itertools/
combinations.rs

1use std::fmt;
2use std::iter::FusedIterator;
3
4use super::lazy_buffer::LazyBuffer;
5use alloc::vec::Vec;
6
7use crate::adaptors::checked_binomial;
8
9/// An iterator to iterate through all the `k`-length combinations in an iterator.
10///
11/// See [`.combinations()`](crate::Itertools::combinations) for more information.
12#[must_use = "iterator adaptors are lazy and do nothing unless consumed"]
13pub struct Combinations<I: Iterator> {
14    indices: Vec<usize>,
15    pool: LazyBuffer<I>,
16    first: bool,
17}
18
19impl<I> Clone for Combinations<I>
20where
21    I: Clone + Iterator,
22    I::Item: Clone,
23{
24    clone_fields!(indices, pool, first);
25}
26
27impl<I> fmt::Debug for Combinations<I>
28where
29    I: Iterator + fmt::Debug,
30    I::Item: fmt::Debug,
31{
32    debug_fmt_fields!(Combinations, indices, pool, first);
33}
34
35/// Create a new `Combinations` from a clonable iterator.
36pub fn combinations<I>(iter: I, k: usize) -> Combinations<I>
37where
38    I: Iterator,
39{
40    Combinations {
41        indices: (0..k).collect(),
42        pool: LazyBuffer::new(iter),
43        first: true,
44    }
45}
46
47impl<I: Iterator> Combinations<I> {
48    /// Returns the length of a combination produced by this iterator.
49    #[inline]
50    pub fn k(&self) -> usize {
51        self.indices.len()
52    }
53
54    /// Returns the (current) length of the pool from which combination elements are
55    /// selected. This value can change between invocations of [`next`](Combinations::next).
56    #[inline]
57    pub fn n(&self) -> usize {
58        self.pool.len()
59    }
60
61    /// Returns a reference to the source pool.
62    #[inline]
63    pub(crate) fn src(&self) -> &LazyBuffer<I> {
64        &self.pool
65    }
66
67    /// Resets this `Combinations` back to an initial state for combinations of length
68    /// `k` over the same pool data source. If `k` is larger than the current length
69    /// of the data pool an attempt is made to prefill the pool so that it holds `k`
70    /// elements.
71    pub(crate) fn reset(&mut self, k: usize) {
72        self.first = true;
73
74        if k < self.indices.len() {
75            self.indices.truncate(k);
76            for i in 0..k {
77                self.indices[i] = i;
78            }
79        } else {
80            for i in 0..self.indices.len() {
81                self.indices[i] = i;
82            }
83            self.indices.extend(self.indices.len()..k);
84            self.pool.prefill(k);
85        }
86    }
87
88    pub(crate) fn n_and_count(self) -> (usize, usize) {
89        let Self {
90            indices,
91            pool,
92            first,
93        } = self;
94        let n = pool.count();
95        (n, remaining_for(n, first, &indices).unwrap())
96    }
97}
98
99impl<I> Iterator for Combinations<I>
100where
101    I: Iterator,
102    I::Item: Clone,
103{
104    type Item = Vec<I::Item>;
105    fn next(&mut self) -> Option<Self::Item> {
106        if self.first {
107            self.pool.prefill(self.k());
108            if self.k() > self.n() {
109                return None;
110            }
111            self.first = false;
112        } else if self.indices.is_empty() {
113            return None;
114        } else {
115            // Scan from the end, looking for an index to increment
116            let mut i: usize = self.indices.len() - 1;
117
118            // Check if we need to consume more from the iterator
119            if self.indices[i] == self.pool.len() - 1 {
120                self.pool.get_next(); // may change pool size
121            }
122
123            while self.indices[i] == i + self.pool.len() - self.indices.len() {
124                if i > 0 {
125                    i -= 1;
126                } else {
127                    // Reached the last combination
128                    return None;
129                }
130            }
131
132            // Increment index, and reset the ones to its right
133            self.indices[i] += 1;
134            for j in i + 1..self.indices.len() {
135                self.indices[j] = self.indices[j - 1] + 1;
136            }
137        }
138
139        // Create result vector based on the indices
140        Some(self.indices.iter().map(|i| self.pool[*i].clone()).collect())
141    }
142
143    fn size_hint(&self) -> (usize, Option<usize>) {
144        let (mut low, mut upp) = self.pool.size_hint();
145        low = remaining_for(low, self.first, &self.indices).unwrap_or(usize::MAX);
146        upp = upp.and_then(|upp| remaining_for(upp, self.first, &self.indices));
147        (low, upp)
148    }
149
150    #[inline]
151    fn count(self) -> usize {
152        self.n_and_count().1
153    }
154}
155
156impl<I> FusedIterator for Combinations<I>
157where
158    I: Iterator,
159    I::Item: Clone,
160{
161}
162
163/// For a given size `n`, return the count of remaining combinations or None if it would overflow.
164fn remaining_for(n: usize, first: bool, indices: &[usize]) -> Option<usize> {
165    let k = indices.len();
166    if n < k {
167        Some(0)
168    } else if first {
169        checked_binomial(n, k)
170    } else {
171        // https://en.wikipedia.org/wiki/Combinatorial_number_system
172        // http://www.site.uottawa.ca/~lucia/courses/5165-09/GenCombObj.pdf
173
174        // The combinations generated after the current one can be counted by counting as follows:
175        // - The subsequent combinations that differ in indices[0]:
176        //   If subsequent combinations differ in indices[0], then their value for indices[0]
177        //   must be at least 1 greater than the current indices[0].
178        //   As indices is strictly monotonically sorted, this means we can effectively choose k values
179        //   from (n - 1 - indices[0]), leading to binomial(n - 1 - indices[0], k) possibilities.
180        // - The subsequent combinations with same indices[0], but differing indices[1]:
181        //   Here we can choose k - 1 values from (n - 1 - indices[1]) values,
182        //   leading to binomial(n - 1 - indices[1], k - 1) possibilities.
183        // - (...)
184        // - The subsequent combinations with same indices[0..=i], but differing indices[i]:
185        //   Here we can choose k - i values from (n - 1 - indices[i]) values: binomial(n - 1 - indices[i], k - i).
186        //   Since subsequent combinations can in any index, we must sum up the aforementioned binomial coefficients.
187
188        // Below, `n0` resembles indices[i].
189        indices.iter().enumerate().try_fold(0usize, |sum, (i, n0)| {
190            sum.checked_add(checked_binomial(n - 1 - *n0, k - i)?)
191        })
192    }
193}