## Naive Integer Factorization

April 9, 2009

After three posts (1, 2, 3) on calculating prime numbers, it is probably worth putting that knowledge to a more useful task. As we will see in a near future, integer factorization, i.e. breaking down a (composite) number into its prime factors is one such task. In purity, factoring a number n is simply decomposing it as the product of two smaller non-trivial, i.e. different from 1 and n itself, divisors. But by repeatedly factoring the divisors one will eventually come up with a unique set of primes which, when multiplied together, render the original number, or so says the fundamental theorem of arithmetic… The point is, we will consider factorization a synonym of prime decomposition, be it formally correct or not.

There are some very sophisticated methods to factor very large numbers, but they use a lot of extremely complex math, so I doubt they will ever find their way onto this blog. So we are going to be left with the naive, straightforward approach as our only option, although I will try to give it an efficiency boost. What is this naive approach? Trial division, of course: given a number n, we know that its smallest factor will be smaller than the square root of n, so we can simply try and see if any of those numbers divide it. No, I will not try to code that yet… If you have read the entries on determining prime numbers, it should come as no surprise that we really do not need to do trial division by all numbers smaller than the square root of n, but only by the primes within. This is a consequence of the fact that, if a composite number divides n, then each of the prime factors of that composite number will also divide n. According to the prime number theorem the number of primes below x is asymptotic to x / log x. So by limiting our trials to prime numbers we can reduce the number of tests from n1/2 to something around 2 n1/2 / log n. If we rescue the `primeListSofE` function from the post on the sieve of Erathostenes, a python implementation of naive factorization could look something like this…

```from time import clock

def factor(n, verbose = False) :
"""
Returns all prime factors of n, using trial division by prime
numbers only. Returns a list of (possibly repeating) prime factors
"""
t = clock()
ret =[]
nn = n
maxFactor = int(n**0.5)
primes = primeListSofE(maxFactor, verbose)
for p in primes :
while nn % p == 0 :
nn //= p
ret += [p]
if nn == 1 :
break
if nn != 1 :
ret += [nn]
t = clock() - t
if verbose :
print "Calculated factors of",n,"in",t,"sec."
return ret
```

While this function will be about as good as we can make it for numbers which are the product of two large prime factors, it will be terribly inefficient for most numbers. Consider, as an extreme example, that we are trying to factor 255 ~ 3.6·1016. We would first calculate all primes up to 1.9·108, a challenging feat in itself with our available tools, only to find out that we only needed the first of those primes, i.e. 2. Taking into account that 50% of all numbers are divisible by 2, 33% are divisible by 3, 20% are divisible by 5… it doesn’t seem wise to disregard the potential time savings. What we can do to profit from this is to do the trial division checks at the same time as we determine the prime numbers, updating the largest prime to test on the fly. This has to be done in two stages, the first while we sieve up to n1/4, the second while we search the rest of the sieve up to n1/2 searching for more primes. The following Franken-code has been written mostly by cut-and-paste from `primeListSofE` and `factor`, which hopefully hasn’t affected its readability much:

```from time import clock

def factorAndSieve(n, verbose = False) :
"""
Returns all prime factors of n, using trial division while sieving
for primes. Returns a list of (possibly repeating) prime factors
"""
t = clock()
ret =[]
nn = n
while nn % 2 == 0 : # remove 2's first, as 2 is not in sieve
nn //= 2
ret += [2]
maxFactor = int(nn**0.5)
maxI = (maxFactor-3) // 2
maxP = int(maxFactor**0.5)
sieve = [True for j in xrange(maxI+1)]
i = 0
for p in xrange(3, maxP+1,2) : # we then sieve as far as needed
if p > maxP :
break
i = (p-3) // 2
if sieve[i] :
while nn % p == 0 :
nn //= p
ret += [p]
maxFactor = int(nn**0.5)
maxI = (maxFactor-3) // 2
maxP = int(maxFactor**0.5)
if nn == 1 :
break
else :
i2 = (p*p - 3) // 2
for k in xrange(i2, maxI+1, p) :
sieve[k] = False
index = i
for i in xrange(index, maxI+1) : # and inspect the rest of the sieve
if i > maxI :
break
if sieve[i] :
p = 2*i + 3
while nn % p == 0 :
nn //= p
ret += [p]
maxFactor = int(nn**0.5)
maxI = (maxFactor-3) // 2
maxP = int(maxFactor**0.5)
if nn == 1 :
break
if nn != 1 :
ret += [nn]
t = clock() - t
if verbose :
print "Calculated factors of",n,"in",t,"sec."
print "Stopped trial division at",2*i+3,"instead of",int(n**0.5)
return ret
```

This new code will very often be much faster than the other one, but at times it will be just about as slow as in the other case, or even slower, since the mixing of both codes introduces some inefficiencies. The most extreme examples of such cases would be a prime number, or the square of a prime number on one side, and a power of 2 on the other one.

The graph above plots times to calculate the factors of numbers between 106 and 106 + 100. Prime numbers in this interval stick out as the red dots among the blue ones: 106 +3, +33, the twin primes +37 and +39, +81 and +99. And numbers with many small prime factors populate the bottom of the red cloud.

If the above graph is not enough to convince you of the benefits of the second approach, maybe this timings for very large numbers will:

``` >>> factor(10**15+37,True) Calculated primes to 31622776 in 6.760 sec. Calculated factors of 1000000000000037 in 8.466 sec. [1000000000000037L] >>> factorAndSieve(10**15+37,True) Calculated factors of 1000000000000037 in 8.666 sec. Stopped trial division at 31622775 instead of 31622776 [1000000000000037L]```

``` ```

```>>> factor(2**55,True) Calculated primes to 189812531 in 42.811 sec. Calculated factors of 36028797018963968 in 43.261 sec. [2, ..., 2] >>> factorAndSieve(2**55,True) Calculated factors of 36028797018963968 in 8.632e-05 sec. Stopped trial division at 3 instead of 189812531 [2, ..., 2]```

## 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…

## Prime Numbers 1. The Wrong Way

March 5, 2009

A prime number is a natural number larger than 1 which is only divisible by 1 and itself. Simple, isn’t it? Euclid already proved quite a while ago that there are infinitely many of them, so lets try and write a function that returns all primes below a given number, using the most straightforward approach. What would that be? Well, there is nothing less sophisticated than to try and divide each number by every integer larger than 1 and smaller than the number itself, and if we never get a zero remainder then the number is prime. Putting it in python:

```from time import time
def primeList1(n, verbose = True) :
t = time()
ret = []
for j in range(2,n+1) :
jIsPrime = True
for k in range(2,j) :
if j % k == 0 :
jIsPrime = False
break
if jIsPrime :
ret += [j]
if verbose :
print "Calculated primes to",n,"in",time()-t,"sec."

return ret<br /></pre><br />
```

Do notice the feeble attempt at code optimization, with the `break` statement that ends the loop as soon as a divisor is found… If you try it out, it will work fine. Use it to calculate all prime numbers below 1,000, and it will probably give you an instant answer. 10,000 delays a second or two, nothing too worrying. If you try it with 100,000, you’ll have time to take your coffee break before it finishes, and adding any more zeroes will require leaving your script running overnight… Not too surprising really, since to determine that, for example, 9973 is indeed a prime number, our program is having to do trial division by every number between 2 and 9972, both inclusive. All in all, the time required grows with n2.

There are several very simple improvements that can speed things up dramatically, and which require a little, but just a little, mathematical insight…

First, we don’t need to test every number between 2 and 9972 to determine that 9973 is prime; we can stop after the largest integer smaller than the square-root of 9973, i.e. 99. Why? Well, if a number is divisible by a number larger than its square-root, it will also be divisible by the quotient of that division, which is necesaarily smaller than the square-root. And since we have already tried (and failed) to divide by those smaller numbers, there is no need to keep going…

The resulting code is very similar to the previous one:

```from time import time
from math import sqrt

def primeList2(n, verbose = True) :
t = time()
ret = []
for j in range(2,n+1) :
jIsPrime = True
kMax = int(sqrt(j))
for k in range(2,kMax+1) :
if j % k == 0 :
jIsPrime = False
break
if jIsPrime :
ret += [j]
if verbose :
print "Calculated primes to",n,"in",time()-t,"sec."

return ret
```

Do try it, and feel the thrill of the new speed of the algorithm. If you are after all primes below 10,000, this new function will probably get them for you… a hundred times faster!!! If you try it for larger numbers, you will find even better speed increases, because the time required by our algorithm now grows with n1.4.

But we are still doing a lot of superfluous checking. Think again of 9973. Once we have tried to divide it by 2 and failed, there is no point in checking with 4, 6, 8 or any other multiple of 2. So yes, we could double the speed of our algorithm by checking only odd numbers, but lets spend a little more time thinking before going back to coding… Because, after failing with 2, once we turn our attention to 3 and again fail, what’s the point in trying 6, 9, 12 or any other multiple of 3? If you keep the reasoning going, it becomes obvious that you only need to do trial division by prime numbers, since by the time we test whether a composite number divides 9973, we have already tested all its prime factors, and if those have failed, this one will as well.

Now wait a minute! Isn’t finding whether a number is prime or not precisely what we are after? How are we going to figure out what numbers to do trial division with? Well, we already saw in our first optimization that our division trials only need to go up to the square root of the number being tested, so we only need to know the first prime number, 2, and afterwards we can use the list of prime numbers that we are building to generate the additional items… The code to do that would look something like this:

```from time import time
from math import sqrt

def primeList3(n, verbose = True) :
t = time()
ret = []
for j in range(2,n+1) :
jIsPrime = True
kMax = sqrt(j)
for k in ret :
if k > kMax :
break
if j % k == 0 :
jIsPrime = False
break
if jIsPrime :
ret += [j]

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

return ret
```

If you test this code you’ll surely notice the new speed increase. This last optimization doesn’t make it a hundred times faster, but on my computer the new code is 3.5 faster than the previous one when calculating all primes below 100,000 and 5.5 times faster when calculating all primes below 1,000,000. And again, what really makes the new code more efficient is not that 4x increase in speed, but the fact that, the larger our input number gets, the better the new code is. It seems that now time required grows with n1.25

This last version holds the fundamental ideas of really fast code, i.e. skipping multiples of primes, but still relies on the slow and time consuming operation of trial division. Getting rid of it is the next optimization, which we will see in the next post.