Which recursive method would return a value of 210 if result(7) was called? Question 1 options: public int result(int a){ if (a == 1){ return a; } else{ return a * result(a - 2); } } public int result(int a){ if (a == 1){ return 1; } else{ return a * result(a - 2); } } public int result(int a){ if (a == 1){ return a + 1; } else{ return a * result(a - 2); } } public int result(int a){ if (a == 1){ return 1; } else{ return a * result(a - 1); } }

Which recursive method would return a value of 210 if result(7) was called? Question 1 options: public int result(int a){ if (a == 1){ return a; } else{ return a * result(a - 2); } } public int result(int a){ if (a == 1){ return 1; } else{ return a * result(a - 2); } } public int result(int a){ if (a == 1){ return a + 1; } else{ return a * result(a - 2); } } public int result(int a){ if (a == 1){ return 1; } else{ return a * result(a - 1); } }

Programming Logic & Design Comprehensive

9th Edition

ISBN:9781337669405

Author:FARRELL

Publisher:FARRELL

Chapter9: Advanced Modularization Techniques

Section: Chapter Questions

Problem 13RQ

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Given the following recursive method: public static int think(int x) { if(x<10) if(x%2!=0) return x; else return 0; else if(x%2!=0) return x%10 + think(x/10); else return think(x/10); } Trace (in details) the calling of the above method with any input from your choice such that the returned value will be 19
Here is a recursive method: public static int mystery(int n, int i) { if(n == 1) return i; else if(n % 2 == 0) return mystery(n/2, i+1); else return mystery(3*n + 1, i+1); } What is the return value of the following call? mystery(3,0);
How to answer the following questions Given the following code, what are the constraints on the input argument? int mystery(int num){ if (num == 0) return 0; else if (num > 100) return -1; else return num + mystery(num – 1); } A. It must be between 0 and 100 inclusive. B. It must be greater than 0. C. It must be less than 0 or greater than 100. D. It must be greater than or equal to 0. The number of recursive calls that a method goes through before returning is called: A. order of growth efficiency. B. the depth of recursion. C. combinatorial recursive count. D. activation stack frame. The following code is supposed to return the sum of the numbers between 1 and n inclusive, for positive n. An analysis of the code using the "Three Question" approach reveals that: int sum(int n){ if (n == 1) return 1; else return (n + sum(n)); } A. it fails the base-case question. B. it fails the smaller-caller question. C. it fails the general-case…
Given the following recursive method:public static int think(int x) {if(x<10)if(x%2!=0)return x;elsereturn 0;elseif(x%2!=0)return x%10 + think(x/10);elsereturn think(x/10);}Trace (in details) the calling of the above method with any input from your choice such that thereturned value will be 19
Java - How many recursive calls will be made if the following method is called with 6? void greeting(int n) { system.out.println(“Hello!”); greeting(n-1); }
I have two recursive methods in Java but I want to also implement a test class to see if both of them perform how can I do that? Method 1: Non Tail Recursive public class NonTailRecursiveFactorial { public static long factorial(int n) {if (n == 0) {return 1;}return n * factorial(n - 1);} public static void main(String[] args) {for (int i = 0; i <= 10; i++) {System.out.println(i + "! = " + factorial(i));}}} Method 2: Tail Recursive Method public class TailRecursiveFactorial { public static long factorial(int n) {return factorial(n, 1);} private static long factorial(int n, long accumulator) {if (n == 0) {return accumulator;}return factorial(n - 1, accumulator * n);} public static void main(String[] args) {for (int i = 0; i <= 10; i++) {System.out.println(i + "! = " + factorial(i));}}
Which of the following recursive methods would return a value of 24 when mystery(3, 2) is called? Question 7 options: public int mystery(int x, int y){ if (x == 1){ return y; } else{ return x * mystery(x - 1, y + 1); }} public int mystery(int x, int y){ if (x == 1){ return x; } else{ return x * mystery(x - 1, y + 1); }} public int mystery(int x, int y){ if (x == 1){ return y; } else{ return x * mystery(x - 1, y); }}
What is the “stopping” case for each (what causes the methods to “end”)? How do you guarantee that the methods will “stop” (infinite loop, infinite recursion)? Which method one is “better”? Why? public static int factorialValue(int n) {int ans = 1;for (int i=1; i <=n;i++) {ans = ans *i;}return ans;} public static int recursiveValue(int n) {if (n == 0) {return 1;}else {return n*factorialValue(n-1);}}
Write a recursive method called digitCount() that takes a positive integer as a parameter and returns the number of digits in the integer. Hint: The number of digits increases by 1 whenever the input number is divided by 10. Use Java. Ex: If the input is: 345 the method digitCount() returns and the program outputs: 3 Ex: If the input is: 0 the method digitCount() returns and the program outputs: 1 import java.util.Scanner; public class LabProgram {/* TODO: Write recursive digitCount() method here. */public static void main(String[] args) {Scanner scnr = new Scanner(System.in);int num, digits;num = scnr.nextInt();digits = digitCount(num);System.out.println(digits);}}
Do the trace in detail and submit The source Code As Well From the following recursive method Trace (in details) the calling of the above method with any input from your choice such that thereturned value will be 19. public static int think(int x) {if(x<10)if(x%2!=0)return x;elsereturn 0;elseif(x%2!=0)return x%10 + think(x/10);elsereturn think(x/10);}
The nth harmonic number is defined non-recursively as: H(n)=1+1/2+1/3+1/4+...+1/n Come up with a recursive definition and use it to guide you to write a method definition for a double-valued method named “harmonic” that accepts an int parameter n and recursively calculates and returns the nth harmonic number. Write a test program that displays the harmonic numbers, H(n)=1,2,3,4...10
The intName method in the "Problem Solving: Stepwise Refinement" section accepted arguments < 1,000. Using a recursive call, extend its range to 999,999. For example an input of 12,345 should return " twelve thousand three hundred forty five". Note that some number names may have extra spaces at the beginning or end. import java.util.Scanner; public class IntegerName { /** Turns a number into its English name. @param number a positive integer < 999,999 @return the name of the number (e.g. "two hundred seventy four") */ public static String intName(int number) { int part = number; // The part that still needs to be converted String name = ""; // The name of the number if (part >= 1000) { // Recursively call intName } if (part >= 100) { name = digitName(part / 100) + " hundred"; part = part % 100; } if (part >= 20) { name = name + " " + tensName(part); part = part % 10; } else if (part >= 10) { name = name + " " + teenName(part); part = 0; } if (part > 0) { name =…

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	public int result(int a){ if (a == 1){ return a; } else{ return a * result(a - 2); } }
	public int result(int a){ if (a == 1){ return 1; } else{ return a * result(a - 2); } }
	public int result(int a){ if (a == 1){ return a + 1; } else{ return a * result(a - 2); } }
	public int result(int a){ if (a == 1){ return 1; } else{ return a * result(a - 1); } }