Baldwin, Doug

Jasinski, James

Klein, Matt

McAlevey, Jack

Parks, Ian

Rohe, Susanna

Wakwella, Praveen

Wilson, John

Wojick, Kevin

English

SUNY Geneseo KnightScholar (State University of New York)

SUNY GeneseoKnightScholarComputability OER Open Educational Resources11-2018Nondeterministic Finite Automata Correspond toDeterministic OnesDoug BaldwinSUNY Geneseo, baldwin@geneseo.eduJames JasinskiSUNY GeneseoMatt KleinSUNY GeneseoJack McAleveySUNY GeneseoIan ParksSUNY GeneseoSee next page for additional authorsFollow this and additional works at: https://knightscholar.geneseo.edu/computability-oerCreative Commons LicenseThis work is licensed under a Creative Commons Attribution 4.0 License.This Open Educational Resource is brought to you for free and open access by the Open Educational Resources at KnightScholar. It has been acceptedfor inclusion in Computability OER by an authorized administrator of KnightScholar. For more information, please contactKnightScholar@geneseo.edu.Recommended CitationBaldwin, Doug; Jasinski, James; Klein, Matt; McAlevey, Jack; Parks, Ian; Rohe, Susanna; Wakwella, Praveen; Wilson, John; andWojick, Kevin, "Nondeterministic Finite Automata Correspond to Deterministic Ones" (2018). Computability OER. 1.https://knightscholar.geneseo.edu/computability-oer/1AuthorsDoug Baldwin, James Jasinski, Matt Klein, Jack McAlevey, Ian Parks, Susanna Rohe, Praveen Wakwella, JohnWilson, and Kevin WojickThis open educational resource is available at KnightScholar: https://knightscholar.geneseo.edu/computability-oer/1Nondeterministic Finite Automata AreEquivalent to Deterministic Finite AutomataDoug Baldwin, James Jasinski, Matt Klein, Jack McAlevey,Ian Parks, Susanna Rohe, Praveen Wakwella,John Wilson, Kevin WojickDepartment of Mathematics, SUNY GeneseoNovember 2018That every nondeterministic finite automaton is equivalent to a deterministicone is a standard theorem in computability theory. Maheshwari and Smid’stext [1] shows how to construct a deterministic automaton from a nondetermin-istic one, but does not show that the constructed automaton accepts the samelanguage as the original. This document provides that proof.Let us first define a notation for the configuration of a finite automaton, as wellas the “yields” relation `. Consider a finite automaton (either deterministicor nondeterministic) with alphabet Σ and transition function δ in some stateq that has an unconsumed string w. We can say that the ordered pair (q, w)indicates the configuration of the automaton, i.e. that the automaton is in stateq and computing on w. If w = av where a ∈ Σ and v is an arbitrary string, thenwe can also say that (q, av) ` (r, v) where r = δ(q, a) for a deterministic finiteautomaton and r ∈ δ(q, a) for a nondeterministic one. For nondeterministicautomata, we also say that (q, u) ` (r, u) when r ∈ δ(q, ). We can furtherdefine the “transitively yields” relation `∗, which describes a configuration thatcan be reached through zero or more “yields” transitions. These notations willbe useful for the following proof.Let N = (Q,Σ, δ, q0, F ) be a nondeterministic finite automaton, and let M =(Q′,Σ, δ′, q′0, F′) be the deterministic finite automaton constructed from N byMaheshwari and Smid’s construction. We show that L(M) = L(N) as a corol-lary to the following pair of theorems:Theorem 1. For every string u over Σ, if there is a state R ∈ Q′ such that(q′0, u) `∗ (R, ) in M , then (q0, u) `∗ (r, ) in N for all states r ∈ R.Copyright ( c©) 2018 by Doug Baldwin, James Jasinski, Matt Klein, Jack McAlevey, IanParks, Susanna Rohe, Praveen Wakwella, John Wilson, and Kevin Wojick. Contact DougBaldwin, baldwin@geneseo.edu, with any questions or comments about this document. Li-censed under a Creative Commons Attribution 4.0 International (CC BY 4.0) license. Seehttps://creativecommons.org/licenses/by/4.0/ for license terms.1Proof. The proof is by induction on |u|, the length of the string u.For the basis step, suppose |u| = 0, i.e., u = . Then (q′0, u) = (q′0, ) `∗ (R, )in M , and we see that R = q′0, since M goes from q′0 to R without consumingany input. From the construction, we know that q′0 is the -closure of q0, whichis exactly the set of states N can reach from q0 without consuming input. Inother words, for every one of the states r in q′0 (which is R), we can write that(q0, u) = (q0, ) `∗ (r, ) in N , and thus we have proven the basis step.For the induction step, we assume that for an integer k = |u|, the existence ofa state Y ∈ Q′ such that (q′0, u) `∗ (Y, ) in M implies that for all states y ∈ Y ,(q0, u) `∗ (y, ) in N .Now, let w be a string over Σ such that w is of length k + 1. We will showthat for a new state R, if (q′0, w) `∗ (R, ) in M , then (q0, w) `∗ (r, ) in N forall r ∈ R. Let w = ua, where a ∈ Σ, and let (q′0, w) = (q′0, ua) `∗ (Y, a) inM . From the inductive hypothesis, we see that this implies that for all statesy ∈ Y , (q0, ua) `∗ (y, a) in N . In M , we can consider the transition functionat Y , δ′(Y, a). Because M is deterministic, we know δ′(Y, a) 6= ∅. Therefore,we can write that δ′(Y, a) equals some state R. As a result, we can write(q′0, w) = (q′0, ua) `∗ (Y, a) ` (R, ). In N , the set of all transitions from (y, a)for all y ∈ Y is the union of the -closure of each state δ(y, a). In other words,the set of all transitions is ⋃y∈YC(δ(y, a))Notice that, because M has been created following the construction laid outin Maheshwari and Smid’s text [1], the above is equal to δ′(Y, a), which is R.Therefore, we can write (q0, w) = (q0, ua) `∗ (y, a) ` (r, ) in N for all statesr ∈ R, and thus we have shown for |w| = k + 1, if (q′0, w) `∗ (R, ) in M forR ∈ Q′, then (q0, w) `∗ (r, ) in N for all states r ∈ R, and the inductive stepis proven.Thus, because we have shown Theorem 1 to be true for the basis step, and truefor each inductive step after, we know that Theorem 1 is true.Figure 1 demonstrates the relationship between the Y and R states in M andthe corresponding y and r states in N .Theorem 2. For every string u over Σ, if R is the set of all states r in Q forwhich (q0, u) `∗ (r, ) in N , then (q′0, u) `∗ (R, ) in M .Proof. Once again, the proof is by induction on the length of u. For the basisstep, suppose |u| = 0, i.e., u = . Then (q0, ) `∗ (r, ) in N where the set of allstates r defines R. Notice that R = C(q0) because N goes from q0 to r withoutconsuming any input. Furthermore, we also know that q′0 = C(q0) from theconstruction in the text [1]. Therefore, R = q′0, and so (q′0, u) = (q′0, ) `∗(q′0, ) = (R, ), and the basis step is proven.2Figure 1: This image illustrates the relationship between the states Y and Rin M and the corresponding states y and r in N . Theorem 1 describes theexistence of the y and r states in N given the existence of the states Y and Rin M , whereas Theorem 2 does the opposite.For the inductive step, we assume that for an integer k = |u|, the existence ofY where Y is the set of all states y ∈ Q for which (q0, u) `∗ (y, ) in N impliesthat (q′0, u) `∗ (Y, ) in M .Now, let w be a string over Σ of length k + 1. We will show that for a new setof states R, if (q0, w) `∗ (r, ) in N for all r ∈ R, then in M (q′0, w) `∗ (R, ).Let w = ua, where a ∈ Σ, and let (q0, w) = (q0, ua) `∗ (y, a) in N where the setof all y is Y . From the inductive hypothesis we can see that this implies that(q′0, ua) `∗ (Y, a) in M . In N , we can consider the set of states r that resultfrom the transition function at each y ∈ Y , δ(y, a). This set is the union ofthe -closure of each transition δ(y, a). Let this set be R, which is described asfollows:R =⋃y∈YC(δ(y, a))By the construction laid out in the text [1], we know that⋃y∈YC(δ(y, a)) = δ′(Y, a)3and so δ′(Y, a) = R. Therefore, we can write (q′0, w) = (q′0, ua) `∗ (Y, a) ` (R, )in M , and thus we have shown for |w| = k + 1, if (q0, w) `∗ (r, ) in N for allr ∈ R, then (q′0, w) `∗ (R, ) in M , and the inductive step is proven.Thus, because we have shown Theorem 2 to be true for a basis step, and truefor each inductive step after, we know that Theorem 2 is true.Again, Figure 1 demonstrates the relationship between the y and r states in Nand the corresponding Y and R states in M .We can now show that a nondeterministic finite automaton and the deterministicautomaton created from it using the construction in the Maheshwari and Smidtext [1] accept the same language.Corollary 1. L(M) = L(N).Proof. We show that L(M) = L(N) by showing that for all strings w over Σ,M accepts w if and only if N does. We prove each direction separately.If. Suppose N accepts w, i.e., (q0, w) `∗ (r, ) for at least one r ∈ F . LetR be the set of all states r such that (q0, w) `∗ (r, ). Then by Theorem 2,(q′0, w) `∗ (R, ) in M , and since at least one member of R is an accepting stateof N , R is an accepting state of M . Thus M accepts w.Only if. Suppose M accepts w, i.e., (q′0, w) `∗ (R, ) for some state R ∈ F ′.Then by Theorem 1, (q0, w) `∗ (r, ) in N for all states r ∈ R. The only way Rcan be an accepting state of M is if at least one such r is an accepting state ofN , so N accepts w.Since we have shown that M accepts w if N does, and N accepts w if M does,then we can conclude that L(M) = L(N).References[1] Anil Maheshwari and Michiel Smid. Introduction to theory of computa-tion. (http://cglab.ca/∼michiel/TheoryOfComputation/), School of Com-puter Science, Carleton University, March 2017.4

Nondeterministic Finite Automata Correspond to Deterministic Ones

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Nondeterministic Finite Automata Correspond to Deterministic Ones

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