Equivalence Two - Automata and Complexity Theory - Lecture Slides, Slides of Theory of Automata

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More on DFA minimization

and DFA equivalence

Example

• Construct a DFA over alphabet {0, 1} that

accepts those strings that end in 111

• This is big, isn’t there a smaller DFA for this?

qε q 0 q 1 q 00 q 10 q 01 q 11 q 000 q 001 q 101 q 111

No smaller DFA!

• Last time we saw that

• In fact, it is also true that

There is no DFA with 3 states for L

This is the unique 4-state DFA for L

q 0 q 1 q 2 q 3

Theorem

• Proof outline

– Suppose there are two minimal DFAs M, M’ for L

– In both M and M’, every pair of states is distinguishable

The DFA M found by the minimization

algorithm is the unique minimal DFA for L

M

q

M’

q0 ’

Proof outline for uniqueness

• Since q

and q

are distinguishable, there is a

string w such that

• But in M’ this cannot happen!

M’ M’’

q 1

q 2

u

v

q’

u , v

w

w

w

Testing equivalence of DFAs

• Two DFAs are called equivalent if they accept

the same language

qε q 0 q 1 q 00 q 10 q 01 q 11

qΑ qΒ qC

Example

• Do the following two NFAs

accept the same language?

q 0 q 1 1

q 0 q 1

Answer

• We cannot run minimization procedure on NFAs

• But we can convert both NFAs to DFA and test

equivalence

regular

expression

NFA DFA