Exercise 2 (25 points) Proof by contradiction Consider a program P that contains two threads of execution. In a simplistic way, it means that two routines Thread and Thread2 in Program P can be running concurrently. The integer variable Turn is initialized to 0 by Program P. The variable Turn is shared by the threads Thread and Thread2. Below are the codes for Thread/ and Thread2. Thread Thread 2 while (0) { while (0) { while (Turn == 0); //loop here while Turn == 0 while (Turn == 1); //loop here while Turn == | Code A Code B Turn = 0; Turn = 1; } } Code A and Code B are blocks of multiple instructions. Pay attention: Code A and Code B are not part of the while (Turn ==..) loops. For example, if (Turn == 1) for Thread I, this while loop keeps looping and Code B will not run unless the variable Turn becomes 0. We assume that Code A and Code B do not modify the variable Turn. Answer the following questions: a) (2 points) When instructions of Code A are running, what is the value of Turn! b) (2 points) When instructions of Code B are running, what is the value of Turn! c) (4 points) Can Code A and Code B be running simultaneously? Answer and justify your answer a. d) Use a proof by contradiction to show that Code A and Code B CANNOT be running simultaneously. For this proof, I suggest to follow these steps: (4 points) What will be your starting assumption? b. (10 points) Can you infer from this assumption some contradiction? C. (3 points) After you show the contradiction, what will be your conclusion