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In this article, we have explored in great detail some of the different factorization algorithms for Integers such as Trial Division, Pollard's rho algorithm, Dixon's algorithm and Quadratic Sieve.
Table of Contents
 Introduction
 Category 1 algorithms
 Trial division
 Pollard's rho algorithm
 Category 2 algorithms
 Dixon's algorithm
 Quadratic sieve
The Time Complexity of the different algorithms for Integer Factorization are:
Algorithm  Time Complexity 

Trial division  O(N) 
Pollard's rho algorithm  O(N^{1/2}) 
Dixon's algorithm  O(e^{2âˆš2 âˆš(logn loglogn)}) 
Quadratic sieve  O(e^{(âˆš(log n log log n))}) 
Let us get started with Integer Factorization Algorithms.
Introduction
In number theory, integer factorization is the decomposition of a composite number into smaller nontrivial divisors, which when multiplied together equals the original integer. There are many different algorithms present to factorize an integer. Depending on the running time of the algorithms, they have been classified into Category 1 and Category 2 which comprises of nonquantum integer factorization algorithms.
Category 1 algorithms
This category of algorithms are also known as special purpose algorithms. The running time of these algorithms depend on certain parameters or unknown factors of the integers like the size, special form, etc.
Some algorithms that come under this category are:
 Trial division
 Pollard's rho algorithm
Let us discuss these algorithms in detail.
Trial division
In trial division, the integer to be factorized (n) is divided by every number less than it. It consists of continually testing if n is divisible by a smaller number.
Further research has shown that the factors of n may be less than or equal to half of it (n/2), but never greater. Implementing this improves the efficiency of this algorithm.Given below is the code for the above
def trial_division(n: int) :
a = []
f = 2
while n > 1:
if n % f == 0:
a.append(f)
n //= f
else:
f += 1
return a
trial_division(12)
The above program gives a list of prime factors of 12 using the trial division.
Output:
[2,2,3]
The Time Complexity of the above algorithm is O(N). Specifically, there are N/2 steps.
Pollard's rho algorithm
In this algorithm, the greatest common divisor of modulus of the difference between two integers and the integer to be factorized (n) is found out. If it is greater than 1, then that is one of the nontrivial factors of n.
This algorithm gets its name 'rho' as the sequence generated will cycle as each value depends on the value before it and the structure of this cycling resembles the Greek letter rho.
The two integers are found by using a function Â f which generates pseudo random numbers. One such function used is â€ƒf = x^{2}+ a mod n where a is generated using a random number generator. We generally start with x_{1}=2 and perform the operations x_{2}=f(x_{1}), x_{3}=f(x_{2}) i.e. the general rule is x_{n+1}=f(x_{n}).
Scheme for the algorithm
x = 2;
while (.. exit criterion .. )
y = f(x);
p = GCD(  y  x  , N);
if ( p > 1)
return "Found factor:";
x = y;
end
return "Failed"
Example:
Let us take n = 55 and f(x) = x^{2} + 2 mod 55.In the table below GCD refers to GCD(x_{n}  x_{(n+1)}, N).
x_{n}  x_{(n+1)}  GCD 

2  6  1 
6  38  1 
38  16  1 
16  36  5 
From the table, 5 is one of the nontrivial factors of 55. The other factor can simply be found by dividing 55 by 5 or continuing the algorithm. Therefore, the factors are 5 * 11.
The Time Complexity of this algorithm will be O(N^{1/2}) where N is the total number of possible values.
Category 2 algorithms
This category of algorithms are also known as general purpose algorithms or Kraitchik family algorithms. The running time of these algorithms depend on only on the size of the integer to be factored. They are used to factor RSA numbers.
Some algorithms that come under this category are:
 Dixon's algorithm
 Quadratic sieve
Let us explore these algorithm in depth.
Dixon's algorithm
Let the number to be factorized be n.

We choose a bound b and a factor base (p) of all primes less than or equal to b.

Look for an integer x such that Â x^{2} mod(n) is bsmooth.
( An integer greater than 0 is called bsmooth if all of its prime factors are lesser than or equal to b).

After we obtain sufficient realtions, we use methods like Gaussian elimination to multiply them together and obtain a final expression in the form Â y^{2}= z^{2} mod(n) .

We then find the factors of the above equation as
GCD(y  z, n), GCD(y + z, n).
The Time Complexity of this method is O(e^{2âˆš2 âˆš(logn loglogn)}) where e = 2.718.
Example
Say we want to factor n = 23449 over p = 2, 3, 5, 7 and b=7.
x = [âˆšn] = 154. Starting here, the first related squares we get are:
970^{2} mod(23449) = 2940 = 2^{2} âˆ— 3 âˆ— 5 âˆ— 7^{2}
8621^{2} mod(23449) = 11760 = 2^{4} âˆ— 3 âˆ— 5 âˆ— 7^{2}
Here, 970 and 8621 are bsmooth numbers whose factors are not greater than 7.
So, (970* 8621)^{2} â‰¡ (2^{3} âˆ— 3 âˆ— 5 âˆ— 7^{2})^{2} (mod 23449)
That is, 14526^{2} â‰¡ 5880^{2} (mod 23449)
Now , we find:
gcd(145265880,23449) = 131
gcd(14526+5880,23449) = 179
Factors are N = 131 * 179
Quadratic sieve
This algorithm is a modification of Dixon's algorithm.
In Dixon's algorithm, we chose 2 from the set of realtios obtained to arrive at the 'congruence of squares modulo n' form. But here, we assign a variable to each of the relation, and work mod 2 so that all the exponents are even. We then obtain a set of reduced relations and then find the values of the variables. These values will tell us which realtions are to be considered so that we get a 'congruence of squares modulo n' form.
The theoretical time and space complexity of this algorithm is O(e^{(âˆš(log n log log n))}).
Example
Let the number to be factorized be n = 539873 and b=19
x = [âˆšn] = 734.6. Starting from 735, we work our way up.
Working mod 539873:
735^{2} â‰¡ 352 â‰¡ 2^{5} âˆ— 11
750^{2} â‰¡ 22627 â‰¡ 11^{3} âˆ— 17
783^{2} â‰¡ 73216 â‰¡ 2^{9} âˆ— 11 âˆ— 13
801^{2} â‰¡ 101728 â‰¡ 2^{5} âˆ— 11 âˆ— 17^{2}
Solving
we get,
working mod 2:
which gives us a non trivial solution x_{1}=1, x_{2}=0, x_{3}=0, x_{4}=1.
Therefore, the first and fourth relations are multiplied to get the 'congruence of squares modulo n' form.
Further proceeding, we have:
By applying the congruence of squares, we will find that the GCD of the difference of the two integers and 539873 will be a factor of 539873. That is:
Hence, 539873 = 1949 * 277
With this article at OpenGenus, you must have the complete idea of Integer Factorization Algorithms.