Im not sure if I am understanding this question correctly. Its a problem without a solution in my text.Starting with 1, and at least 01 or at least 010 or both these substrings.....So a regular expression will be given by:$1(0+1)^*01010(0+1)^* + 1(0+1)^*01001(0+1)^* + 1(0+1)^*010(0+1)^* +  1(0+1)^*01(0+1)^* = M$But then surely $M = 1(0+1)^*01(0+1)^*$ ???And then the NFA can simply be drawn as a three state NFA with starting state joined by edge $1$ to state B then joined by edge 01 to final state. And obviously there will be $0,1$ loops on state B and the final state?Did I miss something? Perhaps the both implies it should be $M = 1(0+1)^*01010(0+1)^* + 1(0+1)^*01001(0+1)^* $Do i need join state B with edges $01,010,01010 and01001$ to the final state?I would draw the graph but I cant find out how to do such things here. Its only a 3 state NFA regardless of the edges. I know that much :)I think im getting confused with these or parts of the question.
NFA that accepts binary strings starting with 1 and with at least 01, or at least 010, or both
automata;finite automata;nondeterminism