--%>

Explain Factorisation by trial division

Factorisation by trial division: The essential idea of factorisation by trial division is straightforward. Let n be a positive integer. We know that n is either prime or has a prime divisor less than or equal to √n. Therefore, if we divide n in turn by the primes 2, 3, 5,..., going possibly as far as [√n], we will either encounter a prime factor of n or otherwise be able to infer that n is prime. Repeating this process as often as necessary we will be able to nd all the prime
factors of n.

We can re fine this idea a little. If we fi nd on division by the prime p that p is a factor of n, then we can recommence trial divisions, but now dividing into the integer n/p rather than n. Also, the divisions can start with the prime p rather than restarting with 2, since we know that n, and hence n/p, has no prime factors smaller than p.

Further, we now need only carry out trial divisions by primes up to [√n/p]. Similarly for later steps.

An obvious difficulty with trial division is that we need either to store or to generate the primes up to [√n], and it may be better simply to divide by all the integers from 2 up to [√n], or, for example, by 2 and then all the odd numbers up to [√n].

Other improvements are possible too, but even with a few improvements the trial division algorithm is inefficient , and the algorithm is unsuitable for all but fairly small initial values of n.

Despite this, the trial division algorithm is in practical use. It is often used as a preliminary phase in a factorisation algorithm to nd the `small' prime factors of a number, and the remaining unfactorised part, containing all the `large' prime factors, is left to later phases.

Most numbers have some small prime factors. For example, it is not hard to show that about 88% of positive integers have a prime factor less than 100 and that about 91% have a prime factor less than 1000, and trial division will be good at finding these factors.

On the other hand, most numbers also have large prime factors. It can be shown (though not so easily) that a random positive integer n has a prime factor greater than √n with probability ln 2, or about 69% of the time, and of course if n is large, then trial division will not be of any help in nding such a factor.

   Related Questions in Mathematics

  • Q : Who firstly discovered mathematical

    Who firstly discovered mathematical theory for random walks, that rediscovered later by Einstein?

  • Q : Set Theory & Model of a Boolean Algebra

    II. Prove that Set Theory is a Model of a Boolean Algebra The three Boolean operations of Set Theory are the three set operations of union (U), intersection (upside down U), and complement ~.  Addition is set

  • Q : Mean and standard deviation of the data

    Below is the amount of rainfall (in cm) every month for the last 3 years in a particular location: 130 172 142 150 144 117 165 182 104 120 190 99 170 205 110 80 196 127 120 175

  • Q : Problem on reduced row-echelon The

    The augmented matrix from a system of linear equations has the following reduced row-echelon form. 280_row echelon method.jpg

  • Q : Explain the work and model proposed by

    Explain the work and model proposed by Richardson.

  • Q : Global And Regional Economic Development

    The Pharmatec Group, a supplier of pharmaceutical equipment, systems and services, has its head office in London and primary production facilities in the US. The company also has a successful subsidiary in South Africa, which was established in 1990. Pharmatec South A

  • Q : Bolzano-Weierstrass property The

    The Bolzano-Weierstrass property does not hold in C[0, ¶] for the infinite set A ={sinnx:n<N} : A is infinite; Show that has no “ limit points”.

  • Q : Elementary Logic Set & Model of a

    Prove that Elementary Logic Set is a Model of a Boolean Algebra The three Boolean operations of Logic are the three logical operations of  OR ( V ), AN

  • Q : Formal logic2 It's a problem set, they

    It's a problem set, they are attached. it's related to Sider's book which is "Logic to philosophy" I attached the book too. I need it on feb22 but feb23 still work

  • Q : Explain Factorisation by Fermats method

    Factorisation by Fermat's method: This method, dating from 1643, depends on a simple and standard algebraic identity. Fermat's observation is that if we wish to nd two factors of n, it is enough if we can express n as the di fference of two squares.