Dixon’s Factorization Method with implementation

Last Updated : 30 Mar, 2020

Dixon’s Factorization method is an integer factorization algorithm. In this article, this method is explained to find the factors of a composite number.

Dixon Factorization is based on the well-known fact of number theory that:

If $x^{2}\equiv y^{2}(mod\;n)$ with $x \neq \pm y(mod\; n)$ it is likely that gcd(x – y, n) will be factor of n.

For example:

if N = 84923,
by starting at 292, the first number greater than √N and counting up
5052 mod 84923 = 256 = 16²
So (505 – 16)(505 + 16) = 0 mod 84923.
Computing the greatest common divisor of 505 – 16 and N using Euclid’s algorithm gives 163, which is a factor of N.

Dixon’s Factorization Algorithm:

Step 1: Choose a bound B and identify the factor base (P) of all primes less than or equal to B.
Step 2: Search for positive integer z, such that $z^{2}mod(n)$ is B-Smooth.
(1) $\begin{equation*}z^{2}\equiv\prod_{p_{i} \in P}p_{i}^{a^{i}} mod(n)\end{equation*}$
B-Smooth: A positive integer is called B-Smooth if none of its prime factors is greater than B.
For example:
720 has prime factorization as 2⁴ * 3² * 5¹
Therefore 720 is 5 smooth because none of its prime factors is greater than 5.
Step 3: After generating enough of these relations (generally few more than the size of P), we use the method of linear algebra (e.g Gaussian Elimination) to multiply together these relations. Repeat this step until we formed a sufficient number of smooth squares.
Step 4: After multiplying all these relations, we get the final equation say:
```
a² = b² mod(N)
```
Step 5: Therefore, the factors of the above equation can be determined as:
```
GCD(a - b, N), GCD(a + b, N)
```

Step by Step execution of the Dixon’s Factorization Algorithm:

Let’s say, we want to factor N = 23449 using bound B = 7. Therefore, the factor base P = {2, 3, 5, 7}.
Here, x = ceil(sqrt(n)) = 154. So, we search randomly for integers between 154 and N whose squares are B-Smooth.
As mentioned before, a positive integer is called B-Smooth if none of its prime factors is greater than B. So, let’s say, we find two numbers which are 970 and 8621 such that their none of their squares have prime factors greater than 7.
Starting here the first related squares we get are:
- 970² % 23499 = 2940 = 2² * 3 * 5 * 7²
- 8621² % 23499 = 11760 = 2⁴ * 3 * 5 * 7²
So, (970 * 8621)² = (2³ * 3 * 5 * 7²)² % 23499. That is:
14256² = 5880² % 23499.

Now, we find:

gcd(14256 - 5880, 23449) = 131
gcd(14256 + 5880, 23449) = 179

Therefore, the factors are:
N = 131 * 179

Below is the implementation of the above approach:

C++

// C++ implementation of Dixon factorization algo
#include<bits/stdc++.h>
using namespace std;
#include<vector>
  
// Function to find the factors of a number
// using the Dixon Factorization Algorithm
void factor(int n)
{
   
    // Factor base for the given number
    int base[4] = {2, 3, 5, 7};
   
    // Starting from the ceil of the root
    // of the given number N
    int start = int(sqrt(n));
   
    // Storing the related squares
    vector<vector<int> >pairs;
      
    // For every number from the square root 
    // Till N
    int len=sizeof(base)/sizeof(base[0]);
    for(int i = start; i < n; i++)
    {
        vector<int> v;
          
        // Finding the related squares 
        for(int j = 0; j < len; j++)
        {
            int lhs = ((int)pow(i,2))% n;
            int rhs = ((int)pow(base[j],2)) % n;
               
            // If the two numbers are the 
            // related squares, then append
            // them to the array 
            if(lhs == rhs)
            {
                v.push_back(i);
                v.push_back(base[j]);
                pairs.push_back(v);
            }
                  
        }
    }
   
    vector<int>newvec;
   
    // For every pair in the array, compute the 
    // GCD such that 
    len = pairs.size();
    for (int i = 0; i < len;i++){
        int factor = __gcd(pairs[i][0] - pairs[i][1], n);
           
        // If we find a factor other than 1, then 
        // appending it to the final factor array
        if(factor != 1)
            newvec.push_back(factor);
   
    }
    set<int>s;
    for (int i = 0; i < newvec.size(); i++)
        s.insert(newvec[i]);
    for(auto i = s.begin(); i != s.end(); i++)
        cout<<(*i)<<" ";
}
   
// Driver Code
int main()
{
    factor(23449);
}
  
// This code is contributed by chitranayal

Python3

# Python 3 implementation of Dixon factorization algo
  
from math import sqrt, gcd
import numpy as np
  
# Function to find the factors of a number
# using the Dixon Factorization Algorithm
def factor(n):
  
    # Factor base for the given number
    base = [2, 3, 5, 7]
  
    # Starting from the ceil of the root
    # of the given number N
    start = int(sqrt(n))
  
    # Storing the related squares
    pairs = []
  
    # For every number from the square root 
    # Till N
    for i in range(start, n):
  
        # Finding the related squares 
        for j in range(len(base)):
            lhs = i**2 % n
            rhs = base[j]**2 % n
              
            # If the two numbers are the 
            # related squares, then append
            # them to the array 
            if(lhs == rhs):
                pairs.append([i, base[j]])
  
    new = []
  
    # For every pair in the array, compute the 
    # GCD such that 
    for i in range(len(pairs)):
        factor = gcd(pairs[i][0] - pairs[i][1], n)
          
        # If we find a factor other than 1, then 
        # appending it to the final factor array
        if(factor != 1):
            new.append(factor)
  
    x = np.array(new)
  
    # Returning the unique factors in the array
    return(np.unique(x))
  
# Driver Code
if __name__ == "__main__":
    print(factor(23449))

Output:

[131 179]

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