4. The Ackermann function is a fast-growing function that takes two nonnegative integers: m and n, and the algorithm is defined as below: A(m, n) = n + 1 if m=0 = A(m-1, 1) if n = 0 = A(m-1, A(m, n-1)), otherwise Write this function in C++ and test the function by calling it from main by using a nested loop, by varying m = 0 to 3, and inside the m loop, use another loop to vary n from 0 to 10 The program should display a table for all Ackermann numbers generated as below: m/n 0 2 3 1 4 5 6 7 8 9 0123 The inside of the table the computed A values must be displayed. 10
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- (Program) Write a program that tests the effectiveness of the rand() library function. Start by initializing 10 counters, such as zerocount, onecount, twocount, and so forth, to 0. Then generate a large number of pseudorandom integers between 0 and 9. Each time 0 occurs, increment zerocount; when 1 occurs, increment onecount; and so on. Finally, display the number of 0s, 1s, 2s, and so on that occurred and the percentage of time they occurred.In C programming Mathematically, given a function f, we recursively define fk(n) as follows: if k = 1, f1(n) = f(n). Otherwise, for k > 1, fk(n) = f(fk-1(n)). Assume that there is an existing function f, which takes in a single integer and returns an integer. Write a recursive function fcomp, which takes in both n and k (k > 0), and returns fk(n). int f(int n);int fcomp(int n, int k){a) Write a non-recursive function in C++/ to multiply all even numbers from 2 to n, where n is an input to the function, and n>=2. (reminder: An even number is divisible by 2 and generates a remainder of 0. for example 2,4,6,... are even numbers). b) Analyze your algorithm in part (a) in the worst-case. Show all your work. Then express the time as Big-O().
- for C++ write a progam for the greatest common divisor of integers x and y is the largest integer that evenly divides both x and y. Write a recursive function gcd that returns the greatest common divisor of x and y. The gcd of x and y is defined recursively as follows. If y is 0 the answer is x; otherwise gcd(x, y) is gcd(y, x%y).The answer must be a Racket Code snapshot with output with the correct answer or a Downvote will be done and the answer Reported to Chegg for investigation Finish the Code bellow for Racket function named (n-list? obj) that takes one argument, returns true if and only if obj is an n-list. n-list? should be mutually recursive with the function num-expr?, which returns true if its argument is a number expression and false if not. (define n-list?(lambda (obj))) in Racket codeWrite a function that adds two polynomials. Ensure that the polynomial produced matches the requirements in HASKELL. addPoly :: (Num a, Eq a) => Poly a -> Poly a -> Poly a For example: > addPoly (P [1]) (P [-1,1])P [0,1]> addPoly (P [17]) (P [0,0,0,1])P [17,0,0,1]>addPoly (P [1,-1]) (P [0,1])P [1] You may find it helpful to write addPoly using two helper functions: one to combine the lists of coefficients, and one to trim trailing zeroes.
- A function k whose domain is the set of positive integers is defined as k(1)=4 and k(n)=k(n-1)-2. Function k was evaluated for several numbers. Select each statement below that is true. A - k(0)=−2 B - k(2)=2 C - k(3)=0 D - k(6)=3 E - k(−1)=−4Write a FRACTION calculator program with c++ that adds, subtracts, multiplies, and divides fractions. Your program should check for the division by 0, have and use the following functions: (a) abs - returns the absolute value of a given integer. (b) min - returns the smallest of two positive integers. (c) gcd - returns the greatest common divisor of two positive integers. (d) reduce - reduces a given fraction. (e) flip - reduces a given fraction and flips the sign if the denominator is negative. (f) add - finds the reduced sum of a pair of given fractions. (g) subtract - finds the reduced difference of a pair of given fractions, by making the second fraction negative then using the add function. (h) multiply - finds the reduced product of a pair of given fractions. (i) divide - finds the reduced quotient of a pair of given fractions by inverting the second fraction then using the multiply function.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.
- In this lab, we will practice: Implementing recursive function printing in a recursive function (before and after the recursion function call) working on setting the base and general cases for the recursion function Instructions Write a recursive function called print_num_pattern() to output the following number pattern. Given two positive integers as input (Ex: 12 3), subtract the second integer (3) from the first one (12) and print the result continually until 0 or a negative value is reached. Then continually print the result of the addition of the second integer to the last value you reached from the subtraction (zero or negative) until the first integer is reached again. Note: For this lab, do not end output with a newline. Ex. If the input is: 12 3 the output is: 12 9 6 3 0 3 6 9 12 Hint You do not need to actually do addition. You can write statements i.e. printing statements after the recursion function call.In Kotlin, Write a higher-order function that takes two arguments and returns a String. The first argument is a String and the second is a function that takes a Char and returns a Boolean. Your function should return a String containing those of the characters in the original String for which the function returns true. Use a tail-recursive helper function. Remember to make sure you get the tail recursion right by using the tailrec annotation. Test the functionWrite a function in C programming language that gets two ints x and y, and returns the sum of all ints between x and y, including them. int sum_interval(int x, int y); For example: - sum_interval(1, 4)returns 1+2+3+4=10. - sum_interval(10, 3)returns 3+4+5+6+7+8+9+10=52. - sum_interval(2, -1)returns (-1)+0+1+2=2. - sum_interval(-1, -1)returns -1. Note that we may have x>=y or x<=y. You may assume that the sum will be within the range of int.