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Let M=(Q,,,q0,F)be a DFA and let be a state of Mcalled its 鈥渉ome鈥. A synchronizing sequence for M and h is a string s鈭埼b垪where(q,s)=hforeveryqQ. (Here we have extended to strings, so that(q,s) equals the state where M ends up when M starts at state q and reads input s .) Say that M is synchronizable if it has a synchronizing sequence for some state h . Prove that if M is a k-state synchronizable DFA, then it has a synchronizing sequence of length at mostk3 . Can you improve upon this bound?

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01

Synchronizable DFA.

A synchronizing word or reset sequence is a word in the input alphabet of the deterministic finite automata that sends any state of the deterministic finite automata to one and the same state. That is, if an ensemble of copies of the deterministic finite automata are each started in different states, and all of the copies process the synchronizing word, they will all end up in the same state. Not every deterministic finite automata has a synchronizing word; for instance, a deterministic finite automata with two states, one for words of even length and one for words of odd length, can never be synchronized.

02

Proof of statement.

Here as from the questionM=Q,,,q0,Fbe a deterministic finite automata and h be a state of M called its 鈥渉ome鈥.

And a synchronizing sequence for M is given.LetM=Q,,,q0,Fbe a deterministic finite automata and let q'be a state of M called its "home".

A Synchronizing sequence for M and q'is a string s* where (q,s)=q'for everys*(We actually have extended to strings so that (q,s)equals the state where M ends up when M starts at state and reads input).

Say that M is Synchronizableif it has a synchronizing sequence for some stateq'.

M' more interested in proving that the synchronized sequence is of length of at most then trying to improve upon this bound. There exists

In which|w|k2wk2sothat:(q1,w)=(q2,w)q1,w=q2,w for two distinct states in M : (thus, M can be read from two states in the automaton and get to the same final state).
If I prove it, I could construct a word M which will be a synchronizing sequence in M and |w|k3 as required.

Consider two states and then claim the following:

If there exists a word ww such that q1,w=q2,w, then there is such a word of length at most k2.

The proof of this a standard shrinking argument: if such a word is longer thank2, then during the runs from two states a pair of states repeats, and we can shrink w.

Now, since you assume the existence of a synchronizing word for all states, now proceed to construct a word that synchronizes all the states, pair by pair

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Most popular questions from this chapter

Give regular expressions generating the languages of Exercise 1.6.

a. {begins with a 1 and ends with a 0}

b. { w|wcontains at least three 1s}

c. { w|wcontains the substring 0101 (i.e., w = x0101y for some x and y)}

d. { w|whas length at least 3 and its third symbol is a 0}

e. { w|wstarts with 0 and has odd length, or starts with 1 and has even length}

f. { w|wdoesn鈥檛 contain the substring 110}

g. { w|the length of wis at most 5}

h. { w|wis any string except 11 and 111}

i. { w|every odd position of w is a 1 }

j. { contains at least two 0s and at most one 1}

k. {,0}

l. { w|wcontains an even number of 0 s, or contains exactly two 1s}

m. The empty set

n. All strings except the empty string

Recall that string x is a prefix of string y if a string z exists where xz=y, and that x is a proper prefix of y if in addition x6=y. In each of the following parts, we define an operation on a language A. Show that the class of regular languages is closed under that operation.

a)NOPREFIXA={wA|noproperprefixofwisamemberofA}.b)NOEXTENDA={wA|wisnottheproperprefixofanystringinA}.

We define the avoids operation for languages A and B to be

AavoidsB={w|wAandwdoesntcontainanystringinBasasubstring}.

Prove that the class of regular languages is closed under the avoids operation.

Let B and C be languages over =0,1. Define

BC={w^IB|forsomey^IC,stringswandycontainequalnumbersof1s}

Show that the class of regular languages is closed under the1operation.

An all- NFAMisa5-tuple(Q,,,q0,F)that accepts x* if every possible state that M could be in after reading input M is a state from F. Note, in contrast, that an ordinary NFA accepts a string if some state among these possible states is an accept state. Prove that all-NFAs recognizes the class of regular languages.

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