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
with 
it is likely that gcd(x – y, n) will be factor of n.
with 
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
is B-Smooth.
(1)
B-Smooth: A positive integer is called B-Smooth if none of its prime factors is greater than B. For example:
is B-Smooth.
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 Algorithmvoid 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 Codeint main(){ factor(23449);}// This code is contributed by chitranayal |
Java
// Java implementation of Dixon factorization algoimport java.util.*;class GFG { static int gcd(int num1, int num2) { int a = Math.abs(num1); int b = Math.abs(num2); while ((a != 0) && (b != 0) && (a != b)) { if (a > b) { a = a - b; } else { b = b - a; }; }; return a | b; } // Function to find the factors of a number // using the Dixon Factorization Algorithm static void factor(int n) { // Factor base for the given number int[] base1 = { 2, 3, 5, 7 }; // Starting from the ceil of the root // of the given number N int start = (int)(Math.sqrt(n)); // For every number from the square root // Till N int len = base1.length; // Storing the related squares ArrayList<ArrayList<Integer> > pairs = new ArrayList<ArrayList<Integer> >(); for (int i = start; i < n; i++) { // Finding the related squares for (int j = 0; j < len; j++) { int lhs = ((int)Math.pow(i, 2)) % n; int rhs = ((int)Math.pow(base1[j], 2)) % n; // If the two numbers are the // related squares, then append // them to the array if (lhs == rhs) { ArrayList<Integer> l1 = new ArrayList<Integer>(); l1.add(i); l1.add(base1[j]); pairs.add(l1); } } } ArrayList<Integer> newvec = new ArrayList<Integer>(); // 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.get(i).get(0) - pairs.get(i).get(1), n); // If we find a factor other than 1, then // appending it to the final factor array if (factor != 1) newvec.add(factor); } HashSet<Integer> s = new HashSet<Integer>(); for (int i = 0; i < newvec.size(); i++) s.add(newvec.get(i)); for (int s1 : s) System.out.print(s1 + " "); } // Driver Code public static void main(String[] args) { factor(23449); }}// This code is contributed by phasing17 |
Python3
# Python 3 implementation of Dixon factorization algofrom math import sqrt, gcdimport numpy as np# Function to find the factors of a number# using the Dixon Factorization Algorithmdef 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 Codeif __name__ == "__main__": print(factor(23449)) |
C#
// C# implementation of Dixon factorization algousing System;using System.Collections.Generic;class GFG{ static int gcd (int num1, int num2) { int a = Math.Abs(num1); int b = Math.Abs(num2); while ( (a != 0) && (b != 0) && (a != b)) { if(a > b){ a = a - b; }else{ b = b - a; }; }; return a | b; } // Function to find the factors of a number // using the Dixon Factorization Algorithm static void factor(int n) { // Factor base for the given number int[] base1 = {2, 3, 5, 7}; // Starting from the ceil of the root // of the given number N int start = (int)(Math.Sqrt(n)); // Storing the related squares List<int []> pairs = new List<int []>(); // For every number from the square root // Till N int len= base1.Length; for(int i = start; i < n; i++) { List<int> v = new List<int>(); // Finding the related squares for(int j = 0; j < len; j++) { int lhs = ((int)Math.Pow(i, 2)) % n; int rhs = ((int)Math.Pow(base1[j], 2)) % n; // If the two numbers are the // related squares, then append // them to the array if(lhs == rhs) { pairs.Add(new [] {i, base1[j]}); } } } List<int> newvec = new List<int>(); // For every pair in the array, compute the // GCD such that len = pairs.Count; 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.Add(factor); } HashSet<int> s = new HashSet<int>(); for (int i = 0; i < newvec.Count; i++) s.Add(newvec[i]); foreach (var s1 in s) Console.Write(s1 + " "); } // Driver Code public static void Main(string[] args) { factor(23449); }}// This code is contributed by phasing17 |
Javascript
// JS implementation of Dixon factorization algoconst gcd = (num1, num2) => { let a = Math.abs(num1); let b = Math.abs(num2); while (a && b && a !== b) { if(a > b){ [a, b] = [a - b, b]; }else{ [a, b] = [a, b - a]; }; }; return a || b;};// Function to find the factors of a number// using the Dixon Factorization Algorithmfunction factor(n){ // Factor base for the given number let base = [2, 3, 5, 7]; // Starting from the ceil of the root // of the given number N let start = Math.floor(Math.sqrt(n)); // Storing the related squares let pairs = []; // For every number from the square root // Till N let len= base.length; for(let i = start; i < n; i++) { let v = []; // Finding the related squares for(let j = 0; j < len; j++) { let lhs = (i ** 2)% n; let rhs = ((base[j] ** 2)) % n; // If the two numbers are the // related squares, then append // them to the array if(lhs == rhs) { pairs.push([i, base[j]]); } } } let newvec = []; // For every pair in the array, compute the // GCD such that len = pairs.length; for (let i = 0; i < len;i++){ let 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(factor); } let s = new Set(); for (let i = 0; i < newvec.length; i++) s.add(newvec[i]); console.log(s)} // Driver Codefactor(23449);// This code is contributed by phasing17 |
Output:
[131 179]
Time Complexity : O(sqrt(N))
Space complexity : O(B)
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