Knapsack algorithm with 2 properties. How to implement that in a 3d array?

Question

I'm having a problem with understanding knapsack problem when there's more than 1 property. When there's 1 property,

I have to write a program that uses knapsack algorithm with a 2 properties. Teacher told us, It has to be done in a 3d array. Wrong implementation will lead to O(2^n) processing time. I can't imagine how would such array look like.

Let's say here's my input:

4 3 4 // number of records below, 1st property of backpack, 2nd property  of backpack
1 1 1 // 1st property, 2nd property, cost
1 2 2 // 1st property, 2nd property, cost
2 3 3 // 1st property, 2nd property, cost
3 4 5 // 1st property, 2nd property, cost

And the output would look like that:

4    // the cheapest sum of costs of 2 records
1 3  // numbers of these 2 records

The explanation of output: 2 sets of records fit's into 1'st line of input:

(1) - record number 1 and record number 3

  1 1 1
+ 2 3 3
-------
  3 4 4

(2) - record number 4

  3 4 5

Because 1st set of the records is the cheapest (4 < 5), we chose it. Not only I'll have to find out whether such set of records exists, I'll also have to find records I've summed.

But for now, I only need to understand, how will 3d array look like. Could some of You help me out with that and show, layer by layer, just like in my image, how would this look like? Thanks.

Answer 1

To convert from the algorithm to solve the 1-constraint knapsack problem with dynamic programming to solving the 2-constraint problem is very easy. For someone to draw a 3d image that shows you what's happening there I imagine is somewhat difficult. The algorithm, on the other hand, is quite easy:

I'll assume you want an exact-fit solution, and that you want to minimize cost, not maximize value (which I derived from your example). I've done neither of these variations before, so I can't guarantee there isn't an error.

1-constraint

The matrix for the 1-constraint knapsack problem is (item x weight) storing the value at each position.

The algorithm is basically:

// WL = backpack weight limit
A[0, 0] = 0
for w = 1, 2, ..., WL // weight
  A[0, w] = INFINITY
for i = 1, 2, ..., n // items
  for w = 0, 1, ..., WL // weight
    // wi and ci are the weight and cost of the item respectively
    // A[i-1, w-wi] + ci = 0 when wi > w
    A[i, w] = min(A[i-1, w], A[i-1, w-wi] + ci)

2-constraint

Now to extend this to the include another constraint is just: (let's say size is the other constraint)

The matrix would be (item x weight x size).

// WL = backpack weight limit
// SL = backpack size limit
for w = 0, 1, ..., WL // weight
  for s = 0, 1, ..., SL // size
    A[0, w, s] = INFINITY
A[0, 0, 0] = 0
for i = 1, 2, ..., n // items
  for w = 0, 1, ..., WL // weight
    for s = 0, 1, ..., SL // size
      // wi, si and ci are the weight, size and cost of the item respectively
      // A[i-1, w-wi, s-si] + ci = 0 when wi > w or si > s
      A[i, w, s] = min(A[i-1, w, s], A[i-1, w-wi, s-si] + ci)