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In theoretical computer science, the continuous knapsack problem (also known as the fractional knapsack problem) is an algorithmic problem in combinatorial optimization in which the goal is to fill a container (the "knapsack") with fractional amounts of different materials chosen to maximize the value of the selected materials.
The bin packing problem can also be seen as a special case of the cutting stock problem. When the number of bins is restricted to 1 and each item is characterized by both a volume and a value, the problem of maximizing the value of items that can fit in the bin is known as the knapsack problem.
The knapsack problem is interesting from the perspective of computer science for many reasons: The decision problem form of the knapsack problem (Can a value of at least V be achieved without exceeding the weight W?) is NP-complete, thus there is no known algorithm that is both correct and fast (polynomial-time) in all cases.
The knapsack problem is one of the most studied problems in combinatorial optimization, with many real-life applications. For this reason, many special cases and generalizations have been examined. For this reason, many special cases and generalizations have been examined.
For example, bin packing is strongly NP-complete while the 0-1 Knapsack problem is only weakly NP-complete. Thus the version of bin packing where the object and bin sizes are integers bounded by a polynomial remains NP-complete, while the corresponding version of the Knapsack problem can be solved in pseudo-polynomial time by dynamic programming.
Many of these problems can be related to real-life packaging, storage and transportation issues. Each packing problem has a dual covering problem, which asks how many of the same objects are required to completely cover every region of the container, where objects are allowed to overlap. In a bin packing problem, people are given:
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Note: consider In the 2-weighted knapsack problem, where each item has two weights and a value, and the goal is to maximize the value such that the sum of squares of the total weights is at most the knapsack capacity: (,) + (,). We could solve it using a similar DP, where each state is (current weight 1, current weight 2, value).