Abstract

We investigate the complexity of algorithms counting ones in different sets of operations. With addition and logical operations (but no shift) O(log⁡2(n))O(\log^2(n)) steps suffice to count ones. Parity can be computed with complexity O(log⁡(n))O(\log(n)), which is the same bound as for methods using shift-operations. If multiplication is available, a solution of time complexity O(log⁡∗(n))O(\log^*(n)) is possible improving the known bound O(log⁡log⁡(n))O(\log\log(n)).

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