Proof by The Pumping Lemma Observation

PumpingLemma

Mindmap

ProofsProof by existence

Proof bycontradiction

ObservationUnary alphabet

Pigeon-hole principle

Binary alphabet

PumpingLemmaPL Game!

Examples

Pumping down

ImplicationsConstant Space

Limitations of the Regular LanguagesThe Pumping Lemma

Dr Kamal Bentahar

School of Computing, Electronics and MathematicsCoventry University

Lecture 4

0 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

PumpingLemma

Theoryof CS

Models ofComputation

DFA/NFA

NFA→DFA

RegExGNFA

TMDecidability

“Haltingproblem”

TuringUnrecog-nizable

Reduction

Complexity

NP-hard

NP-complete

0 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Last week. . .

Regular Languages

The class of regular languages can be:1 Recognized by NFAs. (equiv. GNFA or ε-NFA or NFA or DFA).2 Described using Regular Expressions.

Today:1 See the limit of regular languages.2 How to show a language is not regular.

1 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Types of proofs

We show a language is regular using “proof by existence”:Construct an NFA recognizing it.Write a Regular Expression for it.Using closure under the union, concatenation and star operations.

However, if a languages is not regular then how can we show that?!

2 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Types of proofs

We show a language is regular using “proof by existence”:Construct an NFA recognizing it.Write a Regular Expression for it.Using closure under the union, concatenation and star operations.

However, if a languages is not regular then how can we show that?!

2 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Is it raining now? – example of proof by contradiction

Is it raining now?

Suppose it is.Let us go outside where it is supposed to be raining.

If it is raining then we should get wet.(No umberlla, etc.)

However, we did not get wet!Thus, it is not raining!

3 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Is it raining now?Suppose it is.

Let us go outside where it is supposed to be raining.

3 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Is it raining now?Suppose it is.Let us go outside where it is supposed to be raining.

3 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

3 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

However, we did not get wet!

Thus, it is not raining!

3 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

3 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Eulerian paths – example of proof by contradictionIs it possible to traversethis graph by travellingalong each edgeexactly once?

Suppose it is possible.How many times would each vertex be visited?

Every time a vertex is entered, it is also exited.So, each vertex must have an even number ofneighbours.The starting and ending vertices are exceptions:odd number of neighbours.There can only be 0 or 2 such exceptions.

However, this graph has 4 exceptions!Thus, it is impossible to traverse this graph bytravelling along each path exactly once.

4 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Suppose it is possible.

How many times would each vertex be visited?

4 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

4 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Every time a vertex is entered, it is also exited.

So, each vertex must have an even number ofneighbours.The starting and ending vertices are exceptions:odd number of neighbours.There can only be 0 or 2 such exceptions.

4 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Every time a vertex is entered, it is also exited.So, each vertex must have an even number ofneighbours.

The starting and ending vertices are exceptions:odd number of neighbours.There can only be 0 or 2 such exceptions.

4 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Every time a vertex is entered, it is also exited.So, each vertex must have an even number ofneighbours.The starting and ending vertices are exceptions:odd number of neighbours.

There can only be 0 or 2 such exceptions.

4 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

4 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

However, this graph has 4 exceptions!

Thus, it is impossible to traverse this graph bytravelling along each path exactly once.

4 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

4 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Types of proofs – Proof by contradiction

To prove a language is not regular, we can use proof by contradiction.We need a property that all regular languages must satisfy.Then, if a given language does not satisfy it then it cannot be regular.

Let us try to understand the regular languages (RLs) a bit more. . .Let us examine some examples in the next few slides. . .For each automaton, let us think about the path taken by an acceptedstring – is it “straight” or does it loop?

5 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Types of proofs – Proof by contradiction

To prove a language is not regular, we can use proof by contradiction.We need a property that all regular languages must satisfy.Then, if a given language does not satisfy it then it cannot be regular.

Let us try to understand the regular languages (RLs) a bit more. . .Let us examine some examples in the next few slides. . .For each automaton, let us think about the path taken by an acceptedstring – is it “straight” or does it loop?

5 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Unary alphabet {1} – Strings of length 3,5,7,9, . . .

111 takes a “straight path” to the accept state11111 goes through a loop.Repeating the looped part produces longer strings:11 11 11 1, 11 11 11 11 1, 11 11 11 11 11 1, . . .In fact, we can also omit the 11 loop to get: 111.

We say: we pump the substring 11.

6 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

111 takes a “straight path” to the accept state

11111 goes through a loop.Repeating the looped part produces longer strings:11 11 11 1, 11 11 11 11 1, 11 11 11 11 11 1, . . .In fact, we can also omit the 11 loop to get: 111.

6 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

111 takes a “straight path” to the accept state11111 goes through a loop.

Repeating the looped part produces longer strings:11 11 11 1, 11 11 11 11 1, 11 11 11 11 11 1, . . .In fact, we can also omit the 11 loop to get: 111.

6 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

111 takes a “straight path” to the accept state11111 goes through a loop.Repeating the looped part produces longer strings:11 11 11 1, 11 11 11 11 1, 11 11 11 11 11 1, . . .

In fact, we can also omit the 11 loop to get: 111.

6 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

6 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

We say: we pump the substring 11.6 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

(111)∗+(11111)∗

Set of accepted strings is:{ε,13,16,19, . . .} ∪ {ε,15,110,115, . . .}

111 can be pumped to give:(111)0 = ε,(111)1 = 13,(111)2 = 16,(111)3 = 19, . . .

The shortest string that can bepumped is: 111.3, the length of 111, is called:pumping length.

6 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

(111)∗+(11111)∗

6 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

(111)∗+(11111)∗

6 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

(111)∗+(11111)∗

The shortest string that can bepumped is: 111.

3, the length of 111, is called:pumping length.

6 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

(111)∗+(11111)∗

6 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Unary alphabetLet L be a regular language over a unary alphabet Σ = {1}.The language L is:

either finite, in which case it is regular trivially,or infinite, in which case its DFA will have to loop:

The DFA that recgnizes L has a finite number of states.Any string in L determines a path through the DFA.So any sufficiently long string must visit a state twice.This forms a loop.

This looped part can be repeated any arbitrary number of times to produceother strings in L.

If we put more than n pigeons into n holes then there must be a hole withmore than one pigeon in.

7 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Unary alphabetLet L be a regular language over a unary alphabet Σ = {1}.The language L is:

either finite, in which case it is regular trivially,or infinite, in which case its DFA will have to loop:

The DFA that recgnizes L has a finite number of states.Any string in L determines a path through the DFA.So any sufficiently long string must visit a state twice.This forms a loop.

This looped part can be repeated any arbitrary number of times to produceother strings in L.

If we put more than n pigeons into n holes then there must be a hole withmore than one pigeon in.

7 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

aΣ∗b

8 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

aΣ∗

9 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Σ∗b

10 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Σ∗1Σ

11 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Σ∗aaa

12 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

aabΣ∗ + abaΣ∗

a ba,b

b aa,b

13 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

(Σ11)∗Σ

14 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

(0Σ∗(01 + 10))∗

15 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

A property satisfied by all RLsFinite number of states→ DFA repeats one or more states if the string is long.

S R Fx

When a DFA repeats a state R, divide the string into 3 parts:1 The substring x before the first occurrence of R2 The substring y between the first and last occurrence of R3 The substring z after the last occurrence of R

x , z can be ε but y cannot be ε. (y forms a genuine loop.)Then, if the DFA accepts xyz then it accepts all of:xz, xyz, xyyz, xyyyz, . . .

For any RL L, it is possible to divide an accepted string, that is “long enough”,into 3 substrings xyz, in such a way that xy∗z is a subset of L.

16 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

S R Fx

16 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

S R Fx

16 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

S R Fx

x , z can be ε but y cannot be ε. (y forms a genuine loop.)

Then, if the DFA accepts xyz then it accepts all of:xz, xyz, xyyz, xyyyz, . . .

16 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

S R Fx

16 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

S R Fx

For any RL L, it is possible to divide an accepted string, that is “long enough”,into 3 substrings xyz, in such a way that xy∗z is a subset of L. 16 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

The Pumping Lemma (PL)We will denote a pumping length by p.The precise meaning of “long enough” will be: |w | ≥ p.y has to be in the first p symbols of w .

Pumping Lemma (PL)

Let L be a regular language. Then, there exists a constant p such that everystring w from L, with |w | ≥ p, can be broken into three substrings xyz suchthat

1 y 6= ε (or equivalently: |y | 6= 0 or |y | > 0)2 |xy | ≤ p (y is in the first p symbols of w)3 For any k ≥ 0, the string xykz is also in L (xy∗z ⊂ L)

Its main purpose in practice is to prove that a language is not regular.That is, if we can show that a language L does not have this property, then we conclude that Lcannot be recognized by a DFA/NFA or expressed as a regular expression.

17 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Pumping Lemma (PL)

17 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Pumping Lemma (PL)

17 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

The PL Game!When the PL is used to prove that a language L is not regular, the proof canbe viewed as a “game” between a Prover and a Falsifier as follows:

Ê Prover claims L is regular andfixes a pumping length p.

Ë Falsifier challenges Prover andpicks a string w ∈ L of length atleast p symbols.

Ì Prover writes w = xyz where|xy | ≤ p and y 6= ε.

Í Falsifier wins by finding a valuefor k such that xykz is not in L.

If Falsifier always wins then L is not regular.If Prover always wins then L may be regular.

18 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

18 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

18 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

18 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

18 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

18 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Example (L = {anbn | n ≥ 0})Ê Prover claims L is regular and fixesa pumping length p.

Ë Falsifier challenges Prover andpicks w = apbp ∈ L. (|w | = 2p ≥ p)

Ì Prover tries to split w =a . . . ab . . . b into xyz such that |xy | ≤ p

a . . .x

. . . . . .y

. . . ab . . . . . . bz

Since y must be within the first p sym-bols then y is made of a’s only.

w = a . . . . . . . . . . . .ap symbols

b . . . . . . . . .bp symbols

Í Falsifier now can for example build

xy2z = xyyz = a . . . . . . . . . . . .amore than p symbols

b . . . . . . . . . bstill p symbols

Hence xy2z 6∈ L, and L is not regular.

19 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

a . . .x

. . . . . .y

. . . ab . . . . . . bz

w = a . . . . . . . . . . . .ap symbols

b . . . . . . . . .bp symbols

19 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

a . . .x

. . . . . .y

. . . ab . . . . . . bz

w = a . . . . . . . . . . . .ap symbols

b . . . . . . . . .bp symbols

19 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

a . . .x

. . . . . .y

. . . ab . . . . . . bz

w = a . . . . . . . . . . . .ap symbols

b . . . . . . . . .bp symbols

Hence xy2z 6∈ L, and L is not regular.19 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Example (L = {ww | w ∈ {0,1}∗})Ê Prover claims L is regular and fixesa pumping length p.

Ë Falsifier challenges Prover andChoose w = 0p1 0p1 ∈ L. This haslength |w | = (p + 1) + (p + 1) ≥ p.Ì Prover tries to split w =

0 . . . 01 0 . . . 01 into xyz such that|xy | ≤ p

0 . . .x

. . . . . .y

. . . 010 . . . . . . 01z

Since y must be within the first p sym-bols then y is made of 0’s only.

w = 0 . . . . . . . . . . . .0p symbols

1 0 . . . . . . . . .0p symbols

Í Falsifier pumps y to produce

xy2z = 0 . . . . . . . . . . . .0more than p symbols

1 0 . . . . . . . . .0still p symbols

20 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Ë Falsifier challenges Prover andChoose w = 0p1 0p1 ∈ L. This haslength |w | = (p + 1) + (p + 1) ≥ p.

Ì Prover tries to split w =0 . . . 01 0 . . . 01 into xyz such that|xy | ≤ p

0 . . .x

. . . . . .y

. . . 010 . . . . . . 01z

w = 0 . . . . . . . . . . . .0p symbols

1 0 . . . . . . . . .0p symbols

1 0 . . . . . . . . .0still p symbols

20 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

0 . . .x

. . . . . .y

. . . 010 . . . . . . 01z

w = 0 . . . . . . . . . . . .0p symbols

1 0 . . . . . . . . .0p symbols

1 0 . . . . . . . . .0still p symbols

20 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

0 . . .x

. . . . . .y

. . . 010 . . . . . . 01z

w = 0 . . . . . . . . . . . .0p symbols

1 0 . . . . . . . . .0p symbols

1 0 . . . . . . . . .0still p symbols

20 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Example (L = {aibj | i > j})Ê Prover claims L is regular and fixesa pumping length p.

Ë Falsifier challenges Prover andchooses w = ap+1bp.Here |w | = (p + 1) + p ≥ p.

Ì Prover splits w = a . . . aab . . . b intoxyz:

a . . .x

. . . . . .y

. . . aab . . . . . . bz

So y is made of a’s only.

w = a . . . . . . . . . . . .ap + 1 symbols

b . . . . . . . . .bp symbols

Í Falsifier pumps y down and formsxy0z = xz

xy0z = a . . . . . . . . . . . .aat most p symbols

b . . . . . . . . .bstill p symbols

21 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

a . . .x

. . . . . .y

. . . aab . . . . . . bz

w = a . . . . . . . . . . . .ap + 1 symbols

b . . . . . . . . .bp symbols

21 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

a . . .x

. . . . . .y

. . . aab . . . . . . bz

w = a . . . . . . . . . . . .ap + 1 symbols

b . . . . . . . . .bp symbols

21 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

a . . .x

. . . . . .y

. . . aab . . . . . . bz

w = a . . . . . . . . . . . .ap + 1 symbols

b . . . . . . . . .bp symbols

Hence xy0z 6∈ L, and L is not regular. 21 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Food for thoughtIf “modern computer” = Finite Automaton then:

We can only store a fixed finite amount of data, say1TB = 10244 × 8 = 243 bits of information, i.e. a maximum of2243 ≈ 102,647,887,844,335 states – a finite number still!

So, our “modern computer” is not even able to recognize the (entire)language anbn!

At some point, our “modern computer” can no longer keep track of howmany a’s there are in the input.This occurs when the number of a’s becomes greater than 2243

We have assumed that the input string is not stored in thecomputer. . . (otherwise, it would just run out of memory anyway).However, at 3GHz for example, this would take. . . a length of time soinconceivably huge that the age of the universe would be negligible bycomparison. (So, do we care?)

22 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

We can only store a fixed finite amount of data, say1TB = 10244 × 8 = 243 bits of information, i.e. a maximum of2243 ≈ 102,647,887,844,335 states – a finite number still!So, our “modern computer” is not even able to recognize the (entire)language anbn!

22 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

22 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

We have assumed that the input string is not stored in thecomputer. . . (otherwise, it would just run out of memory anyway).

However, at 3GHz for example, this would take. . . a length of time soinconceivably huge that the age of the universe would be negligible bycomparison. (So, do we care?)

22 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

22 / 23

PumpingLemma

Mindmap

Binary alphabet

Examples

Pumping down

Space Complexity: Constant Space←→ NFAs

Finite Automaton: good model for algorithms which require constantspace (i.e. space used does not grow with respect to the input size).Some languages cannot be recognized by NFAs.Space used in computation must grow with respect to the input size.

We will see a more powerful model of computation next week!

23 / 23

Proof by The Pumping Lemma Observation

Documents

Transcript of Proof by The Pumping Lemma Observation

Optimizing Pumping Schemes In Air‐Conditioningashraeqatar.org/uploads/3/4/5/4/34547927/grundfos_-_ashrae_oryx... · A primary‐secondary pumping scheme divides the ... PRIMARY

THEORETISCHE INFORMATIK · ZHAW/HSR Druckdatum: 05.07.18 FTP_TheoComp Marcel Meschenmoser Dozent: Edgar Lederer / Prof Daniel Mall Seite 5 von 15 Pumping-Lemma Goal Das Pumping Lemma

THE OBSERVATION OF THE WEAK RADIATIVE HYPERON DECAY …ktev.fnal.gov/public/pubs/ktev/caslampig/ping_thesis.pdf · THE OBSERVATION OF THE WEAK RADIATIVE HYPERON DECAY ... friendship,

Wiener Process Ito's Lemma Derivation of Black-Scholes Solving

CLEO PUBLICATIONS - Astrophysics · CLEO PUBLICATIONS May 25, 2004 1. Observation of Three Upsilon States D. Andrews et al., Physical Review Letters 44, 1108 (1980) 2. Observation

Introduction to Szemerédi regularity lemma

Lecture 10 Pumping Lemma A property of regular sets.

The Pumping Lemma · The Pumping Lemma For every regular language L, there is a number ℓ≥ 1 satisfying the pumping lemma property: All w ∈ L with |w| ≥ ℓcan be expressed

Observation of the doubly charmed baryon Xicc++ · EUROPEAN ORGANIZATION FOR NUCLEAR RESEARCH (CERN) CERN-EP-2017-156 LHCb-PAPER-2017-018 12 September 2017 Observation of the doubly

OBSERVATION of RADIO LOUD PULSARS with the Fermi … · OBSERVATION of RADIO LOUD PULSARS with the Fermi-LAT γ-ray telescope Thierry Reposeur ... typically 1000 “pulsed” gammas

Observation of Exclusive γγ Production in CDF (and other exclusive states)

First observation of the magnetic dipole CO2 absorption ...

Observation of Iron Ore Beneficiation within a Spiral ...pure-oai.bham.ac.uk/ws/files/25537751/Boucher_et_al_Observation_of... · 1 Observation of Iron Ore Beneficiation within a

Isolation Lemma for Directed Reachability and NL vs. L

Using the Lovász Local Lemma in the configuration model

Isolation Lemma for Directed Reachability and NL vs. Lnutan/slides/dagstuhl2016.pdf · Isolation Lemma for Directed Reachability and NL vs.L Nutan Limaye (IIT Bombay) Joint work with

The Lopsided Lovász Local Lemma - · PDF fileTheorem (Erd}os, Lov asz 1975) ... Austin Mohr The Lopsided Lov asz Local Lemma. Asymptotics from the Lopsided Local Lemma Let A 1, :::,

Earth Observation Technologies at the European Space ...Earth Observation Technologies at the European Space Agency (ESA) Presenter: Josep.Rosello @ esa.int EO Technology Coordination

5.5: Ekman Pumping - UCD

First observation and Study of the K to π e decay with ...