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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 ...
In the th step, it computes the subarray with the largest sum ending at ; this sum is maintained in variable current_sum. [note 3] Moreover, it computes the subarray with the largest sum anywhere in […], maintained in variable best_sum, [note 4] and easily obtained as the maximum of all values of current_sum seen so far, cf. line 7 of the ...
Let A be the sum of the negative values and B the sum of the positive values; the number of different possible sums is at most B-A, so the total runtime is in (()). For example, if all input values are positive and bounded by some constant C , then B is at most N C , so the time required is O ( N 2 C ) {\displaystyle O(N^{2}C)} .
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.
The top-k data structure at each node is constructed based on the values existing in the subtrees of that node and is meant to answer one-sided range top-k queries. Please note that for a one-dimensional array A {\displaystyle A} , a range tree can be constructed by dividing A {\displaystyle A} into two halves and recursing on both halves ...
The 3-partition problem is similar to the partition problem, in which the goal is to partition S into two subsets with equal sum, and the multiway number partitioning, in which the goal is to partition S into k subsets with equal sum, where k is a fixed parameter. In 3-Partition the goal is to partition S into m = n/3 subsets, not just a fixed ...
Clearly, the new instance has an equal-cardinality equal-sum partition iff the original instance has an equal-sum partition. See also Balanced number partitioning. Product partition is the problem of partitioning a set of integers into two sets with the same product (rather than the same sum). This problem is strongly NP-hard. [14]
The maximum coverage problem is a classical question in computer science, computational complexity theory, and operations research.It is a problem that is widely taught in approximation algorithms.