Midterm Practice Solutions Posted Feb. 16 2005
1. a) xxax b) abxx 2. a) Base case: 0 25 = 0 Recursion clause: (x + 1) 25 = x 25 + 25 This one line () is a complete answer to the question. b) The induction goes most smoothly if you do it on the number of concatenations of elements of {aa} that form the given member of {aa} . Base case: The number of concatenations is 0, then x is the empty string. On the empty string, F1 remains in q0 , which is not a final state of F1 . Induction step: Induction assumption:Say that every string x in {aa} that is formed by n or fewer concatenations of elements of aa is not accepted by F1 . Say that x is formed by the concatenation of n+1 instances of aa. Since 0 < n + 1, there must be strings y {aa} and y = aa, such that x = y y . Since y is formed with n concatenations of aa, it is not accepted by F1 , by the induction assumption. If y is the empty string, then x = aa. F1 enters q1 upon reading a, and then enters q2 upon reading the final a. q2 is not a final state of F1 , so the computation is unsuccessful. y If is not the empty string, then F1 leaves q0 when it reads the first a and there is no transition that can bring it back to q0 . q1 is a final state, and since y is not accepted by F1 , F1 must finish reading y in q2 . Now F1 has only y = aa to read. Since F1 begins reading aa in q2 , it will begin by reading a and moving into q1 , and then reading a again and moving into q2 , terminating in q2 , which is not a final state. So x is not accepted by F1 . This completes the inductive step, and hence the proof. 3. { q0 , a, q1 , q1 , a, q2 , q2, a, q1 } are the transitions, q0 , q1 and q2 are the states, with q0 and q1 th...