- Added solutions by Colin Crain.

4 files changed, 259 insertions, 0 deletions

diff --git a/challenge-137/colin-crain/perl/ch-1.pl b/challenge-137/colin-crain/perl/ch-1.pl
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+#!/Users/colincrain/perl5/perlbrew/perls/perl-5.32.0/bin/perl

+#

+#       .pl

+#

+#       137-1 Long Year

+#         Submitted by: Mohammad S Anwar

+#         Write a script to find all the years between 1900 and 2100 which is a Long Year.

+# 

+#         A year is Long if it has 53 weeks.

+# 

+#         Expected Output

+#         1903, 1908, 1914, 1920, 1925,

+#         1931, 1936, 1942, 1948, 1953,

+#         1959, 1964, 1970, 1976, 1981,

+#         1987, 1992, 1998, 2004, 2009,

+#         2015, 2020, 2026, 2032, 2037,

+#         2043, 2048, 2054, 2060, 2065,

+#         2071, 2076, 2082, 2088, 2093,

+#         2099

+# 

+#         a year, 365 days, divided into 7-day weeks yield 52 and 1/7 or

+#         one extra day. IF it is  leap year, with 366 days, is will have

+#         two. So by my reckoning, every year has 53 weeks worth of days:

+#         52 plus a part or a week. So this challenge hinges on which day

+#         of the week starts the year, and further, which day of the week

+#         starts a given week. Two commonly accepted schemes are for a new

+#         week to either start on Monday, running through Sunday, or for

+#         the week to start on Sunday, running through the following

+#         Saturday. That, I can see, is going to cause some

+#         internationalization problems with our global community. So if we

+#         start our weeks say on Sunday, then in a non-leap year where the

+#         years starts on a Saturday we will roll into Sunday on December

+#         31, and the year will have 53 weeks. For any other day the year

+#         will only have 52 weeks. In the case of a leap year, however,

+#         then the rollover will occur with two start days, either Friday

+#         or Saturday. 

+# 

+#         In light of this sketchiness, the ISO has set up a standard that

+#         the first week of the year is to be considered the week that

+#         contains January 4, which is to say the first week containing at

+#         least 4 days, or put another way the first week where the

+#         majority of days falls within the new year. In this scheme

+#         everything is shifted a few days, but the same precessional logic

+#         applies.

+# 

+#         We could solve this mathematically, by determining the start day

+#         of January 1, 1900, and precessing the day one or two days

+#         depending on the arcane rules for placing leap years, noting when

+#         the start dates properly align.

+# 

+#         But then again, that's why we have computers, to do this dreary,

+#         repetitive work for us. Enter Dave Rolsky, and his DataTime module.

+# 

+#         What we can do, then, is choose the 28th of December in a given 

+#         year, and check its week number. If that value is 53, then the week has

+#         53 ISO weeks. Easy peasy.

+# 

+#         As this result matches the expected output, we will suppose that

+#         interpretation of weeks in a year was what was being requested.

+#         

+# 

+# 

+#

+#       © 2021 colin crain

+## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ##

+

+

+

+use warnings;

+use strict;

+use utf8;

+use feature ":5.26";

+use feature qw(signatures);

+no warnings 'experimental::signatures';

+

+

+use DateTime;

+use DateTime::Duration;

+

+my @years;

+

+my $dt  = DateTime->new (    

+                year    => 1900 ,

+                month   => 12  ,

+                day     => 28   ) ;

+

+my $dur = DateTime::Duration->new (

+                years   => 1    );

+

+for (1..200) {

+    if ($dt->week_number == 53) {

+        push @years, $dt->year;

+    }

+    $dt->add( $dur );

+}

+

+## output phase

+local $" = ', ';

+my @five;

+for (@years) {

+    push @five, $_;

+    if (scalar @five == 5) {

+        say "@five";

+        @five = ();

+    }

+}

+say "@five";

+

diff --git a/challenge-137/colin-crain/perl/ch-2.pl b/challenge-137/colin-crain/perl/ch-2.pl
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+#!/Users/colincrain/perl5/perlbrew/perls/perl-5.32.0/bin/perl

+#

+#       lychrel.pl

+#

+#       137-2 Lychrel Number

+#         Submitted by: Mohammad S Anwar

+#         You are given a number, 10 <= $n <= 1000.

+# 

+#         Write a script to find out if the given number is Lychrel number.

+#         To keep the task simple, we impose the following rules:

+# 

+#         a. Stop if the number of iterations reached 500.

+#         b. Stop if you end up with number >= 10_000_000.

+#         According to wikipedia:

+# 

+#         A Lychrel number is a natural number that cannot form a

+#         palindrome through the iterative process of repeatedly reversing

+#         its digits and adding the resulting numbers.

+# 

+#         Example 1

+# 

+#             Input: $n = 56

+#             Output: 0

+#     

+#             After 1 iteration, we found palindrome number.

+#             56 + 65 = 121

+# 

+#         Example 2

+# 

+#             Input: $n = 57

+#             Output: 0

+#     

+#             After 2 iterations, we found palindrome number.

+#              57 +  75 = 132

+#             132 + 231 = 363

+# 

+#         Example 3

+# 

+#             Input: $n = 59

+#             Output: 0

+#     

+#             After 3 iterations, we found palindrome number.

+#              59 +  95 =  154

+#             154 + 451 =  605

+#             605 + 506 = 1111

+# 

+#       © 2021 colin crain

+## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ##

+

+

+

+use warnings;

+use strict;

+use utf8;

+use feature ":5.26";

+use feature qw(signatures);

+no warnings 'experimental::signatures';

+use integer;

+

+

+for my $num ( 10..1000 ) {

+    say "input:  $num";

+    my @ret = lychrel( $num );

+    say "output: ", $ret[0];

+    ref($ret[1]) eq 'ARRAY' 

+        ?  do {say "steps: (", scalar $ret[1]->@*, ")"; say "\t $_" for $ret[1]->@*} 

+        :  say $ret[1]; 

+    say '';

+}

+

+sub lychrel ( $num ) {

+    my @chain = ($num);

+    return (0, @chain) if $num == reverse $num;

+    for (1..500) {

+        my $revsum = $num + reverse $num;

+        push @chain, $revsum;

+        return (0, \@chain) if $revsum == reverse $revsum;

+        $num = $revsum;

+        return (1, "number too large: $num") if $num > 10_000_000;

+    }

+    say (1, "too many iterations, more than 500");

+}

+

diff --git a/challenge-137/colin-crain/raku/ch-1.raku b/challenge-137/colin-crain/raku/ch-1.raku
new file mode 100755
index 0000000000..a76126e80f
--- /dev/null
+++ b/challenge-137/colin-crain/raku/ch-1.raku
@@ -0,0 +1,24 @@
+#!/usr/bin/env perl6
+#
+#
+#       137-1-long-cat-years.raku
+#
+#
+#
+#       © 2021 colin crain
+## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ##
+
+
+
+unit sub MAIN () ;
+
+my @out = gather {
+    for 1900..2100 {
+        my $d = Date.new( $_, 12, 28 );
+        take $d.year if $d.week-number == 53;
+    }
+}
+
+.join(', ')
+.say        
+    for @out.rotor(5, :partial);
diff --git a/challenge-137/colin-crain/raku/ch-2.raku b/challenge-137/colin-crain/raku/ch-2.raku
new file mode 100755
index 0000000000..b4241aa572
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+++ b/challenge-137/colin-crain/raku/ch-2.raku
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+#!/usr/bin/env perl6
+#
+#
+#       lychrel.raku
+#
+#
+#
+#       © 2021 colin crain
+## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ##
+
+
+
+unit sub MAIN () ;
+
+constant MAX-ITER  = 5000;
+constant MAX-VALUE = 10_000_000_000;
+constant ERR-LARGE = 'number too large - larger than ' ~ MAX-VALUE;
+constant ERR-MANY  = 'too many iterations - more than ' ~ MAX-ITER;
+
+
+for 10..1000 -> $num {
+    my ($out, $res) = lychrel( $num );
+    say qq:to/END/;
+    Input:  $num
+    Output: $out
+            $res
+    END
+}
+
+
+sub lychrel ( $num is copy ) {
+    my @chain = $num;
+    return 1, @chain if $num == $num.flip;
+    for 1..MAX-ITER {
+        my $revsum = $num + $num.flip;
+        @chain.push: $revsum;
+        return 1, @chain if $revsum == $revsum.flip;
+        $num = $revsum;
+        return 0, ERR-LARGE if $num > MAX-VALUE;
+    }
+    return 0, ERR-MANY;
+}
+
+