lecture 20 dpda / review topics given a pda construct a grammar for the language it accepts...

Lecture 20DPDA / Review

Lecture 20DPDA / Review

Topics Topics Given a PDA construct a grammar for the

language it accepts Deterministic PDAs Properties of CFLs

Readings:Readings:

November 5, 2008

CSCE 355 Foundations

– 2 – CSCE 355 Fall 2008

Last Time: Top-down parsing

AlgorithmsLeft-factoring

From Grammars to PDA Section 6.3 Formal PDA Construction from CFL

New: New: When will our parsing algorithm be deterministic?

HomeworkHomework1. …

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When will our parsing algorithm be deterministic?When will our parsing algorithm be deterministic?First(First(αα ) = { x in T such that ) = { x in T such that αα * x* xββ } }

Then if AThen if A αα and A and A β β are two productions we are are two productions we are choosing from and the current input character is a which choosing from and the current input character is a which is in is in First(First(αα) but not in First() but not in First(ββ ) )

Then we choose the production Then we choose the production AA αα..

So if So if First(First(αα) ) ∩∩ First( First(ββ ) = ) = ϕϕ then we can always then we can always deterministically choosedeterministically choose

There is a type of grammar that is LL(1) for which this works There is a type of grammar that is LL(1) for which this works well. A grammar with no well. A grammar with no εε-productions is LL(1) if it satisfies -productions is LL(1) if it satisfies the disjointness property above.the disjointness property above.

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Homework from Last time Homework from Last time

1.1. Given the grammar simple4.y (and simple4.l)Given the grammar simple4.y (and simple4.l)a. Modify the semantic actions so that it produces a trace of

productions involved in the parse.

b. If you look at the sequence do you notice anything?

2.2. 6.2.1b) Strings of 0’s and 1’s such that no prefix has 6.2.1b) Strings of 0’s and 1’s such that no prefix has more 1’s than 0’smore 1’s than 0’s

3.3. 6.2.5b,c6.2.5b,c

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%token INTEGER PLUS TIMES LPAREN RPAREN%token INTEGER PLUS TIMES LPAREN RPAREN

%%%%

line : expr '\n' {printf("Using Prod line -> expr \\n \n");}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");}expr : expr PLUS term {printf("Using Prod expr -> expr PLUS term \n");}

| term | term {printf("Using Prod expr -> term \n");}{printf("Using Prod expr -> term \n");}

;;

term : term TIMES factor {printf("Using Prod term -> term TIMES factor\n");}term : term TIMES factor {printf("Using Prod term -> term TIMES factor\n");}

| factor | factor {printf("Using Prod term -> factor \n");}{printf("Using Prod term -> factor \n");}

;;

factor : LPAREN expr RPAREN {printf("Using Prod factor -> LP expr RP \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");}

;;

%%%%

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cerberus> ./simpleHWcerberus> ./simpleHW

2+3*72+3*7

Using Prod factor -> INTEGER Using Prod factor -> INTEGER

Using Prod term -> factor Using Prod term -> factor

Using Prod expr -> term Using Prod expr -> term


Using Prod term -> factor Using Prod term -> factor


Using Prod term -> term TIMES factorUsing Prod term -> term TIMES factor

Using Prod expr -> expr PLUS term Using Prod expr -> expr PLUS term

Using Prod line -> expr \n Using Prod line -> expr \n

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6.2.1b) no prefix has more 1’s than 0’s6.2.1b) no prefix has more 1’s than 0’s

6.2.1b) Strings of 0’s and 1’s such that no prefix has 6.2.1b) Strings of 0’s and 1’s such that no prefix has more 1’s than 0’smore 1’s than 0’s

q dead

0,Z0/0Z0

0,0/001,0/ε

1,Z0/Z0

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6.2.5b6.2.5bSt Σ Γ NextSt Γ* indx

q0 a Z0 q1 AAZ0 1

q0 b Z0 q2 BZ0 2

q0 ε Z0 f ε 3

q1 a A q1 AAA 4

q1 b A q1 ε 5

q1 ε Z0 q0 Z0 6

q2 a B q3 ε 7

q2 b B q2 BB 8

q2 ε Z0 q0 Z0 9

q3 ε B q2 ε 10

q3 ε Z0 q1 AZ0 11

(q(q00, abb, Z, abb, Z00))├ ├ (rule 1)(rule 1)

(q(q11, bb, AAZ, bb, AAZ00))├ ├ (rule 5)(rule 5)

(q(q11, b, AZ, b, AZ00))├ ├ (rule 5)(rule 5)

(q(q11, , εε, Z, Z00))├ ├ (rule 6)(rule 6)

(q(q00, , εε, Z, Z00))├ ├ (rule 3)(rule 3)

(f, (f, εε, , εε))

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6.2.5c6.2.5cSt Σ Γ NextSt Γ* indx

q0 a Z0 q1 AAZ0 1

q0 b Z0 q2 BZ0 2

q0 ε Z0 f ε 3

q1 a A q1 AAA 4

q1 b A q1 ε 5

q1 ε Z0 q0 Z0 6

q2 a B q3 ε 7

q2 b B q2 BB 8

q2 ε Z0 q0 Z0 9

q3 ε B q2 ε 10

q3 ε Z0 q1 AZ0 11

(q(q00, b, b77aa44, Z, Z00))├ ├ (rule 2)(rule 2)

(q(q22, b, b66aa44, BZ, BZ00))├ ├ (rule 8)(rule 8)

(q(q22, b, b55aa44, BBZ, BBZ00))├ ├ (rule 8)(rule 8)

……

(q(q22, ba, ba44, BBBBBBZ, BBBBBBZ00))├ ├ (r8)(r8)

(q(q22, a, a44, BBBBBBBZ, BBBBBBBZ00))├ ├ (r7)(r7)

(q(q33, a, a33, BBBBBBZ, BBBBBBZ00))├ ├ (rule10)(rule10)

(q(q22, a, a33, BBBBBZ, BBBBBZ00))├ ├ (rule7)(rule7)

(q(q33, a, a22, BBBBZ, BBBBZ00))├ ├ (rule10)(rule10)

(q(q22, a, a22, BBBZ, BBBZ00))├ ├ (r7)(r7)

(q(q33, a, a11, BBZ, BBZ00))├ ├ (rule10)(rule10)

(q(q22, a, a11, BZ, BZ00))├ ├ (rule7)(rule7)

(q(q33, , εε, Z, Z00))├ ├ (rule11)(rule11)

(q(q11, , εε, AZ, AZ00))├ ├ ( No rule!?)( No rule!?)

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Sample Test 2 Minimize DFA problemSample Test 2 Minimize DFA problem

1

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Dump of Distinguished Dump of Distinguished States ArrayStates Array

A B C D E F G H A B C D E F G H

A X A X

B X B X

C X C X

D X D X

E X E X

F X X X X X X X F X X X X X X X

G X G X

H X H X

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Iteration 0 a=0Iteration 0 a=0


Adding dist[A,D] since delta(A, 1) = B, delta(D, 1) = F and dist[B,F]=XAdding 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]=XAdding 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]=XAdding 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]=XAdding 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]=XAdding 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]=XAdding 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]=XAdding 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]=XAdding 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]=XAdding 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]=XAdding 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]=XAdding 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]=XAdding dist[H,B] since delta(H, 1) = H, delta(B, 1) = D and dist[H,D]=X

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Adding dist[A,B] since delta(A, 1) = B, delta(B, 1) = D and dist[B,D]=XAdding 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]=XAdding 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]=XAdding 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]=XAdding 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]=XAdding dist[B,C] since delta(B, 1) = D, delta(C, 1) = E and dist[D,E]=X



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A B C D E F G H

A . X X X X X X X

B X . X X X X X X

C X X . X X X

D X X X . X X X

E X X X . X X

F X X X X X . X X

G X X X X X . X

H X X X X X .

Equivalence ClassesEquivalence Classes

A's equivalence class is: A A's equivalence class is: A

B's equivalence class is: B B's equivalence class is: B

C's equivalence class is: C E H C's equivalence class is: C E H

D's equivalence class is: D G D's equivalence class is: D G

E's equivalence class is: C E H E's equivalence class is: C E H

F's equivalence class is: F F's equivalence class is: F

G's equivalence class is: D G G's equivalence class is: D G

H's equivalence class is: C E HH's equivalence class is: C E H

– 15 – CSCE 355 Fall 2008

Formal PDA Construction from CFLFormal PDA Construction from CFL

Let G = (V, T, Q, S)Let G = (V, T, Q, S)

P = ({q}, T, (V U T), P = ({q}, T, (V U T), δδ, q, S), q, S)

where where δ δ is defined by:is defined by:

1.1. δδ(q, (q, εε, A) = { (q, , A) = { (q, ββ) | A) | A ββ is a production of G}is a production of G}

2.2. δδ(q, a, a) = { (q, (q, a, a) = { (q, εε) }) }

– 16 – CSCE 355 Fall 2008

Homework CFG PDAHomework CFG PDA

Convert the grammarConvert the grammar• S aAA • A aS | bS | a

To a PDA that accepts the same language by empty stackTo a PDA that accepts the same language by empty stack

P = ({q}, {a,b}, {S,A,a,b}, P = ({q}, {a,b}, {S,A,a,b}, δδ, q, S), q, S)

where where δ δ is defined by: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) | Aβ in G}

• δ(q, ε, S) = { (q, aAA) }

• δ(q, ε, A) = { (q, aS), (q, bS), (q, a) }

– 17 – CSCE 355 Fall 2008

HW 10/29 6.1.1b,c PDA trace (tree of IDs)HW 10/29 6.1.1b,c PDA trace (tree of IDs)

1.1. δδ(q, 0, Z(q, 0, Z00) = { (q, XZ) = { (q, XZ00) }) }

2.2. δδ(q, 0, X) = { (q, XX) }(q, 0, X) = { (q, XX) }

3.3. δδ(q, 1, X) = { (q, X) }(q, 1, X) = { (q, X) }

4.4. δδ(q, (q, εε, X) = { (p, , X) = { (p, εε) }) }

5.5. δδ(p, (p, εε, X) = { (p, , X) = { (p, εε) }) }

6.6. δδ(p, 1, X) = { (p, XX) }(p, 1, X) = { (p, XX) }

7.7. δδ(p, 1, Z(p, 1, Z00) = { (p, ) = { (p, εε) }) }

q p

0,Z0/XZ0

0,X/XX1,X/X

ε,X/ε

ε,X/ε1,X/XX1,Z0/ ε

– 18 – CSCE 355 Fall 2008

6.1.1b6.1.1b(q, 0011, Z0)

(q, 011, XZ0) (p, 011, Z0)

(q, 11, XXZ0) (p , 11, XZ0)

(q, 1, XXZ0) (p , 1, XXZ0)

(q, ε, XXZ0) (p , 1, XZ0)

(p, 1, Z0) (p , ε, XZ0)

q p

0,Z0/XZ0

0,X/XX1,X/X

ε,X/ε

ε,X/ε1,X/XX1,Z0/ ε

(p, ε, ε)

(p, 1, Z0) (p , ε, XZ0)

(p, ε, ε)

(p, 1, XZ0)

(p , 11, Z0)

(p , 1, ε)(p , ε, XXXZ0)

(p , ε, XXZ0)

(p , ε, XZ0)

(p , ε, Z0)

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lecture 20 dpda / review topics given a pda construct a grammar for the language it accepts...

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