Question

Suppose we have two DFAs M₁ with 4 total states of which 2 are final and M₂ with 6 total states of which 4 are final. How may final states will the resulting DFA M' = M₁ ∩ M₂ have when using the product construction to create the intersection between these two au- tomatons without any minimization?

Answer 1

The resulting DFA from the intersection of two DFAs using product construction will have 8 final states, determined by the Cartesian product of the final states of the original DFAs, resulting in 2 * 4 = 8 final states without any minimization.

Step-by-step explanation:

The question relates to automata theory, which is a subsection of theoretical computer science and discrete mathematics. Specifically, it concerns constructing a new Deterministic Finite Automaton (DFA) that represents the intersection of two other DFAs using the product construction method. In the product construction, the states of the resulting DFA M' = M₁ ∩ M₂ are formed by the Cartesian product of the states of M₁ and M₂. Therefore, M' will have 4 * 6 = 24 states in total, considering all possible pairs of states from M₁ and M₂.

As for the final states of M', these correspond to the pairs where both elements are final states in their respective automata. Since M₁ has 2 final states and M₂ has 4 final states, the number of final states in M' can be calculated by multiplying these numbers, resulting in 2 * 4 = 8 final states for the resulting DFA without any minimization.

This calculation assumes that all pairs are reachable and contribute to the language recognized by the DFA, keeping in mind that in practice, some pairs might not be reachable or do not contribute to the language, and thus would not be considered in a minimized version of M'. However, the question specifies no minimization, so all possible pairs of final states are included.