This program outputs a downwards facing arrow composed of a rectangle and a right triangle. The arrow dimensions are defined by
user specified arrow base height, arrow base width, and arrow head width.(1) Modify the given program to use a loop to output an arrow base of height arrowBaseHeight. (1 pt)(2) Modify the given program to use a loop to output an arrow base of width arrowBaseWidth. Use a nested loop in which the inner loop draws the *’s, and the outer loop iterates a number of times equal to the height of the arrow base. (1 pt)(3) Modify the given program to use a loop to output an arrow head of width arrowHeadWidth. Use a nested loop in which the inner loop draws the *’s, and the outer loop iterates a number of times equal to the height of the arrow head. (2 pts)(4) Modify the given program to only accept an arrow head width that is larger than the arrow base width. Use a loop to continue prompting the user for an arrow head width until the value is larger than the arrow b
1 answer:
Response:
Here is the JAVA program:
import java.util.Scanner; // to accept user input
public class DrawHalfArrow{ // beginning of the half arrow class
public static void main(String[] args) { // starts main() function
Scanner scnr = new Scanner(System.in); // reads input
int arrowBaseHeight = 0; // variable for arrow base height
int arrowBaseWidth = 0; // variable for arrow base width
int arrowHeadWidth = 0; // variable for arrow head width
// prompts user for arrow base height, width, and head width
System.out.println("Enter arrow base height: ");
arrowBaseHeight = scnr.nextInt(); // scans input as an integer
System.out.println("Enter arrow base width: ");
arrowBaseWidth = scnr.nextInt();
/* while loop to keep asking for arrow head width until it's greater than the arrow base width */
while (arrowHeadWidth <= arrowBaseWidth) {
System.out.println("Enter arrow head width: ");
arrowHeadWidth = scnr.nextInt(); }
// begins the nested loop
// outer loop runs as many times as the arrow base height
for (int i = 0; i < arrowBaseHeight; i++) {
// inner loop prints the asterisks
for (int j = 0; j <arrowBaseWidth; j++) {
System.out.print("*"); } // shows asterisks
System.out.println(); }
// a temporary variable to store the width of the arrowhead
int k = arrowHeadWidth;
// outer loop to iterate the number of times equal to the arrowhead height
for (int i = 1; i <= arrowHeadWidth; i++)
{ for(int j = k; j > 0; j--) {// inner loop to print asterisks
System.out.print("*"); } // shows asterisks
k = k - 1;
System.out.println(); } } } // continues adding more asterisks for a new line
Clarification:
This program requests the user to input the arrow base height, base width, and head width. When the head width is asked, it checks that the width must exceed that of the base width. The while loop keeps prompting until a valid head width is entered.
The loop generates an arrow base matching the specified height. This satisfies the first condition.
A nested loop outputs the base with the specified width. The inner loop draws the asterisks to form the base, while the outer loop iterates according to the base height, thus meeting the second condition.
A temporary variable k stores the initial value of head width during modifications.
The final nested loop produces the arrow head with the specified width. The inner loop handles star printing for the arrow head, fulfilling the third condition.
The temporary variable k decreases by one for subsequent iterations to produce one less asterisk each time.
Attached is the output screenshot.
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Answer:
Below is the JAVA program:
import java.util.Scanner; //to take input from user
public class Main { //name of the class
public static void main(String[] args) { //beginning of main method
Scanner input = new Scanner(System.in); //creates Scanner instance
int integer = input.nextInt(); //defines and collects an integer for the number of words
String stringArray[] = new String[integer]; //initializes an array to hold strings
for (int i = 0; i < integer; i++) { //iterates through the array
stringArray[i] = input.next(); } //collects strings
int frequency[] = new int[integer]; //creates an array for holding frequencies
int count; //initializes variable to calculate frequency of each word
for (int i = 0; i < frequency.length; i++) { //iterates through frequency array
count = 0; //sets count to zero
for (int j = 0; j < frequency.length; j++) { //iterates through array
if (stringArray[i].equals(stringArray[j])) { //checks if element at ith index matches jth index
count++; } } //increments count
frequency[i] = count; } //stores count in the frequency array
for (int i = 0; i < stringArray.length; i++) { //iterates through the string array
System.out.println(stringArray[i] + " " + frequency[i]); } } } //displays each word and its frequency
Explanation:
To illustrate the program, consider:
let integer = 3
for (int i = 0; i < integer; i++) is used to input strings into stringArray.
On the first pass:
i = 0
0<3
stringArray[i] = input.next(); takes a word and saves it to the ith index of stringArray. For example, if the user types "hey", it will be in stringArray[0].
Then, i increments to i = 1
On the second pass:
i = 1
1<3
stringArray[i] = input.next(); takes another word and assigns it to stringArray[1]. If the user enters "hi", then hi will be in stringArray[1]
Next, i becomes i = 2
On the third pass:
i = 2
2<3
stringArray[i] = input.next(); captures a word and places it in stringArray[2]. If the user types "hi", it goes to stringArray[2]
Then, i increments to i = 3
The loop terminates since i<integers is false.
The next outer loop for (int i = 0; i < frequency.length; i++) and inner loop for (int j = 0; j < frequency.length; will check each word and if (stringArray[i].equals(stringArray[j])) identifies any duplicates. Words that repeat will have their count incremented and the frequency array will record these values. For example, hey appears once while hi shows up twice, thus resulting in the final outputs:
hey 1
hi 2
hi 2
The visual representation of the program and its outcome based on the example provided is attached.
