Automata and Languages


 Derick Cole
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1 Automata and Languages Computational Models: An idealized mathematical model of a computer. A computational model may be accurate in some ways but not in others. We shall be defining increasingly powerful models of computation. We begin with the simplest model of computation called the finite state machine or finite automaton. Finite Automata: The finite State automaton (FSA): has a central processing unit of fixed, finite capacity. has no or an extremely limited amount of memory receives its input as a string, it delivered to it on an input tape. delivers no output, except an indication of whether or not the input is acceptable. Various electromechanical devices are built on the FSA models. In addition, FSA are applicable to the design of several types of algorithms and programs. Case of automatic door controller: Automatic doors swing open when sensing that a person is approaching. An automatic door has a pad in front to detect the presence of a person about to walk through the doorway. Another pad is located to the rear of the doorway so that the controller can hold the door open long enough for the person to pass all the way through. The controller is in either of two states: OPEN or CLOSED representing the corresponding condition of the door.
2 Input signal States Neither Front Rear Both Closed Closed Open Closed Closed Open Closed Open Open Open The central concepts of Automata Theory: Alphabets: An alphabet is a finite, nonempty set of symbols. Conventionally, we use the symbol Σ for an alphabet. Common alphabets include: 1. Σ = {0,1}, the binary alphabet 2. Σ = {a, b,,z}, the set of all lower case letters 3. The set of ASCII characters or the set of printable ASCII characters. Strings: A string (or sometimes word) is a finite sequence of symbols chosen from some alphabet. For example is a string from the binary alphabet Σ = {0,1}. The string 111 is another string from this alphabet. The Empty String: The empty string is the string with zero occurrences of symbols. This string denoted ε, is a string that may be chosen from any alphabet whatsoever. Length of a String: If is often useful to classify strings by their length. The length is the number of positions for symbols in the string. For instance, has length 5. The standard notation for the length of a string w is w. For example, 011 = 3 and ε = 0.
3 Powers of an alphabet: If Σ is an alphabet, we can express the set of all strings of a certain length from that alphabet by using exponential notation. We define Σ k to be the set of strings of length k, each of whose symbols is in Σ. Σ 0 = {ε} regardless of the alphabet. That is ε is the only string whose length is 0. if Σ = {0,1}, then Σ 1 = {0,1}, Σ 2 = {00,01,10,11}. What is Σ 3? Confusion between Σ and Σ 1? The former is an alphabet; its members are symbols, The latter is a set of strings, each of which is of length 1. The set of all strings over an alphabet Σ is conventionally denoted Σ *. For instance {0,1} * = {ε, 0, 1, 00, 01, 10, 11, 000,..}. Σ * = Σ 0 Σ 1 Σ 2 Sometimes we wish to exclude the empty string from the set of strings, we then use the set of nonempty strings from alphabet Σ denoted Σ +. Σ + = Σ 1 Σ 2 Σ 3 Σ * = ε Σ + Concatenation of Strings: Let x and y be strings. The xy denotes the concatenation of x and y, that is the string formed by making a copy of x and following it by a copy of y. More precisely, if x is a string composed of i symbols x= a 1 a 2 a 3.a i and y is a string composed of j symbols y = b 1 b 2 b 3 b j, the xy is the string of length i+j: xy= a 1 a 2 a 3.a i b 1 b 2 b 3 b j ε is the identity for the concatenation. Languages: A set of strings chosen from Σ * where Σ is a particular alphabet is called a language. If Σ is an alphabet and L Σ *, the L is a language over Σ.
4 Example of languages: 1. The language of all strings consisting of n 0 s followed by n 1 s, for some n The set of binary numbers whose value is a prime: 3. Σ * is a language for any alphabet Σ 4., the empty language, is a language over any alphabet 5. {ε}, the language consisting of only the empty string. Notice that {ε}, the former has no strings and the latter has one string. Problems: In automata theory, a problem is the question of deciding whether a given string is a member of some particular language. More precisely, if Σ is an alphabet and L is a language over Σ, then the problem is: Given a string w in Σ *, decide whether or not w is in L. Let the following picture depicts a finite automaton called M1: q1 q2 0,1 q3 1,01,11, ,00110? The figure above is called the state diagram of M1: It has three states, labeled q1, q2, and q3. The start state q1 is indicated by the arrow pointing at it from nowhere. The accept state q2 is represented with a double circle The arrows going from one state to another are called transitions. When this automaton receives an input string such as 1101, it processes that string and produces an output. The output is either accept or reject.
5 What language is accepted by the automaton? 1,01,11, ,00110? Formal definition of a finite automaton: A finite automaton is a 5tuple (Q,,δ, q0, F), where: 1. Q is a finite set called the states 2. is a finite set called the alphabet 3. δ : Q x Q is the transition function 4. q0 Q is the start state, and 5. F Q is the set of accept or final states We can describe M1 formally by writing M1= (Q,, δ, q1, F), where 1. Q ={q1, q2, q3} 2. ={0,1} 3. δ is described as: 0 1 q1 q1 q2 q2 q3 q2 q3 q2 q2 4. q1 is the start state, and 5. F= {q2} δ(q1,0)= q1 δ(q2, 0)= q3 δ(q3, 0)= q2 δ(q1, 1) = q2 δ(q2, 1) = q2 δ(q3, 1) = q2 M= {{q1, q2, q3}, {0, 1},δ q1, {q2}} Notations for DFA s There are two preferred notations for describing automata: 1. A transition diagram, which is a graph.
6 2. A transition table, which is a tabular listing of the δ function. Transition Diagrams: A transition diagram for a DFA M= (Q,,δ, q0, F) is a graph defined as follows: 1. For each state in Q there is a node. 2. For each state q in Q and each input symbol a in, Let δ(q,a) =p then the transition diagram has an arc from node q to node p, labeled a. If there are several input symbols that cause transitions from q to p, then the transition diagram can have one arc, labeled by the list of these symbols. 3. There is an arrow into the start q0, labeled Start. This arrow does not originate at any node 4. Nodes corresponding to accepting states (those in F) are marked by a double circle. 5. States not in F have a single circle. Transition Tables: 1. A transition table is a conventional, tabular representation of a function like δ that takes two arguments and returns a value. 2. The rows of the table correspond to the states and the columns correspond to the inputs. 3. The entry for the row corresponding to state q and the column corresponding to input a is the state δ(q,a). Extending the transition function to Strings:
7 The DFA defines a language which is the set of all strings that result in a sequence of state transitions from the start state to an accepting state. Extended Transition function: If δ be the transition function, then the extended transition function is a function that takes a state q and a string w and returns the state p The state that the automaton reaches when starting in state q and processing the sequence of input w. We define δˆ by induction on the length of the input string, as follows: BASIS: δˆ (q, ε) = q. That is, if we are in state q and read no inputs, then we are still in state q. Suppose that δˆ (q, x)= p Induction: Suppose w is a string of the form xa; that is a is the last symbol of w and x is the string consisting of all but the last symbol. Then δˆ (q,w) = δ(δˆ (q,x),a), to compute δˆ (q,w), we first compute δˆ (q,x), suppose that the automaton gets to state p, that isδˆ (q,x)=p. Then δˆ (q,w) is what we get by making a transition from state p on input a, the last symbol of w. Thus δˆ (q,w)= δ(p,a) Language of a DFA If A is the set of all strings that machine M accepts, we say that A is the language of machine M and write L(M) = A that M recognizes A or that M accepts A. L(M)= { w Σ* δˆ (q0, w) F} Non acceptance means that the DFA stops in a non final state so that: L (M ) = { w Σ* δˆ (q0, w) F} A machine may accept several strings, but it always recognizes only one language.
8 If a machine accepts no strings, it still recognizes only one language, namely the empty language φ. The language A defined by the Machine M1 is: A = {w w contains at least one 1 and an even number of 0 s follow the last 1} Regular Languages: Definition: A language L is called regular if and only if there exists some deterministic finite automaton M such that L = L(M). To show that a language is regular, it is necessary and sufficient to build a dfa that accepts it. Closure under the regular operations: Theorem: The class of regular languages is closed under the union operation Theorem: The class of regular languages is closed under the concatenation operation Theorem: The class of regular languages is closed under the star operation. Example: Show that the language L = {abw: w {a, b}* } is regular. The only issue here is that the first two symbols of the string. After they have been processed no further decision are needed. Non deterministic and Deterministic computation:
9 So far, we have seen automata in which every step of a computation follows in a unique way from the preceding step. When the machine is in a given state and reads the next symbol, we know the next state it ll be in, it is determined. In a nondeterministic machine several choices exist for the next state at any point. A nondeterministic finite automaton has the power to be in several states at once. Nondeterminism is a generalization of determinism, so every deterministic finite automaton is automatically a nondeterministic finite automaton. Notation: A deterministic finite state automaton is called DFA and a nondeterministic finite state automaton is called NFA. Like the DFA, and NFA has a finite set of states, a finite set of input symbols, one start state and a set of accepting states. The difference between the DFA and the NFA is in the type of transition. For the NFA, the transition function δ takes a state and an input symbol as arguments and returns 0, 1 or more states, rather than returning exactly one state, as the DFA must. 0, q0 q1 q2 An NFA accepting all strings that end in 01
10 Let s assume that we want to test whether this automaton accepts q0 q0 q0 q0 q0 q0 (stuck) q1 q1 q1 (stuck) q2 q2 (successful computation) (stuck) Draw the computation trees for Computation in an NFA: Suppose we are running an NFA on an input string and come to a state with multiple ways to proceed. For example, say that we are in state q1 in NFA N1, and the next input symbol is 1. After reading that symbol, the machine splits into multiple copies of itself and follows all the possibilities in parallel. Each copy of the machine takes one of the possible ways to proceed and continues as before. If there are subsequent choices, the machine splits again. If the next input symbol doesn t appear on any of the arrows exiting the state occupied by a copy of the machine, that copy along with the branch of computation associated with it, die. Finally, if any one of these copies of the machine is in an accept state at the end of the input, the NFA accepts the input string. Nondeterminism can be viewed as a kind of parallel computation wherein multiple independent processes a or threads can be running concurrently. If at least one of these processes accepts, then the entire computation accepts.
11 Representation of the above automaton: N= ({q0, q1, q2}, {0,1}, δ, q0, {q2}) 0 1 q0 {q0,q1} {q0} q1 {q2} *q2 The extended Transition Function: We need to extend the transition function δ to the function δ that takes a state q and a string of input symbols w and returns the set of states that the NFA is in when it starts at in state q and processes the string w. The language of an NFA: An NFA accepts a string w if it is possible to make any sequence of choices of next state, while reading the characters of w and go from the start state to any accepting state. The fact that other choices using the input symbols of w lead to a non accepting state does not prevent w from being accepted by the NFA as a whole. Formally, if N= (Q, Σ, δ, q0, F) L(N)= { w δ (q0, w) F } Finite Automaton with EpsilonTransition: Another extension of the DFA is to allow transition on ε, the empty string. In effect, this is equivalent to making a transition spontaneously. This new capability does not expand the class of languages that can be accepted by a finite state automaton.
12 Use of εtransitions: We use transition diagrams with ε allowed as a label. However an ε transition contributes nothing to the string along the path. 0,1 0,ε 1 q1 q2 q3 1 q4 0,1 Example: εnfa that accepts decimal numbers consisting of: 1. An optional + or  sign 2. a string of digits 3. a decimal point 4. another string of digits. Either this string or the string (2) can be empty, but at least one of the strings must be nonempty. Formal definition of a nondeterministic finite automaton: The formal definition of a NFA differs from that of a DFA in one essential way: the type of transition function. In a DFA the transition function takes a state and an input symbol and produces the next state. In an NFA the transition function takes a state and an input symbol or the empty string and produces a set of possible states. Definition: We define the power set P(Q) of set Q to be the collection of all subsets of Q. Definition: For any alphabet we write ε = ε
13 Definition: A nondeterministic finite automaton is a 5tuple (Q,, δ, q0, F) where, 1. Q is a finite set of states 2. is a finite alphabet 3. δ: Q x ε P(Q) is the transition function 4. q0 Q is the start state and 5. F Q is the set of accept states Differences between a DFA and a NFA: Every state of a DFA has exactly one exiting transition arrow for each symbol of the alphabet. An NFA violates this rule. in a DFA, labels on the transition arrows are symbols from the alphabet. An NFA may have arrows labeled with members of the alphabet or ε Zero, one or many arrows may exit from each state with the label ε Equivalence of NFAs and DFAs: Deterministic and nondeterministic finite automata recognize the same class of languages. We say that two machines are equivalent if they recognize the same language. Theorem:
14 Every nondeterministic finite automaton has an equivalent deterministic finite automaton. If k is the number of states of the NFA, it has 2 k subsets of states. Each subset corresponds to one of the possibilities that the DFA must remember, so the DFA simulating the NFA will have 2 k states. Proof: Let N =(Q,, δ, q0, F) be the NFA recognizing some language A. We construct a DFA M=(Q,, δ, q0, F ) recognizing A. Case 1: N has no ε arrows. 1. Q = P(Q). Every state of M is a set of states of N. (Recall that P(Q) is the set of subsets of Q). 2. For R Q and a let δ (R, a)= {q Q q δ(r, a) for some r R} if R is a state of M, it is also a set of states of N. When M reads a symbol a in a state R, it shows where a takes each state in R. Because each state may go to a set of states, we take the union of all these sets. Another way to write this expression is: δ ( R, a) = Uδ ( r, a) r R 3. q 0 = {q 0 } M starts in the state corresponding to the collection containing just the start state of N. 4. F = {R Q R contains an accept state of N}. The machine M accepts if one of the possible states that N could be in at this point is an accept state. Case 2: Now we consider the ε arrows: For any state R in M, Let E(R) be the collection of states that can be reached from R by going only along ε arrows, including the members of R themselves. Formally, for R Q let
15 E(R)= {q q can be reached from R by traveling along 0 or more ε arrows}. Thus we modify the transition function of M by placing additional states that can be reached by going along ε arrows. Replacing δ (r, a) by E(δ (r, a)) achieves this effect. δ (R, a)= {q Q q E(δ (r, a)) for some r R} The start state is modified by adding all possible states that can be reached from the start state of N along the ε arrow So q 0 =E({q 0 }) Example: Let us illustrate the procedure we gave for converting an NFA into a DFA by using the machine N4 depicted below: a b 2 1 a a, b ε 3 N4=({1,2,3},{a, b},δ, 1, {1}) To construct a DFA that is equivalent to N4, we first determine D s states, D s states are P(Q) or power set of Q. N4 has three states, so we construct D with 2 3 = 8 states, one of each subset of N4 s states. D s={,{1},{2},{3},{1,3},{1,2},{2,3},{1,2,3}} Next, we determine the start and accept states of D. The start state of N4 is E({1}), the set of states of that are reachable from 1 by traveling along ε arrows, plus 1 itself, so E({1})={1,3}.
16 The new accept states are those containing N4 s accept states; thus {{1},{1,2},{1,3},{1,2,3}}. Finally, we determine D s transition function. Each of D s states goes to one place on input a and one place on input b. In D,  state{2} goes to {2,3} on input a, and to {3} on input b  state{1} goes to on a, because no arrows exit it. It goes to {2} on b.  state{3} goes to {1,3} on a, because in N4, state{3} goes to 1 on a and 1 in turn goes to 3 with an ε arrow.  state {3} on b goes to  state{1,2} on a goes to {2,3} because 1 points at no states with a arrows and 2 points at states {2, 3}  state{1,2} on b goes to {2,3}  state{1,3} on a goes to {1,3}  state{1,3} on b goes to {2}  state{1,2,3} on a goes to {1,2,3}  state{1,2,3} on b goes to {2,3} We may simplify this machine by removing unreachable states such as {1} and {1, 2} (c, d) (a,b)
17 REGULAR EXPRESSIONS: In arithmetic, we can use the operation + and x to build up expressions such as (5+3) x 4. Another example of regular expression is (0 1)*. The value of this expression is the language consisting of all possible strings of 0 s and 1 s. If is an alphabet, the regular expression describes the language consisting of all strings of length 1 over this alphabet and * describes the language consisting of all strings over that alphabet. Similarly *1 is the language that contains all strings ending in a 1. Formal Definition of a Regular Expression: We say that R is a regular expression if R is: 1. a for some a in the alphabet 2. ε (R1 R2), where R1 and R2 are regular expressions 5. (R1 R2), where R1 and R2 are regular expressions or 6. (R1*), where R1 is a regular expression. In items 1 and 2, the regular expressions a and ε represent the languages {a}, and {ε}, respectively. In item 3, the regular expression represents the empty language. In items 4, 5, and 6, the expressions represent the languages obtained by taking the union or concatenation of the languages R1 and R2 or the star of the language R1, respectively. Regular expressions are useful tools in the design of compilers for programming languages.
18 Equivalence with finite automata: Regular expressions and finite automata are equivalent in their descriptive power. That is any regular expression can be converted into a finite automaton that recognizes the language and vice versa. Theorem: A language is regular if and only if some regular expression describes it. Lemma: If a language is described by a regular expression, then it is regular. Proof: (page 67) We consider the six cases in the formal definition of regular expressions. We convert the regular expression (ab a)* to an NFA Lemma: If a language is regular, then it is described by a regular expression Algorithm for converting a DFA into an equivalent regular expression: 1. Add two states to the DFA: a start state called s and an accept state called a. This defines a new automaton called G 2. CONVERT(G): CONVERT(G): 1. Let k be the number of states of G 2. If k=2, then G must consist of a start state, an accept state and a single arrow connecting them and labeled with a regular expression R. return R.
19 3. If k >2, we select any state q rip Q different from s and a and let G = (Q,, δ,s, a), where: Q = Q{q rip } and for any q i Q {a} and any q j Q {s} let: δ (q i, q j ) = (R1)(R2)*(R3) R4 for R1= δ(q i, q rip ), R2= δ( q rip, q rip ), R3= δ( q rip, q j ) and R4= δ(q i, q j ) 4. Compute CONVERT(G ) and return this value Non Regular Languages: Pigeonhole Principle: The pigeonhole Principle states that if n pigeons fly into m pigeonholes and n > m then at least one hole must contain two or more pigeons.
20 Pigeonhole Principle: A function from a finite set to a smaller set cannot be onetoone. There must be at least two elements in the domain that have the same image. Nonregular languages: Because a finite state automaton can assume only a finite number of states and because there are infinitely many input sequences, by the pigeonhole principle, there must be at least one state to which the automaton returns over and over again. This is an essential feature of the automaton. Showing that a language is not regular Example: L= {s * s= 1 n 0 n n 0} If we attempt to find a DFA that recognizes B, we discover that the machine needs to remember how many 0 s have been seen so far as it reads the input. So the machine has to keep track of unlimited number of possibilities, this cannot be done with a finite number of states. In this case, we say that the language B is nonregular. We use the pigeonhole principle to show that B is not regular Proof by contradiction: The Pumping Lemma for regular languages:
21 If A is a regular language, then there is a number p(the pumping length) where, if s is any string in A of length of at least p, then s may be divided into three pieces: 1. for each i 0 xy i z A 2. y > 0 3. xy p Notation: s represents the length of string s. y i means that i copies of y are concatenated together and y 0 equals ε. Proof: Suppose that there is a DFN A that accepts L. [A contradiction will be derived]. Since A has only a finite number of states, these states can be denoted s 1,s 2,.,s n, where n is a positive integer. Consider all inputs strings that consist entirely of 1 s: 1,11,111,1 4, etc.. Now there are infinitely many such strings and only finitely many states. Thus, by the pigeonhole principle, there must be a state s m and two input strings a p and a q with p q such that when either a p or a q are input to A, A goes to state s m. (the pigeons are the strings of a s, the pigeonholes are the states and the correspondences associate each string with the state to which A goes when the string is input). Suppose that A accepts L. Hence A accepts the string 1 p 0 p. This means that after p a s have been input, at which point A is in state s m, inputting p additional b s sends A into an accept state, s q. This also implies that a q b p also sends A to the accept state s q, and so a q b p is accepted by A. Now, we have supposed that L is the language accepted by A. Thus, since s is accepted by A, s L. But by definition of L, L consists of only strings with equal numbers of a s and b s. So, since p q, s L. Hence s L and s L; which is a contradiction. It follows that the supposition is false, and so there is no finitestate automaton that accepts L. Therefore L is not a regular language.
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