Solution to the logic puzzle

solution.pl

use v6;

my $limit = 100;

# before they talk any X*Y where 1<X<Y<X+Y<$limit is allowed as p
my %products;

for 2 .. $limit -> $X {
    for $X+1 .. $limit-$X -> $Y {
        %products{$X * $Y}++;
    }
}

# when P says "I don't know", only products with %products{$p} > 1
# are allowed
my %sums_with_ambiguous_products;

for 2 .. $limit -> $X {
    for $X ^.. $limit - $X -> $Y {
        if %products{ $X * $Y } == 1 {
            %sums_with_ambiguous_products{ $X + $Y } = 1;
        }
    }
}

# when S says "I don't know", only sums that are not in
# %sums_with_ambiguous_products are allowed
my %products_that_cause_S_not_to_know;
for 2 .. $limit -> $s {
    if !%sums_with_ambiguous_products{$s} {
        for 2 .. $s/2 + 1 -> $X {
            my $p = $X * ($s - $X);
            if (%products{$p} // 0) > 1 {
                %products_that_cause_S_not_to_know{$p}++;
            }
        }
    }
}

# only the products that can by split into two pairs (X,Y), where X+Y is
# allowed in only one way, are allowed. i.e.,
# %products_that_cause_S_not_to_know{$p} == 1
my %unambiguous_sums;
for 2 .. $limit -> $s {
    next if %sums_with_ambiguous_products{$s};
    for 2 .. $s / 2 + 1 -> $X {
        my $p = $X * ($s - $X);
        if (%products_that_cause_S_not_to_know{$p} // 0) == 1 {
            %unambiguous_sums{$s}++;
        }
    }
}

for %unambiguous_sums.keys -> $s {
    next unless %unambiguous_sums{$s} == 1;
    for 2 .. $s / 2 + 1 -> $X {
        my $Y = ($s - $X);
        my $p = $X * ($s - $X);
        if %products_that_cause_S_not_to_know{$p} == 1 {
            say "(S, P) = ($s, $p), (X, Y) = ($X, $Y)";
        }
    }
}