2 files changed, 98 insertions, 0 deletions
diff --git a/challenge-120/wanderdoc/perl/ch-1.pl b/challenge-120/wanderdoc/perl/ch-1.pl
new file mode 100644
index 0000000000..306c6013ba
--- /dev/null
+++ b/challenge-120/wanderdoc/perl/ch-1.pl
@@ -0,0 +1,51 @@
+#!perl
+use strict;
+use warnings FATAL => qw(all);
+
+=prompt
+You are given a positive integer $N less than or equal to 255. Write a script to swap the odd positioned bit with even positioned bit and print the decimal equivalent of the new binary representation.
+Example
+
+Input: $N = 101 Output: 154
+Binary representation of the given number is 01 10 01 01.
+The new binary representation after the odd/even swap is 10 01 10 10.
+The decimal equivalent of 10011010 is 154.
+
+Input: $N = 18 Output: 33
+Binary representation of the given number is 00 01 00 10.
+The new binary representation after the odd/even swap is 00 10 00 01.
+The decimal equivalent of 100001 is 33.
+=cut
+
+
+use Scalar::Util qw(looks_like_number);
+
+sub swap_bits
+{
+     my $num = $_[0];
+
+     die "Number in range 1 to 255?$/" 
+          unless looks_like_number($num) and
+          $num == int($num) and
+          ($num > 0 and $num < 256);
+     my $binary = dec_to_bin($num);
+     $binary = length($binary) % 2 ? 0 . $binary : $binary;
+     my @bits = split(//, $binary);
+     @bits[((grep {$_ % 2 == 0} 0 .. $#bits), (grep {$_ % 2 == 1} 0 .. $#bits))] =
+     @bits[((grep {$_ % 2 == 1} 0 .. $#bits), (grep {$_ % 2 == 0} 0 .. $#bits))];
+     $binary = join('', @bits);
+     $num = bin_to_dec($binary);
+     return $num;
+
+
+}
+
+
+
+for my $n ( 1 .. 255 )
+{
+     print join(' -> ', $n, swap_bits($n)), $/;
+}
+
+sub dec_to_bin { return sprintf("%b",$_[0]); }
+sub bin_to_dec { return oct("0b" . $_[0]); }
+\ No newline at end of file
diff --git a/challenge-120/wanderdoc/perl/ch-2.pl b/challenge-120/wanderdoc/perl/ch-2.pl
new file mode 100644
index 0000000000..028e461619
--- /dev/null
+++ b/challenge-120/wanderdoc/perl/ch-2.pl
@@ -0,0 +1,47 @@
+#!perl
+use strict;
+use warnings FATAL => qw(all);
+
+=prompt
+You are given time $T in the format hh:mm. Write a script to find the smaller angle formed by the hands of an analog clock at a given time. HINT: A analog clock is divided up into 12 sectors. One sector represents 30 degree (360/12 = 30).
+Example
+Input: $T = '03:10' Output: 35 degree
+The distance between the 2 and the 3 on the clock is 30 degree. For the 10 minutes i.e. 1/6 of an hour that have passed.
+The hour hand has also moved 1/6 of the distance between the 3 and the 4, which adds 5 degree (1/6 of 30).
+The total measure of the angle is 35 degree.
+
+Input: $T = '04:00' Output: 120 degree
+=cut
+
+
+
+
+
+my %MINUTES;
+@MINUTES{0 .. 59} = map $_ * 6, 0 .. 59;
+
+my %HOURS;
+@HOURS{0 .. 11} = map $_ * 30, 0 .. 11;
+
+
+sub clock_angle
+{
+     my $time = $_[0];
+     my ($hour, $min) = map $_ * 1, split(/:/, $time);
+     $hour = $hour >= 12 ? $hour - 12 : $hour; 
+     my $hour_hand = $HOURS{$hour} + ($min/60) * 30;
+     my $minute_hand = $MINUTES{$min};
+     my $angle = abs($hour_hand - $minute_hand);
+     $angle = $angle > 180 ? 360 - $angle : $angle;
+     return $angle;
+}
+
+
+
+for my $time ( '0000' .. '2359' )
+{
+     next if unpack("xxA2", $time) > 59;
+
+     my $time_formatted = join(':',unpack("A2A2",$time)); 
+     print join(' -> ', $time_formatted, clock_angle($time_formatted)), $/;
+}
+\ No newline at end of file