Define a set X of strings in the symbols 0 and 1 as follows. B. 0 and 1 are in X. R1₁. If x and y are in X, so is xxyy. R₂. If x and y are in X, so is xyx. Explain why the string 01001011 EX using the definition. Build up the string step by step, and justify each step by referring to the appropriate part of the definition. 0 is in X B. B. 1 is in X R2. If 0 and 1 are in X, so is 0 0 R1. If 010 and 1 are in X, so is

What are some real life situations that require a program that is iterative? Include atleast three examples in your answer

C Programming. Could you update my code using the code I provided. I need help with 6.   #include <stdio.h>#include <stdlib.h> #define TRUE 1#define FALSE 0#define NAME 20#define ROW 8#define COL 8#define SPACE ' '#define PLAYER_X 1#define PLAYER_O 2#define ZERO 0#define ONE 1#define TWO 2#define FOUR 4#define INVALID -1 // function prototypesvoid welcomeScreen ();void displayExplicitBoard();void clearScreen();void playGame(); // main functionint main(){    // call function welcomeScreen    welcomeScreen();    // call function clearScreen    clearScreen();     // call function displayExplicitBoard//    displayExplicitBoard();     // call function playGame    playGame();     // program executed successfully    return 0;} // welcomeScreen function displays the Othello logo and rules of the gamevoid welcomeScreen (){    printf ("\t\t OOOO  TTTTTT HH  HH EEEEEE LL      LL      OOOO  \n");    printf ("\t\tOO  OO   TT   HH  HH EE     LL      LL     OO  OO \n");    printf ("\t\tOO…

This code is in Ada code in Ada Please    Not in C code    Please type your solution out in working code don't just give me an example they may just confuse me more    Below is Number_Theory.adb   with Ada.Numerics.Generic_Elementary_Functions;   package body Number_Theory is      -- Instantiate the library for floating point math using Floating_Type.    package Floating_Functions is new Ada.Numerics.Generic_Elementary_Functions(Floating_Type);    use Floating_Functions;    function Factorial(N : in Factorial_Argument_Type) return Positive is    begin       -- TODO: Finish me!       --       -- 0! is 1       -- N! is N * (N-1) * (N-2) * ... * 1       return 1;    end Factorial;    function Is_Prime(N : in Prime_Argument_Type) return Boolean is       Upper_Bound     : Prime_Argument_Type;       Current_Divisor : Prime_Argument_Type;    begin       -- Handle 2 as a special case.       if N = 2 then          return True;       end if;         Upper_Bound := N - 1;       Current_Divisor…