Other answers to lab 4

2.1
    s1.find(s2)> 0 says that s1 is longer than s2. Thus, s2.find(s1)>=0 is never True

2.2
       if B1:
           x = 1
       elif B2:
           x = 2
       elif B3:
           x = 3
       return x

3.1
        dcbabcd

3.2
          N = 0
          for c in S:
               if x==c:
                    N = N+1
          return N

3.3   returns a string that is made up of the even-indexed characters in S. The string is in reverse order.