1. a) Given a recursive algorithm as below: int F(int n) { if (n <= 1) return 1; else if (n % 2 == 0) return F(n-1) +n; else return F(n/2) - 1; } Illustrate how you will find out the value of F(10) using the above algorithm. You need to show all the steps. b) Given a function as below: int recFx(int a, int b) { If (a

1. a) Given a recursive algorithm as below: int F(int n) { if (n <= 1) return 1; else if (n % 2 == 0) return F(n-1) +n; else return F(n/2) - 1; } Illustrate how you will find out the value of F(10) using the above algorithm. You need to show all the steps. b) Given a function as below: int recFx(int a, int b) { If (a

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

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Write a function sum_of_digits(a,b), that takes two arguments a and b, computes a to the power of b, then recursively finds the sum of digits until there is only one digit left. You must PRINT all steps of the process as shown in the below example. sum_of_digits(5,3) will PRINT on the screen the following: 5ˆ3 = 125 = 1 + 2 + 5 = 8 sum_of_digits(2,8) will PRINT on the screen the following: 2ˆ8 = 256 = 2 + 5 + 6 = 13 = 1 + 3 = 4
The following function f uses recursion: def f(n): if n <= 1 return n else return f(n-1) + f(n-2) Let n be a valid input, i.e., a natural number. Which of the following functions returns the same result but without recursion? a) def f(n): a <- 0 b <- 1 if n = 0 return a elsif n = 1 return b else for i in 1..n c <- a + b a <- b b <- c return b b) def f(n): a <- 0 i <- n while i > 0 a <- a + i + (i-1) return a c) def f(n): arr[0] <- 0 arr[1] <- 1 if n <= 1 return arr[n] else for i in 2..n arr[i] <- arr[i-1] + arr[i-2] return arr[n] d) def f(n): arr[0..n] <- [0, ..., n] if n <= 1 return arr[n] else a <- 0 for i in 0..n a <- a + arr[i] return a
Please answer fast Question 1 Answer following questions and write corresponding algorithms. Algorithm: sqrt(n) [main] 1. return sqrtHelper(Arg 1, Arg 2) Algorithm: sqrtHelper(n,m) 1. if m*m <= n 2. return m 3. else 4. return sqrtHelper(Arg 3, Arg 4) 5. endif (a) Write an recursive algorithm called isPrime(n) that takes a positive integer and returns True if it is prime and False otherwise. You must call the function sqrt(n) as a sub- algorithm. Trace your algorithm for isPrime(41). You only need to show the detail in the main algorithm and its helper function (if there is any).
The following function f uses recursion: def f(n): if n <= 1 return n else return f(n-1) + f(n-2) 5 Let n be a valid input, i.e., a natural number. Which of the following functions returns the same result but without recursion? a) def f(n): a <- 0 b <- 1 if n = 0 return a elsif n = 1 return b else for i in 1..n c <- a + b a <- b b <- c return b f(n): a <- 0 i <- n while i > 0 a <- a + i + (i-1) return a f(n): arr[0] <- 0 arr[1] <- 1 if n <= 1 return arr[n] else for i in 2..n arr[i] <- arr[i-1] + arr[i-2] return arr[n] f(n): arr[0..n] <- [0, ..., n] if n <= 1 return arr[n] else a <- 0 for i in 0..n a <- a + arr[i] return a
The first examples of recursion are the mathematical functions factorial and fibonacci. These functions are defined for non-negative integers using the following recursive formulas:factorial(0) = 1factorial(N) = N*factorial(N-1) for N > 0fibonacci(0) = 1fibonacci(1) = 1fibonacci(N) = fibonacci(N-1) + fibonacci(N-2) for N > 1Write recursive functions to compute factorial(N) and fibonacci(N) for a given non-negative integer N, and write a main() routine to test your functions.(In fact, factorial and fibonacci are really not very good examples of recursion, since the most natural way to compute them is to use simple for loops. Furthermore, fibonacci is a particularly bad example, since the natural recursive approach to computing this function is extremely inefficient.)
Write a recursive function that, given a sequence of comparable values, returns the count of elements where the current element is less than the following ( next ) element in the given sequence. See the examples given below. def count_ordered ( seq ) : """ Input : A sequence of comparable elements Output : The number of elements that are less than the following element in the sequence Example : >>> count_ordered ( [ 1 , 2 , 3 , 4 , 5 , 6 ] ) 5 >>> count_ordered ( ( 1 , 12, 7.3 , -2,4 ) ) 2 >>> count_ordered ( 'Python' ) 2 >>> count_ordered ( [ 6 ] ) 0 >>> count_ordered ( [ ] ) 0 """ In the first example above , count_ordered ( [ 1,2,3,4,5,6 ] )the returned answer is 5 because for all the first 5 numbers the current number is less than the next number. In the second example above, count_ordered ( ( 1,12,7.3 , -2,4 ) )the…
What type of recursion is used in the following function? int f(int n){ if (n==1) return 1; else return n+f(n-1); } Tail recursion Multiple recursion Indirect recursion Non-tail recursion
Write a recursive function to implement the recursive algorithm (multiplying two positive integers using repeated addition). Also, write a program to test your function.
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…
is it correct ? Write a recursive function that returns the nth number in a fibonacci sequence when n is passed to function. The fibonacci sequence is like 0,1,1,2,3,5,8,13...... Answer: #include <iostream> using namespace std; int getFibonacci(int n) { if (n == 0 || n == 1) return n; else return getFibonacci(n - 1) + getFibonacci(n - 2); } int main() { int n = 7; int result = getFibonacci(n); cout << result; }