What about implementing David Pearson's algorithm instead? It will find the smallest counterexample in polynomial time O(N^3) when N is the number of coins.
Determining whether or not a given coin system can be done by the greedy algorithm is not NP-complete. Finding optimum change in a given coin system that does not allow the greedy algorithm is NP-complete.
It looks quite easy to implement. I will try it when I get a chance.
Update: Code added below. Update2: Rewrote greedy_change with a simple loop instead of the regular expression one I borrowed from AnonyMonk.
use strict;
use warnings;
sub greedy_change {
my $change = shift;
my @coins = @_;
my @result;
for my $i (0..$#coins) {
my $val = int($change/$coins[$i]);
$change -= $val*$coins[$i];
push @result,$val;
}
return @result;
}
+
+
+
sub pearson_counterexample {
my $i = shift;
my $j = shift;
my @coins = @_;
+
# Let position $i be the highest coin used,
# Let positon $j be the lowest. Then the
# potential counterexample will be:
# 1) Compute the greedy change for the
# next highest coins value - 1.
#
# 2) Copy all coin counts from 0 to $j-1
#
# 3) Add one to the coin count at position $j.
#
# If this combination is more efficient than
# the greedy value, we have a counterexample.
+
my $value = $coins[$i-1] - 1;
my @greedy = greedy_change($value, @coins);
my @example = (0) x @coins;
$example[$_] = $greedy[$_] for (0..$j-1);
$example[$j] = $greedy[$j]+1;
+
# Compute the numeric value and see if it
# represents the amount more efficiently
# than the greedy way.
my $example_value = 0;
$example_value += $coins[$_]*$example[$_] for (0..$#example);
@greedy = greedy_change($example_value, @coins);
my ($example_count, $greedy_count);
$example_count += $example[$_] for (0..$#example);
$greedy_count += $greedy[$_] for (0..$#greedy);
+
# This candidate case did not work.
return if $example_count >= $greedy_count;
return @example;
}
+
# Read the coin system from the command line, in
# decreasing order of size.
my @coins = @ARGV;
if (@coins < 3) {
print "Must have a least three coins\n";
exit(1);
}
my ($i,$j);
my $found = 0;
+
EXAMPLES:
for ($i = 1; $i < @coins; ++$i) {
for ($j = $i; $j < @coins; ++$j) {
my @example = pearson_counterexample($i, $j, @coins);
if (@example) {
print "Counterexample: ",join(" ",@example),"\n";
$found = 1;
last EXAMPLES;
}
}
}
+
print "No counterexamples found. Good coin system\n" unless $found;
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