打印在背包里的物品

Question

Suppose you are a thief and you invaded a house. Inside you found the following items:

A vase that weights 3 pounds and is worth 50 dollars.
A silver nugget that weights 6 pounds and is worth 30 dollars.
A painting that weights 4 pounds and is worth 40 dollars.
A mirror that weights 5 pounds and is worth 10 dollars.

Solution to this Knapsack problem of size 10 pounds is 90 dollars .

Table made from dynamic programming is :-

Now i want to know which elements i put in my sack using this table then how to back track ??

Answer 1

From your DP table we know f[i][w] = the maximum total value of a subset of items 1..i that has total weight less than or equal to w.

We can use the table itself to restore the optimal packing:

def reconstruct(i, w):  # reconstruct subset of items 1..i with weight <= w
                        # and value f[i][w]
  if i == 0: 
      # base case
      return {}
  if f[i][w] > f[i-1][w]:
      # we have to take item i
      return {i} UNION reconstruct(i-1, w - weight_of_item(i))
  else:
      # we don't need item i
      return reconstruct(i-1, w)

Answer 2

Using a loop :

   for (int n = N, w = W; n > 0; n--)
            {
                if (sol[n][w] != 0)
                {
                    selected[n] = 1;
                    w = w - wt[n];
                }
                else
                    selected[n] = 0;
            }
            System.out.print("\nItems with weight ");
            for (int i = 1; i < N + 1; i++)
                if (selected[i] == 1)
                    System.out.print(val[i] +" ");

Answer 3

I have an iterative algorithm inspired by @NiklasB. that works when a recursive algorithm would hit some kind of recursion limit.

def reconstruct(i, w, kp_soln, weight_of_item):
    """
    Reconstruct subset of items i with weights w. The two inputs
    i and w are taken at the point of optimality in the knapsack soln

    In this case I just assume that i is some number from a range
    0,1,2,...n
    """
    recon = set()
    # assuming our kp soln converged, we stopped at the ith item, so
    # start here and work our way backwards through all the items in
    # the list of kp solns. If an item was deemed optimal by kp, then
    # put it in our bag, otherwise skip it.
    for j in range(0, i+1)[::-1]:
        cur_val = kp_soln[j][w]
        prev_val = kp_soln[j-1][w]
        if cur_val > prev_val:
            recon.add(j)
            w = w - weight_of_item[j]
    return recon