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Here is a proof that the asymptotic ratio is at most 2. If there is an FF bin with sum less than 1/2, then the size of all remaining items is more than 1/2, so the sum of all following bins is more than 1/2. Therefore, all FF bins except at most one have sum at least 1/2. All optimal bins have sum at most 1, so the sum of all sizes is at most OPT.
Provided the floating-point arithmetic is correctly rounded to nearest (with ties resolved any way), as is the default in IEEE 754, and provided the sum does not overflow and, if it underflows, underflows gradually, it can be proven that + = +.
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 guillotine cutting problem, both the items and the "bins" are two-dimensional rectangles rather than one-dimensional numbers, and the items have to be cut from the bin using end-to-end cuts. In the selfish bin packing problem, each item is a player who wants to minimize its cost.
In number theory and computer science, the partition problem, or number partitioning, [1] is the task of deciding whether a given multiset S of positive integers can be partitioned into two subsets S 1 and S 2 such that the sum of the numbers in S 1 equals the sum of the numbers in S 2.
A California-based tire company has recalled over 540,000 replacement tires mistakenly labeled as snow traction products, according to an announcement from the National Highway Traffic Safety ...
By formulating MAX-2-SAT as a problem of finding a cut (that is, a partition of the vertices into two subsets) maximizing the number of edges that have one endpoint in the first subset and one endpoint in the second, in a graph related to the implication graph, and applying semidefinite programming methods to this cut problem, it is possible to ...
Currently, Barkley works two days a week for "Inside the NBA," but he doesn't want to work more than that. "I want it in writing how much we're going to be working," Barkley added.