Merge pull request #1671 from fluca1978/pwc59

Pwc59

4 files changed, 86 insertions, 0 deletions

diff --git a/challenge-059/luca-ferrari/blog-1.txt b/challenge-059/luca-ferrari/blog-1.txt
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+https://fluca1978.github.io/2020/05/04/PerlWeeklyChallenge59.html#task1
diff --git a/challenge-059/luca-ferrari/blog-2.txt b/challenge-059/luca-ferrari/blog-2.txt
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+https://fluca1978.github.io/2020/05/04/PerlWeeklyChallenge59.html#task2
diff --git a/challenge-059/luca-ferrari/raku/ch-1.p6 b/challenge-059/luca-ferrari/raku/ch-1.p6
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+#!env raku
+
+
+# You are given a linked list and a value k.
+# Write a script to partition the linked list such that all nodes
+# less than k come before nodes greater than or equal to k.
+# Make sure you preserve the original
+# relative order of the nodes in each of the two partitions.
+#
+#    For example:
+#
+#    Linked List: 1 → 4 → 3 → 2 → 5 → 2
+#
+#    k = 3
+#
+#    Expected Output: 1 → 2 → 2 → 4 → 3 → 5.
+
+
+# example of invocation
+# % raku ch-1.p6 --k=4
+# Index 4 makes 1 4 3 2 5 2 to become 1 3 2 2 4 5
+
+sub MAIN( Int:D :$k = 3 ) {
+
+    my @L = 1, 4, 3, 2, 5, 2 ;
+    my @l = | @L.grep( * < $k ),  | @L.grep( * >= $k );
+    say "Index $k makes { @L } to become { @l }";
+}
diff --git a/challenge-059/luca-ferrari/raku/ch-2.p6 b/challenge-059/luca-ferrari/raku/ch-2.p6
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+#!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";