Final answer:
For every n-state nondeterministic finite automaton, there is an equivalent deterministic finite automaton with at most 2ⁿ states. This can be demonstrated by converting a sample NFA to an equivalent DFA. The DFA will have 2ⁿ states, as determined by the power set of the NFA states. Thus, the statement is true.
Step-by-step explanation:
For every n-state nondeterministic finite automaton, there is an equivalent deterministic finite automaton with at most 2ⁿ states.
To understand this concept, let's consider an example.
Suppose we have a nondeterministic finite automaton (NFA) with n = 2 states, labeled A and B. In the NFA, from state A, there are two possible transitions: one to state B on input 0, and another to both states A and B on input 1. From state B, there is only one transition to state B on input 0.
To convert this NFA to an equivalent deterministic finite automaton (DFA), we start with the initial state as the set of all states reachable from the original initial state, which in this case is {A}. From this set, we determine the transitions on all possible inputs.
In the DFA, the set of possible states will be the power set of the original NFA states. In our example, the power set of {A, B} is {{}, {A}, {B}, {A, B}}. Each subset corresponds to a state in the DFA.
The transition function of the DFA is determined by considering the transitions of the NFA states reachable from each DFA state. We fill in the transition table by checking the transitions of each state in the DFA.
In this example, the DFA will have 2ⁿ = 2² = 4 states, as the power set of {A, B} has four subsets. The equivalent DFA states will be represented by the subsets: {}, {A}, {B}, and {A, B}. The transition table would have four rows, representing the DFA states, and two columns, representing the inputs 0 and 1.
Therefore, the statement is true. For every n-state nondeterministic finite automaton, there is an equivalent deterministic finite automaton having at most 2ⁿ states.