Task2 done.

1 files changed, 56 insertions, 0 deletions

diff --git a/challenge-059/luca-ferrari/raku/ch-2.p6 b/challenge-059/luca-ferrari/raku/ch-2.p6
new file mode 100644
index 0000000000..6e9d5b7ca1
--- /dev/null
+++ b/challenge-059/luca-ferrari/raku/ch-2.p6
@@ -0,0 +1,56 @@
+#!env raku
+
+# Helper Function
+# 
+#     For this task, you will most likely need a function f(a,b)
+#     which returns the count of different bits of binary representation of a and b.
+# 
+#     For example, f(1,3) = 1, since:
+# 
+#     Binary representation of 1 = 01
+# 
+#     Binary representation of 3 = 11
+# 
+#     There is only 1 different bit. Therefore the subroutine should return 1.
+#     Note that if one number is longer than the other in binary,
+#     the most significant bits of the smaller number are padded (i.e., they are assumed to be zeroes).
+#
+#
+#     Script Output
+# 
+#     You script should accept n positive numbers.
+#     Your script should sum the result of f(a,b) for every pair of numbers given:
+# 
+#     For example, given 2, 3, 4, the output would be 6, since f(2,3) + f(2,4) + f(3,4) = 1 + 2 + 3 = 6
+
+
+
+sub f( Int:D $a, Int:D $b ) {
+    my $different-bits = 0;
+
+    my @a-bits = $a.base( 2 ).Str.comb.reverse;
+    my @b-bits = $b.base( 2 ).Str.comb.reverse;
+
+    # find the longest number
+    my $max-length = max( @a-bits.elems, @b-bits.elems );
+    # do the padding with zeros (to the end, the arra)
+    @a-bits.push: 0 for 0 .. ( $max-length - @a-bits.elems  );
+    @b-bits.push: 0 for 0 .. ( $max-length - @b-bits.elems );
+
+
+    # compute the difference
+    for 0 ..^ @a-bits.elems {
+        $different-bits += 1 if ( @a-bits[ $_ ] != @b-bits[ $_ ] );
+    }
+
+    $different-bits;
+}
+
+################
+my $sum = 0;
+for 0 ..^ @*ARGS.elems  -> $first {
+    for $first + 1 ..^ @*ARGS.elems  -> $second {
+        $sum += f( @*ARGS[ $first ].Int, @*ARGS[ $second ].Int );
+    }
+}
+say "Sum is $sum";