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This algorithm finds the greedy (LPT) solution first, but then proceeds to look for better solutions. Several additional heuristics can be used to improve the runtime: [2] In a node in which the current sum-difference is at least the sum of all remaining numbers, the remaining numbers can just be put in the smallest-sum subset.
A solution to EPART consists of two parts, each of which has n/2 elements with a sum of T. It corresponds to an optimal solution of both MSSP variants: two subsets with a sum of (n+1)T, which is the largest possible. Similarly, each optimal solution of MSSP corresponds to a solution to EPART. Any non-optimal solution to MSSP leaves at least one ...
Conversely, given a solution to the SubsetSumZero instance, it must contain the −T (since all integers in S are positive), so to get a sum of zero, it must also contain a subset of S with a sum of +T, which is a solution of the SubsetSumPositive instance. The input integers are positive, and T = sum(S)/2.
The approximation ratio in this context is the smallest sum in the solution returned by the algorithm, divided by the smallest sum in the optimal solution (the ratio is less than 1). For greedy number partitioning, if the numbers are not sorted then the worst-case approximation ratio is 1/k. [11]
Given such an instance, construct an instance of Partition in which the input set contains the original set plus two elements: z 1 and z 2, with z 1 = sum(S) and z 2 = 2T. The sum of this input set is sum(S) + z 1 + z 2 = 2 sum(S) + 2T, so the target sum for Partition is sum(S) + T. Suppose there exists a solution S′ to the SubsetSum instance
Note that this partition is not optimal: in the partition {8,7}, {6,5,4} the sum-difference is 0. However, there is evidence that it provides a "good" partition: If the numbers are uniformly distributed in [0,1], then the expected difference between the two sums is n − Θ ( log ( n ) ) ) {\displaystyle n^{-\Theta (\log(n)))}} .
The two subsets should contain floor(n/2) and ceiling(n/2) items. It is a variant of the partition problem. It is NP-hard to decide whether there exists a partition in which the sums in the two subsets are equal; see [4] problem [SP12]. There are many algorithms that aim to find a balanced partition in which the sum is as nearly-equal as possible.
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.