Knapsack algorithm with an additional property

Question

When there's 1 property, I do understand what's going on in there. I'm having a problem with understanding knapsack problem when there's more than 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. 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

You are trying to do something impossible - that's your problem.

When you are adding to the number of dimensions, your items are getting additional properties. So, instead of a left, column side of a table(prop1), you have rectangle side(prop1 x prop2) or block side (prop1 x prop2 x prop3) and so on. But the existing constraints, that define the upper, row side of the table, should have the same number of dimensions. Not only one dimension!. So, you will never be able to put two-property problem into a 3-dimensional block, you need 4D block instead. 6D block for 3 properties and so on.

Answer 2

Say you are using a three dimension table: A[x][y][z]=k, x: sum 1st property; y: sum 2nd property; z: sum 3rd property; k: minimal cost (maximal reward, which I prefer using reward)

So you iterate over items. Say current item is (p1, p2, p3, reward) (reward = - cost). for each (x,y,z,k), your update formula:

A[x+p1][y+p2][z+p3] = max(A[x+p1][y+p2][z+p3], A[x+p1][y+p2][z+p3] + reward)

If the 1st term on RHS is greater, on slot A[x+p1][y+p2][z+p3], the configuration of knapsack is remain still; otherwise, you update the knapsack by that of A[x+p1][y+p2][z+p3] plus the current item.

Hope this cut things clear.