Puzzle: The Widget Emporium services a wide range of widget needs, and they sell widgets at every exactdollar amount ($1, $2, $3,...). You are shopping for a new widget, so you check in your {wallet,purse} to see what you can afford. You find that it contains an infinite supply of $6 bills, $9 bills, and $20 bills. You have an infinite supply of money, but there's a catch  The Emporium only accepts exact change! What is the most expensive widget you cannot buy using exact change?
The answer for this particular puzzle is $43, but that alone is not very interesting. Of course, we want to generalize the problem to any number of any size denominations. So the challenge is:
Challenge: Given a set of money denominations, what is the largest amount that can not be represented? In other words: write a function, as short as possible, which takes any number of positive integer arguments and returns the largest number that cannot be written as a sum (with repetition) of the arguments.
Rules:
 Don't worry about input validation, assume the function will always be passed 1 or more distinct positive integers.
 Assume the arguments are passed in ascending order
 You can use global variables, but you should be able to call the function several times in the same run and get correct results.
 For some groups of denominations, there are an infinite number of values you cannot represent. Or, when a $1 denomination is used, all values can be represented. In this case, the solution is undefined, so you can return anything you like (or even loop forever).
 Assume any reasonable upper bound on the number of arguments, the size of the arguments, or the size of the answer (but don't use bounds that trivialize the problem). State the bounds you have assumed! If you can, your solution should be coded so that it's possible to increase these upper bounds at the cost of a few more characters. For instance, my golf solutions assume that the answer is less than 1000, but I can increase this to 10,000 by adding another character.
 Don't worry about efficiency.
 Your score is the number of characters inside of sub largest { ... }
 Your code should work within the following framework:
use Test::More 'no_plan';
sub largest {
## your code here
}
is( largest(6,9,20), 43 );
is( largest(3,5), 7 );
is( largest(6,11,20), 27 );
is( largest(17,18,19,20), 101 );
is( largest(2,3,5,10), 1 );
These sample inputs are for your convenience, but of course, your solution should be correct for all inputs.
Think about it for a while, and get golfing. The best I've done so far is 48 characters (two different ways). I'll post my solutions later.
An observation about the problem that might help you out:
If D is any one of the denominations, and you can represent each of X, X+1, ..., X+(D1) using the denominations given, then you can represent all amounts of size X or higher (by taking one of the X+i and adding copies of D). 
I came across this problem during a college radio trivia marathon  only the problem was phrased in terms of blackmarket kidneys that come in packages of 6, 9, or 20. ;) I think it's a very neat problem, as it is has interesting ties to formal language theory. Golfing it down was fun, too, and it even led me to discover a bug in Perl! There are also other fun problems relating to currency and changemaking that I can post, if you're interested.
Update: This problem is also called the Frobenius problem. The largest number not expressible from a set of coins is called the Frobenius number of that set.
Re: Golf: Buying with exact change
by kvale (Monsignor) on Feb 22, 2005 at 01:43 UTC

Fun problem. Here is a solution in 72 characters with a bound of 1000:
sub largest {my$r;$r.='('.5x$_.')*'for@_;for($i=1000;;$i){return$i if(5x$i)!~/^$r$/}} 
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64
sub largest {
# 345678 1 2345678 2 2345678 3 2345678 4 2345678 5 2345678 6 234
my$r;$r.='('.5x$_.')*'for@_;for($i=1000;(5x$i)=~/^$r$/;$i){}$i
}

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59
sub largest {
# 345678 1 2345678 2 2345678 3 2345678 4 2345678 5 23456789
@r=map'('.5x$_.')*',@_;for($=**=2;(5x$=)=~/^@r$/x;$=){}$=
}

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Re: Golf: Buying with exact change
by blokhead (Monsignor) on Feb 22, 2005 at 17:28 UTC

As the other monks observed, the most concise solutions will probably use regular expressions. When looking only at the length of strings, the set of amounts you can represent using $6, $9, and $20 bills is /^(.{6})*(.{9})*(.{20})*$/, or equivalently for our purposes, /^(.{6}.{9}.{20})*$/. Now, to get the function arguments into such a regex and find the longest string that doesn't match is the creative part...
Here are my two 48character solutions:
$"="}1{";$_=1x999;chop while/^(1{@_})*$/;length
$"="}1{";(grep{(1x$_)!~/^(1{@_})*$/}1..999)[1]

And if it weren't for a bug I just found in alternations within negative lookaheads, I'm pretty sure this would be equivalent, giving me 46 characters:
$"="}1{";1x999=~/(1*?)(?!(1{@_})*$)/;length$'

They all use basically the same idea, can you think of a different approach? As I mentioned in the original post, these work for all instances where (a) the solution is 999 or less, and also (b) the arguments are each less than 32k (because they are used in regex quantifiers).
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Re: Golf: Buying with exact change
by fergal (Chaplain) on Feb 22, 2005 at 06:19 UTC

That's very clever but you do appear to have chosen bounds that trivialise the problem. Try largest(1000, 1001, 1003) 100,000 isn't large enough, try 1,000,000 and then go make a cup of tea (actually you could probably grow a cup of tea in the time this will take)
Update Hmm.... the original did say not to worry about efficiency but it's not hard to produce examples with fairly small numbers that will run for years, maybe even centuries.
I've been waiting more than 7 minutes just to find out if 999999 is possible with largest(1000, 1001, 1002, 1003, 1004, 1005) and there's no sign of an answer yet.
Update again I killed it after 100 min of CPU time (1.7Ghz Pentium) and it still hadn't figured out if 999,999 was doable or not.  [reply] [d/l] 

sub largest { 43 }
It works for largest(6,9,20), but one instance is a trivial subset of the problem space. Unreasonable.
The set of instances whose solutions are less than 1000.. 10000.. is that a "reasonable" subset? The idea is to find a concise algorithm to solve this particular problem, and to solve all these instances correctly, (a) you must actually carry out some sort of computation, and (b) your algorithm must probably be correct. To me, that's reasonable enough.
 [reply] [d/l] 

Here's a solution with no limits (beyond the memory used by memoize). It answers largest(1000, 1001, 1002, 1003, 1004, 1005) in 19 secs (the answer is 199,999 by the way). If you turn off memoize it takes much longer, although probably just days rather than centuries, It's based on a previous perlmonks challenge: How to generate restricted partitions of an integer.
It's definitely not golf! Is there a golfish answer that runs in similar time?
I'm going to bed now so I won't explain in detail. Hopefully someone will have posted a more golfish solution by the time I get to work tomorrow and I won't have to bother.
 [reply] [d/l] 

I'm justpointing out that the regex solution is terrible once the numbers start to get large. Efficiency is not the primary concern in golf but a soution should terminate before the sun dies :)
Another problem I have with the regex solution is that you don't know how much is enough for any given input. I'd say twice the largest 2 number would do it but I'm not sure, there will be complications when some of the numbers share factors.
It's an interesting puzzle though, it has a very simple solution when there's only 2 numbers:
largest(m, n) = (m  1) * (n  1)  1 
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Re: Golf: Buying with exact change
by Anonymous Monk on Feb 22, 2005 at 10:28 UTC

Not a golfing solution, but one that doesn't assume an upper bound (although it will loop forever if there's no finite answer):
use strict;
use warnings;
use Test::More 'no_plan';
$ = 1;
my %cache;
sub exact {
my ($amount, @bills) = @_;
return $cache{$amount} if defined $cache{$amount};
return 1 if !$amount;
return 0 if $amount < 0;
foreach my $bill (@bills) {
return $cache{$amount} = 1 if exact($amount  $bill, @bills)
}
return $cache{$amount} = 0;
}
sub largest {
%cache = ();
my @bills = @_;
my $smallest = $bills [0];
return 1 if $smallest == 1;
my $start = 0;
my $prev = 0;
for (my $t = 1; 1; $t++) {
if (exact($t, @bills)) {
if ($prev) {
if ($t  $start + 1 == $smallest) {return $start  1}
}
else {$start = $t}
$prev = 1;
}
else {
$prev = 0;
}
}
}
is (largest(6,9,20), 43);
is (largest(3,5), 7);
is (largest(6,11,20), 27);
is (largest(17,18,19,20), 101);
is (largest(2,3,5,10), 1);
is (largest(1000,1001,1002,1003,1004,1005), 199999);
__END__
ok 1
ok 2
ok 3
ok 4
ok 5
ok 6
1..6
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Golfed, I get you down to 140, assuming the inputs are ordered smallest to largest. Interestingly enough, if you have the definedor patch, I get down to 122. If the inputs aren't ordered, add 13 characters to both solutions.
Without //= patch:
@x=@_;my%c;$e=sub{my$v=pop;exists$c{$v}?$c{$v}:$c{$v}=$v<0?0:$v==0gr
+ep&$e($v$_),@x};$t=$s=0;{&$e(++$t)?$t$s>=$x[0]&&last:($s=$t);redo}$
+s
With //= patch:
@x=@_;my%c;$e=sub{my$v=pop;$c{$v}//=$v<0?0:$v==0grep&$e($v$_),@x};$
+t=$s=0;{&$e(++$t)?$t$s>=$x[0]&&last:($s=$t);redo}$s

Update: Actually, the inputs don't have to be ordered. It just means that the algorithm will take a little longer, but it will get the right results. Also, drop a character by reordering the assignment to $c{$v}. The //= patched version looks like:
@x=@_;my%c;$e=sub{my$v=pop;$c{$v}//=$v==0$v>0&&grep&$e($v$_),@x};$t
+=$s=0;{&$e(++$t)?$t$s>=$x[0]&&last:($s=$t);redo}$s

Update: Rewriting the redoloop as a Cstyle forloop drops to 111 characters for the //= patched version.
@x=@_;my%c;$e=sub{my$v=pop;$c{$v}//=!$v$v>0&&grep&$e($v$_),@x};for(
+$t=$s=0;$t$s<$x[0];&$e(++$t)or$s=$t){}$s

Being right, does not endow the right to be rude; politeness costs nothing. Being unknowing, is not the same as being stupid. Expressing a contrary opinion, whether to the individual or the group, is more often a sign of deeper thought than of cantankerous belligerence. Do not mistake your goals as the only goals; your opinion as the only opinion; your confidence as correctness. Saying you know better is not the same as explaining you know better.
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Re: Golf: Buying with exact change
by Anonymous Monk on Feb 22, 2005 at 09:18 UTC

Or, when a $1 denomination is used, all values can be represented. In this case, the solution is undefined, so you can return anything you like (or even loop forever).
Why is the solution undefined if a $1 denomination is used? I'd say there's an obvious answer: 0.
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Re: Golf: Buying with exact change
by Anonymous Monk on Feb 22, 2005 at 09:14 UTC

The problem was phrased in terms of blackmarket kidneys that come in packages of 6, 9, or 20.
I'm so glad they fixed the organ donor shortage.
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Re: Golf: Buying with exact change
by blazar (Canon) on Feb 23, 2005 at 09:30 UTC

I've not followed this thread in detail so I don't know if it's already been mentioned, but this is indeed a well known mathematical (combinatorial) problem, and a "hard" one, actually called "the money changing problem".
The problem is discussed in some detail e.g. in the beautiful book generatingfunctionology by Prof. Wilf, available for download from http://www.math.upenn.edu/%7Ewilf/DownldGF.html
To quote from Wilf:
There are no general `formulas' for the conductor if M >= 3, and no good algorithms for calculating it if M >= 4.
(M is the number of "changes" and the "conductor" is the smallest quantity N such that all n>=N can be represented as sums of the changes. In this case the latter must be of course coprime.)  [reply] 

I find this claim slightly bizarre. I downloaded the PDF and read Wilf's description of the problem (he's looking for a number 1 bigger than the largest function described above). The following fairly obvious code solves the problem for arbitrary M in worst case time proportional to O(M*n*n) where n is the size of the smallest denomination. (If there are only 2 denominations then this will be O(M*n).)
#! /usr/bin/perl w
use strict;
print largest(@ARGV)."\n";
sub largest {
# Everything works mod $least
my $least = min(@_);
# We start off not knowing how to do anything.
my @first = map {undef} 1..$least;
# For technical reasons this needs to be 0 now,
# and $least later.
$first[0] = 0;
for my $num (@_) {
my @in_process = grep {defined($first[$_])} 0..($least  1);
while (@in_process) {
my $to_process = shift @in_process;
my $value = $first[$to_process] + $num;
my $mod = $value % $least;
if (not defined($first[$mod]) or $value < $first[$mod]) {
$first[$mod] = $value;
print " $value = $mod (mod $least)\n";
push @in_process, $mod;
}
}
}
$first[0] = $least;
if (grep {not defined($_)} @first) {
# We didn't reach one...
return undef;
}
else {
my $worst = max(@first);
return $worst  $least;
}
}
sub min {
my $min = shift;
for (@_) {
$min = $_ if $_ < $min;
}
return $min;
}
sub max {
my $max = shift;
for (@_) {
$max = $_ if $_ > $max;
}
return $max;
}
I can only guess that he's measuring complexity in terms of the number of bits required to represent the numbers in the denomination. So he'd consider my solution as taking exponential time.  [reply] [d/l] 

 For those of us "math challanged" you denominations must not have any common factors. i.e. all even, all multiples of some
N. Otherwise there is an infinite number of prices which you can't buy:(

In a more obscure vein. A closed solution for arbitrary M has been shown to be NPhard.
starbolin
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Re: Golf: Buying with exact change
by aufflick (Deacon) on Feb 23, 2005 at 02:58 UTC

I love the use of black on black as the new rot13!!  [reply] 
Re: Golf: Buying with exact change
by Solo (Deacon) on Feb 25, 2005 at 21:55 UTC

UPDATE: Yuk. Note to self: math at 5PM Friday is a bad idea.
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