1.16 Use the construction given in Theorem 1.39 to convert the following two nonde- terministic finite automata to equivalent deterministic finite automata. b 2 (a) a, b

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hello can you please help me with this problem because I am struggling can you please help me I have add the theorem because I don't understand the theorem at all.

Can you please draw something visual so I can understand it better.

THEOREM 1.39
Every nondeterministic finite automaton has an equivalent deterministic finite
automaton.
PROOF IDEA If a language is recognized by an NFA, then we must show the
existence of a DFA that also recognizes it. The idea is to convert the NFA into an
equivalent DFA that simulates the NFA.
Recall the "reader as automaton" strategy for designing finite automata, How
would you simulate the NFA if you were pretending to be a DFA? What do you
need to keep track of as the input string is processed? In the examples of NFAs,
you kept track of the various branches of the computation by placing a finger
on each state that could be active at given points in the input. You updated the
simulation by moving, adding, and removing fingers according to the way the
NFA operates. All you needed to keep track of was the set of states having fingers
on them.
If k is the number of states of the NFA, it has 2 subsets of states. Each subset
corresponds to one of the possibilities that the DFA must remember, so the DFA
simulating the NFA will have 2" states. Now we need to figure out which will
be the start state and accept states of the DFA, and what will be its transition
function. We can discuss this more easily after setting up some formal notation,
PROOF Let N = (QE, d, go, F) be the NFA recognizing some language A.
We construct a DFA M = (Q',E, 8', qo', F) recognizing A. Before doing the full
construction, let's first consider the easier case wherein N has no e arrows. Later
we take the e arrows into account
1. Q' P(Q).
Every state of M is a set of states of N. Recall that P(Q) is the set of
subsets of Q.
2. For RE Q' and a E, let 6'(R, a) = {g € Q] ged(r, a) for some re K},
If R is a state of M, it is also a set of states of N. When M reads a symbol
a in state R, it shows where a takes each state in R. Because cach state may
go to a set of states, we take the union of all these sets. Another way to
write this expression is
&'(Ra)=
$(r.a).4
TER
3.90 (90).
M starts in the state corresponding to the collection containing just the
start state of N.
4. F(REQ R contains an accept state of N).
The machine M accepts if one of the possible states that I could be in at
this point is an accept state.
The notation (r, a) means: the union of the sets 8(r, a) for each possible r in It.
PGR
Now we need to consider the e arrows. To do so, we set up an extra bit of
notation. For any state R of M, we define E(R) to be the collection of states
that can be reached from members of R by going only along e arrows, including
the members of R themselves. Formally, for RC Q let
E(R) (qq can be reached from R by traveling along 0 or more & arrows).
Then we modify the transition function of M to place additional fingers on all
states that can be reached by going along e arrows after every step. Replacing
s(r,a) by E(8(r, a)) achieves this effect. Thus
(R. a) (ge Ql qE E(S(r, a)) for some r = R}.
8
Additionally, we need to modify the start state of M to move the fingers ini-
tially to all possible states that can be reached from the start state of N along
the e arrows. Changing qo' to be E({0}) achieves this effect. We have now
completed the construction of the DFA & that simulates the NFA N.
The construction of M obviously works correctly. At every step in the com-
putation of M on an input, it dearly enters a state that corresponds to the subset
of states that N could be in at that point. Thus our proof is complete.
Theorem 1.39 states that every NFA can be converted into an equivalent DFA.
Thus nondeterministic finite automata give an alternativo way of characterizing
the regular languages, We state this fact as a corollary of Theorem 1.39.
mondial
Transcribed Image Text:THEOREM 1.39 Every nondeterministic finite automaton has an equivalent deterministic finite automaton. PROOF IDEA If a language is recognized by an NFA, then we must show the existence of a DFA that also recognizes it. The idea is to convert the NFA into an equivalent DFA that simulates the NFA. Recall the "reader as automaton" strategy for designing finite automata, How would you simulate the NFA if you were pretending to be a DFA? What do you need to keep track of as the input string is processed? In the examples of NFAs, you kept track of the various branches of the computation by placing a finger on each state that could be active at given points in the input. You updated the simulation by moving, adding, and removing fingers according to the way the NFA operates. All you needed to keep track of was the set of states having fingers on them. If k is the number of states of the NFA, it has 2 subsets of states. Each subset corresponds to one of the possibilities that the DFA must remember, so the DFA simulating the NFA will have 2" states. Now we need to figure out which will be the start state and accept states of the DFA, and what will be its transition function. We can discuss this more easily after setting up some formal notation, PROOF Let N = (QE, d, go, F) be the NFA recognizing some language A. We construct a DFA M = (Q',E, 8', qo', F) recognizing A. Before doing the full construction, let's first consider the easier case wherein N has no e arrows. Later we take the e arrows into account 1. Q' P(Q). Every state of M is a set of states of N. Recall that P(Q) is the set of subsets of Q. 2. For RE Q' and a E, let 6'(R, a) = {g € Q] ged(r, a) for some re K}, If R is a state of M, it is also a set of states of N. When M reads a symbol a in state R, it shows where a takes each state in R. Because cach state may go to a set of states, we take the union of all these sets. Another way to write this expression is &'(Ra)= $(r.a).4 TER 3.90 (90). M starts in the state corresponding to the collection containing just the start state of N. 4. F(REQ R contains an accept state of N). The machine M accepts if one of the possible states that I could be in at this point is an accept state. The notation (r, a) means: the union of the sets 8(r, a) for each possible r in It. PGR Now we need to consider the e arrows. To do so, we set up an extra bit of notation. For any state R of M, we define E(R) to be the collection of states that can be reached from members of R by going only along e arrows, including the members of R themselves. Formally, for RC Q let E(R) (qq can be reached from R by traveling along 0 or more & arrows). Then we modify the transition function of M to place additional fingers on all states that can be reached by going along e arrows after every step. Replacing s(r,a) by E(8(r, a)) achieves this effect. Thus (R. a) (ge Ql qE E(S(r, a)) for some r = R}. 8 Additionally, we need to modify the start state of M to move the fingers ini- tially to all possible states that can be reached from the start state of N along the e arrows. Changing qo' to be E({0}) achieves this effect. We have now completed the construction of the DFA & that simulates the NFA N. The construction of M obviously works correctly. At every step in the com- putation of M on an input, it dearly enters a state that corresponds to the subset of states that N could be in at that point. Thus our proof is complete. Theorem 1.39 states that every NFA can be converted into an equivalent DFA. Thus nondeterministic finite automata give an alternativo way of characterizing the regular languages, We state this fact as a corollary of Theorem 1.39. mondial
1.16 Use the construction given in Theorem 1.39 to convert the following two nonde-
terministic finite automata to equivalent deterministic finite automata.
1
2
(a)
a, b
Transcribed Image Text:1.16 Use the construction given in Theorem 1.39 to convert the following two nonde- terministic finite automata to equivalent deterministic finite automata. 1 2 (a) a, b
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