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The fact that the Recognition Problem is decidable gives us another algorithm for deciding Emptiness. The pumping lemma tells us that if every string x ∈ L(A) which has length greater than n (where n is the number of states in the minimal DFA recognizing this language and, therefore, no greater than the number of states in this particular DFA) can be split into three components uvw, where |v| > 0 and uviw ∈ L(A) for all i ≥ 0. One consequence of this is that L(A) will be non-empty iff it includes some string of length strictly less than n. To see this, assume (for contradiction) that no string in L(A) was of length less than n. Let x be a minimal length string in L(A), so no string in A is shorter than x. By our assumption |x| ≥ n. Then the pumping lemma applies and x must have the form uvw, etc. But then uw ∈ L(A) also and |uw| < |uvw| contradicting the choice of x as a minimal length string. Hence the shortest string in L(A), whatever it is, must have length strictly less than n. To decide Emptiness, then, all we need to do is to systematically generate all strings in Σ∗ with length less than n (the de?nition of Σ∗ provides the foundation of an algorithm for doing this) and check to see if A accepts any of them. We return "True" iff it accepts at least one. (Thus, the Emptiness Problem reduces to the Recognition Problem.)
Theorem (Finiteness) The Finiteness Problem for Regular Languages is decidable.
PROPERTIES OF Ardens therom
Exercise: Give a construction that converts a strictly 2-local automaton for a language L into one that recognizes the language L r . Justify the correctness of your construction.
Lemma 1 A string w ∈ Σ* is accepted by an LTk automaton iff w is the concatenation of the symbols labeling the edges of a path through the LTk transition graph of A from h?, ∅i to
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Give the Myhill graph of your automaton. (You may use a single node to represent the entire set of symbols of the English alphabet, another to represent the entire set of decima
Construct a Mealy machine that can output EVEN or ODD According to the total no. of 1''s encountered is even or odd.
State and Prove the Arden's theorem for Regular Expression
Theorem The class of recognizable languages is closed under Boolean operations. The construction of the proof of Lemma 3 gives us a DFA that keeps track of whether or not a give
Let L1 and L2 be CGF. We show that L1 ∩ L2 is CFG too. Let M1 be a decider for L1 and M2 be a decider for L2 . Consider a 2-tape TM M: "On input x: 1. copy x on the sec
De?nition Deterministic Finite State Automaton: For any state set Q and alphabet Σ, both ?nite, a ?nite state automaton (FSA) over Q and Σ is a ?ve-tuple (Q,Σ, T, q 0 , F), w
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