2. Consider the following algorithm ALGORITHM Do_Something(int n) //Input: An integer n> 0 if n > 1 do Do_Somehting (n - 1) print n² if n > 1 do Do_Somehting (n - 1)

Given below is a recursive algorithm to compute r". The input r can be any real number. The input n is assumed to be a non-negative integer. Exponent ( r, n) #Input: Real number r and a non-negative integer n. +Output: r If (n-0), return (1) p: Exponent ( ? ) Return ( rp) //The base case 7/The recursive call What missing input values should be used in the recursive call? O (r, n) O (r, n-1) O (r-1, n) O (r-1. n-1)

Using recursion, write a function sum that takes a single argument n and computes the sum of all integers between 0 and n inclusive. Do not write this function using a while or for loop. Assume n is non-negative. def sum(n): """Using recursion, computes the sum of all integers between 1 and n, inclusive. Assume n is positive. >>> sum(1) 1 >>> sum(5) # 1 + 2 + 3+ 4+ 5 15 "*** YOUR CODE HERE ***"

The following code is used to derive Fibonacci algorithm with this sequence: (0, 1, 1, 2, 3, 5, 8, 13, 21, …) where each number is “add” of two previous ones. int fib(int n){ if (n==0) return 0; else if (n == 1) return 1; else return fib(n−1) + fib(n−2); } Explain why the following RISC-V assembly code works, you need to explain all the details using the comments as hints?   # IMPORTANT! Stack pointer must remain a multiple of 16!!!!   addi x10,x10,8                # fib(8), you can change it fib: beq x10, x0, done             # If n==0, return 0 addi x5, x0, 1 beq x10, x5, done             # If n==1, return 1 addi x2, x2, -16              # Allocate 2 words of stack space sd x1, 0(x2)                  # Save the return address sd x10, 8(x2)                 # Save the current n addi x10, x10, -1             # x10 = n-1 jal x1, fib # fib(n-1) ld x5, 8(x2) # Load old n from the stack sd x10, 8(x2) # Push fib(n-1) onto the stack addi x10, x5, -2 # x10 = n-2 jal x1, fib # Call fib(n-2)…