249 Winter 2014 Recursion

Recursion249.java

// --------------------------------------------------------------------------------------------------------
// Recursion249.java
// Written by: Maxwell Mowbray
// For COMP 249 Section PP - Tutorial PB - Winter 2014.
// These are the questions from the COMP 249 final which dealt with recursion
// --------------------------------------------------------------------------------------------------------

public class Recursion249
{
	public static int count = 0;
	
	public static void main(String[] args)
	{
		//magic(6);
		//System.out.println(count);
		
		System.out.println(exp(3,10));
	}

	// Question 1: Analyse this recursive method and predict its output when main() has only:

	/*public static void main(String[] args)
	{
		magic(6);
		System.out.println(count);
	}*/
	
	public static int magic(int n)
	{
		count++;
		int ret = 0;
		
		if (n < 3)
			ret = n;
		else
			ret = magic(n-2) + n + magic(n-3);
		
		System.out.println(ret);
		return ret;
	}

	/*
	Question 2: e^x can be approximated with a Taylor series.

	e^x = 1 + x/1! + x^2/2! + x^3/3! + ... + x^n/n!

	where x is any integer and n is a sufficiently large integer (larger integer -> closer approximation of e^x)

	In simple terms, you can calculate e^7 by computing 1 + 7/1! + 7^2/2! + 7^3/3! + ...

	First, write two recursive functions that will help. The first is the factorial function (factorial(n)) that
	recursively computes n!. Then, a recursive function that computes x^n. This function is pow (x,n).
	*/
	
	public static double factorial(int n) //go to a maximum of ten for n for an int or the number will be too large for an int
	{
		if (n == 0)
			return 1;
		else
			return (n * factorial(n-1));
	}
	
	public static double pow (int x, int n) //x^4 = x * x^3. x^3 = x * x^2. and so on...
	{
		if (n == 0)
			return 1;
		else
			return (x * pow(x, n - 1));
	}
	
	//To get the series computer, put both functions together recursively
	
	public static double exp (int x, int n)
	{
		if (n == 0)
			return 1;
		else
			return ((pow(x,n)) / factorial(n) + exp(x, n - 1));
	}
}