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For the most typical case, a string of bits, this is the number of 1's in the string, or the digit sum of the binary representation of a given number and the ℓ₁ norm of a bit vector. In this binary case, it is also called the population count, [1] popcount, sideways sum, [2] or bit summation. [3]
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.
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)} .
A two-sum formula can be obtained using one of the symmetric formulae for Stirling numbers in conjunction with the explicit formula for Stirling numbers of the second kind. [ n k ] = ∑ j = n 2 n − k ( j − 1 k − 1 ) ( 2 n − k j ) ∑ m = 0 j − n ( − 1 ) m + n − k m j − k m !
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.
If a number is a squarefree positive integer, meaning that it is the product of some number of distinct prime numbers, then gives the number of different multiplicative partitions of . These are factorizations of N {\displaystyle N} into numbers greater than one, treating two factorizations as the same if they have the same factors in a ...
The next number in the sequence (the smallest number of additive persistence 5) is 2 × 10 2×(10 22 − 1)/9 − 1 (that is, 1 followed by 2 222 222 222 222 222 222 222 nines). For any fixed base, the sum of the digits of a number is proportional to its logarithm ; therefore, the additive persistence is proportional to the iterated logarithm .