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- c++ Implement a function that recursively calculates the nth number in the Fibonacci sequence of numbers. The Fibonacci sequence starts at 0 and 1. Each number thereafter is the sum of the two preceding numbers. This gives us the following first ten numbers: 0,1,1,3,5,8,13,21,34, .... Input Expected output0 01 17 139 34Write 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…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; }
- Implement a recursive C++ function which takes a character (ch) and a positive integer (n) and prints thecharacter ch, n times on the screen. The prototype of your function should be:void printChar (char ch, int n)For example, calling printChar('*',5) should display ***** on screen.Note: There should NOT be any loop in your function.Consider the following function: void fun_with_recursion(int x) { printf("%i\n", x); fun_with_recursion(x + 1); } What will happen when this function is called by passing it the value 0?Need some help with this c++ problem In order to compute a power of two, you can take the next-lower power and double it. For example, if you want to compute 211 and you know that 210 = 1024, then 211 = 2 × 1024 = 2048. Based on this observation, write a recursive function int pow2(int n) where n is the exponent. If the exponent is negative, return -1. int pow2(int n) { ..... }
- 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 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.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
- 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 aplease code in python Write a recursive function to add a positive integer b to another number a, add(a, b), where only the unit 1 can be added, For example add(5, 9) will return 14. The pseudocode is: # Base case: if b is 1, you can just return a + 1 # General case: otherwise, return the sum of 1 and what is returned by adding a and b - 1.This is for Java Write a recursive function that takes as a parameter a nonnegative integerand generates the following pattern of stars. If the nonnegative integer is 4,then the pattern generated is:**********Also, write a program that prompts the user to enter the number of lines inthe pattern and uses the recursive function to generate the pattern. Forexample, specifying 4 as the number of lines generates the above pattern. Java please