We know the formula for calculating n! (factorial of n) is:
n! = n * (n-1)!
We can interpret this simple mathematical equation into a Prolog program. To do so, we must determine the basis of the recursion,
0! = 1
We will use two predicates here,
- factorial predicate with one argument N, that will calculate and N!
- factorial predicate with two arguments N and X. This function is recursively used. It will also calculate N!, but store it in the argument X in the process if recursion.
Code snippet given below is a sample program for the purpose.
predicatesfactorial(integer,integer)
factorial(integer)
clauses/* Base Case, 0!=1*/
factorial(0,X):-
X=1./* recursion for factorial */factorial(N,X):-
NN=N-1,factorial(NN,X1),X=X1*N./*One argument function*/factorial(N):-
factorial(N,X),write(X).
Hope you will enjoy!
inside the insight 
very poor
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What is poor?
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I want it by take number from user not in rule form
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You can do that. First take an input from somewhere, them call the top level rule. This SO post has a quick overview on read rule: http://stackoverflow.com/questions/5107745/user-input-how-can-we-do-it
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my friend all the inputs are Out of local stack , thank you first of all but the factorial program you made is not efficient
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Here is a interesting thing: in Prolog, such recursion unfold in a different way than structured programming languages. This code was not written to be efficient, but to be expressive in a similar way to mathematics.
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