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
|
use std
pkg iter =
type permiter(@a) = struct
cmp : (a : @a, b : @a -> std.order)
seq : @a[:]
first : bool
;;
impl iterable permiter(@a) -> @a[:]
generic byperm : (a : @a[:], cmp : (a : @a, b : @a -> std.order) -> permiter(@a))
;;
generic byperm = {a, cmp
/* start off by backwards-sorting a */
std.sort(a, cmp)
-> [.cmp = cmp, .seq = std.sort(a, cmp), .first = true]
}
/*
Generates all permutations of a sequence in place,
mutating the sequence passed in the iterator.
*/
impl iterable permiter(@a) -> @a[:] =
__iternext__ = {itp, valp
var j : std.size = seq.len - 1
var i : std.size = seq.len - 1
var seq : @a[:] = itp.seq
valp# = itp.seq
if itp.first
itp.first = false
-> true
;;
/*
To generate the next permutation, we need
to increase the sequence by as little as
possible. That means that we start by finding
the longest monotonically decreasing trailing
sequence.
*/
j = seq.len - 1
while true
if j == 0
-> false
;;
match itp.cmp(seq[j - 1], seq[j])
| `std.After: j--
| _: break
;;
;;
j--
/*
Find highest i s.t. seq[j] < seq[i]. This
is the index that will maintain the order of
the suffix, while increasing the value of the
element in the 'pivot'.
*/
i = seq.len - 1
while true
if i <= j
-> false
;;
match itp.cmp(seq[j], seq[i])
| `std.Before: break
| _: i--
;;
;;
/*
Then, swap seq[i] and seq[j] and reverse the
sequence. Because the suffix was in decreasing
order, reversing it means that our sequence is
now in increasing order: ie, our most significant
place has the smallest value.
*/
std.swap(&seq[i], &seq[j])
i = 1
while j + i < seq.len - i
std.swap(&seq[j + i], &seq[seq.len - i])
i++
;;
-> true
}
__iterfin__ = {itp, valp
}
;;
|
