computation-theory

This is the DFA i have drawn- Is it correct? I am confused because q4 state has 2 different transitions for same input symbol which violates the rule of DFA , but I can't think of any other solution. Your DFA is not correct. your DFA is completely wrong so I don't comment DFA for RE: 0(1 + 0)*0 + 1(1 + 0)*1 Language Description : if string start with 0 it should end with 0 or if string start with 1 it should end with 1 . hence two final states (state-5, state-4). state-4 : accepts 1(1 + 0)*1 state-5 : accepts 0(1 + 0)*0 state-1 : start state. DFA : EDIT : + Operator in Regular Expression (0 +

I have a little confusion in checking whether the given language is regular or not using pumping lemma. Suppose we have to check whether: L. The language accepting even number of 0 's in regular or not? We know that it is regular because we can construct a DFA for L. But I want to prove this with pumping lemma. Now suppose, I take a String w= "0000" : Now will divide the string as x = 0 , y = 0 , and z = 00 . Now on applying pumping lemma for i = 2 , I will get the string "00000" , which is not present in my language so by pumping lemma its prove that the language is not regular. But it is

My problem may sounds different to you. I am a beginner and I am learning Finite Automata. I am googing over Internet to find the Regular Expression for Finite Automata of Given Machine Below. Can anyone help me to write "Regular Expression for Finite Automata" of above machine Any help will be appreciated Grijesh Chauhan How to write regular expression for a DFA using Arden theorem Lets instead of language symbols 0 , 1 we take Σ = {a, b} and following is new DFA. Notice start state is Q 0 You have not given but In my answer initial state is Q 0 , Where final state is also Q 0 . Language

I know a n b n for n > 0 is not regular by the pumping lemma but I would imagine a*b* to be regular since both a,b don't have to be the same length. Is there a proof for it being regular or not? Grijesh Chauhan Answer to your question: imagine a*b* to be regular, Is there a proof for it being regular or not? No need to imagine, expression a*b* is called r egular e xpression (re), and regular expressions are possible only for regular languages. If a language is not a regular then regular expression is also not possible for that and if a language is regular language then we can always represent

I need help with constructing a left-linear and right-linear grammar for the languages below? a) (0+1)*00(0+1)* b) 0*(1(0+1))* c) (((01+10)*11)*00)* For a) I have the following: Left-linear S --> B00 | S11 B --> B0|B1|011 Right-linear S --> 00B | 11S B --> 0B|1B|0|1 Is this correct? I need help with b & c. Constructing an equivalent Regular Grammar from a Regular Expression First, I start with some simple rules to construct Regular Grammar(RG) from Regular Expression(RE). I am writing rules for Right Linear Grammar (leaving as an exercise to write similar rules for Left Linear Grammar) NOTE: