to enriquto). In the example above, the result is MAX_INT/1 in both cases.

jcranmer 7y ago
An addition operator where adding two positive numbers results in a negative number can hardly be called "the mathematically natural" variant. Saturation arithmetic (which is the norm for floating point, and even most fixed point) is pretty obviously not associative. Also, it's not clear to me that your proposed wraparound addition operator is actually fully associative in all possible cases of overflow.
- saithound 7y ago
  > Also, it's not clear to me that your proposed wraparound addition operator is actually fully associative in all possible cases of overflow. Proof sketch: write out (an/ad + (bn/bd) + cn/cd)) and ((an/ad + bn/bd) + cn/cd) symbolically, and apply the fact that twos complement arithmetic is associative, commutative and distributive in all possible cases of overflow. > An addition operator where adding two positive numbers results in a negative number can hardly be called "the mathematically natural" variant. It's exactly how integer arithmetic works on every CPU designed after 1980, to general satisfaction. In any case, it's mathematically natural because it's exactly how you would construct the field of fractions of an integral domain.
  - enriquto 7y ago
    > write out (an/ad + (bn/bd) + cn/cd)) and ((an/ad + bn/bd) + cn/cd) symbolically, and apply the fact that twos complement arithmetic is associative, commutative and distributive in all possible cases of overflow. I doubt that raku/perl6 uses these rules for rational arithmetic. It would be highly unintuitive to sum two rational numbers close to 0.5 and obtain a negative number. Do you have any reference for raku arithmetic rules? I cannot seem to find them.
    - dragonwriter 7y ago
      Raku Rat falls through to floats if the denominator overflows, and has arbitrary precision numerators, it does not use twos-complement math for normal operations in any way exposed to the user, AFAIK. Most other systems (including Scheme) that I've seen that use rational representation by default use arbitrary precision for both numerator and denominator, so don't have this concern.
  - dnautics 7y ago
    Your proof fails for the odd man out negative integer value.
    - saithound 7y ago
      It does not. Feel free to justify your claim by writing down an explicit counterexample, though.

jcranmer 7y ago

An addition operator where adding two positive numbers results in a negative number can hardly be called "the mathematically natural" variant. Saturation arithmetic (which is the norm for floating point, and even most fixed point) is pretty obviously not associative. Also, it's not clear to me that your proposed wraparound addition operator is actually fully associative in all possible cases of overflow.

saithound 7y ago
> Also, it's not clear to me that your proposed wraparound addition operator is actually fully associative in all possible cases of overflow. Proof sketch: write out (an/ad + (bn/bd) + cn/cd)) and ((an/ad + bn/bd) + cn/cd) symbolically, and apply the fact that twos complement arithmetic is associative, commutative and distributive in all possible cases of overflow. > An addition operator where adding two positive numbers results in a negative number can hardly be called "the mathematically natural" variant. It's exactly how integer arithmetic works on every CPU designed after 1980, to general satisfaction. In any case, it's mathematically natural because it's exactly how you would construct the field of fractions of an integral domain.
- enriquto 7y ago
  > write out (an/ad + (bn/bd) + cn/cd)) and ((an/ad + bn/bd) + cn/cd) symbolically, and apply the fact that twos complement arithmetic is associative, commutative and distributive in all possible cases of overflow. I doubt that raku/perl6 uses these rules for rational arithmetic. It would be highly unintuitive to sum two rational numbers close to 0.5 and obtain a negative number. Do you have any reference for raku arithmetic rules? I cannot seem to find them.
  - dragonwriter 7y ago
    Raku Rat falls through to floats if the denominator overflows, and has arbitrary precision numerators, it does not use twos-complement math for normal operations in any way exposed to the user, AFAIK. Most other systems (including Scheme) that I've seen that use rational representation by default use arbitrary precision for both numerator and denominator, so don't have this concern.
- dnautics 7y ago
  Your proof fails for the odd man out negative integer value.
  - saithound 7y ago
    It does not. Feel free to justify your claim by writing down an explicit counterexample, though.