Re: [boost] [odeint] Iterator semantics

Subject: Re: [boost] [odeint] Iterator semantics
From: Karsten Ahnert (karsten.ahnert_at_[hidden])
Date: 2012-07-31 10:30:45

On 07/31/2012 03:07 PM, Mario Mulansky wrote:
> On Tuesday, July 31, 2012 11:47:55 am Sergey Mitsyn wrote:
>> On 29.07.2012 6:31, Dave Abrahams wrote:
>>> on Thu Jul 12 2012, Karsten Ahnert <karsten.ahnert-AT-ambrosys.de> wrote:
>>>> On 07/12/2012 09:52 AM, Mario Mulansky wrote:
>>>>> On Wed, 2012-07-11 at 23:50 -0700, Jeffrey Lee Hellrung, Jr. wrote:
>>>>>
>>>>> I agree that it1 != it2 should be implemented in the sense of
>>>>> t1+-dt/2 != t2+-dt/2
>>>>
>>>> So you are checking for overlap. This looks good and I think it should
>>>> work.
>>>
>>> If I'm understanding correctly what you're saying, making two iterators
>>> equal if their underlying time ranges overlap would be "evil."  Your
>>> equality operator would not be transitive, and thus not an equivalence
>>> relation, and therefore you would not have implemented a conforming
>>> iterator, and you'd have no right to expect an algorithm that works on
>>> iterators to work on yours.
>>
>> IMHO the values of the iterator's underlying time variable would belong
>> to "almost" discrete {t0 + n*dt +/- epsilon}, n from Z, where epsilon
>> comes from floating point errors and is small. Thus, the distance
>> between neighboring values should be no less than (dt-2*epsilon).> 
>> I would say a test 'distance(t1,t2) < dt/2' would be transitive if
>> epsilon is less than dt/4
>>
>> That's of course true while t0 and dt is the same for all iterators
>> passed to an algorithm.
> 
> dt even changes while iterating when methods with step-size control are used. 
> This case is really ugly and I'm not sure if it's possible to implement a 
> fully compatible iterator there. Maybe one should restrict to fixed-step-size 
> methods where the overlap checking works?

I think already found a solution. The end iterator is marked as end (it
has a flag m_end). Then, two iterators are equal if theirs flags are
equal, or if the flags is not equal the time of the first iterator must
be smaller then the time of the end iterator:

bool equal( iterator other ) const
{
    if( m_end == other.m_end ) return true;
    else
    {
        return ( m_end ) ?
            ( other.m_time < m_time ) :
            ( m_time < other.m_time );
    }
}

This iterator is a single pass iterator, for example it is similar to
the istream_iterator.