Exercise 6 Inclusion and equivalence between NFA ( 1 0 pts bonus 2 0 pts ) In this problem we assume that we are considering NFAs without l o n transitions Given two NFAs N i ( Q i , , i , q 0 i , F i ) , i 1 , 2 , we say that a relation s u b e Q 1 Q 2 is a simulation of N 1 by N 2 , denoted by N 1 N 2 , if the following properties hold ( q 0 1 , q 0 2 ) i n Whenever ( p , q ) i n , for every rin 1 ( p , a ) , there is some s i n 2 ( q , a ) so that ( r , s ) i n , for all ain Whenever ( p , q ) i n , if p i n F 1 then q i n F 2 If N 1 and N 2 are actually DFAs, show that an F map N 1 N 2 of DFAs is a simulation of N 1 by N 2 2 Let N 1 N 2 be a simulation of N 1 by N 2 Prove that for every win , for every q 1 inhat ( ) q ( q 0 1 , w ) , there is some q 2 inhat ( ) 2 ( q 0 2 , w ) so that ( q 1 , q 2 ) i n Conclude that L ( N 1 ) s u b e L ( N 2 ) If N 1 is an NFA and N 2 is a DFA, prove that if L ( N 1 ) s u b e L ( N 2 ) , then there is some simulation N 1 N 2 of N 1 by N 2 Hint Consider the relation ) hat ( ) 1 ( q 0 1 , w ) , q 2 hat ( ) 2 ( q 0 2 , w ) , win Remark If N 1 and N 2 are DFAs and L ( N 1 ) s u b e L ( N 2 ) , then there may not exist any DFA map from N 1 to N 2 , but above shows that there is always a simulation of N 1 by N 2 Give a counter example showing that ( 3 ) is generally false for NFAs, i e , if N 1 and N 2 are both NFAs and L ( N 1 ) s u b e L ( N 2 ) , there may not be any simulation N 1 N 2 In order to salvage ( 3 ) , we modify the conditions of the definition of a simulation we say that N 1 N 2 is a generalized simulation ( or g simulation ) if ( q 0 1 , q 0 2 ) i n Whenever ( p , q ) i n , for all ain , if 1 ( p , a ) O and 2 ( q , a ) O , then for every rin 1 ( p , a ) , there is some s i n 2 ( q , a ) so that ( r , s ) i n For all win with q 1 inhat ( ) 1 ( q 0 1 , w e ) F 1 q 2 inhat ( ) 2 ( q 0 2 , w ) F 2 ( q 1 , q 2 ) i n L ( N 1 ) s u b e L ( N 2 ) g N 1 N 2 N 1 N 2 g N 1 N 2 g N 1 N 2 1 g N 2 N 1

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Question: Exercise 6 : Inclusion and equivalence between NFA ( 1 0 pts + bonus 2 0 pts ) In this problem we assume that we

Exercise

6

: Inclusion and equivalence between NFA

(10

pts

+

bonus

20

pts

)

In this problem we assume that we are considering NFAs without

l o n -

transitions. Given two NFAs

N_{i} = (Q_{i},,_{i}, q_{0 i}, F_{i}), i = 1, 2,

we say that a relation

s u b e Q_{1} Q_{2}

is a simulation of

N_{1}

N_{2},

denoted by

N_{1} N_{2},

if the following properties hold:

(q_{01}, q_{02}) i n .

Whenever

(p, q) i n,

for every rin

_{1} (p, a),

there is some

s i n_{2} (q, a)

so that

(r, s) i n,

for all ain

.

Whenever

(p, q) i n,

p i n F_{1}

then

q i n F_{2} .

N_{1}

and

N_{2}

are actually DFAs, show that an

F -

map

N_{1} N_{2}

of DFAs is a simulation of

N_{1}

N_{2} .

2

Let

N_{1} N_{2}

be a simulation of

N_{1}

N_{2} .

Prove that for every win

^{* *},

for every

q_{1}

inhat

()_{q} (q_{01}, w),

there is some

q_{2}

inhat

()_{2} (q_{02}, w)

so that

(q_{1}, q_{2}) i n .

Conclude that

L (N_{1}) s u b e L (N_{2}) .

N_{1}

is an NFA and

N_{2}

is a DFA, prove that if

L (N_{1}) s u b e L (N_{2}),

then there is some simulation

N_{1} N_{2}

N_{1}

N_{2} .

Hint. Consider the relation

|)

} {

:hat

()_{1} (q_{01}, w), q_{2} =

hat

()_{2} (q_{02}, w),

win

^{* *}} .

Remark. If

N_{1}

and

N_{2}

are DFAs and

L (N_{1}) s u b e L (N_{2}),

then there may not exist any DFA map from

N_{1}

N_{2},

but above shows that there is always a simulation of

N_{1}

N_{2}

Give a counter

-

example showing that

(3)

is generally false for NFAs, i

.

.,

N_{1}

and

N_{2}

are both NFAs and

L (N_{1}) s u b e L (N_{2}),

there may not be any simulation

N_{1} N_{2} .

In order to salvage

(3),

we modify the conditions of the definition of a simulation: we say that

N_{1} N_{2}

is a generalized simulation

(

g -

simulation

)

(q_{01}, q_{02}) i n

Whenever

(p, q) i n,

for all ain

,

_{1} (p, a) \frac{O}{?}

and

_{2} (q, a) \frac{O}{?},

then for every rin

_{1} (p, a),

there is some

s i n_{2} (q, a)

so that

(r, s) i n .

For all win

^{* *}

with

q_{1}

inhat

()_{1} (q_{01}, w e) F_{1} q_{2}

inhat

()_{2} (q_{02}, w) F_{2} (q_{1}, q_{2}) i n L (N_{1}) s u b e L (N_{2}) g

N_{1} N_{2}

N_{1} N_{2} g N_{1} N_{2} g N_{1} N_{2}^{- 1} g N_{2} \frac{N_{1}}{?}

Exercise 6 : Inclusion and equivalence between

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