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  2. Two's complement - Wikipedia

    en.wikipedia.org/wiki/Two's_complement

    The sum of a number and its ones' complement is an N-bit word with all 1 bits, which is (reading as an unsigned binary number) 2 N − 1. Then adding a number to its two's complement results in the N lowest bits set to 0 and the carry bit 1, where the latter has the weight (reading it as an unsigned binary number) of 2 N.

  3. 2Sum - Wikipedia

    en.wikipedia.org/wiki/2Sum

    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 + = +.

  4. Subset sum problem - Wikipedia

    en.wikipedia.org/wiki/Subset_sum_problem

    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)} .

  5. Partition problem - Wikipedia

    en.wikipedia.org/wiki/Partition_problem

    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.

  6. Maximum subarray problem - Wikipedia

    en.wikipedia.org/wiki/Maximum_subarray_problem

    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.

  7. 3-partition problem - Wikipedia

    en.wikipedia.org/wiki/3-partition_problem

    If a 3-set contains two pairing items u ij, u kl and one filler item, then the sum of the two pairing items must be 44T+4 = 4*(5T+6T)+2+2, so they must have matching sizes (a i +a j =a k +a l). Therefore, by replacing as needed, we can assume that the two pairing items are in fact u ij and u ij '.

  8. Musk's X ineffective against surge in US election ... - AOL

    www.aol.com/news/musks-x-ineffective-against...

    Of the 283 misleading X posts that CCDH analyzed, 209, or 74% of the posts, did not show accurate notes to all X users correcting false and misleading claims about the elections, the report said.

  9. Fibonacci sequence - Wikipedia

    en.wikipedia.org/wiki/Fibonacci_sequence

    Since the conversion factor 1.609344 for miles to kilometers is close to the golden ratio, the decomposition of distance in miles into a sum of Fibonacci numbers becomes nearly the kilometer sum when the Fibonacci numbers are replaced by their successors. This method amounts to a radix 2 number register in golden ratio base φ being shifted. To ...