Re: Puzzle Time

And the answer is: 548

(I suppose only positive numbers otherwise multiply it with 2)

The following is just a variation of the code I recently used:

9! = 362880 possible combinationsš are checked in a recursive function.

#!perl
use strict;
use warnings;

my @digits = 1..9;

my @number;
my $maxlevel=10;

my %count_hash;
my $count=0;

append_digit();

print "count: $count\n";

print join "\n", sort { $a <=>$b } keys %count_hash;


sub append_digit {

  my $level = scalar @number;

  my $number= join "", @number;
  
  if ( $level and
       ! grep {$number % $_} @number
     ) {
    $count++;
    print "$count: $number\n" unless $count %100;
    $count_hash{$number} = undef;
  }

  for my $digit (@digits) {
    return if $level > $maxlevel;                 #  for testing only
    next if $digit ~~ @number;

    push @number, $digit;
    append_digit();
    pop @number;
  }
}
[download]

output

Cheers Rolf

š) and of course their substrings from the start, thx Athanasius++ :)

Comment on Re: Puzzle Time Select or Download Code

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Re^2: Puzzle Time
by ikegami (Patriarch) on Dec 24, 2012 at 06:41 UTC

  for my $digit (@digits) {
    next if $digit ~~ @number;
    push @number, $digit;
    append_digit();
    pop @number;
  }
[download]

  for my $pos (0..$#digits) {
    push @number, $digits[$pos];
    local @digits = @digits[0..$pos-1, $pos+1..$#digits];
    append_digit();
    pop @number;
  }
[download]

to cut down the time taken to 2/3 (6s from 9s).

Also, there's reason to use a hash (outside of testing).

Full code:

#!perl
use strict;
use warnings;
use 5.010;

our @digits = 1..9;

my @number;
my @results;
sub append_digit {
  my $number = join "", @number;
  
  push @results, $number
     if @number && !grep $number % $_, @number;

  for my $pos (0..$#digits) {
    push @number, $digits[$pos];
    local @digits = @digits[0..$pos-1, $pos+1..$#digits];
    append_digit();
    pop @number;
  }
}

my $stime = time;
append_digit();
my $etime = time;
say $etime-$stime;

say "count: " . (0+@results);
#say for sort { $a <=> $b } @results;
[download]

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Re^2: Puzzle Time
by Athanasius (Archbishop) on Dec 23, 2012 at 10:21 UTC

++LanX for an excellent answer (way faster than my brute force approach).

Allow me one nit-pick:

9! = 362880 possible combinations are checked in a recursive function.

Throwing in a counter shows that sub append_digit is actually called 986,410 times. Took me a while to work out why...

9! is the number of combinations if every number has exactly 9 digits. But we also allow numbers with 1, 2, ..., 8 digits. So the total number of combinations is:

my $p = 9                                   +   # 1 digit
       (9 * 8)                              +   # 2 digits
       (9 * 8 * 7)                          +   # 3 digits
       (9 * 8 * 7 * 6)                      +   # 4 digits
       (9 * 8 * 7 * 6 * 5)                  +   # 5 digits
       (9 * 8 * 7 * 6 * 5 * 4)              +   # 6 digits
       (9 * 8 * 7 * 6 * 5 * 4 * 3)          +   # 7 digits
       (9 * 8 * 7 * 6 * 5 * 4 * 3 * 2)      +   # 8 digits
       (9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1);     # 9 digits

say $p; # = 986,409
[download]

(That 1 extra sub call is the first one, when @number is empty.)

:-)