1 use std::fmt;
2 use std::iter::FusedIterator;
3 
4 use super::lazy_buffer::LazyBuffer;
5 use alloc::vec::Vec;
6 
7 use 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"]
13 pub struct Combinations<I: Iterator> {
14     indices: Vec<usize>,
15     pool: LazyBuffer<I>,
16     first: bool,
17 }
18 
19 impl<I> Clone for Combinations<I>
20 where
21     I: Clone + Iterator,
22     I::Item: Clone,
23 {
24     clone_fields!(indices, pool, first);
25 }
26 
27 impl<I> fmt::Debug for Combinations<I>
28 where
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.
combinations<I>(iter: I, k: usize) -> Combinations<I> where I: Iterator,36 pub fn combinations<I>(iter: I, k: usize) -> Combinations<I>
37 where
38     I: Iterator,
39 {
40     Combinations {
41         indices: (0..k).collect(),
42         pool: LazyBuffer::new(iter),
43         first: true,
44     }
45 }
46 
47 impl<I: Iterator> Combinations<I> {
48     /// Returns the length of a combination produced by this iterator.
49     #[inline]
k(&self) -> usize50     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]
n(&self) -> usize57     pub fn n(&self) -> usize {
58         self.pool.len()
59     }
60 
61     /// Returns a reference to the source pool.
62     #[inline]
src(&self) -> &LazyBuffer<I>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.
reset(&mut self, k: usize)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 
n_and_count(self) -> (usize, usize)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     /// Initialises the iterator by filling a buffer with elements from the
99     /// iterator. Returns true if there are no combinations, false otherwise.
init(&mut self) -> bool100     fn init(&mut self) -> bool {
101         self.pool.prefill(self.k());
102         let done = self.k() > self.n();
103         if !done {
104             self.first = false;
105         }
106 
107         done
108     }
109 
110     /// Increments indices representing the combination to advance to the next
111     /// (in lexicographic order by increasing sequence) combination. For example
112     /// if we have n=4 & k=2 then `[0, 1] -> [0, 2] -> [0, 3] -> [1, 2] -> ...`
113     ///
114     /// Returns true if we've run out of combinations, false otherwise.
increment_indices(&mut self) -> bool115     fn increment_indices(&mut self) -> bool {
116         if self.indices.is_empty() {
117             return true; // Done
118         }
119 
120         // Scan from the end, looking for an index to increment
121         let mut i: usize = self.indices.len() - 1;
122 
123         // Check if we need to consume more from the iterator
124         if self.indices[i] == self.pool.len() - 1 {
125             self.pool.get_next(); // may change pool size
126         }
127 
128         while self.indices[i] == i + self.pool.len() - self.indices.len() {
129             if i > 0 {
130                 i -= 1;
131             } else {
132                 // Reached the last combination
133                 return true;
134             }
135         }
136 
137         // Increment index, and reset the ones to its right
138         self.indices[i] += 1;
139         for j in i + 1..self.indices.len() {
140             self.indices[j] = self.indices[j - 1] + 1;
141         }
142 
143         // If we've made it this far, we haven't run out of combos
144         false
145     }
146 
147     /// Returns the n-th item or the number of successful steps.
try_nth(&mut self, n: usize) -> Result<<Self as Iterator>::Item, usize> where I::Item: Clone,148     pub(crate) fn try_nth(&mut self, n: usize) -> Result<<Self as Iterator>::Item, usize>
149     where
150         I::Item: Clone,
151     {
152         let done = if self.first {
153             self.init()
154         } else {
155             self.increment_indices()
156         };
157         if done {
158             return Err(0);
159         }
160         for i in 0..n {
161             if self.increment_indices() {
162                 return Err(i + 1);
163             }
164         }
165         Ok(self.pool.get_at(&self.indices))
166     }
167 }
168 
169 impl<I> Iterator for Combinations<I>
170 where
171     I: Iterator,
172     I::Item: Clone,
173 {
174     type Item = Vec<I::Item>;
next(&mut self) -> Option<Self::Item>175     fn next(&mut self) -> Option<Self::Item> {
176         let done = if self.first {
177             self.init()
178         } else {
179             self.increment_indices()
180         };
181 
182         if done {
183             return None;
184         }
185 
186         Some(self.pool.get_at(&self.indices))
187     }
188 
nth(&mut self, n: usize) -> Option<Self::Item>189     fn nth(&mut self, n: usize) -> Option<Self::Item> {
190         self.try_nth(n).ok()
191     }
192 
size_hint(&self) -> (usize, Option<usize>)193     fn size_hint(&self) -> (usize, Option<usize>) {
194         let (mut low, mut upp) = self.pool.size_hint();
195         low = remaining_for(low, self.first, &self.indices).unwrap_or(usize::MAX);
196         upp = upp.and_then(|upp| remaining_for(upp, self.first, &self.indices));
197         (low, upp)
198     }
199 
200     #[inline]
count(self) -> usize201     fn count(self) -> usize {
202         self.n_and_count().1
203     }
204 }
205 
206 impl<I> FusedIterator for Combinations<I>
207 where
208     I: Iterator,
209     I::Item: Clone,
210 {
211 }
212 
213 /// For a given size `n`, return the count of remaining combinations or None if it would overflow.
remaining_for(n: usize, first: bool, indices: &[usize]) -> Option<usize>214 fn remaining_for(n: usize, first: bool, indices: &[usize]) -> Option<usize> {
215     let k = indices.len();
216     if n < k {
217         Some(0)
218     } else if first {
219         checked_binomial(n, k)
220     } else {
221         // https://en.wikipedia.org/wiki/Combinatorial_number_system
222         // http://www.site.uottawa.ca/~lucia/courses/5165-09/GenCombObj.pdf
223 
224         // The combinations generated after the current one can be counted by counting as follows:
225         // - The subsequent combinations that differ in indices[0]:
226         //   If subsequent combinations differ in indices[0], then their value for indices[0]
227         //   must be at least 1 greater than the current indices[0].
228         //   As indices is strictly monotonically sorted, this means we can effectively choose k values
229         //   from (n - 1 - indices[0]), leading to binomial(n - 1 - indices[0], k) possibilities.
230         // - The subsequent combinations with same indices[0], but differing indices[1]:
231         //   Here we can choose k - 1 values from (n - 1 - indices[1]) values,
232         //   leading to binomial(n - 1 - indices[1], k - 1) possibilities.
233         // - (...)
234         // - The subsequent combinations with same indices[0..=i], but differing indices[i]:
235         //   Here we can choose k - i values from (n - 1 - indices[i]) values: binomial(n - 1 - indices[i], k - i).
236         //   Since subsequent combinations can in any index, we must sum up the aforementioned binomial coefficients.
237 
238         // Below, `n0` resembles indices[i].
239         indices.iter().enumerate().try_fold(0usize, |sum, (i, n0)| {
240             sum.checked_add(checked_binomial(n - 1 - *n0, k - i)?)
241         })
242     }
243 }
244