* Exercise #1: 

Question #1:

The main process creates a first child at line 16, and then a second child at line 20. All these processes have then a value of the count variable that's equal to 1, so no more processes are created. 

So the answer is: 2 processes.

Question #2:
 
  The main process creates 2 children during the first iteration.
  At the end of this iteration each process has the following count values:
        - main process: count = 1 (integer division 3/2 = 1)
        - first child:  count = 2 (subtraction 3-1 = 2)
        - second child: count = 2 (subtraction 3-1 = 2)
  Therefore, the main process exists and both children do another iteration of the loop since in their address space count = 2. Each of them will create 2 processes (see Question #1).

  Answer: 2 + 2*2 = 6 processes


Question #3:

   T(0) = 0
   T(1) = 0
   T(2) = 2
   T(3) = 6
   T(n) = 2 + 2*T(n-1) + T(floor(n/2)) 

   At the first iteration, 2 children are created (the first term above).  Then we have
   3 processes.  2 of them re-do the same thing with n-1, and 1 re-does the same thing
   with floor(n/2).

Question #4:

   No need to solve the induction. Instead, we can write a small program to
   compute the values. For instance, in python:

#!/usr/bin/python3
def f(x):
        if (x <= 1):
                return 0
        elif (x == 2):
                return 2
        elif (x == 3):
                return 6
        else:
                return 2 + 2 * f(x-1) + f(x // 2)

for n in range(0,30):
    count = f(n)
    print(f"T({n}) = {count}")
    if count > 10000:
        print(f"Answer: {n}")
        break


   Which gives use the answer: 13    (T(12) = 6118,  T(13) = 12318)


* Exercise #2:

q1: 40
q2: 60
q3: 23
q4: 23
q5: 50
q6: 20
q7: 200
q8: 23
q9: 89, 200, 23