1 Nondeterministic Finite Automata

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1 1 Nondeterministic Finite Automt Suppose in life, whenever you hd choice, you could try oth possiilities nd live your life. At the end, you would go ck nd choose the one tht worked out the est. Then you could decide who to mrry, which jo to ccept, or which nswer to give on n exm knowing the future conseuences. This is the ide of nondeterminism for finite utomt. It s not prcticl to uild devices like this, ut it does help in pplictions of finite utomt to mke use of this concept. In deterministic utomton, in stte s with n input there is one nd only one stte t tht it cn go to in the next time instnt. In nondeterministic utomton, there cn e mny sttes t 1, t 2,..., t n tht it cn go to, or possily no sttes t ll. Also, nondeterministic utomt hve the possiility to go from one stte to nother without reding ny input, if there is n ϵ trnsition etween the two sttes. 1.1 Formlism Formlly, insted of trnsition function δ there is trnsition reltion with K (Σ {ϵ}) K. If (, u, p) then in stte, reding u, the utomton cn go to stte p. Also, u cn e ϵ, in which cse the utomton cn go from stte to stte p without reding nything. Therefore

2 nondeterministic finite utomton cn e represented s uintuple (K, Σ,, s, F ), much s efore. It cn lso e represented s grph; the difference is tht there cn e more thn one rrow lelled with the sme symol coming out of stte, or possily none t ll. The computtion of nondeterministic utomton is similr to tht of deterministic utomton, except tht for configurtions, (, w) M (, w ) iff there is u in Σ {ϵ} such tht w = uw nd (, u, ). Note tht u cn e ϵ. Then M is defined s efore, nd we sy M ccepts w if there is stte F such tht (s, w) M (, e). Also, L(M), the lnguge recognized y M, is the set of strings ccepted y M. Deterministic finite utomt cn e regrded s specil cse of nondeterministic finite utomt, in which there re no ϵ trnsitions nd is function. 1.2 Exmple Here is n exmple: r s Why is this utomton nondeterministic? Wht lnguge does it recognize? Wht re K, Σ,, s, nd F for it? This utomton rejects the strings nd. Consider the computtion of this utomton on the input. There is more thn one possiility:

3 r r s leds to cceptnce. r leds to filure. Becuse there is t lest one computtion seuence leding to cceptnce the string is ccepted y this utomton. To show tht nondeterministic utomton rejects n input string, it is necessry to exmine ll computtion seuences. There could e exponentilly mny possile computtions on given input. This mkes simulting the utomton very expensive. To increse the efficiency, we cn summrize ll computtions of this utomton on this input using sets this wy: Prolem {} {, r} {, r} {r, s} {r, s} Find n euivlent deterministic finite utomton. 1.3 Exmple Here is nondeterministic utomton with ϵ rrows: e e r s

4 Simulte this utomton on inputs nd. The ϵ rrows mke this hrder to do. Give K, Σ,, s, nd F for this utomton. Wht lnguge does it recognize? 1.4 Exmple Note tht nondeterministic utomton my not e le to do nything. Consider the following exmple: r If it receives s input, it cn t do nything. 1.5 Prolem Give nondeterministic finite utomton recognizing the set of strings over {, } hving s sustring, nd mke it s simple s possile. Also, do it for the set of strings over {, } ending with. 1.6 Complexity of Nondeterministic Automt Sometimes nondeterministic utomt cn e much simpler thn ny euivlent deterministic utomton. (We sy two utomt M 1 nd M 2 re euivlent if L(M 1 ) = L(M 2 )). Let Σ e { 1, 2,..., n } nd let L e the set of strings w tht do not contin ll the symols { 1, 2,..., n }. Tht is, t lest one symol must e missing from w. This lnguge cn e recognized y the following n + 1 stte nondeterministic finite utomton. Any euivlent deterministic utomton hs t lest 2 n sttes:

5 e e 1 2 2, 3, 4,..., n 1, 3, 4,..., n s... e n 1, 2,..., [n-1] Prolem Consider this utomton for n = 3 nd using the set ide, simulte this utomton on the inputs nd on egrep nd grep In Unix, grep works directly with nondeterministic utomton nd simultes it using sets of sttes. egrep constructs n euivlent deterministic utomton nd simultes it. Thus egrep cn tke long time on complicted ueries, ecuse the utomton my hve mny sttes. But if the string serched is very long, this cn py off.

6 1.8 Omitting ϵ rrows 0,1 0,1 0,1 r s t 0,1 e e 0,1 0,1 0,1 r s t 0,1 0,1 How cn one eliminte ϵ rrows from the utomton given erlier for the set of strings missing t lest one symol? Theorem 1.1 For ech nondeterministic finite utomton M there is n euivlent nondeterministic finite utomton M without ϵ rrows nd hving the sme numer of sttes. Proof: We otin utomton M from M y pplying two trnsformtion rules s often s possile until they cn e no more pplied. If in M we hve ϵ r s for Σ then in M we hve lso Also, if in M we hve s. ϵ r nd r F then in M, is lso dded to F if it is not lredy there. Let M e M with ll ϵ rrows deleted. Clim: M is euivlent to M, tht is, L(M ) = L(M). Proof of clim: Clerly M is euivlent to M. We know L(M) L(M ) ecuse M hs ll trnsitions tht M hs nd possily more. However, L(M ) L(M) ecuse every ccepting computtion of M cn e simulted y n ccepting computtion of M.

7 Finlly, L(M ) = L(M ). Clerly L(M ) L(M ) ecuse M hs more rrows. But L(M ) L(M ) ecuse ny seuence of moves of M with ϵ rrows cn e simulted y seuence of moves of M. Thus L(M ) = L(M ). For, consider string w tht is ccepted y M. Look t the seuence of sttes tht the utomton goes through in ccepting w, nd in prticulr, t the suseuences consisting entirely of ϵ trnsitions. If s i, s i+1,..., s i+k is seuence of sttes with ϵ trnsitions in etween them in this computtion seuence, then, if s i+k is n ccepting stte, the rules will mke s i n ccepting stte too, so the ϵ trnsitions re not needed in this computtion seuence. If there is some rrow from s i+k to s i+k+1 leled with symol tht is used in the ccepting computtion, then the rules will dd n rrow from s i to s i+k+1 leled with, so the ϵ trnsitions re not needed in this computtion seuence. So w is still ccepted y M. Putting these results together, L(M ) = L(M) Exmple Consider this utomton with ϵ rrows: c e e r s Now, to eliminte the ϵ rrows, first we pply the rules to dd rrow from stte r to stte s, nd mke stte r n ccepting stte. We lso dd n rrow from stte to stte r:

8 c e e r s Now, we cn pply the rules gin. We dd rrow from to s. We lso mke stte n ccepting stte. c e e r s Now no more rules cn e pplied. So we delete the ϵ rrows nd otin the following utomton, hving no ϵ rrows ut euivlent to the strting utomton. c r s Continuing with the proof Now we will show tht every nondeterministic utomton cn e converted to n euivlent deterministic utomton. For exmple, the following nondeterministic utomton M N :

9 , r s cn e converted to this deterministic utomton M D : {} {,r} {,s} How M D is otined from M N : The sttes of M D re sets of sttes of M N ; only those tht re rechle from the strt stte need to e kept. The input lphet of M D is the sme s tht of M N. The strt stte of M D is the set consisting of the strt stte of M N. The ccepting sttes of M D re the sets contining t lest one ccepting stte of M N. If { 1, 2,..., n } is stte of M D nd is in Σ, then there is n rrow in M D from { 1, 2,..., n } to the set of r such tht in M N there is n rrow from some i to r. The euivlence of M N nd M D cn e shown y converting every ccepting computtion seuence for M N into n ccepting computtion seuence for M D, nd vice vers. For exmple, the following computtion for M N : r s

10 corresponds to the following computtion for M D : {} {,r} {,s} {,r} {,s} nd vice vers. Go forwrds (left to right) in the computtion constructing the ottom seuence from the top one, to mp the top seuence into the ottom seuence, showing tht every top stte is n element of the corresponding ottom stte. Thus the finl stte s of the top seuence, which is n ccepting stte of M N, is n element of the finl stte {, s} of the ottom seuence, which mkes {, s} n ccepting stte of M D. Thus the ottom seuence is lso n ccepting seuence. Go ckwrds (right to left) constructing the top seuence from the ottom one, to mp the ottom seuence into the top seuence, gin showing tht the top stte is n element of the ottom stte. Strt with n ccepting stte s of M N tht is n element of the finl stte {, s} of the ottom seuence. Then go ckwrds to some stte r of M N tht cused s to e dded to the finl stte of the ottom seuence. Put this stte in the top seuence. Continue going ck this wy, constructing the top seuence from the ottom one, nd showing tht ech top stte is n element of the ottom stte. Finlly, the first stte of the top seuence is n element of the first stte {} of the ottom seuence, so the first stte of the top seuence hs to e the strt stte.

11 Thus if M N ccepts w then M D ccepts w nd vice vers. This mens tht L(M N ) = L(M D ) so M N nd M D re euivlent. This sme construction cn e used for ny nondeterministic finite utomton, which shows tht for every nondeterministic utomton with n sttes nd no ϵ rrows, there is determistic utomton with t most 2 n sttes. This is stted formlly in the following theorem. Theorem 1.2 Every n stte nondeterministic finite utomton without ϵ rrows cn e converted to n euivlent deterministic finite utomton with t most 2 n sttes. Proof: Suppose M N is (K, Σ,, s, F ). Let M D e (K D, Σ, δ, {s}, F D ) where K D = 2 K F D = {Q K : Q F } δ(q, ) = { K : (,, ) for some Q}. Then ccepting computtions of M N nd M D cn e mpped onto ech other s in the exmple. For the exmple M N given ove, ecuse M N hs three sttes, M D would hve eight sttes. However, only three of these sttes re rechle from the strt stte, so only three of them re shown in the exmple M D given ove. Becuse ϵ rrows cn e eliminted from nondeterministic utomton without incresing the numer of sttes, we otin the following result: Theorem 1.3 Every n stte nondeterministic finite utomton (even if it hs ϵ rrows) cn e converted to n euivlent deterministic finite utomton with t most 2 n sttes.

12 In fct, there is n lgorithm to convert nondeterministic finite utomt to euivlent deterministic finite utomt. Let s pply this construction to the utomton given ove: r Suppose the lphet is {, }. The finl utomton is the following: {} {r}, {}, The stte with the empty set in it rises ecuse some inputs do not led nywhere in the nondeterministic utomton.

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