Chomsky Normal Form

Consider the CFG G

1

:

S → aSb | SS | ε

And the CFG G

2

:

S → aSb | SX | εX → X | S | ε

1. Are G

1

and G

2

equivalent?

(That is, do they generate the

same language?)

2. Is there an advantage of one CFG

over the other?

3. Can the disadvantage of G

2

be

eliminated?

4. How could that procedure be

formalized?

Chomsky Normal Form.

Chomsky Normal Form

A context-free grammar is in Chomsky

Normal Form if every rule is of the form:

S ⟶ εA ⟶ BCA ⟶ a

Where a is any terminal and A, B, and C are

any variables, except that B and C may not

be the start variable.

In this example, S is the start variable.

Implied by the definition:

● In A ⟶ BC, only two nonterminals are allowed in the RHS.

● In A ⟶ a, only one terminal is allowed in the RHS.

● In S ⟶ ε, only the start state may go to ε.● S may not appear on the RHS.

Simplifying and bounding grammars.

Theorem: Any CFG to CNFIdea: Make conversion in stages, systematically

changing or removing rules that do not match

the requirements.

Stages:

1. Add new start S

0

and rule S

0

⟶ S, where S was the original start variable.

(Guarantees S

0

isn’t on RHS.)

2. Eliminate ε-rules. For all A ⟶ ε, where A is not S

0

, then remove the rule and for

each occurrence of A in any RHS, add a

new rule with the occurrence deleted.

(e.g., for every rule R ⟶ uAv, we would add rule R ⟶ uv.)

2. (cont.) Replacement happens for each

occurrence of A, so R ⟶ uAvAw would add R ⟶ uvAw | uAvw | uvw.

If R ⟶ A exists, then add R ⟶ ε (unless this rule has already been removed).

Repeat until all ε-rules are removed.

3. Remove unit rules. For all rules of the

form A ⟶ B, remove the rule and for all B ⟶ u, add the rule A ⟶ u unless this was a unit rule previously removed (u is a string

of variables and terminals).

Repeat until all unit rules removed.

Theorem: Any CFG to CNF (continued)4. Convert all remaining rules into the

proper form.

Reminder: Proper form is

A ⟶ BCA ⟶ a

Replace each rule A ⟶ u1

u

2

…u

k

, where

k ≥ 3 and each u

i

is a variable or terminal,

with rules:

A ⟶ u1

A

1

A

1

⟶ u2

A

2

A

2

⟶ u3

A

3

…

A

k-2

⟶ uk-1

u

k

.

4. (cont.) A

i

’s are new variables.

Replace any terminal u

i

in the preceding

rules with the new variable U

i

and add

the rule U

i

⟶ ui

.

Repeat until all rules are in the proper

form.

Example: G6

to CNF: Step 1Before:

S ⟶ ASA | aBA ⟶ B | SB ⟶ b | ε

After (new starting state):

S

0

⟶ SS ⟶ ASA | aBA ⟶ B | SB ⟶ b | ε

Example: G6

to CNF: Step 2

After (remove ε-rules, B ⟶ ε):S

0

⟶ SS ⟶ ASA | aB | aA ⟶ B | S | εB ⟶ b | ε

After (remove ε-rules, A ⟶ ε):S

0

⟶ SS ⟶ ASA | aB | a | SA | AS | SA ⟶ B | S | εB ⟶ b

Previous:

S

0

⟶ SS ⟶ ASA | aBA ⟶ B | SB ⟶ b | ε

Example: G6

to CNF: Step 3

After (remove unit rules, S ⟶ S):S

0

⟶ SS ⟶ ASA | aB | a | SA | AS | SA ⟶ B | SB ⟶ b

After (remove unit rules, S

0

⟶ S):S

0

⟶ S | ASA | aB | a | SA | ASS ⟶ ASA | aB | a | SA | ASA ⟶ B | SB ⟶ b

Previous:

S

0

⟶ SS ⟶ ASA | aB | a | SA | AS | SA ⟶ B | SB ⟶ b

Example: G6

to CNF: Step 3 (cont.)

After (remove unit rules, A ⟶ B):S

0

⟶ ASA | aB | a | SA | ASS ⟶ ASA | aB | a | SA | ASA ⟶ B | S | bB ⟶ b

After (remove unit rules, A ⟶ S):S

0

⟶ ASA | aB | a | SA | ASS ⟶ ASA | aB | a | SA | ASA ⟶ S | b | ASA | aB | a | SA | ASB ⟶ b

Previous:

S

0

⟶ ASA | aB | a | SA | ASS ⟶ ASA | aB | a | SA | ASA ⟶ B | SB ⟶ b

After (Eliminate aB on RHS):

S

0

⟶ AA1

| aB | UB | a | SA | AS

S ⟶ AA1

| aB | UB | a | SA | AS

A ⟶ b | AA1

| UB | a | SA | AS

A

1

⟶ SAU ⟶ aB ⟶ b

Example: G6

to CNF: Step 4Previous:

S

0

⟶ ASA | aB | a | SA | ASS ⟶ ASA | aB | a | SA | ASA ⟶ b | ASA | aB | a | SA | ASB ⟶ b

After (Eliminate ASA on RHS):

S

0

⟶ ASA | AA1

| aB | a | SA | AS

S ⟶ ASA | AA1

| aB | a | SA | AS

A ⟶ b | ASA | AA1

| aB | a | SA | AS

A

1

⟶ SAB ⟶ b

Previous:

S

0

⟶ AA1

| aB | a | SA | AS

S ⟶ AA1

| aB | a | SA | AS

A ⟶ b | AA1

| UB | a | SA | AS

A

1

⟶ SAB ⟶ b

Example: G6

to CNFBefore:

S ⟶ ASA | aBA ⟶ B | SB ⟶ b | ε

After (Final):

S

0

⟶ AA1

| UB | a | SA | AS

S ⟶ AA1

| UB | a | SA | AS

A ⟶ b | AA1

| UB | a | SA | AS

A

1

⟶ SAU ⟶ aB ⟶ b

Important: Upper size bound is the square of the original Grammar size.

Convert this to CNF:

S ⟶ 0S1 | ε

1. Add new start S

0

2. Eliminate ε-rules.3. Remove unit rules.

4. Convert all remaining rules into

the proper form.

Convert S ⟶ 0S1 | ε to CNFStep 1:

S

0

⟶ SS ⟶ 0S1 | ε

Step 2:

S

0

⟶ S | εS ⟶ 0S1 | 01

Step 3:

S

0

⟶ 0S1 | 01 | εS ⟶ 0S1 | 01

Step 4, part 1:

S

0

⟶ 0A | 01 | εS ⟶ 0A | 01A ⟶ S1

Step 4, part 2:

S

0

⟶ BA | BC | εS ⟶ BA | BCA ⟶ SCB ⟶ 0C ⟶ 1

CNF in Retrospect 1. The book does NOT give a good proof. Why is it a bad example of a proof?2. Wikipedia gives a better overview of the

proof.

https://en.wikipedia.org/wiki/Chomsky_normal_form

3. What benefit does CNF have?

Guaranteed maximum final grammar size.

Guaranteed maximum derivation length.

4. Why do we need CNF?

Proofs! Algorithms! Big-O Analysis!

5. BUT!!!! Does CNF eliminate ambiguity?

No.

https://en.wikipedia.org/wiki/Chomsky_normal_form

Chomsky Language Hierarchies

https://commons.wikimedia.org/wiki/File:Chomsky-hierarchy.svg

Formal Grammar Classifications

Each class is a subset of the class

above it.