How can I make this python function run O(log n) time instead of O(n) time?

Question

def findMax(f,c):
    n=1
    while f(n) <= c:
        n += 1
   return n

This is a higher-order python function that given function f and a maximal count, c returns the largest n such that f(n) ≤ c. This works but not when n gets too large for e.g f(10**6). How can I make this algorithm run O(log n) time so it can facilitate f(10**6) using the function below?

def f(n):
    return math.log(n, 2)

Is the result of f(n) proportionally increasing relative to n? You could just do a binary traversal. — Chris, Mar 21 '17 at 18:49
Is `findMax()` called repeatedly? That is, would it be productive to cache its results? — wallyk, Mar 21 '17 at 18:49
Without more information on the nature of `f`, there's no way to do better. — Kevin, Mar 21 '17 at 18:50
*"returns the largest n such that f(n) ≤ c"*: it seems to return the *smallest* — trincot, Mar 21 '17 at 18:51
This is returning the first positive integer such that `f(n)` is *greater* than `c`, not the largest `n` such that `f(n) <= c`. — user2357112, Mar 21 '17 at 18:53
How are you even supposed to know whether you've found the highest `n`? For all you know, no matter how high you've searched, some higher `n` could have `f(n) <= c`. — user2357112, Mar 21 '17 at 18:54
This is the third time this question has been asked today. http://stackoverflow.com/questions/42935621/algorithm-in-ologn http://stackoverflow.com/questions/42931413/an-higher-order-function-that-returns-the-largest-n-such-that-fn-a-given-ma — Paul Hankin, Mar 21 '17 at 18:56
@Kevin I've provided an example of `f` where `findMax()` takes forever to run. — Wonakee Amos, Mar 22 '17 at 19:13

God Bennett · Answer 1 · 2017-03-30T03:06:47.880

0

Change n += 1 to n *= 2 for logarithmic outcome.

Logarithmic sequences increment in multiples of 2, and are non-linear, thus logarithmic sequences don't increment by 1.

edited Mar 30 '17 at 03:06

answered Mar 30 '17 at 03:01

God Bennett

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score 0 · Answer 2 · edited Dec 18 '20 at 15:59

Use a search algorithm to find the solution faster. This is an implementation using jpe.math.framework.algorythems.brent which is an implementation of the brent search algorithm.

import math
import jpe

import jpe.math.framework.algorythems

def f(x):
    return math.log(x, 2)

value = 9E2
startVal = 1E300

val = int(jpe.math.framework.algorythems.brent(f, a=0, b=startVal, val=value, mode=jpe.math.framework.algorythems.modes.equalVal)[0])#takes 37 iters

print(val)

Alternatively in this scenario with this f:

the result is within 1 of 2**c (c as passedto findMax)

How can I make this python function run O(log n) time instead of O(n) time?

2 Answers2