Time Complexity of a loop that integer divides the loop counter by a constant

问题

I'm trying to calculate the time complexity of a simple algorithm in big O notation, but one part of it is seriously boggling my mind. Here is a simplified version of the algorithm:

int a=n
while(a>0)
{
//for loop with time complexity n^3
a = a/8
}

In this instance, it's integer division, so the while loop will terminate after a's value drops below 8. I'm not sure how to express this in terms of n. I'd also like to know how to tackle future calculations like this, where the number of loops isn't too easy to define.

回答1:

I find it easier to do it the other way around in cases like this. What is the opposite of what you're doing (even approximately)? Something like:

for (a = 1; a <= n; a = a * 8)
{
    ...
}

Now, we've changed the while to a for, which has a fixed number of steps, and decrementation to incrementation, which can be easier to work with.

we have:

1, 8, 8^2, ..., 8^k <= n

8^k <= n | apply logarithm

log (8^k) <= log n

k log 8 <= log n

=> k = O(log n)

So your while loop executes O(log n) times. If you have something that is O(n^3) inside, then your entire sequence will be O(n^3 log n).

回答2:

The key to this is that it's integer division, not Zeno's paradox. Repeated division takes log(n) steps to reduce a to something that will become zero on the next step.

Another way to look at integer division by a power of two is as a bit shift. Shifting a to the right by 3 bits will produce a zero after a number of steps depending on the position of the highest set bit in a. i.e. (sizeof(a)*CHAR_BIT - leading_zero_count(a)) / 3. Bit position is the same thing as log_base2.

来源：https://stackoverflow.com/questions/32490177/time-complexity-of-a-loop-that-integer-divides-the-loop-counter-by-a-constant