I have a number n of size s. What is the computational complexity in big O notation of an algorithm computing n^n? Let's assume I'm using exponentiation by squaring. The result size doubles when we increase n by one bit, so the algorithm computing n^n has exponential computational complexity?

Second question, what is the computational complexity with respect to n of operations on the result n^n, such as multiplying n^n by a number of similar size?

n can be a large number, so I guess multiplication should not be considered O(1), am I wrong?


1 Answer 1


Let M(x) be the time needed to calculate the product of two x-but numbers. n^n is a number with n log n bits. Calculating n^n using repeated squaring, the last product takes M(n log n). All the earlier products needed take M(n/2 log n), M(n/4 log n) etc, and this sums up to roughly 2 M(n log n). A bit more, because we’ll need some multiplications by n, which are quite fast.

So roughly 2 M(n log n), and M(x) is O(x log x), so O(n log^2 n). There may be some small corrections needed.


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