doubling

Given $f(x)$, I want to find $f^n(x)$ fast

• If the width of the definition and value range of f is D and the maximum value of n is N, we can pre-process O(D log N) to main process O(log N)
• pretreatment
  • Since $f^{2k}(x) = f^k(f^k(x))$, we obtain $f^{2k}$ by O(D) when $f^k$ is known
  • Since k=1 is known, we obtain in O(D log N) for k such that [$ k=2^m < N
• main processing
  • We can pick out only the 1's in binary expansion of n from the preprocessing results and synthesize them.
  • The length of the binary expansion is of logarithmic order

This page is auto-translated from /nishio/ダブリング using DeepL. If you looks something interesting but the auto-translated English is not good enough to understand it, feel free to let me know at @nishio_en. I'm very happy to spread my thought to non-Japanese readers.