;;; Solution fsa-3a.lisp
;;;
;;; Modified fsa-3 to accept "the first contestant"

(setf *fsa*
 '((Alphabet      (the first second two three contestant contestants))
   (States        (q0 q1 q2 q3 q4))
   (Initial       q0)
   (Final         (q3 q4))
   (Transitions
                  (((q0 the)         q1)

                   ((q1 first)       q2)
                   ((q1 second)      q2)
                   ((q1 two)         q3)
                   ((q1 three)       q3)
                   ((q1 contestant)  q4)
                   ((q1 contestants) q4)

                   ((q2 two)         q3)
                   ((q2 three)       q3)
                   ((q2 contestant)  q4)

                   ((q3 contestants) q4)))))