Fermat’s Factorization method for large numbers
Given a large number N, the task is to divide this number into a product of two factors, using Fermat’s Factorisation method.
Examples
Input: N = 105327569
Output: 10223, 10303Input: N = 249803
Output: 23, 10861
Fermat Factorization: Fermat’s Factorization method is based on the representation of an odd integer as the difference of two squares.
For an integer N, we want a and b such as:
N = a2 - b2 = (a+b)(a-b) where (a+b) and (a-b) are the factors of the number N.
Approach:
- Get the number as an object of BigInteger class
- Find the square root of N.
- It is guaranteed that the value of a is greater than sqrt(N) and value of b less than sqrt(N).
- Take the value of sqrt(n) as a and increment the number until and unless a number b is found such that N – a^2 is a perfect square.
Below is the implementation of the above approach:
// Java program for Fermat's Factorization // method for large numbers import java.math.*; import java.util.*; class Solution { // Function to find the Floor // of square root of a number public static BigInteger sqrtF(BigInteger x) throws IllegalArgumentException { // if x is less than 0 if (x.compareTo(BigInteger.ZERO) < 0 ) { throw new IllegalArgumentException( "Negative argument." ); } // if x==0 or x==1 if (x.equals(BigInteger.ZERO) || x.equals(BigInteger.ONE)) { return x; } BigInteger two = BigInteger.valueOf(2L); BigInteger y; // run a loop y = x.divide(two); while (y.compareTo(x.divide(y)) > 0 ) y = ((x.divide(y)).add(y)) .divide(two); return y; } // function to find the Ceil // of square root of a number public static BigInteger sqrtC(BigInteger x) throws IllegalArgumentException { BigInteger y = sqrtF(x); if (x.compareTo(y.multiply(y)) == 0 ) { return y; } else { return y.add(BigInteger.ONE); } } // Fermat factorisation static String FermatFactors(BigInteger n) { // constants BigInteger ONE = new BigInteger( "1" ); BigInteger ZERO = new BigInteger( "0" ); BigInteger TWO = new BigInteger( "2" ); // if n%2 ==0 then return the factors if (n.mod(TWO).equals(ZERO)) { return n.divide(TWO) .toString() + ", 2" ; } // find the square root BigInteger a = sqrtC(n); // if the number is a perfect square if (a.multiply(a).equals(n)) { return a.toString() + ", " + a.toString(); } // else perform factorisation BigInteger b; while ( true ) { BigInteger b1 = a.multiply(a) .subtract(n); b = sqrtF(b1); if (b.multiply(b).equals(b1)) break ; else a = a.add(ONE); } return a.subtract(b).toString() + ", " + a.add(b).toString(); } // Driver code public static void main(String args[]) { String N = "105327569" ; System.out.println( FermatFactors( new BigInteger(N))); } } |
Output:
10223, 10303
Performance Analysis:
- Time Complexity: O(sqrt(N))
- Space Complexity: O(1)
Please Login to comment...