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Explanation of Solution
Recursion:
A technique where an
a)
Code:
//class definition
class Main
{
//method definition
public static void mystery3(int n)
{
//if condition
if(n<=0)
{
//display the value
System.out.print("*");
}
//else compare the value of n
else if (n%2==0)
{
//display statement
System.out.print("(");
//function call
mystery3(n-1);
//display statement
System...
Explanation of Solution
Recursion:
A technique where an algorithm is described in terms of itself is called as recursion.
b)
Code:
//class definition
class Main
{
//method definition
public static void mystery3(int n)
{
//if condition
if(n<=0)
{
//display the value
System.out.print("*");
}
//else compare the value of n
else if (n%2==0)
{
//display statement
System.out.print("(");
//function call
mystery3(n-1);
//display statement
System...
Explanation of Solution
c)
Code:
//class definition
class Main
{
//method definition
public static void mystery3(int n)
{
//if condition
if(n<=0)
{
//display the value
System.out.print("*");
}
//else compare the value of n
else if (n%2==0)
{
//display statement
System.out.print("(");
//function call
mystery3(n-1);
//display statement
System.out.print(")");
}
//else part
else
{
//display statement
System...
Explanation of Solution
d)
Code:
//class definition
class Main
{
//method definition
public static void mystery3(int n)
{
//if condition
if(n<=0)
{
//display the value
System.out.print("*");
}
//else compare the value of n
else if (n%2==0)
{
//display statement
System.out.print("(");
//function call
mystery3(n-1);
//display statement
System.out.print(")");
}
//else part
else
{
//display statement
System...
Explanation of Solution
e)
Code:
//class definition
class Main
{
//method definition
public static void mystery3(int n)
{
//if condition
if(n<=0)
{
//display the value
System.out.print("*");
}
//else compare the value of n
else if (n%2==0)
{
//display statement
System.out.print("(");
//function call
mystery3(n-1);
//display statement
System.out.print(")");
}
//else part
else
{
//display statement
System...
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Chapter 12 Solutions
Building Java Programs: A Back To Basics Approach, Loose Leaf Edition (5th Edition)
- EX:[AE00]=fa50h number of ones =1111 1010 0101 0000 Physical address=4AE00h=4000h*10h+AE00h Mov ax,4000 Mov ds,ax; DS=4000h mov ds,4000 X Mov ax,[AE00] ; ax=[ae00]=FA50h Mov cx,10; 16 bit in decimal Mov bl,0 *: Ror ax,1 Jnc ** Inc bl **:Dec cx Jnz * ;LSB⇒CF Cf=1 ; it jump when CF=0, will not jump when CF=1 HW1: rewrite the above example use another wayarrow_forwardEX2: Write a piece of assembly code that can count the number of ones in word stored at 4AE00harrow_forwardWrite a program that simulates a Magic 8 Ball, which is a fortune-telling toy that displays a random response to a yes or no question. In the student sample programs for this book, you will find a text file named 8_ball_responses.txt. The file contains 12 responses, such as “I don’t think so”, “Yes, of course!”, “I’m not sure”, and so forth. The program should read the responses from the file into a list. It should prompt the user to ask a question, then display one of the responses, randomly selected from the list. The program should repeat until the user is ready to quit. Contents of 8_ball_responses.txt: Yes, of course! Without a doubt, yes. You can count on it. For sure! Ask me later. I'm not sure. I can't tell you right now. I'll tell you after my nap. No way! I don't think so. Without a doubt, no. The answer is clearly NO. (You can access the Computer Science Portal at www.pearsonhighered.com/gaddis.)arrow_forward
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