CSE 431531 Algorithms Design And Analysis Fall 2022 Homework

Cse 431531 Algorithms Design And Analysis Fall2022 Homework 4due

Consider the following version of Knapsack. Given are two weight limits W1 and W2, where W1 ≤ W2. Given are also n items (w1, c1), (w2, c2), ..., (wn, cn), where wi is the weight and ci the cost of the i-th item.

We want to find a subset of these items of the highest cost, where the subset weights at least W1 and at most W2. Give an O(nW2) algorithm for this problem. (Recall that the cost (respectively weight) of a subset is the sum of the costs (respectively weights) of the items in the subset.)

Problem For Above instruction

Develop an O(nW2) dynamic programming algorithm to solve the variant of the Knapsack problem described above, where the goal is to maximize the total cost of selected items with total weight between W1 and W2 inclusive. Additionally, provide a proof of correctness for your algorithm and analyze its time complexity.

Answer

Introduction

The classic 0/1 Knapsack problem involves selecting items to maximize total value without exceeding a weight limit. In this problem, the constraints are slightly modified: there are two weight bounds, W1 and W2, with W1 ≤ W2, and the goal is to select a subset of items such that the total weight falls within [W1, W2], and the total cost is maximized. The challenge is to efficiently find such a subset within the given bounds, specifically achieving an O(nW2) runtime.

Algorithm Description

The solution employs a dynamic programming approach that extends the classical Knapsack algorithm to handle the bounds W1 and W2. We define a DP table, dp, where dp[i][w] represents the maximum total cost achievable using the first i items with total weight exactly w.

The transition is as follows: for each item i, and each weight w, we decide whether to include item i or not:

  • If we do not include item i, then dp[i][w] = dp[i-1][w].
  • If we include item i, only if w - wi ≥ 0, then dp[i][w] = dp[i-1][w - wi] + ci.

We initialize dp[0][w] = 0 for all w and fill the table row by row. After processing all items, we look for the maximum cost among all weights w where W1 ≤ w ≤ W2.

Pseudocode

function KnapsackVariant(items, W1, W2):

n = number of items

initialize DP array: dp of size (n+1) x (W2+1) with -infinity

for w in 0 to W2:

dp[0][w] = -infinity

for i in 0 to n:

dp[i][0] = 0 # zero weight, zero cost

for i in 1 to n:

for w in 0 to W2:

Not include item i

dp[i][w] = max(dp[i][w], dp[i-1][w])

Include item i if possible

if w - wi ≥ 0:

dp[i][w] = max(dp[i][w], dp[i-1][w - wi] + ci)

max_cost = -infinity

for w in W1 to W2:

if dp[n][w] > max_cost:

max_cost = dp[n][w]

if max_cost == -infinity:

return "No feasible subset"

else:

return max_cost

Proof of Correctness

The DP approach systematically explores all possible subsets of items considering their cumulative weights and costs. By iterating through all items and updating the dp table, the algorithm ensures that for every possible weight up to W2, the maximum cost achievable is stored. The base case for zero items and zero weight aligns with the initializations, and the update rules properly consider including or excluding each item. Since the final search is over weights within [W1, W2], the algorithm guarantees finding the maximum cost over all feasible subsets within the weight bounds.

Time Complexity Analysis

The DP table has dimensions (n+1) x (W2+1). Filling each cell involves constant time operations, leading to an overall time complexity of O(nW2). This matches the problem's requirement. Additional overhead for initializations is linear in W2, negligible in comparison. Hence, the total time complexity is O(nW2).

Conclusion

This dynamic programming method provides an efficient solution to the constrained Knapsack problem with two weight bounds. It ensures maximum total cost for subsets with weights within the specified interval and operates within the desired time complexity.

References

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