Write an alternative gcd function based on the following observations (arrange so that a > b):  a. gcd(a, b) = 2gcd(a/2, b/2) if a and b are both even.  b. gcd(a, b) = gcd(a/2, b) if a is even and b is odd.  c. gcd(a, b) = gcd(a, b/2) if a is odd and b is even.  d. gcd(a, b) = gcd((a + b)/2, (a − b)/2) if a and b are both odd. Write a main function to test your gcd function.

In c++ please help me answer this question I will give you a good rating :) implement the three versions of the addupto20() function: iterative, recursive, and divide-and-conquer approach test these functions with a few inputs from your main() Iterative solution * check if there exsits two numbers from vector data that add up to 20 e.g., if data=[2,5,3,15], the function returns true, as data [0] +data [3]==20 e.g., if data=[3,4,0,8], the function return false precondition: vector data has been initialized postcondition: if there are two numbers from list add up to 20, return true; otherwise, return false */ bool AddupTo20 (const vector‹int> & data){ } Come up with a recursive solution to the problem, following the hints given below: * check if there exsits two numbers from vector data[first...right] add up to 20 e.g., if data=[2,5,3,15], first=0, last=3, the function returns true, as data [0] +data [3]==20 e.g., if data=[2,5,3,15], first=2, last=3, the function returns…

Given the code: void e(int x, int y) { if(y<2) { cout << x << " "; return; } cout << x << " "; e(x+1, y-1); } 1. Trace the function when x and y are 4.

Problem Description: Implement the described Mancala variant in C++ in the pictures. Player 1 will be a human while the other will be a computer player. The computer player selects a hole following the following decision logic: If there is a hole with enough pieces to finish the turn in the computer player's end zone, select it. Otherwise select the hole on the right to left path with the greatest amount of marbles. If there are no marbles on that path then select the hole closest to the opponent's end zone. It is highly suggested, but not required, to pause the output after the computer makes a move to allow the user to interpret the output before more gets printed out. Look into cin.get(); The program should not allow the player to enter incorrect selections. The player cannot select a nonexistent hole, the end zones, or an empty hole. Look at the example output for an example for how to output the game board formatted in a nice way. You must tell the player what input is valid. For…