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I was thinking,

I wanted to do a variation on the Knapsack Problem.

Imagine the original problem, with items with various weights/value.

My version will, along with having the normal weights/values, contain a "group" value.

eg. Item1[5kg, $600, electronic] Item2[1kg, $50, food]

Now, having a set of items like this, how would I code up the knapsack problem to make sure that a maximum of 1 item from each "group" is selected.

Notes:

  1. You don't need to choose an item from that group
  2. There are multiple items in each group
  3. You're still minimizing weight, maximizing value
  4. The amount of groups are predefined, along with their values.

I'm just writing a draft of the code out at this stage, and I've chosen to use a dynamic approach. I understand the idea behind the dynamic solution for the regular knapsack problem, how do I alter this solution to incorporate these "groups"?

KnapSackVariation(v,w,g,n,W)
{
  for (w = 0 to W)
     V[0,w] = 0;
  for(i = 1 to n)
     for(w = 0 to W)
        if(w[i] <= w)
           V[i,w] = max{V[i-1, w], v[i] + V[i-1, w-w[i]]};
        else
           V[i,w] = V[i-1, w];
     return V[n,W];
}

That's what I have so far, need to add it so that it will remove all corresponding items from the group it is in each time it solves this.

Cœur
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    Add a group item to the state! – flight Oct 07 '11 at 07:05
  • What about solving it as a combinatorial optimization problem? For Each item you choose either an item or no item. You might want to use branch and bound search to solve it. – ysdx Oct 07 '11 at 07:23

2 Answers2

3

just noticed your question trying to find an answer to a question of my own. The problem you've stated is a well-known and well-studied problem called the Multiple Choice Knapsack Problem. If you google that you'll find all sorts of information, and I can also recommend this book: http://www.amazon.co.uk/Knapsack-Problems-Hans-Kellerer/dp/3642073115/ref=sr_1_1?ie=UTF8&qid=1318767496&sr=8-1, which dedicates a whole chapter to the problem. In the classic formulation of MCKP, you have to choose one item from each group. However, you can easily convert that version of the problem to your version by adding a dummy item to each group with profit and weight = 0, and the same algorithms will work. I would caution you against trying to adapt code for the binary knapsack problem to the MCKP with a few tweaks--this approach is likely to lead you to a solution whose performance degrades unacceptably as the number of items in each group increases.

Karen
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0

Assume
c[i] : The category of the ith element
V[i,w,S] : Maximum value of the knapsack such that it contains at max one item from each category in S

Recursive Formulation
V[i,w,S] = max(V[i-1,w,S],V[i,w-w[i],S-{c[i]}] + v[i])
Base Case
V[0,w,S] = -`infinity if w!=0 or S != {}`
Lrrr
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gibraltar
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