in reply to Faster alternative to Math::Combinatorics

Depending on how many elements you have :) 

#!/usr/bin/perl -l

# http://perlmonks.org/?node_id=1198509

use strict;
use warnings;

my @elements = (0, 2, 3);

my ($first, $last) = @elements[0, -1];
my %next;
@next{@elements} = @elements[1 .. @elements];

for my $count (1,2,3,4,7,8)
  {
  print "count=$count";

  $_ = $first x $count;

  1 while print(join ',', /./g),
    s/([^$last])($last*)$/ $next{$1} x length $& /e;
  }

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Outputs: 

count=1
0
2
3
count=2
0,0
0,2
0,3
2,2
2,3
3,3
count=3
0,0,0
0,0,2
0,0,3
0,2,2
0,2,3
0,3,3
2,2,2
2,2,3
2,3,3
3,3,3
count=4
0,0,0,0
0,0,0,2
0,0,0,3
0,0,2,2
0,0,2,3
0,0,3,3
0,2,2,2
0,2,2,3
0,2,3,3
0,3,3,3
2,2,2,2
2,2,2,3
2,2,3,3
2,3,3,3
3,3,3,3
count=7
0,0,0,0,0,0,0
0,0,0,0,0,0,2
0,0,0,0,0,0,3
0,0,0,0,0,2,2
0,0,0,0,0,2,3
0,0,0,0,0,3,3
0,0,0,0,2,2,2
0,0,0,0,2,2,3
0,0,0,0,2,3,3
0,0,0,0,3,3,3
0,0,0,2,2,2,2
0,0,0,2,2,2,3
0,0,0,2,2,3,3
0,0,0,2,3,3,3
0,0,0,3,3,3,3
0,0,2,2,2,2,2
0,0,2,2,2,2,3
0,0,2,2,2,3,3
0,0,2,2,3,3,3
0,0,2,3,3,3,3
0,0,3,3,3,3,3
0,2,2,2,2,2,2
0,2,2,2,2,2,3
0,2,2,2,2,3,3
0,2,2,2,3,3,3
0,2,2,3,3,3,3
0,2,3,3,3,3,3
0,3,3,3,3,3,3
2,2,2,2,2,2,2
2,2,2,2,2,2,3
2,2,2,2,2,3,3
2,2,2,2,3,3,3
2,2,2,3,3,3,3
2,2,3,3,3,3,3
2,3,3,3,3,3,3
3,3,3,3,3,3,3
count=8
0,0,0,0,0,0,0,0
0,0,0,0,0,0,0,2
0,0,0,0,0,0,0,3
0,0,0,0,0,0,2,2
0,0,0,0,0,0,2,3
0,0,0,0,0,0,3,3
0,0,0,0,0,2,2,2
0,0,0,0,0,2,2,3
0,0,0,0,0,2,3,3
0,0,0,0,0,3,3,3
0,0,0,0,2,2,2,2
0,0,0,0,2,2,2,3
0,0,0,0,2,2,3,3
0,0,0,0,2,3,3,3
0,0,0,0,3,3,3,3
0,0,0,2,2,2,2,2
0,0,0,2,2,2,2,3
0,0,0,2,2,2,3,3
0,0,0,2,2,3,3,3
0,0,0,2,3,3,3,3
0,0,0,3,3,3,3,3
0,0,2,2,2,2,2,2
0,0,2,2,2,2,2,3
0,0,2,2,2,2,3,3
0,0,2,2,2,3,3,3
0,0,2,2,3,3,3,3
0,0,2,3,3,3,3,3
0,0,3,3,3,3,3,3
0,2,2,2,2,2,2,2
0,2,2,2,2,2,2,3
0,2,2,2,2,2,3,3
0,2,2,2,2,3,3,3
0,2,2,2,3,3,3,3
0,2,2,3,3,3,3,3
0,2,3,3,3,3,3,3
0,3,3,3,3,3,3,3
2,2,2,2,2,2,2,2
2,2,2,2,2,2,2,3
2,2,2,2,2,2,3,3
2,2,2,2,2,3,3,3
2,2,2,2,3,3,3,3
2,2,2,3,3,3,3,3
2,2,3,3,3,3,3,3
2,3,3,3,3,3,3,3
3,3,3,3,3,3,3,3


real    0m0.023s
user    0m0.018s
sys     0m0.009s

[download]

Replies are listed 'Best First'.
Re^2: Faster alternative to Math::Combinatorics
by AppleFritter (Vicar) on Sep 01, 2017 at 16:46 UTC

    Depending on how many elements you have :)

    Good question! The number could be arbitrarily high in theory, in practice I doubt the number will leave the two-digit range¹. (In case anyone's wondering, BTW, this is related to multistate cellular automata; w is a neighborhood count, and the list I mentioned is a list of (some) states of the CA in question.)

    I'll have to take a long look at your code to really understand it, but wow, those timings look marvellous. Thank you! If this ends up working out I'll definitely owe you a beer.

    Footnotes:

    1. Edited to add: in fact, since the amount of data generated will increase exponentially with the number of CA states, I think realistically, it won't even leave the single-digit range.
[reply]

      Here's a version of the same algorithm working on array elements instead of characters. (Messier, isn't it?)

      With 10 elements (states) it takes 1.5 seconds on my machine, but I think most of the time is taken by the printing.

      How many neighbors can you have? (for code testing purposes.) 

      #!/usr/bin/perl -l

# http://perlmonks.org/?node_id=1198509

use strict;
use warnings;

my @elements = (0, 2, 3);
@elements = 0..9;

my ($first, $last) = @elements[0, -1];
my %next;
@next{@elements} = @elements[1 .. $#elements];

for my $count (1,2,3,4,7,8)
  {
  print "count=$count";

  my @set = ($first) x $count;
  local $, = ',';

  while(1)
    {
    print @set;
    my $i = $#set;
    $i-- while $i >= 0 and $set[$i] eq $last;
    $i < 0 and last;
    @set[$i .. $#set] = ($next{$set[$i]}) x ( @set - $i);
    }
  }

      [download]
[reply]
[d/l]

        How many neighbors can you have? (for code testing purposes.)

        A maximum of eight — the size of the range-1 Moore neighborhood, as I'm not currently considering CAs with larger neighborhoods.

[reply]

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