Add challenge-076's ch-1 solution in Perl

1 files changed, 52 insertions, 0 deletions

diff --git a/challenge-076/andinus/perl/ch-1.pl b/challenge-076/andinus/perl/ch-1.pl
new file mode 100755
index 0000000000..c7660ff559
--- /dev/null
+++ b/challenge-076/andinus/perl/ch-1.pl
@@ -0,0 +1,52 @@
+#!/usr/bin/perl
+
+use strict;
+use warnings;
+use feature 'say';
+
+my $input = shift @ARGV;
+chomp $input;
+
+# Reject invalid inputs.
+die "Invalid input, enter numbers greater than 1.\n" if $input <= 1;
+
+# If it's a prime number then the minimum sum is the number itself.
+say $input and exit 0 if is_prime($input) == 1;
+
+if ($input % 2 == 0) {
+    # If $input is even then we loop from 2 to `$input / 2' & check if
+    # both $i & $diff are primes. If both are primes then we found our
+    # numbers!
+
+    # Eventually we'll find 2 primes to be a sum of even numbers. From
+    # WikiPedia, https://en.wikipedia.org/wiki/Goldbach%27s_conjecture
+    # has been shown to hold for all integers less than `4 × 10^18'.
+    foreach my $i (2 ... $input / 2) {
+        my $diff = $input - $i;
+        say "$i + $diff"
+            if is_prime($i) and is_prime($diff);
+    }
+} elsif (is_prime($input - 2)) {
+    # If $input is odd & `$input - 2' is a prime then $input is sum of
+    # two primes, 2 & `$input -2'.
+    say "2 + $input";
+} else {
+    # If even that doesn't match then the minimum sum will have three
+    # numbers. 3 & then we use the same function as for even numbers
+    # to find the other two primes.
+    foreach my $i (2 ... ($input - 3) / 2) {
+        my $diff = $input - 3 - $i;
+        say "3 + $i + $diff"
+            if is_prime($i) and is_prime($diff);
+    }
+}
+
+sub is_prime {
+    my $num = shift @_;
+    # If $num is divisible by any number between 2 & sqrt($num) then
+    # it's not prime. Checking only till
+    foreach my $i (2 ... sqrt($num)) {
+        return 0 if $num % $i == 0;
+    }
+    return 1;
+}