3. Let D = (Q, E, 6, go, I) be a DFA. Construct a new DFA D' from D such that L(D') = {w E*: w3 € L(D) for some 3 € Σ*}, i.e., L(D') contains every string w that is a prefix of some string in L(D). (Hint: Try to modify the set I of accept states in D.) Prove the correctness of your construction.

Please help me fill in the blanks   The reverse of a string x, denoted rev(x), is the string obtained by writing x backwards. For example, if x = abbaaab then rev(x) = baaabba. If A is a subset of Σ*, the reverse of A, denoted rev(A), is the subset of Σ* consisting of all reverses of strings in A: rev(A)={rev(x) | x∈A} For example, rev({a,ab,aab}) = {a,ba,baa}. Suppose you are given a deterministic finite automaton M accepting a set A. Show how to construct a DFA N accepting rev(A). (Kozen would suggest putting pebbles on the final states of M and moving them backwards along transition edges.) Describe N formally (i.e. in terms of Q, δ, etc.) including the definition of acceptance. Hint: the subset construction will be helpful here. You can start with M = (Q, Σ, δ, s, F) and define the components of N in terms of the components of M. 1. DFA N = (Q_N​, Σ, δ_N​, s_N​, F_N​) where Q_N​ = ___ 2. DFA N = (Q_N​, Σ, δ_N​, s_N​, F_N​) where δ_N​ = ___ 3. DFA N = (Q_N​, Σ, δ_N​, s_N​, F_N​)…

Find all strings in L(ba + ab*a) of length less than four. Find a regular expression for the set { a" bm : n, m are even}.

a) Suppose we are trying to hide a secret message by embedding it in random extra letters. Write a function decode that takes a string called message and a list called key as parameters and returns a new string that represents the decoded text. key contains the index positions of the secret message in message. Example: For a message "KSGIAOELWIERUXOVPQIERCZXIETSQNBY" and key [3, 7, 14, 15, 19 ,21, 27] the function returns "ILOVECS" %3D It is possible that the key contains numbers that are invalid index positions they a greater than the length of message. These extra numbers should be ignored. Example: For a message "OK" and key [0, 1, 2, 3, 4] the function returns "OK" %3D b) Provide the docstring for your function !! + >