Question

Here’s the algorithm in question, in pseudocode:

list.add(2)
for (i = 3 to n)
    isPrime=true
    for each (j in list)
        if ((i mod j) = 0)
            isPrime=false
            break
    if (isPrime)
        list.add(i)
return list

My TA today told me that this algorithm had a complexity of O(n^2), but I am quite sure that is incorrect, especially given that there are approximately n/ln(n) prime numbers up to any n, so the inner loop would not iterate more than n/ln(n) times on its last iteration if n itsself were prime. However I think it’s actually lower than O(n^2/ln(n)) in reality, but I’m not sure how to determine what it actually is. For example, every even number only iterates the second loop twice.

Answer 1

Among n numbers you are checking for being prime there are approximately n / ln(n) actually prime numbers which demands all i / ln(i) already found prime numbers to be tried as divisors. Therefore, the complexity of your algorithm is W(n^2 / ln(n)^2) (W is big-omega) and O(n^2 / ln(n). n^2 / ln(n)^2 growths faster then n * sqrt(n).

For reaching O(n * sqrt(n)) time complexity you should use the following pseudocode:

list.add(2)
for (i = 3 to n)
    isPrime=true
    for each (j in list)
        if j > sqrt(i)
            break
        if ((i mod j) = 0)
            isPrime=false
            break
    if (isPrime)
        list.add(i)
return list