Why is Sieve of Eratosthenes more efficient than the simple "dumb" algorithm?

Question

If you need to generate primes from 1 to N, the "dumb" way to do it would be to iterate through all the numbers from 2 to N and check if the numbers are divisable by any prime number found so far which is less than the square root of the number in question.

As I see it, sieve of Eratosthenes does the same, except other way round - when it finds a prime N, it marks off all the numbers that are multiples of N.

But whether you mark off X when you find N, or you check if X is divisable by N, the fundamental complexity, the big-O stays the same. You still do one constant-time operation per a number-prime pair. In fact, the dumb algorithm breaks off as soon as it finds a prime, but sieve of Eratosthenes marks each number several times - once for every prime it is divisable by. That's a minimum of twice as many operations for every number except primes.

Am I misunderstanding something here?

Answer 1

In the trial division algorithm, the most work that may be needed to determine whether a number n is prime, is testing divisibility by the primes up to about sqrt(n).

That worst case is met when n is a prime or the product of two primes of nearly the same size (including squares of primes). If n has more than two prime factors, or two prime factors of very different size, at least one of them is much smaller than sqrt(n), so even the accumulated work needed for all these numbers (which form the vast majority of all numbers up to N, for sufficiently large N) is relatively insignificant, I shall ignore that and work with the fiction that composite numbers are determined without doing any work (the products of two approximately equal primes are few in number, so although individually they cost as much as a prime of similar size, altogether that's a negligible amount of work).

So, how much work does the testing of the primes up to N take?

By the prime number theorem, the number of primes <= n is (for n sufficiently large), about n/log n (it's n/log n + lower order terms). Conversely, that means the k-th prime is (for k not too small) about k*log k (+ lower order terms).

Hence, testing the k-th prime requires trial division by pi(sqrt(p_k)), approximately 2*sqrt(k/log k), primes. Summing that for k <= pi(N) ~ N/log N yields roughly 4/3*N^(3/2)/(log N)^2 divisions in total. So by ignoring the composites, we found that finding the primes up to N by trial division (using only primes), is Omega(N^1.5 / (log N)^2). Closer analysis of the composites reveals that it's Theta(N^1.5 / (log N)^2). Using a wheel reduces the constant factors, but doesn't change the complexity.

In the sieve, on the other hand, each composite is crossed off as a multiple of at least one prime. Depending on whether you start crossing off at 2*p or at p*p, a composite is crossed off as many times as it has distinct prime factors or distinct prime factors <= sqrt(n). Since any number has at most one prime factor exceeding sqrt(n), the difference isn't so large, it has no influence on complexity, but there are a lot of numbers with only two prime factors (or three with one larger than sqrt(n)), thus it makes a noticeable difference in running time. Anyhow, a number n > 0 has only few distinct prime factors, a trivial estimate shows that the number of distinct prime factors is bounded by lg n (base-2 logarithm), so an upper bound for the number of crossings-off the sieve does is N*lg N.

By counting not how often each composite gets crossed off, but how many multiples of each prime are crossed off, as IVlad already did, one easily finds that the number of crossings-off is in fact Theta(N*log log N). Again, using a wheel doesn't change the complexity but reduces the constant factors. However, here it has a larger influence than for the trial division, so at least skipping the evens should be done (apart from reducing the work, it also reduces storage size, so improves cache locality).

So, even disregarding that division is more expensive than addition and multiplication, we see that the number of operations the sieve requires is much smaller than the number of operations required by trial division (if the limit is not too small).

Summarising:
Trial division does futile work by dividing primes, the sieve does futile work by repeatedly crossing off composites. There are relatively few primes, but many composites, so one might be tempted to think trial division wastes less work.
But: Composites have only few distinct prime factors, while there are many primes below sqrt(p).

Answer 2

In the naive method, you do O(sqrt(num)) operations for each number num you check for primality. Ths is O(n*sqrt(n)) total.

In the sieve method, for each unmarked number from 1 to n you do n / 2 operations when marking multiples of 2, n / 3 when marking those of 3, n / 5 when marking those of 5 etc. This is n*(1/2 + 1/3 + 1/5 + 1/7 + ...), which is O(n log log n). See here for that result.

So the asymptotic complexity is not the same, like you said. Even a naive sieve will beat the naive prime-generation method pretty fast. Optimized versions of the sieve can get much faster, but the big-oh remains unchanged.

The two are not equivalent like you say. For each number, you will check divisibility by the same primes 2, 3, 5, 7, ... in the naive prime-generation algorithm. As you progress, you check divisibility by the same series of numbers (and you keep checking against more and more as you approach your n). For the sieve, you keep checking less and less as you approach n. First you check in increments of 2, then of 3, then 5 and so on. This will hit n and stop much faster.

Answer 3

Because with the sieve method, you stop marking mutiples of the running primes when the running prime reaches the square root of N.

Say, you want to find all primes less than a million.

First you set an array

for i = 2 to 1000000
  primetest[i] = true

Then you iterate

for j=2 to 1000         <--- 1000 is the square root of 10000000
  if primetest[j]                                    <--- if j is prime
    ---mark all multiples of j (except j itself) as "not a prime"
    for k = j^2 to 1000000 step j
      primetest[k] = false

You don't have to check j after 1000, because j*j will be more than a million. And you start from j*j (you don't have to mark multiples of j less than j^2 because they are already marked as multiples of previously found, smaller primes)

So, in the end you have done the loop 1000 times and the if part only for those j's that are primes.

Second reason is that with the sieve, you only do multiplication, not division. If you do it cleverly, you only do addition, not even multiplication.

And division has larger complexity than addition. The usual way to do division has O(n^2) complexity, while addition has O(n).

Answer 4

Explained in this paper: http://www.cs.hmc.edu/~oneill/papers/Sieve-JFP.pdf

I think it's quite readable even without Haskell knowledge.

Answer 5

the first difference is that division is much more expensive than addition. Even if each number is 'marked' several times, it's trivial when compared with the huge number of divisions needed for the 'dumb' algorithm.

Answer 6

A "naive" Sieve of Eratosthenes will mark non-prime numbers multiple times. But, if you have your numbers on a linked list and remove numbers taht are multiples (you will still need to walk the remainder of the list), the work left to do after finding a prime is always smaller than it was before finding the prime.

Answer 7

http://en.wikipedia.org/wiki/Prime_number#Number_of_prime_numbers_below_a_given_number

• the "dumb" algorithm does i/log(i) ~= N/log(N) work for each prime number
• the real algorithm does N/i ~= 1 work for each prime number

Multiply by roughly N/log(N) prime numbers.

Answer 8

It was a time when I was trying to find an efficient way of finding sum of primes less than x:

There I decided to use N by N square table and started checking if numbers with a unit digits in [1,3,7,9]

But Eratosthenes Method of prime made it a little easier: How

1. Let you want to know if N is prime or Not

   You started finding factorization. So you will realize when N is factorized when you divide N with the highest factor quotient will be less.

   So, You take a number: int(sqrt(N)) = K(say) divides N you get somewhat same and close number to K

2. Now let's say you divide N with u<K, but if "U" is not prime and one of the prime factors of U is V(prime) then will obviously be less than U (V<U) and V will also divide N then

   why not divide and check if N is prime or not by DIVIDING 'N' WITH ONLY PRIMES LESS THAN K=int(sqrt(N)) Number of times for which loop Keeps executing = π(√n) This is how the brilliant idea of Eratosthenes starts taking pictures and will start giving you intuition behind this all.

3. Btw using the Sieve of Eratosthenes one can find sum of primes less than a multiple of 10.

because for a given column you just check need to check their unit digits[1,3,7,9] and for how many times a particular unit digit is repeating.

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