确定while循环的迭代次数

Question

I have this code from my computer science class:

int input=15;
while (input < n ) { input = input *3;}

This code has the ceiling of log3(n/15) loops. How can we obtain this result?

Answer 1

I think he is talking about the analytical solution to complexity. I think it's something like this (long time ago i did logaritms):

15 * 3^x = n   // x would be the number of iterations until value reaches n

ln(15*(3^x)) = ln(n)
ln(15) + ln(3^x) = ln(n)
ln(15) + x*ln(3) = ln(n)
x = (ln(n) - ln(15)) / ln(3)
x = ln(n/15) / ln(3)
x = log3(n/15) / log3(3)
x = log3(n/15)

Answer 2

For what values of n does the code loop k times?

It must be that 15 < n, and 15*3 < n and 15*3*3 < n and .... and 15*3^(k-1) < n. Also, it must be that 15*3^k >= n (otherwise the code would do at least one more loop).

That is, 3^k >= n/15 > 3^(k-1), and taking logs (base 3), k >= log3(n/15) > k-1.

Thus k is the smallest integer greater than or equal to log3(n/15), or equivalently: k = ceil(log3(n/15)) as required.