Design a DFA that accepts all strings contain $bbb$
regular expression $(a+b)^*bbb(a+b)^*$
then take complement of DFA such that no string has $3$ consecutive occurrences of the letter $b$.
having regular expression $(a+ba+bba)^*(\epsilon + b+ bb)$
can regular expression be: (a*(b+ ϵ )(b+ ϵ )aa*)* (b+ ϵ )(b+ ϵ )
Hemant Parihar will u
pls share state elimination diagram
"q0= ep + q1a + q2a " sir if we put the value we will get q0 = ep+q0ba+q0bba => q0 = ep+q0(ba+bba)
so we have r=q+rp as r = qp* than q0 = ep(ba+bba)* = (ba+bba)*
as RE is q0+q1+q2 => q0+q0b+q0bb => q0(ep+b+bb)
so final RE is (ba+bba)* (ep+b+bb) but your ans doesn't match
i think we the q0 expression will be q0= ep + q1a + q2a +q0a because we have incoming edge from q0 itself too
then final RE is (a+ba+bba)* (ep+b+bb)
BillyFitt this is regular expression for ur query : a*ba*ba*ba*
In this ,firstly make the dfa of the language which accept all strings from the alphabet (a,b) such that all string contain three consecutive occurrence of the letter b ,then make non final state as final state and final state as non final,initial state will remain same then it become the dfa that accept all string not containing three consecutive b's.
4 stage is a trap state
yeah i have done mistake..sorry for that..if we make transition for 'a' on state q1 and q2 to state q0 then i think it will be correct dfa
whts wrong with this dfa ? @Bikram sir i also got same
From state 2 there must be a transition towards state 1, which is missing in this DFA.
Tht's why it is not correct.
This DFA is for the language accepting strings with less than 3 'b' s . But here they are asking for strings with no 3 consecutive 'b' s .