1.(25)  Write a program that reads in an integer number and prints the following output as long as the number is odd and greater than 0 and less than 10.  For example, if the number entered is 9, print: 

           1 
         3 3 3 
       5 5 5 5 5 
     7 7 7 7 7 7 7 
   9 9 9 9 9 9 9 9 9

Answer:

public class Pyramid {
  public static void main (String args[]){

 	for (int i = 0; i < 10; i++) {
	    if (i%2 != 0) {
		for (int j = 0; j < 20 - i/2; j++){
		System.out.print(" ");}
		for (int k = 0; k < i; k++)
		System.out.print(i);
	     }
	     System.out.println();
	}
  }
	    
}

2.(25) Write a recursive method called computeFN which computes the following function for values of N: 

f(N) = 3*f(N-3) + 4*f(N-2) - 5*f(N-1) where  f(0) = 1, f(1) = 1, f(2) = 3.

Answer:

//import javax.swing.*;

public class Test1 {

public static int function(int n) {
	if (n==0) return 1;
	if (n==1) return 1;
	if (n==2) return 3;
	else return 3*function(n-3) + 4*function(n-2) - 5*function(n-1);
}

public static void main(String []args){

//String a = JOptionPane.showInputDialog("Enter odd number from 1 to 9");

//System.out.println("Enter odd number from 1 to 9");
int num = Integer.parseInt(args[0]);
System.out.println(num);
int b = function(num);

System.out.println("f =" + " " + b);
}
}

The steps of computing of f(5):

1)	check n < 0 ? no, then
2)	check n == 0 ? no, then
3)	check n == 1 ? no, then
4)	check n == 2 ? no, then
5)	compute 3*f(5 – 3) + 4*f(5 – 2) – 5*f(5 – 1)
	Recursive calls: 1 = f(2),
		Compute f(3)
		5.1	step 1, 2, 3, and 4 repeat
		5.2	recursive calls: f(3 – 3) = 1
					 f(3 – 2) = 1
					 f(3 – 1) = 3	
		f(3) = 3*f(3 – 3) + 4*f(3 – 2) – 5*f(3 – 1) = -8

		Compute f(4)
		5.3 	step 1, 2, 3, and 4 repeat
		5.4	recursive calls: f(4 – 3) = 1
					 f(4 – 2) = 3
					 f(4 – 1) = f(3) = -8
		f(4) = 3*f(4 – 3) + 4*f(4 – 2) – 5*f(4 – 1) = 55

	f(5) = 3*f(5 – 3) + 4*f(5 – 2) – 5*f(5 – 1)
		= 3*f(2) +4*f(3) - 5*f(4)
		= 3*3 + 4*(-32) + 5*(-55)
		= -298



3.(20)  Define a Java class, in which the queue abstract data structure (ADT) is implemented.

Answer:


public class As1C implements Queue {
    private static final int N = 10;
    int f;
    int r;
    Object[] q;

    public As1C() {
    	f = 0;
	r = 0;
	q = new Object[N];
    }

    public int size() {
	return (N - f + r) % N;
    }

    public boolean isEmpty() {
    	return (r = f);
    }
    
    public Object front() throws QueueEmptyException {
	if (isEmpty( )) throw new QueueEmptyException("Queue is empty!");
	else { return dequeue(); }
    }

    public void enqueue (Object element) throws QueueFullException {
	if (((N - f + r) % N) == N - 1)
            throw new  QueueFullException("Queue is full!!");
    	else {
            r = (r + 1) % N;
            q[r] = element;
        }
    }

    public Object dequeue() throws QueueEmptyException {
	if (isEmpty) throw new QueueEmptyException("Queue is empty!");
	else {
        	Object temp = q[f];
		f = (f + 1) % N;
            	return temp;
	}
}

    public static void main(String[] args) {
        
        As1C test = new As1C();
        
	test.enqueue("1");
	test.enqueue("2");
	test.enqueue("3");
	test.enqueue("4");
	test.enqueue("5");
	test.enqueue("6");
	test.enqueue("7");
	test.enqueue("8");
	test.enqueue("9");
	test.enqueue("10");
        
	for (int i = 1; i <= 10; i++)
            System.out.println(test.front());
            System.exit(0);
	}
}



4.(30) Using Stack data structure (ArrayStack), write a JAVA program to calculate the following expression:

	1+2+3+4+5+6+7+8+9.

Hints:
1. String s = “1+2+3+4+5+6+7+8+9”;
   Using s.charAt(i) to take a character from s.
   Generate Integer object:
	Integer obj = Integer(s.charAt(i));
   Generate Char object:
	Character obj = Character(s.charAt(i));
5. Since each entry in the stack (ArrayStack) is an instance of Object class, remember type-casting when you take an entry       from the stack.

Answer:

public class As1D {
	public static void main(String [] args){
        String s = "1+2+3+4+5+6+7+8+9";
	char c = 0; int result = 0; Object temp;
	Stack myStack = new ArrayStack();
	myStack.push(new Integer(1));
	for (int i = 1; i < s.length(); i++){
	if (s.charAt(i) == '+')
		myStack.push(new Character(s.charAt(i)));
	else {	temp = myStack.pop();
		result = ((Integer)myStack.pop()).intValue();
	        c = s.charAt(i);
		result = result + (c-'0');
	        myStack.push(new Integer(result));}
	  }
	System.out.println("Total is " + result);
	}
}