CODEWITHSUNDEEP
algorithmdynamic-programmingpseudocodeknapsack-problem
Michalides
Michalides

Reputation: 172

Pseudo code Algorithm for Knapsack Problem with Two Constrains

I'm trying to solve following knapsack problem with two constrains.

What we know:

Constrains:

Can anyone give me some advice about algorithm i should use, pseudo code or good article?

UPDATE:

Important thing I forgot to mention is that I also need to know which items I putted in the bag.

Upvotes: 0

Views: 10204

Answers (2)

Average Enjoyer
Average Enjoyer

Reputation: 3

Non-recursive approach

# N: number of items, W: max weight of knapsack, F: max amount of fragile items
# if item doesn't fit, then skip
# if it fits, check if it's worth to take it

dp[N+1][W+1][F+1] # filled with zeros
benefit, weight, fragility = [] # arrays filled with respective item properties

for n in range(1, N + 1):
   for w in range(1, W + 1):
      for f in range(0, F + 1):
         if weight[n-1] > w or fragility[n-1] > f:
            dp[n][w][f] = dp[n-1][w][f]
         else: 
            dp[n][w][f] = max(dp[n-1][w][f],
                              dp[n-1][w-weight[n-1]][f-fragility[n-1]] + benefit[n-1])

print(dp[N][W][F]) # prints optimal knapsack value

Listing items

knapsack = [] # indexes of picked items
n = N
w = W
f = F

while n > 0:
   if dp[n][w][f] != dp[n-1][w][f]:
      knapsack.append(n-1)
      w -= weight[n-1]
      f -= fragility[n-1]
   n -= 1

print(knapsack) # prints item indexes

Upvotes: 0

Photon
Photon

Reputation: 2777

It looks like a modification to knapsack would solve it.

Let's say we have N items, maximum knapscak weight is W, and max amount of fragile items is F

let's define our dp table as 3-dimensional array dp[N+1][W+1][F+1]

Now dp[n][w][f] stores maximum value we can get if we fill knapsack with some subset of items from first n items, having weight of exacly w and having exacly f fragile items.

frop dp[n][w][f] we can move to states:

  • dp[n+1][w][f] if we skip n+1 th item
  • dp[n+1][w + weight(n+1)][f + isFragile(n+1)] if we take n+1 th item

so pseudocde:

dp[N+1][W+1][F+1] // memo table, initially filled with -1

 int solve(n,w,f)
{
    if(n > N)return 0;
    if(dp[n][w][f] != -1) return dp[n][w][f];

    dp[n][w][f] = solve(n+1,w,f); //skip item
    if(w + weight(n) <= W && f + isFragile(n) <=F)
    dp[n][w][f] = max(dp[n][w][f], value(n) + solve(n+1, w + weight(n), f + isFragile(n)));

    return dp[n][w][f]
}

print(solve(1,0,0))

Getting the actual subset is also not difficult:

vector<int> getSolution(n,w,f)
{   
    int optimalValue = solve(n,w,f);
    vector<int>answer; //just some dynamic array / arrayList

    while(n <= N)
    {
        if(solve(n+1,w,f) == optimalValue)n++; //if after skipping item we can still get optimal answer we just skip it
        else //otherwise we cant so current item must be taken
        {
            int new_w = w + weight(n);
            int new_f = f + isFragile(n);
            answer.push_back(n); //so we just save its index, and update all values
            optimalValue -= value(n);
            n++;
            w = new_w;
            f = new_f;
        }
    }
    return answer;
}

Upvotes: 1

Related Questions