Tutorial Week 2

String num(int value) {

	// Q1.1 Solution
	switch (value) {
		case 1:
			return "one";
		case 2:
			return "two";
		case 3:
			return "three";
		case 4:
			return "four";
		case 5:
			return "five";
		case 6:
			return "six";
		case 7:
			return "seven";
		case 8:
			return "eight";
		case 9:
			return "nine";
		default:
			return "??";
	}

	// Q1.2 -- Work until 9:16
	String[] names = { "one", "two", /* ... */, "nine"};
					//   0      1                 8

	if (value <= 0 || value >= 10) {
		return "??";
	} else {
		// Known: value is between 1 and 9 (incl.)
		// e.g. value == 2  ->  names[2 - 1]  -> "two"
		return names[value - 1]
	}

}


// Q2.1

1, 1, 2, 3, 5, 8, 13, 21, 34 ...

// Q2.2 -- Work until 9:25 on recursive method
/**
 * Returns the nth element of the fibonacci sequence
 */
int fib(int n) {

	// Base case
	if (n <= 1) {
		return 1;
	}

	// Recursive case
	return fib(n - 1) + fib(n - 2);
}

int fib(int n) {

	int prev1_term = 1; // n == 0
	int prev2_term = 1; // n == 1
	int term = 1;

	for (int i = 0; i < n - 1; i++) {
		// Calculate next
		term = prev1_term + prev2_term;

		// Save prior terms
		prev2_term = prev1_term;
		prev1_term = term;
	}

	return term;
}

// Q2.3 
// Inputs: n == 1, n == 5

// fib(1)
//  -> 1
// Result: 1

// fib(5)
//  -> [fib(4) -> 5] + [fib(3) -> 3] (8)
//	     |		          -> [fib(2) -> 2] + [fib(1) -> 1] (3)
//       |                      -> [fib(1) -> 1] + [fib(0) -> 1]  (2)
//       +-> [fib(3) -> 3] + [fib(2) -> 2] (5)
// Result: 8

// n = 4
// i   |  term  |  prev1_term  |  prev2_term 
// 0   |   2    |      2       |      1  
// 1   |   3    |      3       |      2
// 2   |  [5]   |      5       |      3
// Result: 5

// Q3
long ackermann(short m, short n) {

	if (m == 0) {
		return n + 1;
	}

	if (m > 0 && n == 0) {
		return ackermann(m - 1, 1);
	}

	if (m > 0 && n > 0) {
		return ackermann(m - 1, ackermann(m, n - 1));
	}
}