in reply to
Re: Random Derangement Of An Array

in thread Random Derangement Of An Array

*Here's the approach I wanted to code up, but I've been sufficiently distracted.*

After much brain-racking I have been unable to come up with a less silly approach than simply encoding the combinatorial proof that d_(n+1) = n (d_n + d_(n-1)) in Perl, which was essentially your idea.

All of the other ideas in this thread so far produce a biased distribution; the 'obvious' modification to the Fisher-Yates shuffle algorithm looked promising but misses some derangements entirely.

It might be straightforward to do it purely iteratively too, but my brain works recursively so my programs do too!

`#!/usr/bin/perl
use strict;
sub d_l_rec {
# Calculates the pair (d_n, d_{n-1})
#
# Why calculate the pair? It's O(n) instead of O(2^n)
# for the 'obvious' recursive algorithm.
#
# Why recursive? No need to do it iteratively.
my ($n) = @_;
return 1 if $n < 1;
return (0, 1) if $n == 1;
my ($d1, $d2) = d_l_rec($n-1);
my $d = ($n-1) * ($d1 + $d2);
return ($d, $d1);
}
sub random_local_derangement {
# Returns a randomly-chosen local derangement of
# (0..($n-1)). A local derangement is a derangement
# *except* that the last place may be a fixed point.
#
# It's 'local' in the sense that, given $n people and a
# hat of $n tickets you can generate a local derangement
# by each person in turn pulling tickets out of the hat
# until they get one that isn't theirs, ensuring that
# (local to their draw) the permutation looks like it's
# going to be a derangement. This is how 'Secret Santa'
# derangements often seem to be organised, and this
# process sometimes leaves the last person with their
# own ticket.
#
# Unfortunately the 'Secret Santa' approach definitely
# does not give uniform probabilities to each outcome.
# This function, on the other hand, does. [At least,
# it's definitely uniform for $n <= 12 and merely a very
# good approximation for larger $n.]
my ($n) = @_;
if ($n == 0) {
return [];
}
my ($i, $threshold);
# A local $n-derangement is either a full 'total'
# $n-derangement or else it is a ($n-1)-derangement with
# a fixed point added at the end. We must choose between
# these options with appropriate probability weighting.
# Note that this means that l_k = d_k + d_{k-1} where
# l_k is the number of local k-derangements and d_k is
# the number of total k-derangements.
#
# Note that d_{12} < 2^31 < d_{13} so we have to be
# careful of overflow for $n > 13. Fortunately there's a
# good approximation that can be used.
if ($n <= 12) {
# Calculate d_{$n} and d_{$n-1} and a random value
# $i in [0, d_{$n} + d_{$n-1}) to decide which of
# the two options to use.
my ($dn, $dn1) = d_l_rec($n);
$i = int(rand($dn+$dn1));
$threshold = $dn1;
} else {
# If $n is large then d_$n/d_{$n-1} is very close to
# $n. Therefore it'll do to pick a random $i in the
# range [0, $n] and see if it is 0 or not.
$i = int(rand($n+1));
$threshold = 1;
# I think this is ok, but it just might contain an
# off-by-one error. The upshot of such an error
# would be a degree of bias in the results that is
# going to be hard to detect - you may have to run
# it literally trillions of times to pick up a
# statistically significant result.
}
if ($i < $threshold) {
# Case 1 - pick a properly local derangement
my $d = random_derangement($n-1);
push @$d, $n-1;
return $d;
} else {
# Case 2 - pick a total derangement
my $d = random_derangement($n);
return $d;
}
}
sub random_derangement {
# Returns a randomly-chosen (total) derangement of
# (0..($n-1)), uniformly-chosen amongst all possible
# derangements.
my ($n) = @_;
if ($n == 0) {
return [];
}
# There are (n-1) l_{n-1} of them, so pick a (uniformly)
# random local ($n-1)-derangement and a random $m in the
# range [0, $n-1).
my $ld = random_local_derangement($n-1);
my $m = int(rand($n-1));
# If L_k is the set of all local k-derangements and D_k
# is the set of all total k-derangements then the code
# below encodes the proof that (n-1) l_{n-1} = d_n in a
# bijection between [0, $n-1) x L_{n-1} and D_{n}.
#
# Since the pair ($m, $ld) are chosen uniformly, this
# shows that the resulting derangement is also uniformly
# chosen.
if ($n-2 == $ld->[$n-2]) {
# $ld is properly local. Therefore the desired
# derangement swaps the $m'th and last places and
# uses $ld to derange the other places.
my $j = $n-1;
while ($j--) {
my $k = $j < $m ? $j : $j-1;
$ld->[$j]
= $ld->[$k] < $m ? $ld->[$k] : $ld->[$k]+1;
}
$ld->[$n-1] = $m;
$ld->[$m] = $n-1;
return $ld;
} else {
# $ld is total. Therefore put the $m'th entry at the
# end and put $n-1 in the $m'th place.
$ld->[$n-1] = $ld->[$m];
$ld->[$m] = $n-1;
return $ld;
}
}
sub check_derangement {
# Check that we have really generated a derangement
my ($n, $d) = @_;
my $s = join ', ', @$d;
die "Wrong length: $s ($n)" unless ($n == @$d);
for (my $i = 0; $i < $n; $i++) {
die "Not a derangement: $s" if ($i == $d->[$i]);
die "Illegal value: $d->[$i] in $s"
if ($d->[$i] < 0 || $d->[$i] >= $n);
}
eval {
my @check_unique = sort { $a <=> $b || undef } @$d;
};
die "Uniqueness check failed: $s" if $@;
}
my $n = $ARGV[0];
my %f = ();
my $c = 0;
for (my $i = 0; $i < 1e6; $i++) {
my $d = random_derangement($n);
check_derangement($n, $d);
my $s = join ',', @$d;
$f{$s} += 1;
$c += 1;
}
for my $key (sort {$f{$a} <=> $f{$b}} keys %f) {
printf "%s: %0.2f%% (expected %+0.2f%%)\n", $key, 100.0*($f{$key}/
+$c),
100.0*(($f{$key}/$c)-(1.0/(scalar keys %f)));
}
printf "Total %d (%d runs)\n", (scalar keys %f), $c;
`

This produces output like the following

`$ ./derangements.pl 4
2,3,0,1: 11.08% (expected -0.03%)
1,0,3,2: 11.09% (expected -0.02%)
2,3,1,0: 11.10% (expected -0.01%)
1,3,0,2: 11.11% (expected -0.01%)
2,0,3,1: 11.11% (expected -0.00%)
1,2,3,0: 11.12% (expected +0.01%)
3,0,1,2: 11.12% (expected +0.01%)
3,2,1,0: 11.13% (expected +0.02%)
3,2,0,1: 11.14% (expected +0.03%)
Total 9 (1000000 runs)
`

Clearly that's not far off uniform!