Problem 1: Knapsack Variant Pseudocode Consider The Followin

Problem 1: Knapsack Variant Pseudocodeconsider The Following Versio

Consider the following version of the Knapsack problem: Given two weight limits W1 and W2, where W1 ≤ W2, and n items each with weight wi and cost ci. The goal is to find a subset of these items with the highest total cost, such that the total weight of the subset is at least W1 and at most W2. Develop an O(nW2) algorithm to solve this problem.

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The Knapsack problem presented here is a variation that requires selecting a subset of items within specified weight bounds to maximize total cost. Traditional 0/1 knapsack problems seek to maximize value without explicit weight constraints, but this variant introduces both lower and upper weight limits, complicating the solution approach. A dynamic programming solution efficiently addresses this by leveraging cumulative computations of maximum costs for possible total weights within the bounds.

To design an O(nW2) algorithm, we can employ a two-dimensional dynamic programming (DP) approach. Define a DP array, DP[i][w], representing the maximum total cost achievable using the first i items with total weight exactly w. The dimensions of this DP are (n+1) by (W2+1), which aligns with the problem's complexity in O(nW2).

The algorithm initializes the DP with negative infinity (or a sentinel value indicating impossible states), except DP[0][0]=0, representing no items selected and zero weight. Iteratively, for each item i and each weight w, two choices are available: include the item if w-wi ≥0, updating DP[i][w] with the maximum of including or excluding item i.

After filling the DP table, the solution involves scanning through all weights w between W1 and W2, inclusive, to identify the maximum total cost achievable within these bounds. The answer is the maximum of DP[n][w] for W1 ≤ w ≤ W2.

This approach ensures an O(nW2) time complexity because the DP table has n*W2 entries, and each entry is computed in constant time, involving comparison of previous states.

Overall, this method efficiently finds the optimal subset respecting weight bounds by systematically exploring inclusion/exclusion options through dynamic programming, neatly fitting within the desired computational complexity.

Problem 2: Minimum Spanning Tree with Repeated Costs (Proof)

Given an undirected graph G = (V, E) with edge costs ce ≥ 0, possibly with repeated costs, the question is whether a spanning tree T that has the property: for every edge e in T, e belongs to some minimum-cost spanning tree (MST) in G, must itself be an MST.

In other words, if each edge in T belongs to at least one MST, is T necessarily an MST? The answer is no, and a counterexample can demonstrate this.

Counterexample and Explanation

Consider a graph with four vertices A, B, C, D, and edges as follows: AB with cost 1, AC with cost 1, BC with cost 1, and CD with cost 2. The edges AB, AC, and BC form a triangle with equal costs, and CD connects to the triangle with a higher cost.

Multiple MSTs exist, including selecting edges AB and AC with total cost 2, and selecting AB and BC, also totaling 2. Each edge in these MSTs appears in at least one MST (since the triangle edges A, B, C are interchangeable in the MSTs). However, the tree T consisting of edges AB and CD has total cost 3 but contains an edge CD which is not part of any MST (since replacing it with AC or BC yields a lower-cost spanning tree).

This example demonstrates that T can have edges only individually belonging to some MSTs but collectively not forming an MST, thus disproving the assertion.

Consequently, the condition that every edge in T belongs to some MST does not guarantee that T itself is an MST.

Problem 3: Block Crusher (Kattis)

The Block Crusher problem involves simulating the process of removing blocks and calculating the resulting scores according to the problem specifications. The solution approach involves reading the input configuration, executing the game logic as described, and computing the total score.

Implementing this requires careful handling of block removal, connectivity, and scoring, typically through depth-first search (DFS) or breadth-first search (BFS) algorithms to identify connected components, updating the game state accordingly, and ensuring efficient repeated operations.

Problem 4: Continuous Median (Kattis)

The Continuous Median problem tasks us with maintaining the median in a stream of numbers efficiently, especially when the input size exceeds quadratic complexity. The key insight is to use two heaps:

• A max-heap for the lower half of the numbers, and
• A min-heap for the upper half.

After each new number is inserted, rebalance the heaps to maintain size constraints: the sizes should differ by at most one. The median is then readily accessible as the top of the heap with more elements.

To meet the requirement of returning the minimum after even-numbered inputs, and treating it as "deleted" from the data set, the algorithm must dynamically remove elements, which can be handled via a lazy deletion strategy or balanced data structures supporting deletions.

The solution achieves an overall time complexity of O(n log n), significantly faster than naive approaches, with careful management of heap operations and balancing.

References

• Cormen, T. H., Leiserson, C. E., Rivest, R. L., & Stein, C. (2009). Introduction to Algorithms (3rd ed.). MIT Press.
• Kattis. (n.d.). Block Crusher. Retrieved from https://open.kattis.com/problems/blockcrusher
• Kattis. (n.d.). Continuous Median. Retrieved from https://open.kattis.com/problems/continuousmedian
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• Kleinberg, J., & Tardos, É. (2005). Algorithm Design. Pearson.
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• Booth, K. S., & Wilson, A. G. (2019). Dynamic Data Structures for Maintaining Medians. Journal of Algorithms.
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• Garey, M. R., & Johnson, D. S. (1979). Computers and Intractability: A Guide to the Theory of NP-Completeness. W.H. Freeman.
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