Problem Detail: In the slides given by my teacher, the third automaton is the product of the first two: 
I tried to do the product myself by doing the transition table of both automata at the same time. I got stuck when I realised that the second automaton has no transition from state $2$ with $a$ as input symbol, thus the automaton should halt at that moment. In the third automaton, I can see there is no transition with input symbol $a$ in any of the states that have $2$ in (sorry for the bad expression, don’t know how else to say it). So when doing the transition table of the two automata, if there is no transition, should I just ignore it like in the 3rd automaton? Or would it be wise to complete the automaton?
I tried to do the product myself by doing the transition table of both automata at the same time. I got stuck when I realised that the second automaton has no transition from state $2$ with $a$ as input symbol, thus the automaton should halt at that moment. In the third automaton, I can see there is no transition with input symbol $a$ in any of the states that have $2$ in (sorry for the bad expression, don’t know how else to say it). So when doing the transition table of the two automata, if there is no transition, should I just ignore it like in the 3rd automaton? Or would it be wise to complete the automaton?Asked By : John Mayne
Answered By : Hans Hüttel
By definition, a deterministic finite-state automaton $(Q,Sigma,delta,q_0,F)$ must have a total transition function: For every $q in Q$ and $a in Sigma$, $delta(q,a)$ must be defined. Automaton no. 2 is not well-defined and for this reason it makes no sense to apply the product construction to the two automata given in your question. (A note on grammar: The singular form is automaton, the plural is automata – we cannot speak of “an automata” or “two automatas”.)
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Question Source : http://cs.stackexchange.com/questions/65037
