1 (a)
	a = x where x could be 0, 2, 3 or 5 therefore a can be either 0, 2, 3 or 5
	b will always be 6 since it gets set to 6 at the end of T2
	x will always be 5 since it gets set near the end of T1
	y = b where b = 0, y or 6.  since the last write to y is from b, there
		can be a cycle between b and y, therefore y can be any arbitrary value
	z = a where a can be 0, 4 or x, and x can be 0, 2, 3 or 5; therefore z can
		be 0, 2, 3, 4 or 5.
	
1 (b)
	a = x where x can be taken from any stage in T1 or the initial value of x.
		therefore a = 0, 2, 3 or 5.
	b = 6 no matter what since if gets set right at the end of T2 and it never
		gets set in T1
	x = 5 since it's only written to in T1, and the last write is a 5
	y = 0 or 6.  if T1 gets the lock first, then at the end of the critical
					section, y = b where b can only be 0 since T2 didn't write
					to b yet.
				if T2 gets the lock first, y gets b's value which can be 0 or
					6 (from the writes in T2)
	z = a where a = 0, 2, 3, 4 or 5 depending on which value was written to it in
		T2 at the time that T1 reads it.  0 and 4 come from the initial writes to
		a, and the rest come from a = x which could get x from any stage in T1.
	