Merge pull request #4160 from dcw803/master

imported my solutions to this week's challenge..

4 files changed, 141 insertions, 44 deletions

diff --git a/challenge-114/duncan-c-white/README b/challenge-114/duncan-c-white/README
index 6a326449a7..de73e12315 100644
--- a/challenge-114/duncan-c-white/README
+++ b/challenge-114/duncan-c-white/README
@@ -1,52 +1,42 @@
-Task 1: "Represent Integer
+Task 1: "Next Palindrome Number
 
-You are given a positive integer $N and a digit $D.
+You are given a positive integer $N.
 
-Write a script to check if $N can be represented as a sum of positive
-integers having $D at least once. If check passes print 1 otherwise 0.
+Write a script to find out the next Palindrome Number higher than the given integer $N.
 
 Example
 
-Input: $N = 25, $D = 7
-Output: 0 as there are 2 numbers between 1 and 25 having the digit 7
-i.e. 7 and 17. If we add up both we don't get 25.
+  Input: $N = 1234
+  Output: 1331
 
-Input: $N = 24, $D = 7
-Output: 1
+  Input: $N = 999
+  Output: 1001
 "
 
-My notes: sounds very simple.  sum ( grep contains 7 )
+My notes: sounds very simple.  Generate and test for palindromic-ness.
 
 
-Task 2: "Recreate Binary Tree
+Task 2: "Higher Integer Set Bits
 
-You are given a Binary Tree.
+You are given a positive integer $N.
 
-Write a script to replace each node of the tree with the sum of all the remaining nodes.
+Write a script to find the next higher integer having the same number
+of 1 bits in binary representation as $N.
 
 Example
-Input Binary Tree
-
-        1
-       / \
-      2   3
-     /   / \
-    4   5   6
-     \
-      7
-
-Output Binary Tree
-
-        27
-       /  \
-      26  25
-     /   /  \
-    24  23  22
-     \
-     21
+
+Input: $N = 3
+Output: 5
+
+Binary representation of 3 is 011. There are two 1 bits. So the next
+higher integer is 5 having the same the number of 1 bits i.e. 101.
+
+Input: $N = 12
+Output: 17
+
+Binary representation of 12 is 1100. There are two 1 bits. So the next
+higher integer is 17 having the same number of 1 bits i.e. 10001.
 "
 
-My notes: so each node's value becomes sum(allnodes) - that node's value
-Very tempting to view this as an ascii art -> ascii art problem, last time
-we did a similar tree question, I spent a lot of time building the tree
-structure, and getting nice layout from print_tree..
+My notes: also sounds pretty simple.  Generate and test for "having B
+bits set in the binary representation".
diff --git a/challenge-114/duncan-c-white/perl/bintree.eg b/challenge-114/duncan-c-white/perl/bintree.eg
deleted file mode 100644
index 9af8a8a980..0000000000
--- a/challenge-114/duncan-c-white/perl/bintree.eg
+++ /dev/null
@@ -1,8 +0,0 @@
-
-        1
-       / \
-      2   3
-     /   / \
-    4   5   6
-     \
-      7
diff --git a/challenge-114/duncan-c-white/perl/ch-1.pl b/challenge-114/duncan-c-white/perl/ch-1.pl
new file mode 100755
index 0000000000..6f9b316a78
--- /dev/null
+++ b/challenge-114/duncan-c-white/perl/ch-1.pl
@@ -0,0 +1,49 @@
+#!/usr/bin/perl
+# 
+# Task 1: "Next Palindrome Number
+# 
+# You are given a positive integer $N.
+# 
+# Write a script to find out the next Palindrome Number higher than the given integer $N.
+# 
+# Example
+# 
+#   Input: $N = 1234
+#   Output: 1331
+# 
+#   Input: $N = 999
+#   Output: 1001
+# "
+# 
+# My notes: sounds very simple.  Generate and test for palindromic-ness.
+# 
+
+use strict;
+use warnings;
+use feature 'say';
+use Function::Parameters;
+#use Data::Dumper;
+
+#
+# my $ispal = ispalindromic( $n );
+#	Return 1 iff $n is palindromic (in decimal of course)
+#
+fun ispalindromic( $n )
+{
+	my $rev = join( '', reverse split( //, $n ) );
+	return $n==$rev ? 1 : 0;
+}
+
+
+
+
+die "Usage: next-palindromic-number N\n" unless @ARGV==1;
+my $n = shift;
+
+do
+{
+	$n++;
+}
+while( ! ispalindromic($n) );
+
+say $n;
diff --git a/challenge-114/duncan-c-white/perl/ch-2.pl b/challenge-114/duncan-c-white/perl/ch-2.pl
new file mode 100755
index 0000000000..9a836cc748
--- /dev/null
+++ b/challenge-114/duncan-c-white/perl/ch-2.pl
@@ -0,0 +1,66 @@
+#!/usr/bin/perl
+# 
+# Task 2: "Higher Integer Set Bits
+# 
+# You are given a positive integer $N.
+# 
+# Write a script to find the next higher integer having the same number
+# of 1 bits in binary representation as $N.
+# 
+# Example
+# 
+# Input: $N = 3
+# Output: 5
+# 
+# Binary representation of 3 is 011. There are two 1 bits. So the next
+# higher integer is 5 having the same the number of 1 bits i.e. 101.
+# 
+# Input: $N = 12
+# Output: 17
+# 
+# Binary representation of 12 is 1100. There are two 1 bits. So the next
+# higher integer is 17 having the same number of 1 bits i.e. 10001.
+# "
+# 
+# My notes: also sounds pretty simple.  Generate and test for "having B
+# bits set in the binary representation of numbers > N, where B==nbits(N)".
+# 
+
+use strict;
+use warnings;
+use feature 'say';
+use Function::Parameters;
+#use Data::Dumper;
+
+#
+# my $nbits = nbits( $n );
+#	Count up and return the number of 1 bits in the binary representation of $n.
+#
+fun nbits( $n )
+{
+	my $result = 0;
+	while( $n > 0 )
+	{
+		$result++ if $n%2==1;
+		$n >>= 1;
+	}
+	return $result;
+}
+
+
+die "Usage: next-same-no-1-bits N\n" unless @ARGV==1;
+my $n = shift;
+
+my $nbits = nbits( $n );
+
+my $i=$n;
+do {
+	$i++;
+} while( nbits($i) != $nbits );
+
+say "Input: \$N = $n";
+say "Output: $i";
+
+printf "\nBinary representation of $n is %b. There are %d 1 bits. So the next\n",
+	$n, $nbits;
+printf "higher integer is $i having the same number of 1 bits, i.e. %b\n", $i;