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Types of Recursion

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The mathematical combinations operation is a good example of a function that can quickly be implemented as a binary recursive function. The number of combinations, often represented as nCk where we are choosing n elements out of a set of k elements, can be implemented as follows:


int choose(int n, int k)
{
	if (k == 0 || n == k) return(1);
	else return(choose(n-1,k) + choose(n-1,k-1));
}

An exponential recursive function is one that, if you were to draw out a representation of all the function calls, would have an exponential number of calls in relation to the size of the data set (exponential meaning if there were n elements, there would be O(a n) function calls where a is a positive number).

A good example an exponentially recursive function is a function to compute all the permutations of a data set. Let's write a function to take an array of n integers and print out every permutation of it.


void print_array(int arr[], int n)
{
	int i;
	for(i=0; i<n; i) printf("%d ", arr[i]);
	printf("\n");
}

void print_permutations(int arr[], int n, int i)
{
	int j, swap;
	print_array(arr, n);
	for(j=i+1; j<n; j) {
		swap = arr[i]; arr[i] = arr[j]; arr[j] = swap;
		print_permutations(arr, n, i+1);
		swap = arr[i]; arr[i] = arr[j]; arr[j] = swap;
	}
}

To run this function on an array arr of length n, we'd do print_permutations(arr, n, 0) where the 0 tells it to start at the beginning of the array.

In nested recursion, one of the arguments to the recursive function is the recursive function itself! These functions tend to grow extremely fast. A good example is the classic mathematical function, "Ackerman's function. It grows very quickly (even for small values of x and y, Ackermann(x,y) is extremely large) and it cannot be computed with only definite iteration (a completely defined for() loop for example); it requires indefinite iteration (recursion, for example).


Ackerman's function
int ackerman(int m, int n)
{ 
	if (m == 0) return(n+1);
	else if (n == 0) return(ackerman(m-1,1));
	else return(ackerman(m-1,ackerman(m,n-1)));
}

Try computing ackerman(4,2) by hand... have fun!

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