meljun cortes automata theory 5

8/13/2019 MELJUN CORTES Automata Theory 5

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CSC 3130: Automata theory and formal languages

Limitations of finite automata

MELJUN P. CORTES MBA MPA BSCS ACS

MELJUN CORTES


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3/17

Which are regular?

L1= {0n

1m

: n, m 0}L2= {0

n1n: n 0}S= {0, 1}

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

L4= {1n: n is prime}

S= {1}

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

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

S= {0, 1}


4/17

Which are regular?

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

How about:

Lets try to design a DFA for it

it appears that we need infinitely many states!

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


5/17

A non-regular language

Theorem

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

L2

We show something must be wrong with this DFA

In particular, Mmust accept some strings not in L2

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


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What happens when we run Mon input x=

0n+11n+1?

Mbetter accept, because xL2

M

imaginary DFA for L2with nstates

x


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0n+11n+1?

Mbetter accept, because xL2 But since Mhas nstates, it must revisit at least one of

its states while reading 0n+1

Mx 0 0 0 0

0

0r1n+1


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Pigeonhole principle

Here, balls are 0s, bins are states:

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

If you have a DFA with nstates and it reads

n+ 1consecutive 0s, then it must end up inthe same state twice.


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0n+11n+1?

Mbetter accept, because xL2 But since Mhas nstates, it must revisitat least one of

its states while reading 0n+1

But then the DFA must contain a loopwith 0s

Mx 0 0 0 0

0

0r1n+1


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The DFA will then also accept strings that go

around the loop multiple times

But such strings have more 0s than 1s, so theyare not in L2!

M0 0 0 0

0

0r1n+1


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General method for showing non-

regularity

Every regularlanguage Lhas a property:

For every sufficiently long input zin L, there is a

middle part in zthat, even if repeated severaltimes, keeps the input inside L

z a1 ak+1ak an-1

an

an+1am


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Pumping lemma for regular languages

Theorem:For every regular language L

There exists a number nsuch that for every

string zin L, we can write z= u v wwhere

|uv| n|v| 1

For every i 0, the string u vi wis in L.

z

u

v

w


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Proving non-regularity

If Lis regular, then:

So to prove Lis notregular, it is enough to show:

There exists nsuch that for every zin L, we

can write z= u v wwhere |uv| n, |v| 1

andFor every i 0, the string u vi wis in L.

For every nthere exists zin L, such that for

every way of writing z= u v wwhere

|uv| nand|v| 1,the string u vi wis not

in Lfor some i 0.


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Proving non regularity

For every nthere exists zin L, such that forevery way of writing z= u v wwhere

|uv| nand|v| 1,the string u vi wis not

in Lfor some i 0.

This is agame between you and an imagined

adversary

adversary

choose n

write z= uvw(|uv| n,|v| 1)

you

choose zLchoose i

you winif uviwL

1

2


15/17

Proving non-regularity

You need to give a strategythat, regardless ofwhat the adversary does, always wins you the

game

adversary

choose n


you

choose zL

choose i

you winif uviwL

1

2


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Example

L2= {0n1n: n 0} S= {0, 1}

adversary

choose n


you

choose zL

choose i

you win if uviwL

1

2

adversary

choose n


you

z= 0n+11n+1i= 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


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More examples

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

L4= {1n: n is prime}

S= {1}

L5= {x: xhas same number of 0s and 1s}L6= {x: xhas same number of patterns 01 and 10}

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

L8

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

S= {0, 1}

meljun cortes automata theory 5

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