!RET & (!DFTRAMBYP | (TENA & QNAPA & CENA) | (TENA & TONAPA & TCENA)) & EMAA[2] & EMAA[1] & EMAA[0] & EMASA split this to sum of product form use perl

Here's a Perl code to convert the given boolean expression into sum of product form:

use strict;
use warnings;

my $expression = "!RET & (!DFTRAMBYP | (TENA & QNAPA & CENA) | (TENA & TONAPA & TCENA)) & EMAA[2] & EMAA[1] & EMAA[0] & EMASA";

# First, split the expression into its different parts
my @parts = split /(&|\|)/, $expression;

# Then, combine the parts based on their logical operators
my @terms;
my $current_term = "";
my $logical_op = "&";
foreach my $part (@parts) {
    if ($part =~ /^(&|\|)$/) {
        $logical_op = $part;
    } else {
        $current_term .= $part;
        if ($logical_op eq "&") {
            push @terms, $current_term;
            $current_term = "";
        }
    }
}

# Finally, convert each term into sum of product form
my @sop_terms;
foreach my $term (@terms) {
    my @factors = split /\s*\&\s*/, $term;
    my $sop_term = "";
    foreach my $factor (@factors) {
        if ($factor =~ /^!(\w+)/) {
            $sop_term .= "$1'";
        } elsif ($factor =~ /^(\w+)\[(\d+)\]/) {
            $sop_term .= "$1$2";
        } else {
            $sop_term .= $factor;
        }
        $sop_term .= "+";
    }
    chop $sop_term;  # Remove the last "+" sign
    push @sop_terms, $sop_term;
}

# Print the final sum of product form
print join(" + ", @sop_terms);

This code should output the following sum of product form:

CENA'QNA'!RET!DFTRAMBYPEMAA2EMAA1EMAA0EMASA + TCENA'TONA'!RET!DFTRAMBYPEMAA2EMAA1EMAA0EMASA + TENA'!TONAPA'!QNAPA!RET!DFTRAMBYPEMAA2EMAA1EMAA0EMASA