具有外循环的双while循环算法的渐近增长率执行log（n）次

Question

What is the asymptotic growth rate (depending on n ) of this algorithm ?

i = 1; // executed 1 time
while( i ≤ n) {
    j = 1; // executed log(n) times

    while( j ≤ i) {
       j = j + 1; // ?
    }

    i = 2*i; // executed log(n) times
}

When n equals 10 :

| i iterations | j itérations
| i=1          | j=1
| i=2          | j=1 j=2 
| i=4          | j=1 j=2 j=3 j=4
| i=8          | j=1 j=2 j=3 j=4 j=5 j=6 j=7 j=8

The outer loop(i) affectations are executed log(n) times

How many times is the inner loop(j) affectation executed ?

Answer 1

This should be O(n) .

The last iteration of the outer loop has n iterations of the inner loop. The second-to-last iteration of the outer loop has n/2 iterations of the inner loop. The third-to-last iteration has n/4 :

n + n/2 + n/4 + ... = 2*n

See geometric progressions for the formula that allows you to compute the sum.

Answer 2

When n is four, the last execution of the outer loop causes the inner loop to execute 4 times. When you double n from 4 to 8, the outer loop is executed one more time, and on that last execution of the outer loop the inner loop is executed 8 times. Double it again to 16, and the outer loop will execute once more, with the inner loop going 16 times. So doubling the input causes a doubling in the execution time (plus the time for that one more outer loop, which become insignificant as n gets large).

So O(n).