2. Show that if L is regular and M(L) is the FSA for L, then L U {a}, for all a ? ? is regular. Do this by constructing an FSA for L U {a} from M(L). 3. Let L be the language accepted by the automation in below figure. Find a DFA that accepts L². 4. Let L be the language accepted by the FA in the above figure. Find a DFA for the language L² - L. 5. While the language accepted by a given DFA is unique, there are normally many DFA's that accept a language. Find a DFA with exactly six states that accepts the same language as the DFA in the below figure.
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To solve the first problem, we need to find the value of x in the equation 2x + 5 = 15. First, we subtract 5 from both sides of the equation to isolate the term with x: 2x + 5 - 5 = 15 - 5 2x = 10 Next, we divide both sides of the equation by 2 to solve for Show more…
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For each of the following languages state whether it is regular or not. If Li is regular, prove it by drawing a DFA or NFA (your choice) that recognizes it. If the language is not regular, give an argument (in plain language) why there is no DFA or NFA that can recognize it: (a) [4 marks] Lb = {w ∈ {0, 1}* : w represented as a binary integer is a power of 4} (b) [4 marks] La = {wwa^n : w ∈ {a, b}, n > 0} (c) [4 marks] Lc = {wxa : w, x ∈ {a, b}*} (d) [4 marks] Ld = {wwa : w, x ∈ {a, b}*}
Krishna S.
Consider the Deterministic Finite Automata, A = (QA, ΣA, δA, q0A, FA) and B = (QB, ΣB, δB, q0B, FB) with QA ∩ QB = ∅, ΣA ∩ ΣB = ∅ where ∅ stands for the null set. Let LA ⊆ ΣA* and LB ⊆ ΣB* be the languages accepted by A and B respectively and define the interleaved language: LA || LB := {s ∈ (ΣA ∪ ΣB)* | s ↑ A ∈ LA and s ↑ B ∈ LB} where s ↑ A and s ↑ B stand for the projection of s on ΣA and ΣB obtained by erasing all the symbols of s in ΣB and ΣA respectively. (a) Define the interleaving product A || B of A and B as a DFA that accepts the language LA || LB (b) Compute a DFA that accepts the language L = (0.1)* || (a.b)*
Sri K.
THEOREM 1.54 A language is regular if and only if some regular expression describes it. This theorem has two directions. We state and prove each direction as a separate lemma. 2. For each regex R given below, find an NFA equivalent to R by using the technique of our proof of Theorem 1.54. (a) 0(1)* ∪ 00 (b) (0(1)*)* ∪ (0 ∪ 1)*
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