Finite-State Machines with No Output

Finite-State Machines with No Output MSU CSE 260

Outline • Introduction • Set of strings • Concatenation, Kleene Closure • Finite-State Automata • Definition, Example • Extended Transition Function • Language Recognized by FA • Nondeterministic FA • Definition, Example, FA-NFA Equivalence • Exercise 10.3

Introduction • Language recognition • Use final states instead of output • A string is recognized if it takes start state to a final state.

Set of strings - Concatenation • Definition. Let A and B be subsets of V*, where V is a vocabulary. The concatenation of A and B, denoted by AB, is the set of all strings xy | xA  yB. • Example. Let A={0, 11}, B={1, 10, 110}. AB={01, 010, 0110, 111, 1110, 11110} BA={10, 111, 100, 1011, 1100, 11011} • An is recursively defined by: • A0={} • An+1=AnA for n=0, 1, 2, …

Kleene Closure • Definition. Let A be a subset of V. The Kleene closure of A, denoted by A*, is the set consisting of concatenations of arbitrarily many strings from A. A* = k=0Ak • Examples. • Let A={0}. A* = {, 0, 00, 000,…} = {0n | n=0,1,2…}. • Let B={0,1}. B*={0,1}* is the set of all strings of 0s and 1s, including . • Let C={11}. C* = {12n | n=0,1,2…}.

Finite-State Automata • Definition. A finite-state automatonM=(S,I,f,s0,F) consists of: • a finite set S of states, with an initial (orstart) states0, • a finite input alphabetI, • a transition functionf : SI  S, which assigns a next state to every pair of state and input symbol, and • a subset F of S consisting of final (or accepting) states. • FAs can be represented by: • state table, or • state diagram: final states indicated by double circles.

Example - FA M=(S,I,f,s0,F) withS={s0,s1,s2,s3} I={0,1} F={s0,s3} Input and state table and diagram: s1 0 1 1 Start 0 s0 s3 1 0 0,1 s2 • Strings accepted by FA M? , 0, 00, 10, 000, 010, 100, 110, …

Extended Transition Function • The transition function f can be extended: • f: SI*  S • f assigns a next state to every pair of state and a string of symbols. • If s is a state in S and x is a string from I*, then f(s, x) is the state obtained by using each successive symbol of x, from left to right, as input starting with state s.

Language Recognized by FA • Definition. A string x is said to be recognized (or accepted) by the FA M=(S,I,f,s0,F) if it takes the initial state s0 to a final state, that is, f(s0,x)F. • The language recognized or accepted by the machine M, denoted by L(M), is the set of all strings that are recognized by M. • Two FSAs M1, M2 are equivalent if L(M1)=L(M2).

Language of FA - Example • Consider the FA M=(S, I, f, s0, F), with S={s0, s1, s2}, I={0, 1}, F={s2}, and the following state diagram: 1 0 1 0 Start s0 s1 s2 1 0 • Strings accepted by FA M? 10, 010, 110, 0110, … • What is the language accepted by the FA M? • Set of all strings of 0s and 1s that end with 10.

Nondeterministic Finite-State Automata • All FAs discussed so far are deterministic: • A unique next state is assigned to every pair of state and input symbol by the transition function f : SI  S. • The above constraint can be relaxed: • Nondeterministic finite-state automata: NFA • Zero or many possible next states are assigned to every pair of state and input symbol by the transition function f : SI  2S.

NFA - Definition • Definition. A nondeterministic finite-state automatonM=(S,I,f,s0,F) consists of: • a finite set S of states, with an initial state s0, • a finite input alphabet I, • a transition function f : SI  2S, which assigns a set of states (possibly empty) to every pair of state and input symbol, and • a subset F of S consisting of final states. • Starting with s0, on an input string x, an NFA may have multiple transition paths, can be in more than one state, and may get stuck. • x is accepted if at least one of the transition paths normally ends in a final state.

0,1 1 0 s0 s1 s2 Example - NFA • Consider the NFA M=(S, I, f, s0, F), with S={s0, s1, s2}, I={0, 1}, F={s2}, and Input the following state table. Its state diagram is: Start

Example – cont. • Processinginput string: 11010 1 1 0 1 0 start s0 s0 s0 s0 s0 s0 s1 s2 accept s1 s2 stuck s1 stuck • What is the language accepted by the NFA M? • The set of all strings of 1s and 0s that end with 10.

0,1 0,1 0 s0 s1 Another Example • Find an NFA that accepts the set of strings of 0s and 1s which contain at least one 0. • Such an NFA is M=(S, I, f, s0, F), with S={s0, s1}, I={0, 1}, F={s1}, and the following state diagram: Start

NFA-FA Equivalence • Theorem. If a language L is recognized by a nondeterministic NFA M0, then L is also recognized by a deterministic FA M1. • Proof idea:How would an FA simulate an NFA? It needs to keep track of all transition paths by remembering all active states at a given point in the input as NFA operates. If the NFA has n states, there are 2n subsets of states that need to be considered.

Construction of Equiv. FA • For NFA M0=(S0, I, f0 , s0, F0), let’s construct an FA M1=(S1, I, f1, s1, F1), where: • Each state in M1 corresponds to a set of states of M0; • The initial state of M1 is s1={s0}, (s0 initial state of M0); • The input alphabet I of M1is the same as that of M0; • The transition function f1 of M1 is defined as: f1({si1, si2, …, sik}, x) = f0(si1, x)  f0(si2, x) …  f0(sik, x) • The states of M1 are subsets of S0 reachable from {s0}; • The final states of M1 are those sets that contain at least one final state of M0.

Construction of Equiv. FA • For NFA M0=(S0, I, f0 , s0, F0), let’s construct an FA M1=(S1, I, f1, s1, F1), where: • The set of states of M1 is: S12S0; • The initial state of M1 is: s1= {s0}; • The input alphabet I of M1is the same as that of M0; • The transition function f1 of M1 is defined as: sS1aIf1(s, a) = ps f0(p, a) • union of the sets f0(p, a) for each p in s (subset of S0); • The states of M1 are subsets of S0 reachable from {s0}; • The set of final states of M1 is: F1= {s S1 | sF0  } = {s S1 | s contains a final state of M0}.

Proof • To prove that FA M1 accepts the same language as NFA M0, we will prove that:xI* f1(s1,x)=f(s0,x). • Basis step:f1(s1, ) = s1= {s0} = f0(s0, ) • Induction hypothesis:f1(s1, x) = f0(s0, x) • Statement to be shown: aI f1(s1, xa) = f0(s0, xa) f1(s1, xa) = f1(f1(s1, x), a) by def. of f1 for FA = f1(f0(s0, x), a) by induction hypothesis = p f0(s0, x)f0(p, a)by def. of f1 for equiv. FA = f0(s0, xa) by def. of f0 for NFA

0,1 1 0 s0 s1 s2 Equivalent FA of NFA - Example • Consider the NFA M0=(S0, I, f0 , s0, F0), where: S0={s0, s1, s2}, I={0, 1}, F0={s2}, Start • Construct the equivalent FAM1=(S1, I, f1 , s1, F1)

Example – Solution • S1  2S0 = {,{s0},{s1},{s2},{s0,s1},{s0,s2},{s1,s2},{s0,s1,s2}}. • The initial state is s1={s0}. • The input alphabet is I = {0, 1}. • The transition function f1 is defined as: f1(, 0) =  f1(, 1) =  f1({s0},0) = {s0} f1({s0},1) = {s0 , s1} f1({s1},0) = {s2} f1({s1},1) =  f1({s2},0) =  f1({s2},1) =  f1({s0,s1},0) = {s0, s2} f1({s0,s1},1) = {s0, s1} f1({s0,s2},0) = {s0} f1({s0,s2},1) = {s0, s1} f1({s1,s2},0) = {s2} f1({s0,s1},1) =  f1({s0,s1,s2},0) = {s0, s2} f1({s0,s1,s2},1) = {s0, s1}

Example - State Table

Example – State Diagram {s0,s1,s2} 1 1 0 0 Start 0 1 {s0} {s0,s1} {s0,s2} 0,1 1 1 0  {s1,s2} 1 0,1 {s1} {s2} 0 0

1 0 1 0 Start {s0,s1} {s0,s2} {s0} 1 0 Example – cont. • The set of states of M1 is S1={{s0},{s0,s1},{s0,s2}}; • The initial state s1={s0}. • The input alphabet is I = {0, 1}. • The set of final states is F1={{s0,s2}}.

Exercise 10.3

Finite-State Machines with No Output