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Context Free Grammar

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Context Free Grammar

- Why do we want to learn about Context Free Grammars?
- Used in many parsers in compilers
- Yet another compiler-compiler, yacc
- GNU project parser generator, bison

- Web application
- In describing the formats of XML (eXtensible Markup Language) documents

- Used in many parsers in compilers

- G = (V, T, P, S)
- V – a set of variables, e.g. {S, A, B, C, D, E}
- T – a set of terminals, e.g. {a, b, c}
- P – a set of productions rules
- In the form of A , where AV, (VT)* e.g. S aB

- S is a special variable called the start symbol

Construct a CFG for the following language : L1={0n1n | n>0}

Thought: If is a string in L1, so is 01, i.e. 0k+11k+1 = 0(0k1k)1

S 01 | 0S1

Construct a CFG for the following language : L2={0i1j | ij and i, j>0}

Thought: Similar to L1, but at least one more ‘1’ or at least one more ‘0’

S AC | CB

A 0A | 0

B B1 | 1

C 0C1 |

Determine the language generated by the CFG:

S AS |

A A1 | 0A1 | 01

S generates consecutive ‘A’s

A generates 0i1j where i j

Languages: each block of ‘0’s is followed by at least as many ‘1’s

How a string is generated by a grammar G

Consider the grammar G

S AS |

A A1 | 0A1 | 01

S 0110011 ?

S

A

S

A

A

1

S

0

1

0

A

1

0

1

Rightmost Derivation

Leftmost Derivation

Parse Tree or

Derivation Tree

S AS

A1S

011S

011AS

0110A1S

0110011S

0110011

S AS

AAS

AA

A0A1

A0011

A10011

0110011

Each parse tree has one unique leftmost derivation and one unique rightmost derivation.

A grammar is ambiguous if some strings in it have more than one parse trees, i.e., it has more than one leftmost derivations (or more than one rightmost derivations).

- Consider the following grammar G:
- S AS | a | b
- A SS | ab
- A string generated by this grammar can have more than one parse trees. Consider the string abb:

S

S

A

S

A

S

b

S

S

b

a

b

a

b

- As another example, consider the following grammar:
E E + E | E * E | (E) | x | y | z

- There are 2 leftmost derivations for x + y + z:

E E + E

x + E

x + E + E

x + y + E

x + y + z

E E + E

E + E + E

x + E + E

x + y + E

x + y + z

E

E

E

E

E

E

+

+

x

z

y

z

x

y

+

+

- Remove useless variables
- Generating Variables
- Reachable Variables

- Remove -productions, e.g. A
- Remove unit-productions, e.g. AB

- A variable X is useless if:
- X does not generate any string of terminals, or
- The start symbol, S, cannot generate X.
S X

- where , (VT)*, X V and T*

- Mark each production of the form:
X where T*

- Repeat
- Mark X where consists of terminals or variables which are on the left hand side of some marked productions.

- Until no new productions is marked.
- Remove all unmarked productions.

- CFG:
S AB|CAA a

B BC|AB

C aB|b

S CA

A a

C b

- CFG:
S aAa|aB

A aS|bD

B aBa|b

C abb|DD

D aDa

S aAa|aB

A aS

B aBa|b

C abb

- Mark each production of the form:
S where (VT)*

- Repeat
- Mark X where X appears on the right hand side of some marked productions.

- Until no new productions is marked.
- Remove all unmarked productions.

CFG:

S aAa|aB

A aS

B aBa|b

C abb

S aAa|aB

A aS

B aBa|b

CFG:

S Aab|AB

A a

B bD| bB

C A | B

D a

S Aab|AB

A a

B bD | bB

D a

A Bac|bTC|ba

T aB|TB

B aTB|bBC

C TBc|aBC|ac

A ba

C ac

A ba

Step 1: Find nullable variables

Step 2: Remove all -productions by replacing some productions

- Mark all variables A for which there exists a production of the form A .
- Repeat
- Mark X for which there exists X where V* and all symbols in have been marked.

- Until no new variables is marked.

- We can remove all -productions (except S if S is nullable) by rewriting some other productions.
- e.g., If X1 and X3 are nullable, we should replace
- A X1X2X3X4 by:
- A X1X2X3X4 | X2X3X4 | X1X2X4 | X2X4

Example:

S A | B | C

A aAa | B

B bB | bb

C aCaa | D

D baD | abD |

Step1: nullable variables are D, C and S

Step 2:

- eliminate D by replacing:
- D baD | abD into D baD | abD | ba | ab

- eliminate C by replacing:
- C aCaa | D into C aCaa | D | aaa
- S A | B | C into S A | B | C |

- The new grammar:
- SA | B | C |
- AaAa | B
- BbB | bb
- CaCaa | D | aaa
- DbaD | abD | ba | ab

Example:S A | B | C

A aAa | B

B bB | bb

C aCaa | D

D baD | abD |

S ABCD

A a

B

C ED |

D BC

E b

S ABCD|ABC|ABD|ACD|AB|AC|AD|A

- A a
- C ED|E
D BC|B|C

- E b

- Unit production: A B, where A and B V
- 1. Detect cycles of unit productions:
- A1 A2 A3 … A1
- Replaced A1, A2, A3 , …, Ak by any one of them
- 2. Detect paths of unit productions:
- A1 A2 A3 …, Ak , where (VT)*/V
- Add productions A1 , A2 , …, Ak and remove all the unit productions

Example:

S A | B | C

A aa | B

B bb | C

C cc | A

Cycle: A B C A

Remove by replace B,C by A

S A | A | A

A aa | A

A bb | A

A cc | A

Becomes:

S A

A aa | bb | cc

Path: S A

Remove by adding productions

S aa | bb | cc

CFG:

S → A | Bb

A → C | a

B → aBa | b

C → aSa

S → Bb | aSa |a

→ A → a | aSa

B → aBa | b

C → aSa

Substitute

Unit productions of form can be removed immediately

Remove

Substitute

Remove repeated productions

Final grammar

Thank You