60-354, Theory of Computation Fall Asish Mukhopadhyay School of Computer Science University of Windsor

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1 60-354, Theory of Computation Fall 2013 Asish Mukhopadhyay School of Computer Science University of Windsor

2 Pushdown Automata (PDA) PDA = ε-nfa + stack Acceptance ε-nfa enters a final state or Stack is empty Nondeterministic in nature More powerful than its deterministic counterpart

3 An example Design a PDA to accept the language L= {ww r w in {0,1}*} by empty stack Main idea underlying the construction Keep stacking symbols from the input string Guess that the middle has been reached Now start emptying the stack, as long as stacktop and input symbol matches

4 Formal Description of a PDA A PDA is a 6-tuple: (Q,, Λ, δ, q 0, Z 0 ) Q is the set of states of the ε-nfa is the alphabet of the input string Λ is the stack alphabet δ is the transition function: : Q 2 q 0 is the start state of the ε-nfa Z 0 is the bottom of stack marker * Q

5 PDA for the example language Q = {q 0, q 1 } = {0, 1} Λ = {0, 1,Z 0 }

6 PDA for the example language δ(): δ(q 0, a, b) = {(q 0, ab), (q 1,ab)}, a ε{0,1}, bε{0,1,z 0 } δ(q 0, ε, Z 0 ) = {(q 1, Z 0 )} // accept empty string δ(q 1, a, b) = {}, a, b ε {0,1}, a b δ(q 1, a, a) = {(q 1, ε)}, a ε {0,1} δ(q 1, ε, Z 0 ) = {(q 1, ε)} // empty stack

7 Instantaneous Description (ID) An ID is a 3-tuple (q, α, β) describing the state of a PDA q is the current state of the NFA α is the remaining input string β is the contents of the stack

8 Transition (q, α, β) - (q, α, β ) Useful for simulating moves of a PDA on an input string Example: (q 0, 1001, Z 0 ) - (q 0, 001, 1Z 0 ) - (q 1, 01, 01Z 0 ) - (q 1, 1, 1Z 0 ) - (q1, ε, Z 0 ) - (q 1, ε, ε)

9 Language accepted by a PDA L = {w (q 0, w, Z 0 ) -* (q, ε, ε), where q is some state in Q}

10 Context-free grammars An example: S S 0S 0 S 1S1 Productions generate the language L = {ww r w in {0,1}*}

11 Context-free grammars S is a variable, 0 and 1 are terminal symbols A string in L is derived by starting with the symbol S and making a sequence of substitutions, replacing a variable by the right hand side of a production

12 Formal definition A context-free grammar is described by a 4- tuple: (V, T, P, S) V is the set of variables of the grammar T is the set of terminals P is the set of productions S is a special symbol in V, called the start symbol Productions are of the form A -> α, where α is a string over {V U T}* and A is variable in V

13 Derivations Define a relation on {V U T}* thus Let α and β be arbitrary strings over {V U T}* and A -> γ be a production in P Then α A β α γ β Closure of is denoted by *

14 Example of a derivation S 1S1 10S

15 Language generated Let G be a context-free grammar L(G) = {w ε T* S * w}

16 Design a context-free grammar L = {w w ε {a,b}* and is not of the form zz} Claim: The grammar S -> AB BA C A -> aab baa aaa bab B -> abb bba aba bbb A -> a B -> b C -> acb bca aca bcb a b generates L

17 A derivation in G S AB aabb aaaabb aaaabbbb aaaabbbb

18 Another problem Design a context-free grammar for the language over {0,1} that consists of the set S of all strings with twice as many 0 s as 1 s

19 Claim The following grammar generates exactly the strings with this property S SS 10S0 1S 00 0S10 01S 0 00S1 0S01

20 Claim (2) L(G) is contained in S In any derivation every application of a production other than S -> ε introduces two 0 s and a single 1. Since the property is vacuously true for an empty string, the derived string retains this property whenever the production S -> ε is used A formal inductive argument on the number of steps in a derivation can be easily given

21 Claim (3) S is contained in L(G) Any of the sequences 001, 010, 100 can be treated as a balanced pair of parentheses A string with the above property has an adjacent pair of 00 s when the length is more than 3

22 Claim (3) Completing an inductive argument Assume inductively that every sequence of length 3n (n >1) corresponds to a balanced sequence of parentheses Consider a sequence of length 3n + 3 We can remove a sequence 001 or 100 from this sequence The residual sequence corresponds to a balanced sequence of parentheses and into this we can reinsert 001 or 100, each of which corresponds to a balanced parentheses pair

23 Example 20 Design a cfg that generates the language L= { 0 i 1 j 2i = 3j+1, j = 1, 3, 5, } Set j = 2k+1, k=0, 1, 2i = 6k+4 or i =3k+2 The strings are of the form: 0 3k+2 1 2k+1, which can be written as 0 2 (0 3 ) k (1 2 ) k 1

24 Example 20 Grammar: S -> 00B1 ; B -> 000B11 ε

25 Canonical derivations In a derivation, productions can be applied in an arbitrary order In a leftmost (rightmost) derivation, we always replace the leftmost (rightmost) variable by its body in a production

26 Parse Tree Any derivation can be represented by a parse tree

27 Ambiguous grammar A cfg G is ambiguous if there exists more than one parse tree for a string in L(G) In terms of canonical derivations: more than one leftmost or rightmost derivation

28 Decision algorithm Is it decidable if a cfg G is ambiguous? We need considerable infrastructure to answer this question

29 An ambiguous grammar G The productions in G are: E -> E + E E * E (E) I I -> Ia Ib I0 I1 a b 0 1 In G: a + b* a has two leftmost derivations (see courseware)

30 Disambiguating the grammar G The precedence of the operators * and + needs to be defined The new productions that take care of this are: E -> E + T T T -> T * F F F -> (E) I I -> I0 I1 Ia Ib a b 0 1

31 Normal forms Chomsky Normal Form (CNF, for short) All productions are of the form: A -> BC or A -> a Neither B nor C can be the start symbol, S

32 CFG to PDA Let G = (V, T, P, S) be a cfg P = ({q}, T, V U T, δ, q, S) is a PDA that accepts L(G) by empty stack for δ() defined thus: For each variable A in V, δ(q, ε, A ) = { (q, β) A -> β is a production in G} For each terminal symbol a, δ(q, a, a) = {(q, ε)}

33 PDA to CFG (1) Given PDA, A = (Q,, Λ, δ, q 0, Z 0 ) A CFG, G = (V, T, P, S) is constructed thus: V = S U {[pxq] p, q ε Q, X ε Λ} The set of productions P includes S -> [q 0 Z 0 p] for every p ε Q Further, if (r, Y 1 Y 2..Y k ) ε δ(q, a, X), where a ε or a = ε and k 0, then P includes the production [qxr k ] -> a[ry 1 r 1 ][r 1 Y 2 r 2 ] [r k-1 Y k r k ]

34 PDA to CFG (2) When k = 0, the production is [qxr] -> a See Example 22 in courseware

35 Deterministic PDA (DPDA) (1) For each q, a, X, δ(q, a, X) is of size at most 1 When δ(q, a, X) is not empty, δ(q, ε, X) is empty

36 Deterministic PDA (DPDA) (2) Acceptance by empty stack and final state are not equivalent Equivalent under an additional condition is satisfied

37 Prefix language and DPDA L is a prefix language if for a pair of strings x and y in L, neither x nor y is a prefix of the other Theorem A DPDA accepts L by empty stack iff L is a prefix language that is accepted by some DPDA by final state 1

38 DPDA language Languages accepted by DPDAs by final state are called DPDA languages Example of a DPDA language: L wcw r = {wcw r w in {0,1}*}

39 DPDA and CFLs (1) DPDA languages lie strictly between regular languages and context-free languages Given a regular language L, we can construct a DPDA that simulates the action of a DFA that accepts L simply by ignoring the stack Since the language L wcw r is not regular, the inclusion is strict (c is some fixed symbol in )

40 DPDA and CFLs (2) DPDA languages are strictly included in the class of context free languages Example: L ww r (note that this is not a prefix language)

41 DPDA languages and ambiguity A language accepted by a DPDA (by final state or empty stack) has an unambiguous grammar However, not every language that has an unambiguous grammar is accepted by a DPDA Example: L ww r