--%>

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 : Problem on inventory merchandise AB

    AB Department Store expects to generate the following sales figures for the next three months:                            

  • Q : Profit-loss based problems A leather

    A leather wholesaler supplies leather to shoe companies. The manufacturing quantity requirements of leather differ depending upon the amount of leather ordered by the shoe companies to him. Due to the volatility in orders, he is unable to precisely predict what will b

  • Q : Problem on Fermats method A public key

    A public key for RSA is published as n = 17947 and a = 3. (i) Use Fermat’s method to factor n. (ii) Check that this defines a valid system and find the private key X.

    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 : 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 : Problem on Linear equations Anny, Betti

    Anny, Betti and Karol went to their local produce store to bpought some fruit. Anny bought 1 pound of apples and 2 pounds of bananas and paid $2.11.  Betti bought 2 pounds of apples and 1 pound of grapes and paid $4.06.  Karol bought 1 pound of bananas and 2

  • Q : Explain a rigorous theory for Brownian

    Explain a rigorous theory for Brownian motion developed by Wiener Norbert.

  • Q : What is Big-O hierarchy The big-O

    The big-O hierarchy: A few basic facts about the big-O behaviour of some familiar functions are very important. Let p(n) be a polynomial in n (of any degree). Then logbn is O(p(n)) and p(n) is O(an<

  • Q : Statistics math Detailed explanation of

    Detailed explanation of requirements for Part C-1 The assignment states the following requirement for Part 1, which is due at the end of Week 4: “Choose a topic from your field of study. Keep in mind you will need to collect at least [sic] 3- points of data for this project. Construct the sheet y

  • Q : Law of iterated expectations for

     Prove the law of iterated expectations for continuous random variables. 2. Prove that the bounds in Chebyshev's theorem cannot be improved upon. I.e., provide a distribution that satisfies the bounds exactly for k ≥1, show that it satisfies the bounds exactly, and draw its PDF. T