Let L be the set of exactly those strings over the alphabet {a, b, c, g} whose last symbol is either a or g. Language L is accepted by a (noneterministic) finite automaton M = (Q, E, 5, q, F), where: E = {a, b, c, g}, Q= {q, s}, F= {s} and it is your task to complete the transition function 5 (given below) by filling all the editable empty spaces. q, a, [qs11] q, b, [q14] q, c, [q13] q g, [qs12] q, \lambda, [noneq]

2. Let RE be the set of regular expressions over alphabet E = {0, 1}, and for r E RE, let L(r) be the language denoted by r. Consider the claim: VL E P(E*),Vr1, r2 E RE, (|L(r1)| > 1 A L \ L(r2) = L(r1)) = L is regular Either prove this, or find L, rı and r2 that provide a counter-example.

For a Turing machine M, (M) refers to the binary representation of M. For a Turing machine M, L(M) contains the set of all strings accepted by M. For a Turing machine M and an input x € {0,1}*, Steps(M, x) refers to the number of steps taken by M to execute on x before it halts. Here, one step of execution of M on x = one movement (left or right) of the tape head. For a Turing machine M and an input x = {0,1}*, we define the following: ReachCells(M,x) = {i : M reaches ith tape cell when M is executed on x} Informally, it contains all locations on the tape that are visited when M is ecuted on x. The leftmost location on the tape is the first tape cell, the location next to it is the second tape cell, and so on. A string w₁ is an anagram of w2 if w₁ can be obtained by rearranging the alphabets of w2. Formally, if w₁ is an n length string, wê is called an anagram of w₁ if there exists a permutation à on n elements such that π(w₁) = W2.

programm must be solved by python . MY variant number is ( 21 )