Presentation is loading. Please wait.

Presentation is loading. Please wait.

Www.nskinfo.comwww.nskinfo.com && www.nsksofttch.com Department of nskinfo-i educationwww.nsksofttch.com CS2303-THEORY OF COMPUTATION Chapter: Push Down.

Similar presentations


Presentation on theme: "Www.nskinfo.comwww.nskinfo.com && www.nsksofttch.com Department of nskinfo-i educationwww.nsksofttch.com CS2303-THEORY OF COMPUTATION Chapter: Push Down."— Presentation transcript:

1 && Department of nskinfo-i educationwww.nsksofttch.com CS2303-THEORY OF COMPUTATION Chapter: Push Down Automata (PDA) 1

2 PDA - the automata for CFLs What is? FA to Reg Lang, PDA is to CFL PDA == [ -NFA + a stack ] Why a stack? 2 -NFA A stack filled with stack symbols Input string Accept/reject 10 January 2014

3 Pushdown Automata - Definition A PDA P := ( Q,,, δ,q 0,Z 0,F ): Q: states of the -NFA : input alphabet :stack symbols δ:transition function q 0 :start state Z 0 :Initial stack top symbol F:Final/accepting states 310 January 2014

4 δ : The Transition Function δ(q,a,X) = {(p,Y), …} 1. state transition from q to p 2. a is the next input symbol 3. X is the current stack top symbol 4. Y is the replacement for X; it is in * (a string of stack symbols) i. Set Y = for:Pop(X) ii. If Y=X: stack top is unchanged iii. If Y=Z 1 Z 2 …Z k : X is popped and is replaced by Y in reverse order (i.e., Z 1 will be the new stack top) 4 Non-determinism old state Stack top input symb. new state(s) new Stack top(s) δ : Q x x => Q x q a X p Y Y = ?Action i) Y= Pop(X) ii)Y=XPop(X) Push(X) iii)Y=Z 1 Z 2..Z k Pop(X) Push(Z k ) Push(Z k-1 ) … Push(Z 2 ) Push(Z 1 ) 10 January 2014

5 Example Let L wwr = {ww R | w is in (0+1)*} CFG for L wwr : S==> 0S0 | 1S1 | PDA for L wwr : P := ( Q,,, δ,q 0,Z 0,F ) = ( {q 0, q 1, q 2 },{0,1},{0,1,Z 0 },δ,q 0,Z 0,{q 2 }) 510 January 2014

6 PDA for L wwr 1. δ(q 0,0, Z 0 )={(q 0,0Z 0 )} 2. δ(q 0,1, Z 0 )={(q 0,1Z 0 )} 3. δ(q 0,0, 0)={(q 0,00)} 4. δ(q 0,0, 1)={(q 0,01)} 5. δ(q 0,1, 0)={(q 0,10)} 6. δ(q 0,1, 1)={(q 0,11)} 7. δ(q 0,, 0)={(q 1, 0)} 8. δ(q 0,, 1)={(q 1, 1)} 9. δ(q 0,, Z 0 )={(q 1, Z 0 )} 10. δ(q 1,0, 0)={(q 1, )} 11. δ(q 1,1, 1)={(q 1, )} 12. δ(q 1,, Z 0 )={(q 2, Z 0 )} 6 First symbol push on stack Grow the stack by pushing new symbols on top of old (w-part) Switch to popping mode (boundary between w and w R ) Shrink the stack by popping matching symbols (w R -part) Enter acceptance state Z0Z0 Initial state of the PDA: q0q0 Stack top 10 January 2014

7 PDA as a state diagram 7 qiqi qjqj a, X / Y Current state Next input symbol Current state Current stack top Stack Top Replacement (w/ string Y) Next state δ(q i,a, X)={(q j,Y)} 10 January 2014

8 PDA for L wwr : Transition Diagram 8 q0q0 q1q1 q2q2 0, Z 0 /0Z 0 1, Z 0 /1Z 0 0, 0/00 0, 1/01 1, 0/10 1, 1/11 0, 0/ 1, 1/, Z 0 /Z 0, 0/0, 1/1, Z 0 /Z 0 Grow stack Switch to popping mode Pop stack for matching symbols Go to acceptance = {0, 1} = {Z 0, 0, 1} Q = {q 0,q 1,q 2 }, Z 0 /Z 0 This would be a non-deterministic PDA 10 January 2014

9 Example 2: language of balanced paranthesis 9 q0q0 q1q1 q2q2 (, Z 0 / ( Z 0, Z 0 / Z 0 Grow stack Switch to popping mode Pop stack for matching symbols Go to acceptance (by final state) when you see the stack bottom symbol = { (, ) } = {Z 0, ( } Q = {q 0,q 1,q 2 } (, ( / ( ( ), ( / To allow adjacent blocks of nested paranthesis (, ( / ( ( (, Z 0 / ( Z 0, Z 0 / Z 0 10 January 2014

10 Example 2: language of balanced paranthesis (another design) 10 = { (, ) } = {Z 0, ( } Q = {q 0,q 1 } q0q0 (,Z 0 / ( Z 0 (,( / ( ( ), ( / start q1q1,Z 0 / Z 0 10 January 2014

11 PDAs Instantaneous Description (ID) A PDA has a configuration at any given instance: (q,w,y) q - current state w - remainder of the input (i.e., unconsumed part) y - current stack contents as a string from top to bottom of stack If δ(q,a, X)={(p, A)} is a transition, then the following are also true: (q, a, X ) |--- (p,,A) (q, aw, XB ) |--- (p,w,AB) |--- sign is called a turnstile notation and represents one move |---* sign represents a sequence of moves 1110 January 2014

12 How does the PDA for L wwr work on input 1111? 12 (q 0,1111,Z 0 ) (q 0,111,1Z 0 ) (q 0,11,11Z 0 ) (q 0,1,111Z 0 ) (q 0,,1111Z 0 ) (q 1,,1111Z 0 ) (q 1,,11Z 0 ) (q 1,1,111Z 0 ) (q 1,11,11Z 0 ) (q 1,111,1Z 0 ) (q 1,1111,Z 0 ) Path dies… (q 1,1,1Z 0 ) (q 1,,Z 0 ) (q 2,,Z 0 ) Acceptance by final state: = empty input AND final state All moves made by the non-deterministic PDA Path dies… 10 January 2014

13 Principles about IDs Theorem 1: If for a PDA, (q, x, A) |---* (p, y, B), then for any string w * and *, it is also true that: (q, x w, A ) |---* (p, y w, B ) Theorem 2: If for a PDA, (q, x w, A) |---* (p, y w, B), then it is also true that: (q, x, A) |---* (p, y, B) 1310 January 2014

14 Acceptance by… PDAs that accept by final state: For a PDA P, the language accepted by P, denoted by L(P) by final state, is: {w | (q 0,w,Z 0 ) |---* (q,, A) }, s.t., q F PDAs that accept by empty stack: For a PDA P, the language accepted by P, denoted by N(P) by empty stack, is: {w | (q 0,w,Z 0 ) |---* (q,, ) }, for any q Q. 14 Checklist: - input exhausted? - in a final state? Checklist: - input exhausted? - is the stack empty? There are two types of PDAs that one can design: those that accept by final state or by empty stack Q) Does a PDA that accepts by empty stack need any final state specified in the design? 10 January 2014

15 Example: L of balanced parenthesis 15 q0q0 (,Z 0 / ( Z 0 (,( / ( ( ), ( / start q1q1,Z 0 / Z 0 PDA that accepts by final state q0q0 start (,Z 0 / ( Z 0 (, ( / ( ( ), ( /,Z 0 / An equivalent PDA that accepts by empty stack,Z 0 / Z 0 PF:PF: PN:PN: How will these two PDAs work on the input: ( ( ( ) ) ( ) ) ( ) 10 January 2014

16 PDA for L wwr : Proof of correctness Theorem: The PDA for L wwr accepts a string x by final state if and only if x is of the form ww R. Proof: (if-part) If the string is of the form ww R then there exists a sequence of IDs that leads to a final state: (q 0,ww R,Z 0 ) |---* (q 0,w R,wZ 0 ) |---* (q 1,w R,wZ 0 ) |--- * (q 1,,Z 0 ) |---* (q 2,,Z 0 ) (only-if part) Proof by induction on |x| 1610 January 2014

17 PDAs accepting by final state and empty stack are equivalent P F <= PDA accepting by final state P F = (Q F,,, δ F,q 0,Z 0,F) P N <= PDA accepting by empty stack P N = (Q N,,, δ N,q 0,Z 0 ) Theorem: (P N ==> P F ) For every P N, there exists a P F s.t. L(P F )=L(P N ) (P F ==> P N ) For every P F, there exists a P N s.t. L(P F )=L(P N ) 1710 January 2014

18 P N ==> P F construction Whenever P N s stack becomes empty, make P F go to a final state without consuming any addition symbol To detect empty stack in P N : P F pushes a new stack symbol X 0 (not in of P N ) initially before simultating P N 18 q0q0 … pfpf p0p0, X 0 /Z 0 X 0 New start, X 0 / X 0 PNPN PF:PF: P F = (Q N U {p 0,p f },, U {X 0 }, δ F, p 0, X 0, {p f }) PN:PN:, X 0 / X 0 How to convert an empty stack PDA into a final state PDA? 10 January 2014

19 Example: Matching parenthesis ( ) 19 P N :( {q 0 }, {(,)}, {Z 0,Z 1 }, δ N, q 0, Z 0 ) δ N :δ N (q 0,(,Z 0 ) = { (q 0,Z 1 Z 0 ) } δ N (q 0,(,Z 1 ) = { (q 0, Z 1 Z 1 ) } δ N (q 0,),Z 1 ) = { (q 0, ) } δ N (q 0,,Z 0 ) = { (q 0, ) } q0q0 start (,Z 0 /Z 1 Z 0 (,Z 1 /Z 1 Z 1 ),Z 1 /,Z 0 / q0q0 (,Z 0 /Z 1 Z 0 (,Z 1 /Z 1 Z 1 ),Z 1 /,Z 0 / start p0p0 pfpf,X 0 /Z 0 X 0,X 0 / X 0 P f :( {p 0,q 0,p f }, {(,)}, {X 0,Z 0,Z 1 }, δ f, p 0, X 0, p f ) δ f :δ f (p 0,,X 0 ) = { (q 0,Z 0 ) } δ f (q 0,(,Z 0 ) = { (q 0,Z 1 Z 0 ) } δ f (q 0,(,Z 1 ) = { (q 0, Z 1 Z 1 ) } δ f (q 0,),Z 1 ) = { (q 0, ) } δ f (q 0,,Z 0 ) = { (q 0, ) } δ f (p 0,,X 0 ) = { (p f, X 0 ) } Accept by empty stackAccept by final state 10 January 2014

20 P F ==> P N construction Main idea: Whenever P F reaches a final state, just make an -transition into a new end state, clear out the stack and accept Danger: What if P F design is such that it clears the stack midway without entering a final state? to address this, add a new start symbol X 0 (not in of P F ) P N = (Q U {p 0,p e },, U {X 0 }, δ N, p 0, X 0 ) 20 p0p0, X 0 /Z 0 X 0 New start, any/ q0q0 … pepe PFPF PN:PN: How to convert an final state PDA into an empty stack PDA? 10 January 2014

21 Equivalence of PDAs and CFGs 2110 January 2014

22 CFGs == PDAs ==> CFLs 22 CFG PDA by final state PDA by empty stack ? 10 January 2014

23 Converting CFG to PDA Main idea: The PDA simulates the leftmost derivation on a given w, and upon consuming it fully it either arrives at acceptance (by empty stack) or non-acceptance. 23 This is same as: implementing a CFG using a PDA PDA (acceptance by empty stack) CFG w accept reject implements INPUT OUTPUT 10 January 2014

24 Converting a CFG into a PDA Main idea: The PDA simulates the leftmost derivation on a given w, and upon consuming it fully it either arrives at acceptance (by empty stack) or non-acceptance. Steps: 1. Push the right hand side of the production onto the stack, with leftmost symbol at the stack top 2. If stack top is the leftmost variable, then replace it by all its productions (each possible substitution will represent a distinct path taken by the non-deterministic PDA) 3. If stack top has a terminal symbol, and if it matches with the next symbol in the input string, then pop it State is inconsequential (only one state is needed) 24 This is same as: implementing a CFG using a PDA 10 January 2014

25 Formal construction of PDA from CFG Given: G= (V,T,P,S) Output: P N = ({q}, T, V U T, δ, q, S) δ:δ: For all A V, add the following transition(s) in the PDA: δ(q,,A) = { (q, ) | A ==> P} For all a T, add the following transition(s) in the PDA: δ(q,a,a)= { (q, ) } 25 A Before: … a … After: … a … Note: Initial stack symbol (S) same as the start variable in the grammar pop a… 10 January 2014

26 Example: CFG to PDA G = ( {S,A}, {0,1}, P, S) P: S ==> AS | A ==> 0A1 | A1 | 01 PDA = ({q}, {0,1}, {0,1,A,S}, δ, q, S) δ: δ(q,, S) = { (q, AS), (q, )} δ(q,, A) = { (q,0A1), (q,A1), (q,01) } δ(q, 0, 0) = { (q, ) } δ(q, 1, 1) = { (q, ) } 26 How will this new PDA work? Lets simulate string 0011 q,S / S 1,1 / 0,0 /,A / 01,A / A1,A / 0A1,S /,S / AS 10 January 2014

27 Simulating string 0011 on the new PDA … 27 PDA (δ): δ(q,, S) = { (q, AS), (q, )} δ(q,, A) = { (q,0A1), (q,A1), (q,01) } δ(q, 0, 0) = { (q, ) } δ(q, 1, 1) = { (q, ) } S Stack moves (shows only the successful path): S A S 1 A 0 S 1 A 0 S S S 1 1 S 1 Accept by empty stack q,S / S 1,1 / 0,0 /,A / 01,A / A1,A / 0A1,S /,S / AS S => AS => 0A1S => 0011S => 0011 Leftmost deriv.: S =>AS =>0A1S =>0011S => January 2014

28 Proof of correctness for CFG ==> PDA construction Claim: A string is accepted by G iff it is accepted (by empty stack) by the PDA Proof: (only-if part) Prove by induction on the number of derivation steps (if part) If (q, wx, S) |-- * (q,x,B) then S => * lm wB 2810 January 2014

29 Converting a PDA into a CFG Main idea: Reverse engineer the productions from transitions If δ(q,a,Z) => (p, Y 1 Y 2 Y 3 …Y k ): 1. State is changed from q to p; 2. Terminal a is consumed; 3. Stack top symbol Z is popped and replaced with a sequence of k variables. Action: Create a grammar variable called [qZp] which includes the following production: [qZp] => a[pY 1 q 1 ] [q 1 Y 2 q 2 ] [q 2 Y 3 q 3 ]… [q k-1 Y k q k ] Proof discussion (in the book) 2910 January 2014

30 Example: Bracket matching To avoid confusion, we will use b=( and e=) 30 P N :( {q 0 }, {b,e}, {Z 0,Z 1 }, δ, q 0, Z 0 ) 1.δ(q 0,b,Z 0 ) = { (q 0,Z 1 Z 0 ) } 2.δ(q 0,b,Z 1 ) = { (q 0,Z 1 Z 1 ) } 3.δ(q 0,e,Z 1 ) = { (q 0, ) } 4.δ(q 0,,Z 0 ) = { (q 0, ) } 0.S => [q 0 Z 0 q 0 ] 1.[q 0 Z 0 q 0 ] => b [q 0 Z 1 q 0 ] [q 0 Z 0 q 0 ] 2.[q 0 Z 1 q 0 ] => b [q 0 Z 1 q 0 ] [q 0 Z 1 q 0 ] 3.[q 0 Z 1 q 0 ] => e 4.[q 0 Z 0 q 0 ] => Let A=[q 0 Z 0 q 0 ] Let B=[q 0 Z 1 q 0 ] 0.S => A 1.A => b B A 2.B => b B B 3.B => e 4.A => Simplifying, 0.S => b B S | 1.B => b B B | e If you were to directly write a CFG: S => b S e S | 10 January 2014

31 Two ways to build a CFG 31 Build a PDA Construct CFG from PDA Derive CFG directly Derive a CFG Construct PDA from CFG Design a PDA directly Similarly… (indirect) (direct) (indirect) (direct) Two ways to build a PDA 10 January 2014

32 Deterministic PDAs 3210 January 2014

33 This PDA for L wwr is non-deterministic 33 q0q0 q1q1 q2q2 0, Z 0 /0Z 0 1, Z 0 /1Z 0 0, 0/00 0, 1/01 1, 0/10 1, 1/11 0, 0/ 1, 1/, Z 0 /Z 0, 0/0, 1/1, Z 0 /Z 0 Grow stack Switch to popping mode Pop stack for matching symbols Accepts by final state Why does it have to be non- deterministic? To remove guessing, impose the user to insert c in the middle 10 January 2014

34 D-PDA for L wcwr = {wcw R | c is some special symbol not in w} 34 q0q0 q1q1 q2q2 0, Z 0 /0Z 0 1, Z 0 /1Z 0 0, 0/00 0, 1/01 1, 0/10 1, 1/11 0, 0/ 1, 1/ c, Z 0 /Z 0 c, 0/0 c, 1/1, Z 0 /Z 0 Grow stack Switch to popping mode Pop stack for matching symbols Accepts by final state Note: all transitions have become deterministic Note: all transitions have become deterministic Example shows that: Nondeterministic PDAs D-PDAs 10 January 2014

35 Deterministic PDA: Definition A PDA is deterministic if and only if: 1. δ(q,a,X) has at most one member for any a U { } If δ(q,a,X) is non-empty for some a, then δ(q,,X) must be empty January 2014

36 PDA vs DPDA vs Regular languages 36 Regular languages D-PDA non-deterministic PDA L wwr L wcwr 10 January 2014

37 Summary PDAs for CFLs and CFGs Non-deterministic Deterministic PDA acceptance types 1. By final state 2. By empty stack PDA IDs, Transition diagram Equivalence of CFG and PDA CFG => PDA construction PDA => CFG construction 3710 January 2014


Download ppt "Www.nskinfo.comwww.nskinfo.com && www.nsksofttch.com Department of nskinfo-i educationwww.nsksofttch.com CS2303-THEORY OF COMPUTATION Chapter: Push Down."

Similar presentations


Ads by Google