Search results
Results From The WOW.Com Content Network
The maximum sum is 1, attained by giving one agent the item with value 1 and the other agent nothing. But the max-min allocation gives each agent value at least e, so the sum must be at most 3e. Therefore the POF is 1/(3e), which is unbounded. Alice has two items with values 1 and e, for some small e>0. George has two items with value e. The ...
For example, for the array of values [−2, 1, −3, 4, −1, 2, 1, −5, 4], the contiguous subarray with the largest sum is [4, −1, 2, 1], with sum 6. Some properties of this problem are: If the array contains all non-negative numbers, then the problem is trivial; a maximum subarray is the entire array.
Whenever the sum of the current element in the first array and the current element in the second array is more than T, the algorithm moves to the next element in the first array. If it is less than T, the algorithm moves to the next element in the second array. If two elements that sum to T are found, it stops. (The sub-problem for two elements ...
Given a function that accepts an array, a range query (,) on an array = [,..,] takes two indices and and returns the result of when applied to the subarray [, …,].For example, for a function that returns the sum of all values in an array, the range query (,) returns the sum of all values in the range [,].
Therefore, the terms with r, r 2 and r 3 must cancel out, and the terms with r 4 must sum up to 40r 4; so the 4-set must contain a triplet and 3 matching "real" elements, or a triplet and 3 matching "dummy" elements. From the triplets with the 3 matching "real" elements, we construct a valid perfect matching in E.
One can normalize input scores by assuming that the sum is zero (subtract the average: where =), and then the softmax takes the hyperplane of points that sum to zero, =, to the open simplex of positive values that sum to 1 =, analogously to how the exponent takes 0 to 1, = and is positive.
[2] [3] There is an optimization version of the partition problem, which is to partition the multiset S into two subsets S 1, S 2 such that the difference between the sum of elements in S 1 and the sum of elements in S 2 is minimized. The optimization version is NP-hard, but can be solved efficiently in practice. [4]
The algorithm has several stages. First, find a solution using greedy algorithm. In each iteration of the greedy algorithm the tentative solution is added the set which contains the maximum residual weight of elements divided by the residual cost of these elements along with the residual cost of the set.