What is the name of this prime number algorithm?

Question

Does the following recursive algorithm have a name? If so, what is it?

procedure main():
 myFilter = new Filter( myPrime = 2 ) //first prime number
 print 2 //since it would not otherwise be printed
 for each n in 3 to MAX:
  if myFilter.isPrime(n):
   print n

object Filter:
 integer myPrime
 PrimeFilter nextFilter = NULL

 procedure isPrime(integer n):
  if n is multiple of myPrime:
   return FALSE
  else if nextFilter is not NULL:
   return nextFilter.isPrime(n)
  else
   nextFilter = new PrimeFilter(myPrime = n)
   return TRUE

Sample implementation in Java here

This is similar to the Sieve of Eratosthenes though after some discussion in the CS chat, we decided that it is subtly different.

Answer 1

O'Neil [1] call this the "unfaithful sieve". It's much slower than the sieve of Eratosthenes.

For each prime $p$ you do work $\sim p/\log p$ and so the total number of divisions up to $x$ is roughly $x^2/(2\log^2 x)$ if you assume composites are free. (That's essentially true: they take at most $2\sqrt x/\log x$ divisions each for a total of at most $2x^{3/2}/\log x$ divisions.)

Divisions take longer than unit time, so the total bit complexity is about $O(x^2\log\log x/\log x)$.

[1] Melissa E. O’Neill, The Genuine Sieve of Eratosthenes

Answer 2

Let me rephrase your algorithm (starting at a different base case):

Initialize P to be the empty list.
for n from 2 to MAX:
  if no integer in P divides n:
    add n to P
return P

Let $p_1,p_2,\ldots $ be an enumeration of the primes. The probability that $p_1,\ldots,p_i$ do not divide a number $n$ is roughly $$ \prod_{j=1}^i \left(1 - \frac{1}{p_j}\right) \approx e^{-\sum_{j=1}^i p_j} \approx e^{-\sum_{j=1}^i j\log j} \approx e^{-\frac{1}{2} i^2 \log i}.$$ Therefore the inner loop runs for roughly this many iterations: $$ \sum_{i=1}^\infty e^{-\frac{1}{2} i^2\log i} = O(1). $$ In total, the complexity is $O(n)$ divisions. In contrast, the Eratosthenes sieve requires $O(n\log\log n)$ additions.

For a fair comparison, we also need to factor it the computational complexity of operations on large numbers. Assuming that division can be done in time $O(m\log m)$ (which is the conjectured running time), where in our case $m = \log n$, your algorithm has bit complexity $O(n\log n\log\log n)$, matching the bit complexity of the sieve. However, the best known division algorithms are somewhat slower, both asymptotically and in practice, and so I expect your algorithm to be somewhat slower than the sieve.

Atkin's sieve uses only $O(n/\log\log n)$ additions and so is faster than both your algorithm and the Eratosthenes sieve. It also uses only $\tilde{O}(\sqrt{n})$ memory, compared to your $\sum_{p_i \leq n} \log p_i \approx \sum_{m=1}^{n/\log n} \log(m\log m) = \Theta(n)$, also shared by the Eratosthenes sieve.