3 ms·

I think most of us who really like it read it as kids which colors it too much. I distinctly remember this proof that you can't solve the MU-puzzle as the first

• justinator 8y ago
  I read it in art school (swiped from my then-girlfriend's bookshelf), and it was nice to have a connection between my school of art weirdos and putting myself through the same school by teaching myself software engineering. I distinctly remembering writing a Perl script to highlight that the Mu puzzle is unsolvable. #!/usr/bin/perl -w use strict; my %tried = (); main(); sub main { my @rules = ( \&rule_one, \&rule_two, \&rule_three, \&rule_four, ); my $axiom = 'MI'; my @path; my @tries; solve($axiom, \@rules); } sub solve { my @app; my ($axiom, $rules) = @_; my $i = 0; foreach(@$rules){ my $tmp = $_->($axiom); push(@app, $tmp) if $tmp; } foreach(@app){ next if $tried{$_}; next unless (length($_) < 1000); print $_ . "\n"; $tried{$_} = 1; if($_ eq 'MU'){ print "YES!\n\n\n"; exit; } solve($_, $rules); } } sub rule_one { # If any string ends in I, you can append U my $str = shift; if ($str =~ m/I$/){ return $str . 'U'; }else{ return undef; } } sub rule_two { # If any string begins with M, # you can duplicate the string after M, my $str = shift; if($str =~ /^M/){ return 'M' . substr($str, 1) . substr($str, 1); }else{ return undef; } } sub rule_three { # If any string contains III, # you can replace the III with U my $str = shift; if($str =~ /III/){ $str =~ s/III/U/g; return $str; }else{ return undef; } } sub rule_four { #If any string contains UU, # you can delete the UU my $str = shift; if($str =~ /UU/){ $str =~ s/UU//g; return $str; }else{ return undef; } }