Re: Next permutation

[LanX]

You can use Algorithm::Permute for that.

You'll need to split the ciphers into a list and join them again.

There is more to say about it ... but your question is too sparse to go into details of non-repeating and non-module implementation...

[aartist]

Thanks for the answer. I am looking for non-modules solution. I don't want to find all the solution. I just want to find next number in sequence. We should be able to derive from current number. Foe example : ( swap till you hit the largest number as first. If you hit the largest number as first, get the next number on left side and then place remaining number in sorted order ) : 117 => 171 => 711 or : 1234 =>(swap) 12.43 =>(largest 4) : 13.24 =>(swap) 13.42 => (largest) 14.23 =>(swap) :14.32 etc..

[Marshall]

How is what you want to do different than a Bubble Sort where you stop on the first swap? A bubble sort moves the highest element up by one position for each pass (here a digit). What do you want to happen if the highest number is already first in the order?

Update:
Probably a better way to get the index of max value...But is this what you want?
Code should run fairly quickly albeit a bit wordy for Perl.

#!/usr/bin/perl
use strict;
use warnings;


my @array = split '', "117";
print @array, "\n";
print @array,"\n" while rot_max_left(\@array);


@array = split '', "1234";
print "\n",@array,"\n";
print @array,"\n" while rot_max_left(\@array);


sub rot_max_left
{
    my $array_ref = shift;
    
    my $maxi=0;
    return 0 unless ($maxi = get_index_of_max($array_ref));
    
    ($array_ref->[$maxi-1],$array_ref->[$maxi]) = ($array_ref->[$maxi]
+,$array_ref->[$maxi-1]);
    return $maxi;
}

sub get_index_of_max
{
  my $array_ref = shift;
  
  my $maxi = 0;
  foreach (my $i=1;$i<@$array_ref;$i++)
  {
     $maxi = $i if ($array_ref->[$i] > $array_ref->[$maxi])
  }
  return $maxi;
}

__END__
117
171
711

1234
1243
1423
4123

[aartist]

If you hit the highest digit as the first digit, then:
1. Sorting of remaining digits.
2. if you cannot sort any further, (ie.. it is already in reverse order) than sort from the beginning position. ( to return to the original number. 321 -> 123)

[LanX]

Your requirements are cryptic for me.

Please note that you can easily implement any "swap" with array slices, like

@a[0,1]=@a[1,0]

and @a=split//,$number

will initialize it.

Now you know everything needed to implement your ideas.

HTH! :)