## Prime Numbers 3. Wheel Factorization

March 24, 2009

In an earlier post, I described the sieve of Eratosthenes, a very fast algorithm to calculate all primes below a given limit, n. At least when programming it in python, speed is not an issue at all with this algorithm, because we start having memory problems much earlier, due to the fact that we need to keep in memory a list of size n…

This is specially problematic with python, because although all we need is a list of booleans (is prime / is not prime), which could fit in a bit each, I’m afraid python assigns several bytes to each. How many of them is probably machine dependent, but I’m almost certain it is at least 4 of them. So it is very likely that we could improve memory usage by a factor of 32, maybe more, if we could get around this. There are not-so-standard libraries that do just that, but I’ve sworn not to use them, and one day I may expose my failed attempts at using very long ints as bit arrays to the world, but python is definitely not the best language to mess around with low level optimizations, so I’m going to leave this line of thinking aside for the time being.

I also presented a very naive optimization, which simply discards all even numbers, which we already know are not prime, and thus cuts in half the memory requirements. More formally, what was being done there was to partition all numbers into two groups: those of the form 2·n, and those of the form 2·n+1, and then discard the former, since no prime of that form exists, and keep the latter only. This line of thinking is much more promising, as it leads to a very neat way of further reducing the memory required, known as wheel factorization.

For some odd reason that escapes my shallow understanding, the standard description of this algorithm places much emphasis on setting all numbers in circles, and then eliminating what it calls spokes. I personally don’t see any benefit of such a mental image, specially since the number are going to end up arranged in a rectangular array. So for the sake of clarity, I will not honor tradition, and spare you the circles and the spokes.

We begin by choosing a first few prime numbers. Say the first three: 2, 3 and 5. We multiply them together to get M = 30. We are now going to partition the set of all numbers into M groups, of the form M·m + ki, where the ki are the numbers 1 through M, both inclusive. Most of these subsets happen to hold no prime number, and thus need not be considered at all. To find out which of them, remember that M is the product of several prime numbers, M = p1·p2 … pn. Therefore, if ki is a multiple of any of those prime numbers, i.e. ki = pj·k’i, we can always extract the common factor pj from both summands in M·m + ki, which means that all numbers in that subset are composite. These subsets, which we discard, are very closely related to the spokes I was telling you about a while ago.

If M = 30, of the 30 potential ki we are left with just 8 (1, 7, 11, 13, 17, 19, 23 and 29), so we have cut memory required to about 27% of the straightforward implementation of the sieve of Eratosthenes. If we use the first 4 primes, we then have M = 210, and of the 210 potential ki we only need to keep 48. This improves memory use, which now is roughly 23% of the original. But you can probably see a law of diminishing returns at wortk here: the improvement in memory use gets smaller and smaller with every prime you add to the calculation of M. You can find some more data regarding this at the end of this link.

Of course we are still going to have to sieve each of our subsets, in much the same way we did with the sieve of Eratothenes. But sieving all multiples of a prime p when you have several subsets to consider is anything but trivial. Lets see how to tackle it…

The question to answer is, what integer values of m fulfill the equation M·m + ki = p·n? Using modular arithmetic, this equation is equivalent to finding M·m = -ki (mod p). This requires calculating M-1 (mod p), the modular multiplicative inverse of M modulo p, a subject already treated in a previous post. Given that, we can simply write mi = M-1·(-ki) (mod p), and therefore determine all multiples of p in the subset of ki as m = p·j + mi.

Last, but not least, we want to start sieving at the smallest multiple of p in each subset that is larger than or equal to p2, and so we want to increment mi with p·ji, where ji ≥ (p2-ki– M·mi) / (m·p).

Lets try to put all this ideas together in a python script. To make it work, you will need the primeListSofE function from this post, as well as the extEuclideanAlg and modInvEuclid ones from this other post

def primeListWF(n, m, verbose = True) :
"""Returns a list of prime numbers smaller than or equal to n,
using wheel factorization, with a wheel size of the product of
all primes smaller than or equal to m
"""
t = time()
# We use the standard sieve of Eratosthenes first...
primes = primeListSofE(m)
M = 1
for p in primes :
M *= p
if verbose :
print "Size of factorization wheel is",M
# We now sieve out all multiples of these primes,including
# the primes themselves, to get the k[i]
sieve = [True] * M
for p in primes :
sieve[p-1] = False
for j in xrange(p*p,M+1,p) :
sieve[j-1] = False
k = [j+1 for j in range(M) if sieve[j]]
if verbose :
print "Memory use is only",len(k),"/",M,"=",len(k)/M
N = n
if N %M :
N += M
N //= M
maxP = int(sqrt(M*N))
if verbose:
print "Primes will be calculated up to",M*N,"not",n

sieve = [[True]*N for j in range(len(k))]
sieve[0][0] = False # Take care of the non primality of 1
for row in range(N) :
baseVal = M * row
for subset in range(len(k)) :
if sieve[subset][row] :
p = baseVal + k[subset]
primes += [p]
if p > maxP :
continue
# Sieve all multiples of p...
invMp = modInvEuclid(M,p)
for i in range(len(k)) :
mi = invMp * (-k[i] % p)
mi %= p
# ...starting at p**2
ji = (p*p - k[i] - M*mi)
if ji % (M*p) :
ji += M*p
ji //= M*p
mi+= ji*p

for r in range(mi,N,p) :
sieve[j][r] = False
if verbose :
print "Calculated primes to",n,"in",time()-t,"sec."

return primes

You can play around with this algorithm and look at the time required. It probably depends on the value of n, but there seems to be a sweet spot for speed somewhere around m = 7 (and thus M = 210) or m = 11 (M = 2310). This being a general purpose wheel factorization algorithm, it has speeds that are faster than the unoptimized sieve of Eratosthenes code, but it cannot beat the optimized code, which in fact is a particular case of wheel factorization, for m = M = 2.

On the other hand, using m = 7 I have managed to compute all primes up to 108, a feat unreachable for the previous code, at least without finding a workaround the MemoryError they produce. 109 is still out of reach, though…

## Prime Numbers 2. The Sieve of Erathostenes

March 16, 2009

The problem with the algorithms described in the previous post lies on the very large amount of redundant testing that goes on. Because once we have checked the primality of a certain number p, we know that 2p, 3p, 4p, …, np are composites. So what is the point of checking those numbers at all? When you follow this line of thinking to the very end, it turns out you don’t have to do any trial division at all! The resulting algorithm is attributed to Erathostenes, and it employs a technique, known as sieving, which may come in handy in many other problems. But enough for a presentation, lets get going with the Sieve of Erathostenes.

All the information we need to begin with, is that 2 is the first prime. This means, following the previous approach, that no other even number can be prime. To keep track of this information, we begin by writing a list of all the numbers between 2 and the desired limit, say N:

We now draw a circle around 2, because it is prime, and then strike out all even numbers, i.e. all multiples of 2:

Now, there is a good and a bad way to strike out all multiples of a number from a list. The wrong way is to do trial division to find out if it is a multiple. The right way is to use the fact that two consecutive multiples of a number differ by that exact number. Writing it out in python, you can check the tremendous gain for yourself:

from time import time

def strikeOut(mul, n) :
# note that list index i corresponds to number i+2
sieve = [True for j in xrange(2,n+1)]

# The bad way to sieve
t = time()
for j in xrange(2,n+1) :
if j % mul == 0 :
sieve[j-2] = False
t = time() - t
print "Sieved (bad!!!) all multiples of",mul,"below",n,"in",t,"sec."

sieve = [True for j in xrange(2,n+1)]

# The good way to sieve
t = time()
for j in xrange(mul,n+1,mul) :
sieve[j-2] = False
t = time() - t
print "Sieved (good!!!) all multiples of",mul,"below",n,"in",t,"sec."

No, I’m not taking you, dear reader, for an idiot: you can find sieving algorithms on fist pages of google searches that do sieving this very wrong way…

But lets get back to primes. We had our list, with 2 circled and all its multiples crossed out. The first unmarked number we come across is 3. Just a coincidence that 3 is a prime number? Not really: it is unmarked because it is not a multiple of any smaller number. Or to put it another way: it is only divisible by 1 and itself, which is the very definition of prime… So we circle 3 and cross out all its multiples:

I’ve underlined the multiples of three that where already crossed out. The first number which gets crossed out for the first time is 9. Will it be just a coincidence that 9 = 32? Of course not! We have already crossed out all multiples of numbers smaller than 3, and 32 is the first multiple of 3 which is not a multiple of any smaller number.

So we keep going down our list, and find 5 to be the next unmarked number in the list, which of course is prime, so we circle it, and then cross out all of its multiples. But as we saw before, rather than starting at 10, and crossing out the already crossed out 10, 15 and 20, we proceed directly to 52, that is 25.

This goes on until we’ve searched the whole list, circling numbers and crossing out its multiples. But since we start crossing out at the square of the new found prime, we only need to do the crossing out for primes smaller than the square root of N: once we get there, any uncrossed number up to N will be prime. This is what the beginning of our list will look like once we are through:

So lets try and write our very own sieve of Erathostenes in python, with all this square and square root tricks pointed out above…

from time import time
from math import sqrt

def primeListSofE(n, verbose = True) :
t = time()
sieve = [True for j in xrange(2,n+1)]
for j in xrange(2,int(sqrt(n))+1) :
i = j-2
if sieve[i]:
for k in range(j*j,n+1,j) :
sieve[k-2] = False

ret = [j for j in xrange(2,n+1) if sieve[j-2]]

if verbose :
print "Calculated primes to",n,"in",time()-t,"sec."

return ret

Play around with it and once more feel the thrill of speed. On my computer it does all primes below 1,000,000 almost 5 times faster than the best trial division approach from the previous post. But the real beauty is in how the time now scales with n, because you can no longer find a suitable fit of the form nq. We haven’t simply reduced the size of the exponent, but are treading uncharted territory, as it turns out that the time required to calculate all primes below using this sieve requires (n.log(n))(log(log(n))) time. Actually, for the first 1010 numbers, this is basically indistinguishable from linear time, i.e proportional to n.

This speed does come at a memory cost, though: you need to store an array of size n. And while the algorithm could produce all primes below 1010 in just a few minutes, the program will very likely crash giving an out of memory error. Some memory saving can be done by not storing even numbers at all. Without boring you with the details, below is the python function I use to compute primes at work, which does 107 almost three times faster than the previous one, and uses half the memory, but still crashes when asked for all primes below 109

from time import time
from math import sqrt

def primeListSofE(n, verbose = False) :
"""
Returns a list of prime numbers smaller than or equal to n,
using an optimized sieve of Erathostenes algorithm.
"""
t = time()
maxI = (n-3) // 2
maxP = int(sqrt(n))
sieve = [True for j in xrange(maxI+1)]
for p in xrange(3,maxP+1,2) :
i = (p-3) // 2
if sieve[i] :
i2 = (p*p-3) // 2
for k in xrange(i2,maxI+1,p) :
sieve[k] = False
ret = [2]+[2*j+3 for j in xrange(len(sieve)) if sieve[j]]
if verbose :
print "Calculated primes to",n,"in",time()-t,"sec."
return ret

There are slightly faster ways of sieving, and we could memoize precalculated primes, to speed up repeated calls to the function. But this is probably enough for today’s post…