single for-loop runtime explanation problem

Question

I am analyzing some running times of different for-loops, and as I'm getting more knowledge, I'm curious to understand this problem which I have still yet to find out. I have this exercise called "How many stars are printed":

for (int i = N; i > 1; i = i/2) System.out.println("*");

The answers to pick from is A: ~log N B: ~N C: ~N log N D: ~0.5N^2

So the answer should be A and I agree to that, but on the other side.. Let's say N = 500 what would Log N then be? It would be 2.7. So what if we say that N=500 on our exercise above? That would most definitely print more han 2.7 stars? How is that related?

Because it makes sense to say that if the for-loop looked like this:

for (int i = 0; i < N; i++)

it would print N stars.

I hope to find an explanation for this here, maybe I'm interpreting all these things wrong and thinking about it in a bad way. Thanks in advance.

Answer 1

You've overlooked the key characteristic of the logarithm base.

Because i is divided by 2 in each iteration, the running time is logarithmic with base 2. And

log₂(500) ~ 8.9

What you are looking at is

log₁₀(500) ~ 2.7

(logarithm with base 10)

By the way, the reason why the base is often omitted in runtime discussions (and your calculator probably doesn't have a button for log₂) is that due to the mechanisms of logarithmic math, a different base corresponds to a constant factor and thus is not relevant when you're ignoring constant factors anyway. It can be calculated easily:

log_a(x) = log_b(x) / log_b(a)

Answer 2

In addition to Michael Borgwardt's answer, the tilde character in front of each answer should be read as "proportional to". So if you doubled N (say, from 500 to 1000), you would see that the run time (or, in this case, number of stars printed) would increase by a factor which would be equal to (log1000 / log 500), which is independent from which base you use.