Question

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5. (a) Consider the deterministic finite automaton M with states S := {80, 81, 82, 83}, start state so, single accepting state $3, and alphabet E = {0,1}. The following table describes the transition function T:S xHS. State 0 1 So So S1 So S1 S2 So $1 82 S3 S3 82 Draw the transition diagram for M. Let U = {01110,011100}. For each u EU describe the run for input u to M. Does M accept u? (b) Construct a finite-state machine which accepts a string of Os and is if and only if the last 3 input bits read are Os.

Answer 1

5) that, a) given Si= {50, S. 52, 53} Start state = So, final state = S3 state O I So $ 9 SI 92 So S2 S3 5g SA S₂ given, Uz { 01110,011100} bet us take string "O1110" S (S01110) S(51,1110) (84 X1 → [8xos, 2 8 (90,110) [sox 1 so] 8(8.10) (5x3 so] = 8(So, o) [5xo S. = $1 which is not final state so the string "01 110" is not accepted. let us take string Moittoo" $($0,011100) = 8(9444200) CS, X 1 ) [8Xos.] S (50.8 1.00) [8 x L ] [soki so z8( so 100) 90X0 9,3 8 (S0,00) = sco) [5, xo 92 which is not final state ఉం-te so the string "11100" is not accepted. strings in use, olio, 011100} are not accepted by M.

b) != totu* 000 - Regular expresion = Language that accepts atring in which have L- o's 8 bits are O qo NESM This is the i.ee, Non deterministic finise State machine for the given language. DESNT :- (Deterministic Finite State Machine) 5 tuple definition of = { % 4, , , , , } } { Input Alphabet {o, I } S Tramition states a set of states state V 1 quo as Vo ay ato start state & final state
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