Lecture No. 22
6.5.1 0/1 Knapsack Problem: Dynamic Programming Approach
For each i ≤ n and each w ≤ W, solve the knapsack problem for the first i objects when
the capacity is w. Why will this work? Because solutions to larger sub problems can be
built up easily from solutions to smaller ones. We construct a matrix V[0 . . . n, 0 . . .W].
For 1 ≤ i ≤ n, and 0 ≤ j ≤ W, V[i, j] will store the maximum value of any set of objects
{1, 2, . . . , i} that can fit into a knapsack of weight j. V[n,W] will contain the maximum
value of all n objects that can fit into the entire knapsack of weight W.
To compute entries of V we will imply an inductive approach. As a basis, V[0, j] = 0 for
0 ≤ j ≤ W since if we have no items then we have no value. We consider two cases:
Leave object i: If we choose to not take object i, then the optimal value will come about
by considering
how to fill a knapsack of size j with the remaining objects {1, 2, . . . , i - 1}. This is just
V[i - 1, j].
Take object i: If we take object i, then we gain a value of vi. But we use up wi of our
capacity. With the remaining j - wi capacity in the knapsack, we can fill it in the best
possible way with objects {1, 2, . . . , i - 1}. This is vi + V[i - 1, j - wi]. This is only
possible if wi _ j.
This leads to the following recursive formulation:
A naive evaluation of this recursive definition is exponential. So, as usual, we avoid re-
computation by making a table.
Example: The maximum weight the knapsack can hold is W is 11. There are five items
to choose from.
Their weights and values are presented in the following table:
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