BU CS 332 Theory of Computation€¦ · BU CS 332 –Theory of Computation Lecture 7: •More on...

BU CS 332 – Theory of Computation

Lecture 7:

• More on CFGs

• Pushdown Automata

Reading:

Sipser Ch 2.1-2.3

Mark Bun

February 12, 2020

Context-Free Grammar (Formal)

A CFG is a 4-tuple 𝐺 = 𝑉, Σ, 𝑅, 𝑆

• 𝑉 is a finite set of variables

• Σ is a finite set of terminal symbols (disjoint from 𝑉)

• 𝑅 is a finite set of production rules of the form 𝐴 → 𝑤, where 𝐴 ∈ 𝑉 and 𝑤 ∈ (𝑉 ∪ Σ)∗

• 𝑆 ∈ 𝑉 is the start symbol

2/11/2020 CS332 - Theory of Computation 2

Context-Free Grammar

Example Grammar 𝐺 Parse Tree

𝑆 → 0𝑆1𝐴 | 1𝑆0𝐴 | 휀

𝐴 → # | 휀

Derivation


Context-Free Languages


𝐿 is a context-free language if it is the

language of some CFG

Questions about CFLs

1. Which languages are notcontext-free?

2. How do we recognize whether 𝑤 ∈ 𝐿?

3. What are the closure properties of CFLs?

Pumping Lemma for context-free languages


Let 𝐿 be a context-free language.

Then there exists a “pumping length” 𝑝 such that

1. |𝑣𝑦| > 0

2. |𝑣𝑥𝑦| ≤ 𝑝

3. 𝑢𝑣𝑖𝑥𝑦𝑖𝑧 𝐿 for all 𝑖 ≥ 0

For every 𝑤 ∈ 𝐿 where |𝑤| ≥ 𝑝,𝑤 can be split into five parts 𝑤 = 𝑢𝑣𝑥𝑦𝑧 where:

Pumping Lemma example


Claim: 𝐿 = 𝑎𝑛𝑏𝑛𝑐𝑛 𝑛 ≥ 0} is not context-free

Proof: Assume 𝐿 is context-free with pumping length 𝑝

1. Find 𝑤 ∈ 𝐿 with 𝑤 ≥ 𝑝2. Show that 𝑤 cannot be pumped

If 𝑤 = 𝑢𝑣𝑥𝑦𝑧 with |𝑣𝑦| > 0,|𝑣𝑥𝑦| ≤ 𝑝, then…

Case 1: 𝑣, 𝑦 both contain only one kind of symbol

Case 2: Either 𝑣 or 𝑦 contains two kinds of symbols







Case 1: 𝑣, 𝑦 both contain only one kind of symbol







Case 2: Either 𝑣 or 𝑦 contains two kinds of symbols

Pumping Lemma: Proof idea

Let 𝐿 be a context-free language. If 𝑤 ∈ 𝐿 is long enough, then every parse tree for 𝑤 has a repeated variable.


Pumping Lemma Proof

What does “long enough” mean? (How do we choose the pumping length 𝑝?)

• Let 𝐺 be a CFG for 𝐿

• Suppose the right-hand side of every rule in 𝐺 uses at most 𝑏 symbols

• Let 𝑝 = 𝑏 𝑉 +1

Claim: If 𝑤 ∈ 𝐿 with 𝑤 ≥ 𝑝, then the smallest parse tree for 𝑤 has height at least 𝑉 + 1


Pumping Lemma Proof

Claim: If 𝑤 ∈ 𝐿 with 𝑤 ≥ 𝑝, then the smallest parse tree for 𝑤 has height at least 𝑉 + 1

• By the pigeonhole principle, there is a path down the parse tree with a repeated variable 𝑅

• Choose two such occurrences within the bottom 𝑉 + 1 levels


Pushdown Automata



Regular Expressions : Finite Automata ::

Context-Free Languages : ???

Pushdown Automata


Input 𝑎 𝑏 𝑎 𝑎

Finite control

…


Finite control

…

𝑥

𝑥

𝑦 Memory: Infinite Stack

Pushdown Automata (PDAs):Machine with unbounded structured memory in the form of a stack

Finite Automata (FAs):Machine with a finiteamount of unstructured memory

Pushdown Automaton (the idea)

• Nondeterministic finite automaton + stack

• Stack has unlimited size, but machine can only manipulate (push, pop, read) symbol at the top



Finite control

…

𝑥

𝑥

𝑦 Memory: Infinite Stack

Transitions of the form:

𝑎, 𝑥 → 𝑥′𝑝 𝑞

Example: Even Palindromes

𝐿 = {𝑤𝑤𝑅|𝑤 ∈ 0,1 ∗}


Input 𝑎 𝑏 𝑏 𝑏 𝑏 𝑎

Finite control

…

Memory: Infinite Stack


𝐿 = {𝑤𝑤𝑅|𝑤 ∈ 0,1 ∗}


Input 𝑎 𝑏 𝑏 𝑏 𝑏 𝑎

Finite control

…

Memory: Infinite Stack

Algorithmic Description1. Place the marker $ on the stack2. Nondeterministically, either

a) Read a character and push it to the stack, orb) Go to the next step

3. Nondeterministically, eithera) Pop the stack if it matches the next character orb) Go to the next step

4. Accept if the top of the stack is $



휀, 휀 → $ 𝑎, 휀 → 𝑎

휀, 휀 → 휀

𝑎, 𝑎 → 휀휀, $ → 휀

𝑞0 𝑝

𝑞𝑞𝑓

𝑏, 휀 → 𝑏

𝑏, 𝑏 → 휀

Pushdown Automaton (formal)

A PDA is a 6-tuple (sorry) 𝑀 = (𝑄, Σ, Γ, 𝛿, 𝑞0, 𝐹)

• 𝑄 is a finite set of states

• Σ is the input alphabet

• Γ is the stack alphabet

• 𝛿 : 𝑄 × Σ𝜀 × Γ𝜀 → 𝑃(𝑄 × Γ𝜀) is the transition function

• 𝑞0 is the start state

• 𝐹 is the set of final states


𝑀 accepts a string 𝑤 if, starting from 𝑞0 and an empty stack, there exists a path to an accept state that can be followed by reading all of 𝑤.



휀, 휀 → $ 𝑎, 휀 → 𝑎

휀, 휀 → 휀

𝑎, 𝑎 → 휀휀, $ → 휀

𝑞0 𝑝

𝑞𝑞𝑓

𝑏, 휀 → 𝑏

𝑏, 𝑏 → 휀

𝑄 =Σ =Γ =𝐹 =

𝛿 : 𝑄 × Σ𝜀 × Γ𝜀 → 𝑃 𝑄 × Γ𝜀

𝛿 𝑝, 𝑏, 휀 =𝛿 𝑞, 𝑎, 𝑎 =𝛿 𝑞, 𝑎, 𝑏 =

Example𝐿 = {𝑤|𝑤 has an equal number of 𝑎’s and 𝑏’s}


CFGs vs. PDAs

The language 𝐿(𝑀) of a PDA 𝑀 is the set of all strings it accepts.

Theorem (next time): The class of languages recognized by PDAs is exactly the context-free languages.


Closure Properties

• The class of CFLs is closed under the regular operations union, concatenation, star

• Beware: It is not closed under intersection or complement(Find a counterexample!)


Closure under union (Proof 1)

Let 𝐴 be a CFL recognized by PDA 𝑀𝐴 and let 𝐵 be a CFL recognized by PDA 𝑀𝐵

Goal: Construct a PDA recognizing 𝐴 ∪ 𝐵


Closure under union (Proof 2)

Let 𝐴 be a CFL generated by CFG 𝐺𝐴 and let 𝐵 be a CFL recognized by CFG 𝐺𝐵Goal: Construct a CFG 𝐺 recognizing 𝐴 ∪ 𝐵

𝐺𝐴 = (𝑉𝐴, Σ𝐴, 𝑅𝐴, 𝑆𝐴)

𝐺𝐵 = (𝑉𝐵 , Σ𝐵 , 𝑅𝐵 , 𝑆𝐵)

Relabel variables so 𝑉𝐴 and 𝑉𝐵 are disjoint

Let 𝐺 = (𝑉, Σ , 𝑅 , 𝑆 )


BU CS 332 Theory of Computation€¦ · BU CS 332 –Theory of Computation Lecture 7: •More on...

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