1. 1 Module 34 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

2. 2 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(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

3. 3 Visualization CFL LPDA 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

4. 4 Algorithm Specification Input –CFG G Output –PDA M such that L(M) = CFG GPDA M A

5. 5 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

6. 6 Illustration Input Grammar G –V = {S} –  = {a,b} –S = S –P: S --> aSb | What is L(G)? 1. Derive all strings in L(G) on the stack 2. Match the derived string against input (q 0, aabb, Z) /* put S on stack */ (q 1, aabb, SZ) /* derive aabb on stack */ (q 1, aabb, aSbZ) (q 1, aabb, aaSbbZ) (q 1, aabb, aabbZ) /* match stack vs input */ (q 2, aabb, aabbZ) (q 2, abb, abbZ) (q 2, bb, bbZ) (q 2, b, bZ) (q 2,, Z) (q 3,, Z) Illustration of how the PDA might work, though not completely accurate.

7. 7 Difficulty 1. Derive all strings in L(G) on the stack 2. Match the derived string against input (q 0, aabb, Z) /* put S on stack */ (q 1, aabb, SZ) /* derive aabb on stack */ (q 1, aabb, aSbZ) (q 1, aabb, aaSbbZ) (q 1, aabb, aabbZ) /* match stack vs input */ (q 2, aabb, aabbZ) (q 2, abb, abbZ) (q 2, bb, bbZ) (q 2, b, bZ) (q 2,, Z) (q 3,, Z) What is illegal with the computation graph on the left?

8. 8 Construction Input Grammar –G=(V, , S, P) Output PDA –M=(Q, , , q 0, Z, F,  ) –Q = {q 0, q 1, q 2 } –  =  –  = V union  union {Z} –Z = Z –q 0 = q 0 –F = {q 2 }  : –  (q 0,, Z) = (q 1, SZ) –  (q 1,, Z) = (q 2, Z) –For all productions A -->   q 1,, A) = (q 1,  ) –For all a in   q 1, a, a) = (q 1, )

9. 9 Examples

10. 10 Palindromes PALG: –V = {S} –  = {a,b} –S = S –P: S --> aSa | bSb | a | b | Output PDA M=(Q, ,q 0,Z,F,  ) – Q = {q 0, q 1, q 2 } –  = {a,b,S,Z} – q 0 = q 0 – Z = Z – F = {q 2 }  : –  (q 0,, Z) = (q 1, SZ) –  (q 1,, Z) = (q 2, Z) –Production Transitions  q 1,, S) = (q 1, aSa)  q 1,, S) = (q 1, bSb)  q 1,, S) = (q 1, a)  q 1,, S) = (q 1, b)  q 1,, S) = (q 1, ) –Matching transitions  q 1, a, a) = (q 1, )  q 1, b, b) = (q 1, )

11. 11 Palindrome Transition Table Transition Current Input Top of Next Stack Number State Symbol Stack State Update --------------------------------------------------------------------------------- 1 q 0 Z q 1 SZ 2 q 1 Z q 2 Z 3 q 1 S q 1 aSa 4 q 1 S q 1 bSb 5 q 1 S q 1 a 6 q 1 S q 1 b 7 q 1 S q 1 8 q 1 a a q 1 9 q 1 b b q 1

12. 12 Partial Computation Graph (q 0, aba, Z) (q 1, aba, SZ) (q 1, aba, aSaZ) (other branches not shown) (q 1, ba, SaZ) (q 1, ba, baZ) (other branches not shown) (q 1, a, aZ) (q 1,, Z) (q 2,, Z) On your own, draw computation trees for other strings not in the language and see that they are not accepted.

13. 13 {a n b n | n >= 0} Grammar G: –V = {S} –  = {a,b} –S = S –P: S --> aSb | Output PDA M=(Q, ,q 0,Z,F,  ) – Q = {q 0, q 1, q 2 } –  = {a,b,S,Z} – q 0 = q 0 – Z = Z – F = {q 2 }  : –  (q 0,, Z) = (q 1, SZ) –  (q 1,, Z) = (q 2, Z) –Production Transitions –Matching transitions

14. 14 {a n b n | n >= 0} Transition Table Transition Current Input Top of Next Stack Number State Symbol Stack State Update --------------------------------------------------------------------------------- 1 q 0 Z 2 q 1 Z 3 q 1 S 4 q 1 S 5 q 1 a a 6 q 1 b b

15. 15 Partial Computation Graph (q 0, aabb, Z) (q 1, aabb, SZ) (q 1, aabb, aSbZ) (other branch not shown) (q 1, abb, SbZ) (q 1, abb, aSbbZ) (other branch not shown) (q 1, bb, SbbZ) (q 1, bb, bbZ) (other branch not shown) (q 1, b, bZ) (q 1,, Z) (q 2,, Z)

16. 16 {a i b j | 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, ,q 0,Z,F,  ) – Q = {q 0, q 1, q 2 } –  = {a,b,S,T,U,Z} – q 0 = q 0 – Z = Z – F = {q 2 }  –  (q 0,, Z) = (q 1, SZ) –  (q 1,, Z) = (q 2, Z) –Production Transitions –Matching transitions

17. 17 {a i b j | i = j or i = 2j} Transition Table Transition Current Input Top of Next Stack Number State Symbol Stack State Update --------------------------------------------------------------------------------- 1 q 0 Z q 1 SZ 2 q 1 Z q 2 Z 3 q 1 S q 1 T 4 q 1 S q 1 U 5 q 1 T q 1 aTb 6 q 1 T q 1 7 q 1 U q 1 aaUb 8 q 1 U q 1 9 q 1 a a q 1 10 q 1 b b q 1

18. 18 Partial Computation Graph (q 0, aab, Z) (q 1, aab, SZ) (q 1, aab, UZ) (other branch not shown) (q 1, aab, aaUbZ) (other branch not shown) (q 1, ab, aUbZ) (q 1, b, UbZ) (q 1, b, bZ) (other branch not shown) (q 1,, Z) (q 2,, Z)

19. 19 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 answer-preserving input transformations between decision problems about CFL’s.