![]() ![]() Subset sum / knapsack is already a difficult problem - if using it's not going to give you an optimal solution, you may as well use the sorting / greedy approach above. If you just solve subset sum, you're left with 5,5,5 in one bag and 9 in one bag each (for a total of 4 bags), rather than 5,9 in each of 3 bags. Why not? Consider items 5,5,5,9,9,9 with a bag size of 16. For this, you can consider the weight the same as value and solve using a knapsack algorithm, but this won't really work too well for multiple bags. In the case of 1 bag, this is essentially the subset sum problem. The running time will greatly depend on the number of items that can fit into a bag - it will be O(minimumBagsUsed.2 maxItemsPerBag). The linked resource actually considers multiple types, which can occur multiple times - I derived the above solution from that. The array index above is literally a set - think of this as a map of set to value, a bitmap or a multi-dimensional array where each index is either 1 or 0 to indicate whether we include the item corresponding to that dimensional or not. This resource has one option - the basic idea is: D where set2 fits into 1 bag ![]() For this, you can consider the weight the same as value and solve using a knapsack algorithm, but this wont really work too well for multiple bags. In terms of optimal solutions, there isn't a dynamic programming solution that's as well-known as for the knapsack problem. The running time will greatly depend on the number of items that can fit into a bag - it will be O (minimumBagsUsed.2 maxItemsPerBag). ![]() This would easily take O(n²), or possibly O(n log n) with an efficient implementation. Solutions to each sub-problem are stored so that the computation would only need to happen once.This is known as the bin packing problem (which is NP-hard).īy simply sorting the decreasing order by their sizes, and then inserting each item into the first bin in the list with sufficient remaining space, we get 11/9 OPT + 6/9 bins (where OPT is the number of bins used in the optimal solution). This is where dynamic programming techniques can be applied. However, if it is a program, re-computation is not independent and would cause problems. The problem solver only needs to decide whether to take the item or not based on the weight that can still be accepted. This deals with only one item at a time and the current weight still available in the knapsack. ![]() Since an exhaustive search is not possible, one can break the problems into smaller sub-problems and run it recursively. In the knapsack problem, the given items have two attributes at minimum – an item’s value, which affects its importance, and an item’s weight or volume, which is its limitation aspect. It is easily the most important problem in logistics. It also can be found in fields such as applied mathematics, complexity theory, cryptography, combinatorics and computer science. The problem can be found real-world scenarios like resource allocation in financial constraints or even in selecting investments and portfolios. This is a problem that has been studied for more than a century and is a commonly used example problem in combinatorial optimization, where there is a need for an optimal object or finite solution where an exhaustive search is not possible. The knapsack problem is an example of a combinational optimization problem, a topic in mathematics and computer science about finding the optimal object among a set of objects. ![]()
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