Efficient Solutions for the Complement of ww R and the Complement of ww

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1 Effcent Solutons for the Complement of ww R and the Complement of ww Allaoua Refouf Computer Scence Department Unversty of Sétf Algéra allaouarefouf@unv-setfdz Journal of Dgtal Informaton Management ABSTRACT: In ths paper we propose a new approach to tackle the problem of fndng effcent non determnstc solutons for the complement of the language L 1 = { ww R / w {0, 1}*} (the even length non palndromes) and the complement of the type 0 language (recursvely enumerable) L 2 = {ww / w {0, 1}*} The solutons provded are very elegant and make a subtle use of non determnsm We show that these languages are context free languages by desgnng non determnstc pushdown automaton that accepts them Categores and Subject Descrptors I13 Languages and Systems; F42 [Grammars and Other Rewrtng Systems] Grammar types General Terms: Language Analyss, Non-Palndromes Keywords: Context Free Languages, Pushdown Automaton, Non- Determnsm, Even Length Non Palndromes Receved: 18 July 2014, Revsed 3 September 2014, Accepted 12 September Introducton Formalsms descrbng language classes located between determnstc context free languages (CFL) and non determnstc context free languages have been ntensvely studed for many years One of the motvatons was to fnd famles wth acceptable computatonal complextyand suffcent expressveness, as well as havng natural characterzatons by grammars and machne models In what follows we prove that the languages L 1 and L 2 (defned below) are context free by desgnng non determnstc pushdown automaton that accepts them Let L 1 = {ww R / w {0, 1}*} and L 2 ={ww / w {0, 1}*} L 1 s a context free language and L 2 s a language of type 0 accordng to the Chomsky herarchy L 1 s known as the even length palndromes Accordng to the notaton adopted, the language L 1 s the complement of the even length palndrome language, t s also known as the language of even length non palndromes L 2 s the complement of the language L 2 Palndromes are symmetrc strngs that read the same forward and backward, for example the strng radar s a non even length palndrome, whle the strng elle s an even length palndrome Formally, a non-empty strng w s a palndrome f w = w R, where w R denotes the strng w reversed It s convenent to dstngush between even length palndromes that are strngs of the form ww R and odd length palndromes that are strngs of the form waw R, where a s a sngle alphabet symbol The paper s organzed as follows In the followng sectons we present some basc notons and defntons used n formal language theory In chapter II we show that the complement of the language ww R s context free by Journal of Dgtal Informaton Management Volume 12 Number 6 December

2 desgnng a non determnstc fnte pushdown automaton that accepts t In chapter III we prove a smlar result for the complement of the language ww Fnally we conclude our paper and pont out some future drectons of research 11 Prelmnares A word or strng over an alphabet Σ can be any fnte sequence of characters or symbols The set of all words over an alphabet Σ s usually denoted by Σ * (usng the Kleene star notaton, [1]) The length of a word s the number of symbols or letters t s composed of For any alphabet there s only one word of length 0, the empty word, whch s often denoted by ε or λ By concatenaton one can combne two words to form a new word, whose length s the sum of the lengths of the orgnal words A language L s any fnte subset of Σ * Unless stated otherwse we wll use lower case letters (a, b, c) to denote nput symbols, and upper case letters to denote stack symbols (X, Y, Z) Greek letters ndcated strngs of stack symbols Fnally u, v, w denote strngs of nput symbols Defnton 01 The complement L of a language L wth respect to a gven alphabet Σ conssts of all the strngs over the alphabet that do not belong to L It s defned by L = {x Σ / x L} Defnton 02 A non determnstc pushdown automaton or NPDA s a 7-tuple M = (Q, Σ,,, q 0, F) where: Q s a set of fnte states, Σ s the nput alphabet, s the stack alphabet whch may nclude Σ, s the transton functon, q 0 Q s the ntal state, Z 0 s the symbol at the bottom of the stack, whch s used to test whether the stack s empty or not F Q s a set of fnal states The transton functon for a NPDA has the form Q (Σ {λ}) fnte subsets of Q * where s a functon of three arguments The frst one s a state from Q, the second argument s ether λ or the current symbol from the nput alphabet The thrd argument s the current Ysymbol on top of the stack Thus we may wrte the transton functon as: (q, a, Z) = {( p 1, ), (p, ),, (p, )} n n When the nput symbol s from Σ, t s always consumed, that s the nput head moves one step to the rght When the nput symbol s set to λ the nput head does not move; ths move s termed a λ move In the determnstc case, when the functon s appled, the automaton moves to a new state q Q and pushes a new strng of symbols γ * onto the stack Snce we are dealng wth a nondetermnstc pushdown automaton, the result of applyng s a fnte set of (q, x) pars If we were to draw the automaton, each such par would be represented by a sngle arc As wth a fnte state automaton we do not need to specfy for every possble combnaton of arguments; when the transton functon s not explctly defned, t means that the machne enters a dead state and rejects [1, 2] Defnton 03 NPDA moves We dstngush two moves of the NPDA: a regular move when the nput head s advanced, and a λ move when the nput head s not advanced ) The nterpretaton of: where q and p are states; a an nput symbol from Σ, and Z a stack symbol from, s that the NPDA n state q wth the nput symbol a and Z the symbol on top of the stack, can for any enter state p, replace the symbol Z by the strng, and advance the nput head one symbol to the rght We wll adopt the usual conventon that the leftmost symbol of wll be placed on top of the stack Furthermore f: = 1 the top of the stack remans the same (we do not push or pop any symbol) = 0, that s = λ we pop a symbol from the stack, sometmes we wrte pop > 1 we push symbols on the stack )The nterpretaton of: (q, λ, R) = {(p 1, 1 ), (p 2, 2 ),, (p n, n )} s that the NPDA n state q, ndependently of the nput symbol beng scanned, and wth the symbol R on top of the stack, can enter state p, replace R by, for any In ths case the nput head s not advanced Defnton 04 An nstantaneous descrpton (ID) of a pushdown automaton s a trplet (q, w, u), where q s the current state of the automaton, w s the unread part of the nput strng, and u s the stack contents (wrtten as a strng, wth the leftmost symbol lyng at the top of the stack) Let the movement relaton ndcates a move of the NPDA It s related to the transton functon as follows: If (q, a, Z) = {(p 1, y 1 ),(p 2, y 1 ),, (p n, y n )} (p, a, X) = (q, γ) then (p, aw, Xα) (q, W, γα) Where W ndcates the rest of the strng followng the symbol a, and α ndcates the rest of the stack content underneath the symbol X Ths notaton says that n movng from state p to state q, the symbol a s consumed from 350 Journal of Dgtal Informaton Management Volume 12 Number 6 December 2014

3 the nput strng aw, and the strng Xa on the stack ( X s on the top of the stack) s replaced by the strng γa (p, aw, Xα) and (q, W, γa) are known as consecutve nstantaneous descrptons and are often used to defne the current confguraton of the NPDA [2, 3] In essence, a pushdown automaton s nondetermnstc It accepts the nput strng f there s a sequence of actons leadng to the fnal state There are two ways to depct non determnsm n the flowchart: two transtons wth the same label or a transton labeled wth λ (whch does not consume an nput symbol) All possbltes that do not have explct transtons, have mplct transtons that go to an mplct reject state Consderng a computaton of a pushdown automaton we call a sngle step nondetermnstc f the automaton has more than one choce for ts move The branchng of the step s defned to be the number of choces [4, 5]Most of the contextfree languages dffer n the amount of the resource (n ths case, nondetermnsm) that they requre [6, 7] Throughout the paper λ denotes the empty word For a strng x, let x, x [] and x [, j] denote the length of x, the th symbol of x and the factor x [] x [j] respectvely, where 0 < j x Let x R denote the reverse of the word x, that s x R = x [n] x [n 1] x [2] x [1] for x = n We assume throughout the paper that the specal symbol Z 0 can only occur on the bottom of the pushdown automaton, and the nput strng termnates wth the symbol $ 12 Acceptng strngs wth a NPDA We use the relaton to ndcate a sngle move of the NPDA We use the relaton *, the reflexve closure of the relaton to ndcate a sequence of zero or more moves If M = (Q, Σ,,, q 0, F) s an NPDA, then the language accepted by M, L (M), s defned by L (M) = {w Σ : (q 0, w, Z 0 ) * (p, $ ), p F} Ths defnton means that a strng w s accepted f and only f startng from the start state q 0 and an empty stack (depcted by the symbol Z 0 on ts top, we end up, after a sequence of moves, n a fnal state p after consumng entrely the nput strng We assume that the nput strng termnates wth the specal symbol $, and Z 0 s the bottom of the stack The symbol Z 0 s a convenent way to test for the empty stack condton, whch means we are acceptng by empty stack Acceptance by fnal state s the most common mode for acceptance; however proofs of basc theorems are often derved from acceptance by empty stack Henceforth we wll assume throughout ths paper that acceptance s made by empty stack Some results on CFLs can be found n ([2, 3]): A contextfree language or CFL s a language generated by o context free grammar or accepted by a NPDA A determnstc context free language (DCFL) s a language accepted by a determnstc pushdown automaton Throughout the paper we prove that languages are context free by desgnng NPDA that accepts them CFL are not closed under complement, we cannot just reverse fnal and non fnal states to fnd the complement Ths does not work because t dd not work for non determnstc fnte state automaton (NDFA) CFL are closed under unon, however DCFL are not closed under unon DCFL and CFL are not closed under ntersecton In general determnstc machnes are closed under complement Two NPDAs are equvalent f they accept the same language 13 Example of some transtons For ease of understandng the most used transtons are explaned n detal below: 1 (p, a, R) = {(q, ar)} n state p, on nput symbol a and R the top of the stack : move to state q and push a on top of the stack 2 (p, a, R) = {(q, R)} n state p, on nput symbol a and R the top of the stack : move to state q and leave R on top of the stack ( do not push a onto the stack) 3 (p, a, R) = {(q, λ)} n state p, on nput symbol a and R the top of the stack : pop (unstack the symbol R) 4 (p, λ, R) = {(q, R)} ths s a λ move : the nput head s not advanced and the top of the stack remans the same All we do s move to a dfferent state 5 (p, _, R) = {(q, _R)} the anonymous varable _ stands for any nput symbol from Σ (Stack the nput symbol, whatever t s) 6 (p, a, _ ) = {(q, a_)} the anonymous varable _ stands for any stack symbol from Γ 7 (p, _ ) = {(q )} the stack s empty 8 (p, $, _) = {(q, _)} the nput s totally consumed 9 (p, $ ) = {(q )} the nput s totally consumed and the stack s empty In the followng chapter we provde two solutons for the language L 1 : the frst soluton s smple and clear, the second soluton s much more elegant and subtle We frst recall the grammar classfcaton accordng to Chomsky 14 The Chomsky Herarchy of formal grammars Let G = (V, T, P, S) be a formal grammar where V s the non termnal vocabulary, T s the termnal vocabulary, P s the set of the productons, and S s the startng symbol We dstngush four types of grammars dependng on the form of ther productons ([1]) Unrestrcted Grammars: The languages defned by these grammars are accepted by Turng machnes They have rules of the form a b where α and β are arbtrary strngs over a vocabulary V and α # Journal of Dgtal Informaton Management Volume 12 Number 6 December

4 Context-senstve grammars: the languages defned by these grammars are accepted by lnear bounded automata They have rules of the form α β where α β where α, β (V + T) *, and a # Context-free grammars: the languages defned by these grammars are accepted by push-down automata They have rules of the form Α β, where A V and β (V + T) * Regular grammars: the languages defned by these grammars are accepted by fnte state automata There are two knds of regular grammars: 1 Rght-lnear wth rules of the form A wb / w, where A, B V and w T * 2 Left-lnear wth rules of the form A wb / w 2 Non Even Length Palndromes Non determnstc solutons for context free languages have long proven dffcult because the noton of non determnsm n algorthm desgn s not easy to mplement However non determnstc approaches provded an elegant and subtle way to prove that languages are context free Recall that the language known as the non even palndromes s defned as the complement of the language {ww R / w {0, 1}*}, we wll call ths language L 1 In ths secton we prove that L 1 s a context free language Precsely we construct a non determnstc pushdown automaton that accepts L 1 Clam 01: The language L 1 s a context free language We prove the above statement by the constructon of a non determnstc pushdown automaton that accepts the language We begn wth a naïve soluton and then proceed to a much better one The frst soluton s gven to make sure that the average reader understands the desgn of NPDA and also to get used wth the notaton 21 A frst soluton for the language L 1 Consder the followng language: L 1 = {ww R / w {0, 1}*} Ths s the set of strngs n whch each strng begns wth a sequence of one or more 0s or 1s and then mmedately concludes wth an dentcal but reversed sequence In other words, L 1 s the set of symmetrc even-length palndromes over the alphabet {0, 1} Let us construct a NPDA M 1 whch accepts L 1 M 1 s defned by the set M = (Q, Σ,,, q 0, F) where Q ={q 0, q 1, q 2, q 3, q 4, q 5, q 6 }, Σ = [0, 1}, ={Z 0, X, Y}, and F = {q 4, q 6 } 211 Intal part of the algorthm: Frst of all we know that odd length strngs belong to L 1 In ths part all we have to do s to make sure that the nput has an odd length to be accepted From the start state q 0 we make a λ move to state q 5 From q 5 on any nput symbol we move to state q 6, and then loop on state q 5 when readng the next nput symbol Thus the pushdown automaton behaves lke a fnte automaton and we accept f we end up n the fnal state q 6 (q 0, λ ) ={(q 1 ), (q 5 )} make a λ move ether to state q 1 or to state q 5 (q 5, _ ) ={(q 6 )} (q 6, _ ) ={(q 5 )} We move to state q 1 for the second part of the algorthm when the nput has an even length 212 Second part of the algorthm: For the second part of the algorthm whch s the man part of the problem we deal wth even length strngs whch are not palndromes The man dea s to look for msmatches non determnstcally from the frst half of the strng w and the second half of the strng w R, specfcally: 1 In state q 1 push a set of symbols on the stack (readng the symbol 0 push the symbol X, readng the symbol 1 push the symbol Y) 2 Guess that we have reached the mddle of the strng and move to state q 2 3 In state q 2 match the rest of the symbols wth the stack content (match 0 aganst X, and 1 aganst Y) 4 Fnd a msmatch on a 0 or a 1 and move to state q 3 5 Whle n state q 3 pop everythng out of the stack 6 When the nput s totally consumed and the stack becomes empty, move to state q 4 accept and resume In state q 2, we make sure that the symbols we pushed earler n state q 1 match the symbols we popped, whch confrms the fact that the strng s of even length Snce whle readng the nput symbol 0 (respectvely 1) we pushed the symbol X (respectvely Y) on top of the stack A msmatch occurs n state q 2 only f the nput symbol s 0 (respectvely 1) and the symbol on top of the stack s Y (respectvely X) We reach state q 3 f and only f we encounter a msmatch, whch means that the nput strng s not of the form ww R, thus the nput strng belongs to the set L Transtons for M 1 Here are the transtons of the machne:, 0 ) = {(q 1, XZ 0 )} readng the frst nput symbol 0 push X, 1 ) = {(q 1, YZ 0 )} readng the frst nput symbol 1 push Y 352 Journal of Dgtal Informaton Management Volume 12 Number 6 December 2014

5 , 0, T) = {(q 1, XT)} readng the next nput symbol 0 push X, the top of the stack s T ={X, Y},1, T) = {(q 1, YT)} readng the next nput symbol 1 push Y, the top of the stack s T ={X, Y}, λ, R) = {(q 2, R)} At any tme, guess that we have reached the mddle of the strng and move to state q 2 Ths s a l move, the nput head s not advanced and R = {X, Y}, 0, X) = {(q 2, pop)} match the nput symbol 0 wth the symbol X on the stack, 1,Y) = {(q 2, pop)} match the nput symbol 1 wth the symbol Y on the stack, 0, Y) = {(q 3, pop)} guess a msmatch on the symbol 0 when encounterng Y on top of the stack and enter state q 3, 1, X) = {(q 3, pop)} guess a msmatch on the symbol 1 when encounterng X on top of the stack and enter state q 3 (q 3, _, _) = {(q 3, pop)} pop everythng out of the stack (q 3, $ ) = {(q 4 )} move to q 4 and accept 22 A More Effcent Soluton for L 1 Let us consder agan the language L 1, the complement of the language L 1 = { ww R / w {0, 1}*} In what follows we descrbe a more effcent soluton for the even length non palndromes In ths verson we do not need to remember the symbols, all we have to do s count the number of symbols That s the stack s used as a counter Specfcally we push the symbol X ether when readng 0 or 1 We say that two symbols x [] and x [j] from the nput strng msmatch f they hold dfferent values Note that ths noton of msmatch s slghtly dfferent from the one used n secton II A The man ntuton justfyng the conjecture that even length non palndromes are CFL s based on the followng observaton Let us consder the strng x = ww R = x [1] x [2] x [3] x [n 1] x [n] x [n] x [n 1] x [3] x [2] x [1] We can easly note that the symbols x [] and x [2n + 1] are dentcal for 1 n Ths tells us that the strng cannot be a palndrome f the symbols at poston and poston (2n + 1) msmatch That means that all we have to do s make sure that there exts two symbols that msmatch for the strng to be n L 1 Let the msmatchng symbols be x [] and x (2n + 1), from left to rght n the strng The number of symbols that precede x [] s equal to the number of symbols that follow x (2n + 1), whch s exactly ( 1) Moreover all these symbols are dentcal We wll use the stack to make sure that ths condton holds The symbols that le between x [] and x (2n + 1) do not matter much snce the msmatchng symbols are non determnstcally guessed Fgure 01 The man dea s as follows: the msmatchng symbols x [] and x [2n + 1] are ponted by the double arrows n fgure01 Let A 1 be the substrng x [1, 1], A 2 the substrng x [ + 1, 2n ], and A 3 the substrng x [2n + 2, 2n] In zone A 1 we push the symbols onto the stack In part A 2 we just gnore the nput, n part A 3 we pop the symbols makng sure that they match what we have pushed n zone A 1 In fgure 01 the nput strng s represented by the rectangle Substrng A 1 les before the frst vertcal double arrow on the left, substrng A 2 les between the two double arrows, and substrng A 3 les beyond the second double arrow on the rght The vertcal double arrows ndcate the two symbols that must msmatch A 1 and A 3 have the same length The man algorthm for the second soluton s descrbed below 221 Algorthm for L 1 a) Read the nput symbols and push them onto the stack (push the symbol X ether for 0 or 1) b) At any tme we non determnstcally guess the symbol n the frst part of the nput strng (w) that s gong to msmatch a subsequent symbol n the second part of the strng (w R ) In state q 2 we remember the symbol 0, and n state q 3 we remember the symbol 1 c) We remember ths symbol and move across the nput gnorng everythng untl we non determnstcally guess the second symbol for a msmatch From q 2 the msmatchng symbol s 1, from state q 3 the msmatchng symbol s 0 d) From ths pont we make sure that the pushng at the start matches the poppng at the end Precsely the number of symbols (only Xs) pushed n part a) s equal to the symbols remanng on the stack Ths means that the symbols that msmatch are legtmate msmatches as depcted n the fgure 01 above Ths s a hgher level of guessng, where we non determnstcally check that we made the rght guess for a msmatch Ths soluton avods the man loops n the earler verson of the algorthm (secton IIA) and s more subtle and elegant 222Transton functon for L 1, 0 ) ={(q 1, XZ 0 ), (q 2 )} whle readng the frst symbol 0 push X onto the stack or remember the symbol 0 as the msmatch and move to state q 2,1 ) ={(q 1, XZ 0 ), (q 3 )} whle readng the frst symbol 1 push X onto the stack or remember the symbol 1 as the msmatch and move to state q 3 Journal of Dgtal Informaton Management Volume 12 Number 6 December

6 , 0, X) ={(q 1, XX), (q 2, X)} whle readng the next symbol 0 push X onto the stack or remember the symbol 0 as the msmatch and move to state q 2, 1, X) ={(q 1, XX), (q 3, X)} whle readng the next symbol 1 push X onto the stack or remember the symbol 1 as the msmatch and move to state q 3, 0, X) ={(q 2, X)} whle n state q 2 gnore the nput symbol 0, 1, X) ={(q 2, X), (q 4, X)} whle n state q 2 gnore the nput symbol 1 or guess that ths s the msmatch symbol and move to state q 4 (q 3, 1, X) ={(q 3, X)} whle n state q 3 gnore the nput symbol 1 (q 3, 0, X) ={(q 3, X), (q 4, X)} whle n state q 3 gnore the nput symbol 0 or guess ths s the msmatch symbol and move to state q 4 (q 4, 0, X) ={(q 4, pop)} check and pop (q 4, 1, X) ={(q 4, pop)} check and pop (q 4, $ ) ={(q 5 )} f the nput s totally consumed and the stack s empty then accept and resume In state q 2 the msmatch s gong to occur on a 1, snce we remembered the symbol 0 In state q 3 the msmatch s gong to occur on a 0, snce we remembered the symbol 1 We reach state q 4 f and only f we encounter a msmatchng symbol, whch proves that the nput strng belongs to the language L 1 The non determnstc technque s a powerful tool for the desgn of strateges when dealng wth pushdown automaton Example of computatons on the nput The ntal ID of the NPDA s: (q 1, $ ) Each move s represented n a lne and commented (q 1, $ ) read 0 push X (q 1, 01110$, XZ 0 ) guess 0 as a future msmatch, 1110$, XZ 0 ) gnore the symbol 1, 110$, XZ 0 ) gnore the symbol 1, 10$, XZ 0 ) guess 1 as the correspondng msmatchng symbol (q 4, 0$ ) match XZ 0 wth X (q 5, $ ) move to state q 5 (q 5, $ ) accept and resume The nput strng , whch has the form ww R, s not accepted wherever the two msmatchng symbols may le In the next chapter we desgn a NPDA for the language L 2 To the best of our knowledge the solutons provded for L 1 and L 2 are new and have not been dscussed n the lterature 3 The complement of L 2 Let us now consder the complement of the language {ww / w {0, 1}*}, we are about to prove that ths language, labeled L 2 s a context free language Clam 02: the language L 2 s a context free language Proof of Clam 02 To prove that the language L 2 s a CFL we desgn a NDPA that accepts the language As before there are two aspects for strngs n L 2, frst of all odd length words are n L 2 We can take care of ths stuaton by desgnng a fnte automaton that recognzes odd length strngs In what follows we only consder even length words Let the strng x = ww = x [1] x [2] x [3] x [n 1] x [n] x [1] x [2] x [3] x [n 1] x [n] The matchng symbols x [] and x [n + ] are ponted out by the vertcal double arrows n fgure02 Let A 1 be the substrng x [1, 1], A 2 the substrng x [ + 1, n + + 1], and A 3 the substrng x [n + + 2, 2n] We notce that the dentcal symbols x [] and x [n + ] from the frst part of the strng and the second half are separated by exactly n 1 symbols Moreover the number of symbols that le before the frst x [] that s 1, when added to the number of symbols that le after the second x [n + ], that s n, equals n 1 Ths suggests that a strng s n L 2 f and only f the two msmatchng symbols are separated by half the length of the total strng Ths s made clear n the fgure 02 below where the vertcal double arrows show the two msmatchng symbols Ths tells us that we have to make sure that the two msmatchng symbols are separated by half of the total strng length Let us push Xs for a whle, say n 1 Decde that the next nput symbol s the msmatchng symbol, remember t and start to pop symbols from the stack, say n 2 If the stack becomes empty, start pushng agan Xs onto the stack, say n 3 Then arbtrarly guess the next nput symbol s the correspondng msmatchng symbol Now we pop everythng out of the stack, say n 4 symbols At ths pont when the stack emptes then we know that the set consstng of the poppng of the frst set (n 2 ) and the pushng of the second set (n 3 ) s the same as the pushng of the frst set (n 1 ) and the poppng of the second set (n 4 ), that they are equal That s n 2 + n 3 = n 1 + n 4, whch proves that the two msmatchng symbols are legtmate and separated by half of the total strng length Fgure Journal of Dgtal Informaton Management Volume 12 Number 6 December 2014

7 The two msmatchng symbols pctured by the vertcal double arrows n fgure 02 are separated by half the length of the total strng Ths s a neat soluton that requres a sngle counter The whole dea whch s a bt subtle s to make a guess and then later non determnstcally check that our guess was the rght one In fgure 02, as before, substrng A 1 les before the frst vertcal double arrow on the left, substrng A 2 les between the two vertcal double arrows, and substrng A 3 les beyond the second double arrow on the rght Then the length of the strng A 1 added to the length of the strng A 3 equals the length of the strng A 2 If the length of the nput strng ww s 2n then A 1 + A 3 = A 2 = n In fgure 02 the nput strng s represented as a rectangle The vertcal double arrows show where the two msmatchng symbols le They are separated by half the length of the total strng depcted by the horzontal double arrow We descrbe the algorthm n the next secton 31 Algorthm for L 2 a) Push a bunch of symbols on the stack and arbtrarly guess that you are choosng to look at a specfc symbol b) Pop those symbols off the stack, f the stack becomes empty start pushng the symbols agan non determnstcally guessng when to make a msmatch When the stack becomes empty we push a dfferent symbol onto the stack c) Fnd a msmatch, at that pont pop the symbols that you have off the stack d) If the nput strng s entrely consumed and the stack s empty then we know here that the set consstng of the poppng of the frst set and the pushng of the second set was the same as the pushng of the frst set and the poppng of the second set, that they are equal That means that the mddle secton A 2 s half the length of the total nput strng length 32 Transton functon for L 2 The NPDA M 2 s defned by the set M = (Q,,,, q 1, F) where Q = {q 1, q 2, q 3, q 4, q 5, q 6, q 7 }, = {0, 1}, = {Z 0, X, Y}, and F = {q 7 } 1 (q 1,0,T) = {(q 1, XT), (q 2, T)} when readng the symbol 0 push X onto the stack or guess the symbol 0 as a msmatch and move to q 2,T = {Z 0 2 (q 1, 1, T) ={(q 1, XT), (q 3, T)} when readng the symbol 1 push X onto the stack or guess the symbol 1 as a msmatch and move to q 3, T = {Z 0 3 (q 2, 0, X) ={(q 2, pop)} 4 (q 2, 1, X) ={(q 2, pop)} 5 (q 2, λ ) ={(q 4 )} the stack s empty, move to state q 4 6 (q 3, 0, X) ={(q 3, pop)} 7 (q 3, 1,X) ={(q 3, pop)} 8 (q 3, λ ) ={(q 5 )} the stack s empty, move to state q 5 9 (q 4, 0, T) ={(q 4, YT)} when readng the symbol 0 pop agan Ys onto the stack, T = {Z 0 10 (q 4, 1,T) ={(q 4, YT),(q 6, T)} when readng the symbol 1 pop agan Ys onto the stack or guess the msmatch and move to state q 6, T = {Z 0 11 (q 5, 1,T) ={(q 5, YT)} when readng the symbol 1 pop agan Ys onto the stack, T = {Z 0, Y} 12 (q 5, 0,T) ={(q 5, YT),(q 6, T)} when readng the symbol 0 pop agan Ys onto the stack or guess the msmatch and move to state q 6, T = {Z 0 13 (q 6, 0,Y) ={(q 6, pop)} pop the Ys 14 (q 6, 1, Y) ={(q 6, pop)} pop the Ys 15 (q 6, $ ) ={(q 7 )} the nput s entrely consumed and the stack s empty then enter the fnal state q 7 and accept If we assume that the amount of non determnsm that a pushdown automaton requres to recognze an nput strng can be measured by the mnmum number of guesses that t must make to accept the strng then the soluton we provde s effcent as we only make two (2) guesses ([4]) 33 Some Input computatons Here are some computatons for the nput The ntal ID s (q 1, $ ) (q 1, $ ) read 0 push X (q 1, 01011$, XZ 0 ) guess 0 as the msmatch, 1011$, XZ 0 ) pop, 011$ ) stack empty move to q 4 (q 4, 011$ ) push Y (q 4, 11$, YZ 0 ) guess 1 as the second msmatch and move to q 6 (q 6,1$,YZ 0 ) pop (q 6, $ ) stack s empty move to q 7 ( q 7, $ ) accept and resume The msmachng symbols occur n the second and the ffth symbol Here are some computatons for the nput (q 1, $ ) (q 1, 1000$, XXZ 0 ) (q 3, 000$, XXZ 0 ) (q 3,0$,Z 0 ) (q 5, 0$ ) (q 6, $ ) (q 7, $ ) The msmachng symbols occur n the thrd and the sxth symbol If we try the strng , t s clear that t s Journal of Dgtal Informaton Management Volume 12 Number 6 December

8 mpossble to fnd two symbols to msmatch that are separated by half the strng length, because ths example s of the form ww, so t s rejected The man results of ths paper are that the complement of the language L 1 = { ww R / w {0, 1}*} and the complement of the type 0 language L 2 ={ww / w {0, 1}*} are context free We have shown these results by desgnng non determnstc pushdown automaton that accept these languages 4 Concluson In ths paper we have proved that the complement of the language L 1 = { ww R / w {0, 1}*} and the complement of the language L 2 ={ww / w {0, 1}*} are context free by desgnng non determnstc pushdown automatons that recognze these languages The solutons provded are elegant and show that non determnsm s a powerful tool for desgnng strateges to solve hard problems Both languages can be characterzed by the presence of two nput symbols that msmatch The man feature of the procedures s to determne the poston of the symbols that have to msmatch n the nput strng At some pont we make a guess and then non determnstcally prove that our guess was legtmate The NPDA for the language L 2 uses the stack only as a counter; the value of the nput symbol (0 or 1) does not matter To the best of our knowledge the solutons provded are state of the art and have not been publshed n the lterature Future drectons of research mght deal wth the refnement of the non determnsm measure, n other words how to reduce the branchng factor (non determnstc moves) n the desgn of pushdown machnes It s also mportant to consder the relaton between complexty measures, n terms of the sze of nput strngs, and non determnstc moves when desgnng effcent algorthms References 1 Hopcroft, J, Ullman, J (1979) Introducton to Automata Theory, Languages and Computaton Addson-Wesley, Readng, MA 2 Harrson, M (1978) Introducton to Formal Language Theory Addson-Wesley, Readng, MA 3 Drobot, V (1989) Formal Languages and Automata Theory Computer Scence Press, Rockvlle, MD 4 Goldstne, J, Leung, H, Wotschke, D, (2005) Measurng nondetermnsm n pushdown automata, Journal of Computer and System Scences Kutrb M Martn, Malcher, A (2007) Context-dependent nondetermnsm for pushdown automata Theoretcal Computer Scence Jang, T, Salomaa, A, Salomaa, K, Yu, S (1995) Decson problems for patterns Journal of Computer and System Scences 50 (1) Kutrb M, Malcher A, Werlen, L (2009) Regulated nondetermnsm n pushdown automata Theoretcal Computer Scence Bography The author s currently a Professor n computer scence at the Unversty of Setf, Algera He graduated from the Unversty of Colorado at Boulder, USA (Master of Scences n 1980) and from the Unversty of Sheffeld, Unted Kngdom (PhD n Artfcal Intellgence n 1990) Hs areas of research ncludes formal language theory, natural language processng and semantc web technologes The author may be reached at allaouarefouf@unv-setfdz 356 Journal of Dgtal Informaton Management Volume 12 Number 6 December 2014