Deterministic Finite Automata

Transcription

1 Deterministic Finite Automt Bkhdyr Khoussinov Computer Science Deprtment, The University of Aucklnd, New Zelnd In this lecture we introduce deterministic finite utomt, one of the foundtionl concepts in computing sciences. Finite utomt re the simplest mthemticl model of computers. Informlly, finite utomton is system tht consists of sttes nd trnsitions. Ech stte represents finite mount of informtion gthered from the strt of the system to the present moment. Trnsitions represent stte chnges descried y the system rules. Prcticl pplictions of finite utomt include digitl circuits, lnguge design nd implementtions, imge processing, modeling nd uilding relile softwre, nd theoreticl computing. 1 Strings nd lnguges Recll tht n lphet is finite set Σ of symols. In most cses, our lphets contin symols,, c, d,.... If Σ contins k letters then we sy tht Σ is k-letter lphet. A 1-letter lphet is clled unry lphet, nd 2-letter lphet is inry lphet. We will often del with the inry lphet {, }. Let Σ e n lphet. The elements of the lphet re clled input symols. A finite sequence of symols from Σ is clled string of the lphet. Sometimes strings re lso clled words. Thus, ech string is of the form σ 1 σ 2... σ n, where ech σ i is symol of the lphet. The length of given string is the numer of symols it hs. Thus, the length of strings,, nd re 4, 6, nd 2, respectively. There is specil string whose length is 0 clled the empty string. We denote this string y λ. Using mthemticl induction, it is esy to show tht the numer of strings of length n of k-letter lphet Σ is equl to k n. Let Σ e n lphet. The set of ll strings over the lphet Σ is denoted y Σ. Thus, Σ = {σ 1 σ 2... σ m σ 1, σ 2,...,σ m Σ, m N}. Note tht when m = 0, we hve the empty string λ nd therefore λ Σ. We denote the strings of the lphet y the letters u, v, w,.... Let u nd v e strings. Then the conctention of these two strings, denoted y u v, is otined y writing down u followed y v. For exmple, produces the string. It is cler tht the conctention opertion on words stisfies the equlity u (v w) = (u v) w for ll words u, v, w. Note tht λ u = u λ for ny string u. Often, insted of writing u v we my omit the dot sign nd write uv insted. Let u e string. The nottion u n represents the string otined y writing down u exctly n times. Thus, u n is the string otined y conctenting u to itself n times. For exmple, () 3 =. When n = 0 then u n is the empty string λ for ny string u.

2 Let u nd w e strings. We sy tht w is sustring of u if w occurs in u. More formlly, w is sustring of u if u = u 1 wu 2 for some strings u 1 nd u 2. For exmple, is sustring of while is not. Clerly, every string u is sustring of itself. We sy tht w is prefix of u if u cn e written s wu 1. For exmple, the prefixes of the string re λ,,,,, nd. Now we give n importnt definition: Definition 1. A lnguge is suset of Σ, where Σ is n lphet. When we use the term lnguge we lwys ssume, implicitly or explicitly, tht we re given n lphet. Here re some exmples of lnguges:, Σ, {λ,,,,,,...}, {, }, {w is sustring of w}. We denote lnguges y cpitl letters U, V, W, L, etc. We now define severl opertions on lnguges. Given two lnguges U nd V, their union is U V ; their intersection is U V ; nd the complement of U is Σ \ U. Below we discuss two dditionl opertions on lnguges. The conctention opertion. Let U nd W e lnguges. The conctention of U nd W is the lnguge: {u w u U, w W }. We denote this lnguge y U W. For exmple, if U = {, } nd W = {, } then U W = {,,, }. Here is nother exmple. Let U = {}. Then for ny lnguge L, U L = {u u L}. The str opertion. This opertion is lso often clled the Kleene s str opertor. Let U e the lnguge. Consider the following sequence of lnguges: U 0, U 1, U 2, U 3, U 4,... where U n with n N is defined recursively s follows: U 0 = {λ}, U 1 = U, U 2 = U U, nd U n = U n 1 U. We cn tke the union of ll these lnguges nd denote the resulting lnguge y U. Thus, U = U 0 U 1 U 2 U The lnguge U is clled the str of the lnguge U. More informlly, U is the set of strings otined y finite numer of conctentions pplied to strings from the lnguge U. For exmple, if U = {} then U = {λ,,,,,...}. Note tht the str of every lnguge contins λ. The lnguge U is lwys n infinite lnguge prt from the cses U = nd U = {λ}. 2 Deterministic finite utomt Let Σ e n lphet nd U e lnguge of Σ. A typicl prolem tht we wnt to solve is the following. Design n lgorithm tht, given string w, determines whether or not w is in U.

3 Here is n exmple. Let U e the lnguge consisting of ll strings w over the lphet {, } such tht w contins the sustring. We wnt to design n lgorithm tht, given string w, determines whether or not w contins sustring. Here is the Find-(w) lgorithm. Let w e the string w = σ 1... σ n. 1. Initilize vriles i nd stte of integer type s i = 1 nd stte = 0 2. While i n do () If stte = 0 nd σ i = then set stte = 0. () If stte = 0 nd σ i = then set stte = 1. (c) If stte = 1 nd σ i = then set stte = 1. (d) If stte = 1 nd σ i = then set stte = 2. (e) If stte = 2 nd σ i = then set stte = 3. (f) If stte = 2 nd σ i = then set stte = 0. (g) If stte = 3 nd σ i {, } then stte = 3. (h) Increment i y one. 3. If stte = 3 then output ccept. Otherwise output reject. The importnt fctor in this lgorithm is the vlues of the vrile stte. These vlues re re 0, 1, 2, nd 3. Cll these the sttes of the progrm. The progrm mkes its trnsitions from one stte to nother depending on the input σ i it reds. Thus, we hve trnsition function ssocited with the progrm. The tle of the trnsition function is presented in Tle Tle 1. Trnsition function for the Find-(w) The stte 0 is the initil stte of the lgorithm. The stte 3 is the ccepting stte s w contins when stte = 3. Finlly, if the lgorithm outputs reject then the stte vrile hs vlue 0, 1, or 2. Thus, we hve given finite stte nlysis of the Find-(w) lgorithm. Now we strct from this exmple nd give the following definition: Definition 2. A deterministic finite utomton is 5-tuple (S, q 0, T, F, Σ), where 1. S is the set of sttes. 2. q 0 is the initil stte nd q 0 S. 3. T is the trnsition function T : S Σ S. 4. F is suset of S clled the set of ccepting sttes. 5. Σ is n lphet.

4 We revite deterministic finite utomton s DFA. We often write (S, q 0, TF) insted of (S, q 0, T, F, Σ). We use letters A, B,... to denote finite utomt. Thus, the stte nlysis of the Find-(w) lgorithm ove gives us the utomton (S, 0, T, F), where: S = {0, 1, 2, 3}, 0 is the initil stte, T is presented in Tle 1 ove, nd F = {3}. As for trnsition functions from the previous lecture, we cn visulize finite utomt s leled grphs. Let (S, q 0, T, F) e DFA. The sttes of the DFA re represented s vertices of the grph. We put n edge from stte s to stte q if there is n input signl σ such tht T(s, σ) = q. The edges leled with σ re clled σ-trnsitions. The initil stte is presented s vertex with n ingoing rrow without source. The finl sttes re vertices tht re douly circled. We cll this visul presenttion trnsition digrm of the utomton. For exmple, the trnsition digrm of the utomton for the Find-(w) lgorithm is given in Figure Fig. 1. Trnsition digrm for DFA for the F ind-(w) lgorithm Another exmple of DFA is presented in Figure 2. Fig.2. An exmple of trnsition digrm

5 Let M = (S, q 0, T, F, Σ) e DFA. Tke ny string w = σ 1... σ n of the lphet Σ. The run of the utomton on this string is the sequence of sttes s 1, s 2,..., s n, s n+1 such tht T(s i, σ i ) = s i+1 for ll i = 1,...,n nd tht s 1 is the initil stte. Thus, the run of M on string w = σ 1...σ n cn e thought of s the execution the following lgorithm Run(M, w): 1. Initilize s = q 0 nd i = Print s. 3. While i n do () Set σ = σ i. () Set s = T(s, σ). (c) Print s. (d) Increment i This lgorithm outputs the sttes of the DFA M s it reds through the string w. First, the lgorithm prints the initil stte. It then reds the first input σ 1, mkes trnsition from the initil stte to stte T(q 0, σ 1 ) nd outputs T(q 0, σ 1 ), reds the next symol σ 2, mkes trnsition, nd so on. Note tht every run of M is pth strting from the initil stte q 0. One cn lso visulize the run s pth leled y the string w nd strting with q 0. For exmple, the pth 0, 0, 0, 1, 1, 2, 3, 3 is the run of the utomton in Figure 1 on the string. On the sme input string the utomton on Figure 1 produces the run 0, 0, 0, 1, 4, 7, 5, 2. Definition 3. Let M = (S, q 0, T, F) e DFA. We sy tht M ccepts the string w = σ 1...σ n if the run of M s 1,...,s n, s n+1 on w is such tht the lst stte s n+1 is in F. We cll such run n ccepting run. Note tht the run of M on w is lwys unique. Therefore, the run is either ccepting or rejecting ut not oth. For exmple, the utomton in Figure 1 ccepts those strings tht contin s sustring nd rejects ll other strings. The utomton in Figure 2 ccepts ll the strings tht hve in the third position from the lst nd rejects ll other strings. Thus, we now define the following importnt concept: Definition 4. Let M = (S, q 0, T, F, Σ) e DFA. The lnguge ccepted y M, denoted y L(M), is the following lnguge: {w the utomton M ccepts w}. A lnguge L Σ is DFA recognizle if there exists DFA M such tht L = L(M). Now we give severl simple exmples: 1. Consider DFA with exctly one stte. If the stte is the ccepting stte then the utomton ccepts the lnguge Σ. Otherwise, if the stte is not ccepting, the utomton ccepts the empty lnguge.

6 2. Consider the lnguge {w} consisting of one word w = σ 1... σ n. This lnguge is DFA recognizle. The utomton recognizing this lnguge just rememers the entire string. Indeed, the following DFA (S, 0, T, F) recognizes the lnguge: () S = {0, 1, 2, 3, 4,..., n + 1}. () 0 is the initil stte. (c) For ll i n 1, T(i, σ i+1 ) = i + 1. In ll other cses T(s, σ) = n + 1. (d) The ccepting stte is n. The trnsition digrm of this utomton when w = is in Figure Fig.3. A DFA recognizing the lnguge {}. 3. The lnguge L = {w w {, } } is DFA recognizle. The trnsition digrm of DFA tht recognizes the lnguge is in Figure 4. Formlly, here is the utomton M = (S, q 0, T, F) recognizing L: () S = {0, 1, 2} () The initil stte is 0. (c) The trnsition tle is defined s follows: T(s, σ) = 1 for s = 0 nd σ =, T(s, σ) = 1 for s = 1 nd σ {, }, nd T(s, σ) = 2 in ll other cses. (d) 1 is the ccepting stte Fig.4. A DFA recognizing the lnguge {w w {, } }.

7 3 Constructing finite utomt Suppose tht we re given lnguge L nd re sked to design DFA tht recognizes the lnguge L or rgue tht such DFA does not exist. A very helpful ide for solving this type of prolem is to pretend tht you re the mchine nd see how you would go out solving the prolem. We explin this in the following two exmples. Assume tht we wnt to construct DFA tht recognizes the lnguge L = {w w {} nd w contins n odd numer of s nd n even numer of s}. Let s pretend tht we re the mchine. We strt reding from left to right the input string w, nd see symols or. Do we need to rememer the entire string we hve seen so fr in order to tell whether we hve pssed through n odd numer of s nd n even numer of s? Our nswer is no ecuse of the following oservtion. We keep two coins oth hving two colored sides lue nd red. We ssocite the first coin with the symol nd the second with the symol. When we strt, oth coins re on their lue sides. Reding w, when we see we flip the first coin nd when we see we flip the second coin. Thus, if the first coin is on its lue side thn we hve seen n even numer of s, nd if the coin is on its red side then we hve seen n odd numer of s. The sme holds true for the symol. Therefore, t t ny given time, our stte is determined y the current colors of the coins. There re four possile sttes only: 1. The first coin is lue nd the coin is lue. Code this stte s The first coin is red nd the coin is lue. Code this stte s The first coin is lue nd the coin is red. Code this stte s The first coin is red nd the coin is red. Code this stte s 3. When we finish reding the whole input string w we ccept the string if the first coin is on its red side nd the second coin is in its lue side. In ll other cses we reject the string. Thus, ll this resoning helps us to uild DFA tht recognizes L. The Figure 5 is the trnsition digrm of the utomton we hve uilt Fig.5. A DFA recognizing the lnguge {w w hs n odd numer of s nd n even numer of s}.

8 Consider the following lnguge L = { n n n N}. Let s pretend to e mchine tht recognizes L. Given n input string w, we hve the following constrints s dictted y the definition of run of DFA. We must red the string w from left to right. We re not llowed to come ck to ny position in w which we hve pssed. Assume w strts with. We then reject w. Assume tht w strts with n. We red nd rememer tht we hve red one. Informlly, this tells us tht we hve to crete stte, let s denote it y s 1, tht rememers tht w strts with n. If is the next nd the lst symol then we ccept w. Otherwise, we reject w. If the second symol is, then we rememer tht we hve red two s. Informlly, this tells us tht we hve to crete stte, let s denote it y s 2, tht rememers tht w strts with. We must keep s 1 nd s 2 seprte. Indeed, if we declre s 1 = s 2 then we hve to ccept the string which is clerly not desirle. Lets continue this on. Assume tht w is of the form n u, where n 1, nd we hve creted sttes s 1, s 2,..., s n. Stte s i detects tht the input string strts with i. Consider the first symol of u. If it is then we cn use our sttes s n, s n 1,..., s 1 nd control the next n symols of w. If ll the symols of u re nd there re exctly n of them, we ccept w nd otherwise reject. However, if the first symol of u is, then we need to rememer tht we hve red exctly n + 1 symols of. This tells us tht we hve to crete stte, let s denote it y s n+1, tht rememers tht w strts with n+1. We must keep s n+1 seprte from ll the sttes s 1, s 2,..., s n we hve uilt so fr. Thus, we see tht in order to recognize L, our informl nlysis tells us tht we need to hve infinitely mny sttes. This intuitive resoning sed on pretending to e mchine for recognizing L gives us reson to elieve tht no DFA recognizes L. Indeed, our intuition here is correct s will e shown in Lecture 13. Constructing utomt for the union opertion. Here is design prolem we wnt to solve. Assume we re given two DFA M 1 = (S 1, q (1) 0, T 1, F 1 ) nd M 2 = (S 2, q (2) 0, T 2, F 2 ). These two DFA recognize lnguges L 1 = L(M 1 ) nd L 2 = L(M 2 ). We wnt to design DFA M = (S, q 0, T, F) tht recognizes the union L 1 L 2. We use the method of pretending to e mchine tht recognizes L 1 L 2. The first ttempt for uilding the desired M is this. For n input string w, we simulte the first mchine M 1. If M 1 ccepts w then w L 1 L 2. If M 1 rejects w then we run the second mchine M 2 on w. If M 2 ccepts w then w L 1 L 2. Otherwise, we reject w. The prolem with this ide is tht we cn not red the input w twice. If we re not llowed to run M 1 nd M 2 on w turn y turn then why don t we run M 1 nd M 2 on w in prllel? We tke the following pproch. On input string w, we run M 1 nd M 2 simultneously s follows. Initilly, we rememer the initil sttes q (1) 0 nd q (2) 0. We red the first input symol σ 1 of w, mke simultneous trnsitions on M 1 nd M 2, nd rememer the sttes s (1) 1 = T 1 (q (1) 0, σ 1) nd s (2) 1 = T 2 (q (2) 0, σ 1). We then repet this process y mking simultneous trnsitions from s (1) 1 to s (1) 2 = T 1 (s (1) 1, σ 2) on M 1, nd from s (2) 1 to s (2) 2 = T 2 (s (2) 1, σ 2) on M 2, where σ 2 is the second letter of w. We continue this on. Once we finish the simultneous runs of M 1 nd M 2 on w, we look t the resulting end sttes of these two runs. If one of these sttes is ccepting then we ccept w nd otherwise we reject w. Thus, t ny given stge of the two runs we rememer pir (p, q), where p S 1 nd q S 2. Our trnsition on this pir on input σ is simply the simultneous trnsitions from p to T 1 (p, σ) of M 1 nd from q to T 2 (q, σ) of M 2. We cn e in t most S 1 S 2 sttes.

9 Now we formlly define the DFA M = (S, q 0, T, F) tht recognizes the lnguge L 1 L 2 : 1. The set S of sttes is S 1 S The initil stte is the pir (q (1) 0, q(2) 0 ). 3. The trnsition function T is the product of the trnsition functions T 1 nd T 2, tht is: T((p, q), σ) = (T 1 (p, σ), T 2 (q, σ)), where p S 1, q S 2, nd σ Σ. 4. The set F of finl sttes consists of ll pirs (p, q) such tht either p F 1 or q F 2. The proof tht M is DFA recognizing the union L 1 L 2 is informlly explined s follows. We first show tht, nn the one hnd, if w is in L 1 L 2 then the utomton M ccepts w. Indeed, if w L 1 L 2 then either M 1 ccepts w or M 2 ccepts w. In either cse, since M simultes oth M 1 nd M 2, the string w must e ccepted y M. On the other hnd, if w is ccepted y M then the run of M on w cn e split into two runs: one is the run of M 1 on w nd the other is the run of M 2 on w. Since M ccepts w, it must e the cse tht one of the runs is ccepting. The nottion for the utomton uilt is this: M 1 M 2. Constructing utomt for the intersection opertion. Assume we re given two DFA M 1 = (S 1, q (1) 0, T 1, F 1 ) nd M 2 = (S 2, q (2) 0, T 2, F 2 ). These two DFA recognize lnguges L 1 = L(M 1 ) nd L 2 = L(M 2 ). We wnt to design DFA M = (S, q 0, T, F) tht recognizes the intersection L 1 L 2. We use the ide of constructing the DFA for the union of lnguges. Given n input w, we run M 1 nd M 2 on w simultneously s we explined for the union opertion. Once we finish the runs of M 1 nd of M 2 on w, we look t the resulting end sttes of these two runs. If oth end sttes re ccepting then we ccept w nd otherwise we reject w. Formlly, we define the DFA M = (S, q 0, T, F) tht recognizes the lnguge L 1 L 2 s follows: 1. The set S of sttes is S 1 S The initil stte is the pir (q (1) 0, q(2) 0 ). 3. The trnsition function T is the product of the trnsition functions T 1 nd T 2, tht is: T((p, q), σ) = (T 1 (p, σ), T 2 (q, σ)), where p S 1, q S 2, nd σ Σ. 4. The set F of finl sttes consists of ll pirs (p, q) such tht p F 1 nd q F 2. It is not difficult to prove tht M constructed recognizes L 1 L 2. The nottion for the utomton uilt is this: M 1 M 2. Constructing utomt for the complementtion opertion. Given DFA M = (S, q 0, T, F) we wnt to design DFA tht recognizes the complement of L(M). This is indeed very simple prolem s for this we just keep the sme sttes, the initil stte, nd the trnsition

10 function T. The only chnge we mke is tht we swp the ccepting sttes with the non-ccepting sttes. Thus, the utomton tht recognizes the lnguge Σ \L(M) is M (c) = (S, q 0, T, S \F). The most importnt prt in this construction is the following oservtion. On every input w, the DFA M (nd hence M (c) ) hs unique run. Therefore, M ccepts w if nd only if M (c) rejects w. 4 Minimiztion lgorithm Suppose tht we re given DFA A. The numer of sttes of A mesures the complexity of A. If A hs mny sttes then this could e due to two resons. One is tht the lnguge recognized y A cn e complex. Therefore there could e no wy to reduce the numer of sttes of A without chnging the lnguge recognized. The second is tht A could e implemented inefficiently nd therefore cn hve mny redundnt sttes. In this cse we would like to remove those redundnt sttes nd mke the utomton A smller. To mke ll these things more precise we give the following definition: Definition 5. We sy tht DFA A is equivlent to DFA B if L(A) = L(B). Thus, if A nd B re equivlent DFA, then for ny string w, A ccepts w if nd only if B ccepts w. Consider the two utomt, A 1 nd A 2, in Figure 6: A 1 is the utomton with three sttes nd A 2 with two sttes. Both A 1 nd A 2 re equivlent nd recognize the lnguge L = {w w {, } nd ech in w is followed y n lter on fter the }. There does not exist one stte utomton recognizing L. Therefore the two stte utomton A 2 hs the fewest possile sttes recognizing L. Fig.6. Two equivlent utomt Our gol is now to provide method tht reduces the redundnt sttes from given utomton. We give the following definition. Definition 6. We sy tht DFA A is miniml if no DFA B equivlent to A hs fewer sttes thn A itself.

11 For exmple, the two stte utomton A 2 in Figure 6 is miniml. Note tht, y dfeinition, for every DFA A there exists miniml utomton equivlent to A. Let A = (S, q 0, T, F) e DFA. Tke stte s S nd string w. Strting t stte s, run the utomton on string w. We denote the lst stte of this run y T(s, w). Formlly, the stte T(s, w) is defined inductively s follows: T(s, λ) = s, T(s, wσ) = T(T(s, w), σ), where w Σ. We now give the following importnt definition tht is used in constructing miniml utomt. Definition 7. Let p nd q e two sttes of the DFA A. We sy tht p nd q re indistinguishle if for ll strings w Σ, T(p, w) F if nd only if T(q, w) F. Otherwise, we sy tht the sttes p nd q re distinguishle For exmple, consider the utomton A 1 with three sttes in Figure 6. The two finl sttes re indistinguishle. If sttes p nd q re distinguishle then this is equivlent to sying the following. There exists string w such tht either T(p, w) F nd T(q, w) F or T(p, w) F nd T(q, w) F. We cll ny string w tht stisfies one of these two properties witness tht distinguishes p nd q. As n exmple, ny ccepting stte p is distinguishle from non-ccepting stte q. The string λ is witness tht distinguishes p nd q ecuse T(p, λ) F nd T(q, λ) F. Here is now the prolem we would like to solve. Design n lgorithm, tht given DFA A = (S, q 0, T, F, Σ), produces miniml DFA equivlent to A. In order to solve this prolem, let s nlyze the DFA A presented in Figure Fig. 7. A DFA for minimiztion The utomton A hs 9 sttes. We wnt to reduce the numer of sttes of this utomton nd uild smller utomton equivlent to it. First of ll, we remove ll the sttes tht re

12 not rechle from the initil stte. There is only one such stte which is 8. So we tke the stte 8 out nd the resulting utomton is still equivlent to the given one. From now on we ssume tht stte 8 does not exist. Secondly, the ide is this. For ech stte i we wnt to collpse ll sttes j indistinguishle from i into one stte denoted y [i]. Thus, the stte [i] is the set {j j is indistinguishle from i}. Clerly, every stte i elongs to [i]. Therefore [i]. This ide is implemented s follows. We form 8 8 tle whose rows nd columns re nmed y the sttes. The cell (i, j) in the tle represents the cell on row i nd column j. Initilly we hve the Tle Tle 2. Our gol is to put X s into those cells (i, j) such tht sttes i nd j re distinguishle. We know tht if one of the sttes i, j is ccepting nd the other is not then i nd j re distinguishle. Thus, we put X into ll cells (i, j) such tht either i or j is in F ut not oth. This produces Tle X X 1 X X 2 X X X X X X 3 X X 4 X X 5 X X 6 X X X X X X 7 X X Tle 3. We process Tle 3 s follows. Tke ny cell (p, q) tht is not mrked with n X. Put n X into cell (p, q) if there is symol σ {, } such tht the cell (r, s) is mrked with n X, where r = T(p, σ) nd s = T(q, σ). This process produces Tle 4.

13 We then process Tle 4 s follows in the sme wy s processed the previous tle. Nmely, tke ny cell (p, q) without n X. Put n X into the cell (p, q) if there is symol σ {, } such tht the cell (r, s) is mrked with n X, where r = T(p, σ) nd s = T(q, σ). This process produces Tle 5. Finlly, we process Tle 5 in the sme wy s ove, nd produce the finl Tle 6. We stop when there exists no cell such tht the cell is without n X nd cn still get n X with the method descried X X X X X 1 X X X X X 2 X X X X X X 3 X X X X X 4 X X X X X X X 5 X X X X X 6 X X X X 7 X X X X X X Tle X X X X X X 1 X X X X X 2 X X X X X X 3 X X X X X X 4 X X X X X X X 5 X X X X X X X 6 X X X X X 7 X X X X X X Tle 5. Consider the finl Tle 6. Tke ny stte i of the utomton, go through the ith row of the tle, collpse ll the sttes j such tht the cell (i, j) is not mrked with n X into one stte denoted y [i]. Thus, we hve [0] = {0, 3}, [1] = {1, 7}, [2] = {2, 6}, [4] = {4}, nd [5] = {5}. Thus we hve uilt new utomton A new with sttes [0], [1], [2], [4], nd [5] presented in Figure elow. Keeping the ove exmple in mind, we now present the minimiztion lgorithm nd prove its correctness. Let A = (S, q 0, T, F) e DFA. Here is the Minimiztion(A) lgorithm:

14 X X X X X X 1 X X X X X X 2 X X X X X X 3 X X X X X X 4 X X X X X X X 5 X X X X X X X 6 X X X X X X 7 X X X X X X Tle 6. [0] [1] [2] [4] [5] Fig.8. The DFA A new 1. Remove sttes tht re not rechle from the initil stte of A. (Comment: From now on we ssume tht ll the sttes of A re rechle from the initil stte). 2. Let 0, 1,..., n 1 e ll the sttes of A, where 0 is the initil stte. Mke n n n tle whose rows nd columns re leled with sttes. 3. Initilize the tle y mrking with n X ll the cells (i, j) such tht one of the sttes i, j is n ccepting stte nd the other is not. 4. Repet the following until no cell (i, j) cn e mrked with n X: () Mrk cell (p, q) with n X if (p, q) does not hve mrk, nd there is σ Σ such tht the cell (T(p, σ), T(q, σ) hs mrk. 5. For ech stte i define [i] to e the set {j the cell (i, j) is not mrked}, where i = 0,...,n Construct the following new DFA: () Declre ech [i] to e stte of the new DFA. () The initil stte is [0]. (c) Put trnsitions from ll [i] to [T(i, σ)] nd lel the trnsition y σ.

15 (d) Declre [i] to e n ccepting stte if i is n ccepting stte. The first step of the lgorithm produces the utomton tht is equivlent to the given one. Note tht Step 6 of the lgorithm defines sttes, the initil stte, trnsition, nd ccepting sttes of the new utomton. Our gol is to show tht the new utomton is correctly defined nd is miniml. To do this we now nlyze the lgorithm. There re n 2 cells in the initil tle defined t Step 2, the numer of mrked cells in the tle increses with ech itertion in Step 4, nd hence the lgorithm must terminte. Fct 1If cell (p, q) is mrked with n X then p nd q re distinguishle. If (p, q) is mrked t Step 2 of the lgorithm then p nd q re clerly distinguishle. The witness tht distinguishes these sttes is λ. Consider the loop of the lgorithm in Step 4. Assume (p, q) is mrked with n X ecuse there exists σ Σ such tht (T(p, σ), T(q, σ)) hs lredy een mrked. By inductive ssumption the sttes p = T(p, σ) nd q = T(q, σ) re distinguishle. Let w e the string distinguishing p nd q. Then σw distinguishes p nd q. Indeed, in this cse we hve T(p, σw) = T(T(p, σ), w) = T(p, w) nd T(q, σw) = T(T(q, σ), w) = T(q, w). This explins the fct. Fct 2.If cell (p, q) does not hve n X then p nd q re indistinguishle. Assume string w = σ 1... σ k distinguishes p nd q. Consider the following two sequences p 0 = p, p 1 = T(p 0, σ 1 ), p 2 = T(p 1, σ 2 ),...,p n = T(p n 1, σ n ) nd q 0 = q, q 1 = T(q 0, σ 1 ), q 2 = T(q 1, σ 2 ),..., q n = T(q n 1, σ n ). The pir (p n, q n ) gets n X y Step 2 of the lgorithm since w distinguishes p nd q. Therefore the pir (p n1, q n 1 ) must get mrk. Continue this on we see tht the pir (p 0, q 0 ) lso gets mrk X. This contrdicts the ssumption of the lemm. This explins the fct. For ech stte i S, the lgorithm considers the set [i] = {j the cell (i, j) is not mrked} nd declres [i] to e stte of the new DFA. These sttes stisfy the following properties (the reder, check ll these properties!): Property 1. Every stte i of the originl utomton is in [i]. Property 2. For ny two sttes [i] nd [j] either [i] = [j] or [i] [j] =. Property 3. If i nd j re indistinguishle then T(i, σ) nd T(j, σ) re lso indistinguishle. In the M inimiztion(a) lgorithm Step 5 constructs the deterministic finite utomton A new = (S new, q new, T new, F new ) s follows: 1. The set S new consists of ll the sttes [i]. 2. The initil stte q new is [0]. 3. T neq ([i], σ) = [T(i, σ)] for ll [i] S new nd σ Σ. 4. F new = {[i] i F }.

16 By properties 1 nd 2, ove the numer of sttes in A is not smller thn the numer of sttes in S new. Property 3 ove tells us tht this definition of T new is correct. In other words, T new is function from S new Σ to S new. Fct 3The DFA utomt A new nd A re equivlent. The proof of this fct follows from how we defined A new nd the previous fcts. Indeed, ssume tht w is ccepted y A. This mens tht T(0, w) F. By the definition of T new we then must hve T new ([0], w) = [T(0, w)]. Since T(0, w) F it must e the cse tht [T(0, w)] F new. Thus, w is ccepted y A new. Now ssume tht A new ccepts w. Then T new ([0], w) F new which mens [T(0, w)] F new. Therefore, T(0, w) F nd hence A ccepts w. Now we wnt to show tht the DFA A new is miniml DFA equivlent to the given DFA A. For this we need to show tht ny miniml DFA equivlent to A does not hve fewer sttes thn A new. Thus, let A = (S, q 0, T, F ) e miniml utomton equivlent to A. We wnt to show tht A new nd A hve the sme numer of sttes. We reson recursively tht defines function f from sttes of A into the sttes of A new. Bsis: In this cse we mp q 0 to [0]. Thus we hve prtil function f : S S new such tht no two sttes in S re mpped into the sme stte in S new. Inductive step. Our hypothesis is the following. We hve uilt prtil function f : S S new such tht no two sttes in S re mpped into the sme stte in S new. Moreover, for ny s Domin(f) the following property is stisfied. The sttes s nd f(s) re indistinguishle in the sense tht for ll w Σ, T(s, w) F if nd only if T new (f(s), w) F new. Assume tht f is not totl. Tke p Domin(f) nd σ Σ such tht T(p, σ) = q nd q Domin(f). Such p nd σ must exist s f is not totl. Define the vlue of f on q s follows. Declre the vlue of f on q to e T new (f(p ), σ). This vlue T new (f(p ), σ) of f cn not e equl to ny other vlue s S new from the rnge of f defined in the previous steps of the construction of f. Otherwise, s nd T new (f(p ), σ) would e indistinguishle. Continuing this construction of f, when the construction termintes we hve totl ijective function from S to S new. This shows tht A new hs the sme numer of sttes s the miniml the miniml utomton A. Thus, we hve proved the lgorithm Minimiztion(A) is correct. In other words, given DFA A the lgorithm produces miniml DFA equivlent to A.