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Count combinations with replacement #737

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Philippe-Cholet
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@Philippe-Cholet Philippe-Cholet commented Aug 26, 2023

@phimuemue
Maybe a comment in remaining_for to suggest to see the big comment in combinations.rs.
I have a test combinations_with_replacement_inexact_size_hints if you want.

Only two differences with combinations *without* replacement:
- this `count` instead of `checked_binomial` ;
- `k` can be bigger than `n`.
For a fast test: `n` and `k` are only up to 7.
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@phimuemue What do you think?

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Hi @Philippe-Cholet, thanks again!

Please comment how remaining_for works - I believe that you did your homework there, but I'd really appreciate if you could clear things up for the reader.

Comment on lines 114 to 128
let count = |n: usize, k: usize| checked_binomial((n + k).saturating_sub(1), k);
let k = indices.len();
if first {
count(n, k)
} else {
indices
.iter()
.enumerate()
// TODO: Once the MSRV hits 1.37.0, we can sum options instead:
// .map(|(i, n0)| count(n - 1 - *n0, k - i))
// .sum()
.fold(Some(0), |sum, (i, n0)| {
sum.and_then(|s| s.checked_add(count(n - 1 - *n0, k - i)?))
})
}
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I believe you that this is correct, but I have a hard time comprehending this without comments. Is it the "stars and sticks" thing that leads to binomial(n+k-1, k)?

Could you comment the loop similar to what we did in the other cases?

In addition: Could n+k overflow? And if so, shouldn't we return None here instead of panicking?

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I chose the expression "stars and bars" from wikipedia, expression I did not know. I added a comment.
Good catch n+k overflowing, handled.
And I copied/updated the previous comment about the loop.

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@phimuemue
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bors r+

Thanks @Philippe-Cholet.

One thing: I'm constantly annoyed by the first field. Do you happen to know if there's a way to avoid this? (No need to work on it, I was just interested if you have insights into that.)

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bors bot commented Sep 4, 2023

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@bors bors bot merged commit c7a038e into rust-itertools:master Sep 4, 2023
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Philippe-Cholet commented Sep 4, 2023

@phimuemue I was thinking it could be indices: Option<Vec<usize>> to remove this field too.
Without replacement, the lazy buffer length would give us k but there is reset for powerset so it would be a real pain I think if we could even make it work.
With replacement, we don't prefill the lazy buffer so its length does not give us k.
So basically to know k without storing it (which is bigger than a boolean), we kinda need indices to be a vector of length k, not an optional one.

And with indices: Vec<usize> and not first we would skip the first combination with the current "next" algo. I suppose we could update the "next" algo: get the next element from current unchanged indices then only then update indices for the next time instead of updating indices first but that's quite some change for such a small benefit. And it would offset what we did here (count remaining elements based on indices) by 1. And when getting the last element, indices should become what: None? (representing the indices for the next element when there is no next element).

Another better alternative I guess would be our own alternative to Option<Vec<usize>> to know k anyway: enum State { Start { k: usize }, Ongoing { indices: Vec<usize> } }. That would be my choice.

PS: I was also thinking about merging Powerset, Combinations and CombinationsWithReplacement but I'm unsure it would be beneficial.

EDIT: Because of reset where we update indices rather than ditch it, the above State enum is not as good as our current first: bool, indices: Vec<usize>.

@Philippe-Cholet Philippe-Cholet deleted the count_combinations_with_replacement branch September 4, 2023 16:08
@jswrenn jswrenn added this to the next milestone Nov 14, 2023
@jswrenn jswrenn mentioned this pull request Nov 14, 2023
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3 participants