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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.

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Presentation Transcript
slide2

Presentation Outline

  • Introduction
  • Chomsky normal form
  • Greibach Normal Form
    • Algorithm ( with Example)
  • Summary
slide3

Introduction

  • Grammar: G = (V, T, P, S)

T = { a, b }

  • Terminals

V = A, B, C

  • Variables

S

  • Start Symbol

P = S → A

  • Production
slide4

Presentation Outline

  • Introduction
  • Chomsky normal form
  • Greibach Normal Form
    • Algorithm ( with Example)
  • Summary
slide5

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
slide6

Presentation Outline

  • Introduction
  • Chomsky normal form
  • Greibach Normal Form
    • Algorithm ( with Example)
  • Summary
slide7

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.
slide8

Presentation Outline

  • Introduction
  • Chomsky normal form
  • Greibach Normal Form
    • Algorithm ( with Example)
  • Summary
conversion
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
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 recursion1
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 recursion2
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
Transform A  uBv rules
  • Before

A  uBb

B  w1 | w1 |…| wn

  • After

Add A  uw1b | uw1b |…| uwnb

DeleteA  uBb

conversion step 1
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 11
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
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
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
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
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
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
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
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 rules1
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
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
Step 2: Fix starting symbol
  • Rules S, A, B and C don’t have direct left recursion, and RHS variables are of higher number
  • All C rules start with terminal symbol
  • 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
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
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
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
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
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
slide31

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

slide32

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

slide33

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

slide34

Greibach Normal Form

Example:

A1 → A2A3 | A4A4

A4 → bA3A4 | A4A4A4 | b

A2 → b

A3 → a

Second Step

Eliminate Left Recursions

A4 → A4A4A4

slide35

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

slide36

Greibach Normal Form

Example:

A1 → A2A3 | A4A4

A4 → bA3A4 | b | bA3A4Z | bZ

Z → A4A4 | A4A4 Z

A2 → b

A3 → a

A → αX

GNF

slide37

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

slide38

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

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