Question

Prove that for every NFA with an arbitrary number of final states, there is an equivalent NFA with only one final state. Can we make a similar claim for DFAs? Provide a proof and discussion.

Answer 1

Every NFA can be converted into an equivalent NFA with only one final state by adding an epsilon-transition to a new sole final state from each existing final state. However, a similar claim cannot be made for DFAs since DFAs do not permit epsilon-transitions and require unique transitions for each symbol of the alphabet.

Step-by-step explanation:

The question asks us to prove that for every Non-deterministic Finite Automaton (NFA) with an arbitrary number of final states, there is an equivalent NFA with only one final state. The proof is quite straightforward. One can construct a new state that is not originally in the NFA and make it the only final state. Then, for every transition that leads to a final state in the original NFA, add an epsilon-transition (a transition that consumes no input symbol) from that state to the new, sole final state. This new NFA has the same language as the original one but only one final state. For Deterministic Finite Automatons (DFA), however, this claim does not hold because DFAs do not allow epsilon-transitions, and each state must have exactly one transition for each symbol in the alphabet, so we cannot simply converge multiple final states into one without potentially altering the language the DFA recognizes.