Theory of Computation

Beyond Regularity:Proofs by Contradiction,Pumping Lemma Proofs,

Pumping Lemma & Finite Languages

Vladimir Kulyukin

Outline

● Proofs by Contradiction● Detailed Pumping Lemma Proofs● Pumping Lemma & Finite Languages

Beyond Regularity

● Are there languages that cannot be recognized by finite state automata?● Yes, there are: anbn is one of the languages that cannot be recognized

by any FSA ● In general, any language whose strings have symbolic dependencies

(the number of some symbol depends on the number of some other symbol) cannot be recognized by an FSA

Proof by Contradiction● Pumping Lemma applications are examples of a proof technique known

as proof by contradiction● Proof by contradiction is used to prove statements of the form

IF A THEN B● A typical use of this technique consists of two steps:

Assume that A is true and B is false

Find a contradiction, i.e., a statement that contradicts another statement known to be true; that statement can be A itself, which is assumed to be true

Proof by Contradiction

● Why does proof by contradiction works?● It works because the statement IF A THEN B is true in all cases except

when A is true and B is false● Proof by contradiction allows us to rule out that case by finding a

contradiction● It is this contradiction that proves the original statement IF A THEN B

because it excludes the only case under which this statement can be false

Three Questions about Proof by Contradiction● What contradictions can/should we look for?

No general answer but experience, intuition, and insight● How can we use the assumptions that A is true and B is false to

reach a contradiction? Work forward from both assumptions and see what shakes

loose● When can/should we use proof by contradiction?

A reliable indicator is the word not in the statement B

Proof by Contradiction: Example

If r is a real number such that r2 = 2, then r is not rational (irrational)

A: r is a real number such that r2 = 2B: r is not rational (irrational)

Proof by Contradiction: Example

Assume A and NOT B:A: r is a real number such that r2 = 2NOT B: r is rationalNow we have to find a contradiction

Proof by Contradiction: Example

If r is rational (remember that NOT B is assumed to be true), then r = p/q, q != 0

We can assume that p and q have no common divisors (aka mutually prime); if they had, they could be divided out.

Proof by Contradiction: Example

Since r = p/q, r2 = (p/q)2.Since r2 = 2, then 2 = (p/q)2.Since 2 = (p/q)2, 2q2 = p2.Since p2 = 2q2, p2 is even.Since p2 is even, p is even.

Proof by Contradiction: Example

Since p is even, p = 2k, where k is an integer.

Since 2q2 = p2, 2q2 = (2k)2 = 4k2.

Since 2q2 = (2k)2 = 4k2, q2 = 2k2.

Since q2 = 2k2, then q is even.

But then, since both p and q are even,

they have at least one common divisor.

We found a contradiction because p and q were assumed to have no common divisors.

Proof by Contradiction: Example

Since p is even, p = 2k, where k is an integer.

Since 2q2 = p2, 2q2 = (2k)2 = 4k2.

Since 2q2 = (2k)2 = 4k2, q2 = 2k2.

Since q2 = 2k2, then q is even.

But then, since both p and q are even,

they have at least one common divisor.

We found a contradiction because p and q were assumed to have no common divisors.

Pumping Lemma Examples

Review: Pumping Lemma

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Pumping Lemma Use

● Pumping Lemma is not used to prove that languages are regular● Pumping Lemma states that if a language is regular, then its

strings have a specific property (pumping property)● Pumping Lemma is used to show that languages are not

regular: assume that a language is regular and then show that a specific string does not satisfy the pumping property

Pumping Lemma Example: Proof 1

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Pumping Lemma & Finite Languages

● Any finite language is regular● We can construct a finite automaton for each individual

string (that automaton will accept that string and only that string)

● The individual automaton can then be compiled into one union automaton that accepts all strings in that language

Pumping Lemma & Finite Languages● Does the Pumping Lemma contradict the regularity of finite

languages?● No, because for any finite language we can find the length of the

longest string (say n)● The Pumping Lemma's statement is vacuously true for all strings

whose length is greater than n, because there are no such strings in the language

● So we do not get any contradiction out of the Pumping Lemma for finite languages

References & Reading Suggestions

● Davis, Weyuker, Sigal. Computability, Complexity, and Languages, 2nd Edition, Academic Press

● Brooks Webber. Formal Language: A Practical Introduction, Franklin, Beedle & Associates, Inc

● Hopcroft and Ullman. Introduction to Automata Theory, Languages, and Computation, Narosa Publishing House

● D. Solow. How to Read & Do Proofs, Wiley & Sons