Use math first before programming.

Note that you can factor out a1 and b1 from the formula, you'll get 

(a1 + b1) * (a2 * a3 * a4 + a2 * a3 * b4 + ...)

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In the second factor, you can use the same technique to factor out a2 + b2, and so on. In the end, you'll get 
(a1 + b1) * (a2 + b2) * (a3 + b3) * (a4 + b4)

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So, the solution is 

$sum = 1;
for my $i (0 .. 3) {
    $sum *= $A[$i] + $B[$i];
}
say $sum;

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And it gives the same result as 

my $sum = 0;
for my $pattern (map unpack('b4', pack 's', $_), 0 .. 15) {
    my $prod = 1;
    for my $i (0 .. 3) {
        $prod *= (\@A, \@B)[substr $pattern, $i, 1][$i];
    }
    $sum += $prod;
}
say $sum;

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map{substr$_->[0],$_->[1]||0,1}[\*||{},3],[[]],[ref qr-1,-,-1],[{}],[sub{}^*ARGV,3]

In reply to Re: adding n-tuples by choroba
in thread adding n-tuples by kwatts59

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