Solution 1
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        1) No pointer  information is missing. The  pointer information 
           (x -> p) is propagated after call to function f.             

        2) (x->p)  after function  f, gets  killed because  of statement
           x=&q. Therefore, information x->q is propagated beyond second
           call of function f.

        3) The  information  x->p is  propagated  beyond  first call  to
           function f1. But the statement after first call to f (x = &q)
           kills the  pointer information x->p and  generates x->q which
           is propagated beyond second call to function f.