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 ...
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)} .
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 ...
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 ...
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.
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 [,].
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 algorithm performs summation with two accumulators: sum holds the sum, and c accumulates the parts not assimilated into sum, to nudge the low-order part of sum the next time around. Thus the summation proceeds with "guard digits" in c , which is better than not having any, but is not as good as performing the calculations with double the ...