Question

I have written the following script that creates a string that is of length 3 or 4 units for all possible strings that contain the 26 letters of the alphabet. I’d like to add a 27th character. Call it ?, but it does not matter which one. I cannot think how I should do this. Any help would be much welcomed.

Here is the code that does the 26 letters:

 #!/usr/bin/perl

 use 5.10.0;

 my $str = 'AAA';
 while ( $str ne 'ZZZZ' ) {
    say $str;
    $str++;
  }

How would I transform a number into base 27?

Answer 1

perldoc perlop:

The auto-increment operator has a little extra builtin magic to it. If you increment a variable that is numeric, or that has ever been used in a numeric context, you get a normal increment. If, however, the variable has been used in only string contexts since it was set, and has a value that is not the empty string and matches the pattern /^[a-zA-Z]*[0-9]*\z/, the increment is done as a string, preserving each character within its range, with carry …

Algorithm::Combinatorics to the rescue!

#!/usr/bin/env perl

use strict;
use warnings;
use feature 'say';

use Algorithm::Combinatorics qw(variations_with_repetition);

my $digits = ['A' .. 'Z', '?'];

print_sequence($digits, 3);

sub print_sequence {
    my ($digits, $n) = @_;

    my $it = variations_with_repetition($digits, $n);
    while (my $v = $it->next) {
        say join '', @$v;
    }

    return;
}