Chapter 4: Problem 31

Show that the worst-case number of entries computed by the refined dynamic programming algorithm for the \(0-1\) Knapsack Problem is in \(\Omega\left(2^{n}\right) .\) Do this by considering the instance where \(W=2^{n}-2\) and \(w_{i}=2^{i-1}\) for \(1 \leq t \leq n\).

Short Answer

Expert verified

By considering a specific instance where \(W=2^{n}-2\) and \(w_{i}=2^{i-1}\) for \(1 \leq t \leq n\), it is clear that there are \(2^{n}\) different possible binary representations and thus \(2^{n}\) different possible sums of weights, each corresponding to a different subset of items that can be included in the knapsack. Each of these possibilities has to be computed and stored in the table by the dynamic programming algorithm, leading to at least \(2^{n}\) operations in the worst-case, showing that the algorithm is in \(\Omega\left(2^{n}\right)\).

Step by step solution

Understand the instance

We are considering a specific instance where the total weight \(W=2^{n}-2\) and each item has a weight \(w_{i}=2^{i-1}\) for \(1 \leq i \leq n\). In binary representation, it means each item corresponds to a unique bit in the binary representation of \(W\).

Determine possible sums of weights

Each item can either be included in the knapsack or not, thus we can get any sum of weights from 0 to \(W\) by deciding whether to include each item. There're \(2^{n}\) different ways, corresponding to \(2^{n}\) different binary representations from \(000...00\) to \(111...11\).

Explain how it's related to the algorithm

The dynamic programming algorithm computes and stores values for every possible sum of weights (every distinct subset of items), thus it has to compute at least \(2^{n}\) entries in the worst-case scenario. The size of the table to be computed is determined by exactly these possibilities.

Conclude the proof

Considering the specific nature of our given instance and the function of the dynamic programming algorithm, the algorithm will run \(2^{n}\) operations in the worst case. Therefore, the refined dynamic programming solution for the \(0-1\) Knapsack Problem is in \(\Omega\left(2^{n}\right)\).

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Show that the worst-case number of entries computed by the refined dynamic programming algorithm for the \(0-1\) Knapsack Problem is in \(\Omega\left(2^{n}\right) .\) Do this by considering the instance where \(W=2^{n}-2\) and \(w_{i}=2^{i-1}\) for \(1 \leq t \leq n\).

Short Answer

Step by step solution

Understand the instance

Determine possible sums of weights

Explain how it's related to the algorithm

Conclude the proof

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