intro to theory of computation

04/19/2023

Sofya Raskhodnikova

Intro to Theory of Computation

LECTURE 6 Theory of Computation• Proving that a language is

not regular• Pushdown automaton

(PDA)• Context-free grammar

(CFG)

Sofya Raskhodnikova; based on slides by Nick Hopper 6.1

THE PUMPING LEMMA

Let L be a regular language with |L| =

Then there exists a length p such that

1. |y| > 0

2. |xy| ≤ p

3. xyiz L for all i ≥ 0

if w L and |w| ≥ p thenw can be split into three parts w=xyz where:

Sofya Raskhodnikova; based on slides by Nick Hopper04/19/2023 6.2

GENERAL STRATEGY

Proof by contradiction – assume L is regular

Find a string w 2 L with |w| ¸ p

Show that no matter how an adversarychooses xyz, your w cannot be pumped!

Conclude that L is not regular

Then there is a pumping length p

Sofya Raskhodnikova; based on slides by Nick Hopper L6.304/19/2023

PUMPING DOWNProve C = { 0i1j | i > j ¸ 0} is not regular.Proof: Assume … pumping length p

Find a w 2 C longer than p

Show that … w cannot be pumped:0P+11P

00…0011…11

p+1 p

w=

y

xyyz = 00…00011…11

> p+1P

xz = 0…0011…11

· p PSofya Raskhodnikova; based on slides by Nick Hopper L6.4

USING THE PUMPING LEMMA:

04/19/2023

PUMPING DOWN

Prove C = { 0i1j | i > j ¸ 0} is not regular.Proof: Assume … pumping length p

Find a w 2 C longer than p

If w = xyz with |xy| · p theny = 0J for some J¸ 1.Then xy0z = xz = 0P+1-J1P C

0P+11P

xz = 0…0011…11

Sofya Raskhodnikova; based on slides by Nick Hopper L6.5

Contradiction!

04/19/2023

For every p, w L with |w| ≥ p

1. |y| > 0there exist xyz=w, where


2. |xy| ≤ p

WHAT THE PUMPING LEMMADOESN’T SAY - I

Let L = 1*. L is regular.The string 10 cannot be pumped with p=1.

You need to show that no matter what p is chosen,L has a string w, |w| ≥ p, that cannot be pumped.


WHAT THE PUMPING LEMMADOESN’T SAY - II

For every xyz=w, where:

Example: Let L = 10*1*. Pickingw = 10P1P; x=ε; y=10P; z=1P

does not contradict pumping lemma!

(You can pick x=1, y=0, z=0P-11P.)

1. |y| > 0


2. |xy| ≤ p


CHOOSING WISELY

Let BALANCED = { w | w has an equal # of 1sand 0s }

Assume … there is a p

Find a w 2 BALANCED longer than p

(01)P

Show that w cannot be pumped:

If w = xyz with |xy| · p theny = 0J for some J>0.

Then xyyz = 0P+J1P BALANCED

0P1P


REUSING A PROOF

Pumping a language can be lots of work…

Let’s try to reuse that work!

{0n1n : n ¸ 0} = BALANCED ∩ 0*1*

If BALANCED is regular then so is {0n1n: n ¸ 0}


USING CLOSURE

A C B∩ =

(Not regular)

Prove: A is not regular

or any of {¢, [, ∩, :, R, *}

If A is regular, then A ∩ C (= B) is regular.

(regular)

But B is not regular so neither is A.


Prove A = {0i1j : i j} is not regular using B = {0n1n : n¸0}

B [ { strings that mix 0s and 1s }

:A ∩ 0*1* = B

:A =

USING CLOSURE: example


PUMPINGNON-REGULAR LANGUAGES

Let F = {aibjck | i, j, k ¸ 0 and i=1 ) j=k}

F has pumping length 2!

HW3: prove that F is not regular.

i = 0i = 1i = 2

A non-regular language may still satisfy the pumping lemma.


464 so far

MODEL OF A PROGRAM: DFA

MODEL OF A PROBLEM: LANGUAGE

EQUIVALENT MODELS: NFA, REGEXP

PROBLEMS THAT A DFA CAN’T SOLVE

ARE WE DONE?


• Σ = {0, 1}, L = { 0n1n | n ≥ 0 }

• Σ = {a, b, c, …, z}, L = { w | w = wR }

• Σ = { (, ) }, L = { balanced strings of parens }

NONE OF THESE ARE REGULAR

We can write a SCHEME or JAVA program for any of them!


PUSHDOWN AUTOMATA (PDA)

FINITE STATE CONTROL

STACK(Last in, first out)

INPUT


ε,ε → $0,ε → 0

1,0 → ε

1,0 → εε,$ → ε

string pop push

0011

STACK $

0011011

$0

11

$0

1

ε,ε → $0,ε → 0

1,0 → ε

1,0 → εε,$ → ε

string pop push

001

STACK $ $0 $0

101001

PDA to recognize L = { 0n1n | n ≥ 0 }

Formal Definition

04/19/2023

Q is a finite set of statesΣ is the alphabet

q0 Q is the start stateF Q is the set of accept states

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

Sofya Raskhodnikova; based on slides by Nick Hopper

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

P(Q Γε) is the set of subsets of Q Γε and Σε = Σ {ε}

Γ is the stack alphabet

Note: A PDA is defined to be nondeterministic.

6.18

Formal Definition

04/19/2023

Q is a finite set of statesΣ is the alphabet

q0 Q is the start stateF Q is the set of accept states

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

Sofya Raskhodnikova; based on slides by Nick Hopper

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

Γ is the stack alphabet

A PDA starts with an empty stack. It accepts a string if at least one of its computational branches reads all the input and gets into an accept state at the end of it.

6.19

ε,ε → $0,ε → 0

1,0 → ε

1,0 → εε,$ → ε

q0 q1

q2q3

Q = {q0, q1, q2, q3} Γ =Σ =

: Q Σε Γε → P(Q Γε)

{0,1} {$,0}

(q1,1,0) = { (q2,ε) } (q2,1,1) =

EVEN-LENGTH PALINDROMES

Σ = {a, b, c, …, z}

ε,ε → $

ε,ε → ε

, → εε,$ → ε

q0 q1

q2q3

,ε →

Build a PDA to recognize L = { aibjck | i, j, k ≥ 0 and (i = j or i = k) }

ε,ε → $

b,a → ε

ε,$ → εq0

q5q1

q3

a,ε → a

q2

q4

ε,ε →

ε

q6ε,ε → ε ε,ε → ε ε,$ → ε

b,ε → ε c,a → ε

c,ε → ε

A → 0A1A → BB → #

CONTEXT-FREE GRAMMARS (CFGs)

A

variablesterminals

productionrulesstart variable

0A1 00A11 00B11 00#11

04/19/2023 Sofya Raskhodnikova; based on slides by Nick Hopper 6.23

<PHRASE> → <START><END>

<PHRASE> → <FILLER><PHRASE>

<FILLER> → LIKE

<FILLER> → UMM

<START> → YOU KNOW

<END> → GAG ME WITH A SPOON

<START> → ε

<END> → AS IF

<END> → WHATEVER

VALLEY GIRL GRAMMAR

<END> → LOSER

04/19/2023 Sofya Raskhodnikova; based on slides by Nick Hopper 6.24

intro to theory of computation

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