1. Lecture 20 DPDA / Review Topics Given a PDA construct a grammar for the language it accepts Deterministic PDAs Properties of CFLsReadings: November 5, 2008 CSCE 355 Foundations

2. – 2 – CSCE 355 Fall 2008 Last Time: Top-down parsing Algorithms Left-factoring From Grammars to PDA Section 6.3 Formal PDA Construction from CFLNew: When will our parsing algorithm be deterministic?Homework 1. …

3. – 3 – CSCE 355 Fall 2008 When will our parsing algorithm be deterministic? First(α ) = { x in T such that α  * x β } Then if A  α and A  β are two productions we are choosing from and the current input character is a which is in First(α) but not in First( β ) Then we choose the production A  α. So if First(α) ∩ First( β ) = ϕ then we can always deterministically choose There is a type of grammar that is LL(1) for which this works well. A grammar with no ε-productions is LL(1) if it satisfies the disjointness property above.

4. – 4 – CSCE 355 Fall 2008 Homework from Last time 1.Given the grammar simple4.y (and simple4.l)  Modify the semantic actions so that it produces a trace of productions involved in the parse.  If you look at the sequence do you notice anything? 2.6.2.1b) Strings of 0’s and 1’s such that no prefix has more 1’s than 0’s 3.6.2.5b,c

5. – 5 – CSCE 355 Fall 2008 %token INTEGER PLUS TIMES LPAREN RPAREN % line : expr '\n' {printf("Using Prod line -> expr \\n \n");} | line expr '\n' {printf("Using Prod line -> line expr \\n \n");} | line expr '\n' {printf("Using Prod line -> line expr \\n \n");} ; expr : expr PLUS term {printf("Using Prod expr -> expr PLUS term \n");} | term {printf("Using Prod expr -> term \n");} | term {printf("Using Prod expr -> term \n");} ; term : term TIMES factor {printf("Using Prod term -> term TIMES factor\n");} | factor {printf("Using Prod term -> factor \n");} | factor {printf("Using Prod term -> factor \n");} ; factor : LPAREN expr RPAREN {printf("Using Prod factor -> LP expr RP \n");} | INTEGER {printf("Using Prod factor -> INTEGER \n");} | INTEGER {printf("Using Prod factor -> INTEGER \n");} ;%

6. – 6 – CSCE 355 Fall 2008 cerberus>./simpleHW 2+3*7 Using Prod factor -> INTEGER Using Prod term -> factor Using Prod expr -> term Using Prod factor -> INTEGER Using Prod term -> factor Using Prod factor -> INTEGER Using Prod term -> term TIMES factor Using Prod expr -> expr PLUS term Using Prod line -> expr \n

7. – 7 – CSCE 355 Fall 2008 6.2.1b) no prefix has more 1’s than 0’s 6.2.1b) Strings of 0’s and 1’s such that no prefix has more 1’s than 0’s qdead 0,Z 0 /0Z 0 0,0/00 1,0/ε 1,Z 0 /Z 0

8. – 8 – CSCE 355 Fall 2008 6.2.5b St ΣΓ NextSt Γ*Γ* indx q0q0 aZ0Z0 q1q1 AAZ 0 1 q0q0 bZ0Z0 q2q2 BZ 0 2 q0q0 εZ0Z0 fε3 q1q1 aAq1q1 AAA4 q1q1 bAq1q1 ε5 q1q1 εZ0Z0 q0q0 Z0Z0 6 q2q2 aBq3q3 ε7 q2q2 bBq2q2 BB8 q2q2 εZ0Z0 q0q0 Z0Z0 9 q3q3 εBq2q2 ε10 q3q3 εZ0Z0 q1q1 AZ 0 11 (q 0, abb, Z 0 ) ├ (rule 1) (q 1, bb, AAZ 0 ) ├ (rule 5) (q 1, b, AZ 0 ) ├ (rule 5) (q 1, ε, Z 0 ) ├ (rule 6) (q 0, ε, Z 0 ) ├ (rule 3) (f, ε, ε)

9. – 9 – CSCE 355 Fall 2008 6.2.5c St ΣΓ NextSt Γ*Γ* indx q0q0 aZ0Z0 q1q1 AAZ 0 1 q0q0 bZ0Z0 q2q2 BZ 0 2 q0q0 εZ0Z0 fε3 q1q1 aAq1q1 AAA4 q1q1 bAq1q1 ε5 q1q1 εZ0Z0 q0q0 Z0Z0 6 q2q2 aBq3q3 ε7 q2q2 bBq2q2 BB8 q2q2 εZ0Z0 q0q0 Z0Z0 9 q3q3 εBq2q2 ε10 q3q3 εZ0Z0 q1q1 AZ 0 11 (q 0, b 7 a 4, Z 0 ) ├ (rule 2) (q 2, b 6 a 4, BZ 0 ) ├ (rule 8) (q 2, b 5 a 4, BBZ 0 ) ├ (rule 8) … (q 2, ba 4, BBBBBBZ 0 ) ├ (r8) (q 2, a 4, BBBBBBBZ 0 ) ├ (r7) (q 3, a 3, BBBBBBZ 0 ) ├ (rule10) (q 2, a 3, BBBBBZ 0 ) ├ (rule7) (q 3, a 2, BBBBZ 0 ) ├ (rule10) (q 2, a 2, BBBZ 0 ) ├ (r7) (q 3, a 1, BBZ 0 ) ├ (rule10) (q 2, a 1, BZ 0 ) ├ (rule7) (q 3, ε, Z 0 ) ├ (rule11) (q 1, ε, AZ 0 ) ├ ( No rule!?)

10. – 10 – CSCE 355 Fall 2008 Sample Test 2 Minimize DFA problem 1

11. – 11 – CSCE 355 Fall 2008 Dump of Distinguished States Array A B C D E F G H A B C D E F G H A X B X C X D X E X F X X X X X X X G X H X

12. – 12 – CSCE 355 Fall 2008 Iteration 0 a=0 Iteration 0 a=1 Adding dist[A,D] since delta(A, 1) = B, delta(D, 1) = F and dist[B,F]=X Adding dist[A,G] since delta(A, 1) = B, delta(G, 1) = F and dist[B,F]=X Adding dist[B,D] since delta(B, 1) = D, delta(D, 1) = F and dist[D,F]=X Adding dist[B,G] since delta(B, 1) = D, delta(G, 1) = F and dist[D,F]=X Adding dist[C,D] since delta(C, 1) = E, delta(D, 1) = F and dist[E,F]=X Adding dist[C,G] since delta(C, 1) = E, delta(G, 1) = F and dist[E,F]=X Adding dist[D,E] since delta(D, 1) = F, delta(E, 1) = H and dist[F,H]=X Adding dist[D,H] since delta(D, 1) = F, delta(H, 1) = H and dist[F,H]=X Adding dist[E,B] since delta(E, 1) = H, delta(B, 1) = D and dist[H,D]=X Adding dist[E,G] since delta(E, 1) = H, delta(G, 1) = F and dist[H,F]=X Adding dist[G,H] since delta(G, 1) = F, delta(H, 1) = H and dist[F,H]=X Adding dist[H,B] since delta(H, 1) = H, delta(B, 1) = D and dist[H,D]=X

13. – 13 – CSCE 355 Fall 2008 Iteration 1 a=0 Iteration 1 a=1 Adding dist[A,B] since delta(A, 1) = B, delta(B, 1) = D and dist[B,D]=X Adding dist[A,C] since delta(A, 1) = B, delta(C, 1) = E and dist[B,E]=X Adding dist[A,E] since delta(A, 1) = B, delta(E, 1) = H and dist[B,H]=X Adding dist[A,H] since delta(A, 1) = B, delta(H, 1) = H and dist[B,H]=X Adding dist[B,C] since delta(B, 1) = D, delta(C, 1) = E and dist[D,E]=X Iteration 2 a=0 Iteration 2 a=1

14. – 14 – CSCE 355 Fall 2008 ABCDEFGH A.XXXXXXX BX.XXXXXX CXX.XXX DXXX.XXX EXXX.XX FXXXXX.XX GXXXXX.X HXXXXX. Equivalence Classes A's equivalence class is: A B's equivalence class is: B C's equivalence class is: C E H D's equivalence class is: D G E's equivalence class is: C E H F's equivalence class is: F G's equivalence class is: D G H's equivalence class is: C E H

15. – 15 – CSCE 355 Fall 2008 Formal PDA Construction from CFL Let G = (V, T, Q, S) P = ({q}, T, (V U T), δ, q, S) where δ is defined by: 1.δ (q, ε, A) = { (q, β ) | A  β is a production of G} 2.δ (q, a, a) = { (q, ε) }

16. – 16 – CSCE 355 Fall 2008 Homework CFG  PDA Convert the grammar S  aAA A  aS | bS | a To a PDA that accepts the same language by empty stack P = ({q}, {a,b}, {S,A,a,b}, δ, q, S) where δ is defined by: δ (q, a, a) = { (q, ε) } for all a in Σ 1.δ (q, a, a) = { (q, ε) } 2.δ (q, b, b) = { (q, ε) } δ (q, ε, A) = { (q, β ) | A  β in G} 1.δ (q, ε, S) = { (q, aAA) } 2.δ (q, ε, A) = { (q, aS), (q, bS), (q, a) }

17. – 17 – CSCE 355 Fall 2008 HW 10/29 6.1.1b,c PDA trace (tree of IDs) 1.δ(q, 0, Z 0 ) = { (q, XZ 0 ) } 2.δ(q, 0, X) = { (q, XX) } 3.δ(q, 1, X) = { (q, X) } 4.δ(q, ε, X) = { (p, ε) } 5.δ(p, ε, X) = { (p, ε) } 6.δ(p, 1, X) = { (p, XX) } 7.δ(p, 1, Z 0 ) = { (p, ε) } qp 0,Z 0 /XZ 0 0,X/XX 1,X/X ε,X/ε 1,X/XX 1,Z 0 / ε

18. – 18 – CSCE 355 Fall 2008 6.1.1b (q, 0011, Z 0 ) (q, 011, XZ 0 )(p, 011, Z 0 ) (q, 11, XXZ 0 )(p, 11, XZ 0 ) (q, 1, XXZ 0 )(p, 1, XXZ 0 ) (q, ε, XXZ 0 )(p, 1, XZ 0 ) (p, 1, Z 0 )(p, ε, XZ 0 ) qp 0,Z 0 /XZ 0 0,X/XX 1,X/X ε,X/ε 1,X/XX 1,Z 0 / ε (p, ε, ε) (p, 1, Z 0 )(p, ε, XZ 0 ) (p, ε, ε) (p, 1, XZ 0 ) (p, 11, Z 0 ) (p, 1, ε) (p, ε, XXXZ 0 ) (p, ε, XXZ 0 ) (p, ε, XZ 0 ) (p, ε, Z 0 )

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