Automata Theory - Automata and Complexity Theory - Lecture Slides, Slides of Theory of Automata

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Formal Languages and Automata Theory

Outline

• Context-free Languages, Context-free

grammars (CFG), Push-down Automata (PDA)

• Grammar Transformations

– Remove ε- production;

– Remove unit-production;

– Covert to Chomsky-Normal-Form (CNF)

• Cocke-Younger-Kasami (CYK) algorithm

Example (I)

• Given the following CFG

S  X | Y

X  aXb | aX | a

Y  aYb | Yb | b

• (1) L(G) =?

• (2) Design an equivalent PDA for it.

Σ={a, b}

Example (I) --- solution: L(S)

S  X | Y

X  aXb | aX | a

Y  aYb | Yb | b

Try to write some strings generated by it: SXaXbaaXbbaaaXbbaaaabb SYaYbaYbbaaYbbbaabbbb

more a ’s than b ’s more b ’s than a ’s

Observations:

• Start from S, we can enter two States X & Y, and X, Y are “independent”;
• In X state, always more a are generated;
• In Y state, always more b are generated.

Ls = Lx U Ly

Lx = { a ib j^ ; i>j } Lx = { a ib j^ ; i<j }

L(S) =

{ aibj; i≠j }

Example (II)

• Given the following language:

• (1) design a CFG for it;

• (2) design a PDA for it.

L = {0 i 1 j : i ≤ j ≤ 2 i, i=0,1,… }, Σ = {0, 1}

Example (II) -- solution: CFG

L = {0 i 1 j : i ≤ j ≤ 2 i, i=0,1,… }, Σ = {0, 1}

Consider two extreme cases:

(a). if j = i, then L 1 = { 0i 1 j: i=j }; (b). if j = 2i, then L 2 = { 0i 1 j: 2i=j }.

S  0S

S  ε

S  0S

S  ε If i ≤ j ≤ 2 i , then randomly choose “ red- rule ” or “ blue-rule ” in the generation.

“ red-rule ” “ blue-rule ”

S  0S

S  0S

S  ε

Example (II) -- solution: PDA

L = {0 i 1 j : i ≤ j ≤ 2 i, i=0,1,… }, Σ = {0, 1}

Similar idea: “randomly choose two extreme cases”

ε,ε/# q (^0)

0,ε/X ε,ε/ε q^1

q (^2)

ε,#/ε

1,X/ε

1,X/X 1,X/ε

q (^3)

Example (III)

• Given the following language:

• (1). Design a CFG for it;

• (2). Design a PDA for it.

L = { a i b j c k d l : i , j,k,l=0,1 ,…; i+k=j+l },

where the alphabet Σ= {a, b, c, d}

Example (III) – solution: PDA

L = { a i b j c k d l : i , j,k,l=0,1 ,…; i+k=j+l },

Main idea: (1) use X to record an a or c ; use Y to record an b or d. (2) Compare # X and # Y : by cancellation.

How to realize the comparison by cancellation? Action1: Push an X, when a or c was read; Action2: Pop an X (if any, otherwise push a Y), when b or d was read. Action3: Pop an Y (if any, otherwise push an X), when a or c was read.

Example (III) – solution: PDA

L = { a i b j c k d l : i , j,k,l=0,1 ,…; i+k=j+l },

Action1: Push an X, when a or c was read; Action2: Pop an X (if any, otherwise push a Y), when b or d was read. Action3: Pop an Y (if any, otherwise push an X), when a or c was read.

ε,ε/# (^) q (^1) q (^5)

a,ε/X ε,ε/ε

b,#/Y#

q 2 ε,ε/ε

c,X/XX

q 3 ε,ε/ε^ q 4 ε,#/ε

b,X/ε

b,Y/YY

c,#/X#

c,Y/ε

d,X/ε

d,#/Y# d,Y/YY

Remove ε-production

• Example:

Remove ε-productions of the

following grammar G:

S → ABaC

A → BC

B → b |ε

C → D |ε

D → d

Remove ε-production

S → ABaC | BaC

| AaC | aC

| ABa | Ba | Aa | a

A → BC | C | B

B → b |ε

C → D |ε

D → d

Nullable variables: A, B, C Add Delete For A : S  BaC For B : S  AaC | aC A  C

B → ε

For C : S  ABa | Ba | Aa | a A  B

C → ε

For i = 1 to k

For every production of the form A → αN i β,

add another production A → αβ

If N i → ε is a production, remove it Docsity.com

Remove unit-productions

Add Delete

S  T

For T TF (^) S  TF For T F  a (^) S  a

S → S+T | T

| T*F | a | ( S )

T → T*F | F

| a | ( S )

F → ( S ) | a

For T F  (S) (^) S  (S) T  F For F  a (^) S  a For F  (S) (^) S  (S)

A 1 → A 2 → ... → A k → A 1

A 1 → A 2 → ... → A k → α A 1 →^ α,... , Ak →^ α

delete it and replace

Rule 1: everything with A 1

Rule 2:

Convert CFG to Chomsky-Normal-Form

(CNF)

• Example: Convert the following CFG to CNF:

S  0AB

A  0D | 1AD

B  0

D  1