Re: Puzzle Time

[LanX]

in reply to 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:

(I love branch and bound! :)

9! = 362880 possible combinations¹ 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;
  }
}

output

<Reveal this spoiler or all in this thread>

¹) and of course their substrings from the start, thx Athanasius++ :)

[Anonymous Monk]

You are a legend, your help is greatly appreciated :)

[Athanasius]

++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

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

:-)

Read more... (1067 Bytes)

[LanX]

> Allow me one nit-pick:

Yeah I noticed this, but was too tired to correct it. =)

Anyway my guess was wrong (9!+8!+7!+...) you got it right.

> (way faster than my brute force approach).

If it's about speed you can limit the $maxlevel, because the longest number can't have more than 7 digits:

1. Evidently 0 is excluded!

2. Any number this long includes even ciphers. So 5 is excluded! But any number divisible by 5 and 2 must end with a 0

3. An number from the remaining 8 digits would include 9 and 3, but the digit sum would be 40, which is impossible.¹

Even your approach with a brute force loop could compete when only considering 7 digits, cause you don't have the overhead of 1 million function calls.

¹) and therefor a 7 digit number excludes 4 to be divisible by 9.

[ikegami]

Your method of getting the next digit is expensive! Replace

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

with

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

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;