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The multiple subset sum problem is an optimization problem in computer science and operations research.It is a generalization of the subset sum problem.The input to the problem is a multiset of n integers and a positive integer m representing the number of subsets.
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)} .
In computational complexity theory, the 3SUM problem asks if a given set of real numbers contains three elements that sum to zero. A generalized version, k-SUM, asks the same question on k elements, rather than simply 3. 3SUM can be easily solved in () time, and matching (⌈ / ⌉) lower bounds are known in some specialized models of computation (Erickson 1999).
Karatsuba's basic step works for any base B and any m, but the recursive algorithm is most efficient when m is equal to n/2, rounded up. In particular, if n is 2 k , for some integer k , and the recursion stops only when n is 1, then the number of single-digit multiplications is 3 k , which is n c where c = log 2 3.
It may be used to prove Nicomachus's theorem that the sum of the first cubes equals the square of the sum of the first positive integers. [2] Summation by parts is frequently used to prove Abel's theorem and Dirichlet's test.
G(3) is at least 4 (since cubes are congruent to 0, 1 or −1 mod 9); for numbers less than 1.3 × 10 9, 1 290 740 is the last to require 6 cubes, and the number of numbers between N and 2N requiring 5 cubes drops off with increasing N at sufficient speed to have people believe that G(3) = 4; [17] the largest number now known not to be a sum of ...
For example, if all the values are O(n k) for some k, then the time required is O(n k+2). This algorithm is easily modified to return the subset with sum 0 if there is one. This solution does not count as polynomial time in complexity theory because P-N is not polynomial in the size of the problem, which is the number of bits used to represent it.