MELJUN CORTES Automata Theory (Automata5)

CSC 3130: Automata theory and formal languages

Limitations of finite automata

MELJUN P. CORTES, MBA,MPA,BSCS,ACSMELJUN P. CORTES, MBA,MPA,BSCS,ACS

MELJUN CORTESMELJUN CORTES

Many examples of regular languages

q0 q1

0 1

1

0

1 0 1

0, 1

q q q q

((0 + 1)0 + (0 + 1)(0 + 1)(0 + 1))*

all strings not containing pattern 010 followed by 101

Is every language regular?

Which are regular?

L1 = {0n1m: n, m ≥ 0}

L2 = {0n1n: n ≥ 0}

= {0, 1}

L3 = {1n: n is divisible by 3}

L4 = {1n: n is prime}

= {1}

L5 = {x: x has same number of 0s and 1s}

L6 = {x: x has same number of patterns 01 and 10}

= {0, 1}

Which are regular?

L1 = {0n1m: n, m ≥ 0}= 0*1*, so regular

• How about:

• Let’s try to design a DFA for it

… it appears that we need infinitely many states!

L2 = {0n1n: n ≥ 0} = {, 01, 0011, 000111, …}

A non-regular language

• Theorem

• To prove this, we argue by contradiction:– Suppose we have managed to construct a DFA

M for L2

– We show something must be wrong with this DFA

– In particular, M must accept some strings not in L2

The language L2 = {0n1n: n ≥ 0} is not regular.


• What happens when we run M on input x = 0n+11n+1?– M better accept, because x L2

M

imaginary DFA for L2 with n states

x



– But since M has n states, it must revisit at least one of its states while reading 0n+1

Mx 0 0 0 0

0

0r1n+1

Pigeonhole principle

• Here, balls are 0s, bins are states:

Suppose you are tossing n + 1 balls into n bins. Then two balls end up in the same bin.

If you have a DFA with n states and it reads n + 1 consecutive 0s, then it must end up in the same state twice.



– But since M has n states, it must revisit at least one of its states while reading 0n+1

– But then the DFA must contain a loop with 0s

Mx 0 0 0 0

0

0r1n+1


• The DFA will then also accept strings that go around the loop multiple times

• But such strings have more 0s than 1s, so they are not in L2!

M0 0 0 0

0

0r1n+1

General method for showing non-regularity• Every regular language L has a property:

• For every sufficiently long input z in L, there is a “middle part” in z that, even if repeated several times, keeps the input inside L

z a1 ak+1ak …… an-1

an

an+1…am

Pumping lemma for regular languages• Theorem: For every regular language L

There exists a number n such that for every string z in L, we can write z = u v w where |uv| ≤ n |v| ≥ 1 For every i ≥ 0, the string u vi w is in L.

z ……

u

v

w

Proving non-regularity

• If L is regular, then:

• So to prove L is not regular, it is enough to show:

There exists n such that for every z in L, we can write z = u v w where |uv| ≤ n, |v| ≥ 1 and For every i ≥ 0, the string u vi w is in L.

For every n there exists z in L, such that for every way of writing z = u v w where|uv| ≤ n and |v| ≥ 1, the string u vi

w is not in L for some i ≥ 0.

Proving non regularity

For every n there exists z in L, such that for every way of writing z = u v w where|uv| ≤ n and |v| ≥ 1, the string u vi

w is not in L for some i ≥ 0.• This is a game between you and an

imagined adversary

adversary

choose nwrite z = uvw (|uv| ≤ n,|v| ≥ 1)

you

choose z Lchoose iyou win if uviw L

1

2

Proving non-regularity

• You need to give a strategy that, regardless of what the adversary does, always wins you the game

adversary


you


1

2

Example

• L2 = {0n1n: n ≥ 0} = {0, 1}

adversary


you


1

2

adversary


you

z = 0n+11n+1

i = 2

uviw = 0j+2k+l1n+1

= 0n+1+k1n+1

L

1

2

u = 0j, v = 0k, w = 0l1n+1

j + k + l = n + 1

More examples

L3 = {1n: n is divisible by 3}

L4 = {1n: n is prime}

= {1}

L5 = {x: x has same number of 0s and 1s}

L6 = {x: x has same number of patterns 01 and 10}

L7 = {x: x has more 0s than 1s}

L8 = {x: x has different number of 0s and 1s}

= {0, 1}

MELJUN CORTES Automata Theory (Automata5)

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