Finding the best combination of a complex collection of items?

Question

I have the following objects:

enum Slot
{
   HANDS, LEGS, CHEST, HEAD, FEET;
}

class Clothing
{
   // The slot this piece of clothing is worn on.
   Slot s;
   // The color of the clothing, used for `gradeOutfit`
   Color c;
}

class Person
{
   Map<Slot, Clothing> body;

   // look through his outfit and give a score
   // for how well he looks
   int gradeOutfit()
   {
      return ...
   }
}

I have one Person object and a collection of Clothing. This collection has many Clothing objects of each Slot. For example, it might look like this:

MyCloset = { GREEN_HAT, RED_VEST, BLACK_VEST, 
BLUE_JEANS, BROWN_PANTS, RED_SHOES, BLACK_HAT, BLUE_GLOVES, PURPLE_VEST }

In the reality of my program, there are a lot more items than just these but this is just a simplified example.

Problem:

I need to find a combination of these clothes that lead to the highest gradeOutfit score. That means my Person will have to make sure he tries on every Clothing item with every other Clothing item (within limits, ex. it's impossible for two hats to be worn because both are for HEAD Slot). A Person cannot have their gradeOutfit called until they are wearing a Clothing item for every Slot.

I was thinking recursion is the best way to do this but then I think I'd get a stack overflow very fast if I had a decent amount of items. I tried doing it iteratively but I cannot seem to find a good easy way to loop through everything. My program basically looks like

Person p = new Person();
for (Clothing i : MyCloset)
{
    for (Clothing h : MyCloset)
    {
        if (i == h) continue;

        if (!p.isWearing(h.slot())
        {
            p.wear(h);
        }
    }

    int score = p.gradeOutfit();
}

But I know this is just a terrible approach. In order to ensure that every clothing item has been paired up with every other Clothing item, I would need so much more looping logic than just this. No matter what I try, it turns into spaghetti code. I also need to avoid looping over the same outfit twice and make sure that no outfit combination is forgotten about.

What is the best way to approach something like this?

Answer 1

This is an example of a mathematical optimization problem. You seem to already have the objective function (the function that calculates the gradeOutfit score - taking as an input five clothings, one per slot) and you need some constraints (e.g. each clothing in a combination of 5 belongs to a different slot). You need a Java solver library to do this. If your objective function is linear, a linear solver will do. As I have only experience with commercial solvers, I cannot recommend an open-source one, a list of options can be found here.

A simpler (but not extremely elegant) way, without a solver:

1. Create 5 sets of Clothing objects, one per slot (you can use Java HashSet for this).
2. Iterate over all combinations, each time taking one item from each of the 5 sets. You need n1 x n2 x n3 x n4 x n5 combinations, where ni is the number of clothing instances per slot.

It also seems to me that the gradeOutfit function should not be part of the Person class - as it is actually the property of an outfit, not a person (i.e. two persons with the same outfit have exactly the same scores). I 'd prefer to have an Outfit class and put it there.

Answer 2

You have very poorly created the data structure.

enum Slot
{
   HANDS, LEGS, CHEST, HEAD, FEET;
   numbers = new int[values.length()]
}

enum COLOR
{
   RED,BLUE,...;
}
enum Clothing {
    GREEN_HAT(HEAD,GREEN), ...; 

    Slot slot;
    Color color;
    public static Clothing (Slot slot, Color color){...}
}
class Outfit extends Map <Slot, Clothing> {
    countScore(){};
    public static Outfit(){
        //foreach slot this.put(slot, Clothing.values().get(0));
    }
}
... 
int n=slot.values.length()-1;
Outfit currentOutfit = new Outfit();
Outfit bestOutfit = new Outfit();
int currentActiveSlot = 0;
// make a cycle for combination of all Outfits

Answer 3

for an enum , you have to use the method "values()" to loop on it:

 For (clothe c: clothes.values())