Merge pull request #496 from dmanto/branch-for-challenge-020

my proposed solutions for challenge-020, p5 1 & 2

2 files changed, 94 insertions, 0 deletions

diff --git a/challenge-020/daniel-mantovani/perl5/ch-1.pl b/challenge-020/daniel-mantovani/perl5/ch-1.pl
new file mode 100644
index 0000000000..e3f689ee34
--- /dev/null
+++ b/challenge-020/daniel-mantovani/perl5/ch-1.pl
@@ -0,0 +1,32 @@
+# Write a script to accept a string from command line and split it
+# on change of character. For example, if the string is “ABBCDEEF”,
+# then it should split like “A”, “BB”, “C”, “D”, “EE”, “F”.
+
+use strict;
+use warnings;
+use v5.10;
+use utf8;
+use Encode;
+use open qw(:std :utf8);    # for other languages
+
+# we start by reading string from command line
+
+my $string = Encode::decode 'utf8', $ARGV[0];
+die "Usage: $0 <string>" unless $string;
+
+my $current;                # this to store current char to see if changed
+
+while ($string) {
+    my $char = $1
+      if $string =~ s/(.)//;   # there are maaaany ways to do this, but the
+                               # idea is to take the first char from $string and
+                               # store it in $char
+    print "\n"
+      unless !defined $current
+      || $char eq $current;    # new line on change of character
+        # note that on first pass $current will not be defined,
+        # so we don't get a newline before first char
+    $current = $char;    # keep current char for next pass
+    print $char;
+}
+print "\n";              # to end last line
diff --git a/challenge-020/daniel-mantovani/perl5/ch-2.pl b/challenge-020/daniel-mantovani/perl5/ch-2.pl
new file mode 100644
index 0000000000..b81a4fbe0d
--- /dev/null
+++ b/challenge-020/daniel-mantovani/perl5/ch-2.pl
@@ -0,0 +1,62 @@
+# Write a script to print the smallest pair of Amicable Numbers.
+# For more information, please checkout wikipedia page.
+
+use strict;
+use warnings;
+use v5.10;
+
+# we start by defining a sub that sums of all divisors of a given number:
+
+sub sum_divisors {
+    my $x   = shift;
+    my $sum = 0;
+    for my $d ( 1 .. $x - 1 ) {
+        $sum += $d unless $x % $d;
+    }
+    return $sum;
+}
+
+# now we will start calculating sum of divisors for each integer,
+# starting by integer 2.
+# we will name that integer $candidate, to show that is a potential
+# amicable number, and we will try to get the corresponding pair
+# and see if they accomplish the given criteria
+
+my $candidate = 2;
+
+while (1) {
+    my $candidate2 = sum_divisors $candidate;
+
+    # we will not consider cases where candidates are the same number
+    # (that wouldn't be a pair, right?).
+    # (Note also that this candidate would be a perfect number, something
+    # we saw on challenge 008-1).
+    # Also we don't care for candidates that are lower than the one
+    # we are checking, as those would have been found before
+    if ( $candidate < $candidate2 && $candidate == sum_divisors($candidate2) ) {
+        say "Found amicable numbers: $candidate & $candidate2";
+        last;    # challenge ask to stop on first match (i.e. smallest pair)
+    }
+    $candidate++;
+}
+
+# commenting out line # 38 above will get you a list of first
+# amicable numbers. You can check that the first 10 pairs on that
+# list match the ones listed on the wiki page:
+#
+# $> perl ch-2.pl
+# Found amicable numbers: 220 & 284
+# Found amicable numbers: 1184 & 1210
+# Found amicable numbers: 2620 & 2924
+# Found amicable numbers: 5020 & 5564
+# Found amicable numbers: 6232 & 6368
+# Found amicable numbers: 10744 & 10856
+# Found amicable numbers: 12285 & 14595
+# Found amicable numbers: 17296 & 18416
+# Found amicable numbers: 63020 & 76084
+# Found amicable numbers: 66928 & 66992
+# Found amicable numbers: 67095 & 71145
+# Found amicable numbers: 69615 & 87633
+# Found amicable numbers: 79750 & 88730
+# ^C
+# $>