Knapsack

Fatma Kahveci
Coding
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  • For given weights and values of $n$ items, we have to choose sets of weights that can fill up the maximum capacity $w$ in a bag of capacity $W$.

  • This set should contain a maximum number of items.

  • The expected output is an integer with a count of a maximum number of items.

  • Applications

    • Resource allocation

  • Approaches

    • Greedy
    • DP
    • Brute force

0-1 Knapsack

# O(2^n) as there are redundant subproblems
def knapSack(W, wt, val, n):

    if n == 0 or W == 0:
        return 0
 
    if wt[n-1] > W:
        return knapSack(W, wt, val, n-1)
    
    # pick or don't pick
    return max( knapSack(W - wt[n-1], wt, val, n-1) + val[n-1],
                knapSack(W, wt, val, n-1) )
 
val = [60, 100, 120]
wt = [10, 20, 30]
W = 50
n = len(val)
print(knapSack(W, wt, val, n))

0-1 Knapsack DP

  • You cannot break an item
    • pick OR
    • don’t pick
def knapSack(W, w, val):
    num_items = len(val)
    m = {}
    for c in range(W+1):
        m[(0,c)] = 0

    for i in range(1, num_items+1):
        for c in range(W+1):
            if w[i-1] <= c:
                m[(i,c)] = max(m[i-1,c], val[i-1]+m[(i-1,c-w[i-1])])
            else:
                m[(i,c)] = m[(i-1,c)]

    return m[(num_items,W)]

print(knapSack(W=50, w=[10, 20, 30], val=[60, 100, 120]))