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How to fill a knapsack table when using recursive dynamic programming

* NOT HOMEWORK *

I have implemented the knapsack in python and am successfully getting the best value however I would like to expand the problem to fill a table with all appropriate values for a knapsack table of all weights and items.

I've implemented it in python which I'm new to so please advise me if theres anything I could improve upon however the concepts should work in any language.

values, weights, table = [], [], [[]]

def knapsack(i, W):
    global weights, values, table, counter
    if (i < 0):
        # Base case
        return 0
    if (weights[i] > W):
        # Recursion
        return knapsack(i - 1, W)
    else:
        # Recursion
        return max(knapsack(i - 1, W), values[i] + knapsack(i - 1, W - weights[i]))

def main():
    global values, weights, table
    W = int(input())
    values = list(map(int, input().split()))
    weights = list(map(int, input().split()))
    # initalise table with 0's
    table = [[0 for i in range(W)] for i in range(len(values))]
    for i in range(len(values)):
        for j in range(W):
            table[i][j] = 0
    # Fill table
    print("Result: {}".format(knapsack(len(values) - 1, W)))
    printKnapsack(W)

if __name__ == '__main__':
    main()

I also have this print table method which is unrelated but just so you can see what I'm outputting it as:

def printLine(W):
    print(" ",end="")
    for i in range(W + 1):
        print("-----",end="")
    print("")

def printKnapsack(W):
    global table
    print("\nKnapsack Table:")
    printLine(W)
    print("| k\w |", end="")
    for i in range(W):
        print("{0: >3} |".format(i + 1), end="")
    print("")
    printLine(W)
    for i in range(len(values)):
        print("|  {}  |".format(i+1), end="")
        for j in range(W):
            print("{0: >3} |".format(table[i][j]), end="")
        print("")
        printLine(W)

This is the sample input:

10
18 9 12 25
5 2 4 6

This is what is should output:

Result: 37

Knapsack Table:
 -------------------------------------------------------
| k\w |  1 |  2 |  3 |  4 |  5 |  6 |  7 |  8 |  9 | 10 |
 -------------------------------------------------------
|  1  |  0 |  0 |  0 |  0 | 18 | 18 | 18 | 18 | 18 | 18 |
 -------------------------------------------------------
|  2  |  0 |  9 |  9 |  9 | 18 | 18 | 27 | 27 | 27 | 27 |
 -------------------------------------------------------
|  3  |  0 |  9 |  9 | 12 | 18 | 21 | 27 | 27 | 30 | 30 |
 -------------------------------------------------------
|  4  |  0 |  9 |  9 | 12 | 18 | 25 | 27 | 34 | 34 | 37 |
 -------------------------------------------------------

I have tried multiple different lines in the knapsack(i, W) function to add elements to the table and I've drawn it out but I can't understand how the recursion is working well enough to figure out what indexes to put in to add the unravelled recursive call values to.

This is the method I have to fix.

def knapsack(i, W):
    global weights, values, table, counter
    if (i < 0):
        # Base case
        return 0
    if (weights[i] > W):
        # Recursion
        table[?][?] = ?
        return knapsack(i - 1, W)
    else:
        # Recursion
        table[?][?] = ?
        return max(knapsack(i - 1, W), values[i] + knapsack(i - 1, W - weights[i]))
like image 897
joshuatvernon Avatar asked Oct 19 '22 02:10

joshuatvernon


1 Answers

In your recursive algorithm you just can't get full filled table, because this step skip a lot:

return max(knapsack(i - 1, W), values[i] + knapsack(i - 1, W - weights[i]))

I can sudgest you this solution:

def knapsack(i, W):
    global weights, values, table, counter
    if (i < 0):
        # Base case
        return 0
    if (weights[i] > W):
        # Recursion
        table[i][W-1] = knapsack(i - 1, W)
    else:
        # Recursion
        table[i][W-1] = max(knapsack(i - 1, W), values[i] + knapsack(i - 1, W - weights[i]))
    return table[i][W-1]

In resulted table non-zero cells means that your algorithm step through here and got this intermediate solution. Also you can run your algorithm more than once with different input values and got full-filled table.

Hope this helped

like image 196
A. Chubarov Avatar answered Nov 01 '22 09:11

A. Chubarov