lecture 4 – outline - tau › ~bchor › cm › compute4.pdf · lecture 4 – outline so far we...

Lecture 4 – OutlineSo far we saw

finite automata,regular languages,regular expressions,pumping lemma for regular languages.

Today, we introduce stronger machines andlanguages with more expressive power:

pushdown automata,context-free languages,context-free grammars,pumping lemma for context-free languages.

Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. – p.1

finite automata,

regular languages,regular expressions,pumping lemma for regular languages.

finite automata,regular languages,

regular expressions,pumping lemma for regular languages.

finite automata,regular languages,regular expressions,

pumping lemma for regular languages.

pushdown automata,

context-free languages,context-free grammars,pumping lemma for context-free languages.

pushdown automata,context-free languages,

context-free grammars,pumping lemma for context-free languages.

pushdown automata,context-free languages,context-free grammars,

pumping lemma for context-free languages.

Context-Free GrammarsThis is an example of a context free grammer, G1:

A → 0A1

A → B

B → #

Terminology:

Each line is a substitution rule or production.

Each rule has the form: symbol → string.The left-hand symbol is a variable(usually upper-case).

A string consists of variables and terminals.

One variable is the start variable.

A → 0A1

A → B

B → #

Terminology:

A → 0A1

A → B

B → #

Terminology:

A → 0A1

A → B

B → #

Terminology:

Rules for Generating Strings

Write down the start variable (lhs of top rule).

Pick a variable written down in current string anda derivation that starts with that variable.

Replace that variable with right-hand side of thatderivation.

Repeat until no variables remain.

Return final string (concatenation of terminals).

Example

Grammar G1:

A → 0A1

A → B

B → #

Derivation with G1:

A ⇒ 0A1

⇒ 00A11

⇒ 000A111

⇒ 000B111

⇒ 000#111

Example

Grammar G1:

A → 0A1

A → B

B → #

Derivation with G1:

A ⇒ 0A1

⇒ 00A11

⇒ 000A111

⇒ 000B111

⇒ 000#111

A Parse TreeA

#0 10 0 1 1

Question: What strings can be generated in this wayfrom the grammar G1?

Answer: Exactly those of the form 0n#1n (n ≥ 0).

A Parse TreeA

#0 10 0 1 1

A Parse TreeA

#0 10 0 1 1

Context-Free Languages

The language generated in this way is the language ofthe grammar.

For example, L(G1) is {0n#1n|n ≥ 0}.

Any language generated by a context-free grammar is

called a context-free language.

A Useful AbbreviationRules with same variable on left hand side

A → 0A1

A → B

are written as:

A → 0A1 | B

A Useful AbbreviationRules with same variable on left hand side

A → 0A1

A → B

are written as:

A → 0A1 | B

English-like Sentences

A grammar G2 to describe a few English sentences:

< SENTENCE > → < NOUN-PHRASE >< VERB >

< NOUN-PHRASE > → < ARTICLE >< NOUN >

< NOUN > → boy | girl | flower

< ARTICLE > → a | the

< VERB > → touches | likes | sees

Deriving English-like Sentences

A specific derivation in G2:

< SENTENCE > ⇒ < NOUN-PHRASE >< VERB >

⇒ < ARTICLE >< NOUN >< VERB >

⇒ a < NOUN >< VERB >

⇒ a boy < VERB >

⇒ a boy sees

More strings in G2:

a flower sees

the girl touches

Deriving English-like Sentences

A specific derivation in G2:

⇒ a boy < VERB >

⇒ a boy sees

More strings in G2:

a flower sees

the girl touches

Derivation and Parse Tree

⇒ a boy < VERB >

⇒ a boy sees

SENTENCE

NOUN-PHRASE VERB

ARTICLE NOUN

a boy seesSlides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. – p.10

Formal DefinitionsA context-free grammar is a 4-tuple (V, Σ, R, S)where

V is a finite set of variables,

Σ is a finite set of terminals,

R is a finite set of rules: each rule is a variableand a finite string of variables and terminals.

S is the start symbol.

Formal Definitions

If u and v are strings of variables and terminals,

and A → w is a rule of the grammar, then

we say uAv yields uwv, written uAv ⇒ uwv.

We write u∗⇒ v if u = v or

u ⇒ u1 ⇒ . . . ⇒ uk ⇒ v.

for some sequence u1, u2, . . . , uk.

Definition: The language of the grammar is{

w | S ∗⇒ w}

Formal Definitions

u ⇒ u1 ⇒ . . . ⇒ uk ⇒ v.

w | S ∗⇒ w}

Formal Definitions

u ⇒ u1 ⇒ . . . ⇒ uk ⇒ v.

w | S ∗⇒ w}

Formal Definitions

u ⇒ u1 ⇒ . . . ⇒ uk ⇒ v.

w | S ∗⇒ w}

Formal Definitions

u ⇒ u1 ⇒ . . . ⇒ uk ⇒ v.

w | S ∗⇒ w}

Example

Consider G4 = (V, {a, b} , R, S).

R (Rules): S → aSb | SS | ε .

Some words in the language: aabb, aababb.

Q.: But what is this language?

Hint: Think of parentheses.

Example

Arythmetic Example

Consider (V, Σ, R,E) where

V = {E, T, F}Σ = {a, +,×, (, )}

Rules:E → E + T | TT → T × F | FF → (E) | a

Strings generated by the grammer:a + a× a and (a + a)× a.

What is the language of this grammer?Hint: arithmetic expressions.

E = expression, T = term, F = factor.

Arythmetic Example

V = {E, T, F}Σ = {a, +,×, (, )}

E = expression, T = term, F = factor.

Arythmetic Example

V = {E, T, F}Σ = {a, +,×, (, )}

E = expression, T = term, F = factor.Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. – p.14

Parse Tree for a + a× a

E → E + T |TT → T × F | FF → (E) | a

Parse Tree for a + a× a

E → E + T |TT → T × F | FF → (E) | a

Parse Tree for (a + a)× a

E → E + T | TT → T × F | FF → (E) | a

( a + aX)a

Parse Tree for (a + a)× a

E → E + T | TT → T × F | FF → (E) | a

( a + aX)a

Designing Context-Free Grammars

No recipe in general, but few rules-of-thumb

If CFG is the union of several CFGs, renamevariables (not terminals) so they are disjoint, andadd new rule S → S1 | S2 | . . . | Si.

To construct CFG for a regular language,“follow” a DFA for the language. For initial stateq0, make R0 the start variable. For state transitionδ(qi, a) = qj add rule Ri → aRj to grammer. Foreach final state qf , add rule Rf → ε to grammer.

For languages with linked substrings (like{0n#1n|n ≥ 0} ), a rule of form R → uRv maybe helpful, forcing desired relation betweensubstrings.

Closure Properties

Regular languages are closed under

unionconcatenationstar

Context-Free Languages are closed underunion : S → S1 | S2

concatenation S → S1S2

star S → ε | SS

Closure Properties

Regular languages are closed underunion

concatenationstar

star S → ε | SS

Closure Properties

Regular languages are closed underunionconcatenation

star S → ε | SS

Closure Properties

Regular languages are closed underunionconcatenationstar

star S → ε | SS

Closure Properties

Context-Free Languages are closed under

union : S → S1 | S2

star S → ε | SS

Closure Properties

star S → ε | SS

Closure Properties

star S → ε | SS

Closure Properties

star S → ε | SS

More Closure Properties

Regular languages are also closed under

complement (reverse accept/non-accept statesof DFA)

intersection(

L1 ∩ L2 = L1 ∪ L2

What about complement and intersection ofcontext-free languages?

Not clear . . .

Regular languages are also closed undercomplement (reverse accept/non-accept statesof DFA)

intersection(

L1 ∩ L2 = L1 ∪ L2

Not clear . . .

intersection(

L1 ∩ L2 = L1 ∪ L2

Not clear . . .

intersection(

L1 ∩ L2 = L1 ∪ L2

Not clear . . .

intersection(

L1 ∩ L2 = L1 ∪ L2

Not clear . . .

Ambiguity

Grammar: E → E+E | E×E | (E) | a

Ambiguity

We say that a string w is derived ambiguously fromgrammer G if w has two or more parse trees thatgenerate it from G.

Ambiguity is usually not only a syntactic notion butalso a semantic one, implying multiple meanings forthe same string.

It is sometime possible to eliminate ambiguity byfinding a different context free grammer generatingthe same language. This is true for the grammerabove, which can be replaced by unambiguousgrammer from slide (14).

Some languages (e.g. {1i2j3k | i = j or j = k} are in-

herentrly ambigous.

Ambiguity

herentrly ambigous.

Ambiguity

herentrly ambigous.

Ambiguity

herentrly ambigous.Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. – p.21

Chomsky Normal Form

A simplified, canonical form of context freegrammers.Every rule has the form

A → BC

A → a

S → ε

where S is the start symbol, A, B and C are any vari-

able, except B and C not the start symbol.

TheoremTheorem: Any context-free language is generated bya context-free grammar in Chomsky normal form.Basic idea:

Add new start symbol S0.

Eliminate all ε rules of the form A → ε.

Eliminate all “unit” rules of the form A → B.

Patch up rules so that grammar generates thesame language.

Convert remaining long rules to proper form.

ProofAdd new start symbol S0 and rule S0 → S.

Guarantees that new start symbol does not appear on

right-hand-side of a rule.

ProofEliminating ε rules.

Repeat:

remove some A → ε.

for each R → uAv, add rule R → uv.

and so on: for R → uAvAw add R → uvAw,R → uAvw, and R → uvw.

for R → A add R → ε, except if R → ε hasalready been removed.

until all ε-rules not involving the start variable have

been removed.

Repeat:

been removed.

Repeat:

been removed.

Repeat:

been removed.

Repeat:

been removed.

ProofEliminate unit rules.

Repeat:

remove some A → B.

for each B → u, add rule A → u, unless this ispreviously removed unit rule. (u is a string ofvariables and terminals.)

until all unit rules have been removed.

Repeat:

ProofFinally, convert long rules.To replace each A → u1u2 . . . uk (for k ≥ 3),introduce new non-terminals

N1, N2, . . . , Nk−1

and rules

A → u1N1

N1 → u2N2

...Nk−3 → uk−2Nk−2

Nk−2 → uk−1uk ♠

ProofFinally, convert long rules.To replace each A → u1u2 . . . uk (for k ≥ 3),introduce new non-terminals

N1, N2, . . . , Nk−1

and rules

A → u1N1

N1 → u2N2

...Nk−3 → uk−2Nk−2

Nk−2 → uk−1uk ♠

Conversion Example

Initial Grammar:

S → ASA | aB

A → B | SB → b | ε

(1) Add new start state:

S0 → S

S → ASA | aB

A → B | SB → b | ε

Conversion Example (2)

S0 → S

S → ASA | aB

A → B | SB → b | ε

(2) Remove ε-rule B → ε:

S0 → S

S → ASA | aB | aA → B | S | εB → b | ε

S0 → S

S → ASA | aB | aA → B | S | εB → b

(3) Remove ε-rule A → ε:

S0 → S

S → ASA | aB | a | AS | SA | SA → B | S | εB → b

S0 → S

S → ASA | aB | a | AS | SA | SA → B | SB → b

(4) Remove unit rule S → S

S0 → S

S → ASA | aB | a | AS | SA | SA → B | SB → b

S0 → S

S → ASA | aB | a | AS | SA

A → B | SB → b

(5) Remove unit rule S0 → S:

S0 → S | ASA | aB | a | AS | SA

A → B | SB → b

S0 → ASA | aB | a | AS | SA

A → B | SB → b

(6) Remove unit rule A → B:

A → B | S | bB → b

A → S | bB → b

Remove unit rule A → S:

A → S | b | ASA | aB | a | AS | SA

B → bSlides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. – p.34

A → b | ASA | aB | a | AS | SA

B → b

(8) Final simplification – treat long rules:

S0 → AA1 | UB | a | SA | AS

S → AA1 | UB | a | SA | AS

A → b | AA1 | UB | a | SA | AS

A1 → SA

U → a

B → b√

lecture 4 – outline - tau › ~bchor › cm › compute4.pdf · lecture 4 – outline so far we...

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