Solutions to challenge 30 task 1 and 2 in Perl 6 by Noud

2 files changed, 31 insertions, 0 deletions

diff --git a/challenge-030/noud/perl6/ch-1.p6 b/challenge-030/noud/perl6/ch-1.p6
new file mode 100644
index 0000000000..838e53d537
--- /dev/null
+++ b/challenge-030/noud/perl6/ch-1.p6
@@ -0,0 +1,8 @@
+# Write a script to list dates for Sunday Christmas between 2019 and 2100. For
+# example, 25 Dec 2022 is Sunday.
+
+for 2019 .. 2100 -> $year {
+    if (Date.new($year, 12, 25).day-of-week == 7) {
+        $year.say;
+    }
+}
diff --git a/challenge-030/noud/perl6/ch-2.p6 b/challenge-030/noud/perl6/ch-2.p6
new file mode 100644
index 0000000000..45efa3f42b
--- /dev/null
+++ b/challenge-030/noud/perl6/ch-2.p6
@@ -0,0 +1,23 @@
+# Write a script to print all possible series of 3 positive numbers, where in
+# each series at least one of the number is even and sum of the three numbers
+# is always 12. For example, 3,4,5.
+
+# This should not be too hard. First notice that if (i, j, k) is a triple that
+# sums up to 12, we have symetric tuples (i, k, j), (j, i, k), (j, k, i),
+# (k, i, j), and (k, j, i) that also sum up to 12. We make use of this
+# symmetry property. Next, if i and j are given in tuple (i, j, k), then we
+# know that k = 12 - i - j. So we don't have to loop over k. Last, one of the
+# integers needs to be even. Due to symmetry arguments we can always take i to
+# be even.
+
+my $total = 12;
+my @triplets = [];
+for 2, 4 ... $total -> $i {
+    for 1 .. ($total - $i - 1) -> $j {
+        @triplets = [|(@triplets), |([$i, $j, $total - $i - $j].permutations)];
+    }
+}
+
+# Because we use permutations we have duplicates. Ex. (10, 1, 1) with j <-> k
+# gives (10, 1, 1).
+say @triplets.unique;