Dating problem cryptography
A similar dynamic programming solution for the 0/1 knapsack problem also runs in pseudo-polynomial time. From Definition A, we can know that there is no need for computing all the weights when the number of items and the items themselves that we chose are fixed.That is to say, the program above computes more than expected because that the weight changes from 0 to W all the time.Of the possible subsets of problems whose total point values add up to 100, a knapsack algorithm would determine which subset gives each student the highest possible score. Informally, the problem is to maximize the sum of the values of the items in the knapsack so that the sum of the weights is less than or equal to the knapsack's capacity.The bounded knapsack problem (BKP) removes the restriction that there is only one of each item, but restricts the number One example of the unbounded knapsack problem is given using the figure shown at the beginning of this article and the text "if any number of each box is available" in the caption of that figure.
A multiple constrained problem could consider both the weight and volume of the boxes.(Solution: if any number of each box is available, then three yellow boxes and three grey boxes; if only the shown boxes are available, then all but the green box.) The knapsack problem or rucksack problem is a problem in combinatorial optimization: Given a set of items, each with a weight and a value, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible.It derives its name from the problem faced by someone who is constrained by a fixed-size knapsack and must fill it with the most valuable items.Another algorithm for 0-1 knapsack, discovered in 1974 and sometimes called "meet-in-the-middle" due to parallels to a similarly named algorithm in cryptography, is exponential in the number of different items but may be preferable to the DP algorithm when input: a set of items with weights and values output: the greatest combined value of a subset partition the set into two sets A and B of approximately equal size compute the weights and values of all subsets of each set for each subset of A find the subset of B of greatest value such that the combined weight is less than W keep track of the greatest combined value seen so far space, and efficient implementations of step 3 (for instance, sorting the subsets of B by weight, discarding subsets of B which weigh more than other subsets of B of greater or equal value, and using binary search to find the best match) result in a runtime of ), at the cost of using exponential rather than constant space (see also baby-step giant-step).As for most NP-complete problems, it may be enough to find workable solutions even if they are not optimal.
Besides, we can break the recursion and convert it into a tree.