For function log, write the missing base case condition and the recursive call. This function computes the log of n to the base b. As an example: log 8 to the base 2 equals 3 since 8 = 2*2*2. We can find this by dividing 8 by 2 until we reach 1, and we count the number of divisions we make. You should assume that n is exactly b to some integer power. Examples: log(2, 4) -> 2 and log(10, 100) -> 2 public int log(int b, int n ) { if <> { return 0; } else { return <> } }

For function log, write the missing base case condition and the recursive call. This function computes the log of n to the base b. As an example: log 8 to the base 2 equals 3 since 8 = 222. We can find this by dividing 8 by 2 until we reach 1, and we count the number of divisions we make. You should assume that n is exactly b to some integer power. Examples: log(2, 4) -> 2 and log(10, 100) -> 2 public int log(int b, int n ) { if <> { return 0; } else { return <> } }

C++ Programming: From Problem Analysis to Program Design

8th Edition

ISBN:9781337102087

Author:D. S. Malik

Publisher:D. S. Malik

Chapter15: Recursion

Section: Chapter Questions

Problem 8SA

See similar textbooks

Similar questions

Write a recursive function for int powerOfTwo (int k). The function determines the value of 2k. (Note: k is a positive integer). Example, when k=0 the function returns 1 and when k-3 the function returns 8. To show that your code is correct, give the recursive trace for powerOfTwo (2) that returns 4.
Consider the following recursive function: if b = 0, if 6 > a > 0, a f(b, a) f (b, 2.(a mod b)) otherwise. f(a, b) = Estimate the number of recursive applications required to compute f(a, b).
For function addOdd(n) write the missing recursive call. This function should return the sum of all postive odd numbers less than or equal to n. Examples: addOdd(1) -> 1addOdd(2) -> 1addOdd(3) -> 4addOdd(7) -> 16 public int addOdd(int n) { if (n <= 0) { return 0; } if (n % 2 != 0) { // Odd value return <<Missing a Recursive call>> } else { // Even value return addOdd(n - 1); }}
Write a recursive fibonacci (n) function with an expression body. The function should return an Int; it will be too slow to deal with inputs whose fibonacci numbers are too large anyway. You do *not* need to use memoization. Note that: fibonacci(0) = 0 fibonacci(1) = 1 fibonacci(n, where n is greater than 1) = fibonacci(n-2) + fibonacci(n - 1)
Write a recursive function called that takes a string of single names separated by spaces and prints out all possible combinations (permutations), each combination on a new line. When the input is: Alice Bob Charlie then the output is: Alice Bob Charlie Alice Charlie Bob Bob Alice Charlie Bob Charlie Alice Charlie Alice Bob Charlie Bob Alice Here is my original code that needs to be fixed: def all_permutations(permList, nameList): # TODO: Implement method to create and output all permutations of the list of names. if nameList == len(permList) - 1: return nameList else: for x in range(permList, len(nameList)): permList[nameList], permList[x] = permList[x], permList[name_List] return all_permutations(permList, nameList + 1) permList[nameList], permList[x] = permList[x], permList[name_List] if __name__ == "main": nameList = input().split(' ') permList = [] all_permutations(permList, nameList)
Write a recursive function for Euclid's algorithm to find the greatest common divisor (gcd) of two positive integers. gcd is the largest integer that divides evenly into both of them. For example, the gcd(102, 68) = 34. You may recall learning about the greatest common divisor when you learned to reduce fractions. For example, we can simplify 68/102 to 2/3 by dividing both numerator and denominator by 34, their gcd. Finding the gcd of huge numbers is an important problem that arises in many commercial applications. We can efficiently compute the gcd using the following property, which holds for positive integers p and q: If p > q, the gcd of p and q is the same as the gcd of q and p % q.
Write a recursive function called that takes a string of single names separated by spaces and prints out all possible combinations (permutations), each combination on a new line. When the input is: Alice Bob Charlie then the output is: Alice Bob Charlie Alice Charlie Bob Bob Alice Charlie Bob Charlie Alice Charlie Alice Bob Charlie Bob Alice Here is the original code in Python: def all_permutations(permList, nameList): # TODO: Implement method to create and output all permutations of the list of names. pass if __name__ == "__main__": nameList = input().split(' ') permList = [] all_permutations(permList, nameList)
For function decToBinary, write the missing parts of the recursion case. This function should return a string that stores the binary equivalent for int variable num. Example: The binary equivalent of 13 may be found by repeatedly dividing 13 by 2. So, 13 in base 2 is represented by the string "1101". Examples: decToBinary(13) -> "1101" public String decToBinary (int num) { if (num < 2) return Integer.toString(num); else return <<Missing recursive call>> + <<Missing calculation>>;}
Recursive Palindrome! Recall that a palindrome is a string that reads the same forward and backward. Write a recursive function is_palindrome (s:str) -> bool to check whether a string s is a palindrome. Here's a hint: think about how you can use is_palindrome(t), where t is a substring of s, to help you decide whether s is a palindrome. Your Answer: 1 # Put your answer here 2 Submit
Consider a recursive function, called f, that computes powers of 3 using only the + operator. Assume n > = 0. int f(int n) { if (n == 0) return 1; return f(n-1) + f(n-1) + f(n-1); } Give an optimized version of f, called g, where we save the result of the recursive call to a temporary variable t, then return t+t+t. i got int g(int n) { if (n == 0) return 1; int t = g(n - 1); return t+t+t; } so now Write a recurrence relation for T(n), the number addition operations performed by g(n) in terms of n.
Compute f(6) for the recursive function below. def f(n): if n == 0: return 1 if n == 1: return 2 else: return f(n-1)+n*f(n-2)-n
Recursion can be direct or indirect. It is direct when a function calls itself and it is indirect recursion when a function calls another function that then calls the first function. To illustrate solving a problem using recursion, consider the Fibonacci series: - 1,1,2,3,5,8,13,21,34...The way to solve this problem is to examine the series carefully. The first two numbers are 1. Each subsequent number is the sum of the previous two numbers. Thus, the seventh number is the sum of the sixth and fifth numbers. More generally, the nth number is the sum of n - 2 and n - 1, as long as n > 2.Recursive functions need a stop condition. Something must happen to cause the program to stop recursing, or it will never end. In the Fibonacci series, n < 3 is a stop condition. The algorithm to use is this: 1. Ask the user for a position in the series.2. Call the fib () function with that position, passing in the value the user entered.3. The fib () function examines the argument (n). If n < 3…