--%>

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 : Pig Game Using the PairOfDice class

    Using the PairOfDice class design and implement a class to play a game called Pig. In this game the user competes against the computer. On each turn the player rolls a pair of dice and adds up his or her points. Whoever reaches 100 points first, wins. If a player rolls a 1, he or she loses all point

  • Q : Logic and math The homework is attached

    The homework is attached in the first two files, it's is related to Sider's book, which is "Logic for philosophy" I attached this book too, it's the third file.

  • Q : Elasticity of Demand For the demand

    For the demand function D(p)=410-0.2p(^2), find the maximum revenue.

  • Q : Econ For every value of real GDP,

    For every value of real GDP, actual investment equals

  • Q : Breakfast program if the average is

    if the average is 0.27 and we have $500 how much break fastest will we serve by 2 weeks

  • Q : Linear programming model of a Cabinet

    A cabinet company produces cabinets used in mobile and motor homes. Cabinets produced for motor homes are smaller and made from less expensive materials than those for mobile homes. The home office in Dayton Ohio has just distributed to its individual manufacturing ce

  • Q : Theorem-Group is unique and has unique

    Let (G; o) be a group. Then the identity of the group is unique and each element of the group has a unique inverse.In this proof, we will argue completely formally, including all the parentheses and all the occurrences of the group operation o. As we proce

  • Q : Uniform scaling what is uniform scaling

    what is uniform scaling in computer graphic

  • Q : Row-echelon matrix Determine into which

    Determine into which of the following 3 kinds (A), (B) and (C) the matrices (a) to (e) beneath can be categorized:       Type (A): The matrix is in both reduced row-echelon form and row-echelon form. Type (B): The matrix

  • 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.