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Re: Balanced division of iterator range without LegacyRandomAccessIterator trait

From: Arthur O'Dwyer <arthur.j.odwyer_at_[hidden]>
Date: Sun, 6 Sep 2020 09:13:24 -0400
On Sun, Sep 6, 2020 at 2:55 AM Andreas Ringlstetter via Std-Proposals <
std-proposals_at_[hidden]> wrote:

> Follow up on
> https://stackoverflow.com/questions/62306580/balanced-partition-of-iterator-range-without-legacyrandomaccessiterator

Interesting problem and relevant motivation!

On SO, darune writes:
> All associate containers are already inherently sorted and already
provides specialized accessors and api's.

But this isn't really true for your use-case, is it? I mean, suppose I
have a std::set. It is a balanced binary tree, but (unless I'm missing
something) there is no specialized accessor that would give me an iterator
to the "midpoint/root" element in anything less than O(n) time. And a very
slow and pointer-chasey O(n) at that!
So (unless I'm missing something) darune's dismissal is a bit premature.

However, I also don't see any clean way to implement what you're asking
for. Consider what the compiler sees for

    std::set<int> s = {1,2,3,4,5,6,7,8,9,10,11};
    auto first = s.begin(), last = s.end();
    auto it1 = std2::divide_in_half(first, last);
    assert(*it1 == 6);
    std::advance(first, 2); std::advance(last, -4);
    auto it2 = std2::divide_in_half(first, last);
    assert(*it2 == 5);

The iterators `first` and `last` don't know anything about the container
that they're part of. They don't have any way to get at the "root" node of
that tree, except to traverse over to it. (It is true AFAIK that an
iterator would be able to tell *whether* it were pointing directly at the
root node, but that doesn't help us here AFAIK.)
And the type-system doesn't see any difference between the first call to
`std2::divide_in_half` there and the second call. The first one we'd like
to quickly return the root node; the second one we do *not *want to return
the root node, and must do the slow O(n) traversal.

NOTE: For the purposes of this thread, I'd like to completely handwave away
the practical difference between "the root node of the red-black tree" and
"the n/2th node in sorted order." I'd like to just assume that we can solve
any potential off-by-1 or off-by-2 errors that might arise, and not go down
that nerdsnipe rabbit hole at all.


> If they provide only the `LegacyForwardIterator` trait, only thing we
> can do in constant time is:
> std::set<int> foo;
> ... // fill it
> auto begin = foo.begin();
> auto end = foo.end();
> auto middle = (begin != end) ? ++(begin) : begin;
> No chance of processing the iterator range in an even remotely
> balanced way without a linear scan.

FWIW, `std::for_each` is allowed to take O(n) time to do its job. If it
spends 1/Kth of that time doing an O(n) traversal of the list to find its
actual midpoint, then that's fine, AFAIK. There's nothing in the formal
standard AFAIK that says it *must* use a dumb O(1) algorithm to partition
the work. If you're seeing that dumb algorithm being used in
implementations today, I think that's likely for software-engineering
reasons, not standard-conformance reasons.


> For reference, the following is about on-par with what the existing
> STL implementations do:
> // Generic implementation
> template<typename Iter, ~~~>
> std::pair<std::pair<Iter, Iter>, std::pair<Iter, Iter>>
> divide(const Iter &begin, const Iter &end)
> {
> if constexpr (std::is_convertible_v<typename
> std::iterator_traits<Iter>::iterator_category,
> std::random_access_iterator_tag>)
> {
> // Ideal bisection
> const Iter middle = end - std::distance(begin, end) / 2;
> return { {begin, middle}, {begin, middle} };

Here and below, that should be {{begin, middle}, {middle, end}}, right?

Finally, I think you have at least two different problems here:
(1) Bisect (trisect,...) a range into contiguous subranges of the same
iterator-type and *exactly* the right sizes. I think range-v3 would be
capable of this task; something with std::ranges::size and then repeatedly
views::slice. Someone better at Ranges than me might try it.
(2) Given a range, split it "quick and dirty" into subranges of possibly
different iterator-type, and possibly not exactly the right sizes, but
close enough to get speedup on parallel tasks.

You mention that unordered_set would probably be very good at #2 — but
horrible at #1.
Another kind of data structure that might be good at #2 would be
skiplists, whose iterators would know how to take "chunks" easily.


> A heuristic alternative to "std::distance". The exact form we already
> have does not have an efficient implementation for anything except
> dense, compact allocations where it can be reduced to pointer
> arithmetic.

Pedantry: No, std::distance uses iterator arithmetic, so it works
efficiently on any random-access iterator type. std::distance of std::deque
iterators is constant-time, for example, even though deque is not
C++20 adds "contiguous iterator" as a refinement of "random-access


Received on 2020-09-06 08:17:06