Chomsky &amp; Greibach Normal Forms

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# Chomsky & Greibach Normal Forms - PowerPoint PPT Presentation

Chomsky &amp; Greibach Normal Forms. Presentation Outline. Introduction Chomsky normal form Greibach Normal Form Algorithm ( with Example) Summary. Introduction. Grammar: G = (V, T, P, S). T = { a, b }. Terminals. V = A, B, C. Variables. S. Start Symbol. P = S → A. Production.

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### Chomsky & Greibach Normal Forms

Presentation Outline

• Introduction
• Chomsky normal form
• Greibach Normal Form
• Algorithm ( with Example)
• Summary

Introduction

• Grammar: G = (V, T, P, S)

T = { a, b }

• Terminals

V = A, B, C

• Variables

S

• Start Symbol

P = S → A

• Production

Presentation Outline

• Introduction
• Chomsky normal form
• Greibach Normal Form
• Algorithm ( with Example)
• Summary

Chomsky Normal Form

A context free grammar is said to be in Chomsky Normal Form if all productions are in the following form:

A → BC

A → α

• A, B and C are non terminal symbols
• α is a terminal symbol

Presentation Outline

• Introduction
• Chomsky normal form
• Greibach Normal Form
• Algorithm ( with Example)
• Summary

Greibach Normal Form

A context free grammar is said to be in Greibach Normal Form if all productions are in the following form:

A → αX

• A is a non terminal symbols
• α is a terminal symbol
• X is a sequence of non terminal symbols.
• It may be empty.

Presentation Outline

• Introduction
• Chomsky normal form
• Greibach Normal Form
• Algorithm ( with Example)
• Summary
Conversion
• Convert from Chomsky to Greibach in two steps:
• From Chomsky to intermediate grammar
• Eliminate direct left recursion
• Use A  uBv rules transformations to improve references (explained later)
• From intermediate grammar into Greibach
Eliminate direct left recursion
• Before

A  Aa| b

• After

A  bZ | b

Z  aZ | a

• Remove the rule with direct left recursion, and create a new one with recursion on the right
Eliminate direct left recursion
• Before

A  Aa| Ab| b| c

• After

A  bZ| cZ| b| c

Z  aZ | bZ | a| b

• Remove the rules with direct left recursion, and create new ones with recursion on the right
Eliminate direct left recursion
• Before

A  AB| BA| a

B  b | c

• After

A  BAZ| aZ| BA| a

Z  BZ | B

B  b | c

Transform A  uBv rules
• Before

A  uBb

B  w1 | w1 |…| wn

• After

Add A  uw1b | uw1b |…| uwnb

DeleteA  uBb

Conversion: Step 1
• Goal: construct intermediate grammar in this format
• A  aw
• A  Bw
• S  

where w  V* and B comes after A

Conversion: Step 1
• Assign a number to all variables starting with S, which gets 1
• Transform each rule following the order according to given number from lowest to highest
• Eliminate direct left recursion
• If RHS of rule starts with variable with lower order, apply A  uBb transformation to fix it
Conversion: Step 2
• Goal: construct Greibach grammar out of intermediate grammar from step 1
• Fix A  Bw rules into A  aw format
• After step 1, last original variable should have all its rules starting with a terminal
• Working from bottom to top, fix all original variables using A  uBb transformation technique, so all rules become A  aw
• Fix introduced recursive rules same way
Conversion Example
• Convert the following grammar from Chomsky normal form, into Greibach normal form
• S AB | 
• A  AB | CB | a
• B  AB | b
• C  AC | c
Conversion Strategy
• Goal: transform all rules which RHS does not start with a terminal
• Apply two steps conversion
• Work rules in sequence, eliminating direct left recursion, and enforcing variable reference to higher given number
• Fix all original rules, then new ones
Step 1: S rules
• Starting with S since it has a value to of 1
• S AB | 
• S rules comply with two required conditions
• There is no direct left recursion
• Referenced rules A and B have a given number higher than 1. A corresponds to 2 and B to 3.
Step 1: A rules
• A  AB | CB | a
• Direct left recursive rule A  AB needs to be fixed. Other A rules are fine
• Apply direct left recursion transformation

A  CBR1 | aR1 | CB | a

R1  BR1 | B

Step 1: B rules
• B  AB | b
• B  AB rule needs to be fixed since B corresponds to 3 and A to 2. B rules can only have on their RHS variables with number equal or higher. Use A  uBb transformation technique
• B  CBR1B | aR1B | CBB | aB| b
Step 1: C rules
• C  AC | c
• C  AC rule needs to be fixed since C corresponds to 4 and A to 2. Use same A  uBb transformation technique
• C  CBR1C | aR1C | CBC | aC| c
• Now variable references are fine according to given number, but we introduced direct left recursion in two rules…
Step 1: C rules
• C  CBR1C | aR1C |CBC | aC| c
• Eliminate direct left recursion

C  aR1CR2 | aCR2| cR2 | aR1C | aC| c

R2  BR1CR2 | BCR2 | BR1C | BC

Step 1: Intermediate grammar
• S AB | 
• A  CBR1 | aR1 | CB | a
• B  CBR1B | aR1B | CBB | aB | b
• C  aR1CR2 | aCR2| cR2 | aR1C | aC | c
• R1  BR1 | B
• R2  BR1CR2 | BCR2 | BR1C | BC
Step 2: Fix starting symbol
• Rules S, A, B and C don’t have direct left recursion, and RHS variables are of higher number
• Proceed to fix rules B, A and S in bottom-up order, so they start with terminal symbol.
• Use A  uBb transformation technique
Step 2: Fixing B rules
• Before

B  CBR1B | aR1B | CBB | aB | b

• After

B  aR1B | aB | b

B  aR1CR2BR1B | aCR2BR1B| cR2BR1B | aR1CBR1B | aCBR1B| cBR1B

B  aR1CR2BB | aCR2BB| cR2BB | aR1CBB | aCBB| cBB

Step 2: Fixing A rules
• Before

A  CBR1 | aR1 | CB | a

• After

A  aR1| a

A  aR1CR2BR1 | aCR2BR1| cR2BR1 | aR1CBR1 | aCBR1| cBR1

A  aR1CR2B | aCR2B| cR2B | aR1CB | aCB| cB

Step 2: Fixing S rules
• Before

SAB | 

• After

S 

S  aR1B| aB

S  aR1CR2BR1B | aCR2BR1B| cR2BR1B | aR1CBR1B | aCBR1B| cBR1B

S  aR1CR2BB | aCR2BB| cR2BB | aR1CBB | aCBB| cBB

Step 2: Complete conversion
• All original rules S, A, B and C are fully converted now
• New recursive rules need to be converted next

R1  BR1 | B

R2  BR1CR2 | BCR2 | BR1C | BC

• Use same A  uBb transformation technique replacing starting variable B
Conclusions
• After conversion, since B has 15 rules, and R1 references B twice, R1 ends with 30 rules
• Similar for R2which referencesB four times. Therefore, R2 ends with 60 rules
• All rules start with a terminal symbol (with the exception of S  )
• Parsing algorithms top-down or bottom-up would complete on a grammar converted to Greibach normal form

Greibach Normal Form

Example:

S → XA | BB

B → b | SB

X → b

A → a

S = A1

X = A2

A = A3

B = A4

A1 → A2A3 | A4A4

A4 → b | A1A4

A2 → b

A3 → a

CNF

New Labels

Updated CNF

Greibach Normal Form

Example:

Ai → AjXkj >= i

A1 → A2A3 | A4A4

A4 → b | A1A4

A2 → b

A3 → a

First Step

Xk is a string of zero or more variables

A4 → A1A4

Greibach Normal Form

Example:

Ai → AjXk j > i

First Step

| A4A4A4

| b

A4 → bA3A4

A4 → A1A4

A1 → A2A3 | A4A4

A4 → b | A1A4

A2 → b

A3 → a

| A4A4A4

| b

A4 → A2A3A4

Greibach Normal Form

Example:

A1 → A2A3 | A4A4

A4 → bA3A4 | A4A4A4 | b

A2 → b

A3 → a

Second Step

Eliminate Left Recursions

A4 → A4A4A4

Greibach Normal Form

Example:

Second Step

Eliminate Left Recursions

A4 → bA3A4 | b

| bA3A4Z| bZ

A1 → A2A3 | A4A4

A4 → bA3A4 | A4A4A4 | b

A2 → b

A3 → a

Z → A4A4

| A4A4Z

Greibach Normal Form

Example:

A1 → A2A3 | A4A4

A4 → bA3A4 | b | bA3A4Z | bZ

Z → A4A4 | A4A4 Z

A2 → b

A3 → a

A → αX

GNF

Greibach Normal Form

Example:

A1 → A2A3 | A4A4

A4 → bA3A4 | b | bA3A4Z | bZ

Z → A4A4 | A4A4 Z

A2 → b

A3 → a

A1 → bA3

| bA3A4A4

| bA4

| bA3A4ZA4

| bZA4

Z → bA3A4A4 | bA4 | bA3A4ZA4 | bZA4 | bA3A4A4Z| bA4Z| bA3A4ZA4 Z| bZA4 Z

Greibach Normal Form

Example:

A1 → bA3 |bA3A4A4 | bA4 | bA3A4ZA4 | bZA4

A4 → bA3A4 | b | bA3A4Z | bZ

Z → bA3A4A4 | bA4 | bA3A4ZA4 | bZA4 | bA3A4A4Z| bA4Z| bA3A4ZA4 Z|

bZA4 Z

A2 → b

A3 → a

Grammar in Greibach Normal Form