Solving the NP-complete problem in XKCD

Answer 1

Read up on the Knapsack Problem.

Answer 2

It gives all the permutations of the solutions, but I think I'm beating everyone else for code size.

item(X) :- member(X,[215, 275, 335, 355, 420, 580]).
solution([X|Y], Z) :- item(X), plus(S, X, Z), Z >= 0, solution(Y, S).
solution([], 0).

Solution with swiprolog:

?- solution(X, 1505).

X = [215, 215, 215, 215, 215, 215, 215] ;

X = [215, 355, 355, 580] ;

X = [215, 355, 580, 355] ;

X = [215, 580, 355, 355] ;

X = [355, 215, 355, 580] ;

X = [355, 215, 580, 355] ;

X = [355, 355, 215, 580] ;

X = [355, 355, 580, 215] ;

X = [355, 580, 215, 355] ;

X = [355, 580, 355, 215] ;

X = [580, 215, 355, 355] ;

X = [580, 355, 215, 355] ;

X = [580, 355, 355, 215] ;

No

Answer 3

Here's a solution with F#:

#light

type Appetizer = { name : string; cost : int }

let menu = [
    {name="fruit"; cost=215}
    {name="fries"; cost=275}
    {name="salad"; cost=335}
    {name="wings"; cost=355}
    {name="moz sticks"; cost=420}
    {name="sampler"; cost=580}
    ] 

// Choose: list<Appetizer> -> list<Appetizer> -> int -> list<list<Appetizer>>
let rec Choose allowedMenu pickedSoFar remainingMoney =
    if remainingMoney = 0 then
        // solved it, return this solution
        [ pickedSoFar ]
    else
        // there's more to spend
        [match allowedMenu with
         | [] -> yield! []  // no more items to choose, no solutions this branch
         | item :: rest -> 
            if item.cost <= remainingMoney then
                // if first allowed is within budget, pick it and recurse
                yield! Choose allowedMenu (item :: pickedSoFar) (remainingMoney - item.cost)
            // regardless, also skip ever picking more of that first item and recurse
            yield! Choose rest pickedSoFar remainingMoney]

let solutions = Choose menu [] 1505

printfn "%d solutions:" solutions.Length 
solutions |> List.iter (fun solution ->
    solution |> List.iter (fun item -> printf "%s, " item.name)
    printfn ""
)

(*
2 solutions:
fruit, fruit, fruit, fruit, fruit, fruit, fruit,
sampler, wings, wings, fruit,
*)

Answer 4

You've got all the correct combinations now, but you're still checking many more than you need to (as evidenced by the many permutations your result shows). Also, you're omitting the last item that hits the 15.05 mark.

I have a PHP version that does 209 iterations of the recursive call (it does 2012 if I get all permutations). You can reduce your count if right before the end of your loop, you pull out the item you just checked.

I don't know CF syntax, but it will be something like this:

        <cfelse>
            <!--- less than 15.50 --->
            <!--<cfoutput>#arguments.currentCombo# < 15.05 (traversing)</cfoutput><br />-->
            <cfset found = testCombo(CC, CT, arguments.apps) />
        ------- remove the item from the apps array that was just checked here ------
    </cfif>
</cfloop>

EDIT: For reference, here's my PHP version:

<?
  function rc($total, $string, $m) {
    global $c;

    $m2 = $m;
    $c++;

    foreach($m as $i=>$p) {
      if ($total-$p == 0) {
        print "$string $i\n";
        return;
      }
      if ($total-$p > 0) {
        rc($total-$p, $string . " " . $i, $m2);
      }
      unset($m2[$i]);
    }
  }

  $c = 0;

  $m = array("mf"=>215, "ff"=>275, "ss"=>335, "hw"=>355, "ms"=>420, "sp"=>580);
  rc(1505, "", $m);
  print $c;
?>

Output

 mf mf mf mf mf mf mf
 mf hw hw sp
209

EDIT 2:

Since explaining why you can remove the elements will take a little more than I could fit in a comment, I'm adding it here.

Basically, each recursion will find all combinations that include the currently search element (e.g., the first step will find everything including at least one mixed fruit). The easiest way to understand it is to trace the execution, but since that will take a lot of space, I'll do it as if the target was 6.45.

MF (2.15)
  MF (4.30)
    MF (6.45) *
    FF (7.05) X
    SS (7.65) X
    ...
  [MF removed for depth 2]
  FF (4.90)
    [checking MF now would be redundant since we checked MF/MF/FF previously]
    FF (7.65) X
    ...
  [FF removed for depth 2]
  SS (5.50)
  ...
[MF removed for depth 1]

At this point, we've checked every combination that includes any mixed fruit, so there's no need to check for mixed fruit again. You can use the same logic to prune the array at each of the deeper recursions as well.

Tracing through it like this actually suggested another slight time saver -- knowing the prices are sorted from low to high means that we don't need to keep checking items once we go over the target.

Answer 5

In python.
I had some problems with "global vars" so I put the function as method of an object. It is recursive and it calls itself 29 times for the question in the comic, stopping at the first successful match

class Solver(object):
    def __init__(self):
        self.solved = False
        self.total = 0
    def solve(s, p, pl, curList = []):
        poss = [i for i in sorted(pl, reverse = True) if i <= p]
        if len(poss) == 0 or s.solved:
            s.total += 1
            return curList
        if abs(poss[0]-p) < 0.00001:
            s.solved = True # Solved it!!!
            s.total += 1
            return curList + [poss[0]]
        ml,md = [], 10**8
        for j in [s.solve(p-i, pl, [i]) for i in poss]:
            if abs(sum(j)-p)<md: ml,md = j, abs(sum(j)-p)
        s.total += 1
        return ml + curList


priceList = [5.8, 4.2, 3.55, 3.35, 2.75, 2.15]
appetizers = ['Sampler Plate', 'Mozzarella Sticks', \
              'Hot wings', 'Side salad', 'French Fries', 'Mixed Fruit']

menu = zip(priceList, appetizers)

sol = Solver()
q = sol.solve(15.05, priceList)
print 'Total time it runned: ', sol.total
print '-'*30
order = [(m, q.count(m[0])) for m in menu if m[0] in q]
for o in order:
    print '%d x %s \t\t\t (%.2f)' % (o[1],o[0][1],o[0][0])

print '-'*30
ts = 'Total: %.2f' % sum(q)
print ' '*(30-len(ts)-1),ts

Output:

Total time it runned:  29
------------------------------
1 x Sampler Plate   (5.80)
2 x Hot wings       (3.55)
1 x Mixed Fruit       (2.15)
------------------------------
               Total: 15.05

Answer 6

Learning from @rcar's answer, and another refactoring later, I've got the following.

As with so many things I code, I've refactored from CFML to CFScript, but the code is basically the same.

I added in his suggestion of a dynamic start point in the array (instead of passing the array by value and changing its value for future recursions), which brought me to the same stats he gets (209 recursions, 571 combination price checks (loop iterations)), and then improved on that by assuming that the array will be sorted by cost -- because it is -- and breaking as soon as we go over the target price. With the break, we're down to 209 recursions and 376 loop iterations.

What other improvements could be made to the algorithm?

function testCombo(minIndex, currentCombo, currentTotal){
    var a = 0;
    var CC = "";
    var CT = 0;
    var found = false;

    tries += 1;
    for (a=arguments.minIndex; a <= arrayLen(apps); a++){
        combos += 1;
        CC = listAppend(arguments.currentCombo, apps[a].name);
        CT = arguments.currentTotal + apps[a].cost;
        if (CT eq 15.05){
            //print current combo
            WriteOutput("<strong>#CC# = 15.05</strong><br />");
            return(true);
        }else if (CT gt 15.05){
            //since we know the array is sorted by cost (asc),
            //and we've already gone over the price limit,
            //we can ignore anything else in the array
            break; 
        }else{
            //less than 15.50, try adding something else
            found = testCombo(a, CC, CT);
        }
    }
    return(found);
}

mf = {name="mixed fruit", cost=2.15};
ff = {name="french fries", cost=2.75};
ss = {name="side salad", cost=3.35};
hw = {name="hot wings", cost=3.55};
ms = {name="mozarella sticks", cost=4.20};
sp = {name="sampler plate", cost=5.80};
apps = [ mf, ff, ss, hw, ms, sp ];

tries = 0;
combos = 0;

testCombo(1, "", 0);

WriteOutput("<br />tries: #tries#<br />combos: #combos#");