# recursive

abalfazl hossein

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posted 5 years ago

- 0

http://cse.unl.edu/~dsadofs/RecursionTutorial/index.php?s=algorithms#recfactalg

In this tutorial , it is said that a recursive function divide to the two part, Base and inductive

Now, look at this function:

In this function, If n>0 is base, and rem=n%10;

s += rem;

sum(n/10)

Is inductive?

Is it right that said every recursive function divide to the two parts?

Well, as stated in the description of a recursive definition, a base case and inductive rule were needed. In this case, fact(0)=1 is the base case. This is a logical place to start, seeing as how the n in fact(n) must be larger than or equal to 1 (only positive integers). The other half is the inductive rule of the problem, which in this case is represented by “fact(n)=fact(n-1)*n”. We can see the logic in this, as the factorial of any number, is simply the factorial of the number one smaller than it, times the number you’re finding the factorial of.

In this tutorial , it is said that a recursive function divide to the two part, Base and inductive

Now, look at this function:

In this function, If n>0 is base, and rem=n%10;

s += rem;

sum(n/10)

Is inductive?

Is it right that said every recursive function divide to the two parts?

posted 5 years ago

- 0

What language is that second function in? It's not Java, it's not C, it looks like some kind of pseudo-code, and it doesn't look complete. What happens in the second function when n == 0 or n < 0? That (the missing part) would be the base case. The inductive case is what's between the { and } of the if-statement; that's the part that does the recursive call.

Recursive functions always need an if-statement in some form. If there is no terminating condition (the base case), you would have an infinite recursive loop, and the program would end with a stack overflow error.

Recursive functions always need an if-statement in some form. If there is no terminating condition (the base case), you would have an infinite recursive loop, and the program would end with a stack overflow error.

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