complexity for nested loops

Question

I am trying to figure out the complexity of a for loop using Big O notation. I have done this before in my other classes, but this one is more rigorous than the others because it is on the actual algorithm. The code is as follows:

for(i=n ; i>1 ; i/=2) //for any size n
{
    for(j = 1; j < i; j++)
    {
      x+=a
    }
}

and

for(i=1 ; i<=n;i++,x=1) //for any size n
{
    for(j = 1; j <= i; j++)
    {
      for(k = 1; k <= j; x+=a,k*=a)
      {

      }
    }
}

I have arrived that the first loop is of O(n) complexity because it is going through the list n times. As for the second loop I am a little lost! Thank you for the help in the analysis. Each loop is in its own space, they are not together.

Is your comment in the last paragraph describing your thoughts on the first problem or the second one? — hatchet - done with SOverflow, May 24 '13 at 20:58
There is no list in your code, so I assume by "going through the list" you simply mean that `x += a` is executed `n` times, right? If so, how did you come to that conclusion? — sepp2k, May 24 '13 at 21:31
This may help http://cs.stackexchange.com/questions/4590/big-o-nested-for-loop-with-dependence — hatchet - done with SOverflow, May 24 '13 at 22:01

score 2 · Accepted Answer · answered May 25 '13 at 19:20

Consider the first code fragment,

for(i=n ; i>1 ; i/=2) //for any size n
{
    for(j = 1; j < i; j++)
    {
      x+=a
    }
}

The instruction x+=a is executed for a total of n + n/2 + n/4 + ... + 1 times.

Sum of the first log₂n terms of a G.P. with starting term n and common ratio 1/2 is, (n (1-(1/2)^log₂n))/(1/2). Thus the complexity of the first code fragment is O(n).

Now consider the second code fragment,

for(i=1 ; i<=n; i++,x=1)
{
    for(j = 1; j <= i; j++)
    {
      for(k = 1; k <= j; x+=a,k*=a)
      {

      }
    }
}

The two outer loops together call the innermost loop a total of n(n+1)/2 times. The innermost loop is executed at most log<sub>a</sub>n times. Thus the total time complexity of the second code fragment is O(n²log_an).

Mohamed Ennahdi El Idrissi · Answer 2 · 2014-06-09T00:22:17.990

1

You may formally proceed like the following:

Fragment 1:

enter image description here

Fragment 2 (Pochhammer, G-Function, and Stirling's Approximation):

enter image description here

With log(G(n)).

[UPDATE of Fragment 2]:

With some enhancements from "DISCRETE LOOPS AND WORST CASE PERFORMANCE" publication, by Dr. Johann Blieberger (All cases verified for a = 2):

enter image description here

Where:

enter image description here

Therefore,

enter image description here

edited Jun 09 '14 at 00:22

answered Apr 09 '14 at 20:03

Mohamed Ennahdi El Idrissi

2,931
2
19
32

qwwqwwq · Answer 3 · 2013-05-24T21:39:56.500

0

EDIT: I agree the first code block is O( n )

You decrement the outer loop i by diving by 2, and in the inner loop you run i times, so the number of iterations will be a sum over all the powers of two less than or equal to N but greater than 0, which is n^log(n)+1 - 1, so O(n).

The second code block is O(log_a(n)n²) assuming a is a constant.

The two outermost loops equate to a sum of all the numbers less than or equal to n, which is n(n-1)/2, so O(n²). Finally the inner loop is the powers of a less than an upper bound of n, which is O(log_an).

edited May 24 '13 at 21:39

answered May 24 '13 at 21:18

qwwqwwq

6,999
2
26
49

can you please more expling why O(2log(n)) – Hassan Bendouj May 24 '13 at 21:23
`2^log_2(n)` is just `n`, but I don't think you're right here. I'm pretty sure it actually is `O(n)`. – sepp2k May 24 '13 at 21:23

complexity for nested loops

3 Answers3

Linked