http://www.perlmonks.org?node_id=1197354

Saw this idea recently. Wondered how susceptible it would be to a brute force approach.

Given a square grid size N, and a list of numbers 2*N**2 2*N, find a fillin (like a crossword), and report the digits in the major diagonal (as an easy proof of solution).

The reference in the spoiler took an hour or two to find the solution. I won't post the solution here, you'll have to do the work yourself.

-QM
--
Quantum Mechanics: The dreams stuff is made of

Replies are listed 'Best First'.
Re: Number Grid Fillin
by tybalt89 (Curate) on Aug 14, 2017 at 12:05 UTC

    Fun problem!

    I call this "Smart Brute Force".
    It fills in one column at a time with each remaining number in turn, then checks to see if the leading parts of each row are possible from left over numbers.

    #!/usr/bin/perl -l # http://perlmonks.org/?node_id=1197354 use strict; use warnings; @ARGV or @ARGV = qw( 113443 143132 241131 321422 323132 331222 341114 412433 414422 431331 443112 444313 ); my $half = @ARGV / 2; my $steps = 0; my @stack = [ "\n" x $half, join '', map "$_\n", @ARGV ]; NEXT: while( @stack ) { my ($have, $rest) = @{ pop @stack }; $steps++; my %lefts; # validate legal so far $lefts{$_}++ for $have =~ /^(.+)\n/gm; for my $head (keys %lefts) { $lefts{$head} <= ( () = $rest =~ /^$head/gm ) or goto NEXT; } if( $rest =~ tr/\n// == $half ) # half left means completed { print "answer in $steps steps\n\n$have"; print "diagonal ", $have =~ /(\d)(?:..{$half})?/gs; exit; } while( $rest =~ /^(.+)\n/gm ) # try each number remaining { my ($before, $after, @digits) = ($`, $', split //, $1); push @stack, [ $have =~ s/(?=\n)/ shift @digits /ger, $before . $after ]; } } print "failed to find solution in $steps steps\n";

    Output:

    answer in 35 steps 412433 414422 431331 341114 143132 331222 diagonal 411132

    Computes the answer in less than 0.1 seconds.

Re: Number Grid Fillin
by hdb (Monsignor) on Aug 15, 2017 at 12:40 UTC

    Thanks for posting this inspiring problem. Similar to what tybalt89 has posted, one can significantly reduce the search space by checking whether for a given number in a given position the remaining numbers would still fit. Below some code to do that based on positions counting from 0 to 5 and each can be vertical or horizontal. Given the output it is nearly trivial to solve the puzzle manually.

    use strict; use warnings; my @numbers = qw( 113443 143132 241131 321422 323132 331222 341114 412433 414422 431331 443112 444313 ); # find possible positions my %positions; for my $n (@numbers) { $positions{$n} = []; # frequency of digits in current number my @nfreq = (0) x 5; $nfreq[$_]++ for split //, $n; # check which position is possible for my $p (0..5) { # frequency of digits in current position w/o current number my @freq = (0) x 5; $freq[substr( $_, $p, 1 )]++ for @numbers; $freq[substr( $n, $p, 1 )]--; # check if position is feasible # ie enough of each digit available my $possible = 1; $freq[$_]<$nfreq[$_] and $possible = 0 for 1..4; push @{$positions{$n}}, $p if $possible; } } for my $n (sort { scalar(@{$positions{$a}}) <=> scalar(@{$positions{$b +}}) } @numbers) { print "Number $n can be at positions @{$positions{$n}}.\n"; }
Re: Number Grid Fillin
by LanX (Bishop) on Aug 14, 2017 at 15:56 UTC