Just for fun :)

no guarantees whatsoever...

use strict;
use warnings;

my @equations;

while(<DATA>){
  chomp;
  next unless $_;               # ignore empty
  next if $_ =~ /^\s*#/;        # ignore comment
  s/(\w+)/\$$1/g;               # variables

  # ops by decending precedence
  s/'/^1/g;                     # negation by xor 1
  s/\*/ and /g;                 # and
  s/\+/ or /g;                  # or
  s/=(.*)$/= ( $1 )/;           # assignment after ( RHS )

  push @equations,"$_;\n";
}

my $format = "%s | %s %s %s %s\n";
$\="\n";
#print @equations;


for my $eq (@equations) {
  print "\n $eq";
  printf  $format, qw/Y A B C D/;
  for my $A (0,1) {
    for my $B (0,1) {
      for my $C (0,1) {
        for my $D (0,1) {
          my $Y=undef;
          eval $eq;
          printf $format, $Y, $A, $B, $C, $D; 
        }
      }
    }
  }
}
__DATA__
#Y=A+B'
#Y=A*B'
#Y=A'
#Y=A*B
#Y=A+B

Y = A + (B*C)
Y = A + (B' + (C*D)')
Y = A*(B*(C'+D)')
[download]

 $Y = (  $A  or  ($B and $C) );

Y | A B C D
0 | 0 0 0 0
0 | 0 0 0 1
0 | 0 0 1 0
0 | 0 0 1 1
0 | 0 1 0 0
0 | 0 1 0 1
1 | 0 1 1 0
1 | 0 1 1 1
1 | 1 0 0 0
1 | 1 0 0 1
1 | 1 0 1 0
1 | 1 0 1 1
1 | 1 1 0 0
1 | 1 1 0 1
1 | 1 1 1 0
1 | 1 1 1 1

 $Y = (  $A  or  ($B^1  or  ($C and $D)^1) );

Y | A B C D
1 | 0 0 0 0
1 | 0 0 0 1
1 | 0 0 1 0
1 | 0 0 1 1
1 | 0 1 0 0
1 | 0 1 0 1
1 | 0 1 1 0
0 | 0 1 1 1
1 | 1 0 0 0
1 | 1 0 0 1
1 | 1 0 1 0
1 | 1 0 1 1
1 | 1 1 0 0
1 | 1 1 0 1
1 | 1 1 1 0
1 | 1 1 1 1

 $Y = (  $A and ($B and ($C^1 or $D)^1) );

Y | A B C D
0 | 0 0 0 0
0 | 0 0 0 1
0 | 0 0 1 0
0 | 0 0 1 1
0 | 0 1 0 0
0 | 0 1 0 1
0 | 0 1 1 0
0 | 0 1 1 1
0 | 1 0 0 0
0 | 1 0 0 1
0 | 1 0 1 0
0 | 1 0 1 1
0 | 1 1 0 0
0 | 1 1 0 1
1 | 1 1 1 0
0 | 1 1 1 1
[download]

update

wait there is a precedence problem ... = before and

Solved

---

In reply to Re: Parsing Boolean expressions (hack) by LanX
in thread Parsing Boolean expressions by Anonymous Monk

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