Merge pull request #1448 from dcw803/master

imported my solutions for challenge 52..

3 files changed, 223 insertions, 18 deletions

diff --git a/challenge-052/duncan-c-white/README b/challenge-052/duncan-c-white/README
index 1c6e36e8ae..557a59563d 100644
--- a/challenge-052/duncan-c-white/README
+++ b/challenge-052/duncan-c-white/README
@@ -1,30 +1,49 @@
-Task 1: "3 Sum
+Task 1: "Stepping Numbers
 
-Given an array @L of integers, and a target T. Write a script to find
-all unique triplets such that a + b + c is same as the target T. Also
-make sure a <= b <= c.
+Write a script to accept two numbers between 100 and 999. It should then
+print all Stepping Numbers between them.
 
-Example:
-
-    @L = (-25, -10, -7, -3, 2, 4, 8, 10);
-
-One such triplet for target 0 i.e. -10 + 2 + 8 = 0.
+    A number is called a stepping number if the adjacent digits have a
+    difference of 1. For example, 456 is a stepping number but 129 is not.
 "
 
-My notes: Fine.
+My notes: Sounds simple.
 
 
-Task #2: "Colourful Number
+Task #2: "Lucky Winner
 
-Write a script to display all Colorful Number with 3 digits.
+Suppose there are following coins arranged on a table in a line in
+random order.
 
-    A Colorful Number is an integer where all the products of consecutive
-    subsets of digit are different.
+    £1, 50p, 1p, 10p, 5p, 20p, £2, 2p
 
-For example, 263 is a Colorful Number since 2, 6, 3, 2x6, 6x3, 2x6x3 are unique.
+Suppose you are playing against the computer. Player can only pick one
+coin at a time from either ends. Find out the lucky winner, who has the
+larger amounts in total?
 "
 
-My notes: Interesting.
+My notes: umm.. with this wording, it wasn't very clear what exactly we're
+being asked to do, so the wording I've chosen is:
+
+  "Suppose the following coins are arranged in a line, but in
+   a different random order each game:
+
+    £1, 50p, 1p, 10p, 5p, 20p, £2, 2p
+
+   A human is playing against the computer. Pick a random player to go
+   first, after that players play alternative moves.  Each move, either pick
+   the FIRST coin or the LAST coin.  The winner is the player with the greatest
+   amount of money.  Write a program to play the whole game, with a computer
+   move choice method with the greatest choice of winning.
+  "
+
+Given this understanding, and those specific coin values, then whoever picks
+up the £2 coin will win.  obviously if the £2 coin is initially at one end,
+picking it is the correct move, otherwise both players should ensure they do
+not give the other player the opportunity to pick the £2 coin next move.
+ie. avoid leaving the £2 coin at either end..
 
-Having written this, assuming that it's correct, I find 328 Colourful 3-digit
-numbers.
+In practice: the first player alway wins, so rather a dull game.
+If playing second, the best the computer can do is pounce on a human
+mistake.
+#
diff --git a/challenge-052/duncan-c-white/perl/ch-1.pl b/challenge-052/duncan-c-white/perl/ch-1.pl
new file mode 100755
index 0000000000..7918222184
--- /dev/null
+++ b/challenge-052/duncan-c-white/perl/ch-1.pl
@@ -0,0 +1,47 @@
+#!/usr/bin/perl
+#
+# Task 1: "Stepping Numbers
+# 
+# Write a script to accept two numbers between 100 and 999. It should then
+# print all Stepping Numbers between them.
+# 
+#     A number is called a stepping number if the adjacent digits have a
+#     difference of 1. For example, 456 is a stepping number but 129 is not.
+# "
+#
+# My notes: Sounds simple.
+# 
+
+use feature 'say';
+use strict;
+use warnings;
+use Function::Parameters;
+#use Data::Dumper;
+
+die "Usage: stepping-nos [from [to]]\n" if @ARGV > 2;
+my $from = shift // 100;
+my $to   = shift // 999;
+
+
+#
+# my $isstepping = stepping( $x );
+#	Return true iff $x is a stepping number in the sense of the question,
+#	ie. one where the adjacent digits have a difference of 1. eg 456.
+#
+fun stepping( $x )
+{
+	my @digits = split(//,$x);
+	my $prev = shift @digits;
+	foreach my $next (@digits)
+	{
+		return 0 unless $next == $prev+1;
+		$prev = $next;
+	}
+	return 1;
+}
+
+
+foreach my $n ($from..$to)
+{
+	say $n if stepping($n);
+}
diff --git a/challenge-052/duncan-c-white/perl/ch-2.pl b/challenge-052/duncan-c-white/perl/ch-2.pl
new file mode 100755
index 0000000000..bd03820f26
--- /dev/null
+++ b/challenge-052/duncan-c-white/perl/ch-2.pl
@@ -0,0 +1,139 @@
+#!/usr/bin/perl
+#
+# Task #2: "Lucky Winner
+#
+# Suppose the following coins are arranged in a line, but in
+# a different random order each game:
+# 
+#     £1, 50p, 1p, 10p, 5p, 20p, £2, 2p
+# 
+# A human is playing against the computer. Pick a random player to go
+# first, after that players play alternative moves.  Each move, either pick
+# the FIRST coin or the LAST coin.  The winner is the player with the greatest
+# amount of money.  Write a program to play the whole game, with a computer
+# move choice method with the greatest choice of winning.
+# "
+# 
+# My notes: umm.. with the original wording, it wasn't very clear what exactly
+# we're being asked to do, so I've changed the wording above to clarify what
+# I think was meant.
+# Given those coin values, whoever picks up the £2 coin will win.  obviously
+# if the £2 coin is initially at one end,. picking it immediately is the
+# correct move, otherwise both players should ensure they do not give the
+# other player the opportunity to pick the £2 coin next move.  ie. avoid
+# leaving the £2 coin at either end..
+# In practice: the first player alway wins, so rather a dull game.
+# If playing second, the best the computer can do is pounce on a human
+# mistake.
+#
+
+use strict;
+use warnings;
+use feature 'say';
+use Function::Parameters;
+use Data::Dumper;
+use List::Util qw(max);
+
+my @coins = ( 100, 50, 1, 10, 5, 20, 200, 2 );
+
+my $biggest = max(@coins);
+
+srand( $$ ^ time() );
+
+# randomise their order..
+@coins = sort { int rand(3) - 1 } @coins;
+
+#die Dumper \@coins;
+
+#
+# my $choice = pick_first_or_last( @coins );
+#	Given an array of coin values @coins (expressed in pennies),
+#	pick either the "first" coin (return 'first') or the "last" coin
+#	(return 'last'), depending on which is a better move to make.
+#	
+#	The $biggest coin (with these coins the £2 coin) is critical: pick
+#	it if it's at either end, otherwise prevent it from getting to either
+#	end..  if it's already been picked, calculate the biggest remaining
+#	and apply the same strategy to that value..
+#
+fun pick_first_or_last( @coins )
+{
+	my $firstc = $coins[0];
+	return 'first' if $firstc == $biggest;
+	my $lastc  = $coins[$#coins];
+	return 'last' if $lastc == $biggest;
+
+	# find position of biggest (if it's still here)
+	my @bigpos = grep { $coins[$_] == $biggest } 0..$#coins;
+	# if not here.. change biggest to the biggest that is still here
+	if( @bigpos == 0 )
+	{
+		$biggest = max(@coins);
+		# find the position of that new biggest
+		@bigpos = grep { $coins[$_] == $biggest } 0..$#coins;
+	}
+	# now: @bigpos==1, $bigpos[0] is the position of that biggest.
+	my $nbp = @bigpos;
+	die "logic error, bigpos array has $nbp elements, should be 1\n"
+		unless $nbp==1;
+	my $bigpos = shift @bigpos;
+
+	return 'last' if $bigpos == 1;		# biggest very close to front
+	return 'first' if $bigpos == $#coins-1;	# biggest very close to back
+
+	# pick bigger
+	return 'last' if $coins[$#coins] > $coins[0];
+	return 'first';
+}
+
+my $humtot = 0;
+my $comptot = 0;
+
+# randomise who plays first
+my $player = int(rand(2));	# 0 is human, 1 is computer
+my @who = qw(You I);
+
+say "$who[$player] play first";
+
+while( @coins )
+{
+	say "coins: ", join(',',@coins);
+	if( $player == 0 )
+	{
+		my $choice = 'f';
+		if( @coins > 1 )
+		{
+			print " pick first coin (f) or last coin (l)? ";
+			$choice = <STDIN>;
+			chomp $choice;
+			$choice = lc($choice);
+		}
+
+		my $coin = ($choice eq 'f')? shift @coins : pop @coins;
+		$humtot += $coin;
+		say " you picked $coin, your total is now $humtot";
+	} else
+	{
+		my $choice = pick_first_or_last( @coins );
+		say " I pick $choice coin";
+
+		my $coin = ($choice eq 'first')? shift @coins : pop @coins;
+		$comptot += $coin;
+		say " I picked $coin, my total is now $comptot";
+	}
+
+	# switch players
+	$player = 1-$player;
+}
+
+say "you scored $humtot, I scored $comptot";
+if( $humtot > $comptot )
+{
+	say "well done, you win";
+} elsif( $humtot < $comptot )
+{
+	say "yippee, I win";
+} else
+{
+	say "it's a draw!";
+}