I would like to find the most efficient way to fill a vehicle, basically the different combinations to satisfy the vehicles capacity

Question

I am trying to write an algorithm that will 'fill' a vehicle to it's capacity based on different input combinations. The problem is solved, but it is far too slow to be usable for a greater amount of combinations.

For example: I have a vehicle with a capacity of 10, I have different combinations of types(attributes) of seats that can fill the vehicle differently(ambulatory, or normal seat capacity of 1 is default). Also, each combination that is not equal to 10(greater than 10 is removed) will be filled in with ambulatory capacity, or just a normal seat. Like so:

const a = [ {
 name: 'wheelchair',
 capacity: 0,
 total: 3
}, {
  name: 'walker',
  capacity: 2,
  total: 5
  }, {
  name: 'service animal',
  capacity: 2,
  total: 5
}];

It is also worth noting in the example above that the wheelchair can only be added 3 times per combination because it has a total of 3(max). The 0 capacity for the wheelchair is an example of a designated spot for this particular attribute that doesn't take up any of the other ambulatory seats.

I have tried a few different approaches for this and my algorithm works fine for this particular combination, or even if I add a few more. But, if I add an attribute with a total of 10 and capacity of one, this will increase the total possibilities by an order of magnitude and slow down the algorithm immensely. In my approach I am finding the different permutations and then filtering out the duplicates to find the combinations, if there were a way to find only the combinations, perhaps it would reduce the calculation load, but I can't think of a way. I have a specific way the output needs to look, which is the output at the bottom however, I have control over the input and can change if necessary. Any ideas or help is greatly appreciated.

This code was modified from this answer https://stackoverflow.com/a/21640840/6025994

  // the power set of [] is [[]]
  if(arr.length === 0) {
      return [[]];
  }

  // remove and remember the last element of the array
  var lastElement = arr.pop();

  // take the powerset of the rest of the array
  var restPowerset = powerSet(arr);


  // for each set in the power set of arr minus its last element,
  // include that set in the powerset of arr both with and without
  // the last element of arr
  var powerset = [];
  for(var i = 0, len = restPowerset.length; i < len; i++) {

      var set = restPowerset[i];

      // without last element
      powerset.push(set);

      // with last element
      set = set.slice(); // create a new array that's a copy of set
      set.push(lastElement);
      powerset.push(set);
  }

  return powerset;
};

var subsetsLessThan = function (arr, number) {
  // all subsets of arr
  var powerset = powerSet(arr);

  // subsets summing less than or equal to number
  var subsets = new Set();
  for(var i = 0, len = powerset.length; i < len; i++) {

      var subset = powerset[i];

      var sum = 0;
      const newObject = {};
      for(var j = 0, len2 = subset.length; j < len2; j++) {
          if (newObject[subset[j].name]) {
            newObject[subset[j].name]++;
          } else {
            newObject[subset[j].name] = 1;
          }
          sum += subset[j].seat;
      }
      const difference = number - sum;

      newObject.ambulatory = difference;

      if(sum <= number) {
          subsets.add(JSON.stringify(newObject));
      }
  }

  return [...subsets].map(subset => JSON.parse(subset));
};

const a = [{
  name: 'grocery',
  capacity: 2,
  total: 5
}, {
  name: 'wheelchair',
  capacity: 0,
  total: 3
}];
const hrStart = process.hrtime();
const array = [];

for (let i = 0, len = a.length; i < len; i++) {
  for (let tot = 0, len2 = a[i].total; tot < len2; tot++) {
    array.push({
      name: a[i].name,
      seat: a[i].capacity
    });
  }
}
const combinations = subsetsLessThan(array, 10);
const hrEnd = process.hrtime(hrStart);
// for (const combination of combinations) {
//   console.log(combination);
// }

console.info('Execution time (hr): %ds %dms', hrEnd[0], hrEnd[1] / 1000000)

Expected results are all combinations of results passed in to be less than vehicle capacity, so it is a combination less than sum algorithm essentially. For instance, the expected result of the code I have posted would be -->

[{"ambulatory":10},{"wheelchair":1,"ambulatory":10},{"wheelchair":2,"ambulatory":10},{"wheelchair":3,"ambulatory":10},{"grocery":1,"ambulatory":8},{"grocery":1,"wheelchair":1,"ambulatory":8},{"grocery":1,"wheelchair":2,"ambulatory":8},{"grocery":1,"wheelchair":3,"ambulatory":8},{"grocery":2,"ambulatory":6},{"grocery":2,"wheelchair":1,"ambulatory":6},{"grocery":2,"wheelchair":2,"ambulatory":6},{"grocery":2,"wheelchair":3,"ambulatory":6},{"grocery":3,"ambulatory":4},{"grocery":3,"wheelchair":1,"ambulatory":4},{"grocery":3,"wheelchair":2,"ambulatory":4},{"grocery":3,"wheelchair":3,"ambulatory":4},{"grocery":4,"ambulatory":2},{"grocery":4,"wheelchair":1,"ambulatory":2},{"grocery":4,"wheelchair":2,"ambulatory":2},{"grocery":4,"wheelchair":3,"ambulatory":2},{"grocery":5,"ambulatory":0},{"grocery":5,"wheelchair":1,"ambulatory":0},{"grocery":5,"wheelchair":2,"ambulatory":0},{"grocery":5,"wheelchair":3,"ambulatory":0}]

Answer 1

There is one trick you can use to improve your algorithm, called backtracking: If you reach an impossible path, e.g. 5 -> 6, then you don't have to continue searching there, as 5 + 6 is already greater 10. Through that you can eliminate a lot of combinations.

   function* combineMax([current, ...rest], max, previous = {}) {
     // Base Case:  if there are no items left to place, end the recursion here
     if(!current) { 
       // If the maximum is reached exactly, then this a valid solution, yield it up
       if(!max) yield previous; 
       return;
     }

     // if the "max" left is e.g. 8, then the grocery with "seat" being 2 can only fit in 4 times at max, therefore loop from 0 to 4
     for(let amount = 0; (!current.seat || amount <= max / current.seat) && amount <= current.total; amount++) {
       // The recursive call
       yield* combineMax(
        rest, // exclude the current item as that was  used already
        max - amount * current.seat, // e.g. max was 10, we take "seat: 2" 3 times, then the max left is "10 - 2 * 3"
        { ...previous, [current.name]: amount } // add the current amount
       );
     }
   }

   const result = [...combineMax([
    { name: 'grocery', seat: 2, total: Infinity }, 
    { name: 'wheelchair', seat: 0, total: 3 },
    { name: 'ambulatory equipment', seat: 1, total: Infinity },
     //...
   ], 10)];