The code for implementation of Dixon's Algorithm is as follows:import java.util.Scanner;class Exe{ public static void main(String args[]) Scanner ob=new Scanner(System.in); System.out.print("n------------------n!! ENTER THE NO. !! : "); double n=ob.nextDouble(); System.out.println("n ---------------------------------------------"); dixon(n); public static double gcd(double a,double b) if(b==0) return a; else if(b>a) return gcd(b,a); else return gcd(b,a%b); public static int checkprime(double n) { int f=0; for(int i=2;i
In number theory, Dixon's factorization method (also Dixon's random squares method[1] or Dixon's algorithm) is a general-purpose integer factorization algorithm; it is the prototypical factor base method. Unlike for other factor base methods, its run-time bound comes with a rigorous proof that does not rely on conjectures about the smoothness properties of the values taken by polynomial.
Dixon's implementation in java for prime number factorization
Dixon's method replaces the condition "is the square of an integer" with the much weaker one "has only small prime factors"; for example, there are 292 squares smaller than 84923; 662 numbers smaller than 84923 whose prime factors are only 2,3,5 or 7; and 4767 whose prime factors are all less than 30. (Such numbers are called B-smooth with respect to some bound B.)
Suppose the composite number N is being factored. Bound B is chosen, and the factor base is identified (which is called P), the set of all primes less than or equal to B. Next, positive integers z are sought such that z2 mod N is B-smooth. Therefore it can be written, for suitable exponents ai,
When enough of these relations have been generated (it is generally sufficient that the number of relations be a few more than the size of P), the methods of linear algebra can be used (for example, Gaussian elimination) to multiply together these various relations in such a way that the exponents of the primes on the right-hand side are all even:
Not all factoring problems are difficult; we can spot an even number instantly and know that one of its factors is 2. In fact, there are specific criteria for choosing numbers that are difficult to factor, but the basic idea is to choose the product of two large prime numbers.
Notice, if the number that you want to factorize is actually a prime number, most of the algorithms, especially Fermat's factorization algorithm, Pollard's p-1, Pollard's rho algorithm will run very slow.So it makes sense to perform a probabilistic (or a fast deterministic) primality test before trying to factorize the number.
We divide by each possible divisor $d$.We can notice, that it is impossible that all prime factors of a composite number $n$ are bigger than $\sqrtn$.Therefore, we only need to test the divisors $2 \le d \le \sqrtn$, which gives us the prime factorization in $O(\sqrtn)$.(This is pseudo-polynomial time, i.e. polynomial in the value of the input but exponential in the number of bits of the input.)
The smallest divisor has to be a prime number.We remove the factor from the number, and repeat the process.If we cannot find any divisor in the range $[2; \sqrtn]$, then the number itself has to be prime.
This method can be extended.If the number is not divisible by 3, we can also ignore all other multiples of 3 in the future computations.So we only need to check the numbers $5, 7, 11, 13, 17, 19, 23, \dots$.We can observe a pattern of these remaining numbers.We need to check all numbers with $d \bmod 6 = 1$ and $d \bmod 6 = 5$.So this leaves us with only $33.3\%$ percent of the numbers to check.We can implement this by checking the primes 2 and 3 first, and then start checking with 5 and alternatively skip 1 or 3 numbers.
Extending the wheel factorization with more and more primes will leave exactly the primes to check.So a good way of checking is just to precompute all prime numbers with the Sieve of Eratosthenes until $\sqrtn$ and test them individually.
Fermat's factorization method tries to exploit the fact, by guessing the first square $a^2$, and check if the remaining part $b^2 = a^2 - n$ is also a square number.If it is, then we have found the factors $a - b$ and $a + b$ of $n$.
It is very likely that at least one factor of a number is $B$-powersmooth for small $B$.$B$-powersmooth means that every prime power $d^k$ that divides $p-1$ is at most $B$.E.g. the prime factorization of $4817191$ is $1303 \cdot 3697$.And the factors are $31$-powersmooth and $16$-powersmooth respectably, because $1303 - 1 = 2 \cdot 3 \cdot 7 \cdot 31$ and $3697 - 1 = 2^4 \cdot 3 \cdot 7 \cdot 11$.In 1974 John Pollard invented a method to extracts $B$-powersmooth factors from a composite number.
Let the prime factorization of a number be $n = p q$.The algorithm looks at a pseudo-random sequence $\x_i\ = \x_0,f(x_0),f(f(x_0)),\dots\$ where $f$ is a polynomial function, usually $f(x) = (x^2 + c) \bmod n$ is chosen with $c = 1$.
In number theory, integer factorization is the decomposition of a composite number into smaller non-trivial 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 non-quantum integer factorization algorithms.
Many algorithms have been devised for determining the prime factors of a given number (a process called prime factorization). They vary quite a bit in sophistication and complexity. It is very difficult to build a general-purpose algorithm for this computationally "hard" problem, so any additional information that is known about the number in question or its factors can often be used to save a large amount of time.
The simplest method of finding factors is so-called "direct search factorization" (a.k.a. trial division). In this method, all possible factors are systematically tested using trial division to see if they actually divide the given number. It is practical only for very small numbers. 2ff7e9595c
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