Write a Pseudocode for a function that takes an integer n ≥ 1 and returns f(n) where f(n) = 3 (f(n-1))2 +2 f(n-2) and f(0) = f(1) = 1
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- The following function f uses recursion:def f(n):if n <= 1return nelsereturn f(n-1) + f(n-2)5Let 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 <- 0b <- 1if n = 0return aelsif n = 1return belsefor i in 1..nc <- a + ba <- bb <- creturn b b) def f(n):a <- 0i <- nwhile i > 0a <- a + i + (i-1)return a c) def f(n):arr[0] <- 0arr[1] <- 1if n <= 1return arr[n]elsefor i in 2..narr[i] <- arr[i-1] + arr[i-2]return arr[n] d) def f(n):arr[0..n] <- [0, ..., n]if n <= 1return arr[n]elsea <- 0for i in 0..na <- a + arr[i]return aWhat does the function return for any positive values of x and n ? The answer should be a general formula in terms of x and n. int what_do_I_do ( int x, int n ) { if ( n = = 1 ) return x; else return x + what_do_I_do ( x, n – 1 ); } C++ be quick pleaseThe following function f uses recursion:def f(n):if n <= 1return nelse 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 <- 0b <- 1 if n = 0return aelsif n = 1 return belsefor i in 1..nc <- a + b a <- b b <- c return bb) def f(n):a <- 0i <- n while i > 0 a <- a + i + (i-1) return ac) def f(n): arr[0] <- 0 arr[1] <- 1 if n <= 1return arr[n]elsefor 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 <= 1return arr[n]elsea <- 0 for i in 0..n a <- a + arr[i]return a
- int functionC (int n) { int i, j, sumC = 0; for (i=n; i > 0; i=i-5) for (j=1; j 0) { if (functionC(n) % 2 == 0) { for (i=m; i > 0; i=i/3) sumE++; } else (10) Asymptotic runtime of functionE { for (i=m; i > 0; i=i-3) sumE--; } n--; } return sumE;Q11/ A function f(t) is said to be even if: Oa) f(t) = f(-t) Ob) f(t) = -f(-t) Oc) f(t) # f(-t) %3D Od) None of theseConsider the following function fun: int fun(int n) { int s = 0; while(n > 1) { n= n/2; s++; } return s; } O A) O(n log n) O B) O(n) C) O(n²) O D) O(log n)
- Code in Python Write a function, print_integers_less_than(n), which takes an integer parameter n and prints each integer k which is at least 0 and is less than n, in ascending order. Hint: use a simple for loop. For example: Test Result print_integers_less_than(2) 0 1 print_integers_less_than(5) 0 1 2 3 4 print_integers_less_than(-3)497 Bytes Write a recursive function that returns the sum of the digits of a given integer.Input format :Integer NOutput format :Sum of digits of NConstraints :0 <= N <= 10^9Sample Input 1 :12345Sample Output 1 :15Sample Input 2 :9Sample Output 2 :9 Solution:///////////// public class solution { public static int sumOfDigits(int input){ int sum; if(input<10){ return input; } sum = (input % 10) + sumOfDigits(input / 10); return sum; }}. .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
- Write a recursive function that finds n-th power of number m. Ex:m=3 n=4 Ans=81 WRITE IN PYTHON PLEASECompute 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)-nO(nlgn) means that there is function f(n) that is O(nlgn) which is an upper bound for the running time at large n Select one: True False
