1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
use alloc::vec::Vec;
use std::fmt;
use std::iter::FusedIterator;
use super::lazy_buffer::LazyBuffer;
use crate::adaptors::checked_binomial;
/// An iterator to iterate through all the `n`-length combinations in an iterator, with replacement.
///
/// See [`.combinations_with_replacement()`](crate::Itertools::combinations_with_replacement)
/// for more information.
#[derive(Clone)]
#[must_use = "iterator adaptors are lazy and do nothing unless consumed"]
pub struct CombinationsWithReplacement<I>
where
I: Iterator,
I::Item: Clone,
{
indices: Vec<usize>,
pool: LazyBuffer<I>,
first: bool,
}
impl<I> fmt::Debug for CombinationsWithReplacement<I>
where
I: Iterator + fmt::Debug,
I::Item: fmt::Debug + Clone,
{
debug_fmt_fields!(CombinationsWithReplacement, indices, pool, first);
}
impl<I> CombinationsWithReplacement<I>
where
I: Iterator,
I::Item: Clone,
{
/// Map the current mask over the pool to get an output combination
fn current(&self) -> Vec<I::Item> {
self.indices.iter().map(|i| self.pool[*i].clone()).collect()
}
}
/// Create a new `CombinationsWithReplacement` from a clonable iterator.
pub fn combinations_with_replacement<I>(iter: I, k: usize) -> CombinationsWithReplacement<I>
where
I: Iterator,
I::Item: Clone,
{
let indices: Vec<usize> = alloc::vec![0; k];
let pool: LazyBuffer<I> = LazyBuffer::new(iter);
CombinationsWithReplacement {
indices,
pool,
first: true,
}
}
impl<I> Iterator for CombinationsWithReplacement<I>
where
I: Iterator,
I::Item: Clone,
{
type Item = Vec<I::Item>;
fn next(&mut self) -> Option<Self::Item> {
// If this is the first iteration, return early
if self.first {
// In empty edge cases, stop iterating immediately
return if !(self.indices.is_empty() || self.pool.get_next()) {
None
// Otherwise, yield the initial state
} else {
self.first = false;
Some(self.current())
};
}
// Check if we need to consume more from the iterator
// This will run while we increment our first index digit
self.pool.get_next();
// Work out where we need to update our indices
let mut increment: Option<(usize, usize)> = None;
for (i, indices_int) in self.indices.iter().enumerate().rev() {
if *indices_int < self.pool.len() - 1 {
increment = Some((i, indices_int + 1));
break;
}
}
match increment {
// If we can update the indices further
Some((increment_from, increment_value)) => {
// We need to update the rightmost non-max value
// and all those to the right
for indices_index in increment_from..self.indices.len() {
self.indices[indices_index] = increment_value;
}
Some(self.current())
}
// Otherwise, we're done
None => None,
}
}
fn size_hint(&self) -> (usize, Option<usize>) {
let (mut low, mut upp) = self.pool.size_hint();
low = remaining_for(low, self.first, &self.indices).unwrap_or(usize::MAX);
upp = upp.and_then(|upp| remaining_for(upp, self.first, &self.indices));
(low, upp)
}
fn count(self) -> usize {
let Self {
indices,
pool,
first,
} = self;
let n = pool.count();
remaining_for(n, first, &indices).unwrap()
}
}
impl<I> FusedIterator for CombinationsWithReplacement<I>
where
I: Iterator,
I::Item: Clone,
{
}
/// For a given size `n`, return the count of remaining combinations with replacement or None if it would overflow.
fn remaining_for(n: usize, first: bool, indices: &[usize]) -> Option<usize> {
// With a "stars and bars" representation, choose k values with replacement from n values is
// like choosing k out of k + n − 1 positions (hence binomial(k + n - 1, k) possibilities)
// to place k stars and therefore n - 1 bars.
// Example (n=4, k=6): ***|*||** represents [0,0,0,1,3,3].
let count = |n: usize, k: usize| {
let positions = if n == 0 {
k.saturating_sub(1)
} else {
(n - 1).checked_add(k)?
};
checked_binomial(positions, k)
};
let k = indices.len();
if first {
count(n, k)
} else {
// The algorithm is similar to the one for combinations *without replacement*,
// except we choose values *with replacement* and indices are *non-strictly* monotonically sorted.
// The combinations generated after the current one can be counted by counting as follows:
// - The subsequent combinations that differ in indices[0]:
// If subsequent combinations differ in indices[0], then their value for indices[0]
// must be at least 1 greater than the current indices[0].
// As indices is monotonically sorted, this means we can effectively choose k values with
// replacement from (n - 1 - indices[0]), leading to count(n - 1 - indices[0], k) possibilities.
// - The subsequent combinations with same indices[0], but differing indices[1]:
// Here we can choose k - 1 values with replacement from (n - 1 - indices[1]) values,
// leading to count(n - 1 - indices[1], k - 1) possibilities.
// - (...)
// - The subsequent combinations with same indices[0..=i], but differing indices[i]:
// Here we can choose k - i values with replacement from (n - 1 - indices[i]) values: count(n - 1 - indices[i], k - i).
// Since subsequent combinations can in any index, we must sum up the aforementioned binomial coefficients.
// Below, `n0` resembles indices[i].
indices.iter().enumerate().try_fold(0usize, |sum, (i, n0)| {
sum.checked_add(count(n - 1 - *n0, k - i)?)
})
}
}