Time complexity of Sieve of Eratosthenes algorithm

From Wikipedia:

> The complexity of the algorithm is > O(n(logn)(loglogn)) bit operations.

How do you arrive at that?

That the complexity includes the loglogn term tells me that there is a sqrt(n) somewhere.

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Suppose I am running the sieve on the first 100 numbers (n = 100), assuming that marking the numbers as composite takes constant time (array implementation), the number of times we use mark_composite() would be something like

n/2 + n/3 + n/5 + n/7 + ... + n/97        =      O(n^2)                         

And to find the next prime number (for example to jump to 7 after crossing out all the numbers that are multiples of 5), the number of operations would be O(n).

So, the complexity would be O(n^3). Do you agree?

1. Your n/2 + n/3 + n/5 + … n/97 is not O(n), because the number of terms is not constant. [Edit after your edit: O(n²) is too loose an upper bound.] A loose upper-bound is n(1+1/2+1/3+1/4+1/5+1/6+…1/n) (sum of reciprocals of all numbers up to n), which is O(n log n): see Harmonic number. A more proper upper-bound is n(1/2 + 1/3 + 1/5 + 1/7 + …), that is sum of reciprocals of primes up to n, which is O(n log log n). (See here or here.)

2. The "find the next prime number" bit is only O(n) overall, amortized — you will move ahead to find the next number only n times in total, not per step. So this whole part of the algorithm takes only O(n).

So using these two you get an upper bound of O(n log log n) + O(n) = O(n log log n) arithmetic operations. If you count bit operations, since you're dealing with numbers up to n, they have about log n bits, which is where the factor of log n comes in, giving O(n log n log log n) bit operations.

> That the complexity includes the loglogn term tells me that there is a sqrt(n) somewhere.

Keep in mind that when you find a prime number P while sieving, you don't start crossing off numbers at your current position + P; you actually start crossing off numbers at P^2. All multiples of P less than P^2 will have been crossed off by previous prime numbers.

1. The inner loop does n/i steps, where i is prime => the whole complexity is sum(n/i) = n * sum(1/i). According to prime harmonic series, the sum (1/i) where i is prime is log (log n). In total, O(n*log(log n)).
2. I think the upper loop can be optimized by replacing n with sqrt(n) so overall time complexity will O(sqrt(n)loglog(n)):

void isPrime(int n){
    int prime[n],i,j,count1=0;
    for(i=0; i < n; i++){
        prime[i] = 1;
    }
    prime[0] = prime[1] = 0;
    for(i=2; i <= n; i++){
        if(prime[i] == 1){
            printf("%d ",i);
            for(j=2; (i*j) <= n; j++)
                prime[i*j] = 0;
        }
    }    
}

int n = 100;
int[] arr = new int[n+1];  
for(int i=2;i<Math.sqrt(n)+1;i++) {
  if(arr[i] == 0) {
    int maxJ = (n/i) + 1;
    for(int j=2;j<maxJ;j++)
    {
      arr[i*j]= 1;
    }
  }
}
for(int i=2;i<=n;i++) {
  if(arr[i]==0) {
    System.out.println(i);
  }
}

For all i>2, Ti = sqrt(i) * (n/i) => Tk = sqrt(k) * (n/k) => Tk = n/sqrt(k)

Loop stops when k=sqrt(n) => n[ 1/sqrt(2) + 1/sqrt(3) + ...] = n * log(log(n)) => O(nloglogn)

see take the above explanation the inner loop is harmonic sum of all prime numbers up to sqrt(n). So, the actual complexity of is O(sqrt(n)*log(log(sqrt(n))))