Merge pull request #5175 from dcw803/master

imported my solutions to this week's tasks..  task 2 was a bit strange

3 files changed, 228 insertions, 50 deletions

diff --git a/challenge-137/duncan-c-white/README b/challenge-137/duncan-c-white/README
index c41627048b..a9b09917c6 100644
--- a/challenge-137/duncan-c-white/README
+++ b/challenge-137/duncan-c-white/README
@@ -1,77 +1,73 @@
-TASK #1 - Two Friendly
+TASK #1 - Long Year
 
-You are given 2 positive numbers, $m and $n.
+Write a script to find all the years between 1900 and 2100 which is a Long Year.
 
-Write a script to find out if the given two numbers are Two Friendly.
+    A year is Long if it has 53 weeks.
 
-Two positive numbers, m and n are two friendly when gcd(m, n) = 2 ^
-p where p > 0. The greatest common divisor (gcd) of a set of numbers
-is the largest positive number that divides all the numbers in the set
-without remainder.
+For more information about Long Year, please refer to
 
-Example 1
-
-    Input: $m = 8, $n = 24
-    Output: 1
-
-    Reason: gcd(8,24) = 8 => 2 ^ 3
-
-Example 2
-
-    Input: $m = 26, $n = 39
-    Output: 0
+https://en.wikipedia.org/wiki/ISO_week_date#Weeks_per_year
 
-    Reason: gcd(26,39) = 13
+Expected Output
 
-Example 3
+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
 
-    Input: $m = 4, $n = 10
-    Output: 1
+MY NOTES: Pretty easy.  Especially using the p(year) and weeks(year) functions given,
+won't even need a date manipulation module..
 
-    Reason: gcd(4,10) = 2 => 2 ^ 1
 
-MY NOTES: Pretty easy.  Euclid's GCD algorithm, then check if the result is a power of 2.
+TASK #2 - Lychrel Number
 
+You are given a number, 10 <= $n <= 1000.
 
-TASK #2 - Fibonacci Sequence
+Write a script to find out if the given number is Lychrel number;
+output 1 if it is, and zero if it is not.  To keep the task simple, we impose the following rules:
 
-You are given a positive number $n.
+a. Stop and return 1 if the number of iterations reached 500.
+b. Stop and return 1 if you end up with number >= 10_000_000.
 
-Write a script to find how many different sequences you can create using
-Fibonacci numbers where the sum of unique numbers in each sequence are
-the same as the given number.
+According to wikipedia:
 
-    Fibonacci Numbers: 1,2,3,5,8,13,21,34,55,89,..
+    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 = 16
-  Output: 4
+	Input: $n = 56
+	Output: 0
 
-  Reason: There are 4 possible sequences that can be created using Fibonacci numbers
-  i.e. (3 + 13), (1 + 2 + 13), (3 + 5 + 8) and (1 + 2 + 5 + 8).
+	After 1 iteration, we found palindrome number.
+	56 + 65 = 121
 
 Example 2
 
-  Input:  $n = 9
-  Output: 2
+	Input: $n = 57
+	Output: 0
 
-  Reason: There are 2 possible sequences that can be created using Fibonacci numbers
-  i.e. (1 + 3 + 5) and (1 + 8).
+	After 2 iterations, we found palindrome number.
+	 57 +  75 = 132
+	132 + 231 = 363
 
 Example 3
 
-  Input:  $n = 15
-  Output: 2
-
-  Reason: There are 2 possible sequences that can be created using Fibonacci numbers
-  i.e. (2 + 5 + 8) and (2 + 13).
+	Input: $n = 59
+	Output: 0
 
-MY NOTES: Pretty easy - find subsets of the first few Fibonacci numbers that sum to N.
-How many Fibonacci numbers should we consider?  Those <= N.  Then, how do we sum subsets
-of them?  Two obvious ways:
+	After 3 iterations, we found palindrome number.
+	 59 +  95 =  154
+	154 + 451 =  605
+	605 + 506 = 1111
 
-1. recursive function, iterating over the Fibonacci values: each value is either in the
-   subset or not, try both possibilities.
-2. binary-counting method, iterate over every combination C from 1 .. 2**(number values)-1,
-   then sum up only the elements where the corresponding bit in C is true.
+MY NOTES: Sounds pretty easy.  The strangest thing about this question
+is that the Wikipedia page has that (without the artificial constraints)
+noone found a definite Lychrel base 10 number.  So any "Output: 1" entries
+we can find (such as 89) are caused by sequences violating one or other
+of the constraints.
diff --git a/challenge-137/duncan-c-white/perl/ch-1.pl b/challenge-137/duncan-c-white/perl/ch-1.pl
new file mode 100755
index 0000000000..a0adda2685
--- /dev/null
+++ b/challenge-137/duncan-c-white/perl/ch-1.pl
@@ -0,0 +1,82 @@
+#!/usr/bin/perl
+# 
+# TASK #1 - Long Year
+# 
+# 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.
+# 
+# For more information about Long Year, please refer to
+# 
+# https://en.wikipedia.org/wiki/ISO_week_date#Weeks_per_year
+# 
+# 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
+# 
+# MY NOTES: Pretty easy.  Especially using the p(year) and weeks(year) functions given,
+# won't even need a date manipulation module..
+# 
+
+use strict;
+use warnings;
+use feature 'say';
+use Getopt::Long;
+use Function::Parameters;
+#use Data::Dumper;
+
+
+my $debug=0;
+die "Usage: long-year\n" unless
+	GetOptions( "debug"=>\$debug ) && @ARGV==0;
+
+#
+# my $dow = dow_31stdec( $year );
+#	Return the Dow (0..6, 0 is Sunday, 6 is Saturday) of 31st Dec in $year.
+#	Wikipedia's P() function:
+#
+fun dow_31stdec( $year )
+{
+	my $x = $year + int($year/4) - int($year/100) + int($year/400);
+	return $x % 7;
+}
+
+
+#
+# my $weeks = weeksinyear( $year );
+#	How many weeks does year $year have?  52 or 53..
+#
+fun weeksinyear( $year )
+{
+	my $result = 52;
+	$result++ if dow_31stdec( $year ) == 4	 # current year ends on Thursday
+		  || dow_31stdec( $year-1) == 3; # previous year ends on Wednesday
+	return $result;
+}
+
+#my @dow = qw(Sun Mon Tue Wed Thu Fri Sat);
+
+my @result;
+foreach my $y (1900..2100)
+{
+	#my $dow = dow_31stdec( $y );
+	#say "31st dec $y is a $dow[$dow]";
+	my $weeks = weeksinyear( $y );
+	#say "$y has $weeks weeks";
+	next if $weeks == 52;
+	push @result, $y;
+}
+
+while( @result > 4 )
+{
+	say join( ', ', @result[0..4] ).',';
+	splice( @result, 0, 5 );
+}
+say join( ', ', @result );
diff --git a/challenge-137/duncan-c-white/perl/ch-2.pl b/challenge-137/duncan-c-white/perl/ch-2.pl
new file mode 100755
index 0000000000..546e0a3abe
--- /dev/null
+++ b/challenge-137/duncan-c-white/perl/ch-2.pl
@@ -0,0 +1,100 @@
+#!/usr/bin/perl
+# 
+# TASK #2 - Lychrel Number
+# 
+# You are given a number, 10 <= $n <= 1000.
+# 
+# Write a script to find out if the given number is Lychrel number;
+# output 1 if it is, and zero if it is not.  To keep the task simple, we impose the following rules:
+# 
+# a. Stop and return 1 if the number of iterations reached 500.
+# b. Stop and return 1 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
+# 
+# MY NOTES: Sounds pretty easy.
+# 
+
+use strict;
+use warnings;
+use feature 'say';
+use Function::Parameters;
+use Getopt::Long;
+use Data::Dumper;
+
+my $debug = 0;
+
+die "Usage: is-lychrel [-d|--debug] N\n"
+	unless GetOptions( "debug"=>\$debug ) && @ARGV==1;
+my $n = shift @ARGV;
+die "$n is out of range 10..1000\n" if $n < 10 || $n > 1000;
+
+
+#
+# my @steps = get_to_palindrome( $n );
+#	Try to find a sequence of n + reverse(n) values leading to a palindrome,
+#	if you find one in less than 500 steps, with no intermediate value
+#	exceeding 10,000,000, then return the sequence of steps.
+#	If you can't find such a sequence, return ().
+#
+fun get_to_palindrome( $n )
+{
+	my @result;
+	for( my $nit = 0; $nit<500; $nit++ )
+	{
+		return () if $n > 10000000;		# intermediate value too big
+		my $r = reverse($n);
+		my $msg = sprintf( "%4d + %4d = %d", $n, $r, $n+$r );
+		push @result, $msg;
+		say "debug: $msg" if $debug;
+		$n += $r;
+		return @result if $n == reverse($n);	# found a palindrome
+	}
+	return ();					# too many iterations
+}
+
+
+my @seq = get_to_palindrome( $n );
+say "Input:  $n";
+print "Output: ";
+if( @seq )
+{
+	say 0;
+	my $steps = @seq;
+	say "\nAfter $steps iterations, we found a palindrome:";
+	say for @seq;
+} else
+{
+	say 1;
+}