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CS 111: Lab 8 main content

Student Learning Objectives

Demonstrate the ability to

Lab 8 Assignment

Based on Chapter 12, Programming Challenges 1 (Recursive Multiplication), 7 (Recursive Power Method), and 8 (Sum of Numbers).

Create a new Lab 8 project and copy and paste the code below into your Driver class. Complete the recursive methods multiply, power, and sumOfNumbers

Driver Class

   import java.util.Scanner;

   public class Driver
   {
      public static void main(String[] args)
      {
         Scanner keyboard = new Scanner(System.in);
         
         System.out.print("Recursive Multiplication - Enter two numbers: ");
         int x = keyboard.nextInt();
         int y = keyboard.nextInt();
         System.out.println(x + " times " + y + 
                           " equals " + multiply(x, y));  
                           
         System.out.print("\nRecursive power - Enter two numbers: ");
         x = keyboard.nextInt();
         y = keyboard.nextInt();
         System.out.println(x + " raised to the power of " + y +
                           " is " + power(x, y));
      	
         System.out.print("\nSum of Numbers - Enter a number: ");
         int number = keyboard.nextInt();
      
         System.out.println("The sum of the integers 1 through " 
                             + number + " is " + sumOfNumbers(number));
      }
      
   	
      public static double multiply(double x, double y)
      {
         
      }
      
      public static double power(double x, double y)
      {
         
      }
   
      public static int sumOfNumbers(int max)
      {
         
      }
   }

Sample Session

Recursive Multiplication - Enter two numbers: 5 3
5 times 3 equals 15.0

Recursive power - Enter two numbers: 2 4
2 raised to the power of 4 is 16.0

Sum of Numbers - Enter a number: 6
The sum of the integers 1 through 6 is 21

To Receive Credit

Programming Challenge

If you finish early, get yourself checked off first! As a programming challenge (not required but recommended):

Do Programming Challenge 9 (Ackermann's Function) as well

ackermann(0, 0) = 1
ackermann(0, 1) = 2
ackermann(1, 1) = 3
ackermann(1, 2) = 4
ackermann(1, 3) = 5
ackermann(2, 2) = 7
ackermann(3, 2) = 29