Anna Savarin
HW#2

Question 1.
Pg. 273, Exercise 7.1.

Q: The first known correct software solution to the critical-section problem 
   for two threads was developed by Dekker; it is shown in Figure 7.43.  The 
   two threads, T0 and T1, coordinate activity sharing an object of class Dek-
   ker.  Show that the algorithm satisfies all three requirements for the
   critical section problem.

A: The three criteria referred to by the book are mutual exclusion, progress,
   and bounded waiting.  (1) Mutual exclusion is achieved by the use of two
   variables, flag and turn.  If T0 has its flag set to true, and T1's flag
   is false, then T0 has no problem entering its critical section.  If both
   flags are true, then the turn will be checked to see who is up next, T0 
   or T1.  If T1 is up next, then T0 will wait its turn; otherwise, T0 will
   go while T1 waits.  So, in the possible situation where both processes are
   interested in entering the critical section, the process whose turn it is 
   will go and, once it is done, the turn will be handed to the other process
   who will now have the entrance to the critical section.  (2) This solution
   guarantees progress because all possible flag-turn combinations allow the
   interested processes to proceed.  In the case of both processes with their
   flags being true, the turn variable will determine who gets to go and the
   turn variable can only equal to one value at a time.  Once a process comp-
   letes its turn, it will give the turn to the other process, allowing it to
   go.  The only way for the processes to not enter their critical section
   is when they are not interested. (3) Bounded waiting is guaranteed in the
   following way.  If T1 has just exited its critical section and attempts to
   gain entry again while T0's flag is true, then, upon T1's exit, the turn 
   will be given to T0.  In this case, T0 will have the turn and the flag, 
   which will cause T0 to proceed and T1 to clear its flag and wait.  


Question 2.

Q: Peterson's solution to MUTEX was for two processes.  
   (a) Generalize Peterson's solution to an arbitrary number of processes.  
       Please write the solution in a conceptual formulation.
   (b) Show that your solution achieves mutual exclusion.
   (c) Show that no process can be locked out in your solution.
   (d) Show that your solution is fair.

A: (a) For processes 0 through n:

       Loop Forever
         Process i wants = true;
	 turn = ( i + 1 ) % n; (in other words, give the turn to the neighbor)
	 while ( neighbor wants and it is neighbor's turn );
	 keep giving turn to the next process after neighbor and waiting if the
	      process wants to go until turn comes back to you
	       Critical Section only if you checked on everyone and made sure
			that everyone who wanted to go before you went;
			hold on to your turn while in CS
	 Process i wants = false;
	 turn released;
		 Non-Critical Section

       
       The generalization here is the following.  Assign turns to processes
       0 through n consecutively and let them go if they want to go and it is
       their turn.  Otherwise, proceed to your critical section but HOLD ON
       TO YOUR TURN.  It is analogous to keeping, say, a token in your pocket
       that says, "It is my turn and I will do what I have to do with this
       token in my pocket.  I will pass on this token once I am done."

   (b) This solution achieves mutual exclusion because the process that has
       the turn will go.  All other processes have to wait for the turn to
       come to them and will not be able to enter the critical section
       without the turn "token."

   (c) A process can only block itself while it waits for its neighbors to take
       their turn.  The turn is guaranteed to come back to the process and
       allow the process to go since it will want to go and have the turn
       allow it to do so.

   (d) No single process will have to wait longer than any other assuming
       all processes want to go.  All processes will have to let everyone
       go once before proceeding.  The waiting is bounded, however, to at
       most the amount of time it takes everyone else to skip you.