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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 + = +. [1] [6] [2]
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
The subset sum problem is a special case of the decision and 0-1 problems where each kind of item, the weight equals the value: =. In the field of cryptography, the term knapsack problem is often used to refer specifically to the subset sum problem. The subset sum problem is one of Karp's 21 NP-complete problems. [2]
Best-fit is an online algorithm for bin packing.Its input is a list of items of different sizes. Its output is a packing - a partition of the items into bins of fixed capacity, such that the sum of sizes of items in each bin is at most the capacity.
A Fenwick tree or binary indexed tree (BIT) is a data structure that stores an array of values and can efficiently compute prefix sums of the values and update the values. It also supports an efficient rank-search operation for finding the longest prefix whose sum is no more than a specified value.
For example, to solve a system of n equations for n unknowns by performing row operations on the matrix until it is in echelon form, and then solving for each unknown in reverse order, requires n(n + 1)/2 divisions, (2n 3 + 3n 2 − 5n)/6 multiplications, and (2n 3 + 3n 2 − 5n)/6 subtractions, [10] for a total of approximately 2n 3 /3 operations.
After processing n input elements, the input sequence can be partitioned into (n−c) / 2 pairs of unequal elements, and c copies of m left over. This is a proof by induction; it is trivially true when n = c = 0, and is maintained every time an element x is added: If x = m, add it to the set of c copies of m (and increment c).