Computational Models Lecture 5

Transcription

1 Computational Models Lecture 5 Chomsky Normal Form of CFGs Closure Properties of CFLs Push-down Automata (PDAs) Equivalence of PDAs and CFLs Algorithmic Aspects of PDAs and CFLs DFAs and PDAs: Perspectives Sipser s book, 2.1, 2.2 & 2.3 Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 1

2 Chomsky Normal Form A simplified, canonical form of context free grammars. Elegant by itself, useful (but not crucial) in proving equivalence theorem. Can also be used to slightly simplify proof of pumping lemma. All rules are of the form (1)A (2)A BC (3)S ε where S is the start symbol, A, B and C are any variable, except B and C not the start symbol, and A can be the start symbol. a Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 2

3 Theorem: Any context-free language is generated by a context-free grammar in Chomsky normal form. Basic idea: Add new start symbol S 0. Eliminate all ε rules of the form A ε. Eliminate all unit rules of the form A B. At each step, patch up rules so that grammar generates the same language. Convert remaining long rules to proper form. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 3

4 Proof Add new start symbol S 0 and rule S 0 S. Guarantees that new start symbol does not appear on right-hand-side of a rule. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 4

5 Proof Eliminating ε rules. Repeat: remove some A ε. for each R uav, add rule R uv. and so on: for R uavaw add R uvaw, R uavw, and R uvw. for R A add R ε, except if R ε has already been removed. until all ε-rules not involving the original start variable have been removed. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 5

6 Proof Eliminate unit rules. Repeat: remove some A B. for each B u, add rule A u, unless this is previously removed unit rule. (u is a string of variables and terminals.) until all unit rules have been removed. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 6

7 Proof Finally, convert long rules. To replace each A u 1 u 2...u k (for k 3), introduce new non-terminals N 1,N 2,...,N k 1 and rules A u 1 N 1 N 1 u 2 N 2. N k 3 u k 2 N k 2 N k 2 u k 1 u k Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 7

8 Conversion Example Initial Grammar: S A B ASA ab B S b ε (1) Add new start state: S 0 S S ASA ab A B S B b ε Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 8

9 Conversion Example (2) S 0 S S ASA ab A B S B b ε (2) Remove ε-rule B ε: S 0 S S ASA ab a A B S ε B b ε Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 9

10 Conversion Example (3) S 0 S S ASA ab a A B S ε B b (3) Remove ε-rule A ε: S 0 S S ASA ab a AS SA S A B S ε B b Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 10

11 Conversion Example (4) S 0 S S ASA ab a AS SA S A B S B b (4) Remove unit rule S S S 0 S S ASA ab a AS SA S A B S B b Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 11

12 Conversion Example (5) S 0 S S ASA ab a AS SA A B S B b (5) Remove unit rule S 0 S: S 0 S ASA ab a AS SA S ASA ab a AS SA A B S B b Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 12

13 Conversion Example (6) S 0 ASA ab a AS SA S ASA ab a AS SA A B S B b (6) Remove unit rule A B: S 0 ASA ab a AS SA S ASA ab a AS SA A B S b B b Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 13

14 Conversion Example (7) S 0 ASA ab a AS SA S ASA ab a AS SA A S b B b Remove unit rule A S: S 0 ASA ab a AS SA S ASA ab a AS SA A S b ASA ab a AS SA B b Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 14

15 Conversion Example (8) S 0 ASA ab a AS SA S ASA ab a AS SA A b ASA ab a AS SA B b (8) Final simplification treat long rules: S 0 AA 1 UB a SA AS S AA 1 UB a SA AS A b AA 1 UB a SA AS A 1 SA U a B b Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 15

16 Why is CNF Important? Theorem: If G is a CFG in Chomsky normal form, then for any w L(G), such that w = n, there is a derivation of size 2n 1 of w in G. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 16

17 CFL Closure Properties We saw that Context-Free Languages are closed under union, concatenation, and star. It is time we resolve closure with respect to complementation and intersection. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 17

18 CFL Closure Properties Are the context free languages closed under intersection? Suggested approach: Can we intersect two context free languages to get 0 n 1 n 2 n? Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 18

19 CFL Closure Properties Are the context free languages closed under intersection? S 1 A 1 B 1 S 2 A 2 B 2 A 1 0A A 2 0A 2 ε B 1 2B 1 ε B 2 1B L 1 = 0 n 1 n 2 L 2 = 0 1 n 2 n L 1 L 2 = 0 n 1 n 2 n L 1 is a context free language, L 2 is a context free language, but L 1 L 2 is not! Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 19

20 CFL Closure Properties The fact that CFLs are not closed under intersection but are closed under union implies they are not closed under complementation, as L 1 L 2 = L 1 L 2. Example: L = {ww w {0, 1} }. For any y L, either y s length is odd. y s length is even, 2l, and there is an i 1 such that y i y l+i. It can be shown that L is context-free. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 20

21 CFL Closure Properties Context free languages are closed under intersection with a regular language. That is, if L 1 is a context free language, and L 2 is regular, L 1 L 2 is context-free. Details - in the tutorial... Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 21

22 CFL Closure Properties: Example Is L = {(0 1 2) : # of 0 s = # of 1 s = # of 2 s } context free? Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 22

23 CFL Closure Properties L {(0 1 2) : # 0 s = # 1 s = # 2 s } Is L context free? L ={0 n 1 n 2 n : n > 0} which is not context free. Context free languages intersected with a regular languages are context free is regular. So L is not a context free language! Could this be proven directly, using pumping? Probably, but we prefer to work less (provided the proofs are correct :-). Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 23

24 String Generators and String Acceptors Regular expressions are string generators they tell us how to generate all strings in a language L Finite Automata (DFA, NFA) are string acceptors they tell us if a specific string w is in L CFGs are string generators Are there string acceptors for Context-Free languages? YES! Push-down automata Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 24

25 A Finite Automaton Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 25

26 A PushDown Automaton Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 26

27 A PushDown Automaton We add a memory device with restricted access: A stack (last in, first out; push/pop). Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 27

28 An Example Recall that the language {0 n 1 n n 0} is not regular. Consider the following PDA: Read input symbols, one by one For each 0, push it on the stack As soon as a 1 is seen, pop a 0 for each 1 read Accept if stack is empty when last symbol read Reject if Stack is non-empty when end of input symbol read Stack is empty but input symbol(s) still exist, 0 is read after 1. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 28

29 Comparing PDA and Finite Automata PDA may be deterministic or non-deterministic. Unlike finite automata, non-determinism adds power: There are some languages accepted only by non-deterministic PDAs. Transition function δ looks different than DFA or NFA cases, reflecting stack functionality. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 29

30 The Transition Function Denote input alphabet by Σ and stack alphabet by Γ. the domain of the transition function δ is current state: Q next input, if any: Σ ε = Σ {ε} stack symbol popped, if any: Γ ε = Γ {ε} and its range is new state: Q stack symbol pushed, if any: Γ ε non-determinism: P(of two components above) δ : Q Σ ε Γ ε P(Q Γ ε ) Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 30

31 Formal Definitions A pushdown automaton (PDA) is a 6-tuple (Q, Σ, Γ,δ,q 0,F), where Q is a finite set called the states, Σ is a finite set called the input alphabet, Γ is a finite set called the stack alphabet, δ : Q Σ ε Γ ε P(Q Γ ε ) is the transition function, q 0 Q is the start state, and F Q is the set of accept states. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 31

32 Meaning of ε-transitions δ(q,ε,a) = {(q 1,b), (q 2,c)} - don t read input δ(q,b,ε) = {(q 1,b), (q 2,c)} - don t pop δ(q,b,a) = {(q 1,ε), (q 2,c)} - don t push Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 32

33 How to define acceptance of a PDA? Two options: Accept when reaching a final state. Accept when the stack is empty. These two notions of acceptance are equivalent in the sense that for every PDA accepting L by a final state, there is a PDA accepting L by empty stack, and vice versa. To show this equivalence, it will be convenient to assume that at the beginning of the run the stack of a PDA contains a special beginning marker X. So we extend the definition of a PDA from M = (Q, Σ, Γ,δ,q 0,F) to M = (Q, Σ, Γ,δ,q 0,F,X), where X Γ is a marker on top of stack of M at the beginning of the run. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 33

34 Formal Definition - Acceptance by Final State A PDA M = (Q, Σ, Γ,δ,q 0,F,X) accepts input w if w = w 1 w 2...w m, where each w i Σ ε, and sequences of r 0,r 1,...,r m Q and strings s 0,s 1,...,s m Γ exist that satisfy the following conditions: r 0 = q 0 and s 0 = X (M starts in start state with X on top of stack). For i = 0,...,m 1, we have (r i+1,b) δ(r i,w i+1,a), where s i = at and s i+1 = bt for some a,b Γ ε and t Γ (M moves properly according to the state, stack and next input symbol). r m F (Acceptance by final state). The language L(M) is the set of all words accepted by M by final state. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 34

35 Formal Definition - Acceptance by Empty Stack A PDA M = (Q, Σ, Γ,δ,q 0,F,X) accepts input w if w = w 1 w 2...w m, where each w i Σ ε, and sequences of r 0,r 1,...,r m Q and strings s 0,s 1,...,s m Γ exist that satisfy the following conditions: r 0 = q 0 and s 0 = X (M starts in start state with X on top of stack). For i = 0,...,m 1, we have (r i+1,b) δ(r i,w i+1,a), where s i = at and s i+1 = bt for some a,b Γ ε and t Γ (M moves properly according to the state, stack and next input symbol). s m = ε (Acceptance by empty stack). The language E(M) is the set of all words accepted by M by empty stack. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 35

36 Equivalence of Acceptance - 1 If a language L is L(M 2 ) for some PDA M 2, then L is E(M 1 ) for some PDA M 1. Proof Idea: M 1 should simulate M 2, with the option for M 1 to erase its stack whenever M 2 enters a final state. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 36

37 Equivalence of Acceptance - The Proof If a language L is L(M 2 ) for some PDA M 2, then L is E(M 1 ) for some PDA M 1. Let M 2 = (Q, Σ, Γ,δ,q 0,F,Z 0 ). Construct M 1 as follows: M 1 = (Q {q e,q 0}, Σ, Γ {X 0 },δ,q 0,,X 0 ) δ (q 0,ǫ,ǫ) = {(q 0,Z 0 )}. For all q Q, a Σ ε and b Γ ε, δ (q,a,b) = δ(q,a,b). For all q F,(q e,ε) δ (q,ε,ε). For all a Γ ε {X 0 }, (q e,ε) δ (q e,ε,a). Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 37

38 Equivalence of Acceptance - 2 If a language L is E(M 1 ) for some PDA M 1, then L is L(M 2 ) for some PDA M 2. Proof Idea: M 2 should simulate M 1, and detect when M 1 empties its stack. M 2 enters a final state when and only when this occurs. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 38

39 Equivalence of Acceptance - Proof If a language L is E(M 1 ) for some PDA M 1, then L is L(M 2 ) for some PDA M 2. Let M 1 = (Q, Σ, Γ,δ,q 0,,Z 0 ). Construct M 2 as follows: M 2 = (Q {q 0,q f }, Σ, Γ {X 0 },δ,q 0, {q f },X 0 ) δ (q 0,ε,ε) = {(q 0,Z 0 )}. For all q Q, a Σ ε and b Γ ε, δ (q,a,b) = δ(q,a,b). For all q Q, (q f,ε) δ (q,ε,x 0 ). From now on we will use the more standard definition of acceptance by final state. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 39

40 Notation Transition a,b c means if read a from input and pop b from stack then push c onto stack Meaning of ε transitions: if a = ε, don t read inputs if b = ε, don t pop any symbols if c = ε, don t push any symbols Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 40

41 Example The PDA accepts {0 n 1 n n 1}. Does it also accept the empty string? Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 41

42 Another Example A PDA that accepts { } a i b j c k $ i,j,k > 0 and i = j or i = k Informally: read and push a s either pop and match with b s or else ignore b s. Pop and match with c s a non-deterministic choice! Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 42

43 Another Example (cont.) This PDA accepts { } a i b j c k $ i,j,k > 0 and i = j or i = k Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 43

44 Non-determinism Adds Power to PDAs Recall that all regular languages are accepted by deterministic finite automata. For finite automata, non-determinism does not add power. But there are context free languages that are accepted only by non-deterministic push down automata. Examples: { a i b j c k i,j,k 0 and i = j or i = k } { ww R w {0, 1} } Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 44

45 PDA Languages The Push-Down Automata Languages, L PDA, is the set of all languages that can be described by some PDA: L PDA = {L : PDA M L[M] = L} We already know L PDA L DFA, since every DFA is just a PDA that ignores the stack. L CFG L PDA? L PDA L CFG? Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 45

46 CFL/PDA Equivalence Theorem Theorem: A language is context free if and only if some pushdown automaton accepts it. We will present a high level view of the proof (not all details). Full details in pp of Sipser s book. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 46

47 If Part Theorem: If a language is context free, then some pushdown automaton accepts it. Let A be a context-free language. By definition, A has a context-free grammar G generating it. On input w, the PDA P should figure out if there is a derivation of w using G. Question: How does P figure out which substitution to make? Answer: It guesses. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 47

48 CFL Implies PDA (cont.) Where do we keep the intermediate string? can t put it all on the stack only strings whose first letter is a variable are kept on stack Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 48

49 CFL Implies PDA Informally, on input string w Σ : P pushes start variable S on stack keeps making substitutions when popping a terminal, P checks equality with current input string rejects if not equal when popping a variable, P pushes to top of stack a right hand side of some rule corresponding to variable (zero, one, or two symbols). if EOI reached when stack is empty, accept. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 49

50 CFL Implies PDA (cont.) Informal description of P : Place the marker symbol $ and the start variable on stack Repeat the following steps: if top of stack is variable A, non-deterministically select rule and substitute. if top of stack is terminal a, read next input symbol and compare. If they differ, reject. if top of stack is $, enter accept state. (Namely accepts only if input has all been read and stack is empty!). Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 50

51 CFL Implies PDA (cont.) Need shorthand to push strings of length > 1 onto stack. For example, suppose S ABC is a derivation of the CFG. And suppose that we would like the PDA to go from state q to state r, on input a popping S and pushing the string ABC to stack: δ(q,a,s) = {(r,abc),...} This cannot be done in one move, so we need to add intermediate states q 1 and q 2, so that: δ(q,a,s) = {(q 1,C),..., } δ(q 1,ε,ε) = {(q 1,B)} δ(q 1,ε,ε) = {(r,a),..., } Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 51

52 CFL Implies PDA (cont.) States of P are start state q s accept state q a loop state q l q e states, needed for shorthand of right hand sides of rules Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 52

53 Transition Function Initialize stack δ(q s,ε,ε) = {q l,s$} Top of stack is variable (shorthand for a number of transitions) δ(q l,ε,a) = {(q l,w) where A w is a rule } Top of stack is terminal δ(q l,a,a) = {(q l,ε)} End of Stack δ(q l,ε, $) = {(q a,ε)} Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 53

54 Example S T atb b Ta ε Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 54

55 Only If Part Theorem: If a PDA accepts a language, L, then L is context-free. For each pair of states p and q in P, we will have a variable A pq in the grammar G. This variable, A pq, generates all strings that take P from p with an empty stack to q with empty stack. Same string also takes p with any stack to q with same stack! Start variable is A q0,q a (assuming a single accept state q a ). Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 55

56 PDA Implies CFL To make things easier, we slightly modify P Has single accept state q a. It empties stack before accepting. Each transition either pushes a symbol on stack, or pops a symbol from stack, but not both. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 56

57 PDA Implies CFL (2) Modify P to make things easier single accept state q a ε,ε ε,ε ε ε empties stack before accepting each transition pushes or pops, but not both. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 57

58 PDA Implies CFL (3) Modify P to make things easier single accept state q a empties stack before accepting ε,ε ε ε,ε ε ε,ε ε ε,a ε ε,ε ε ε,ε ε Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 58

59 PDA Implies CFL (4) Modify P to make things easier single accept state q a empties stack before accepting transition either pushes or pops, but not both a,b c a,b ε c ε,ε Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 59

60 Proof Idea Suppose string x takes P from p with empty stack to q with empty stack. First move that touches the stack must be a push, last must be a pop. In between, two possibilities: Stack is empty only at start and finish, but not in middle. Stack was also empty at some point in between. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 60

61 Proof Idea (2) Suppose string x takes P from p with empty stack to q with empty stack. First move that touches the stack must be a push, last must be a pop. In between, two possibilities: Stack is empty only at start and finish, but not in middle. Simulate by: A pq aa rs b, where a,b are first and last symbols in x (shorter x will be taken care of too), r follows p, and s precedes q. Stack was also empty at some point in between. Simulate by: A pq A pr A rq, r is intermediate state where P has empty stack. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 61

62 Details of Simulating Grammar Given PDA P = (Q, Σ, Γ,δ,q 0, {q a }), construct grammar G. Variables are {A pq p,q Q}. Start variable is A q0 q a. Rules: For p,q,r,s Q, t Γ, and a,b Σ, if (r,t) δ(p,a,ε) and (q,ε) δ(s,b,t), add rule A pq aa rs b. for every p,q,r Q, add rule A pq A pr A rq. for each p Q, add rule A pp ε. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 62

63 Overall Structure Should now prove Claim: A pq generates x if and only if x brings P from p with empty stack to q with empty stack. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 63

64 Only If Part Theorem: If a PDA accepts a language, L, then L is context-free. Proof: After constructing the grammar G, should prove it generates exactly the same language accepted by the PDA. This is done by induction on the length of any computation of P on any input string x. The induction argument is a bit lengthy and tedious, and we ll skip it. Diehards are welcome to consult pp in Sipser s book. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 64

65 Algorithmic Questions for NFAs Q.: Given an NFA, N, and a string s, is s L(N)? Answer: Construct the DFA equivalent to N and run it on w. Q.: Is L(N) =? Answer: This is a reachability question in graphs: Is there a path in the states graph of N from the start state to some accepting state. There are simple, efficient algorithms for this task. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 65

66 More Algorithmic Questions for NFAs Q.: Is L(N) = Σ? Answer: Check if L(N) =. Q.: Given N 1 and N 2, is L(N 1 ) L(N 2 )? Answer: Check if L(N 2 ) L(N 1 ) =. Q.: Given N 1 and N 2, is L(N 1 ) = L(N 2 )? Answer: Check if L(N 1 ) L(N 2 ) and L(N 2 ) L(N 1 ). Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 66

67 Algorithmic Questions Regarding DFAs Given a regular expression, R, find the smallest DFA (minimum number of states) that accepts L(R). Initial Idea: Use the algorithm described in class to transform R into an NFA. Then transform this NFA into a DFA, M. That s very nice, but how do we know M is minimal? It need not be! Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 67

68 Algorithmic Questions for DFAs (2) Given a regular expression, R, find the smallest DFA that accepts L(R) (minimum number of states). We can enumerate all DFAs that are strictly smaller than M. For each such M i, test if L(M i ) = L(M) (we saw an algorithm for this). Take the smallest such M i. Algorithm is very inefficient. More efficient algorithm is known, using the Myhill-Nerode theorem. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 68

69 Algorithmic Questions Regarding CFGs Given a CFG, G, and a string w, does G generate w? Initial Idea: Design an algorithm that tries all derivations. Problem: If G does not generate w, we ll never stop. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 69

70 Algorithmic Questions for CFGs (2) Recall: If G is in Chomsky normal form, w = n, and w is generated by G, then w has a derivation of length 2n 1 or less. Modified algorithm s idea: First, convert G to Chomsky normal form. Now need only consider a finite number of derivations those of length 2n 1 or less. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 70

71 Algorithmic Questions for CFGs (3) Theorem: There is an algorithm (that halts on every inputs) A, that on inputs G and w, decides if G generates w. On input G,w, where G is a grammar and w a string, 1. Convert G to Chomsky normal form. 2. List all derivations with 2n 1 steps, were n = w. 3. If any generates w, accept, otherwise reject. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 71

72 Emptiness of CFGs Given a CFG, G, is L(G) =? In other words, is there any string w, such that G generates w? Theorem: There is an algorithm that solves this problem (and always halts). Possible approaches for a proof: Bad Idea: We know how to test whether w L(G) for any string w, so just try it for each w. (criticize this...) Better Idea: Can the start variable generate a string of terminals? Even Better Idea: Can a particular variable generate a string of terminals? Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 72

73 CFG Emptiness (2) Algorithm: On input G (a CFG), 1. Mark all terminal symbols in G. 2. Repeat until no new variables become marked. 3. Mark any A where A U 1 U 2...U k and all U i have already been marked. 4. If start symbol marked, accept, else reject. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 73

74 CFGs Fullness Given a CFG, G, is L(G) = Σ? We just saw an algorithm to determine, given a CFG, G, if L(G) = L(G) = Σ iff L(G) =. Why not modify the algorithm so it determines emptiness of the complement? Unfortunately, CFGs are not closed under complement. Fact: There is no algorithm to solve this problem. We are not prepared to prove this remarkable fact (yet). Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 74

75 When Are Two CFGs equivalent? Given two CFGs, G,H, is L(G) = L(H)? Hey, we did this already for equivalence of DFAs! We constructed C from A and B: ( ) L(C) = L(A) L(B) and tested whether L(C) is empty. ( ) L(A) L(B). Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 75

76 When Are Two CFGs equivalent? This approach was fine for DFAs, but not for CFLs! The class of context-free languages is not closed under complementation or intersection. Fact: There is no algorithm to solve this problem. We are not prepared to prove this remarkable fact (yet). Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 76

77 A Short Summary Regular Languages Finite Automata. Context Free Languages Push Down Automata. Closure properties of regular languages and of CFLs. Most algorithmic problems for finite automata are solvable. Some algorithmic problems for finite automata are not solvable. Pumping lemmata for both classes of languages. There are additional languages out there. Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 77

78 The View Over The Horizon enumerable decidable context free regular Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. p. 78