# Formal Languages and Automata Theory - Regular Expressions and Finite Automata -

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1 Formal Languages and Automata Theory - Regular Expressions and Finite Automata - Samarjit Chakraborty Computer Engineering and Networks Laboratory Swiss Federal Institute of Technology (ETH) Zürich March 7, 23 Contents Why should you read this? 2 A word about notation 2 3 Languages 2 4 Regular Expressions and the Corresponding Languages 3 5 Deterministic Finite Automata 5 6 From Regular Expression to an FA via NFA- and NFA 7 7 Nondeterministic Finite Automata 7 8 Euivalence of FAs, NFAs, and NFA-s 9 9 There exists an FA for every Regular Expression 3 There exists a Regular Expression for every FA 5 References 7 Why should you read this? If you already know what a regular expression is and what a finite state machine (or a finite automaton) is, then you probably also know that given a regular expression many times it is very difficult to directly construct the corresponding finite automaton. This is because there are two intermediate steps that you should go through before you can come up with the finite automaton. This note will explain to you what these two steps are. Most of the proofs given here are constructive in nature, i.e. they will help you to come up with an algorithm. All the proofs are to a large extent informal, and the aim has been to explain the basic underlying concepts. If you understand these, then you should be able to come up with more formal proofs on your own.

2 2 A word about notation There is a slight deviation in notation from what has been presented to you during the lecture, and what you are going to read here. Be assured that both the notations mean the same thing. First of all, we use the term finite automaton (FA). In many books this is also called a finite state machine. Both are exactly the same. In the script and in the lecture this was referred to as the Endlicher Automat. We denote an FA by the 5-tuple Q E δ A, where Q is a set of states, E is an alphabet, is the starting state, δ is a transition function, and A is the set of accepting states. In the script (and the slides presented in the lecture), this 5-tuple is denoted by E X f x F. E corresponds to our E, X corresponds to our Q (set of states), f corresponds to our δ (transition function), x corresponds to our (initial state), and lastly F corresponds to our A (set of accepting states). 3 Languages To define what we mean by language, we first have to define what an alphabet is. An alphabet is a finite set of symbols which are used to form words in a language. An example of an alphabet might be a set like a b. In this note we will denote an alphabet using the symbol E. A string over E is some number of elements of E (possibly none) placed in order. So if E a b then strings over E can be a ab bbaa abab aaaabbaab and so on and so forth. A very important string, which is always a string over E, no matter what E is, is the null string denoted by. This is the string with no symbols. If x is a string over E when we will use x to denote the length of x. Hence aba 3 and a, and. Note also that strings of length one over E are the same as elements of E. For some alphabet E, we use E to denote the set of all possible strings over E. Applying to some set is called the closure operation. So if E a b then a b a b aa ab ba bb aaa aab aba abb baa A language over E will be a set of strings over E, so it will be some subset of E. So examples of languages might be: a aa bb x a b x 3 x a b x 4 x a b x has eual number of as and bs x a b x always ends with an a It is possible to give many more examples of languages, but hopefully the idea is clear to you by now. Since languages are simply sets of strings, it is possible to generate new languages by applying standard operations on sets. So, for example, if L and L 2 are languages over E then L L 2 L L 2 L L 2, etc. are also languages over E. We use L to denote the complement of L. So L E L. 2

3 In addition to the standard set operations, we can also use operations on strings like concatenation to generate new languages. If x and y are strings over E then the concatenation of x and y, denoted by xy is a new string formed by writing the symbols of x followed by the symbols of y. So if x aa and y bbb then xy aabbb. If L and L 2 are two languages then we can generate a new language L L 2 which is defined as follows. L L 2 xy x L and y L 2 We can extend this notion of concatenation as follows. If L is a language over E then for any integer k we can define a language L k LL L k times L So L k is the set of strings obtained by concatenating k elements of L and L. Hence from the above discussion you might have already started feeling that there are principally two ways of specifying languages. In the first way, we define how a language can be constructed. An example of this can be, if L and L 2 are two languages, then we can construct a new language L L L 2. In the second approach, we define some means by which the strings that belong to the language (which we are trying to specify) can be recognized. An example of this might be L x a b x always ends with an a Clearly, there is no hard line distinguishing the two approaches. As you read this note you will learn more ways of generating languages, which will be more sophisticated than the way the example language that we gave before as an example of constructing languages. You will also read more about recognizing languages, which is more or less what is note is all about. 4 Regular Expressions and the Corresponding Languages Here we will see how new languages can be constructed from existing ones by using three simple operations union (denoted by ), concatenation, and the closure operation (denoted by ). If we start with the simplest possible languages those that consist of a single string which is either of length one, or is the null string and then apply any combination of the three operators, then the resulting languages are called regular languages. Such languages can be described by explicit formulas called regular expressions and they consist of the three operations mentioned, i.e. union, concatenation, and the closure operation. Example Language Regular Expression representing the language i.e. To explain our definition of regular languages a bit more clearly, we might say that they are languages containing only or a string of length one, together with those 3

4 which can be obtained from such languages by a finite seuence of steps, where each step consists of applying one of the three operations to the languages that were obtained at an earlier step. Example 2 is a regular expression representing the language consisting of all strings which consist of the substring preceded by arbitrarily many s. Applying our step by step procedure, we can obtain this by:. Apply concatenation to, yielding 2. Apply to, yielding 3. Apply concatenation to and, yielding Now we are ready to give formal definitions. Definition (Regular Expression) A regular expression over the alphabet E is defined as follows:. / is a regular expression corresponding to the empty language /. 2. is a regular expression corresponding to the language. 3. For each symbol a E, a is a regular expression corresponding to the language a. 4. For any regular expression r and s over E, corresponding to the languages L r and L s respectively, each of the following is a regular expression corresponding to the language indicated (a) rs corresponding to the language L r L s (b) r s corresponding to L r L s (c) r corresponding to the language L r 5. Only those formulas that can be produced by the application of rules -4 are regular expressions over E. Definition 2 (Regular Language) A language over the alphabet E is a regular language if there is some regular expression over E corresponding to it. Example 3 Let L be the language consisting of all strings of s and s that have even length. Note that since is of length zero, and zero is even, is in L. Note that L can be thought of as consisting of a number, possibly zero, of strings of length 2 concatenated. Hence the regular expression corresponding to L can be given as. Exercise What would be the regular expression corresponding to the language which consist of all strings of s and s that have odd length? Exercise 2 Prove that every finite language is regular. (Hint: Use mathematical induction) 4

6 L N, L Figure : Recognizing the language in Example 4 state after the entire input string has been read, then this indicates that the string is in the language. Observe that at the heart of the machine is a set of states and a function which receives a state and a symbol as input and outputs the next state. The crucial property of the machine is the finiteness of the set of states. So while recognizing a string, we need not remember the entire substring that has been read so far, but only which of the n (n 4 in our case) states the substring belongs to. These set of states is precisely the memory aspect that we talked about at the beginning of this section. Now we will give a formal definition of machines like the one we just described. We call such a machine a (deterministic) finite automaton or a finite-state machine. Definition 3 (Deterministic Finite Automaton) A finite automaton (FA) is a 5-tuple Q E δ A where Q is a finite set of states, E is a finite set of input symbols, Q is the initial state, and δ is a function from Q E to Q (known as the state transition function), and lastly A Q is the set of accepting states. Note that it is possible to extend our definition where the transition function takes as an input not a single symbol, but a string of symbols. Let δ denote this new transition function. Then if for any string y E and symbol a E, and state Q, δ ya δ δ y a. The rest of the functionally of the FA remains the same. Now let us state a theorem, whose proof we will work out later. Theorem A language L E is regular if and only if there is a FA that recognizes L. Using the above theorem, we can state another theorem describing an important property about regular languages. Theorem 2 If L and L 2 are regular languages in E, then L L 2 L L 2 L L 2 and L (complement of L ), are all regular languages. We will now try to sketch the proof of the above theorem. Let L and L 2 be recognized by the FAs M P E p δ A and M 2 Q E δ 2 A 2 respectively. Suppose that we wish to find an FA M R E r δ A to recognize L L 2. If we can find M then by Theorem we prove that L L 2 is regular. Note that the FA M while processing a string needs to keep track of the status of both M and M 2 simultaneously, and M accepts when either of M or M 2 accepts. How can we do this? Basically, M should remember the state it is in and M 2 should remember the state it is in, and M needs to remember the states of both M and M 2. For 6

7 this to happen, we construct the states of M to be pairs p where p P and Q. So R P Q. Initially M is in the state p and M 2 is in the state, and so M should be in the state p. If M is in the state p and receives input a, then it should go to the state δ p a δ 2 a, since δ p a and δ 2 a are the states to which M and M 2 would respectively go. Since M has to accept a string whenever M or M 2 does, the accepting states of M should be pairs p for which either p A or A 2. That leads to the fact that M accepts L L 2. By now you probably came to understand that exactly the same strategy works for the cases L L 2 and L L 2, except for the definition of accepting states. In the first case p should be an accepting state if both p A and A 2, and in the second case p is an accepting state if p A and A 2. For the last case i.e. L, note that we can reduce it to the third case by noting that L E L. That solves this case. However we can solve it in an even more simple way by noting that the M which accepts L will be the same as M accepting L with the only difference that the accepting states of M will be P A, i.e. exactly those states which are not the accepting states in M. Exercise 3 Let L and L 2 be the following subsets of : L x is not a substring of x L 2 x x ends with Construct the FAs accepting L, L 2, L L 2, and L L 2. 6 From Regular Expression to an FA via NFA- and NFA Given a regular expression, in most cases you will see that it can be broken into natural subparts. For each such subpart, it will be easy to construct an automaton, and then join these together to obtain a finite automaton corresponding to the given regular expression. However, you will see that it will be very natural to come up with an automaton for which there are multiple transitions from each state for a single input symbol. Moreover, there will also be states from which it will be possible to jump to other states with only the null string () as input. These are what we will call nondeterministic finite automaton with -transitions (NFA-). If there are no -transitions in any state then the automaton is simply a nondeterministic finite automaton. Given the NFA- obtained by joining together the NFA-s corresponding to the subparts of the given regular expression, it is possible to first convert this to a nondeterministic finite automaton without the -transitions and then convert this to a deterministic finite automaton. How to do all this is the subject of the remaining sections. 7 Nondeterministic Finite Automata Now we will consider machines similar to the ones described in Section 5, except that we will introduce some element of nondeterminism into them. These new machines will be capable of recognizing exactly the same languages as those recognized by deterministic finite automata (FA), but often these machines will be simpler to construct and will have fewer states than the corresponding FA. 7

8 , A B Figure 2: FA corresponding to the language in Example 5,, A B Figure 3: Nondeterministic finite automaton corresponding to the language in Example 5 The only difference these machines will have is that from each state there might be multiple transitions possible for any given input symbol, whereas in FAs for a given state and input symbol there was exactly one next state. Example 5 Consider the language L corresponding to the regular expression: Try to convince yourself that the FA shown in Figure 2 corresponds to the language L. The automaton shown in Figure 3 is similar to the one in Figure 2, but is not a FA since in state A, the input offers possible transitions to two different states. It is unclear as to which path should be taken if the input occurs in state A. However, if we decide on the rule that this automaton accepts strings for which there exists some path to an accepting state, then note that it reflects the structure in the regular expression much more than the FA in Figure 2 does. For any string in the language L it is easy to describe a path that leads to the accepting state B: start at A and continue looping back to A until the first occurrence of either or, use these to go to the accepting state B, and continue looping back to B for each of the remaining symbols. Note that this is exactly what the regular expression in Example 5 also says. Further, also note that for any string not in L, there doesn t exist any path which starts in A and leads to B. Now let us formally define machines like that described in Figure 3, which we will call nondeterministic finite automaton (NFA). 8

9 Definition 4 (Nondeterministic Finite Automaton) A nondeterministic finite automaton (NFA) is a 5-tuple Q E δ A where Q E and A are exactly the same as described in the case of a FA. The function is δ is different and is defined from Q E to 2 Q (i.e. the set of all possible subsets of Q). The above definition simply says that an NFA is exactly the same as a FA (or a Deterministic FA) except that from each state there can be multiple different transitions for any given input symbol. Note that it is also possible to define δ as a relation instead of a function, in which case δ Q E Q. The two definitions are clearly euivalent. As in the case of FA, it is also possible to extend the definition of δ to δ so that it accepts strings of symbols. So if for a string y E, symbol a E, and state p Q, δ p ya δ δ p y a. Lastly, we can generalize our definition of an NFA even further by including transitions i.e. state transitions which reuire only the null string () as input. We will see that this additional extension will simplify the process of finding an abstract machine for recognizing a given language. Therefore we can define an NFA with - transitions (denoted by NFA-) to be the same as an NFA, except that the transition function is defined as δ : Q E 2 Q. 8 Euivalence of FAs, NFAs, and NFA-s In an NFA- there are states from which it is possible to go to other states with only -transitions. Given any such state we need to compute all the states to which we can go from this state via -transitions. We will see shortly why this is important, but first we need to formalize this idea. Definition 5 (-closure) Let M -closure of S is the subset S Q E δ A be an NFA-. For a subset S of Q, the Q which can be defined as follows:. Every element of S is an element of S. 2. For any S, every element of δ is an element of S. 3. No elements of Q are in S unless they can be obtained from the above rules and 2. So the -closure of a set S is simply the set of states that can be reached from the elements of S using only -transitions. Algorithm to calculate S : Begin with T S, and at each step add to T the union of all the sets δ for T. Stop when the set T doesn t change any more. S is the final value of T. Now note that an FA can be considered to be a special case of an NFA, since a function from Q E to Q can in an obvious way be identified with a function from Q E to 2 Q, whose values are all sets with one element. Similarly, an NFA can be considered to be a special case of NFA-, one in which for each Q, δ /. Therefore, any language that is recognized by an FA can be recognized by an NFA, and any language that is recognized by an NFA can be recognized by an NFA-. The euivalence of FA, NFA, and NFA- in terms of the class of languages that they recognize, is proved with the additional fact that any language which is recognized by 9

10 an NFA- can be recognized by an FA. This completes the loop, and proves that allowing nondeterminism doesn t enlarge the class of languages that FAs can recognize. Theorem 3 Let L E, and suppose L is recognized by the NFA- M Q E δ A. There is an FA M 2 Q 2 E 2 δ 2 A 2 recognizing L. We will not give a formal proof of the above theorem, but we will sketch what the proof would look like. We will do this in two parts. First we will find an NFA M (without -transitions) recognizing L, and second we will find an FA euivalent to this NFA. PART I. Finding an NFA M Q E δ A recognizing L (a) Defining M (Eliminating the -transitions) We let Q Q and. The only thing that we have to do is define the transition function δ such that, if in the machine M we can get from a state p to a state using certain symbols together with -transitions, then in the machine M we should be able to get from p to using only those symbols, without the the -transitions. If in the machine M the initial state is not an accepting state, but it is possible to get from to an accepting state using only -transitions, then clearly the state in M should also be labeled as an accepting state. Hence, for any Q and a E, we define the transition function in M as δ a δ a Note that δ a is the set of all states that can be reached from, using the input symbol a but allowing -transitions both before and after. Therefore, the way we have defined δ, if M can move from p to using the input symbol a together with - transitions, then M can move from p to using the input a alone. Finally, as we mentioned before, it might be necessary to make an accepting state in M. For this, we define A A if A / in M and A otherwise (b) For any x E and Q, if x then δ x δ x This can be proved using induction on x. Note that the basis step, with x follows from our definition of δ above. So δ x δ x, and we have already defined δ x to be δ x. The rest of the proof is eually easy and we omit it here. (c) M recognizes L Since M recognizes L, a string x is in L if and only if δ x A /. On the other hand, x is accepted by M if and only if δ x A /. First let us consider the case when A / in M. Here A is defined to be A. If x then from part (b) above, δ x δ x. So x is in L if and only if x is accepted by M. If x then x is not accepted by either M or M. Hence it follows that M recognizes L. In the second case, when A, then A A. is accepted by both M and M. If x, δ x δ x. Either this set contains an element of A (in which case both M and M accept x), or this set contains neither nor any elements of A (in which case both M and M reject x). It is impossible for this set to contain

11 and no elements of A since if δ x, then since δ x is the -closure of another set (by definition), δ x contains and thus contains an element of A. Hence M and M accept exactly the same strings. PART II. For any NFA M Q E δ A recognizing L, there is an FA M 2 Q 2 E 2 δ 2 A 2 recognizing L (a) Definition of M 2 We are trying to eliminate the nondeterminism present in M, which means that in M 2 each combination of state and input symbol should result in exactly one state. Note that the transition function δ takes a pair a and returns a set of elements of Q. Now suppose we define our notion of state to be a set of elements from Q. Let s be such a subset of Q. For any element p s, there is a set δ p a of (possibly several) elements of Q to which M may go on input a. But for a single subset of elements s of Q there is a single subset of elements of Q in which M may end up this is the union of the sets δ p a for all elements p s. So for each state-input pair (with our new notion of state), there is one and only one state. The machine obtained in this way clearly simulates in a natural way the action of the original machine M, provided the initial and the final states are defined correctly. Hence we have eliminated nondeterminism by what might be called as subset construction: states in Q 2 are subsets of Q. From the above discussion, we can now define M 2 Q 2 E 2 δ 2 A 2 as follows. Q 2 the set of all subsets of Q 2 δ 2 a δ p a for Q 2 and a E p A 2 Q 2 A / The definition of A 2 follows from the fact that for a string to be accepted in M, starting in, the machine can end up only in sets of states which contain an element of A. (b) M 2 recognizes L Clearly this will follow if we can show that δ 2 2 x δ x (for every x) This can be proved using mathematical induction on x. At this point, note that the above proof gives us two algorithms. The first one to find an NFA euivalent to a given NFA-, and the second one allows us to find an FA euivalent to a given NFA. Example 6 Consider the NFA- shown in Figure 4(a). Figure 4(b) shows the corresponding NFA and Figure 4(c) the corresponding FA. Let us consider a few steps in obtaining the NFA and the FA in the above example. Let M Q E δ A be the NFA-, M Q E δ A be the NFA, and M 2 Q 2 E 2 δ 2 A 2 the FA.

12 2 3, 2 3 (a) (b) ,, φ (c) Figure 4: Obtaining an FA from a given NFA-. (a) denotes an NFA-, (b) the euivalent NFA, and (c) the euivalent FA 2

13 For finding δ, for example, we use δ δ p δ p The steps involved are: calculating, finding δ p for each p in this set, taking the union of the δ p s, and calculating the -closure of the result. We can see that from the state with input, M can move to state (using a -transition to first), or to state 3 (first to 2 with, and then from there to 3 with ). There are no - transitions from or 3, and so δ 3. A similar procedure will give the remaining values of δ. Finally since does not contain 3, A A 3. Figure 4(b) shows the NFA. For the DFA shown in Figure 4(c), first note that it has only seven states and not sixteen, which is the number of subsets of 2 3. This is because we create a state only when it is needed, during the construction process of the FA. From the two states that can be reached with one symbol are 3 and 2 3. To compute δ 2 2 3, we proceed as follows. δ 2 3 δ δ 3 / Since this state didn t appear yet during our construction, we add this state. Proceeding in this fashion we reach a point where for each state that is drawn so far, δ 2 and δ 2 are both already drawn, and this indicates that we have all the reachable states. Lastly, note that δ 2 3 δ 2 3 /. To complete the FA we add an extra state φ for handling this case. The resulting FA is the one shown in Figure 4(c). 9 There exists an FA for every Regular Expression Here we want to show that if R is a regular expression over the alphabet E, and L is the language in E corresponding to R, then there is a finite automaton M recognizing L. We will rely on the result of the last section where we proved the euivalence between an NFA- and an FA, and show that for a regular expression corresponding to a language L, we can construct an NFA- for recognizing L. Recall the definition of a regular expression that was given by Definition. We will give an algorithm for constructing an NFA-. Towards this we will show how to construct an NFA- corresponding to the rules 4(a) (c) in Definition. The NFA-s corresponding to rules 3 in Definition are really easy and are shown in Figure 5. Figure 5(a) shows the NFA- corresponding to the empty language /, (b) corresponding to the regular expression (which results in the language ), and (c) corresponding to the regular expression a (which results in the language a ). Now we will show how to construct an NFA- corresponding to 4(a) (c) in Definition. Note that this will give us all the essential ingredients for a proof of the fact that for every regular expression there is an NFA- (and hence an FA). The proof will use mathematical induction. The basis step was just shown above that there is an NFA- corresponding to the regular expressions /,, and a E. The induction step will show that if r and s are regular expressions for which there exist NFA-s then for the regular expressions obtained by applying rules 4(a) (c) of Definition we can also obtain NFA-s. 3

14 (a) (b) a (c) Figure 5: NFA-s corresponding to (a) /, (b), and (c) a E M r M s r s Accepting states in M s Accepting states in M r Figure 6: NFA- construction corresponding to concatenation Case. Suppose R rs. Let L r and L s be the languages corresponding to r and s respectively and let L L r L s correspond to R. Suppose the NFA-s M r Q r E r δ r A r and M s Q s E s δ s A s recognize L r and L s. We assume that Q r Q s / (by renaming states if necessary). We want to construct M Q E δ A which will recognize L L r L s. For this we use a very simple idea: we make the initial state of M i.e. r, the accepting states of M i.e. A A s, and we add transitions from every element of A r to s. Strings in the language L r L s will correspond to paths from r to an element in A r, then jump to s using the -transition, and finally finish in some state in A s. Figure 6 shows this construction. Try to convince yourself that if x L r L s then x is accepted by M, and for all strings x accepted by M, x x r x s where x r L r and x s L s. Case 2. Suppose R r s. L r and L s are recognized by M r Q r E r δ r A r and M s Q s E s δ s A s, where as before Q r Q s /. Let L L r L s and let M Q E δ A recognize L. The way we will construct M is, we take M r and M s, add a new initial state, and add -transitions from to the initial states of M r and M s (i.e. r and s ). The construction is shown in Figure 7 If x L r then M with x as input moves from to r by a -transition, and then from r to an element of A r (and therefore to an element of A, since A A r A s ). Hence x x is accepted by M. An exactly similar argument works for x L s. Clearly, the converse is also true, i.e. if x is accepted by M then x either belongs to L r or to L s, and hen x L r L s. Case 3. Suppose R r. Let M r Q r E r δ r A r recognize L r, and L L r. We want to construct M Q E δ A for recognizing L. This we do as follows. We 4

15 M r Accepting states in M r r M s Accepting states in M s s Figure 7: NFA- construction corresponding to union assume that is a state not in Q r. for Q Q Q r A for Q and a E δ a δ r a if Q / if δ δ r if Q r A r δ r if A r r if Now suppose x L r. If x then clearly x is accepted by M. For some m, let x x x 2 x m, where each x i L r. M moves from to r by a -transition. Then for each i, M moves from r to an element of A r by a seuence of transitions corresponding to those that happen in M r for x i, and finally M moves back to by a -transition. It follows that x x 2 x m x is accepted by M. The construction of M and this process of accepting x is illustrated in Figure 8 Clearly, for every x accepted by M there is a seuence of transitions begining and ending at, and it follows that x can be decomposed to the form x x 2 x m where each x i L r. There exists a Regular Expression for every FA Given an FA M Q E δ A recognizing a language L, we want to show that there exists a regular expression over E corresponding to L. 5

16 M r r Accepting states in M r (note that these are not accepting states in M) Figure 8: NFA- construction corresponding to closure For elements p and belonging to Q let L p x E δ p x L p is the set of strings that allow M to reach state if it begins in state p. If we can show that each language L p correspond to a regular expression, then since the language recognized by M is a union of the languages L for all A, a regular expression for L can be obtained by combining these individual regular expressions using. We will give an inductive proof that each language L p is regular. On what shall we base our induction? For this, consider the elements of Q to be labeled with integers from to N. Next we formalize the idea of a path going through a state s. If x E, we say x represents a path from p to through s if x can be written in the form x yz for some y and z with y z, δ p y s and δ s z. For any J we define the set L p J as follows. L p J x E x corresponds to a path from p to that goes through no state numbered higher than J. Note that L p N L p since N is the highest numbered state in the FA. Thus it will be sufficient to show that L p N is regular, and the way we shall show this is by proving that L p J is regular for each J, J N. We will use mathematical induction over J. For the basis step we have to show that L p is regular. Now a path from p to can go from p to without going to any state numbered higher than (i.e. without going through any other state) only if it corresponds to a single symbol, or if p and the path corresponds to the string. Thus L p is a subset of E and is regular. Now we want to show that if the language L p K is regular then L p K is also regular for p N, and K N. A string x L p K represents a path from p to that goes through no state numbered higher than K. There are two possible ways in which this can happen: the path could bypass the state K altogether, in which case x L p K and is regular, or the path could go from p to K, possibly looping back to K several times and then go from K to, never going to any states higher than K. In the later case, we can write x in the form x yx 2, where δ p x K δ K y K δ K x 2 6

17 If we include all the looping from K back to itself within the string y, then x L p K K and x 2 L K K. This simply says that before arriving at K for the first time and after leaving K for the last time, the path goes through no state numbered higher than K. Further, if y, and each separate loop in the path is represented by a string y i, then y y y 2 y l and each y i is an element of L K K K. Therefore y L K K K. Hence it follows from the above discussion that L p K L p K L p K K L K K K L K K This proves that L p K is regular. The results in this and the last two sections together show that a language is regular if and only if it is accepted by a finite automaton. This result is known as Kleene s theorem. References. Introduction to Automata Theory, Languages, and Computation, by John Hopcroft, Rajeev Motwani, and Jeffrey D. Ullman, Addison-Wesley, Reading, Massachusetts, USA, Elements of the Theory of Computation, by Harry R. Lewis and Christos H. Papadimitriou, Prentice-Hall International, Introduction to Languages and the Theory of Computation, by John C. Martin, McGraw-Hill Inc., 99. 7

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