Re: [boost] boost interval arithmetic

Subject: Re: [boost] boost interval arithmetic
From: Marc Glisse (marc.glisse_at_[hidden])
Date: 2015-01-14 16:58:17

On Thu, 15 Jan 2015, Павел Кудан wrote:

>> I tried the following program:
>>
>> #include <boost/numeric/interval.hpp>
>> #include <iostream>
>>
>> typedef boost::numeric::interval<double> I;
>> int main(){
>> Â Â Â I a(1,2);
>> Â Â Â I b(-1,1);
>> Â Â Â std::cout << lower(a) << ',' << upper(a) << std::endl;
>> Â Â Â std::cout << lower(b) << ',' << upper(b) << std::endl;
>> Â Â Â a/=b;
>> Â Â Â std::cout << lower(a) << ',' << upper(a) << std::endl;
>> }
>>
>> and it prints:
>>
>> 1,2
>> -1,1
>> -inf,inf
>>
>> and not -inf,-1 as you claim. If you have code showing a different result, 
>> please post it. I can't see anything in the documentation that supports 
>> your claim.
>>
>> [-inf,inf] is the smallest interval containing all possible values for the 
>> quotient, so it makes a lot of sense to return that. Now, if I understand 
>> correctly, you would like to add a policy so that anytime an operation has 
>> a disconnected image, instead of bridging the gap, boost would throw. That 
>> would be useless in my applications (well, maybe a shortcut since once I 
>> get [-inf,inf] somewhere it tends to propagate, eventually to a comparison 
>> that will throw), but maybe it makes sense for yours... In any case, it 
>> doesn't look like it would hurt boost to add such an optional policy. So 
>> if you want to prepare a patch, someone might be willing to review and 
>> pull it.
>
> In that case [-inf, inf] should be a result of any operation in the 
> world :) As it is universal :)))

Note the word "smallest" in my explanation.

> 2 + 2 == [-inf, inf], is it true? :) 
>
> Of course - as it is somewhere between :)
>
> Not true? Of course, not. 
>
> 2 + 2 !=Â Â [-inf, inf]
> and [-inf, inf] != [-inf, 1] U [1, inf]

[-inf,inf] is the smallest interval containing both [-inf, 1] and [1, 
inf]. It isn't possible to return a smaller interval that still contains 
all possible values.

> Let's see presentation about boost interval and interval arithmetic.
>
> https://www.lri.fr/~melquion/doc/03-rnc5-expose.pdf
>
> On page 11:
> Functions can also be used to compute a pair of intervals: [1, 2] ÷ 
> [−1, 1] = [−∞, −1] ∪ [1, +∞]. (that is not boost interval operator 
> / yet, it is basic arithmetic operation for interval. That is what 
> should be a result)
>
> also, on page 5:
> [a, b] × [c, d] = [min(ac, bc, ad, bd), max(ac, bc, ad, bd)],
> [a, b] ÷ [c, d] = [a, b] × [1 ÷ d, 1 ÷ c] if 0 NOT ∈ [c, d]
>
> (that is not boost interval operator / yet, but it is the way its result 
> is calculated when zero is not inside of [c, d] interval (by the way, 
> typo here - must be 'if 0 NOT ∈ (c, d)).
> And here official boost interval documentation
>
> http://www.boost.org/doc/libs/1_37_0/libs/numeric/interval/doc/interval.htm
>
> The operators / and /=  will try to produce an empty interval if the 
> denominator is exactly zero. If the denominator contains zero (but not 
> only zero), the result will be the smallest interval containing the set 
> of division results; so one of its bound will be infinite, but it may 
> not be the whole interval.
>
>
>
> So, official documented result of / operator for that case is [-inf, 1] 
> as it is smallest interval (the result of division_part1() function of 
> boost interval class). And this result is not correct as it is not full.

Language barrier. You are grossly misunderstanding the documentation. "The 
set of division results" is [-inf,-1] union [1,inf]. "The smallest 
interval containing" this set is [-inf,inf]. It is NOT the same thing as 
the smallest connected component of the set.

-- 
Marc Glisse