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Subject: Re: [boost] Interest in some Algorithms
From: Phil Endecott (spam_from_boost_dev_at_[hidden])
Date: 2016-01-19 12:31:00

Steven Watanabe wrote:
> On 01/18/2016 05:13 AM, Dean Michael Berris wrote:
>> I wanted to ask whether anybody's interested in a couple of
>> algorithms, to be made part of the Boost.Algorithm library.
>> One is tentatively named ?set_inplace_difference' which has
>> the following synopsis:
>> template <class I1, class I2>
>> I1 set_inplace_difference(I1 f1, I1 l1, I2 f2, I2 l2);
>> template <class I1, class I2, class C>
>> I1 set_inplace_difference(I1 f1, I1 l1, I2 f2, I2 l2, C comp);

(I've fixed an obvious typo above...)
>> It's essentially a set_difference which re-uses the storage of
>> the range [f1, l1). It takes elements from [f1, l1) that are in
>> [f2, l2) and moves them to [p, l1) where p is the partition point
>> (or the new 'end'). This is best used in conjunction with 'erase'
>> on vectors and/or std::deque. These algorithms return p.
>> Another is tentatively named 'set_inplace_intersection' which
>> instead of the above which does the difference, does an intersection
>> instead.
>> Requirements on I1 and I2 are that they are RandomAccess iterators,
>> and that they are sorted. I haven't figured out whether
>> partially-sorted is a sufficient condition. I also haven't figured
>> out whether it would work for weaker iterator classes, but so far
>> my implementation relies on std::lower_bound and std::upper_bound.
> Why do you need lower/upper_bound? set_difference
> can be implemented easily with a straight linear
> scan through the two sequences.

If you do that, the complexity is at least O(N) even for the best
case where the ranges don't overlap.

Here is my attempt at this algorithm, which might be O(N log N) worst
case if I'm thinking straight. It is O(1) when the ranges don't overlap.
Dean, is this anything like what you've implemented?

template <typename I1, typename I2>
I1 set_inplace_difference(I1 f1, I1 l1, I2 f2, I2 l2)
  // If either range is empty, there is nothing to do:
  if (f1 == l1 || f2 == l2) return l1;

  // If the ranges are both singletons, compare the values:
  if (l1-f1 == 1 && l2-f2 == 1) {
    if (*f1 == *f2) return f1; else return l1;

  // If ranges don't overlap, there is nothing to do:
  if (*(l2-1) < *f1 || *f2 > *(l1-1)) return l1;

  // Recurse, splitting the larger range:
  if (l2-f2 > l1-f1) {
    // Second range is larger. Split it at its midpoint and
    // subtract each half in turn.
    I2 m2 = f2 + (l2-f2)/2;
    I1 p = set_inplace_difference(f1,l1, m2,l2);
    return set_inplace_difference(f1,p, f2,m2);

  } else {
    // First range is larger. Split it at its midpoint and
    // subtract from each half in turn.
    I1 m1 = f1 + (l1-f1)/2;
    I1 p = set_inplace_difference(f1,m1, f2,l2);
    I1 q = set_inplace_difference(m1,l1, f2,l2);
    // Now merge the two results.
    return std::rotate(p,m1,q);

Regards, Phil.

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