The following YACC snippet is used to implement a grammar of an expression. The expressions are : E -> E+E E-> E'E E-> id program: program expr '\n' { printf("Xd\n", S2); } expr: { SS - $1; } { SS - $1 + $3; } { $S - $1 - $3; } INTEGER I expr + expr I expr '-' expr Modify the above code to handle the following grammar rules: E-> Add E E E -> Sub EE E-> Mul EE E-> Div EE E-> id, where Add, Sub, Mul, and Div are used to implement Addition, subtraction, multiplication, and division operations respectively.

Theory of computation: Write a program using c++ and pseudo code to check if the entering value following the grammar or not without checking if the grammar is left recursion or not: E ---> E + T | T T---> T * F | F F---> (E) | i Thanks in advance ( I hope you solve it fast with the right answer if you can't solve it leave it please! )

Assignment 16: Iteration Control Structure - Hands on practice Hint: Consider numbers 1 to 10, for each number find the odd multiples. Example, for 1, it should be 1, 3, 5, 7, 9 Objective: Given a real world problem, implement the logic and solve the problem using appropriate constructs (sequen- tial, selection, iteration) using an object oriented program- ming language (Java) for 2, it should be 2, 6, 10, 14, 18 Estimated time: 50 minutes Summary of this assignment: In this assignment, you have learnt the implementation of iteration control structures. Problem Description: Write simple java programs to implement the following prob- lems. 1.Find the sum of first 5 numbers (1 to 5) using for loop End of document statement 2.Find the product of first 5 numbers (1 to 5) using while loop statement 3. Generate the multiplication table of 10 upto multiple value 500 4.Calculate the square of first 5 numbers (1 to 5) using do..while statement 5.Find the odd multiples among the first 10…

4- Write a C program to read an integer number num and find it's factorial. Note: the factorial number is all the number multiplied from 1 to the number itself. n! = n(n - 1)(n – 2) ... (2)(1) Example: 5! = (5)(4)(3)(2)(1) = 120