20081211, 10:09  #1 
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
3·419 Posts 
Implementing Factoring Algorithms
In my leisure time, I can read about factoring algorithms and try to implement them. So far, I have implemented Trial Division and Pollard's Rho, and including a SPRP test for the cofactor in a JAVA program. The source code is attached.
When SPRP test finds that the cofactor is composite, and trial division or Pollard's Rho is unable to find out a factor, it asks for the factor from the user as input. Perhaps that I should implement p1 and Quadratic Sieve as soon as possible such that the frequency of input becomes lesser, and they should not be as hard as to implement when compared to ECM or the Number Field Sieve. The code is organized when compared to my older version, such that it is easy to add new factoring algorithms to the program. Just the function "Factor" has to be modified for it. The factors are stored in a stack, and the function "display_factors" displays them in the usual manner. The code is organized into various functions for each algorithm: Trial Divide, Pollard's Rho, GCD, SPRP, Modular Exponentiation, Sum of factors, etc. It may be necessary in the near future to add algorithms for: Legendre Symbol, Modular Square Root, Euler's phi function, etc. For learning purposes, as a first step, I am using BigInteger class in Java. Please don't criticize me for using that now, I will later on switch over to some other useful and faster library in C, or write my own Big Integer class. Perhaps, Trial Division and Pollard's Rho take far less time for execution when compared to, when I implement ECM or Quadratic Sieve. So, Big Integer may be avoided in such advanced algorithms. 
20081211, 10:40  #2  
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
3×419 Posts 
p1
Suppose that I want to implement p1 as my next algorithm. Stage 1.
Quote:
1) Suppose that first I calculate LCM of all numbers to B and store it in a variable, and then do exponentiation by squaring, say recursively Code:
a^{L} mod p if L = 1 then return a if L mod 2 = 0 then return a^{2 (L/2)} mod p if L mod 2 = 1 then return a × a^{2 (L1)/2} mod p 2) On the other hand, if I do exponentiation by squaring for every prime, (and for small primes, their powers below B). To the maximum, log_{2} B squarings are needed so, for every prime. If there are Pi(B) primes below B, then the number of squarings that will be needed so, will thus be, O(Pi(B) log_{2} B). Which way is more efficient, or are there any more optimization methods? If so, let me know about them up. 

20081211, 12:00  #3 
(loop (#_fork))
Feb 2006
Cambridge, England
3×19×113 Posts 
Dan Bernstein, whose expertise in optimisation is terrifying, recommends computing L first if you're going for bestpossible performance; you need an efficient multiplication algorithm and I'm not sure whether java BigIntegers use FFTbased multiply. Then compute the product of the primes by sticking them into a priority queue and successively taking the two smallest elements, multiplying them, and putting that back into the queue.
There are various 'window methods' (I can't find a good article about them) are a bit more efficient than basic exponentiationbysquaring; you can't avoid O(log_2 L) squarings, but you can get the number of multiplications by a to be much smaller at the price of a bit of precomputation. The easiest one is just to work in base 4 or 8, precompute a^(2..7) and use a lookup table. You probably want to use Montgomery representation for the modular arithmetic, which is another interesting exercise. NB if I read http://docjar.net/html/api/java/math...eger.java.html correctly, BigInteger's modpow algorithm uses both windowing and Montgomery representation. If you're implementing things for fun, I strongly recommend FFTbased multiplication: it's one of the really magical algorithms. 
20081211, 12:58  #4 
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
1257_{10} Posts 
Sorry to question about FFT without reading about it.
But is it possible to understand about Fast Fourier Transform algorithm without knowing about ordinary Fourier Transforms or Laplace Transforms? I think that the latter two are not related to number theory, right? How many times, the speed will increase when using FFT, than without using FFT? Is it too difficult to understand it, or to implement it up? I am implementing these algorithms to learn about them. 
20081211, 16:41  #5 
Tribal Bullet
Oct 2004
3×1,181 Posts 
See the Handbook of Applied Cryptography online for more on efficient modular exponentiation.
I think most of the FFT references on the web assume you are interested in signal processing and not number theory. Multiplying two numbers using an FFT involves converting each number into a "signal" and performing a convolution on the two signals using FFTs. This adds a significant amount of overhead, but the asymptotic complexity is so much better than other convolution methods that beyond a certain size of input you have no choice but to use transform arithmetic. The difference is really between 'impossible' and 'a few seconds' once the inputs become large enough. 
20081211, 16:53  #6  
(loop (#_fork))
Feb 2006
Cambridge, England
3·19·113 Posts 
Quote:
All you need to know for the FFT is that you're computing out[j] = sum_{i=0}^{N1} omega^{i*j} in[i] where omega is an Nth root of unity, and that you can play about with this to get it as a combination of two similar things involving the even and oddnumbered elements of 'in'; I strongly recommend working out a length8 Fourier transform symbolically with pencil and paper to get what the expression looks like. Quote:


20081211, 19:40  #7 
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
3×419 Posts 
Sieve
Hmmm... Looks like FFT is the key boost to p1 and ECM, and until I understand about that, I cannot implement these algorithms.
Getting to sieving, what is the best way to implement sieve algorithms? Take Quadratic Sieve: Suppose that there are B primes in the factor base, such that N is a quadratic residue mod p, for every prime p in the factor base. Should I need to create an array for storing x^{2}kN, for (say M) consecutive values of x, starting with the sqrt(kN)? In Java, it may take time for the program at run time to allocate M indices of storage for an array or a dynamic Vector. In C, it can be simply done so using int* a=int* malloc(sizeof(int)*m); or in C++ and Java, int a[]=new int[m]; Then, I can check for every prime in the factor base, divides x^{2}kN if and only if x=sqrt(N) mod p. So, divide by the prime p and mark it as a divisor, for the values of x=Kp+sqrt(N), where sqrt(N) can have two values if the Legendre Symbol (p/N)=1, and no solution if 1. What about the Sieve of Eratosthenes? For listing all the primes till (say so) 10000, should I need to create a boolean array, for all values till 10000, and flag it for multiple of every prime till 100? How much is it better than Trial Division, in which for every (odd) number, you test by all the primes starting with 2, till a factor is found out, or till that number is completely being factored so. In trial division, you test by all the primes for each number. In the Sieve of Eratosthenes, for every prime, you flag all its multiples till 10000. Just to me, one of them is a horizontal process and the other is a vertical process. And the Sieve of Eratosthenes, occupies so more space for the boolean array. 
20081211, 21:10  #8 
Tribal Bullet
Oct 2004
3·1,181 Posts 
If P1 or ECM are attempted on large numbers, then arithmetic modulo those large numbers would benefit from using FFTs; but the numbers have to be really large.
The P1 and ECM algorithms themselves don't care whether you use fast transform arithmetic, unless you are performing stage 2 using the advanced techniques in packages like GMPECM. Even then you cannot use floating point FFTs, for technical reasons. 
20081211, 21:46  #9  
"Ben"
Feb 2007
2×1,789 Posts 
Quote:
Sieving can be sped up by cache blocking, bit packing (not sure how the booleans in java are actually represented in hardware... I'm betting they use 8 bits) and the presieving talked about above. There is a good explanation of some of the possible optimizations here, which has some source code to study as well. Quote:


20081213, 18:56  #10 
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
3·419 Posts 
Quadratic Sieve
Have a look at my brief implementation of Quadratic Sieve.
I take 5 arguments, N > The number to be factored start > The number whose square to start sieving k > The multiplier for x^{2}kN it > The number of iterations to sieve FB > The factor base bound Further value[] holds the value of x^{2}kN and residue[] holds the residues after sieving. Code:
For i=0 to it, i=i+1 { value[i] = (start+i)^{2}kN residue[i]=value[i] } Sieving,p is the prime p=2, i=0 if residue[0] mod 2=0 then i=1 else i=0 while i<it { Divide residue[i] by 2 till it is even i=i+2 } Other primes For p=3 to FB, p=p+1 { If p is composite or Legendre symbol (p/kN)=1, then continue i1 = sqrt(kN) mod p // Modular square root is a special function that is being written so separately i2 = pi1 phase = start mod p // This gives the phase of the starting sieving value, mod p start1=(i1phase) mod p start2=(i2phase) mod p // These give the starting index of the two arithmetic progressions For j=start1 to it, j=j+p { Divide residue[i] by p till it is divisible by p } For j=start2 to it, j=j+p { Divide residue[i] by p till it is divisible by p } } For i=0 to it, i=i+1 { If residue[i]=1, then Factor residue[i], and thus display its factors } Here is a sample output: N: 11111111111 start: 105410 k: 1 Number of iterations is 10000 Factor base limit being 100 Output: Code:
3319514 = 2 * 11^3 * 29 * 43 11966045 = 5 * 7^2 * 13^2 * 17^2 17872925 = 5^2 * 7 * 41 * 47 * 53 83798525 = 5^2 * 17 * 37 * 73^2 157135993 = 7^2 * 31^2 * 47 * 71 272498525 = 5^2 * 13 * 17 * 31 * 37 * 43 410549810 = 2 * 5 * 11 * 29 * 41 * 43 * 73 478703225 = 5^2 * 7 * 11^2 * 13 * 37 * 47 512316233 = 11 * 13^3 * 17 * 29 * 43 516413450 = 2 * 5^2 * 7^2 * 41 * 53 * 97 516629113 = 7 * 11 * 13^2 * 29 * 37^2 558721618 = 2 * 7^3 * 13 * 31 * 43 * 47 572553170 = 2 * 5 * 7 * 37 * 43 * 53 * 97 629911625 = 5^3 * 7 * 17^2 * 47 * 53 774031250 = 2 * 5^6 * 17 * 31 * 47 827073533 = 11 * 13 * 29 * 53^2 * 71 872350850 = 2 * 5^2 * 7 * 31 * 37 * 41 * 53 876730010 = 2 * 5 * 13 * 43 * 47^2 * 71 889215005 = 5 * 7^2 * 17 * 31 * 71 * 97 953714489 = 7^2 * 13^2 * 41 * 53^2 1017065273 = 7 * 17^2 * 71 * 73 * 97 1435353010 = 2 * 5 * 7 * 13 * 17 * 31 * 41 * 73 1904276114 = 2 * 13^4 * 17 * 37 * 53 2175831250 = 2 * 5^5 * 37 * 97^2 
20081216, 13:57  #11 
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
3×419 Posts 
I have made notes of the above algorithm, with one trivial and one nontrivial linear dependency.
If anybody has spare time, please check whether all of the notes belong to the Dixon's Factorization Method, or some can be transferred to the Quadratic Sieve. Also let me know if there exist any extra concepts to be known for the Quadratic Sieve, like the Multiple Polynomial Quadratic Sieve, and the like, to optimize the algorithm. The next step will be to implement an algorithm for Linear Algebra, to get dependencies. I can implement Gaussian Elimination with my knowledge, but is Block Lanczos more efficient for sparse matrices? I have absolutely no idea on how it works, although I have read that it is based on continued fraction analysis... I have thought of two optimizations to the code for sieving: 1) Instant storage of prime factors, on division by a prime. Perhaps it may consume lots of additional memory space because representation of prime factorization will take larger space when compared to the number itself. 2) If any residue reaches 1, immediately I can store that index in a hash table, and the lookup process for the residues will be a lot easier, even in constant time, obviously. Rather than searching for all of the array whether the residue is 1 or not. Perhaps, I should try to allow partial relations, with one or more large prime, that is not in the factor base. Of course, it may help to collect more relations in lesser amounts of time, although the probability of the relations merging to a perfect squares may go down a little. If more than one large prime is allowed, then Pollard's Rho might be a suitable algorithm indeed, for verifying whether the cofactor is prime, and then, splitting it up, thus. 
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