--%>

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 : Calculus I need it within 4 hours. Due

    I need it within 4 hours. Due time March 15, 2014. 3PM Pacific Time. (Los Angeles, CA)

  • Q : How do it? integral e^(-t)*e^(tz) t

    integral e^(-t)*e^(tz) t between 0 and infinity for Re(z)<1

  • 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 : Problem on budgeted cash collections

    XYZ Company collects 20% of a month's sales in the month of sale, 70% in the month following sale, and 5% in the second month following sale. The remainder is not collectible. Budgeted sales for the subsequent four months are:     

  • Q : Who firstly use the finite-difference

    Who firstly use the finite-difference method?

  • Q : Who independently developed

    Who independently developed a model for simply pricing risky assets?

  • Q : Examples of groups Examples of groups:

    Examples of groups: We now start to survey a wide range of examples of groups (labelled by (A), (B), (C), . . . ). Most of these come from number theory. In all cases, the group axioms should be checked. This is easy for almost all of the examples, an

  • Q : Formal logic 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 : What is the definition of a group Group

    Group: Let G be a set. When we say that o is a binary operation on G, we mean that o is a function from GxG into G. Informally, o takes pairs of elements of G as input and produces single elements of G as output. Examples are the operations + and x of

  • Q : State Fermat algorithm The basic Fermat

    The basic Fermat algorithm is as follows: Assume that n is an odd positive integer. Set c = [√n] (`ceiling of √n '). Then we consider in turn the numbers c2 - n; (c+1)2 - n; (c+2)2 - n..... until a perfect square is found. If th