Pushdown Automata

1. Pushdown Automata Chapter 12

2. Recognizing Context-Free Languages Two notions of recognition: (1) Say yes or no, just like with FSMs (2) Say yes or no, AND if yes, describe the structure a + b * c

3. Just Recognizing • We need a device similar to an FSM except that it needs more power. • The insight: Precisely what it needs is a stack, which gives it an unlimited amount of memory with a restricted structure. • Example: Bal (the balanced parentheses language) • (((()))

4. Definition of a Pushdown Automaton • M = (K, , , , s, A), where: • K is a finite set of states •  is the input alphabet •  is the stack alphabet • sK is the initial state • AK is the set of accepting states, and •  is the transition relation. It is a finite subset of • (K ( {}) *)  (K*) • state input or  string of state string of • symbols symbols • to pop to push • from top on top • of stack of stack

5. Definition of a Pushdown Automaton • A configuration of M is an element of K* *. • The initial configuration of M is (s, w, ).

6. Manipulating the Stack • c will be written as cab • a • b • If c1c2…cn is pushed onto the stack: • c1 • c2 • cn • c • a • b • c1c2…cncab

7. Yields Let c be any element of  {}, Let 1, 2 and  be any elements of *, and Let w be any element of *. Then: (q1, cw, 1) |-M (q2, w, 2) iff ((q1, c, 1), (q2, 2)) . Let |-M* be the reflexive, transitive closure of |-M. C1yields configuration C2 iff C1 |-M*C2

8. Computations A computation by M is a finite sequence of configurations C0, C1, …, Cn for some n 0 such that: ● C0 is an initial configuration, ● Cn is of the form (q, , ), for some state qKM and some string  in *, and ● C0 |-MC1 |-MC2 |-M … |-MCn.

9. Nondeterminism If M is in some configuration (q1, s, ) it is possible that: ● contains exactly one transition that matches. ● contains more than one transition that matches. ● contains no transition that matches.

10. Accepting A computation C of M is an accepting computation iff: ● C = (s, w, ) |-M* (q, , ), and ● qA. Maccepts a string w iff at least one of its computations accepts. Other paths may: ● Read all the input and halt in a nonaccepting state, ● Read all the input and halt in an accepting state with the stack not empty, ● Loop forever and never finish reading the input, or ● Reach a dead end where no more input can be read. The language accepted byM, denoted L(M), is the set of all strings accepted by M.

11. Rejecting A computation C of M is a rejecting computation iff: ● C = (s, w, ) |-M* (q, w, ), ● C is not an accepting computation, and ● M has no moves that it can make from (q, , ). Mrejects a string w iff all of its computations reject. So note that it is possible that, on input w, M neither accepts nor rejects.

12. A PDA for Balanced Parentheses

13. A PDA for Balanced Parentheses M = (K, , , , s, A), where: K = {s} the states  = {(, )} the input alphabet  = {(} the stack alphabet A = {s}  contains: ((s, (, **), (s, ( )) ((s, ), ( ), (s, )) **Important: This does not mean that the stack is empty

14. A PDA for AnBn = {anbn: n 0}

15. A PDA for AnBn = {anbn: n 0}

16. A PDA for {wcwR: w {a, b}*}

17. A PDA for {wcwR: w {a, b}*} M = (K, , , , s, A), where: K = {s, f} the states  = {a, b, c} the input alphabet  = {a, b} the stack alphabet A = {f} the accepting states  contains: ((s, a, ), (s, a)) ((s, b, ), (s, b)) ((s, c, ), (f, )) ((f, a, a), (f, )) ((f, b, b), (f, ))

18. A PDA for {anb2n: n 0}

19. A PDA for {anb2n: n 0}

20. Exploiting Nondeterminism A PDA M is deterministic iff: ● M contains no pairs of transitions that compete with each other, and ● Whenever M is in an accepting configuration it has no available moves. But many useful PDAs are not deterministic.

21. A PDA for PalEven ={wwR: w {a, b}*} S SaSa SbSb A PDA:

22. A PDA for PalEven ={wwR: w {a, b}*} S SaSa SbSb A PDA:

23. A PDA for {w {a, b}* : #a(w) = #b(w)}

24. A PDA for {w {a, b}* : #a(w) = #b(w)}

25. More on NondeterminismAccepting Mismatches L = {ambn: mn; m, n > 0} Start with the case where n = m: b/a/ a//a b/a/ 1 2

26. More on NondeterminismAccepting Mismatches L = {ambn: mn; m, n > 0} Start with the case where n = m: b/a/ a//a b/a/ 1 2 ● If stack and input are empty, halt and reject. ● If input is empty but stack is not (m > n) (accept): ● If stack is empty but input is not (m < n) (accept):

27. More on NondeterminismAccepting Mismatches L = {ambn: mn; m, n > 0} b/a/ a//a b/a/ 2 1 ● If input is empty but stack is not (m < n) (accept): b/a/ a//a /a/ /a/ b/a/ 2 1 3

28. More on NondeterminismAccepting Mismatches L = {ambn: mn; m, n > 0} b/a/ a//a b/a/ 2 1 ● If stack is empty but input is not (m > n) (accept): b// b/a/ a//a b// b/a/ 2 4 1

29. Putting It Together L = {ambn: mn; m, n > 0} ● Jumping to the input clearing state 4: Need to detect bottom of stack. ● Jumping to the stack clearing state 3: Need to detect end of input.

30. The Power of Nondeterminism Consider AnBnCn = {anbncn: n 0}. PDA for it?

31. The Power of Nondeterminism Consider AnBnCn = {anbncn: n 0}. Now consider L =  AnBnCn. L is the union of two languages: 1. {w {a, b, c}* : the letters are out of order}, and 2. {aibjck: i, j, k 0 and (ij or jk)} (in other words, unequal numbers of a’s, b’s, and c’s).

32. A PDA for L = AnBnCn

33. Are the Context-Free Languages Closed Under Complement? AnBnCn is context free. If the CF languages were closed under complement, then AnBnCn = AnBnCn would also be context-free. But we will prove that it is not.

34. L = {anbmcp: n, m, p 0 and nm or mp} SNC /* nm, then arbitrary c's SQP /* arbitrary a's, then pm NA /* more a's than b's NB /* more b's than a's Aa AaA AaAb Bb BBb BaBb C | cC /* add any number of c's PB' /* more b's than c's PC' /* more c's than b's B'b B'bB' B'bB'c C'c | C'c C'C'c C'bC'c Q | aQ /* prefix with any number of a's

35. Reducing Nondeterminism ● Jumping to the input clearing state 4: Need to detect bottom of stack, so push # onto the stack before we start. ● Jumping to the stack clearing state 3: Need to detect end of input. Add to L a termination character (e.g., $)

36. Reducing Nondeterminism ● Jumping to the input clearing state 4:

37. Reducing Nondeterminism ● Jumping to the stack clearing state 3:

38. More on PDAs A PDA for {wwR : w {a, b}*}: What about a PDA to accept {ww : w {a, b}*}?

39. PDAs and Context-Free Grammars Theorem: The class of languages accepted by PDAs is exactly the class of context-free languages. Recall: context-free languages are languages that can be defined with context-free grammars. Restate theorem: Can describe with context-free grammar Can accept by PDA

40. Going One Way • Lemma: Each context-free language is accepted by some PDA. • Proof (by construction): • The idea: Let the stack do the work. • Two approaches: • Top down • Bottom up

41. Top Down The idea: Let the stack keep track of expectations. Example: Arithmetic expressions EE + T ET TTF TF F (E) Fid (1) (q, , E), (q, E+T) (7) (q, id, id), (q, ) (2) (q, , E), (q, T) (8) (q, (, ( ), (q, ) (3) (q, , T), (q, T*F) (9) (q, ), ) ), (q, ) (4) (q, , T), (q, F) (10) (q, +, +), (q, ) (5) (q, , F), (q, (E) ) (11) (q, , ), (q, ) (6) (q, , F), (q, id)

42. A Top-Down Parser The outline of M is: M = ({p, q}, , V, , p, {q}), where  contains: ● The start-up transition ((p, , ), (q, S)). ● For each rule Xs1s2…sn. in R, the transition: ((q, , X), (q, s1s2…sn)). ● For each character c, the transition: ((q, c, c), (q, )).

43. Example of the Construction L = {anb*an} 0 (p, , ), (q, S) (1) S  * 1 (q, , S), (q, ) (2) S  B 2 (q, , S), (q, B) (3) S aSa 3 (q, , S), (q, aSa) (4) B  4 (q, , B), (q, ) (5) B bB 5 (q, , B), (q, bB) 6 (q, a, a), (q, ) input = a a b b a a 7 (q, b, b), (q, ) trans state unread input stack p a a b b a a 0 q a a b b a a S 3 q a a b b a aaSa 6 q a b b a a Sa 3 q a b b a aaSaa 6 q b b a a Saa 2 q b b a a Baa 5 q b b a abBaa 7 q b a a Baa 5 q b a abBaa 7 q a a Baa 4 q a aaa 6 q aa 6 q 

44. Another Example L = {anbmcpdq : m + n = p + q}

45. Another Example L = {anbmcpdq : m + n = p + q} (1) SaSd (2) ST (3) SU (4) TaTc (5) TV (6) U bUd (7) UV (8) VbVc (9) V input = a a b c d d

46. Another Example L = {anbmcpdq : m + n = p + q} 0 (p, , ), (q, S) (1) SaSd 1 (q, , S), (q, aSd) (2) ST 2 (q, , S), (q, T) (3) SU 3 (q, , S), (q, U) (4) TaTc 4 (q, , T), (q, aTc) (5) TV 5 (q, , T), (q, V) (6) U bUd 6 (q, , U), (q, bUd) (7) UV 7 (q, , U), (q, V) (8) VbVc 8 (q, , V), (q, bVc) (9) V 9 (q, , V), (q, ) 10 (q, a, a), (q, ) 11 (q, b, b), (q, ) input = a a b c d d 12 (q, c, c), (q, ) 13 (q, d, d), (q, ) trans state unread input stack

47. The Other Way to Build a PDA - Directly L = {anbmcpdq : m + n = p + q} (1) SaSd(6) UbUd (2) ST (7) UV (3) SU (8) VbVc (4) TaTc(9) V (5) TV input = a a b c d d

48. The Other Way to Build a PDA - Directly L = {anbmcpdq : m + n = p + q} (1) SaSd (6) UbUd (2) ST (7) UV (3) SU (8) VbVc (4) TaTc (9) V (5) TV input = a a b c d d a//a b//a c/a/ d/a/ b//a c/a/ d/a/ 1 2 3 4 c/a/ d/a/ d/a/

49. Notice Nondeterminism Machines constructed with the algorithm are often nondeterministic, even when they needn't be. This happens even with trivial languages. Example: AnBn = {anbn: n 0} A grammar for AnBn is: A PDA M for AnBn is: (0) ((p, , ), (q, S)) [1] SaSb (1) ((q, , S), (q, aSb)) [2] S (2) ((q, , S), (q, )) (3) ((q, a, a), (q, )) (4) ((q, b, b), (q, )) But transitions 1 and 2 make M nondeterministic. A directly constructed machine for AnBn:

50. Bottom-Up The idea: Let the stack keep track of what has been found. (1) EE + T (2) ET (3) TTF (4) TF (5) F (E) (6) Fid Reduce Transitions: (1) (p, , T + E), (p, E) (2) (p, , T), (p, E) (3) (p, , FT), (p, T) (4) (p, , F), (p, T) (5) (p, , )E( ), (p, F) (6) (p, , id), (p, F) Shift Transitions (7) (p, id, ), (p, id) (8) (p, (, ), (p, () (9) (p, ), ), (p, )) (10) (p, +, ), (p, +) (11) (p, , ), (p, )