ref: ed2bcb605e43881ebe854040f1e3e6f918ed9b26
dir: /lib/iter/perm.myr/
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
;;
j--
match itp.cmp(seq[j], seq[j + 1])
| `std.Before: break
| _:
;;
;;
/*
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
/*
By the previous loop, we know that
seq[j] < seg[j + 1], thus, in this loop,
we always have j + 1 <= i, and the array
accesses are always in bounds.
*/
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
}
;;