Most Complex Regular Ideal Languages

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1 Discrt Mthmtics nd Thorticl Computr Scinc DMTCS vol. 18:3, 2016, #15 Most Complx Rgulr Idl Lngugs Jnusz Brzozowski 1 Sylvi Dvis 2 Bo Yng Victor Liu 1 rxiv: v3 [cs.fl] 13 Oct Dvid R. Chriton School of Computr Scinc, Univrsity of Wtrloo 2 Dprtmnt of Pur Mthmtics, Univrsity of Wtrloo rcivd 3 rd Dc. 2015, rvisd 4 th Oct. 2016, ccptd 11 th Oct A right idl (lft idl, two-sidd idl) is non-mpty lngug L ovr n lpht Σ such thtl = LΣ (L = Σ L, L = Σ LΣ ). Lt k = 3 for right idls, 4 for lft idls nd 5 for two-sidd idls. W show tht thr xist squncs (L n n k) of right, lft, nd two-sidd rgulr idls, whrl n hs quotint complxity (stt complxity) n, such tht L n is most complx in its clss undr th following msurs of complxity: th siz of th syntctic smigroup, th quotint complxitis of th lft quotints of L n, th numr of toms (intrsctions of complmntd nd uncomplmntd lft quotints), th quotint complxitis of th toms, nd th quotint complxitis of rvrsl, str, product (conctntion), nd ll inry ooln oprtions. In tht sns, ths idls r most complx lngugs in thir clsss, or univrsl witnsss to th complxity of th vrious oprtions. Kywords: tom, sic oprtions, idl, most complx, quotint, rgulr lngug, stt complxity, syntctic smigroup, univrsl witnss 1 Introduction W gin informlly, postponing dfinitions until Sction 2. In [4] Brzozowski introducd list of conditions tht rgulr lngug should stisfy in ordr to clld most complx, nd found squnc(u n n 3) of rgulr lngugs with quotint/stt complxity n tht hv th smllst possil lpht nd mt ll of ths conditions [4]. Nmly, th lngugs U n mt th uppr ounds for th siz of th syntctic smigroup, th quotint complxitis of lft quotints, th numr of toms (intrsctions of complmntd nd uncomplmntd lft quotints), th quotint complxitis of th toms, nd th quotint complxitis of th following oprtions: rvrsl, str, product (conctntion), nd ll inry ooln oprtions. In this sns th lngugs in this squnc r most complx whn comprd to othr rgulr lngugs of th sm quotint complxity. Howvr, ths univrsl witnsss cnnot usd whn studying th complxitis listd ov in suclsss of rgulr lngugs, sinc thy gnrlly lck th proprtis of thos clsss. This ppr is prt of n ongoing projct to invstigt whthr th pproch usd for gnrl rgulr lngugs cn xtndd to suclsss. W prsnt squncs of most complx lngugs for th clsss of right, lft, nd two-sidd rgulr idls. Right idls wr chosn s n initil tst cs for this projct du to thir simpl structur; w wr l to otin squnc of most complx right idls y mking smll modifictions to th squnc (U n n 3). A prliminry vrsion of our rsults out right idls pprd in [6]. Th squncs of witnsss for lft nd two-sidd idls r mor complictd, nd first pprd in [15] whr thy wr conjcturd to hv syntctic smigroups of mximl siz; this ws ltr This work ws supportd y th Nturl Scincs nd Enginring Rsrch Council of Cnd undr grnt No. OGP ISSN c 2016 y th uthor(s) Distriutd undr Crtiv Commons Attriution 4.0 Intrntionl Licns

2 2 Jnusz Brzozowski, Sylvi Dvis, Bo Yng Victor Liu provd in [12]. Our min nw rsult is dmonstrtion tht ths squncs r in fct most complx. It hs n shown in [11] tht most complx squnc dos not xist for th clss of suffix-fr lngugs. Hving singl squnc of witnsss for ll th complxity msurs is usful whn on nds to tst systms tht prform oprtions on rgulr lngugs nd finit utomt: to dtrmin th sizs of th lrgst utomt tht cn hndld y th systm, on cn us th sm squnc of witnsss for ll th oprtions. For furthr discussion of rgulr idls s [6, 8, 10, 12, 15]. It ws pointd out in [8] tht idls dsrv to studid for svrl rsons. Thy r fundmntl ojcts in smigroup thory. Thy ppr in th thorticl computr scinc litrtur s rly s 1965, nd continu to of intrst; n ovrviw of historicl nd rcnt work on idl lngugs is givn in [8]. Bsids ing of thorticl intrst, idls lso ply rol in lgorithms for pttrn mtching: For xmpl, whn srching for ll words nding in word from som stl, on is looking for ll th words of th lft idlσ L. Additionl xmpls of th us of idls in pplictions cn found in [1, 16, 17, 24]. 2 Bckground A dtrministic finit utomton (DFA)D = (Q,Σ,δ,q 1,F) consists of finit non-mpty stqof stts, finit non-mpty lphtσ, trnsition functionδ: Q Σ Q, n initil sttq 1 Q, nd st F Q of finl stts. Th trnsition function is xtndd to functions δ : Q Σ Q nd δ : 2 Q Σ 2 Q s usul, ut ths xtnsions r lso dnotd y δ. Stt q Q is rchl if thr is word w Σ such tht δ(q 1,w) = q. Th lngug ccptd y D is L(D) = {w Σ δ(q 1,w) F}. Two DFAs r quivlnt if thir lngugs r qul. Th lngug of stt q is th lngug ccptd y th DFA (Q,Σ,δ,q,F). Two stts r quivlnt if thir lngugs r qul; othrwis, thy r distinguishl y som word tht is in th lngug of on of th stts, ut not th othr. A DFA is miniml if ll of its stts r rchl nd no two stts r quivlnt. A stt is mpty if its lngug is mpty. A nondtrministic finit utomton (NFA) is tupln = (Q,Σ,η,Q I,F), whrq,σ, ndf r s in DFA, η: Q Σ 2 Q is th trnsition function ndq I Q is th st of initil stts. An ε-nfa hs ll th fturs of n NFA ut its trnsition functionη: Q (Σ {ε}) 2 Q llows lso trnsitions undr th mpty wordε. Th lngug ccptd y n NFA or n ε-nfa is th st of wordsw for which thr xists squnc of trnsitions such tht th conctntion of th symols cusing th trnsitions is w, nd this squnc lds from stt inq I to stt inf. Two NFAs r quivlnt if thy ccpt th sm lngug. Without loss of gnrlity w us th st Q n = {1,2,...,n} s th st of stts for our utomt. A trnsformtion ofq n is mpping ofq n into itslf. W dnot th img of sttq undr trnsformtion t yqt. An ritrry trnsformtion ofq n cn writtn s ( ) 1 2 n 1 n t =, p 1 p 2 p n 1 p n whrp q = qt,1 q n, ndp q Q n. Th img of st P Q n is Pt = {pt p P}. Th idntity trnsformtion1 mps ch lmnt to itslf, tht is, q1 = q forq = 1,...,n. For k 2, trnsformtiont of st P = {p 1,...,p k } is k-cycl if thr xist pirwis diffrnt lmntsp 1,...,p k such tht p 1 t = p 2,p 2 t = p 3,...,p k 1 t = p k, p k t = p 1, nd ll othr lmnts of Q n r mppd to thmslvs. A k-cycl is dnotd y (p 1,p 2,...,p k ). A trnsposition is 2-cycl (p,q). A trnsformtion tht chngs only on lmnt p to n lmnt q p is dnotd y (p q). A trnsformtion mpping sustp ofq n to singl lmntq nd cting s th idntity onq n \P is dnotd y(p q). A prmuttion ofq n is mpping ofq n onto itslf. Th st of ll prmuttions of stq n ofnlmnts is group, clld th symmtric group of dgr n. This group hs siz n!. It is wll known tht two

3 Most Complx Rgulr Idl Lngugs 3 gnrtors r ncssry nd sufficint to gnrt th symmtric group of dgrn; in prticulr, th pirs {(1,2,...,n),(1,2)} nd{(1,2,...,n),(2,3,...,n)} gnrts n. Th st T Qn of ll trnsformtions of Q n is smigroup, in fct monoid with 1 s th idntity. It is wll known tht thr trnsformtions r ncssry nd sufficint to gnrtt Qn ; in prticulr, th tripls {(1,2,...,n),(1,2),(n 1)} nd{(1,2,...,n),(2,3,...,n),(n 1)} gnrtt Qn. Lt D = (Q,Σ,δ,q 0,F) DFA. For ch word w Σ, th trnsition function inducs trnsformtion δ w of Q y w: for ll q Q, qδ w = δ(q,w). Th st T D of ll such trnsformtions y non-mpty words forms smigroup of trnsformtions clld th trnsition smigroup of D [23]. Convrsly, w cn us st {δ Σ} of trnsformtions to dfin δ, nd so th DFA D. W writ : t, whr t is trnsformtion ofq, to mn tht th trnsformtionδ inducd yis t. Th Myhill congrunc [20] L of lngugl Σ is dfind onσ + s follows: Forx,y Σ +, x L y if nd only if wxz L wyz L for ll w,z Σ. This congrunc is lso known s th syntctic congrunc ofl. Th quotint st Σ + / L of quivlnc clsss of th rltion L is smigroup clld th syntctic smigroup of L. If D is miniml DFA of L, thn T D is isomorphic to th syntctic smigroup T L of L [23], nd w rprsnt lmnts of T L y trnsformtions in T D. Th siz of th syntctic smigroup hs n usd s msur of complxity for rgulr lngugs [4, 15, 18], nd is dnotd yσ(l). Th Nrod right congrunc [21] of lngugl Σ is dfind onσ s follows: Forx,y Σ, x L y if nd only ifxz L yz L for llz Σ. Th (lft) quotint of L Σ y word w Σ is th lngugw 1 L = {x Σ wx L}. Thus two wordsxndy r in th sm clss of th Nrod right congrunc if thy dfin th sm quotint, tht is, if x 1 L = y 1 L, nd th numr of quivlnc clsss of L is th numr of quotints, which is clld th quotint complxity [3] of L. An quivlnt concpt is th stt complxity of rgulr lngug [25] L, which is th numr of stts in miniml DFA with lpht Σ tht rcognizs L. This ppr uss th trm complxity for oth of ths quivlnt notions. W dnot th (quotint/stt) complxity of rgulr lngug L y κ(l). Atoms of rgulr lngugs wr studid in [14], nd thir complxitis in [6, 13, 19]. Considr th lft congrunc dfind s follows: Forx,y Σ +,x L y if nd only ifwx L wy L for ll w Σ. Thusx L y ifx w 1 L if nd only ify w 1 L. An quivlnc clss of this rltion is clld n tom. It follows tht toms r intrsctions of complmntd nd uncomplmntd quotints. In prticulr, if th quotints of L r K 1,...,K n, thn for ch tom A thr is uniqu st S {K 1,...,K n } such tht A = K S K K S (Σ \K). In [4] it ws rgud tht for rgulr lngug to considrd most complx whn comprd with othr lngugs of th sm (quotint/stt) complxity, it should hv th mximl possil numr of toms nd ch tom should hv mximl complxity. Th complxity of toms of idls ws studid in [6, 7], nd w shll stt th rsults otind thr without proofs. Most of th rsults in th litrtur concntrt on th (quotint/stt) complxity of oprtions on rgulr lngugs. Th complxity of n oprtion is th mximl complxity of th lngug rsulting from th oprtion s function of th complxitis of th rgumnts. It is gnrlly ssumd whn studying complxity of inry oprtions tht oth rgumnts r lngugs ovr th sm lpht, sinc if thy hv diffrnt lphts, w cn just viw thm s lngugs ovr

4 4 Jnusz Brzozowski, Sylvi Dvis, Bo Yng Victor Liu th union of th lphts. Howvr, in 2016, Brzozowski dmonstrtd tht this viwpoint lds to incorrct complxity rsults for oprtions on lngugs ovr diffrnt lphts [5]. H introducd notion of unrstrictd (quotint/stt) complxity of oprtions, which givs corrct rsults whn lngugs hv diffrnt lphts. Th trditionl notion of complxity, in which lngugs r ssumd to hv th sm lpht, is rfrrd to s rstrictd (quotint/stt) complxity of oprtions. As this ppr ws writtn wll for [5], ll our rsults r in trms of rstrictd complxity. A study of unrstrictd complxity of inry oprtions on idls cn found in [9]. Thr r two prts to th procss of stlishing th complxity of n oprtion. First, on must find n uppr ound on th complxity of th rsult of th oprtion y using quotint computtions or utomton constructions. Scond, on must find witnsss tht mt this uppr ound. On usully dfins squnc (L n n k) of lngugs, whrk is som smll positiv intgr (th ound my not pply for smll vlus of n). This squnc is clld strm. Th lngugs in strm usully diffr only in th prmtr n. For xmpl, on might study unry lngugs ({ n } n 1) tht hv zro s modulo n. A unry oprtion thn tks its rgumnt from strm (L n n k). For inry oprtion, on dds s th scond rgumnt strm(l m m k). Somtims on considrs th cs whr th inputs to th oprtions r rstrictd to som suclss of th clss of rgulr lngugs. In this stting, typiclly th uppr ounds on complxity r diffrnt nd diffrnt witnsss must found. Th complxity of oprtions on rgulr idl lngugs ws studid in [8]. Whil witnss strms r normlly diffrnt for diffrnt oprtions, th min rsult of this ppr shows tht for th suclsss of right, lft nd two-sidd idls, th complxity ounds for ll sic oprtions (thos mntiond in th introduction) cn mt y singl strm of lngugs long with strm of dilcts, which r slightly modifid vrsions of th lngugs. Svrl typs of dilcts wr introducd in [4]; in this ppr w considr only dilcts dfind s follows: Lt Σ = { 1,..., k } n lpht; w ssum tht its lmnts r ordrd s shown. Lt π prtil prmuttion of Σ, tht is, prtil function π: Σ Γ whr Γ Σ, for which thr xists Σ such tht π is ijctiv whn rstrictd to nd undfind on Σ\. W dnot undfind vlus of π y th symol. If L is lngug ovr Σ, w dnot it y L( 1,..., k ) to strss its dpndnc on Σ. If π is prtil prmuttion, lt s π th lngug sustitution dfind s follows: for Σ, {π()} whn π() is dfind, nd whnπ() is not dfind. For xmpl, ifσ = {,,c},l(,,c) = {,,c} {,cc}, nd π() = c, π() =, nd π(c) =, thn s π (L) = {,c} {c}. In othr words, th lttr c plys th rol of, nd plys th rol of c. A prmuttionl dilct of L( 1,..., k ) is lngug of th form s π (L( 1,..., k )), whrπ is prtil prmuttion ofσ; this dilct is dnotd yl(π( 1 ),...,π( k )). If th ordr onσis undrstood, w usl(σ) forl( 1,..., k ) ndl(π(σ)) forl(π( 1 ),...,π( k )). LtΣ = { 1,..., k }, nd lt D = (Q,Σ,δ,q 1,F) DFA; w dnot it yd( 1,..., k ) to strss its dpndnc onσ. Ifπ is prtil prmuttion, thn th prmuttionl dilct D(π( 1 ),...,π( k )) of D( 1,..., k ) is otind y chnging th lpht of D from Σ to π(σ), nd modifying δ so tht in th modifid DFA π( i ) inducs th trnsformtion inducd y i in th originl DFA; thus π( i ) plys th rol of i. On vrifis tht if th lngug L( 1,..., k ) is ccptd y DFA D( 1,..., k ), thn L(π( 1 ),...,π( k )) is ccptd yd(π( 1 ),...,π( k )). In th squl w rfr to prmuttionl dilcts simply s dilcts. Exmpl 1. SupposD = D(,,c) = ({1,2,3},{,,c},δ,q 1,F), whrδ is dfind y th trnsformtions : (1,2,3), : (3 1), nd c: (1,2). Lt L = L(,,c) th lngug of this DFA. Considr th

5 Most Complx Rgulr Idl Lngugs 5 prtil prmuttion π() =, π() =, nd π(c) =. In th dilct ssocitd with π, th lttr plys th rol of, ndplys th rol ofc. ThusD(,,) is th DFA ({1,2,3},{,},δ π,q 1,F), whrδ π is dfind y: (1,2) nd: (1,2,3). Th lngug ofd(,,) is th dilctl(,,) ofl. Th notion of most complx strm of rgulr lngugs ws introducd informlly in [4]. A most complx strm is on whos lngugs togthr with thir dilcts mt ll th uppr ounds for th complxity msurs dscrid in th introduction. W now mk this notion prcis. First, howvr, w rcll proprty of ooln functions. Lt th truth vlus of propositions 1 (tru) nd 0 (fls). Lt : {0,1} {0,1} {0,1} inry ooln function. Extnd to function : 2 Σ 2 Σ 2 Σ : If w Σ ndl,l Σ, thnw (L L ) (w L) (w L ). A inry ooln function is propr if it is not constnt nd not function of on vril only. Dfinition 1. LtC clss of lngugs nd ltc n th suclss ofc tht consists of ll th lngugs ofc tht hv (quotint/stt) complxityn. LtΣ = { 1,..., k }, nd lt(l n (Σ) n k) strm of lngugs, whr L n C n for ll n k. Thn (L n (Σ) n k) is most complx in clss C if it stisfis ll of th following conditions: 1. Th syntctic smigroup ofl n (Σ) hs mximl crdinlity for chn k. 2. Ech quotint ofl n (Σ) hs mximl complxity for chn k. 3. L n (Σ) hs th mximl possil numr of toms for chn k. 4. Ech tom ofl n (Σ) hs mximl complxity for chn k. 5. Th rvrs ofl n (Σ) hs mximl complxity for chn k. 6. Th str ofl n (Σ) hs mximl complxity for chn k. 7. Th productl m (Σ)L n (Σ) hs mximl complxity for ll m,n k. 8. Thr xists dilct L n (π(σ)) such tht ch propr inry ooln function L m (Σ) L n (π(σ)) hs mximl complxity for ll m,n k. A most complx strm (L n n 3) for th clss of rgulr lngugs ws introducd in [4]. W giv th dfinition ofl n low. Dfinition 2. Forn 3, ltd n = D n (,,c) = (Q n,σ,δ n,1,{n}), whrσ = {,,c}, ndδ n is dfind y th trnsformtions: (1,...,n), : (1,2), c: (n 1). Lt L n = L n (,,c) th lngug ccptd yd n. Th structur ofd n (,,c) is shown in Figur 1. c c,c,c,c, n 2 n 1 n,c Fig. 1: Miniml DFA D n(,,c) of Dfinition 2. Our min contriutions in this ppr r most complx strms for th clsss of right, lft, nd two-sidd rgulr idls.

6 6 Jnusz Brzozowski, Sylvi Dvis, Bo Yng Victor Liu 3 Right Idls A strm of right idls tht is most complx ws dfind nd studid in [6]. For compltnss w includ th rsults from tht ppr; th proof of Thorm 5 did not ppr in [6]. Dfinition 3. For n 3, lt D n = D n (,,c,d) = (Q n,σ,δ n,1,{n}), whr Σ = {,,c,d}, nd δ n is dfind y th trnsformtions: (1,...,n 1), : (2,...,n 1), c: (n 1 1) nd d: (n 1 n). Not tht is th idntity whn n = 3. Lt L n = L n (,,c,d) th lngug ccptd y D n. Th structur ofd n (,,c,d) is shown in Figur 2.,c,d c,d c,d c,d,,c,d,,,, d n 2 n 1 n,c Fig. 2: Miniml DFA D n(,,c,d) of Dfinition 3. Th DFA of Figur 2 hs similr structur to th DFA of Figur 1. Mor prcisly, DFA D n of Figur 2 is constructd y tking DFA D n 1 of Figur 1, dding nw stt n nd nw input d: (n 1 n), mking n th only finl stt, nd hving induc th cyclic prmuttion (2,...,n 1), rthr thn th trnsposition(1,2). Th nw stt nd input d nsur tht L n is right idl. Th nw trnsformtion y is ncssry sinc, ifinducs(1,2) in D n, thnl n dos not mt th ound for product. Thorm 1 (Right Idls [6]). For ch n 3, th DFA D n (,,c,d) of Dfinition 3 is miniml nd its lngug L n (,,c,d) is right idl of complxity n. Th strm (L n (,,c,d) n 3) with dilct strm(l n (,,c,d) n 3) is most complx in th clss of rgulr right idls. In prticulr, this strm mts ll th complxity ounds listd low, which r mximl for right idls. In svrl css th ounds cn mt with rstrictd lphts, s shown low. 1. Th syntctic smigroup of L n (,,c,d) hs crdinlity n n 1. Morovr, fwr thn four inputs do not suffic to mt this ound. 2. Th quotints of L n (,,,d) hv complxity n, xcpt for th quotint {,d}, which hs complxity L n (,,,d) hs2 n 1 toms. 4. For ch toma S of L n (,,c,d), th complxityκ(a S ) stisfis: { 2 n 1, if S = Q n ; κ(a S ) = 1+ S n S ( n 1 )( n x x=1 y=1 x 1 y 1), if S Qn. 5. Th rvrs ofl n (,,,d) hs complxity2 n Th str ofl n (,,,d) hs complxityn+1.

7 Most Complx Rgulr Idl Lngugs 7 7. Th productl m (,,,d)l n (,,,d) hs complxitym+2 n For ny propr inry ooln function, th complxity ofl m (,,,d) L n (,,,d) is mximl. In prticulr, () L m (,,,d) L n (,,,d) ndl m (,,,d) L n (,,,d) hv complxitymn. () L m (,,,d)\l n (,,,d) hs complxitymn (m 1). (c) L m (,,,d) L n (,,,d) hs complxitymn (m+n 2). (d) If m n, th ounds r mt y L m (,,,d) ndl n (,,,d). Proof: For 1 q n 1, non-finl stt q ccpts n 1 q d nd no othr non-finl stt ccpts this word. All non-finl stts r distinguishl from th finl stt n. Hnc D n (,,,d) is miniml nd L n (,,,d) hs n quotints. Sinc D(,,c,d) hs only on finl stt nd tht stt ccpts Σ, it is right idl. 1. Th cs n = 3 is sily chckd. For n 4, lt D n = (Q n,σ,δ n,1,{n}), whr Σ = {,,c,d}, nd : (1,...,n 1), : (1,2), c: (n 1 1) nd d: (n 1 n). It ws provd in [15] tht th trnsition smigroup of miniml DFA ccpting right idl hs t mostn n 1 trnsformtions, nd tht th trnsition smigroup ofd n hs crdinlityn n 1. Sinc forn 3,(1,2) is inducd y n 2 in D n, ll th trnsformtions of D n cn inducd in D n, nd th clim follows. Morovr, it ws provd in [12] tht n lpht of t lst four lttrs is rquird to mt this ound. 2. Ech quotint of L n (,,,d), xcpt {,d}, hs complxity n, sinc stts 1,...,n 1 r strongly connctd. Ech right idl must hv finl stt tht ccpts Σ (for L n (,,,d) this is stt n), nd so th complxity of th quotint corrsponding to this finl stt is 1. Hnc th complxitis of th quotints r mximl for right idls. 3. It ws provd in [13] tht th numr of toms of ny rgulr lngug L is qul to th complxity of th rvrs ofl. IfLis right idl of complxityn, th mximl complxity of th rvrsl R is 2 n 1 [8]. For n = 3, it is sily chckd tht our witnss mts this ound. For n > 3, it ws provd in [15] tht th rvrs ofl n (,,,d) rchs this ound. 4. This ws provd in [7]. 5. S Itm 3 ov. 6. S Thorm S Thorm S Thorms 4 nd 5.

8 8 Jnusz Brzozowski, Sylvi Dvis, Bo Yng Victor Liu 3.1 Str Thorm 2 (Right Idls: Str [6]). Forn 3 th complxity of th str of L n (,,,d) isn+1. Proof: IfLis right idl, thnl = L {ε}. Considr th DFA for str constructd s follows. To ddε tol, th initil stt1of our witnss cnnot md finl sinc this would dd othr words to th lngug, for xmpl, n 1. Thus n dditionl stt, sy 0, is rquird; this stt is oth initil nd finl in th DFA for L, its outgoing trnsitions r th sm s thos of th initil stt 1 of th DFA for L, nd it hs no incoming trnsitions. Sinc this DFA ccpts L {ε}, n + 1 stts r sufficint. All th stts r pirwis distinguishl, sinc 0 rjcts d, whil n ccpts it, nd ll th non-finl stts r distinguishl y words in d. 3.2 Product W us th DFAs D m(,,,d) = (Q m,σ,δ m,1,{m }) ndd n (,,,d) = (Q n,σ,δ n,1,{n}) shown in Figur 3 for m = 4 nd n = 5, whr Σ = {,,d} nd th stts of th first DFA r primd to distinguish thm from thos of th scond DFA. W show tht th complxity of th product ofl m (,,,d)l n (,,,d) rchs th mximum possil ound m +2 n 2 drivd in [8]. Dfin th ε-nfa P = (Q m Q n,σ,δ P,{1 },{n}), whr δ P (p,) = {δ (p,)} if p Q m, Σ, δ P(q,) = {δ(q,)} if q Q n, Σ, nd δ P (m,ε) = {1}. This ε-nfa ccptsl m L n, nd is illustrtd in Figur 3.,d d,,d 1, 2 3 d 4 ε,d d d,,d,, d Fig. 3: ε-nfa for product with m = 4, n = 5. Thorm 3 (Right Idls: Product [6]). For m,n 3 th complxity of th product of L m (,,,d) nd L n (,,,d) ism+2 n 2. Proof: It ws shown in [8] tht m+2 n 2 is n uppr ound on th complxity of th product of two right idls. To prov this ound is mt, w pply th sust construction to P to otin DFA D for L m L n. Th stts of D r susts of Q m Q n. W prov tht ll stts of th form {p }, p = 1,...,m 1 nd ll stts of th form{m,1} S, whrs Q n \{1,n}, nd stt {m,1,n} r rchl, for totl of m+2 n 2 stts. Stt {1 } is th initil stt, nd{p } is rchd y p 1 forp = 2,...,m 1. Also,{m,1} is rchd y m 2 d, nd stts m nd 1 r prsnt in vry sust rchl from {m,1}. By pplying q 2 to {m,1} w rch{m,1,q}; hnc ll susts{m,1} S with S = 1 r rchl. Assum now tht w cn rch ll sts {m,1} S with S = k, nd suppos tht w wnt to rch {m,1} T witht = {q 0,q 1,...,q k } with2 q 0 < q 1 < < q k n 1. This cn don y strting with S = {q 1 q 0 + 1,...,q k q 0 + 1} nd pplying q0 2. Finlly, to rch {m,1,n}, pply d to {m,1,n 1}.

9 Most Complx Rgulr Idl Lngugs 9 If 1 p < q m 1, thn stt {p } is distinguishl from {q } y m 1 q d n 1 d. Also, stt p Q n with 2 p n 1 ccpts n 1 p d nd no othr stt q Q n with 2 q n 1 ccpts this word. Hnc, ifs,t Q n \{1,n} nds T, thn{m,1} S nd{m,1} T r distinguishl. Stt {p } with 2 p m 1 is distinguishl from stt {m,1} S cus thr is word with singl d tht is ccptd from {m,1} S ut no such word is ccptd y {q }. Hnc ll th non-finl stts r distinguishl, nd{m,1,n} is th only finl stt. 3.3 Booln Oprtions W rstrict our ttntion to th four ooln oprtions,,\,, sinc th complxity of ny othr propr inry ooln oprtion cn otind from ths four. For xmpl, w hvκ(l L) = κ(l L) = κ(l L) = κ(l\l ). Tight uppr ounds for ooln oprtions on right idls [8] r mn for intrsction nd symmtric diffrnc,mn (m 1) for diffrnc, ndmn (m+n 2) for union. SincL n L n = L n L n = L n, ndl n \L n = L n L n =, two diffrnt lngugs must usd to rch th ounds if m = n. W us th DFAs D m(,,,d) = (Q m,σ,δ m,1,{m }) nd D n (,,,d) = (Q n,σ,δ n,1,{n}) shown in Figur 4 form = 4 ndn = 5.,d d,,d 1, 2 3 d 4,d d d,,d,, d Fig. 4: Right-idl witnsss for ooln oprtions with m = 4, n = 5. LtD m,n = D m D n. Dpnding on th ssignmnt of finl stts, this DFA rcognizs diffrnt ooln oprtions onl m ndl n. Th dirct product ofd 4 (,,,d) ndd 5 (,,,d) is in Figur 5. Lt S n dnot th symmtric group of dgr n. A sis [22] of S n is n ordrd pir (s,t) of distinct trnsformtions of Q n = {1,...,n} tht gnrt S n. A DFA hs sis (t,t ) for S n if it hs lttrs, Σ such tht inducs t nd inducs t. In th cs of our right idl D m (D n), th trnsitions δ nd δ (δ nd δ ) rstrictd to {1,...,(m 1) }({1,...,n 1}), constitut sis fors m 1 (S n 1 ). Considr DFAs A m 1 = (Q m 1,Σ,δ m 1,1,{(m 1) }) nd A n 1 = (Q n 1,Σ,δ n 1,1,{n 1}), whr Σ = {,} nd th trnsitions of A m 1 (A n 1) r th sm s thos of D m (D n) rstrictd to Q m 1 (Q n 1). By [2, Thorm 1] ll th stts in th dirct product ofa m 1 nda n 1 r rchl nd distinguishl form 1,n 1 2,(m 1,n 1) {(2,2),(3,4),(4,3),(4,4)}. W shll us this rsult to simplify our proof of th nxt thorm. Thorm 4 (Right Idls: Booln Oprtions [6]). Ifm,n 3, thn 1. Th complxity ofl m (,,,d) L n (,,,d) ismn. 2. Th complxity ofl m (,,,d) L n (,,,d) ismn. 3. Th complxity ofl m (,,,d)\l n (,,,d) is mn (m 1).

10 10 Jnusz Brzozowski, Sylvi Dvis, Bo Yng Victor Liu 1,1 1,2 1,3 1 d,4 1,5 2,1 2,2 2,3 2,4 d 2,5, 3,1 d 4,1 3,2 3,3 3,4 d d d 4,,2 4,,3 4 d,4 3,5 d 4,5 Fig. 5: Dirct product for ooln oprtions. Unlld trnsitions undr r solid, undr r dshd, nd undr oth ndr thick. Slf-loops r omittd. 4. Th complxity ofl m (,,,d) L n (,,,d) ismn (m+n 2). Proof: In th css whr (m,n) {(3,3),(4,5),(5,4),(5,5)}, w cnnot pply th rsult from [2, Thorm 1], ut w hv vrifid computtionlly tht th ounds r mt. Thus w ssum tht (m,n) {(3,3),(4,5),(5,4),(5,5)}. Our first tsk is to show tht ll mn stts of D m,n r rchl. By [2, Thorm 1], ll stts in th st S = {(p,q) 1 p m 1,1 q n 1} r rchl. Th rmining stts r th ons in th lst row or lst column (tht is, Row m or Columnn) of th dirct product. For 1 q n 2, from stt ((m 1),q) w cn rch (m,q) y d. From stt (m,n 2) w cn rch(m,n 1) y. From stt((m 1),n 1) w cn rch(m,n) yd. Hnc ll stts in Rowm r rchl. For 1 p m 2, from stt (p,n 1) w cn rch (p,n) y d. From stt ((m 2),n) w cn rch((m 1),n) y. Hnc ll stts in Columnnr rchl, nd thus ll mn stts r rchl. W now count th numr of distinguishl stts for ch oprtion. Lt H = {(m,q) 1 q n} th st of stts in th lst row nd lt V = {(p,n) 1 p m} th st of stts in th lst column. If {,,\, }, thnl m (,,,d) L n (,,,d) is rcognizd yd m,n, whr th st of finl stts is tkn to H V. Lt H = {((m 1),q) 1 q n 1} nd lt V = {(p,n 1) 1 p m 1}. Tht is, H is th scond lst row of stts, ndv is th scond lst column, rstrictd to S. By [2, Thorm 1], ll stts ins r distinguishl with rspct toh V, for ch ooln oprtion {,,\, }. W clim tht thy r lso distinguishl with rspct to H V for {,,\, }. To s this, on vrifis th following sttmnt: for ch {,,\, } nd ch stt (p,q) S, w hv(p,q) H V if nd only if(p,q)d H V. (This only pplis to stts(p,q) S; for xmpl, in Figur 5 w s tht (4,4)d H V ut (4,4) H V.) Sinc stts in S r distinguishl with rspct to H V, it follows tht for ny pir of stts (p,q),(r,s) S thr is word w with (p,q)w H V nd(r,s) H V. Thn y th sttmnt, th wordwd snds(p,q) intoh V nd (r,s) outsid ofh V,thus distinguishing th two stts with rspct toh V.

11 Most Complx Rgulr Idl Lngugs 11 Thus for ch ooln oprtion, ll stts in S r distinguishl from ch othr with rspct to th finl stt st H V. Nxt, w prov tht th stts in S r distinguishl from th rst of th stts (thos inh V) with rspct to th finl stt st H V. Sinc ll stts ins r non-finl, it suffics to distinguish stts ins from stts inx = H V \(H V), th st of non-finl stts in H V. If =, thn X contins no stts nd thr is nothing to don. If =, th only stt in X is (m,n), which is mpty, ut ll stts in S r non-mpty. If = \, thn ll stts in X r mpty ut no stts in S r. Finlly, if =, osrv tht ch stt in X ccpts word contining singld, whil stts ins\{((m 1),n 1)} ccpt only words with t lst two occurrncs of d. To distinguish((m 1),n 1) from(p,q) X, pply(which mps((m 1),n 1) to (1,2)) nd thn pply word which distinguishs(1,2) from(p,q). W hv shown tht ch stt in S is distinguishl from vry othr stt in th dirct product D m,n, with rspct to ch of th four finl stt stsh V with {,,\, }. Sinc thr r(m 1)(n 1)= mn m n+1 stts ins, thr r t lst tht mny distinguishl stts for ch oprtion. To show tht th complxity ounds r rchd yl m (,,,d) L n (,,,d), it suffics to considr how mny of th rmining m +n 1 stts in H V r distinguishl with rspct to H V. W considr ch oprtion in turn. Intrsction: Hr th st of finl stts is H V = {(m,n)}. Stt (m,n) is th only finl stt nd hnc is distinguishl from ll th othr stts. Any two stts inh (V ) r distinguishd y words in d ( d). Stt (m,1) ccpts n 2 d, whil (1,n) rjcts it. For 2 q n 1, (m,q) is snt to (m,1) y n q, whil stt (1,n) is not chngd y tht word. Hnc (m,q) is distinguishl from (1,n). By symmtric rgumnt,(p,n) is distinguishl from(m,1) for 2 p m 1. For 2 q n 1 nd 2 p m 1,(m,q) is distinguishd from(p,n) cus n q snds th formr to(m,1) nd th lttr to stt of th form(r,n), whr2 r m 1. Hnc ll pirs of stts fromh V r distinguishl. Sinc thr r m+n 1 stts in H V, it follows thr r (mn m n+1)+(m+n 1) = mn distinguishl stts. Symmtric Diffrnc: Hr th st of finl stts is H V, tht is, ll stts in th lst row nd column xcpt(m,n), which is th only mpty stt. This sitution is complmntry to tht for intrsction. Thus vry two stts fromh V r distinguishl y th sm word s for intrsction. Hnc thr r mn distinguishl stts. Diffrnc: Hr th st of finl stts is H \V, tht is, ll stts in th lst row H xcpt (m,n), which is mpty. All othr stts in th lst column V r lso mpty. Th m mpty stts in V r ll quivlnt, nd th n 1 finl stts in H \ V r distinguishd in th sm wy s for intrsction. Hnc thr r (n 1)+1 = n distinguishl stts inh\v. It follows thr r(mn m n+1)+n = mn (m 1) distinguishl stts. Union: Hr th st of finl stts is H V. From stt in H V it is possil to rch only othr stts inh V, nd ll ths stts r finl; so vry stt inh V ccptsσ. Thus ll th stts inh V r quivlnt, nd so thr r(mn m n+1)+1 = mn (m+n 2) distinguishl stts. Although it is impossil for th strm (L n (,,,d) n 3) to mt th ounds for ooln oprtions whnm = n, this strm is s complx s it could possily in viw of th following thorm: Thorm 5 (Right Idls: Booln Oprtions,m n). Supposm,n 3 ndm n. 1. Th complxity ofl m (,,,d) L n (,,,d) ismn. 2. Th complxity ofl m (,,,d) L n (,,,d) ismn. 3. Th complxity ofl m (,,,d)\l n (,,,d) is mn (m 1).

12 12 Jnusz Brzozowski, Sylvi Dvis, Bo Yng Victor Liu 4. Th complxity ofl m (,,,d) L n (,,,d) ismn (m+n 2). Proof: Lt D m = D m(,,,d), D n = D n (,,,d), nd D m,n = D m D n th dirct product utomton. If (m, n) {(4, 5),(5, 4)}, w hv vrifid computtionlly tht th ounds r mt. If (m,n) {(4,5),(5,4)}, w cn pply [2, Thorm 1]. Thus y th rgumnts usd in th proof of Thorm 4, ll stts ofd m,n r rchl. Most of th distinguishility rgumnts crry ovr s wll. If H nd V r th lst row nd column of stts in D m.n rspctivly, nd S is th st of stts lying outsid of H V, w cn us nrly idnticl rgumnts s in th proof of Thorm 4 to show tht for {,,\, }, vry stt ins is distinguishl from vry othr stt ind m,n with rspct toh V. It rmins to count th numr of stts inh V tht r distinguishl with rspct toh V. Intrsction: Hr th st of finl stts is H V = {(m,n)}. Sinc (m,n) is th only finl stt, it is distinguishl from ll othr stts. Any two stts oth in H (or oth in V ) r distinguishd y words in d. Suppos m < n. Thn m 1 snds (m,1) to (m,m) nd fixs (1,n). Words in cn snd (m,m) to (m,q) for 2 q n 1, nd thy fix (1,n). For 2 q n 1, (m,q) ccpts n 1 q d, whil (1,n) rmins fixd. Hnc (m,q) is distinguishl from(1,n) for ll q. For 2 p m 1 nd 2 q n 1, (m,q) is distinguishd from (p,n) cus m p snds (p,n) to (1,n) nd (m,q) to som stt tht is distinguishl from (1,n). Hnc ll pirs of stts from H V r distinguishl if m < n. A symmtric rgumnt works for m > n. Thus ll mn stts r distinguishl. Symmtric Diffrnc, Diffrnc, nd Union: Th rgumnts r similr to thos usd in th proof of Thorm 4. Rmrk 1. For ch clss of lngugs w studid in this ppr, our gol ws to find singl DFA strm tht mts th uppr ounds (for tht clss) on ll of our complxity msurs. For rgulr right idls, four-lttr lpht ws ncssry to chiv this, cus fwr thn four lttrs r not sufficint for th siz of th syntctic smigroup to mximl. Hving found such DFA, w thn osrvd tht th lpht of this DFA cn rducd for svrl oprtions. On th othr hnd, if on wishs to minimiz th lpht for on prticulr oprtion only, it is possil to find witnsss ovr vn smllr lphts. W list hr ch oprtion with th siz of th smllst known lpht (first ntry) long with our lpht siz (scond ntry): rvrsl (2/2), str (2/2), product (2/3), union (2/3), intrsction (2/3), symmtric diffrnc (2/3), nd diffrnc (2/3). As n xmpl, considr th two inry witnsss for th product oprtion tht r usd in [8]: D m = (Q m,{,},δ m,1,{m }), whr,: ((m 1) m )((m 2) (m 1) ) (1 2 ), nd D n = (Q n,{,},δ n,1,{n}), whr : (1,2,...,n 1), : (n 1 n)(n 2 n 1) (2 3). Not tht, lthough only two inputs r usd, thy induc thr diffrnt trnsformtions. Thus on cn rgu tht ths witnsss r not simplr thn ours. 4 Lft Idls Th following strm of lft idls ws dfind in [15]: Dfinition 4. For n 4, lt D n = D n (,,c,d,) = (Q n,σ,δ n,1,{n}), whr Σ = {,,c,d,}, nd δ n is dfind y trnsformtions: (2,...,n), : (2,3), c: (n 2), d: (n 1), : (Q n 2). Lt L n = L n (,,c,d,) th lngug ccptd yd n. Th structur ofd n (,,c,d,) is shown in Figur 6. This strm of lngugs is closly rltd to th strm of Figur 1. Th DFAD n (,,c,d,) of Dfinition 4 is constructd y tkingd n 1 (,,c) of Figur 1 with stts rlld2,...,n, dding nw stt1 nd nw inputsd: (n 1) nd: (Q n 2).

13 Most Complx Rgulr Idl Lngugs 13,,c,d c,d, c,d,c,d,c,d, 1 2, n 1 n d,c, Fig. 6: Miniml DFAD n(,,c,d,) of Dfinition 4. Thorm 6 (Lft Idls). For ch n 4, th DFA D n (,,c,d,) of Dfinition 4 is miniml nd its lngugl n (,,c,d,) is lft idl of complxity n. Th strm (L n (,,c,d,) n 4) with dilct strm(l n (,,,d,c) n 4) is most complx in th clss of rgulr lft idls. In prticulr, this strm mts ll th complxity ounds listd low, which r mximl for lft idls. In svrl css th ounds cn mt with rstrictd lphts, s shown low. 1. Th syntctic smigroup of L n (,,c,d,) hs crdinlity n n 1 +n 1. Morovr, fwr thn fiv inputs do not suffic to mt this ound. 2. All quotints of L n (,,,d,) hv complxityn. 3. L n (,,c,d,) hs2 n 1 +1 toms. 4. For ch toma S of L n (,,c,d,), th complxityκ(a S ) stisfis: n, if S = Q n ; κ(a S ) = 2 n 1, if S = ; 1+ S ) x=1, othrwis. n S y=1 ( n 1 x )( n x 1 y 1 5. Th rvrs ofl n (,,c,d,) hs complxity2 n Th str ofl n (,,,,) hs complxityn Th productl m (,,,,)L n (,,,,) hs complxitym+n For ny propr inry ooln function, th complxity of L m (,,c,,) L n (,,,,c) ismn. Proof: DFA D n (,,,d,) is miniml sinc only stt 1 dos not ccpt ny word in, whrs vry othr stt p > 1 ccpts n p nd no stt q with 1 < q p ccpts this word. It ws provd in [15] tht D n (,,c,d,) ccpts lft idl. 1. It ws shown in [10] tht th syntctic smigroup of lft idl of complxityn hs crdinlity t most n n 1 +n 1, nd in [15] tht th syntctic smigroup ofl n (,,c,d,) hs crdinlityn n 1 +n 1. Morovr, it ws provd in [12] tht n lpht of t lst fiv lttrs is rquird to mt this ound.

14 14 Jnusz Brzozowski, Sylvi Dvis, Bo Yng Victor Liu 2. Ech quotint ofl n (,,,d,) hs complxityn, sinc its miniml DFA is strongly connctd. 3. Th numr of toms of ny rgulr lngug L is qul to th complxity of th rvrs of L [13]. It ws provd in [15] tht th complxity of th rvrs ofl n (,,c,d,) is 2 n This ws provd in [7]. 5. S Itm 3 ov. 6. Th rgumnt is th sm s for th str of right idls. 7. S Thorm S Thorm Product W now show tht th complxity of th product of D m(,,,,) with D n (,,,,) rchs th mximum possil ound m + n 1 drivd in [8]. As in [8] w us th following construction: Dfin DFA D from DFAsD m ndd n y omitting th finl stt ofd m nd ll th trnsitions from th finl stt, nd rdircting ll th trnsitions tht go from non-finl stt of D m to th finl stt of D m to th initil stt ofd n. It ws provd in [8] thtd ccptsl m L n. Th construction is illustrtd in Figur 7 form = 4 ndn = , Fig. 7: Product of lft-idl witnsss with m = 4, n = 5. Thorm 7 (Lft Idls: Product). For m,n 4, th complxity of th product of L m (,,,,) nd L n (,,,,) ism+n 1. Proof: By constructiond hsm+n 1 stts. It is lso clr tht th shortst word ccptd y stt 1 is m 2 n 2, whrs for stt p with 2 p m 1 it is m p n 2, for stt 1 it is n 2, nd for ny stt q with2 q n it is n q. Hnc ll th stts r distinguishl y thir shortst words.

15 Most Complx Rgulr Idl Lngugs 15,c c, c ,c, c, c, c c 5,c, 4.2 Booln Oprtions Fig. 8: Lft-idl witnsss for ooln oprtions with m = 4, n = 5. As pointd out rlir, two diffrnt lngugs hv to usd to rch th mximl complxity for ooln oprtions. Lt D m = D m(,,c,,),d n = D n (,,,,c), ndd m,n = D m D n. Figur 8 shows DFAsD 4 (,,c,,) ndd 5(,,,,c). Thorm 8 (Lft Idls: Booln Oprtions). If m,n 4 nd is ny propr inry ooln function, thn th complxity ofl m (,,c,,) L n (,,,,c) ismn. Proof: Our first tsk is to show tht llmn stts ofd m,n r rchl. Stt(1,1) is th initil stt. For q = 2,...,n,(1,q) is rchl yc q 2, nd forp = 2,...,m,(p,1) is rchl y p 2. Nxt,(p,2) is rchd from(p,1) ycforp = 2,...,m 1, nd(2,q) is rchd from(1,q) yforq = 2,...,n 1. Lt g = gcd(m 1,n 1) nd l = lcm(m 1,n 1); thn g l = (m 1)(n 1). Not tht is prmuttion on th st S = {(p,q) 2 p m,2 q n}. Sinc is cyclic prmuttion of ordr m 1 on Q m \{1 }, nd is lso cyclic prmuttion of ordr n 1 on Q n \{1}, is prmuttion of ordrl = lcm(m 1,n 1) ons. Lt(p 1,q 1 ) nd(p 2,q 2 ) lmnts ofs, such thtp 1 q 1 p 2 q 2 mod g. Thnp 2 p 1 q 2 q 1 mod g. Sincm 1 nd n 1 g r coprim, y th Chins Rmindr Thorm th quivlncsk p 2 p 1 mod (m 1) nd k q 2 q 1 mod n 1 g hv uniqu solution modulo(m 1) n 1 g = l for k. Sinc k p 2 p 1 mod (m 1), w hvk p 2 p 1 q 2 q 1 mod g. Comind withk q 2 q 1 mod n 1 g, this givsk q 2 q 1 mod (n 1). Applying k to(p 1,q 1 ), givs th stt(p 1+k mod (m 1),q 1 +k mod (n 1)) = (p 2,q 2). So for vry pir (p 1,q 1) nd (p 2,q 2) such tht p 1 q 1 p 2 q 2 mod g, (p 2,q 2 ) is rchl from(p 1,q 1 ) y som numr of s. If (p 1,q 1) S is rchl, ll (p,q) S such tht p 1 q 1 p q mod g r lso rchl. Sinc (p 1,2) is rchl forp 1 = 2,...,m 1, ll (p,q) S such tht p q {0,1,2,...,(m 3) mod g} r rchl. Sinc (2,3) is rchl, ll (p,q) S such tht p q 1 (m 2) mod g r rchl. Sinc ll rmindrs modulog hv t lst on rprsnttiv, ll ofs is rchl. W prov distinguishility using numr of clims: 1. (2,2) is distinguishl from vry othr stt in Column 2. () If th oprtion is intrsction (symmtric diffrnc), thn (m,n) is distinguishl from ll othr stts in Columnn, sinc(m,n) is th only finl (non-finl) stt in this column. Not tht m 1 ( n 1 ) cts s th idntity on th st Q m \{1 } (Q n \{1}). Hnc lcm(m 1,n 1) is th idntity onq m\{1} Q n \{1}, ndx = lcm(m 1,n 1) 1 mps(2,2) to(m,n). If two stts r in th sm column, thn so r thir succssors ftr l is pplid, for nyl 0. Thrfor pplyingxto(p,2) lds to stt in Columnn; so (2,2) nd(p,2) r distinguishl yx.

16 16 Jnusz Brzozowski, Sylvi Dvis, Bo Yng Victor Liu () If th oprtion is union (diffrnc), thn (m,n 1) is distinguishl from ny othr stt in Column n 1. Considr (2,2) nd (p,2); pplying c rsults in (3,2) nd (r,2), whr r 3. Following this y lcm(m 1,n 1) 2 yilds (m,n 1) nd (s,n 1), whr s m. Hncc lcm(m 1,n 1) 2 distinguishs(2,2) from(p,2). 2. (2,2) is distinguishl from vry othr stt in Row 2. Th rgumnt is symmtric to tht for Clim 1, whn th oprtion is intrsction, symmtric diffrnc, or union. If th oprtion is diffrnc, th stts cn distinguishd y lcm(m 1,n 1) 1 sinc this mps(2,2) to(m,n), which is distinguishl from ll othr stts in Row m. 3. For ny two stts in th sm column thr is word mpping xctly on of thm to(2,2). Lt th two stts (p,q) nd(r,q). If {p,r} = {2,m}, lt s = 1; othrwis, lt s = 0. Applying s rsults in (p 1,q 1 ) nd (r 1,q 1 ), whr {p 1,r 1 } {2,m}, sinc m 4. Thus w cn ssum tht th pir of stts to distinguishd is (p,q) nd (r,q), whr {p,r} {2,m}. Now c tks ths stts to(p 1,2) nd(r 1,2), ndp 1 r 1, sincccn mp two stts ofq m\{1 } to th sm stt only if ths stts r2 ndm. Osrv thtc cycliclly prmuts stts{(p,2) 2 p m 1}. So pplyingc sufficint numr of tims mps xctly on of th two stts to (2,2). 4. For ny two stts in th sm row thr is word mpping xctly on of thm to(2,2). Th proof is symmtric to tht for Clim 3, if w intrchng rows nd columns nd rplcc y. 5. For ny pir of stts, thr xists word tht mps on of th stts to (2,2) nd th othr to stt(p,2),p 2, or(2,q), q 2. Thr r svrl css: () If th stts r in th sm column or row, thn th rsult holds y Clims 3 nd 4. Hnc ssum thy r not in th sm column or row. () If oth stts li outsid of Row m, thn c tks oth stts to Column 2 ut it tks ch stt to diffrnt row. Th rsult now follows y Clim 3. (c) If oth stts li outsid of Columnn, thntks oth stts to Row2ut it tks ch stt to diffrnt column. Th rsult now follows y Clim 4. (d) If on of th stts is(m,n), thntks it to(2,2). If Cs () dos not pply, th othr stt must hv n tkn to Row m. If Cs (c) lso dos not pply, it must lso hv n tkn to Columnn. Thus th othr stt must hv n tkn to (m,n). Applying gin rsults in (2,2) nd(3,3), nd this rducs to Cs (). () If th stts r(m,n 1) nd((m 1),n), thn pplying 2 rsults in(3,2) nd(2,3), nd Cs () pplis. (f) In th rmining cs, on stt is in Row m nd th othr stt is in Column n; furthrmor, nithr stt is th stt (m,n) from Cs (d), nd th pir of stts ing considrd is not th pir(m,n 1) nd((m 1),n) from Cs (). If th stt in Rowmis not(m,n 1), pplying snds it to stt tht is not in Columnn, nd th othr stt to Column 2; so Cs (c) pplis. Othrwis, th stt in Row m is (m,n 1) nd th stt in Columnnis not((m 1),n). Applyingsnds th first stt to Row 2 nd th othr stt to stt not in Rowm; so Cs () pplis. W hv shown tht for ny pir of stts, thr xists word tht tks on of th stts to (2,2) nd th othr stt not to (2,2) ut to ithr Row 2 or Column 2 y Clim 5. By Clims 1 nd 2, thos stts r distinguishl. Thrfor th originl stts r lso distinguishl.

17 Most Complx Rgulr Idl Lngugs 17 Rmrk 2. For rgulr lft idls, th miniml lpht rquird to mt ll th ounds hs fiv lttrs. As ws th cs with right idls, it is possil to rduc th lpht for som oprtions [8]. Th sizs r s follows: rvrsl (3/4), str (2/2), product (1/2), union(4/3), intrsction (2/3), symmtric diffrnc (2/3), nd diffrnc (3/3). Not tht th prviously known witnss for union usd four-lttr lpht, whil ours only uss thr lttrs. 5 Two-Sidd Idls Th following strm of two-sidd idls ws dfind in [15]: Dfinition 5. For n 5, lt D n = D n (,,c,d,,f) = (Q n,σ,δ,1,{n}), whr Σ = {,,c,d,,f}, nd δ n is dfind y th trnsformtions: (2,3,...,n 1), : (2,3), c: (n 1 2), d: (n 1 1), : (Q n 1 2), ndf: (2 n). Th structur ofd n (,,c,d,,f) is shown in Figur 9.,,c,d,,f,c, n,,c,d,f f c,d,f,c,d,f,c,d,f,f, , n 2 n 1 c,d, Fig. 9: Miniml DFAD n(,,c,d,,f) of Dfinition 5. This strm of lngugs is closly rltd to th strm of Figur 1. Th DFA D n (,,c,d,,f) of Dfinition 5 is constructd y tking D n 2 (,,c) of Figur 1 with stts rlld 2,...,n 1, dding nw stts1ndn, nd nw inputsd: (n 1 1), : (Q n 1 2), ndf: (2 n). Thorm 9 (Two-Sidd Idls). For chn 5, th DFAD n (,,c,d,,f) of Dfinition 5 is miniml nd its lngugl n (,,c,d,,f) is two-sidd idl of complxityn. Th strm(l n (,,c,d,,f) n 5) with dilct strm (L n (,,c,d,,f) n 5) is most complx in th clss of rgulr two-sidd idls. In prticulr, this strm mts ll th complxity ounds listd low, which r mximl for two-sidd idls. In svrl css th ounds cn mt with rstrictd lphts, s shown low. d 1. Th syntctic smigroup of L n (,,c,d,,f) hs crdinlity n n 2 + (n 2)2 n Morovr, fwr thn six inputs do not suffic to mt this ound. 2. All quotints of L n (,,,d,,f) hv complxity n, xcpt th quotint corrsponding to stt n, which hs complxity L n (,,,d,,f) hs2 n 2 +1 toms.

18 18 Jnusz Brzozowski, Sylvi Dvis, Bo Yng Victor Liu 4. For ch toma S of L n (,,c,d,,f), th complxityκ(a S ) stisfis: n, if S = Q n ; κ(a S ) = 2 n 2 +n 1, if S = Q n \{1}; 1+ S )( n x 1 ) x=1, othrwis. n S y=1 ( n 2 x 1 y 1 5. Th rvrs ofl n (,,,d,,f) hs complxity2 n Th str ofl n (,,,,,f) hs complxityn Th productl m (,,,,,f)l n (,,,,,f) hs complxitym+n For ny propr inry ooln function, th complxity of L m (,,,d,,f) L n (,,,d,,f) is mximl. In prticulr, () L m (,,,d,,f) L n (,,,d,,f) hs complxitymn, s dos L m (,,,d,,f) L n (,,,d,,f). () L m (,,,d,,f)\l n (,,,d,,f) hs complxitymn (m 1). (c) L m (,,,d,,f) L n (,,,d,,f) hs complxitymn (m+n 2). (d) If m n, th ounds r mt y L m (,,,d,,f) ndl n (,,,d,,f). Proof: Notic tht inputs, nd f r ndd to mk ll th stts rchl. It ws provd in [15] tht D n (,,c,d,,f) is miniml nd ccpts two-sidd idl. 1. It ws shown in [10] tht th syntctic smigroup of two-sidd idl of complxity n hs crdinlity t most n n 2 + (n 2)2 n 2 +1, nd in [15] tht th syntctic smigroup of L n (,,c,d,,f) hs tht crdinlity. Morovr, it ws provd in [12] tht n lpht of t lst six lttrs is rquird to mt this ound. 2. This follows from Dfinition Th numr of toms of ny rgulr lngug L is qul to th complxity of th rvrs of L [13]. It ws provd in [15] tht th complxity of th rvrs ofl n (,,,d,,f) is2 n This ws provd in [7]. 5. S Itm 3 ov. 6. Th rgumnt is th sm s for th str of right idls. 7. S Thorm S Thorms 11 nd 12.

19 Most Complx Rgulr Idl Lngugs Product W show tht th complxity of th product of th DFA D m (,,,,,f) with D n(,,,,,f) mts th uppr ound m + n 1 drivd in [8]. W us th sm construction s for lft idls for th product DFA D. Th construction is illustrtd in Figur 10 for m = n = 5. Thorm 10 (Two-Sidd Idls: Product). Form,n 5, th product of th lngugl m (,,,,,f) with L n (,,,,,f) hs complxitym+n 1. Proof: By constructiond hsm+n 1 stts. W know tht ll th stts ind m r pirwis distinguishl. Hnc in D, for ch pir thr xists word tht tks on stt to stt 1 nd th othr to stt in Q m \{m }. Also, ll pirs of distinct stts in D n r distinguishl. To distinguish stt p from stt q in D, not tht vry word ccptd from p contins two f s, whrs thr r words ccptd from q tht contin only on f.,, f f f,f, f f 4 5,,f Fig. 10: Product of two-sidd-idl witnsss with m = 5, n = Booln Oprtions Thorm 11 (Two-Sidd Idls: Booln Oprtions). If m,n 5, thn 1. Th complxity ofl m (,,,d,,f) L n (,,,d,,f) is mn. 2. Th complxity ofl m (,,,d,,f) L n (,,,d,,f) ismn. 3. Th complxity ofl m (,,,d,,f)\l n (,,,d,,f) ismn (m 1). 4. Th complxity ofl m (,,,d,,f) L n (,,,d,,f) is mn (m+n 2). Proof: As for, w tk th dirct product DFA D m,n = D m (,,,d,,f) D n(,,,d,,f) nd count rchl nd distinguishl stts. First w prov ll stts r rchl, s illustrtd in Figur 11. Stt(1,1) is th initil stt, nd(2,2) is rchd from(1,1) y. Sinc{,} gnrts ll prmuttions ofq m\{1,m } ndq n \{1,n}, y [2, Thorm 1] ll stts in(q m\{1,m }) (Q n \{1,n}) r rchl, unlss(m 2,n 2) {(2,2),(3,4),(4,3),(4,4)}; th cs(2,2) dos not occur sincm,n 5, nd w hv vrifid th othr spcil css computtionlly. Thus it rmins to show tht ll stts in Rows 1 nd m nd Columns 1 nd n r rchl. For Row 1, first not tht(1,2) is rchl sinc((m 1),2)d = (1,2). Thn(1,q) = (1,q+1) for 2 q n 2, so w cn rch(1,3),...,(1,n 1). Finlly, w cn rch(1,n) sinc(1,2)f = (1,n).

20 20 Jnusz Brzozowski, Sylvi Dvis, Bo Yng Victor Liu A symmtric rgumnt pplis to Column 1. For Row m, not (m,1) is rchl sinc it is in Column 1. Thn w hv (m,1) = (m,2), nd (m,q) = (m,q + 1), for 2 q n 2, nd finlly (m,2)f = (m,n). A symmtric rgumnt pplis to Columnn. f 1,1 1,2 1,3 1,4 d 1,5 1,6 2,1 2,2 2,3 2,4 2,5 2,6 d 3,1 3,2 3,3 3,4 3,5 3,6 f 4,1 4,2 4,3 4,4 4,5 4,6 5,1 5,2 5,3 5,4 5,5 5,6 f Fig. 11: Prtil dirct product for ooln oprtions on two-sidd idls. Bfor w gin th distinguishility proofs, w mk fw osrvtions. Lt(p,q) nd(r,s) stts with1 p < r m; not thtr > If r < m, th word m r snds (r,s) to stt in Row 2. It snds (p,q) to ithr Row 1 (if p = 1) or Row p+m r (ifp 2); in ithr cs(p,q) is not snt to Row2or Row m. 2. Ifr < m, th word m r f snds(r,s) to Rowm, nd it snds(p,q) to th sm row s m r. 3. Ifr < m, thn m r f m r = m r f snds(r,s) to Row m, nd(p,q) to row othr thn Row m. If r = m, thn m r f m r = ε, th stt (r,s) = (m,s) is in Row m, nd th stt (p,q) is not in Rowm. Thus th word m r f m r will snd ny stt in Rowr to Rowm, nd ny stt not in Rowr to row othr thn Row m. By symmtric rgumnt, th word n s f n s snds stts in Column s to Column n, nd stts not in Columnsto diffrnt column, for 1 q < s n. W us this fct frquntly to distinguish stts. For ch ooln oprtion w now prov tht th numr of distinguishl stts mts th rlvnt uppr ound. 1. Intrsction/Symmtric Diffrnc. For intrsction, th only finl stt is (m,n). For symmtric diffrnc, vry stt in Row m or Column n xcpt (m,n) is finl, nd (m,n) is th only mpty stt. W cn us th sm distinguishing words in oth css.

21 Most Complx Rgulr Idl Lngugs 21 () Stts in distinct rows, p < r m: Stts (p,q) nd (r,s), q,s Q n, r distinguishl y first pplying m r f m r. This snds on of th stts to Rowm, nd th othr to diffrnt row. W thn pply 2 f: pplying snds th stt in Row m to (m,2) or (m,n), nd th stt not in Rowmto(2,2) or(2,n). Thn 2 fixs(m,2) or(m,n) nd snds th othr stt to (4,2) or(4,n). Finlly, pplyingf snds on stt to (m,n), nd th othr stt to(4,n). () Stts in distinct columns,q < s n: Th rgumnt is symmtric if w intrchng nd. Sinc ll stts r distinguishl, th complxity ismn. 2. Diffrnc. Hr th finl stts r thos which r in Row m, ut not Column n. Stts in Column n r indistinguishl nd mpty, sinc no word cn tk ny of ths stts out of Column n, nd tht column contins only non-finl stts. Hnc thr r t most mn (m 1) distinguishl stts. () Stts in Row m, 1 q < s n: Stts (m,q) nd (m,s) r distinguishl y n s f n s, sinc (m,n) is th only non-finl stt in Row m. () Stts in distinct rows, nd on stt is in Column n: Any stt outsid of Columnnccpts 2 f, nd thus ll of ths stts r non-mpty. It follows tht ll stts outsid of Column n r distinguishl from th mpty stts in Columnn. (c) Stts in distinct rows, nd nithr stt is in Column n: Stts (p,q) nd (r,s), with p < r m nd q,s n 1, r distinguishl y m r f m r unlss this word snds (r,s) to (m,n). This occurs only if r < m nd m r snds (r,s) to (2,2). In this cs, pplying 2 ftr m r snds(2,2) to(2,4), nd dos not ffct th row numrs; thus on of thm will in Row 2 nd th othr in nithr Row 2 nor Rowm. Thnf distinguishs th stts. (d) Stts in distinct columns: Th rrngmnt of finl nd non-finl stts in Row m mtchs tht of symmtric diffrnc. Thus th rgumnt usd for intrsction/symmtric diffrnc pplis hr lso. Sinc ll stts inq m (Q n\{n}) r distinguishl, th complxity of diffrnc ismn (m 1). 3. Union. Th finl stts r thos in Rowmnd Columnn. All finl stts r indistinguishl, sinc thy ll ccptσ. Thrfor thr r t most1+(m 1)(n 1)= mn (m+n 2) distinguishl stts. () Non-finl stts in distinct rows: Two non-finl stts (p,q) nd (r,s), with p < r < m nd q,s < n, r distinguishl y m r f, unlss this word snds(p,q) to Columnn(sinc it lso snds (r,s) to Row m). This occurs only if m r snds (p,q) to Column 2. In this cs w cn us m r 2 f to distinguish th stts; this is similr to Cs (c) of th diffrnc oprtion. () Non-finl stts in distinct columns: Ths stts r distinguishl y symmtric rgumnt. Thus th complxity of union is mn (m+n 2). If m n, th complxity ounds for ooln oprtions cn mt y using lngugs from th strm (L n (,,,d,,f) n 5) for oth rgumnts. Tht is, w cn mt th ounds for ooln oprtions without th dilct strm(l n (,,,d,,f) n 5).

22 22 Jnusz Brzozowski, Sylvi Dvis, Bo Yng Victor Liu Thorm 12 (Two-Sidd Idls: Booln Oprtions,m n). Supposm,n 5 ndm n. 1. Th complxity ofl m (,,,d,,f) L n (,,,d,,f) is mn. 2. Th complxity ofl m (,,,d,,f) L n (,,,d,,f) ismn. 3. Th complxity ofl m (,,,d,,f)\l n (,,,d,,f) ismn (m 1). 4. Th complxity ofl m (,,,d,,f) L n (,,,d,,f) is mn (m+n 2). Proof: Hr w considr th DFA D m,n = D m(,,,d,,f) D n (,,,d,,f). Th proof tht ll stts r rchl is idnticl to th proof ov, xcpt stt (1, n 1) is rchl from(1, n 2) y instd of, nd whn pplying [2, Thorm 1] th spcil css w must vrify r only(m 2,n 2) {(3,4),(4,3)}, sinc w r ssuming m n. Also, for stts (p,q) nd (r,s) with p < r, th sm rmrk out th word m r f m r pplis, i.., this word snds (r,s) to Row m nd (p,q) to diffrnt row. For (p,q) nd (r,s) with q < s, th word n s f n s snds (r,s) to Column n nd (p,q) to diffrnt column (prviously w usd n s f n s for this purpos). For ch ooln oprtion w now prov tht th numr of distinguishl stts mts th rlvnt uppr ound. 1. Intrsction/Symmtric Diffrnc. As for, th sm distinguishing words cn usd for intrsction nd symmtric diffrnc. () Stts in Column n: Stts (p,n) nd (r,n), with p < r m, r distinguishl y m r f m r. () Stts (2,2) nd (m,2) r distinguishl s follows. Sinc ll stts in (Q m \ {1,m }) (Q n \ {1,n}) r rchl from (2,2) using words in {,}, thr is word w {,} which snds(2,2) to(3,2). If w viww s prmuttion onq n \{1,n} lon, it must fix2. Thusw snds(m,2) to(m,2). Thn(3,2) nd(m,2) r distinguishd yf. (c) Stts in Column q < n: Stts (p,q) nd (r,q), with p < r m, r distinguishl sinc th word m r f m r rducs this cs to Cs () or (). (d) By symmtry, stts in th sm row r distinguishl. () Stts in distinct columns: (p,q) nd (r,s), q < s n, r distinguishl y first pplying n s f n s. This snds(p,q) to som stt in Column 2 nd(r,s) to som stt in Columnn. If ths succssor stts r in th sm row, thn this cs rducs to Cs (d). Othrwis, sinc ws pplid, th only possil stts r (2,2) nd (m,n), or (2,n) nd (m,2). In ithr cs, ths stts r distinguishd y min(m,n) 2 f. (f) By symmtry, stts in distinct rows r distinguishl. Sinc ll stts r distinguishl, th complxity of intrsction ismn. 2. Diffrnc. Stts in Column n r mpty nd thus indistinguishl. () Stts in distinct columns, nd on stt is in Column n: All stts (p,q), q < n r nonmpty, nd thus distinguishl from thos in Columnn. To s this, osrv tht ifp = m, thn (p,q) is finl stt nd thus is non-mpty. Ifp<m thnsnds(p,q) to(2,2). Sinc vry stt in (Q m \{1,m }) (Q n \{1,n}) is rchl from(2,2), thr is word w {,} tht snds(2,2) to (2,3). Thnf snds(2,3) to th finl stt (m,3).

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