Lecture 35

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Lecture 35

• CFG --> PDA construction– Shows that for any CFL L, there exists a PDA

M such that L(M) = L– The reverse is true as well, but we do not prove

that here

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CFL subset LPDA

• Let L be an arbitrary CFL

• Let G be the CFG such that L(G) = L– G exists by definition of L is CF

• Construct a PDA M such that L(M) = L– M is constructed from CFG G

• Argue L(M) = L

• There exists a PDA M such that L(M) = L

• L is in LPDA– By definition of L in LPDA

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Visualization

CFLLPDA

CFG’s PDA’s

L L

G M

•Let L be an arbitrary CFL•Let G be the CFG such that L(G) = L

•G exists by definition of L is CF•Construct a PDA M such that L(M) = L

•M is constructed from CFG G•Argue L(M) = L•There exists a PDA M such that L(M) = L•L is in LPDA

•By definition of L in LPDA

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Algorithm Specification

• Input– CFG G

• Output– PDA M such that L(M) = L(G)

CFG G PDA MA

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Construction Idea

• The basic idea is to have a 2-phase PDA– Phase 1:

• Derive all strings in L(G) on the stack nondeterministically

• Do not process any input while we are deriving the string on the stack

– Phase 2: • Match the input string against the derived string on the stack

– This is a deterministic process

• Move to an accepting state only when the stack is empty

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Illustration• Input Grammar G

– V = {S}– = {a,b}– S = S– P:

S --> aSb |

• What is L(G)?– {anbn | n >= 0}

1. Derive all strings in L(G) on the stack2. Match the derived string against input

(q0, aabb, Z)/* put S on stack */

(q1, aabb, SZ)/* derive aabb on stack */

(q1, aabb, aSbZ)(q1, aabb, aaSbbZ)(q1, aabb, aabbZ)

/* match stack vs input */(q2, aabb, aabbZ)(q2, abb, abbZ)(q2, bb, bbZ)(q2, b, bZ)(q2,, Z)

(q3, , Z)

Illustration of how the PDA might work,though not completely accurate.

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Difficulty 1. Derive all strings in L(G) on the stack2. Match the derived string against input

(q0, aabb, Z)/* put S on stack */

(q1, aabb, SZ)/* derive aabb on stack */

(q1, aabb, aSbZ)(q1, aabb, aaSbbZ)(q1, aabb, aabbZ)

/* match stack vs input */(q2, aabb, aabbZ)(q2, abb, abbZ)(q2, bb, bbZ)(q2, b, bZ)(q2,, Z)

(q3, , Z)

What is wrong with the computation graph on the left?

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Construction• Input Grammar

– G=(V,, S, P)

• Output PDA – M=(Q, , , q0, Z, F, )

– Q = {q0, q1, q2}

– = – = V union union {Z}

– Z = Z

– q0 = q0

– F = {q2}

• :– (q0, , Z) = (q1, SZ)

– (q1, , Z) = (q2, Z)

– For all productions A --> • q1, , A) = (q1, )

– For all a in • q1, a, a) = (q1, )

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Examples

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Palindromes• PALG:

– V = {S}

– = {a,b}

– S = S

– P:• S --> aSa | bSb | a | b |

• Output PDA M=(Q,,q0,Z,F,)

– Q = {q0, q1, q2}

– = {a,b,S,Z}

– q0 = q0

– Z = Z

– F = {q2}

• :

– (q0, , Z) = (q1, SZ)

– (q1, , Z) = (q2, Z)

– Production Transitions• q1, , S) = (q1, aSa)

• q1, , S) = (q1, bSb)

• q1, , S) = (q1, a)

• q1, , S) = (q1, b)

• q1, , S) = (q1, )

– Matching transitions• q1, a, a) = (q1, )

• q1, b, b) = (q1, )

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Palindrome Transition Table

Transition Current Input Top of Next StackNumber State Symbol Stack State Update--------------------------------------------------------------------------------- 1 q0 Z q1 SZ 2 q1 Z q2 Z 3 q1 S q1 aSa 4 q1 S q1 bSb 5 q1 S q1 a 6 q1 S q1 b 7 q1 S q1 8 q1 a a q1 9 q1 b b q1

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Partial Computation Graph(q0, aba, Z) (q1, aba, SZ) (q1, aba, aSaZ) (other branches not shown)(q1, ba, SaZ)(q1, ba, baZ) (other branches not shown)(q1, a, aZ)(q1, , Z)(q2, , Z)

On your own, draw computation trees for other stringsnot in the language and see that they are not accepted.

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{anbn | n >= 0}• Grammar G:

– V = {S}

– = {a,b}

– S = S

– P:• S --> aSb |

• Output PDA M=(Q,,q0,Z,F,)

– Q = {q0, q1, q2}

– = {a,b,S,Z}

– q0 = q0

– Z = Z

– F = {q2}

• :

– (q0, , Z) = (q1, SZ)

– (q1, , Z) = (q2, Z)

– Production Transitions• q1, , S) = (q1, aSb)

• q1, , S) = (q1, )

– Matching transitions• q1, a, a) = (q1, )

• q1, b, b) = (q1, )

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{anbn | n >= 0} Transition Table

Transition Current Input Top of Next StackNumber State Symbol Stack State Update--------------------------------------------------------------------------------- 1 q0 Z q1 SZ 2 q1 Z q2 Z 3 q1 S q1 aSb 4 q1 S q1 5 q1 a a q1 6 q1 b b q1

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Partial Computation Graph

(q0, aabb, Z) (q1, aabb, SZ) (q1, aabb, aSbZ) (other branch not shown)(q1, abb, SbZ)(q1, abb, aSbbZ) (other branch not shown)(q1, bb, SbbZ)(q1, bb, bbZ) (other branch not shown)(q1, b, bZ)(q1, , Z) (q2, , Z)

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{aibj | i = j or i = 2j}• Grammar G:

– V = {S,T,U}

– = {a,b}

– S = S

– P:• S --> T | U

• T --> aTb | • U --> aaUb |

• Output PDA M=(Q,,q0,Z,F,)

– Q = {q0, q1, q2}

– = {a,b,S,T,U,Z}

– q0 = q0

– Z = Z

– F = {q2}

• – (q0, , Z) = (q1, SZ)

– (q1, , Z) = (q2, Z)

– Production Transitions• q1, , S) = (q1, T)

• q1, , S) = (q1, U)

• q1, , T) = (q1, aTb)

• q1, , T) = (q1, )

• q1, , U) = (q1, aaUb)

• q1, , U) = (q1, )

– Matching transitions• q1, a, a) = (q1, )

• q1, b, b) = (q1, )

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{aibj | i = j or i = 2j} Transition Table

Transition Current Input Top of Next StackNumber State Symbol Stack State Update--------------------------------------------------------------------------------- 1 q0 Z q1 SZ 2 q1 Z q2 Z 3 q1 S q1 T 4 q1 S q1 U 5 q1 T q1 aTb 6 q1 T q1 7 q1 U q1 aaUb 8 q1 U q1 9 q1 a a q1 10 q1 b b q1

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Partial Computation Graph

(q0, aab, Z) (q1, aab, SZ) (q1, aab, UZ) (other branch not shown)(q1, aab, aaUbZ) (other branch not shown)(q1, ab, aUbZ)(q1, b, UbZ)(q1, b, bZ) (other branch not shown)(q1, , Z) (q2, , Z)

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Things you should be able to do

• You should be able to execute this algorithm– Given any CFG, construct an equivalent PDA

• You should understand the idea behind this algorithm– Derive string on stack and then match it against input

• You should understand how this construction can help you design PDA’s

• You should understand that it can be used in many-one reductions between decision problems about CFL’s.