1. Consider the following recursive function: def foo(n): if (n == 0): return 0 return n + foo(n - 1) a. What is foo(5) b. What is foo(10) c. Suppose that n + foo(n - 1) is changed to n * foo(n - 1). What is foo(5) now? d. What happens if foo(-1) is called?
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1.
Consider the following recursive function:
def foo(n):
if (n == 0):
return 0
return n + foo(n - 1)
a. What is foo(5)
b. What is foo(10)
c. Suppose that n + foo(n - 1) is changed to n * foo(n - 1). What is foo(5) now?
d. What happens if foo(-1) is called?
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- Consider the following recursive function: def rec1(aList,first,last): if first==last: return aList[first] else: return aList[last]*rec1(aList,first,last-1) Show the output of the following call (show your work and all recursive calls): print(rec1([1,2,3,4,5,6,7],3,5))? describe the task of the rec1 function.Which is the base case of the following recursion function: def mult3(n): if n == 1: return 3 else: return mult3(n-1) + 3 else n == 1 mult3(n) return mult3(n-1) + 3Consider the recursive procedure which computes the nth Fibonacci number is the one below. procedure Fl (n) //a function which returns the nth Fibonacci number.// if n < 2 then return(n) else return (F2(2,n,1,1)) endif end Fl procedure F2(i,n,x,y) if i
- 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 <<Missing base case condition>> { return 0; } else { return <<Missing a Recursive case action>> }}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 aThe 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
- 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…Using recursion, write a Python function def before(k,A) which takes an integer k and an array A of integers as inputs and returns a new array consisting of all the integers in A which come before the last occurrence of k in A, in the same order they are in A. For example, if A is [1,2,3,6,7,2,3,4] then before(3,A) will return [1,2,3,6,7,2]. If k does not occur in A, the function should return None.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; }
- 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)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 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