structured/0707.0430.structured.json

{"paper_meta":{"paper_id":"arxiv:0707.0430","title":"0707.0430","authors":[],"year":null,"published":null,"updated":null,"venue_or_source":"arxiv-papers-shard","primary_category":"cs","secondary_categories":[],"doi":null,"license":"unknown","source_type":"pdf+shard","source_url":null,"pdf_url":null},"abstract":{"raw":"","cleaned":"","offsets":[]},"full_paper_text":{"raw_ordered_text":"arXiv:0707.0430v1 [cs.CC] 3 Jul 2007\nAssisted Problem Solving and Decompositions of\nFinite Automata∗\nPeter Gaˇzi\nBranislav Rovan\nDepartment of Computer Science, Comenius University\nMlynsk ́a dolina, 842 48, Bratislava, Slovakia\n{gazi,rovan}@dcs.fmph.uniba.sk\nAbstract\nA study of assisted problem solving formalized via decompositions of\ndeterministic finite automata is initiated. The landscape of new types of\ndecompositions of finite automata this study uncovered is presented. Lan-\nguages with various degrees of decomposability between undecomposable\nand perfectly decomposable are shown to exist.\n1\nIntroduction\nIn the present paper we initiate the study of assisted problem solving. We intend\nto model and study situations, where solution to the problem can be sought\nbased on some additional a priori information about the inputs. One can expect\nto obtain simpler solution in such case. There are similar approaches known\nin the literature, most notably the notions of advice functions [1], where the\nadditional information is based on the length of the input word and the notion\nof promise problems [2], where the set of inputs is separated into three classes\n– those with “yes” answer, those with “no” answer and those where we do not\ncare about the outcome. By considering the simplest case where the “problem\nsolving” machinery is the deterministic finite automaton (DFA) we obtain a new\nmotivation for studying new types of finite automata decompositions.\nIn this paper we shall thus consider the case where solving a problem shall\nmean constructing an automaton for a given language L. The “assistance” shall\nbe given by additional information about the input, e.g., that we can assume the\ninputs shall be restricted to words from a particular regular language L′. Thus,\ninstead of looking for an automaton A such that L = L(A) we can look for a\n(possibly simpler) automaton B such that L = L(B)∩L′. We can then say that\nB accepts L with the assistance of L′. We shall call L′ (or the corresponding\nautomaton A′ such that L′ = L(A′)) an advisor to B. In this case the advisor\nA′ provides assistance to the solver B by guaranteeing that A′ accepts the given\ninput word.\nWe shall also study a case where the assistance provides more\ndetailed information about the outcome of the computation of A′ on the input\nword (e.g., the state reached). Clearly the advisor can be considered useful\nonly if it enables B to be simpler than A and at the same time A′ is not more\n∗This work was supported in part by the grant VEGA 1/3106/06.\n1\n\ncomplicated than A. The measure of complexity we shall consider is the number\nof states of the deterministic finite automaton. This measure of complexity was\nused quite often recently due to renewed interest in finite automata prompted\nby applications such as model checking (see e.g. [3] for a recent survey). (Note\nthat results complementary to ours, namely results on complexity of automata\nfor the intersection of regular sets were studied in [4].)\nThe contribution of our paper is twofold. First, we can interpret the ‘solver’\nand the ‘advisor’ as two parallel processes each performing a different task and\njointly solving a problem. Since our approach lends itself to a generalisation to\nk advisors it may stimulate new parallel solutions to problems (the traditional\nones usually using parallel processes to perform essentially the same task). Sec-\nond, the choice of finite automata as the simplest problem solving machinery\nbrought about new types of decompositions motivated by the information the\n‘advisor’ can provide to the ‘solver’. Our results provide a complete picture of\nthe landscape of these decompositions.\nThe problem within this scenario we shall address in this paper is the exis-\ntence of a useful advisor for a given automaton A. We shall compare the power\nof several types of advisors, and investigate the effect of the advisor on the\ncomplexity of the assisted solver B. We can formulate this also as a problem of\ndecomposition of deterministic finite state automata – given DFA A find DFA\nA1 (a solver) and A2 (an advisor) such that w ∈L(A) can be determined from\nthe computations of A1 and A2. We shall study several new types of decompo-\nsitions of DFA, one of them is analogous to the state behavior decomposition of\nfinite state transducers studied in [5]. In Sect. 3 we prove relations among these\ndecompositions. For each type of decomposition there are automata which are\nundecomposable and automata for which there is a decomposition that is the\nbest possible. In Sect. 4 we consider the space between these extreme points\nand study the degree of decomposability.\n2\nDefinitions and Notation\nWe shall use standard notions of the theory of formal languages (see e.g. [6]).\nOur notation shall be as follows.\nΣ∗denotes the set of all words over the\nalphabet Σ, the length of a word w is denoted by |w|, ε denotes the empty\nword, and for a language L we shall denote by ΣL the minimal alphabet such\nthat L ⊆Σ∗\nL. The number of occurrences of a given letter a in a word w is\ndenoted by #a(w). Throughout this paper we shall consider deterministic finite\nautomata only.\nA deterministic finite automaton (DFA) is a quintuple (K, Σ, δ, q0, F), such\nthat K is a finite set of states, Σ is a finite input alphabet, q0 ∈K is the initial\nstate, F ⊆K is the set of accepting states and δ: K × Σ →K is a transition\nfunction. As usual, we shall denote by δ also the standard extension of δ to\nwords, i.e., δ: K × Σ∗→K. We shall denote by |K| the number of states in K.\nFormalizing the notions of assisted problem solving from the Introduction\nwe shall now define several types of decompositions of DFA A into two (sim-\npler) DFAs A1 and A2 (a solver and an advisor) so that the membership of\nan input word w in L(A) can be determined based on the information on the\ncomputations of A1 and A2 on w.\nWe first introduce an acceptance-identifying decomposition of deterministic\n2\n\nfinite automata.\nDefinition 2.1. A pair of DFAs (A1, A2), where A1 = (K1, Σ, δ1, q1, F1) and\nA2 = (K2, Σ, δ2, q2, F2), forms an acceptance-identifying decomposition (AI-\ndecomposition) of a DFA A = (K, Σ, δ, q0, F), if L(A) = L(A1) ∩L(A2). This\ndecomposition is nontrivial if |K1| < |K| and |K2| < |K|.\nBy decomposing A in this manner, one of the decomposed automata (say\nA2) can act as an advisor and narrow down the set of input words for the other\none (say A1), whose task to recognize the words of L(A) may become easier.\nAnother requirement we could pose on a decomposition is to identify the\nfinal state of any computation of the original automaton by only knowing the\nfinal states of both corresponding computations of the automata forming the\ndecomposition. This requirement can be formalized as follows.\nDefinition 2.2. A pair of DFAs (A1, A2), where A1 = (K1, Σ, δ1, q1, F1) and\nA2 = (K2, Σ, δ2, q2, F2), forms a state-identifying decomposition (SI-decomposition)\nof a DFA A = (K, Σ, δ, q0, F), if there exists a mapping β : K1 × K2 →K, such\nthat it holds β(δ1(q1, w), δ2(q2, w)) = δ(q0, w) for all w ∈Σ∗. This decomposi-\ntion is nontrivial if |K1| < |K| and |K2| < |K|.\nThe third – and the weakest – requirement we pose on a decomposition of a\nDFA is to require that there must exist a way to determine whether the original\nautomaton would accept some given input word based on knowing the states in\nwhich the computations of both decomposition automata have finished.\nDefinition 2.3. A pair of DFAs (A1, A2), where A1 = (K1, Σ, δ1, q1, F1) and\nA2 = (K2, Σ, δ2, q2, F2), forms a weak acceptance-identifying decomposition\n(wAI-decomposition) of a DFA A = (K, Σ, δ, q0, F), if there exists a relation\nR ⊆K1 × K2 such that it holds R(δ1(q1, w), δ2(q2, w)) ⇔w ∈L(A) for all\nw ∈Σ∗. This decomposition is nontrivial if |K1| < |K| and |K2| < |K|.\nNote that in the last two definitions, the sets of accepting states of A1 and\nA2 are irrelevant.\nBy a decomposability of a regular language L in some way, we shall mean\nthe decomposability of the corresponding minimal automaton over ΣL.\nTo be able to compare these new types of decomposition to the parallel\ndecompositions of state behavior introduced for sequential machines in [5], we\nshall redefine them for DFAs.\nDefinition 2.4. A DFA A′ = (K′, Σ, δ′, q′\n0, F ′) is said to realize the state\nbehavior of a DFA A = (K, Σ, δ, q0, F) if there exists an injective mapping\nα: K →K′ such that\n(i) (∀a ∈Σ)(∀q ∈K); δ′(α(q), a) = α(δ(q, a)),\n(ii) α(q0) = q′\n0.\nMoreover, A′ is said to realize the state and acceptance behavior of A, if in\naddition the following property holds:\n(iii) (∀q ∈K); α(q) ∈F ′ ⇔q ∈F.\n3\n\nDefinition 2.5. The parallel connection of two DFA A1 = (K1, Σ, δ1, q1, F1)\nand A2 = (K2, Σ, δ2, q2, F2) is the DFA A = A1||A2 = (K1 × K2, Σ, δ, (q1, q2),\nF1 × F2) such that δ((p1, p2), a) = (δ1(p1, a), δ2(p2, a)).\nDefinition 2.6. A pair of DFAs (A1, A2) is a state behavior (SB-) decomposi-\ntion of a DFA A if A1||A2 realizes the state behavior of A. The pair (A1, A2) is\nan acceptance and state behavior (ASB-) decomposition of A if A1||A2 realizes\nthe state and acceptance behavior of A. This decomposition is nontrivial if both\nA1 and A2 have fewer states than A.\nWe have modified the definitions to fit the formalism and purpose of deter-\nministic finite automata (i.e., to accept formal languages) without loosing the\nconnection to the strongly related and useful concept of S.P.partitions, exhibited\nbelow.\nWe shall use the following notation and properties of S.P. partitions from [5].\nA partition π on a set of states of a DFA A = (K, Σ, δ, q0, F) has substitution\nproperty (S.P.), if it holds ∀p, q ∈K;\np ≡π q ⇒(∀a ∈Σ; δ(p, a) ≡π δ(q, a)). If\nπ1 and π2 are partitions on a given set M, then\n(i) π1 · π2 is a partition on M such that a ≡π1·π2 b ⇔a ≡π1 b ∧a ≡π2 b,\n(ii) π1 +π2 is a partition on M such that a ≡π1+π2 b iffthere exists a sequence\na = a0, a1, a2, . . . , an = b, such that ai ≡π1 ai+1 ∨ai ≡π2 ai+1 for all\ni ∈{0, . . ., n −1},\n(iii) π1 ⪯π2 if it holds (∀x, y ∈M);\nx ≡π1 y ⇒x ≡π2 y.\nThe set of all partitions on a given set (with the partial order ⪯, join realized\nby + and meet realized by .) forms a lattice. The set of all S.P. partitions on\nthe set of states of a given DFA forms a sublattice of the lattice of all partitions\non this set. The trivial partitions {{q0}, {q1}, . . . , {qn}} and {{q0, q1, . . . , qn}}\nshall be denoted by symbols 0 and 1, respectively. The block of a partition π\ncontaining the state q shall be denoted by [q]π. In addition, we shall use the\nfollowing separation notion.\nDefinition 2.7. The partitions π1 = {R1, . . . , Rk} and π2 = {S1, . . . , Sl} on a\nset of states of a DFA A = (K, Σ, δ, q0, F) are said to separate the final states\nof A if there exist indices i1, . . . , ir and j1, . . . , js such that it holds (Ri1 ∪. . . ∪\nRir) ∩(Sj1 ∪. . . ∪Sjs) = F.\n3\nRelations Between Types of Decompositions\nThe concept of partitions separating the final states allows us to derive a neces-\nsary and sufficient condition for the existence of SB- and ASB-decompositions\nsimilar to the one stated in [5].\nTheorem 3.1. A DFA A = (K, Σ, δ, q0, F) has a nontrivial SB-decomposition\niffthere exist two nontrivial S.P. partitions π1 and π2 on the set of states of A\nsuch that π1 · π2 = 0. This decomposition is an ASB-decomposition if and only\nif these partitions separate the final states of A.\nProof. The proof is analogous to that in [5] but had to be extended for the\nASB-decomposition. We omit it due to space constraints.\n4\n\nFor the other decompositions, we can derive the following sufficient condi-\ntions that exploit the concept of S.P. partitions.\nTheorem 3.2. Let A = (K, Σ, δ, q0, F) be a deterministic finite automaton, let\nπ1 and π2 be nontrivial S.P. partitions on the set of states of A, such that they\nseparate the final states of A. Then A has a nontrivial AI-decomposition.\nProof. Since π1 and π2 separate the final states of A, there exist blocks B1, . . . , Bk\nand C1, . . . , Cl of the partitions π1 and π2 respectively, such that (B1 ∪. . . ∪\nBk) ∩(C1 ∪. . . ∪Cl) = F. We shall construct two automata A1 and A2 having\nstates corresponding to blocks of these partitions and show that (A1, A2) is a\nnontrivial AI-decomposition of A. Let A1 = (π1, Σ, δ1, [q0]π1, {B1, . . . , Bk}) and\nA2 = (π2, Σ, δ2, [q0]π2, {C1, . . . , Cl}) be DFAs with δi defined by δi([q]πi, a) =\n[δ(q, a)]πi, i ∈{1, 2} (this definition does not depend on the choice of q since πi\nis an S.P. partition). We now need to prove that L(A) = L(A1) ∩L(A2).\nLet w ∈L(A). Suppose that the computation of A on the word w ends\nin some accepting state qf ∈F. Then, from the construction of A1 and A2 it\nfollows that the computation of Ai on the word w ends in the state corresponding\nto the block [qf]πi of the partition πi. Since qf ∈F, it must hold [qf]π1 ∈\n{B1, . . . , Bk} and [qf]π2 ∈{C1, . . . , Cl}, hence from the construction of Ai,\nthese blocks correspond to the accepting states in the respective automata.\nThus w ∈L(Ai) for i ∈{1, 2}, therefore L(A) ⊆L(A1) ∩L(A2).\nNow suppose w ∈L(A1) ∩L(A2), Thus the computation of A1 on w ends\nin one of the states B1, . . . , Bk, which means that the computation of A on w\nwould end in a state from the union of blocks B1 ∪. . . ∪Bk. Using the same\nargument for A2, we get that the computation of A on w would end in a state\nfrom C1 ∪. . . ∪Cl.\nSince (B1 ∪. . . ∪Bk) ∩(C1 ∪. . . ∪Cl) = F we obtain\nthat the computation of A ends in an accepting state, hence w ∈L(A) and\nL(A1) ∩L(A2) ⊆L(A).\nSince both partitions are nontrivial, so is the AI-decomposition obtained.\nTheorem 3.3. Let A = (K, Σ, δ, q0, F) be a deterministic finite automaton,\nlet π1 and π2 be nontrivial S.P. partitions on the set of states of A, such that\nπ1 · π2 ⪯{F, K −F}. Then A has a nontrivial wAI-decomposition.\nProof. We shall construct A1 and A2 corresponding to the S.P. partitions π1\nand π2 as follows: Ai = (πi, Σ, δi, [q0]πi, ∅), where δi([q]πi, a) = [δ(q, a)]πi and\ni ∈{1, 2}. To show that (A1, A2) is a wAI-decomposition of A, we define the\nrelation R ⊆π1 × π2 by the equivalence R(D1, D2) ⇔(D1 ∩D2 ⊆F),where\nDi is some block of the partition πi. Now we need to prove that ∀w ∈Σ∗;\nw ∈L(A) ⇔R(δ1([q0]π1, w), δ2([q0]π2, w)).\nLet the computation of A on w end in some state p ∈K. It follows that the\ncomputation of Ai on the word w ends in the state corresponding to the block\n[p]πi, i ∈{1, 2}. Thus R(δ1([q0]π1, w), δ2([q0]π2, w)) ⇔R([p]π1, [p]π2) and by the\ndefinition of R, we have R(δ1([q0]π1, w), δ2([q0]π2, w)) ⇔[p]π1 ∩[p]π2 ⊆F. Since\np ∈[p]π1 ∩[p]π2, [p]π1 ∩[p]π2 is a block of the partition π1 · π2 and π1 · π2 ⪯\n{F, K −F}, it must hold that either [p]π1 ∩[p]π2 ⊆F or [p]π1 ∩[p]π2 ⊆K −F.\nTherefore R(δ1([q0]π1, w), δ2([q0]π2, w)) ⇔p ∈F and the proof is complete.\nIt follows directly from the definitions, that each SI-decomposition is also a\nwAI-decomposition, and so is each AI-decomposition. Also, each ASB-decomposition\nis an AI-decomposition, which is a consequence of the definition of acceptance\n5\n\nand state behavior realization. For minimal automata, a relationship between\nAI- and SI-decompositions can be obtained.\nTheorem 3.4. Let A = (K, Σ, δ, q0, F) be a minimal DFA, let (A1, A2) be its\nAI-decomposition. Then (A1, A2) is also an SI-decomposition of A.\nProof. Since (A1, A2) is an AI-decomposition of A, L(A) = L(A1) ∩L(A2).\nTherefore if we use the well-known Cartesian product construction, we obtain\nthe automaton A1||A2 such that L(A1||A2) = L(A). Since A is the minimal\nautomaton accepting the language L(A), there exists a mapping β : K′ →K\nsuch that it holds (∀w ∈Σ∗);\nβ(δ′(q′\n0, w)) = δ(β(q′\n0), w), where δ′ is the\ntransition function of A1||A2, K′ is its set of states and q′\n0 is its initial state.\nSince A1||A2 is a parallel connection (i.e., K′ = K1 ×K2, q′\n0 is the pair of initial\nstates of A1 and A2), it is easy to see that β is in fact exactly the mapping\nrequired by the definition of the SI-decomposition.\nThe ASB-decomposition is a combination of the SB-decomposition and the\nAI-decomposition, as the next theorem shows.\nTheorem 3.5. Let A be a DFA without unreachable states.\n(A1, A2) is an\nASB-decomposition of A iff(A1, A2) is both an SB-decomposition and an AI-\ndecomposition of A.\nProof. The first implication clearly follows from the definitions, Theorem 3.1\nand Theorem 3.2. Now let (A1, A2) be an SB- and AI-decomposition of A =\n(K, Σ, δ, q0, F). Let α be the mapping given by the definition of SB-decomposition.\nWe need to prove that for all states q of A, q ∈F iffα(q) ∈F1 × F2, where Fi is\nthe set of accepting states of Ai, i ∈{1, 2}. Let q ∈K and let w be a word such\nthat δ(q0, w) = q. Then q ∈F ⇔w ∈L(A) ⇔w ∈L(A1) ∩L(A2) ⇔α(q) ∈\nF1 × F2, where the first equivalence is implied by the choice of w, the second\nholds because (A1, A2) is an AI-decomposition and the third is a consequence\nof the properties of α guaranteed by the SB-decomposition definition.\nThere is also a relationship between SB- and SI-decompositions, in fact SB-\nis a stronger version of the state-identifying decomposition, as the following two\ntheorems show. We need the notion of reachability on pairs of states.\nDefinition 3.1. Let A1 = (K1, Σ, δ1, p1, F1) and A2 = (K2, Σ, δ2, p2, F2) be\nDFAs. We shall call a pair of states (q, r) ∈K1 × K2 reachable, if there exists\na word w ∈Σ∗such that δ1(p1, w) = q and δ2(p2, w) = r.\nTheorem 3.6. Let A = (K, Σ, δ, q0, F) be a DFA and let (A1, A2) be its SB-\ndecomposition. Then (A1, A2) also forms an SI-decomposition of A.\nProof. Let Ai = (Ki, Σ, δi, qi, Fi), i ∈{1, 2}. Since (A1, A2) is an SB-decomposition\nof A, there exists an injective mapping α: K →K1 × K2 such that it holds\nα(q0) = (q1, q2) and (∀a ∈Σ)(∀p ∈K); α(δ(p, a)) = (δ1(p1, a), δ2(p2, a)), where\nα(p) = (p1, p2). Let us define a new mapping β : K1 × K2 →K by\nβ(p1, p2) =\n \np\nif ∃p ∈K, α(p) = (p1, p2)\nq0\notherwise.\n(1)\nSince α is injective, there exists at most one such p and this definition is correct.\n6\n\nWe now need to prove that β satisfies the condition from the definition of\nSI-decomposition, i.e., that (∀w ∈Σ∗); β(δ1(q1, w), δ2(q2, w)) = δ(q0, w). Since\nα(q0) = (q1, q2) and all the pairs of states we encounter in the computation of\nA1||A2 are thus reachable, this follows from the definition of α and (1) by an\neasy induction.\nLemma 3.7. Let A be a DFA without unreachable states and let (A1, A2) be\nits SI-decomposition, with β being the corresponding mapping. Then (A1, A2) is\nan SB-decomposition of A if and only if β is injective on all reachable pairs of\nstates.\nProof. Let (A1, A2) be an SB-decomposition of A. It clearly follows from Def-\ninition 2.2, that the corresponding β satisfies the equation (1) in the proof of\nTheorem 3.6 on all reachable pairs of states. Since the mapping α is a bijection\nbetween the set of states of A and the set of all reachable pairs of states of A1\nand A2, β defined as its inverse on the set of reachable pairs of states will be\ninjective on this set.\nFor the other implication, let (A1, A2) be an SI-decomposition of A and let\nβ be injective on the set of reachable pairs of states, let βr denote the mapping\nβ restricted onto the set of all reachable pairs of states of A1, A2. Since A has\nno unreachable states, βr is also surjective, thus we can define a new mapping\nα: K →K1 × K2 by the equation α(q) = β−1\nr (q). Since β maps the initial state\nonto the initial state, so does α, and since β satisfies the condition from the\nDefinition 2.2, it implies that also α satisfies the condition (i) from the definition\nof realization of state behavior. Therefore (A1, A2) is an SB-decomposition of\nA, with the corresponding mapping α.\nThe converse of Theorem 3.6 does not hold. The minimal automaton for the\nlanguage L = {a4kb4l|k ≥0, l ≥1} gives a counterexample. Inspecting its S.P.\npartitions shows that it has no nontrivial SB-decomposition, but it can be AI-\ndecomposed into minimal automata for languages L1 = {a4kbl|k ≥0, l ≥1} and\nL2 = {w|#b(w) = 4l; l ≥0}. According to Theorem 3.4, this AI-decomposition\nis also state-identifying.\nEach ASB-decomposition is obviously also an SB-decomposition. On the\nother hand, there exist SB-decomposable automata, that are ASB-undecomposable.\nFor example, the minimal automaton for the language\nL1\n=\n{w ∈{a, b, c}∗|#a(w)\nmod 3 = 0 ∧#b(w)\nmod 5 = 0} ∪\n∪\n{w ∈{a, b, c}∗|#a(w)\nmod 3 = 2 ∧#b(w)\nmod 5 = 4}\nhas this property, because the corresponding S.P. partitions on the set of its\nstates do not separate the final states in the sense of Definition 2.7.\nIt is also not so difficult to see that for any non-minimal automaton A without\nunreachable states, there exists a nontrivial AI- and wAI-decomposition (A1, A2)\nsuch that A1 is the minimal automaton equivalent to A and A2 has only one\nstate. This decomposition is obviously not state-identifying.\nFigure 1 summarizes all the relationships among the decomposition types\nthat we have shown so far.\nNow we show that for the case of so-called perfect decompositions, some of\nthe types of decomposition mentioned coincide.\n7\n\nASB\n{vvvvvvvvv\n#G\nG\nG\nG\nG\nG\nG\nG\nAI\n×\nv\nv\nv\nv\n;v\nv\nv\nv\nmin\nH\nH\nH\nH\n#H\nH\nH\nH\n \n×\n#\nSB\n{wwwwwwww\n×GGGG\ncGGGG\nSI\n{vvvvvvvvv\n×\nwwww\n;wwww\nwAI\n×\n;\nDescription:\nA\n/ B : every A-decomposition is also a B-\ndecomposition\nA\n×\n/ B : not every A-decomposition is also\na B-decomposition\nA\n×\n/ B : there exists a DFA that has a non-\ntrivial A-decomposition but does not have\na nontrivial B-decomposition\nFigure 1: Relationships between decomposition types of DFA\nDefinition 3.2. Let t be a type of decomposition, t ∈{ASB, SB, AI, SI, wAI}.\nLet A be a DFA having n states, let A1 and A2 be DFAs having k and l states,\nrespectively. We shall call the pair (A1, A2) a perfect t-decomposition of A, if\nit forms a t-decomposition of A and n = k · l.\nTheorem 3.8. Let A be a DFA with no unreachable states and let (A1, A2) be a\npair of DFAs. Then (A1, A2) forms a perfect SI-decomposition of A iff(A1, A2)\nforms a perfect SB-decomposition of A.\nProof. One of the implications is a consequence of Theorem 3.6.\nAs to the\nsecond one, since (A1, A2) forms a perfect SI-decomposition of A, each of the\npairs of states of A1 and A2 is reachable and each pair has to correspond to a\ndifferent state of A in the mapping β, therefore β is bijective and the theorem\nfollows from Theorem 3.7.\nCorollary 3.9. Let A be a minimal DFA and let (A1, A2) be a pair of DFAs.\nThen (A1, A2) forms a perfect AI-decomposition of A iff(A1, A2) forms a perfect\nASB-decomposition of A.\nProof. The claim follows from Theorem 3.5, Theorem 3.4 and Theorem 3.8.\nAs a consequence of these facts, we can use the necessary and sufficient\nconditions stated in Theorem 3.1 to look for perfect AI- and SI-decompositions.\nNow, let us inspect the relationship between decompositions of an automaton\nand the decompositions of the corresponding minimal automaton.\nTheorem 3.10. Let A = (K, Σ, δ, q0, F) be a DFA and let Amin be a min-\nimal DFA such that L(A) = L(Amin).\nLet (A1, A2) be an SI-decomposition\n(AI-decomposition, wAI-decomposition) of A, then (A1, A2) also forms a de-\ncomposition of Amin of the same type.\nProof. First, note that this theorem does not state that any of the decomposi-\ntions is nontrivial. To prove the statement for SI-decompositions, suppose that\n(A1, A2) is an SI-decomposition of A, thus there exists a mapping α: K1×K2 →\nK such that it holds (∀w ∈Σ∗); α(δ1(q1, w), δ2(q2, w)) = δ(q0, w), where δi and\nqi are the transition function and the initial state of the automaton Ai. Since\nAmin is the minimal automaton corresponding to A, there exists some mapping\nβ : K →Kmin such that (∀w ∈Σ∗); β(δ(q0, w)) = δmin(β(q0), w), where δmin is\n8\n\na1\nb\n/\na\n \nR\na0\nb\n/\na\nH\nb1\nb\n)\na\nO\nb0\nb\ni\na\nm\na1\nb\n/\na\n \nR1\nb\n* R0\nb\nj\na0\nb\n/\na\nH\nb1\nb\n)\na\nO\nb0\nb\ni\na\nO\nFigure 2: Transition functions of Amin and A′.\nthe transition function of Amin and Kmin is the set of states of Amin. By the com-\nposition of these mappings we obtain the mapping β◦α: K1×K2 →Kmin, which\ncombines A1 and A2 into Amin in the way that the definition of SI-decomposition\nrequires. For both the AI- and the wAI-decomposition, this statement is trivial,\nsince L(A) = L(Amin).\nBased on the above theorem it thus suffices to inspect the SI- (AI-, wAI-)\ndecomposability of the minimal automaton accepting a given language, and if\nwe show its undecomposability, we know that the recognition of this language\ncannot be simplified using an advisor of the respective type.\nHowever, this\ndoes not hold for SB- and ASB-decompositions, as exhibited by the following\nexample.\nExample 3.1. Let us consider the language L = {a2kb2l|k ≥0, l ≥1}. The\nminimal automaton Amin = (K, ΣL, δ, a0, {a0, b0}) has its transition function\ndefined by the first transition diagram in Fig.2. We can easily show that this\nautomaton does not have any nontrivial SB- (and thus neither ASB-) decompo-\nsition by enumerating its S.P. partitions.\nNow let us examine the automaton A′ = (K′, ΣL, δ′, a0, {a0, b0}) with the\ntransition function δ′ defined by the second transition diagram in Fig.2. Clearly,\nL(A′) = L(Amin), but by inspecting the lattice of S.P. partitions of A′, we can\nfind the pair π1 = {{a0}, {a1}, {b0, b1}, {R0, R1}} and π2 = {{a0, a1, b0, R0}, {b1, R1}}\nsuch that π1 · π2 = 0 and they separate the final states of A′. By Theorem 3.1\nwe can use these partitions to construct a nontrivial ASB- (and thus also SB-)\ndecomposition of A′ formed by the automata A1 and A2 having two and four\nstates, respectively. Note that both A1 and A2 have less states than Amin.\nIn the following theorem (inspired by a similar theorem in [5]) we state a\ncondition, under which the situation from the last example cannot occur, i.e.,\nunder which any SB-decomposition of a DFA implies a (maybe simpler) SB-\ndecomposition of the equivalent minimal DFA.\nTheorem 3.11. Let A = (K, Σ, δ, q0, F) be a deterministic finite automa-\nton and let Amin = (Kmin, Σ, δmin, qmin, Fmin) be the minimal DFA such that\nL(A) = L(Amin). Let (A1, A2) be a nontrivial SB-decomposition of A consisting\nof automata having k and l states. If the lattice of S.P. partitions of A is dis-\ntributive, then there exists an SB-decomposition of Amin consisting of automata\nhaving k′ and l′ states, such that k′ ≤k and l′ ≤l.\nProof. Since Amin is the minimal DFA such that L(A) = L(Amin), there exists\na mapping f : K →Kmin such that (∀w ∈Σ∗); f(δ(q0, w)) = δmin(qmin, w).\nUsing the mapping f, let us define a partition ρ on the set of states of A by\np ≡ρ q ⇔f(p) = f(q). Clearly, ρ is an S.P. partition.\n9\n\nSince (A1, A2) is a nontrivial SB-decomposition of A, we can use it to obtain\nS.P. partitions π1 and π2 on the set of states of A such that π1 · π2 = 0. Let us\ndefine new partitions π′\n1 and π′\n2 on the set of states of Amin by f(p) ≡π′\ni f(q) ⇔\np ≡ρ+πi q. Since it holds that ρ + πi ⪯ρ, this definition does not depend on\nthe choice of the states p and q. It holds that |π′\ni| = |ρ + πi| ≤|πi|, therefore if\nwe prove that π′\n1 and π′\n2 are S.P. partitions and π′\n1 · π′\n2 = 0, we can use them to\nconstruct the desired decomposition.\nThe fact that π′\ni is an S.P. partition on the set of states of Amin is a trivial\nconsequence of the fact that ρ + πi is an S.P. partition on the set of states of\nA. We need to prove that π′\n1 · π′\n2 = 0. Let us assume that p′ and q′ are states\nof Amin such that p′ ≡π′\n1·π′\n2 q′ and p, q are some states of A such that f(p) = p′\nand f(q) = q′. Then p′ ≡π′\n1 q′ and p′ ≡π′\n2 q′, and by definition of π′\ni we get\np ≡ρ+π1 q and p ≡ρ+π2 q, which is equivalent to p ≡(ρ+π1)·(ρ+π2) q. Since the\nlattice of all S.P. partitions of A is distributive, we have (ρ + π1) · (ρ + π2) =\nρ + (π1 · π2) = ρ + 0 = ρ, therefore p ≡ρ q, which by definition of ρ implies that\nf(p) = f(q), in other words p′ = q′. Hence π′\n1 · π′\n2 = 0.\n4\nDegrees of Decomposability\nIt is easy to see that for each type of decomposition, there exist undecomposable\nregular languages (e.g. L(n) = {ak|k ≥n −1} is wAI-undecomposable for each\nn ∈N). There also exist regular languages, that are perfectly decomposable in\neach way (e.g. L(k,l) = {w ∈{a, b}∗|#a(w) mod k = 0 ∧#b(w) mod l = 0} has\na perfect ASB-decomposition for all k, l ≥2). We shall now investigate whether\nall values between these two limits can be achieved.\nDefinition 4.1. Let A be a DFA, let (A1, A2) be its nontrivial SB- (ASB-)\ndecomposition with the corresponding S.P. partitions π1 and π2. We shall call\nthis decomposition redundant, if there exist S.P. partitions π′\n1 ⪰π1 and π′\n2 ⪰π2\nsuch that at least one of these inequalities is strict, but it still holds π′\n1 · π′\n2 = 0\n(and π′\n1 and π′\n2 separate the final states of A).\nLemma 4.1. For each r, s ∈N, r, s ≥2, there exists a minimal DFA A consist-\ning of r.s states and having only one nontrivial nonredundant SB-decomposition\n(ASB-decomposition) up to the order of automata, consisting of automata hav-\ning r and s states.\nProof. Let us study the minimal automaton Ar,s = (K, Σ, δ, q0,0, F) defined\nby K = {qi,j|i ∈{0, . . . , r −1}, j ∈{0, . . ., s −1}}, F = {qr−1,s−1} and the\ntransition function δ defined by\nδ(qi,j, a)\n=\nqi+1,j for i ∈{0, . . ., r −2}, j ∈{0, . . ., s −1}\nδ(qr−1,j, a)\n=\nqr−1,j for j ∈{0, . . . , s −1}\nδ(qi,j, b)\n=\nqi,j+1 for i ∈{0, . . ., r −1}, j ∈{0, . . ., s −2}\nδ(qi,s−1, b)\n=\nqi,s−1 for i ∈{0, . . . , r −1}.\nTo inspect the SB-decompositions of Ar,s, let us study the S.P. partitions on\nthe set of its states. From the method for generating all S.P. partitions of an\nautomaton that is described in [5], we know that each nontrivial S.P. partition\ncan be obtained as a sum of some partitions πm\np,t, where πm\np,t denotes the minimal\n10\n\nS.P. partition such that it does not distinguish between states p and t, i.e., they\nbelong into the same block. Let us determine πm\np,t for various states p and t of\nAr,s.\nFirst, let us consider the case of πm\np,t such that p = qi,j, t = qi′,j′ and both\ninequalities i < i′ and j < j′ hold. Since qi,j ≡π qi′,j′, δ(qi,j, ai′−ibj′−j) = qi′,j′\nand δ(qi′,j′, ai′−ibj′−j) = q2i′−i,2j′−j (if 2i′ −i < r and 2j′ −j < s), as a\nconsequence of the substitution property of π, we obtain qi,j ≡π q2i′−i,2j′−j.\nBy applying this argument a finite number of times (keeping in mind the con-\nstruction of Ar,s), we obtain qi,j ≡π qr−1,s−1. Now let k ∈{i, . . ., r −1} and\nlet l ∈{i, . . . , s −1}.\nThen δ(qi,j, ak−ibl−j) = qk,l and δ(qi′,j′, ak−ibl−j) =\nqk+i′−i,l+j′−j (if such states exist), therefore qk,l ≡π qk+i′−i,l+j′−j. Again, we\ncan use the same argument to show that qk,l ≡π qr−1,s−1. Therefore, for this\ntype of π = πm\np,t, we have qk,l ≡π qk′,l′ for all k, l, k′, l′ such that i ≤k, k′ < r\nand j ≤l, l′ < s.\nNow let us consider the case of πm\np,t such that p = qi,j, t = qi′,j′ and it holds\ni > i′ and j < j′. Since qi,j ≡π qi′,j′, δ(qi,j, ar−1−ibs−1−j′) = qr−1,s−1−(j′−j)\nand δ(qi′,j′, ar−1−ibs−1−j′) = qr−1−(i−i′),s−1, as a consequence of the substitu-\ntion property of π, we have qr−1,s−1−(j′−j) ≡π qr−1−(i−i′),s−1. By exploiting\nthe substitution property again on this equivalence, using the words ai−i′−1,\nbj′−j−1 and bj′−j, we obtain qr−2,s−1 ≡π qr−1,s−1 ≡π qr−2,s−2. Therefore in\nthis case, no such πm\np,t partition can distinguish between states qr−2,s−1, qr−1,s−1\nand qr−2,s−2.\nThe last case to consider is the case of πm\np,t such that p = qi,j, t = qi′,j′ and\nit holds i = i′ (the case j = j′ is analogous). Without loss of generality, we\ncan assume that j < j′. Now, using the same arguments as in the first case, we\ncan show that qi,l ≡π qi,l′ for all l, l′ such that j ≤l, l′ < s. Therefore for each\ngiven k such that i ≤k < r, it holds that qk,l ≡π qk,l′ and all of the states not\nmentioned in this equivalence form separate blocks of πm\np,t.\nIt is easy to verify that one nontrivial ASB-decomposition of Ar,s is given\nby the S.P. partitions\nπ1\n=\n{{q0,0, . . . , q0,s−1}, {q1,0, . . . , q1,s−1}, . . . , {qr−1,0, . . . , qr−1,s−1}} and\nπ2\n=\n{{q0,0, . . . , qr−1,0}, {q0,1, . . . , qr−1,1}, . . . , {q0,s−1, . . . , qr−1,s−1}}\nNow we show that any other SB-decomposition of Ar,s is given by S.P. partitions\npreceding to π1 and π2 in the partial order ⪯and therefore is redundant.\nIndeed, notice that none of the πm\np,t partitions of the first and the second\ndiscussed type can distinguish between any of the states qr−2,s−1, qr−1,s−1 and\nqr−2,s−2, therefore no sum of them can, either. For the partitions of the third\ntype, it holds either qr−2,s−1 ≡π qr−1,s−1 or qr−1,s−1 ≡π qr−2,s−2, therefore it\nwill take two partitions to distinguish between these three states. Hence any\nnontrivial SB-decomposition is determined by two S.P. partitions, both of which\nmust be of the third type. But it is easy to see that for any partition π of this\ntype it holds either π ⪯π1 or π ⪯π2.\nDefinition 4.2. Let A = (K, Σ, δ, q0, F) be a deterministic finite automaton,\nlet K ∩{p0, p1, . . . , pk−1} = ∅and let c be a new symbol not included in Σ. We\nshall define a k-extension A′ of the automaton A by the following construction:\nA′ = (K ∪{p0, p1, . . . , pk−1}, Σ ∪{c}, δ′, p0, F), where the transition function δ′\n11\n\nis defined as follows:\n(∀q ∈K) (∀a ∈Σ);\nδ′(q, a)\n=\nδ(q, a)\n(∀q ∈K);\nδ′(q, c)\n=\nq\n(∀p ∈{p0, p1, . . . , pk−1}) (∀a ∈Σ);\nδ′(p, a)\n=\np\n(∀i ∈{0, 1, . . ., k −2});\nδ′(pi, c)\n=\npi+1\nδ′(pk−1, c)\n=\nq0.\nLemma 4.2. Let A be a DFA consisting of n states, all of which are reach-\nable. Let A′ be its k-extension. Then A has a nontrivial nonredundant SB-\ndecomposition (ASB-decomposition) consisting of automata having r and s states\niffA′ has a nontrivial nonredundant decomposition of the same type, consisting\nof automata having k + r and k + s states.\nProof. We will try to inspect S.P. partitions on the set of states of A′, using\nthe notation from Definition 4.2. Let us assume that π′ is an S.P. partition\non the set of states of A′ such that pi and pj are in the same block of π′;\ni, j ∈{0, 1, . . ., k −1}. As a consequence of the S.P. property, if i, j < k −1\nthen also pi+1 and pj+1 are in the same block of π′, because δ′(pi, c) = pi+1\nand δ′(pj, c) = pj+1. By applying this argument a finite number of times, we\ncan show that there exists some l ∈{0, 1, . . ., k −2} such that pl ≡π′ pk−1,\nand using the argument once more, we obtain pl+1 ≡π′ q0. However, it holds\nδ′(pl, a) = pl for all a ∈Σ, hence pl ≡π′ δ′(q0, w) for all w ∈Σ∗. Since all of the\nstates of A′ are reachable, we have pl ≡π′ q for all q ∈K. Thus such a partition\ncannot distinguish between the original states of the automaton A.\nNow let us suppose that π′ is an S.P. partition on the set of states of A′ such\nthat for some i ∈{0, 1, . . ., k −1}, pi ≡π′ q for some q in K. Then it also holds\nthat pi ≡π′ pi+1, because δ(pi, c) = pi+1 and δ(q, c) = q. But we have already\nshown that pi ≡π′ pi+1 implies that all of the states in K are equivalent modulo\nπ′, thus this S.P. partition cannot distinguish between the states of A, either.\n¿From these observations it follows that if π′ is any S.P. partition on the set\nof states of A′ such that the states of A are not all equivalent modulo π′, then\nπ′ must also contain k blocks, each of which contains only one state pi, where\ni ∈{0, 1, . . ., k −1}. Now we can prove the equivalence stated in the theorem.\nLet A have an SB-decomposition consisting of r and s states. Then there\nexist S.P. partitions π1 and π2 on the set of states of A having r and s blocks,\nsuch that π1 · π2 = 0.\nLet us now construct new partitions π′\n1 and π′\n2 on\nthe set of states of A′ by π′\n1 = π1 ∪{{p0}, {p1}, . . . , {pk−1}} and π′\n2 = π2 ∪\n{{p0}, {p1}, . . . , {pk−1}}. Obviously, π′\n1 and π′\n2 have substitution property, be-\ncause for the states in K this property is inherited from π1 and π2, and the new\nstates p0, p1, . . . , pk−1 cannot violate this property either, because each of these\nstates belongs to a separate block in π′\n1 and π′\n2, making the substitution prop-\nerty hold trivially. Neither do the new c-moves defined on the states from K\nviolate the substitution property. Finally, it holds that π′\n1 · π′\n2 = 0. To see this,\nnote that for a state q ∈K, it holds [q]π′\n1·π′\n2 = [q]π1·π2 = {q}, since π1 · π2 = 0.\nFor a state q ∈K′ −K, [q]π′\ni = {q} for i ∈{1, 2} thus [q]π′\n1·π′\n2 = {q}, too. Hence\neach state of A′ belongs to a separate block of π′\n1 · π′\n2, which implies π′\n1 · π′\n2 = 0.\nTherefore π′\n1 and π′\n2 induce an SB-decomposition of A′. It is also easy to see\nthat if π1 and π2 separate the final states of A, then also π′\n1 and π′\n2 separate the\nfinal states of A′, making the induced decomposition an ASB-decomposition.\n12\n\nOn the other hand, let us now assume that A′ has an SB-decomposition and\nπ′\n1 and π′\n2 are the S.P. partitions on K′ that induce this decomposition, thus\nπ′\n1 · π′\n2 = 0. From the observations made in the beginning of this proof, we know\nthat any S.P. partition that can distinguish between the states in K in any way,\nmust contain each of the states p0, p1 . . . pk−1 in a separate block containing only\nthis state. As π′\n1 ·π′\n2 = 0, for all q1, q2 ∈K, at least one of these partitions must\ndistinguish between these states, i.e., [q1]π′\ni ̸= [q2]π′\ni. If one of the partitions\ndistinguished between all such pairs, it would imply that this partition must\ncontain a separate block for each one of the states in K′, thus becoming a trivial\npartition 0, resulting in a trivial decomposition. Therefore both π′\n1 and π′\n2 have\nto distinguish between some pair of states from K, which implies that they both\ncontain a separate block for each of the states p0, p1 . . . pk−1 containing no other\nstate. By removing these k blocks from π′\n1 and π′\n2, we obtain new partitions π1\nand π2 on the set K, such that π1 = π′\n1−{{p0}, {p1}, . . . , {pk−1}} and π2 = π′\n2−\n{{p0}, {p1}, . . . , {pk−1}}. These partitions preserve the substitution property,\nsince (∀a ∈Σ)(∀q ∈K): δ(q, a) ∈K and π′\n1 and π′\n2 were S.P. partitions. It also\nholds π1 ·π2 = 0, as for all q1, q2 ∈K, q1 ≡π1·π2 q2 implies q1 ≡π′\n1·π′\n2 q2 and that\nimplies q1 = q2. So π1 and π2 induce an SB-decomposition of A. As π′\n1 and π′\n2\nwere nontrivial, so are π1 and π2 and the obtained decomposition. It is again\neasy to see that if π′\n1 and π′\n2 separate the final states of A′, then also π1 and π2\nmust separate the final states of A.\nThe described relationship between the S.P. partitions on the set of states\nof A and the corresponding S.P. partitions on A′ also implies, that each de-\ncomposition of A is nonredundant iffthe corresponding decomposition of A′ is\nnonredundant, too.\nSince a k-extension of a minimal DFA is again a minimal DFA, we can\ncombine the lemmas to obtain the following theorem.\nTheorem 4.3. Let n ∈N be such that n = k + r.s, where r, s, k ∈N, r, s ≥2.\nThen there exists a minimal DFA A consisting of n states, such that it has only\none nontrivial nonredundant SB-decomposition (ASB-decomposition) up to the\norder of the automata in the decomposition, and this decomposition consists of\nautomata with k + r and k + s states.\nReferences\n[1] J. L. Balcazar, J. Diaz, J. Gabarro, Structural Complexity I., Springer-Verlag\nNew York, 1988\n[2] S. Even, A. L. Selman, Y. Yacobi, The Complexity of Promise Problems\nwith Applications to Public-Key Cryptography, Information and Control 61(2):\n159-173 (1984)\n[3] S. Yu, State Complexity: Recent Results and Open Problems, Fundamenta\nInformaticae 64: 471-480 (2005)\n[4] J. C. Birget, Intersection and Union of Regular Languages and State Com-\nplexity, Information Processing Letters 43: 185-190 (1992)\n[5] J. Hartmanis and R. E. Stearns, Algebraic Structure Theory of Sequential\nMachines, Prentice-Hall, 1966\n13\n\n[6] J. E. Hopcroft and J. D. Ullman, Introduction to Automata Theory, Lan-\nguages, and Computation, Addison-Wesley, 1979\n14","paragraphs":[{"paragraph_id":"p1","order":1,"text":"arXiv:0707.0430v1 [cs.CC] 3 Jul 2007\nAssisted Problem Solving and Decompositions of\nFinite Automata∗\nPeter Gaˇzi\nBranislav Rovan\nDepartment of Computer Science, Comenius University\nMlynsk ́a dolina, 842 48, Bratislava, Slovakia\n{gazi,rovan}@dcs.fmph.uniba.sk\nAbstract\nA study of assisted problem solving formalized via decompositions of\ndeterministic finite automata is initiated. The landscape of new types of\ndecompositions of finite automata this study uncovered is presented. Lan-\nguages with various degrees of decomposability between undecomposable\nand perfectly decomposable are shown to exist.\n1\nIntroduction\nIn the present paper we initiate the study of assisted problem solving. We intend\nto model and study situations, where solution to the problem can be sought\nbased on some additional a priori information about the inputs. One can expect\nto obtain simpler solution in such case. There are similar approaches known\nin the literature, most notably the notions of advice functions [1], where the\nadditional information is based on the length of the input word and the notion\nof promise problems [2], where the set of inputs is separated into three classes\n– those with “yes” answer, those with “no” answer and those where we do not\ncare about the outcome. By considering the simplest case where the “problem\nsolving” machinery is the deterministic finite automaton (DFA) we obtain a new\nmotivation for studying new types of finite automata decompositions.\nIn this paper we shall thus consider the case where solving a problem shall\nmean constructing an automaton for a given language L. The “assistance” shall\nbe given by additional information about the input, e.g., that we can assume the\ninputs shall be restricted to words from a particular regular language L′. Thus,\ninstead of looking for an automaton A such that L = L(A) we can look for a\n(possibly simpler) automaton B such that L = L(B)∩L′. We can then say that\nB accepts L with the assistance of L′. We shall call L′ (or the corresponding\nautomaton A′ such that L′ = L(A′)) an advisor to B. In this case the advisor\nA′ provides assistance to the solver B by guaranteeing that A′ accepts the given\ninput word.\nWe shall also study a case where the assistance provides more\ndetailed information about the outcome of the computation of A′ on the input\nword (e.g., the state reached). Clearly the advisor can be considered useful\nonly if it enables B to be simpler than A and at the same time A′ is not more\n∗This work was supported in part by the grant VEGA 1/3106/06.\n1"},{"paragraph_id":"p2","order":2,"text":"complicated than A. The measure of complexity we shall consider is the number\nof states of the deterministic finite automaton. This measure of complexity was\nused quite often recently due to renewed interest in finite automata prompted\nby applications such as model checking (see e.g. [3] for a recent survey). (Note\nthat results complementary to ours, namely results on complexity of automata\nfor the intersection of regular sets were studied in [4].)\nThe contribution of our paper is twofold. First, we can interpret the ‘solver’\nand the ‘advisor’ as two parallel processes each performing a different task and\njointly solving a problem. Since our approach lends itself to a generalisation to\nk advisors it may stimulate new parallel solutions to problems (the traditional\nones usually using parallel processes to perform essentially the same task). Sec-\nond, the choice of finite automata as the simplest problem solving machinery\nbrought about new types of decompositions motivated by the information the\n‘advisor’ can provide to the ‘solver’. Our results provide a complete picture of\nthe landscape of these decompositions.\nThe problem within this scenario we shall address in this paper is the exis-\ntence of a useful advisor for a given automaton A. We shall compare the power\nof several types of advisors, and investigate the effect of the advisor on the\ncomplexity of the assisted solver B. We can formulate this also as a problem of\ndecomposition of deterministic finite state automata – given DFA A find DFA\nA1 (a solver) and A2 (an advisor) such that w ∈L(A) can be determined from\nthe computations of A1 and A2. We shall study several new types of decompo-\nsitions of DFA, one of them is analogous to the state behavior decomposition of\nfinite state transducers studied in [5]. In Sect. 3 we prove relations among these\ndecompositions. For each type of decomposition there are automata which are\nundecomposable and automata for which there is a decomposition that is the\nbest possible. In Sect. 4 we consider the space between these extreme points\nand study the degree of decomposability.\n2\nDefinitions and Notation\nWe shall use standard notions of the theory of formal languages (see e.g. [6]).\nOur notation shall be as follows.\nΣ∗denotes the set of all words over the\nalphabet Σ, the length of a word w is denoted by |w|, ε denotes the empty\nword, and for a language L we shall denote by ΣL the minimal alphabet such\nthat L ⊆Σ∗\nL. The number of occurrences of a given letter a in a word w is\ndenoted by #a(w). Throughout this paper we shall consider deterministic finite\nautomata only.\nA deterministic finite automaton (DFA) is a quintuple (K, Σ, δ, q0, F), such\nthat K is a finite set of states, Σ is a finite input alphabet, q0 ∈K is the initial\nstate, F ⊆K is the set of accepting states and δ: K × Σ →K is a transition\nfunction. As usual, we shall denote by δ also the standard extension of δ to\nwords, i.e., δ: K × Σ∗→K. We shall denote by |K| the number of states in K.\nFormalizing the notions of assisted problem solving from the Introduction\nwe shall now define several types of decompositions of DFA A into two (sim-\npler) DFAs A1 and A2 (a solver and an advisor) so that the membership of\nan input word w in L(A) can be determined based on the information on the\ncomputations of A1 and A2 on w.\nWe first introduce an acceptance-identifying decomposition of deterministic\n2"},{"paragraph_id":"p3","order":3,"text":"finite automata.\nDefinition 2.1. A pair of DFAs (A1, A2), where A1 = (K1, Σ, δ1, q1, F1) and\nA2 = (K2, Σ, δ2, q2, F2), forms an acceptance-identifying decomposition (AI-\ndecomposition) of a DFA A = (K, Σ, δ, q0, F), if L(A) = L(A1) ∩L(A2). This\ndecomposition is nontrivial if |K1| < |K| and |K2| < |K|.\nBy decomposing A in this manner, one of the decomposed automata (say\nA2) can act as an advisor and narrow down the set of input words for the other\none (say A1), whose task to recognize the words of L(A) may become easier.\nAnother requirement we could pose on a decomposition is to identify the\nfinal state of any computation of the original automaton by only knowing the\nfinal states of both corresponding computations of the automata forming the\ndecomposition. This requirement can be formalized as follows.\nDefinition 2.2. A pair of DFAs (A1, A2), where A1 = (K1, Σ, δ1, q1, F1) and\nA2 = (K2, Σ, δ2, q2, F2), forms a state-identifying decomposition (SI-decomposition)\nof a DFA A = (K, Σ, δ, q0, F), if there exists a mapping β : K1 × K2 →K, such\nthat it holds β(δ1(q1, w), δ2(q2, w)) = δ(q0, w) for all w ∈Σ∗. This decomposi-\ntion is nontrivial if |K1| < |K| and |K2| < |K|.\nThe third – and the weakest – requirement we pose on a decomposition of a\nDFA is to require that there must exist a way to determine whether the original\nautomaton would accept some given input word based on knowing the states in\nwhich the computations of both decomposition automata have finished.\nDefinition 2.3. A pair of DFAs (A1, A2), where A1 = (K1, Σ, δ1, q1, F1) and\nA2 = (K2, Σ, δ2, q2, F2), forms a weak acceptance-identifying decomposition\n(wAI-decomposition) of a DFA A = (K, Σ, δ, q0, F), if there exists a relation\nR ⊆K1 × K2 such that it holds R(δ1(q1, w), δ2(q2, w)) ⇔w ∈L(A) for all\nw ∈Σ∗. This decomposition is nontrivial if |K1| < |K| and |K2| < |K|.\nNote that in the last two definitions, the sets of accepting states of A1 and\nA2 are irrelevant.\nBy a decomposability of a regular language L in some way, we shall mean\nthe decomposability of the corresponding minimal automaton over ΣL.\nTo be able to compare these new types of decomposition to the parallel\ndecompositions of state behavior introduced for sequential machines in [5], we\nshall redefine them for DFAs.\nDefinition 2.4. A DFA A′ = (K′, Σ, δ′, q′\n0, F ′) is said to realize the state\nbehavior of a DFA A = (K, Σ, δ, q0, F) if there exists an injective mapping\nα: K →K′ such that\n(i) (∀a ∈Σ)(∀q ∈K); δ′(α(q), a) = α(δ(q, a)),\n(ii) α(q0) = q′\n0.\nMoreover, A′ is said to realize the state and acceptance behavior of A, if in\naddition the following property holds:\n(iii) (∀q ∈K); α(q) ∈F ′ ⇔q ∈F.\n3"},{"paragraph_id":"p4","order":4,"text":"Definition 2.5. The parallel connection of two DFA A1 = (K1, Σ, δ1, q1, F1)\nand A2 = (K2, Σ, δ2, q2, F2) is the DFA A = A1||A2 = (K1 × K2, Σ, δ, (q1, q2),\nF1 × F2) such that δ((p1, p2), a) = (δ1(p1, a), δ2(p2, a)).\nDefinition 2.6. A pair of DFAs (A1, A2) is a state behavior (SB-) decomposi-\ntion of a DFA A if A1||A2 realizes the state behavior of A. The pair (A1, A2) is\nan acceptance and state behavior (ASB-) decomposition of A if A1||A2 realizes\nthe state and acceptance behavior of A. This decomposition is nontrivial if both\nA1 and A2 have fewer states than A.\nWe have modified the definitions to fit the formalism and purpose of deter-\nministic finite automata (i.e., to accept formal languages) without loosing the\nconnection to the strongly related and useful concept of S.P.partitions, exhibited\nbelow.\nWe shall use the following notation and properties of S.P. partitions from [5].\nA partition π on a set of states of a DFA A = (K, Σ, δ, q0, F) has substitution\nproperty (S.P.), if it holds ∀p, q ∈K;\np ≡π q ⇒(∀a ∈Σ; δ(p, a) ≡π δ(q, a)). If\nπ1 and π2 are partitions on a given set M, then\n(i) π1 · π2 is a partition on M such that a ≡π1·π2 b ⇔a ≡π1 b ∧a ≡π2 b,\n(ii) π1 +π2 is a partition on M such that a ≡π1+π2 b iffthere exists a sequence\na = a0, a1, a2, . . . , an = b, such that ai ≡π1 ai+1 ∨ai ≡π2 ai+1 for all\ni ∈{0, . . ., n −1},\n(iii) π1 ⪯π2 if it holds (∀x, y ∈M);\nx ≡π1 y ⇒x ≡π2 y.\nThe set of all partitions on a given set (with the partial order ⪯, join realized\nby + and meet realized by .) forms a lattice. The set of all S.P. partitions on\nthe set of states of a given DFA forms a sublattice of the lattice of all partitions\non this set. The trivial partitions {{q0}, {q1}, . . . , {qn}} and {{q0, q1, . . . , qn}}\nshall be denoted by symbols 0 and 1, respectively. The block of a partition π\ncontaining the state q shall be denoted by [q]π. In addition, we shall use the\nfollowing separation notion.\nDefinition 2.7. The partitions π1 = {R1, . . . , Rk} and π2 = {S1, . . . , Sl} on a\nset of states of a DFA A = (K, Σ, δ, q0, F) are said to separate the final states\nof A if there exist indices i1, . . . , ir and j1, . . . , js such that it holds (Ri1 ∪. . . ∪\nRir) ∩(Sj1 ∪. . . ∪Sjs) = F.\n3\nRelations Between Types of Decompositions\nThe concept of partitions separating the final states allows us to derive a neces-\nsary and sufficient condition for the existence of SB- and ASB-decompositions\nsimilar to the one stated in [5].\nTheorem 3.1. A DFA A = (K, Σ, δ, q0, F) has a nontrivial SB-decomposition\niffthere exist two nontrivial S.P. partitions π1 and π2 on the set of states of A\nsuch that π1 · π2 = 0. This decomposition is an ASB-decomposition if and only\nif these partitions separate the final states of A.\nProof. The proof is analogous to that in [5] but had to be extended for the\nASB-decomposition. We omit it due to space constraints.\n4"},{"paragraph_id":"p5","order":5,"text":"For the other decompositions, we can derive the following sufficient condi-\ntions that exploit the concept of S.P. partitions.\nTheorem 3.2. Let A = (K, Σ, δ, q0, F) be a deterministic finite automaton, let\nπ1 and π2 be nontrivial S.P. partitions on the set of states of A, such that they\nseparate the final states of A. Then A has a nontrivial AI-decomposition.\nProof. Since π1 and π2 separate the final states of A, there exist blocks B1, . . . , Bk\nand C1, . . . , Cl of the partitions π1 and π2 respectively, such that (B1 ∪. . . ∪\nBk) ∩(C1 ∪. . . ∪Cl) = F. We shall construct two automata A1 and A2 having\nstates corresponding to blocks of these partitions and show that (A1, A2) is a\nnontrivial AI-decomposition of A. Let A1 = (π1, Σ, δ1, [q0]π1, {B1, . . . , Bk}) and\nA2 = (π2, Σ, δ2, [q0]π2, {C1, . . . , Cl}) be DFAs with δi defined by δi([q]πi, a) =\n[δ(q, a)]πi, i ∈{1, 2} (this definition does not depend on the choice of q since πi\nis an S.P. partition). We now need to prove that L(A) = L(A1) ∩L(A2).\nLet w ∈L(A). Suppose that the computation of A on the word w ends\nin some accepting state qf ∈F. Then, from the construction of A1 and A2 it\nfollows that the computation of Ai on the word w ends in the state corresponding\nto the block [qf]πi of the partition πi. Since qf ∈F, it must hold [qf]π1 ∈\n{B1, . . . , Bk} and [qf]π2 ∈{C1, . . . , Cl}, hence from the construction of Ai,\nthese blocks correspond to the accepting states in the respective automata.\nThus w ∈L(Ai) for i ∈{1, 2}, therefore L(A) ⊆L(A1) ∩L(A2).\nNow suppose w ∈L(A1) ∩L(A2), Thus the computation of A1 on w ends\nin one of the states B1, . . . , Bk, which means that the computation of A on w\nwould end in a state from the union of blocks B1 ∪. . . ∪Bk. Using the same\nargument for A2, we get that the computation of A on w would end in a state\nfrom C1 ∪. . . ∪Cl.\nSince (B1 ∪. . . ∪Bk) ∩(C1 ∪. . . ∪Cl) = F we obtain\nthat the computation of A ends in an accepting state, hence w ∈L(A) and\nL(A1) ∩L(A2) ⊆L(A).\nSince both partitions are nontrivial, so is the AI-decomposition obtained.\nTheorem 3.3. Let A = (K, Σ, δ, q0, F) be a deterministic finite automaton,\nlet π1 and π2 be nontrivial S.P. partitions on the set of states of A, such that\nπ1 · π2 ⪯{F, K −F}. Then A has a nontrivial wAI-decomposition.\nProof. We shall construct A1 and A2 corresponding to the S.P. partitions π1\nand π2 as follows: Ai = (πi, Σ, δi, [q0]πi, ∅), where δi([q]πi, a) = [δ(q, a)]πi and\ni ∈{1, 2}. To show that (A1, A2) is a wAI-decomposition of A, we define the\nrelation R ⊆π1 × π2 by the equivalence R(D1, D2) ⇔(D1 ∩D2 ⊆F),where\nDi is some block of the partition πi. Now we need to prove that ∀w ∈Σ∗;\nw ∈L(A) ⇔R(δ1([q0]π1, w), δ2([q0]π2, w)).\nLet the computation of A on w end in some state p ∈K. It follows that the\ncomputation of Ai on the word w ends in the state corresponding to the block\n[p]πi, i ∈{1, 2}. Thus R(δ1([q0]π1, w), δ2([q0]π2, w)) ⇔R([p]π1, [p]π2) and by the\ndefinition of R, we have R(δ1([q0]π1, w), δ2([q0]π2, w)) ⇔[p]π1 ∩[p]π2 ⊆F. Since\np ∈[p]π1 ∩[p]π2, [p]π1 ∩[p]π2 is a block of the partition π1 · π2 and π1 · π2 ⪯\n{F, K −F}, it must hold that either [p]π1 ∩[p]π2 ⊆F or [p]π1 ∩[p]π2 ⊆K −F.\nTherefore R(δ1([q0]π1, w), δ2([q0]π2, w)) ⇔p ∈F and the proof is complete.\nIt follows directly from the definitions, that each SI-decomposition is also a\nwAI-decomposition, and so is each AI-decomposition. Also, each ASB-decomposition\nis an AI-decomposition, which is a consequence of the definition of acceptance\n5"},{"paragraph_id":"p6","order":6,"text":"and state behavior realization. For minimal automata, a relationship between\nAI- and SI-decompositions can be obtained.\nTheorem 3.4. Let A = (K, Σ, δ, q0, F) be a minimal DFA, let (A1, A2) be its\nAI-decomposition. Then (A1, A2) is also an SI-decomposition of A.\nProof. Since (A1, A2) is an AI-decomposition of A, L(A) = L(A1) ∩L(A2).\nTherefore if we use the well-known Cartesian product construction, we obtain\nthe automaton A1||A2 such that L(A1||A2) = L(A). Since A is the minimal\nautomaton accepting the language L(A), there exists a mapping β : K′ →K\nsuch that it holds (∀w ∈Σ∗);\nβ(δ′(q′\n0, w)) = δ(β(q′\n0), w), where δ′ is the\ntransition function of A1||A2, K′ is its set of states and q′\n0 is its initial state.\nSince A1||A2 is a parallel connection (i.e., K′ = K1 ×K2, q′\n0 is the pair of initial\nstates of A1 and A2), it is easy to see that β is in fact exactly the mapping\nrequired by the definition of the SI-decomposition.\nThe ASB-decomposition is a combination of the SB-decomposition and the\nAI-decomposition, as the next theorem shows.\nTheorem 3.5. Let A be a DFA without unreachable states.\n(A1, A2) is an\nASB-decomposition of A iff(A1, A2) is both an SB-decomposition and an AI-\ndecomposition of A.\nProof. The first implication clearly follows from the definitions, Theorem 3.1\nand Theorem 3.2. Now let (A1, A2) be an SB- and AI-decomposition of A =\n(K, Σ, δ, q0, F). Let α be the mapping given by the definition of SB-decomposition.\nWe need to prove that for all states q of A, q ∈F iffα(q) ∈F1 × F2, where Fi is\nthe set of accepting states of Ai, i ∈{1, 2}. Let q ∈K and let w be a word such\nthat δ(q0, w) = q. Then q ∈F ⇔w ∈L(A) ⇔w ∈L(A1) ∩L(A2) ⇔α(q) ∈\nF1 × F2, where the first equivalence is implied by the choice of w, the second\nholds because (A1, A2) is an AI-decomposition and the third is a consequence\nof the properties of α guaranteed by the SB-decomposition definition.\nThere is also a relationship between SB- and SI-decompositions, in fact SB-\nis a stronger version of the state-identifying decomposition, as the following two\ntheorems show. We need the notion of reachability on pairs of states.\nDefinition 3.1. Let A1 = (K1, Σ, δ1, p1, F1) and A2 = (K2, Σ, δ2, p2, F2) be\nDFAs. We shall call a pair of states (q, r) ∈K1 × K2 reachable, if there exists\na word w ∈Σ∗such that δ1(p1, w) = q and δ2(p2, w) = r.\nTheorem 3.6. Let A = (K, Σ, δ, q0, F) be a DFA and let (A1, A2) be its SB-\ndecomposition. Then (A1, A2) also forms an SI-decomposition of A.\nProof. Let Ai = (Ki, Σ, δi, qi, Fi), i ∈{1, 2}. Since (A1, A2) is an SB-decomposition\nof A, there exists an injective mapping α: K →K1 × K2 such that it holds\nα(q0) = (q1, q2) and (∀a ∈Σ)(∀p ∈K); α(δ(p, a)) = (δ1(p1, a), δ2(p2, a)), where\nα(p) = (p1, p2). Let us define a new mapping β : K1 × K2 →K by\nβ(p1, p2) ="},{"paragraph_id":"p7","order":7,"text":"p\nif ∃p ∈K, α(p) = (p1, p2)\nq0\notherwise.\n(1)\nSince α is injective, there exists at most one such p and this definition is correct.\n6"},{"paragraph_id":"p8","order":8,"text":"We now need to prove that β satisfies the condition from the definition of\nSI-decomposition, i.e., that (∀w ∈Σ∗); β(δ1(q1, w), δ2(q2, w)) = δ(q0, w). Since\nα(q0) = (q1, q2) and all the pairs of states we encounter in the computation of\nA1||A2 are thus reachable, this follows from the definition of α and (1) by an\neasy induction.\nLemma 3.7. Let A be a DFA without unreachable states and let (A1, A2) be\nits SI-decomposition, with β being the corresponding mapping. Then (A1, A2) is\nan SB-decomposition of A if and only if β is injective on all reachable pairs of\nstates.\nProof. Let (A1, A2) be an SB-decomposition of A. It clearly follows from Def-\ninition 2.2, that the corresponding β satisfies the equation (1) in the proof of\nTheorem 3.6 on all reachable pairs of states. Since the mapping α is a bijection\nbetween the set of states of A and the set of all reachable pairs of states of A1\nand A2, β defined as its inverse on the set of reachable pairs of states will be\ninjective on this set.\nFor the other implication, let (A1, A2) be an SI-decomposition of A and let\nβ be injective on the set of reachable pairs of states, let βr denote the mapping\nβ restricted onto the set of all reachable pairs of states of A1, A2. Since A has\nno unreachable states, βr is also surjective, thus we can define a new mapping\nα: K →K1 × K2 by the equation α(q) = β−1\nr (q). Since β maps the initial state\nonto the initial state, so does α, and since β satisfies the condition from the\nDefinition 2.2, it implies that also α satisfies the condition (i) from the definition\nof realization of state behavior. Therefore (A1, A2) is an SB-decomposition of\nA, with the corresponding mapping α.\nThe converse of Theorem 3.6 does not hold. The minimal automaton for the\nlanguage L = {a4kb4l|k ≥0, l ≥1} gives a counterexample. Inspecting its S.P.\npartitions shows that it has no nontrivial SB-decomposition, but it can be AI-\ndecomposed into minimal automata for languages L1 = {a4kbl|k ≥0, l ≥1} and\nL2 = {w|#b(w) = 4l; l ≥0}. According to Theorem 3.4, this AI-decomposition\nis also state-identifying.\nEach ASB-decomposition is obviously also an SB-decomposition. On the\nother hand, there exist SB-decomposable automata, that are ASB-undecomposable.\nFor example, the minimal automaton for the language\nL1\n=\n{w ∈{a, b, c}∗|#a(w)\nmod 3 = 0 ∧#b(w)\nmod 5 = 0} ∪\n∪\n{w ∈{a, b, c}∗|#a(w)\nmod 3 = 2 ∧#b(w)\nmod 5 = 4}\nhas this property, because the corresponding S.P. partitions on the set of its\nstates do not separate the final states in the sense of Definition 2.7.\nIt is also not so difficult to see that for any non-minimal automaton A without\nunreachable states, there exists a nontrivial AI- and wAI-decomposition (A1, A2)\nsuch that A1 is the minimal automaton equivalent to A and A2 has only one\nstate. This decomposition is obviously not state-identifying.\nFigure 1 summarizes all the relationships among the decomposition types\nthat we have shown so far.\nNow we show that for the case of so-called perfect decompositions, some of\nthe types of decomposition mentioned coincide.\n7"},{"paragraph_id":"p9","order":9,"text":"ASB\n{vvvvvvvvv\n#G\nG\nG\nG\nG\nG\nG\nG\nAI\n×\nv\nv\nv\nv\n;v\nv\nv\nv\nmin\nH\nH\nH\nH\n#H\nH\nH\nH"},{"paragraph_id":"p10","order":10,"text":"×\n#\nSB\n{wwwwwwww\n×GGGG\ncGGGG\nSI\n{vvvvvvvvv\n×\nwwww\n;wwww\nwAI\n×\n;\nDescription:\nA\n/ B : every A-decomposition is also a B-\ndecomposition\nA\n×\n/ B : not every A-decomposition is also\na B-decomposition\nA\n×\n/ B : there exists a DFA that has a non-\ntrivial A-decomposition but does not have\na nontrivial B-decomposition\nFigure 1: Relationships between decomposition types of DFA\nDefinition 3.2. Let t be a type of decomposition, t ∈{ASB, SB, AI, SI, wAI}.\nLet A be a DFA having n states, let A1 and A2 be DFAs having k and l states,\nrespectively. We shall call the pair (A1, A2) a perfect t-decomposition of A, if\nit forms a t-decomposition of A and n = k · l.\nTheorem 3.8. Let A be a DFA with no unreachable states and let (A1, A2) be a\npair of DFAs. Then (A1, A2) forms a perfect SI-decomposition of A iff(A1, A2)\nforms a perfect SB-decomposition of A.\nProof. One of the implications is a consequence of Theorem 3.6.\nAs to the\nsecond one, since (A1, A2) forms a perfect SI-decomposition of A, each of the\npairs of states of A1 and A2 is reachable and each pair has to correspond to a\ndifferent state of A in the mapping β, therefore β is bijective and the theorem\nfollows from Theorem 3.7.\nCorollary 3.9. Let A be a minimal DFA and let (A1, A2) be a pair of DFAs.\nThen (A1, A2) forms a perfect AI-decomposition of A iff(A1, A2) forms a perfect\nASB-decomposition of A.\nProof. The claim follows from Theorem 3.5, Theorem 3.4 and Theorem 3.8.\nAs a consequence of these facts, we can use the necessary and sufficient\nconditions stated in Theorem 3.1 to look for perfect AI- and SI-decompositions.\nNow, let us inspect the relationship between decompositions of an automaton\nand the decompositions of the corresponding minimal automaton.\nTheorem 3.10. Let A = (K, Σ, δ, q0, F) be a DFA and let Amin be a min-\nimal DFA such that L(A) = L(Amin).\nLet (A1, A2) be an SI-decomposition\n(AI-decomposition, wAI-decomposition) of A, then (A1, A2) also forms a de-\ncomposition of Amin of the same type.\nProof. First, note that this theorem does not state that any of the decomposi-\ntions is nontrivial. To prove the statement for SI-decompositions, suppose that\n(A1, A2) is an SI-decomposition of A, thus there exists a mapping α: K1×K2 →\nK such that it holds (∀w ∈Σ∗); α(δ1(q1, w), δ2(q2, w)) = δ(q0, w), where δi and\nqi are the transition function and the initial state of the automaton Ai. Since\nAmin is the minimal automaton corresponding to A, there exists some mapping\nβ : K →Kmin such that (∀w ∈Σ∗); β(δ(q0, w)) = δmin(β(q0), w), where δmin is\n8"},{"paragraph_id":"p11","order":11,"text":"a1\nb\n/\na"},{"paragraph_id":"p12","order":12,"text":"R\na0\nb\n/\na\nH\nb1\nb\n)\na\nO\nb0\nb\ni\na\nm\na1\nb\n/\na"},{"paragraph_id":"p13","order":13,"text":"R1\nb\n* R0\nb\nj\na0\nb\n/\na\nH\nb1\nb\n)\na\nO\nb0\nb\ni\na\nO\nFigure 2: Transition functions of Amin and A′.\nthe transition function of Amin and Kmin is the set of states of Amin. By the com-\nposition of these mappings we obtain the mapping β◦α: K1×K2 →Kmin, which\ncombines A1 and A2 into Amin in the way that the definition of SI-decomposition\nrequires. For both the AI- and the wAI-decomposition, this statement is trivial,\nsince L(A) = L(Amin).\nBased on the above theorem it thus suffices to inspect the SI- (AI-, wAI-)\ndecomposability of the minimal automaton accepting a given language, and if\nwe show its undecomposability, we know that the recognition of this language\ncannot be simplified using an advisor of the respective type.\nHowever, this\ndoes not hold for SB- and ASB-decompositions, as exhibited by the following\nexample.\nExample 3.1. Let us consider the language L = {a2kb2l|k ≥0, l ≥1}. The\nminimal automaton Amin = (K, ΣL, δ, a0, {a0, b0}) has its transition function\ndefined by the first transition diagram in Fig.2. We can easily show that this\nautomaton does not have any nontrivial SB- (and thus neither ASB-) decompo-\nsition by enumerating its S.P. partitions.\nNow let us examine the automaton A′ = (K′, ΣL, δ′, a0, {a0, b0}) with the\ntransition function δ′ defined by the second transition diagram in Fig.2. Clearly,\nL(A′) = L(Amin), but by inspecting the lattice of S.P. partitions of A′, we can\nfind the pair π1 = {{a0}, {a1}, {b0, b1}, {R0, R1}} and π2 = {{a0, a1, b0, R0}, {b1, R1}}\nsuch that π1 · π2 = 0 and they separate the final states of A′. By Theorem 3.1\nwe can use these partitions to construct a nontrivial ASB- (and thus also SB-)\ndecomposition of A′ formed by the automata A1 and A2 having two and four\nstates, respectively. Note that both A1 and A2 have less states than Amin.\nIn the following theorem (inspired by a similar theorem in [5]) we state a\ncondition, under which the situation from the last example cannot occur, i.e.,\nunder which any SB-decomposition of a DFA implies a (maybe simpler) SB-\ndecomposition of the equivalent minimal DFA.\nTheorem 3.11. Let A = (K, Σ, δ, q0, F) be a deterministic finite automa-\nton and let Amin = (Kmin, Σ, δmin, qmin, Fmin) be the minimal DFA such that\nL(A) = L(Amin). Let (A1, A2) be a nontrivial SB-decomposition of A consisting\nof automata having k and l states. If the lattice of S.P. partitions of A is dis-\ntributive, then there exists an SB-decomposition of Amin consisting of automata\nhaving k′ and l′ states, such that k′ ≤k and l′ ≤l.\nProof. Since Amin is the minimal DFA such that L(A) = L(Amin), there exists\na mapping f : K →Kmin such that (∀w ∈Σ∗); f(δ(q0, w)) = δmin(qmin, w).\nUsing the mapping f, let us define a partition ρ on the set of states of A by\np ≡ρ q ⇔f(p) = f(q). Clearly, ρ is an S.P. partition.\n9"},{"paragraph_id":"p14","order":14,"text":"Since (A1, A2) is a nontrivial SB-decomposition of A, we can use it to obtain\nS.P. partitions π1 and π2 on the set of states of A such that π1 · π2 = 0. Let us\ndefine new partitions π′\n1 and π′\n2 on the set of states of Amin by f(p) ≡π′\ni f(q) ⇔\np ≡ρ+πi q. Since it holds that ρ + πi ⪯ρ, this definition does not depend on\nthe choice of the states p and q. It holds that |π′\ni| = |ρ + πi| ≤|πi|, therefore if\nwe prove that π′\n1 and π′\n2 are S.P. partitions and π′\n1 · π′\n2 = 0, we can use them to\nconstruct the desired decomposition.\nThe fact that π′\ni is an S.P. partition on the set of states of Amin is a trivial\nconsequence of the fact that ρ + πi is an S.P. partition on the set of states of\nA. We need to prove that π′\n1 · π′\n2 = 0. Let us assume that p′ and q′ are states\nof Amin such that p′ ≡π′\n1·π′\n2 q′ and p, q are some states of A such that f(p) = p′\nand f(q) = q′. Then p′ ≡π′\n1 q′ and p′ ≡π′\n2 q′, and by definition of π′\ni we get\np ≡ρ+π1 q and p ≡ρ+π2 q, which is equivalent to p ≡(ρ+π1)·(ρ+π2) q. Since the\nlattice of all S.P. partitions of A is distributive, we have (ρ + π1) · (ρ + π2) =\nρ + (π1 · π2) = ρ + 0 = ρ, therefore p ≡ρ q, which by definition of ρ implies that\nf(p) = f(q), in other words p′ = q′. Hence π′\n1 · π′\n2 = 0.\n4\nDegrees of Decomposability\nIt is easy to see that for each type of decomposition, there exist undecomposable\nregular languages (e.g. L(n) = {ak|k ≥n −1} is wAI-undecomposable for each\nn ∈N). There also exist regular languages, that are perfectly decomposable in\neach way (e.g. L(k,l) = {w ∈{a, b}∗|#a(w) mod k = 0 ∧#b(w) mod l = 0} has\na perfect ASB-decomposition for all k, l ≥2). We shall now investigate whether\nall values between these two limits can be achieved.\nDefinition 4.1. Let A be a DFA, let (A1, A2) be its nontrivial SB- (ASB-)\ndecomposition with the corresponding S.P. partitions π1 and π2. We shall call\nthis decomposition redundant, if there exist S.P. partitions π′\n1 ⪰π1 and π′\n2 ⪰π2\nsuch that at least one of these inequalities is strict, but it still holds π′\n1 · π′\n2 = 0\n(and π′\n1 and π′\n2 separate the final states of A).\nLemma 4.1. For each r, s ∈N, r, s ≥2, there exists a minimal DFA A consist-\ning of r.s states and having only one nontrivial nonredundant SB-decomposition\n(ASB-decomposition) up to the order of automata, consisting of automata hav-\ning r and s states.\nProof. Let us study the minimal automaton Ar,s = (K, Σ, δ, q0,0, F) defined\nby K = {qi,j|i ∈{0, . . . , r −1}, j ∈{0, . . ., s −1}}, F = {qr−1,s−1} and the\ntransition function δ defined by\nδ(qi,j, a)\n=\nqi+1,j for i ∈{0, . . ., r −2}, j ∈{0, . . ., s −1}\nδ(qr−1,j, a)\n=\nqr−1,j for j ∈{0, . . . , s −1}\nδ(qi,j, b)\n=\nqi,j+1 for i ∈{0, . . ., r −1}, j ∈{0, . . ., s −2}\nδ(qi,s−1, b)\n=\nqi,s−1 for i ∈{0, . . . , r −1}.\nTo inspect the SB-decompositions of Ar,s, let us study the S.P. partitions on\nthe set of its states. From the method for generating all S.P. partitions of an\nautomaton that is described in [5], we know that each nontrivial S.P. partition\ncan be obtained as a sum of some partitions πm\np,t, where πm\np,t denotes the minimal\n10"},{"paragraph_id":"p15","order":15,"text":"S.P. partition such that it does not distinguish between states p and t, i.e., they\nbelong into the same block. Let us determine πm\np,t for various states p and t of\nAr,s.\nFirst, let us consider the case of πm\np,t such that p = qi,j, t = qi′,j′ and both\ninequalities i < i′ and j < j′ hold. Since qi,j ≡π qi′,j′, δ(qi,j, ai′−ibj′−j) = qi′,j′\nand δ(qi′,j′, ai′−ibj′−j) = q2i′−i,2j′−j (if 2i′ −i < r and 2j′ −j < s), as a\nconsequence of the substitution property of π, we obtain qi,j ≡π q2i′−i,2j′−j.\nBy applying this argument a finite number of times (keeping in mind the con-\nstruction of Ar,s), we obtain qi,j ≡π qr−1,s−1. Now let k ∈{i, . . ., r −1} and\nlet l ∈{i, . . . , s −1}.\nThen δ(qi,j, ak−ibl−j) = qk,l and δ(qi′,j′, ak−ibl−j) =\nqk+i′−i,l+j′−j (if such states exist), therefore qk,l ≡π qk+i′−i,l+j′−j. Again, we\ncan use the same argument to show that qk,l ≡π qr−1,s−1. Therefore, for this\ntype of π = πm\np,t, we have qk,l ≡π qk′,l′ for all k, l, k′, l′ such that i ≤k, k′ < r\nand j ≤l, l′ < s.\nNow let us consider the case of πm\np,t such that p = qi,j, t = qi′,j′ and it holds\ni > i′ and j < j′. Since qi,j ≡π qi′,j′, δ(qi,j, ar−1−ibs−1−j′) = qr−1,s−1−(j′−j)\nand δ(qi′,j′, ar−1−ibs−1−j′) = qr−1−(i−i′),s−1, as a consequence of the substitu-\ntion property of π, we have qr−1,s−1−(j′−j) ≡π qr−1−(i−i′),s−1. By exploiting\nthe substitution property again on this equivalence, using the words ai−i′−1,\nbj′−j−1 and bj′−j, we obtain qr−2,s−1 ≡π qr−1,s−1 ≡π qr−2,s−2. Therefore in\nthis case, no such πm\np,t partition can distinguish between states qr−2,s−1, qr−1,s−1\nand qr−2,s−2.\nThe last case to consider is the case of πm\np,t such that p = qi,j, t = qi′,j′ and\nit holds i = i′ (the case j = j′ is analogous). Without loss of generality, we\ncan assume that j < j′. Now, using the same arguments as in the first case, we\ncan show that qi,l ≡π qi,l′ for all l, l′ such that j ≤l, l′ < s. Therefore for each\ngiven k such that i ≤k < r, it holds that qk,l ≡π qk,l′ and all of the states not\nmentioned in this equivalence form separate blocks of πm\np,t.\nIt is easy to verify that one nontrivial ASB-decomposition of Ar,s is given\nby the S.P. partitions\nπ1\n=\n{{q0,0, . . . , q0,s−1}, {q1,0, . . . , q1,s−1}, . . . , {qr−1,0, . . . , qr−1,s−1}} and\nπ2\n=\n{{q0,0, . . . , qr−1,0}, {q0,1, . . . , qr−1,1}, . . . , {q0,s−1, . . . , qr−1,s−1}}\nNow we show that any other SB-decomposition of Ar,s is given by S.P. partitions\npreceding to π1 and π2 in the partial order ⪯and therefore is redundant.\nIndeed, notice that none of the πm\np,t partitions of the first and the second\ndiscussed type can distinguish between any of the states qr−2,s−1, qr−1,s−1 and\nqr−2,s−2, therefore no sum of them can, either. For the partitions of the third\ntype, it holds either qr−2,s−1 ≡π qr−1,s−1 or qr−1,s−1 ≡π qr−2,s−2, therefore it\nwill take two partitions to distinguish between these three states. Hence any\nnontrivial SB-decomposition is determined by two S.P. partitions, both of which\nmust be of the third type. But it is easy to see that for any partition π of this\ntype it holds either π ⪯π1 or π ⪯π2.\nDefinition 4.2. Let A = (K, Σ, δ, q0, F) be a deterministic finite automaton,\nlet K ∩{p0, p1, . . . , pk−1} = ∅and let c be a new symbol not included in Σ. We\nshall define a k-extension A′ of the automaton A by the following construction:\nA′ = (K ∪{p0, p1, . . . , pk−1}, Σ ∪{c}, δ′, p0, F), where the transition function δ′\n11"},{"paragraph_id":"p16","order":16,"text":"is defined as follows:\n(∀q ∈K) (∀a ∈Σ);\nδ′(q, a)\n=\nδ(q, a)\n(∀q ∈K);\nδ′(q, c)\n=\nq\n(∀p ∈{p0, p1, . . . , pk−1}) (∀a ∈Σ);\nδ′(p, a)\n=\np\n(∀i ∈{0, 1, . . ., k −2});\nδ′(pi, c)\n=\npi+1\nδ′(pk−1, c)\n=\nq0.\nLemma 4.2. Let A be a DFA consisting of n states, all of which are reach-\nable. Let A′ be its k-extension. Then A has a nontrivial nonredundant SB-\ndecomposition (ASB-decomposition) consisting of automata having r and s states\niffA′ has a nontrivial nonredundant decomposition of the same type, consisting\nof automata having k + r and k + s states.\nProof. We will try to inspect S.P. partitions on the set of states of A′, using\nthe notation from Definition 4.2. Let us assume that π′ is an S.P. partition\non the set of states of A′ such that pi and pj are in the same block of π′;\ni, j ∈{0, 1, . . ., k −1}. As a consequence of the S.P. property, if i, j < k −1\nthen also pi+1 and pj+1 are in the same block of π′, because δ′(pi, c) = pi+1\nand δ′(pj, c) = pj+1. By applying this argument a finite number of times, we\ncan show that there exists some l ∈{0, 1, . . ., k −2} such that pl ≡π′ pk−1,\nand using the argument once more, we obtain pl+1 ≡π′ q0. However, it holds\nδ′(pl, a) = pl for all a ∈Σ, hence pl ≡π′ δ′(q0, w) for all w ∈Σ∗. Since all of the\nstates of A′ are reachable, we have pl ≡π′ q for all q ∈K. Thus such a partition\ncannot distinguish between the original states of the automaton A.\nNow let us suppose that π′ is an S.P. partition on the set of states of A′ such\nthat for some i ∈{0, 1, . . ., k −1}, pi ≡π′ q for some q in K. Then it also holds\nthat pi ≡π′ pi+1, because δ(pi, c) = pi+1 and δ(q, c) = q. But we have already\nshown that pi ≡π′ pi+1 implies that all of the states in K are equivalent modulo\nπ′, thus this S.P. partition cannot distinguish between the states of A, either.\n¿From these observations it follows that if π′ is any S.P. partition on the set\nof states of A′ such that the states of A are not all equivalent modulo π′, then\nπ′ must also contain k blocks, each of which contains only one state pi, where\ni ∈{0, 1, . . ., k −1}. Now we can prove the equivalence stated in the theorem.\nLet A have an SB-decomposition consisting of r and s states. Then there\nexist S.P. partitions π1 and π2 on the set of states of A having r and s blocks,\nsuch that π1 · π2 = 0.\nLet us now construct new partitions π′\n1 and π′\n2 on\nthe set of states of A′ by π′\n1 = π1 ∪{{p0}, {p1}, . . . , {pk−1}} and π′\n2 = π2 ∪\n{{p0}, {p1}, . . . , {pk−1}}. Obviously, π′\n1 and π′\n2 have substitution property, be-\ncause for the states in K this property is inherited from π1 and π2, and the new\nstates p0, p1, . . . , pk−1 cannot violate this property either, because each of these\nstates belongs to a separate block in π′\n1 and π′\n2, making the substitution prop-\nerty hold trivially. Neither do the new c-moves defined on the states from K\nviolate the substitution property. Finally, it holds that π′\n1 · π′\n2 = 0. To see this,\nnote that for a state q ∈K, it holds [q]π′\n1·π′\n2 = [q]π1·π2 = {q}, since π1 · π2 = 0.\nFor a state q ∈K′ −K, [q]π′\ni = {q} for i ∈{1, 2} thus [q]π′\n1·π′\n2 = {q}, too. Hence\neach state of A′ belongs to a separate block of π′\n1 · π′\n2, which implies π′\n1 · π′\n2 = 0.\nTherefore π′\n1 and π′\n2 induce an SB-decomposition of A′. It is also easy to see\nthat if π1 and π2 separate the final states of A, then also π′\n1 and π′\n2 separate the\nfinal states of A′, making the induced decomposition an ASB-decomposition.\n12"},{"paragraph_id":"p17","order":17,"text":"On the other hand, let us now assume that A′ has an SB-decomposition and\nπ′\n1 and π′\n2 are the S.P. partitions on K′ that induce this decomposition, thus\nπ′\n1 · π′\n2 = 0. From the observations made in the beginning of this proof, we know\nthat any S.P. partition that can distinguish between the states in K in any way,\nmust contain each of the states p0, p1 . . . pk−1 in a separate block containing only\nthis state. As π′\n1 ·π′\n2 = 0, for all q1, q2 ∈K, at least one of these partitions must\ndistinguish between these states, i.e., [q1]π′\ni ̸= [q2]π′\ni. If one of the partitions\ndistinguished between all such pairs, it would imply that this partition must\ncontain a separate block for each one of the states in K′, thus becoming a trivial\npartition 0, resulting in a trivial decomposition. Therefore both π′\n1 and π′\n2 have\nto distinguish between some pair of states from K, which implies that they both\ncontain a separate block for each of the states p0, p1 . . . pk−1 containing no other\nstate. By removing these k blocks from π′\n1 and π′\n2, we obtain new partitions π1\nand π2 on the set K, such that π1 = π′\n1−{{p0}, {p1}, . . . , {pk−1}} and π2 = π′\n2−\n{{p0}, {p1}, . . . , {pk−1}}. These partitions preserve the substitution property,\nsince (∀a ∈Σ)(∀q ∈K): δ(q, a) ∈K and π′\n1 and π′\n2 were S.P. partitions. It also\nholds π1 ·π2 = 0, as for all q1, q2 ∈K, q1 ≡π1·π2 q2 implies q1 ≡π′\n1·π′\n2 q2 and that\nimplies q1 = q2. So π1 and π2 induce an SB-decomposition of A. As π′\n1 and π′\n2\nwere nontrivial, so are π1 and π2 and the obtained decomposition. It is again\neasy to see that if π′\n1 and π′\n2 separate the final states of A′, then also π1 and π2\nmust separate the final states of A.\nThe described relationship between the S.P. partitions on the set of states\nof A and the corresponding S.P. partitions on A′ also implies, that each de-\ncomposition of A is nonredundant iffthe corresponding decomposition of A′ is\nnonredundant, too.\nSince a k-extension of a minimal DFA is again a minimal DFA, we can\ncombine the lemmas to obtain the following theorem.\nTheorem 4.3. Let n ∈N be such that n = k + r.s, where r, s, k ∈N, r, s ≥2.\nThen there exists a minimal DFA A consisting of n states, such that it has only\none nontrivial nonredundant SB-decomposition (ASB-decomposition) up to the\norder of the automata in the decomposition, and this decomposition consists of\nautomata with k + r and k + s states.\nReferences\n[1] J. L. Balcazar, J. Diaz, J. Gabarro, Structural Complexity I., Springer-Verlag\nNew York, 1988\n[2] S. Even, A. L. Selman, Y. Yacobi, The Complexity of Promise Problems\nwith Applications to Public-Key Cryptography, Information and Control 61(2):\n159-173 (1984)\n[3] S. Yu, State Complexity: Recent Results and Open Problems, Fundamenta\nInformaticae 64: 471-480 (2005)\n[4] J. C. Birget, Intersection and Union of Regular Languages and State Com-\nplexity, Information Processing Letters 43: 185-190 (1992)\n[5] J. Hartmanis and R. E. Stearns, Algebraic Structure Theory of Sequential\nMachines, Prentice-Hall, 1966\n13"},{"paragraph_id":"p18","order":18,"text":"[6] J. E. Hopcroft and J. D. Ullman, Introduction to Automata Theory, Lan-\nguages, and Computation, Addison-Wesley, 1979\n14"}],"pages":[{"page":1,"text":"arXiv:0707.0430v1 [cs.CC] 3 Jul 2007\nAssisted Problem Solving and Decompositions of\nFinite Automata∗\nPeter Gaˇzi\nBranislav Rovan\nDepartment of Computer Science, Comenius University\nMlynsk ́a dolina, 842 48, Bratislava, Slovakia\n{gazi,rovan}@dcs.fmph.uniba.sk\nAbstract\nA study of assisted problem solving formalized via decompositions of\ndeterministic finite automata is initiated. The landscape of new types of\ndecompositions of finite automata this study uncovered is presented. Lan-\nguages with various degrees of decomposability between undecomposable\nand perfectly decomposable are shown to exist.\n1\nIntroduction\nIn the present paper we initiate the study of assisted problem solving. We intend\nto model and study situations, where solution to the problem can be sought\nbased on some additional a priori information about the inputs. One can expect\nto obtain simpler solution in such case. There are similar approaches known\nin the literature, most notably the notions of advice functions [1], where the\nadditional information is based on the length of the input word and the notion\nof promise problems [2], where the set of inputs is separated into three classes\n– those with “yes” answer, those with “no” answer and those where we do not\ncare about the outcome. By considering the simplest case where the “problem\nsolving” machinery is the deterministic finite automaton (DFA) we obtain a new\nmotivation for studying new types of finite automata decompositions.\nIn this paper we shall thus consider the case where solving a problem shall\nmean constructing an automaton for a given language L. The “assistance” shall\nbe given by additional information about the input, e.g., that we can assume the\ninputs shall be restricted to words from a particular regular language L′. Thus,\ninstead of looking for an automaton A such that L = L(A) we can look for a\n(possibly simpler) automaton B such that L = L(B)∩L′. We can then say that\nB accepts L with the assistance of L′. We shall call L′ (or the corresponding\nautomaton A′ such that L′ = L(A′)) an advisor to B. In this case the advisor\nA′ provides assistance to the solver B by guaranteeing that A′ accepts the given\ninput word.\nWe shall also study a case where the assistance provides more\ndetailed information about the outcome of the computation of A′ on the input\nword (e.g., the state reached). Clearly the advisor can be considered useful\nonly if it enables B to be simpler than A and at the same time A′ is not more\n∗This work was supported in part by the grant VEGA 1/3106/06.\n1"},{"page":2,"text":"complicated than A. The measure of complexity we shall consider is the number\nof states of the deterministic finite automaton. This measure of complexity was\nused quite often recently due to renewed interest in finite automata prompted\nby applications such as model checking (see e.g. [3] for a recent survey). (Note\nthat results complementary to ours, namely results on complexity of automata\nfor the intersection of regular sets were studied in [4].)\nThe contribution of our paper is twofold. First, we can interpret the ‘solver’\nand the ‘advisor’ as two parallel processes each performing a different task and\njointly solving a problem. Since our approach lends itself to a generalisation to\nk advisors it may stimulate new parallel solutions to problems (the traditional\nones usually using parallel processes to perform essentially the same task). Sec-\nond, the choice of finite automata as the simplest problem solving machinery\nbrought about new types of decompositions motivated by the information the\n‘advisor’ can provide to the ‘solver’. Our results provide a complete picture of\nthe landscape of these decompositions.\nThe problem within this scenario we shall address in this paper is the exis-\ntence of a useful advisor for a given automaton A. We shall compare the power\nof several types of advisors, and investigate the effect of the advisor on the\ncomplexity of the assisted solver B. We can formulate this also as a problem of\ndecomposition of deterministic finite state automata – given DFA A find DFA\nA1 (a solver) and A2 (an advisor) such that w ∈L(A) can be determined from\nthe computations of A1 and A2. We shall study several new types of decompo-\nsitions of DFA, one of them is analogous to the state behavior decomposition of\nfinite state transducers studied in [5]. In Sect. 3 we prove relations among these\ndecompositions. For each type of decomposition there are automata which are\nundecomposable and automata for which there is a decomposition that is the\nbest possible. In Sect. 4 we consider the space between these extreme points\nand study the degree of decomposability.\n2\nDefinitions and Notation\nWe shall use standard notions of the theory of formal languages (see e.g. [6]).\nOur notation shall be as follows.\nΣ∗denotes the set of all words over the\nalphabet Σ, the length of a word w is denoted by |w|, ε denotes the empty\nword, and for a language L we shall denote by ΣL the minimal alphabet such\nthat L ⊆Σ∗\nL. The number of occurrences of a given letter a in a word w is\ndenoted by #a(w). Throughout this paper we shall consider deterministic finite\nautomata only.\nA deterministic finite automaton (DFA) is a quintuple (K, Σ, δ, q0, F), such\nthat K is a finite set of states, Σ is a finite input alphabet, q0 ∈K is the initial\nstate, F ⊆K is the set of accepting states and δ: K × Σ →K is a transition\nfunction. As usual, we shall denote by δ also the standard extension of δ to\nwords, i.e., δ: K × Σ∗→K. We shall denote by |K| the number of states in K.\nFormalizing the notions of assisted problem solving from the Introduction\nwe shall now define several types of decompositions of DFA A into two (sim-\npler) DFAs A1 and A2 (a solver and an advisor) so that the membership of\nan input word w in L(A) can be determined based on the information on the\ncomputations of A1 and A2 on w.\nWe first introduce an acceptance-identifying decomposition of deterministic\n2"},{"page":3,"text":"finite automata.\nDefinition 2.1. A pair of DFAs (A1, A2), where A1 = (K1, Σ, δ1, q1, F1) and\nA2 = (K2, Σ, δ2, q2, F2), forms an acceptance-identifying decomposition (AI-\ndecomposition) of a DFA A = (K, Σ, δ, q0, F), if L(A) = L(A1) ∩L(A2). This\ndecomposition is nontrivial if |K1| < |K| and |K2| < |K|.\nBy decomposing A in this manner, one of the decomposed automata (say\nA2) can act as an advisor and narrow down the set of input words for the other\none (say A1), whose task to recognize the words of L(A) may become easier.\nAnother requirement we could pose on a decomposition is to identify the\nfinal state of any computation of the original automaton by only knowing the\nfinal states of both corresponding computations of the automata forming the\ndecomposition. This requirement can be formalized as follows.\nDefinition 2.2. A pair of DFAs (A1, A2), where A1 = (K1, Σ, δ1, q1, F1) and\nA2 = (K2, Σ, δ2, q2, F2), forms a state-identifying decomposition (SI-decomposition)\nof a DFA A = (K, Σ, δ, q0, F), if there exists a mapping β : K1 × K2 →K, such\nthat it holds β(δ1(q1, w), δ2(q2, w)) = δ(q0, w) for all w ∈Σ∗. This decomposi-\ntion is nontrivial if |K1| < |K| and |K2| < |K|.\nThe third – and the weakest – requirement we pose on a decomposition of a\nDFA is to require that there must exist a way to determine whether the original\nautomaton would accept some given input word based on knowing the states in\nwhich the computations of both decomposition automata have finished.\nDefinition 2.3. A pair of DFAs (A1, A2), where A1 = (K1, Σ, δ1, q1, F1) and\nA2 = (K2, Σ, δ2, q2, F2), forms a weak acceptance-identifying decomposition\n(wAI-decomposition) of a DFA A = (K, Σ, δ, q0, F), if there exists a relation\nR ⊆K1 × K2 such that it holds R(δ1(q1, w), δ2(q2, w)) ⇔w ∈L(A) for all\nw ∈Σ∗. This decomposition is nontrivial if |K1| < |K| and |K2| < |K|.\nNote that in the last two definitions, the sets of accepting states of A1 and\nA2 are irrelevant.\nBy a decomposability of a regular language L in some way, we shall mean\nthe decomposability of the corresponding minimal automaton over ΣL.\nTo be able to compare these new types of decomposition to the parallel\ndecompositions of state behavior introduced for sequential machines in [5], we\nshall redefine them for DFAs.\nDefinition 2.4. A DFA A′ = (K′, Σ, δ′, q′\n0, F ′) is said to realize the state\nbehavior of a DFA A = (K, Σ, δ, q0, F) if there exists an injective mapping\nα: K →K′ such that\n(i) (∀a ∈Σ)(∀q ∈K); δ′(α(q), a) = α(δ(q, a)),\n(ii) α(q0) = q′\n0.\nMoreover, A′ is said to realize the state and acceptance behavior of A, if in\naddition the following property holds:\n(iii) (∀q ∈K); α(q) ∈F ′ ⇔q ∈F.\n3"},{"page":4,"text":"Definition 2.5. The parallel connection of two DFA A1 = (K1, Σ, δ1, q1, F1)\nand A2 = (K2, Σ, δ2, q2, F2) is the DFA A = A1||A2 = (K1 × K2, Σ, δ, (q1, q2),\nF1 × F2) such that δ((p1, p2), a) = (δ1(p1, a), δ2(p2, a)).\nDefinition 2.6. A pair of DFAs (A1, A2) is a state behavior (SB-) decomposi-\ntion of a DFA A if A1||A2 realizes the state behavior of A. The pair (A1, A2) is\nan acceptance and state behavior (ASB-) decomposition of A if A1||A2 realizes\nthe state and acceptance behavior of A. This decomposition is nontrivial if both\nA1 and A2 have fewer states than A.\nWe have modified the definitions to fit the formalism and purpose of deter-\nministic finite automata (i.e., to accept formal languages) without loosing the\nconnection to the strongly related and useful concept of S.P.partitions, exhibited\nbelow.\nWe shall use the following notation and properties of S.P. partitions from [5].\nA partition π on a set of states of a DFA A = (K, Σ, δ, q0, F) has substitution\nproperty (S.P.), if it holds ∀p, q ∈K;\np ≡π q ⇒(∀a ∈Σ; δ(p, a) ≡π δ(q, a)). If\nπ1 and π2 are partitions on a given set M, then\n(i) π1 · π2 is a partition on M such that a ≡π1·π2 b ⇔a ≡π1 b ∧a ≡π2 b,\n(ii) π1 +π2 is a partition on M such that a ≡π1+π2 b iffthere exists a sequence\na = a0, a1, a2, . . . , an = b, such that ai ≡π1 ai+1 ∨ai ≡π2 ai+1 for all\ni ∈{0, . . ., n −1},\n(iii) π1 ⪯π2 if it holds (∀x, y ∈M);\nx ≡π1 y ⇒x ≡π2 y.\nThe set of all partitions on a given set (with the partial order ⪯, join realized\nby + and meet realized by .) forms a lattice. The set of all S.P. partitions on\nthe set of states of a given DFA forms a sublattice of the lattice of all partitions\non this set. The trivial partitions {{q0}, {q1}, . . . , {qn}} and {{q0, q1, . . . , qn}}\nshall be denoted by symbols 0 and 1, respectively. The block of a partition π\ncontaining the state q shall be denoted by [q]π. In addition, we shall use the\nfollowing separation notion.\nDefinition 2.7. The partitions π1 = {R1, . . . , Rk} and π2 = {S1, . . . , Sl} on a\nset of states of a DFA A = (K, Σ, δ, q0, F) are said to separate the final states\nof A if there exist indices i1, . . . , ir and j1, . . . , js such that it holds (Ri1 ∪. . . ∪\nRir) ∩(Sj1 ∪. . . ∪Sjs) = F.\n3\nRelations Between Types of Decompositions\nThe concept of partitions separating the final states allows us to derive a neces-\nsary and sufficient condition for the existence of SB- and ASB-decompositions\nsimilar to the one stated in [5].\nTheorem 3.1. A DFA A = (K, Σ, δ, q0, F) has a nontrivial SB-decomposition\niffthere exist two nontrivial S.P. partitions π1 and π2 on the set of states of A\nsuch that π1 · π2 = 0. This decomposition is an ASB-decomposition if and only\nif these partitions separate the final states of A.\nProof. The proof is analogous to that in [5] but had to be extended for the\nASB-decomposition. We omit it due to space constraints.\n4"},{"page":5,"text":"For the other decompositions, we can derive the following sufficient condi-\ntions that exploit the concept of S.P. partitions.\nTheorem 3.2. Let A = (K, Σ, δ, q0, F) be a deterministic finite automaton, let\nπ1 and π2 be nontrivial S.P. partitions on the set of states of A, such that they\nseparate the final states of A. Then A has a nontrivial AI-decomposition.\nProof. Since π1 and π2 separate the final states of A, there exist blocks B1, . . . , Bk\nand C1, . . . , Cl of the partitions π1 and π2 respectively, such that (B1 ∪. . . ∪\nBk) ∩(C1 ∪. . . ∪Cl) = F. We shall construct two automata A1 and A2 having\nstates corresponding to blocks of these partitions and show that (A1, A2) is a\nnontrivial AI-decomposition of A. Let A1 = (π1, Σ, δ1, [q0]π1, {B1, . . . , Bk}) and\nA2 = (π2, Σ, δ2, [q0]π2, {C1, . . . , Cl}) be DFAs with δi defined by δi([q]πi, a) =\n[δ(q, a)]πi, i ∈{1, 2} (this definition does not depend on the choice of q since πi\nis an S.P. partition). We now need to prove that L(A) = L(A1) ∩L(A2).\nLet w ∈L(A). Suppose that the computation of A on the word w ends\nin some accepting state qf ∈F. Then, from the construction of A1 and A2 it\nfollows that the computation of Ai on the word w ends in the state corresponding\nto the block [qf]πi of the partition πi. Since qf ∈F, it must hold [qf]π1 ∈\n{B1, . . . , Bk} and [qf]π2 ∈{C1, . . . , Cl}, hence from the construction of Ai,\nthese blocks correspond to the accepting states in the respective automata.\nThus w ∈L(Ai) for i ∈{1, 2}, therefore L(A) ⊆L(A1) ∩L(A2).\nNow suppose w ∈L(A1) ∩L(A2), Thus the computation of A1 on w ends\nin one of the states B1, . . . , Bk, which means that the computation of A on w\nwould end in a state from the union of blocks B1 ∪. . . ∪Bk. Using the same\nargument for A2, we get that the computation of A on w would end in a state\nfrom C1 ∪. . . ∪Cl.\nSince (B1 ∪. . . ∪Bk) ∩(C1 ∪. . . ∪Cl) = F we obtain\nthat the computation of A ends in an accepting state, hence w ∈L(A) and\nL(A1) ∩L(A2) ⊆L(A).\nSince both partitions are nontrivial, so is the AI-decomposition obtained.\nTheorem 3.3. Let A = (K, Σ, δ, q0, F) be a deterministic finite automaton,\nlet π1 and π2 be nontrivial S.P. partitions on the set of states of A, such that\nπ1 · π2 ⪯{F, K −F}. Then A has a nontrivial wAI-decomposition.\nProof. We shall construct A1 and A2 corresponding to the S.P. partitions π1\nand π2 as follows: Ai = (πi, Σ, δi, [q0]πi, ∅), where δi([q]πi, a) = [δ(q, a)]πi and\ni ∈{1, 2}. To show that (A1, A2) is a wAI-decomposition of A, we define the\nrelation R ⊆π1 × π2 by the equivalence R(D1, D2) ⇔(D1 ∩D2 ⊆F),where\nDi is some block of the partition πi. Now we need to prove that ∀w ∈Σ∗;\nw ∈L(A) ⇔R(δ1([q0]π1, w), δ2([q0]π2, w)).\nLet the computation of A on w end in some state p ∈K. It follows that the\ncomputation of Ai on the word w ends in the state corresponding to the block\n[p]πi, i ∈{1, 2}. Thus R(δ1([q0]π1, w), δ2([q0]π2, w)) ⇔R([p]π1, [p]π2) and by the\ndefinition of R, we have R(δ1([q0]π1, w), δ2([q0]π2, w)) ⇔[p]π1 ∩[p]π2 ⊆F. Since\np ∈[p]π1 ∩[p]π2, [p]π1 ∩[p]π2 is a block of the partition π1 · π2 and π1 · π2 ⪯\n{F, K −F}, it must hold that either [p]π1 ∩[p]π2 ⊆F or [p]π1 ∩[p]π2 ⊆K −F.\nTherefore R(δ1([q0]π1, w), δ2([q0]π2, w)) ⇔p ∈F and the proof is complete.\nIt follows directly from the definitions, that each SI-decomposition is also a\nwAI-decomposition, and so is each AI-decomposition. Also, each ASB-decomposition\nis an AI-decomposition, which is a consequence of the definition of acceptance\n5"},{"page":6,"text":"and state behavior realization. For minimal automata, a relationship between\nAI- and SI-decompositions can be obtained.\nTheorem 3.4. Let A = (K, Σ, δ, q0, F) be a minimal DFA, let (A1, A2) be its\nAI-decomposition. Then (A1, A2) is also an SI-decomposition of A.\nProof. Since (A1, A2) is an AI-decomposition of A, L(A) = L(A1) ∩L(A2).\nTherefore if we use the well-known Cartesian product construction, we obtain\nthe automaton A1||A2 such that L(A1||A2) = L(A). Since A is the minimal\nautomaton accepting the language L(A), there exists a mapping β : K′ →K\nsuch that it holds (∀w ∈Σ∗);\nβ(δ′(q′\n0, w)) = δ(β(q′\n0), w), where δ′ is the\ntransition function of A1||A2, K′ is its set of states and q′\n0 is its initial state.\nSince A1||A2 is a parallel connection (i.e., K′ = K1 ×K2, q′\n0 is the pair of initial\nstates of A1 and A2), it is easy to see that β is in fact exactly the mapping\nrequired by the definition of the SI-decomposition.\nThe ASB-decomposition is a combination of the SB-decomposition and the\nAI-decomposition, as the next theorem shows.\nTheorem 3.5. Let A be a DFA without unreachable states.\n(A1, A2) is an\nASB-decomposition of A iff(A1, A2) is both an SB-decomposition and an AI-\ndecomposition of A.\nProof. The first implication clearly follows from the definitions, Theorem 3.1\nand Theorem 3.2. Now let (A1, A2) be an SB- and AI-decomposition of A =\n(K, Σ, δ, q0, F). Let α be the mapping given by the definition of SB-decomposition.\nWe need to prove that for all states q of A, q ∈F iffα(q) ∈F1 × F2, where Fi is\nthe set of accepting states of Ai, i ∈{1, 2}. Let q ∈K and let w be a word such\nthat δ(q0, w) = q. Then q ∈F ⇔w ∈L(A) ⇔w ∈L(A1) ∩L(A2) ⇔α(q) ∈\nF1 × F2, where the first equivalence is implied by the choice of w, the second\nholds because (A1, A2) is an AI-decomposition and the third is a consequence\nof the properties of α guaranteed by the SB-decomposition definition.\nThere is also a relationship between SB- and SI-decompositions, in fact SB-\nis a stronger version of the state-identifying decomposition, as the following two\ntheorems show. We need the notion of reachability on pairs of states.\nDefinition 3.1. Let A1 = (K1, Σ, δ1, p1, F1) and A2 = (K2, Σ, δ2, p2, F2) be\nDFAs. We shall call a pair of states (q, r) ∈K1 × K2 reachable, if there exists\na word w ∈Σ∗such that δ1(p1, w) = q and δ2(p2, w) = r.\nTheorem 3.6. Let A = (K, Σ, δ, q0, F) be a DFA and let (A1, A2) be its SB-\ndecomposition. Then (A1, A2) also forms an SI-decomposition of A.\nProof. Let Ai = (Ki, Σ, δi, qi, Fi), i ∈{1, 2}. Since (A1, A2) is an SB-decomposition\nof A, there exists an injective mapping α: K →K1 × K2 such that it holds\nα(q0) = (q1, q2) and (∀a ∈Σ)(∀p ∈K); α(δ(p, a)) = (δ1(p1, a), δ2(p2, a)), where\nα(p) = (p1, p2). Let us define a new mapping β : K1 × K2 →K by\nβ(p1, p2) =\n \np\nif ∃p ∈K, α(p) = (p1, p2)\nq0\notherwise.\n(1)\nSince α is injective, there exists at most one such p and this definition is correct.\n6"},{"page":7,"text":"We now need to prove that β satisfies the condition from the definition of\nSI-decomposition, i.e., that (∀w ∈Σ∗); β(δ1(q1, w), δ2(q2, w)) = δ(q0, w). Since\nα(q0) = (q1, q2) and all the pairs of states we encounter in the computation of\nA1||A2 are thus reachable, this follows from the definition of α and (1) by an\neasy induction.\nLemma 3.7. Let A be a DFA without unreachable states and let (A1, A2) be\nits SI-decomposition, with β being the corresponding mapping. Then (A1, A2) is\nan SB-decomposition of A if and only if β is injective on all reachable pairs of\nstates.\nProof. Let (A1, A2) be an SB-decomposition of A. It clearly follows from Def-\ninition 2.2, that the corresponding β satisfies the equation (1) in the proof of\nTheorem 3.6 on all reachable pairs of states. Since the mapping α is a bijection\nbetween the set of states of A and the set of all reachable pairs of states of A1\nand A2, β defined as its inverse on the set of reachable pairs of states will be\ninjective on this set.\nFor the other implication, let (A1, A2) be an SI-decomposition of A and let\nβ be injective on the set of reachable pairs of states, let βr denote the mapping\nβ restricted onto the set of all reachable pairs of states of A1, A2. Since A has\nno unreachable states, βr is also surjective, thus we can define a new mapping\nα: K →K1 × K2 by the equation α(q) = β−1\nr (q). Since β maps the initial state\nonto the initial state, so does α, and since β satisfies the condition from the\nDefinition 2.2, it implies that also α satisfies the condition (i) from the definition\nof realization of state behavior. Therefore (A1, A2) is an SB-decomposition of\nA, with the corresponding mapping α.\nThe converse of Theorem 3.6 does not hold. The minimal automaton for the\nlanguage L = {a4kb4l|k ≥0, l ≥1} gives a counterexample. Inspecting its S.P.\npartitions shows that it has no nontrivial SB-decomposition, but it can be AI-\ndecomposed into minimal automata for languages L1 = {a4kbl|k ≥0, l ≥1} and\nL2 = {w|#b(w) = 4l; l ≥0}. According to Theorem 3.4, this AI-decomposition\nis also state-identifying.\nEach ASB-decomposition is obviously also an SB-decomposition. On the\nother hand, there exist SB-decomposable automata, that are ASB-undecomposable.\nFor example, the minimal automaton for the language\nL1\n=\n{w ∈{a, b, c}∗|#a(w)\nmod 3 = 0 ∧#b(w)\nmod 5 = 0} ∪\n∪\n{w ∈{a, b, c}∗|#a(w)\nmod 3 = 2 ∧#b(w)\nmod 5 = 4}\nhas this property, because the corresponding S.P. partitions on the set of its\nstates do not separate the final states in the sense of Definition 2.7.\nIt is also not so difficult to see that for any non-minimal automaton A without\nunreachable states, there exists a nontrivial AI- and wAI-decomposition (A1, A2)\nsuch that A1 is the minimal automaton equivalent to A and A2 has only one\nstate. This decomposition is obviously not state-identifying.\nFigure 1 summarizes all the relationships among the decomposition types\nthat we have shown so far.\nNow we show that for the case of so-called perfect decompositions, some of\nthe types of decomposition mentioned coincide.\n7"},{"page":8,"text":"ASB\n{vvvvvvvvv\n#G\nG\nG\nG\nG\nG\nG\nG\nAI\n×\nv\nv\nv\nv\n;v\nv\nv\nv\nmin\nH\nH\nH\nH\n#H\nH\nH\nH\n \n×\n#\nSB\n{wwwwwwww\n×GGGG\ncGGGG\nSI\n{vvvvvvvvv\n×\nwwww\n;wwww\nwAI\n×\n;\nDescription:\nA\n/ B : every A-decomposition is also a B-\ndecomposition\nA\n×\n/ B : not every A-decomposition is also\na B-decomposition\nA\n×\n/ B : there exists a DFA that has a non-\ntrivial A-decomposition but does not have\na nontrivial B-decomposition\nFigure 1: Relationships between decomposition types of DFA\nDefinition 3.2. Let t be a type of decomposition, t ∈{ASB, SB, AI, SI, wAI}.\nLet A be a DFA having n states, let A1 and A2 be DFAs having k and l states,\nrespectively. We shall call the pair (A1, A2) a perfect t-decomposition of A, if\nit forms a t-decomposition of A and n = k · l.\nTheorem 3.8. Let A be a DFA with no unreachable states and let (A1, A2) be a\npair of DFAs. Then (A1, A2) forms a perfect SI-decomposition of A iff(A1, A2)\nforms a perfect SB-decomposition of A.\nProof. One of the implications is a consequence of Theorem 3.6.\nAs to the\nsecond one, since (A1, A2) forms a perfect SI-decomposition of A, each of the\npairs of states of A1 and A2 is reachable and each pair has to correspond to a\ndifferent state of A in the mapping β, therefore β is bijective and the theorem\nfollows from Theorem 3.7.\nCorollary 3.9. Let A be a minimal DFA and let (A1, A2) be a pair of DFAs.\nThen (A1, A2) forms a perfect AI-decomposition of A iff(A1, A2) forms a perfect\nASB-decomposition of A.\nProof. The claim follows from Theorem 3.5, Theorem 3.4 and Theorem 3.8.\nAs a consequence of these facts, we can use the necessary and sufficient\nconditions stated in Theorem 3.1 to look for perfect AI- and SI-decompositions.\nNow, let us inspect the relationship between decompositions of an automaton\nand the decompositions of the corresponding minimal automaton.\nTheorem 3.10. Let A = (K, Σ, δ, q0, F) be a DFA and let Amin be a min-\nimal DFA such that L(A) = L(Amin).\nLet (A1, A2) be an SI-decomposition\n(AI-decomposition, wAI-decomposition) of A, then (A1, A2) also forms a de-\ncomposition of Amin of the same type.\nProof. First, note that this theorem does not state that any of the decomposi-\ntions is nontrivial. To prove the statement for SI-decompositions, suppose that\n(A1, A2) is an SI-decomposition of A, thus there exists a mapping α: K1×K2 →\nK such that it holds (∀w ∈Σ∗); α(δ1(q1, w), δ2(q2, w)) = δ(q0, w), where δi and\nqi are the transition function and the initial state of the automaton Ai. Since\nAmin is the minimal automaton corresponding to A, there exists some mapping\nβ : K →Kmin such that (∀w ∈Σ∗); β(δ(q0, w)) = δmin(β(q0), w), where δmin is\n8"},{"page":9,"text":"a1\nb\n/\na\n \nR\na0\nb\n/\na\nH\nb1\nb\n)\na\nO\nb0\nb\ni\na\nm\na1\nb\n/\na\n \nR1\nb\n* R0\nb\nj\na0\nb\n/\na\nH\nb1\nb\n)\na\nO\nb0\nb\ni\na\nO\nFigure 2: Transition functions of Amin and A′.\nthe transition function of Amin and Kmin is the set of states of Amin. By the com-\nposition of these mappings we obtain the mapping β◦α: K1×K2 →Kmin, which\ncombines A1 and A2 into Amin in the way that the definition of SI-decomposition\nrequires. For both the AI- and the wAI-decomposition, this statement is trivial,\nsince L(A) = L(Amin).\nBased on the above theorem it thus suffices to inspect the SI- (AI-, wAI-)\ndecomposability of the minimal automaton accepting a given language, and if\nwe show its undecomposability, we know that the recognition of this language\ncannot be simplified using an advisor of the respective type.\nHowever, this\ndoes not hold for SB- and ASB-decompositions, as exhibited by the following\nexample.\nExample 3.1. Let us consider the language L = {a2kb2l|k ≥0, l ≥1}. The\nminimal automaton Amin = (K, ΣL, δ, a0, {a0, b0}) has its transition function\ndefined by the first transition diagram in Fig.2. We can easily show that this\nautomaton does not have any nontrivial SB- (and thus neither ASB-) decompo-\nsition by enumerating its S.P. partitions.\nNow let us examine the automaton A′ = (K′, ΣL, δ′, a0, {a0, b0}) with the\ntransition function δ′ defined by the second transition diagram in Fig.2. Clearly,\nL(A′) = L(Amin), but by inspecting the lattice of S.P. partitions of A′, we can\nfind the pair π1 = {{a0}, {a1}, {b0, b1}, {R0, R1}} and π2 = {{a0, a1, b0, R0}, {b1, R1}}\nsuch that π1 · π2 = 0 and they separate the final states of A′. By Theorem 3.1\nwe can use these partitions to construct a nontrivial ASB- (and thus also SB-)\ndecomposition of A′ formed by the automata A1 and A2 having two and four\nstates, respectively. Note that both A1 and A2 have less states than Amin.\nIn the following theorem (inspired by a similar theorem in [5]) we state a\ncondition, under which the situation from the last example cannot occur, i.e.,\nunder which any SB-decomposition of a DFA implies a (maybe simpler) SB-\ndecomposition of the equivalent minimal DFA.\nTheorem 3.11. Let A = (K, Σ, δ, q0, F) be a deterministic finite automa-\nton and let Amin = (Kmin, Σ, δmin, qmin, Fmin) be the minimal DFA such that\nL(A) = L(Amin). Let (A1, A2) be a nontrivial SB-decomposition of A consisting\nof automata having k and l states. If the lattice of S.P. partitions of A is dis-\ntributive, then there exists an SB-decomposition of Amin consisting of automata\nhaving k′ and l′ states, such that k′ ≤k and l′ ≤l.\nProof. Since Amin is the minimal DFA such that L(A) = L(Amin), there exists\na mapping f : K →Kmin such that (∀w ∈Σ∗); f(δ(q0, w)) = δmin(qmin, w).\nUsing the mapping f, let us define a partition ρ on the set of states of A by\np ≡ρ q ⇔f(p) = f(q). Clearly, ρ is an S.P. partition.\n9"},{"page":10,"text":"Since (A1, A2) is a nontrivial SB-decomposition of A, we can use it to obtain\nS.P. partitions π1 and π2 on the set of states of A such that π1 · π2 = 0. Let us\ndefine new partitions π′\n1 and π′\n2 on the set of states of Amin by f(p) ≡π′\ni f(q) ⇔\np ≡ρ+πi q. Since it holds that ρ + πi ⪯ρ, this definition does not depend on\nthe choice of the states p and q. It holds that |π′\ni| = |ρ + πi| ≤|πi|, therefore if\nwe prove that π′\n1 and π′\n2 are S.P. partitions and π′\n1 · π′\n2 = 0, we can use them to\nconstruct the desired decomposition.\nThe fact that π′\ni is an S.P. partition on the set of states of Amin is a trivial\nconsequence of the fact that ρ + πi is an S.P. partition on the set of states of\nA. We need to prove that π′\n1 · π′\n2 = 0. Let us assume that p′ and q′ are states\nof Amin such that p′ ≡π′\n1·π′\n2 q′ and p, q are some states of A such that f(p) = p′\nand f(q) = q′. Then p′ ≡π′\n1 q′ and p′ ≡π′\n2 q′, and by definition of π′\ni we get\np ≡ρ+π1 q and p ≡ρ+π2 q, which is equivalent to p ≡(ρ+π1)·(ρ+π2) q. Since the\nlattice of all S.P. partitions of A is distributive, we have (ρ + π1) · (ρ + π2) =\nρ + (π1 · π2) = ρ + 0 = ρ, therefore p ≡ρ q, which by definition of ρ implies that\nf(p) = f(q), in other words p′ = q′. Hence π′\n1 · π′\n2 = 0.\n4\nDegrees of Decomposability\nIt is easy to see that for each type of decomposition, there exist undecomposable\nregular languages (e.g. L(n) = {ak|k ≥n −1} is wAI-undecomposable for each\nn ∈N). There also exist regular languages, that are perfectly decomposable in\neach way (e.g. L(k,l) = {w ∈{a, b}∗|#a(w) mod k = 0 ∧#b(w) mod l = 0} has\na perfect ASB-decomposition for all k, l ≥2). We shall now investigate whether\nall values between these two limits can be achieved.\nDefinition 4.1. Let A be a DFA, let (A1, A2) be its nontrivial SB- (ASB-)\ndecomposition with the corresponding S.P. partitions π1 and π2. We shall call\nthis decomposition redundant, if there exist S.P. partitions π′\n1 ⪰π1 and π′\n2 ⪰π2\nsuch that at least one of these inequalities is strict, but it still holds π′\n1 · π′\n2 = 0\n(and π′\n1 and π′\n2 separate the final states of A).\nLemma 4.1. For each r, s ∈N, r, s ≥2, there exists a minimal DFA A consist-\ning of r.s states and having only one nontrivial nonredundant SB-decomposition\n(ASB-decomposition) up to the order of automata, consisting of automata hav-\ning r and s states.\nProof. Let us study the minimal automaton Ar,s = (K, Σ, δ, q0,0, F) defined\nby K = {qi,j|i ∈{0, . . . , r −1}, j ∈{0, . . ., s −1}}, F = {qr−1,s−1} and the\ntransition function δ defined by\nδ(qi,j, a)\n=\nqi+1,j for i ∈{0, . . ., r −2}, j ∈{0, . . ., s −1}\nδ(qr−1,j, a)\n=\nqr−1,j for j ∈{0, . . . , s −1}\nδ(qi,j, b)\n=\nqi,j+1 for i ∈{0, . . ., r −1}, j ∈{0, . . ., s −2}\nδ(qi,s−1, b)\n=\nqi,s−1 for i ∈{0, . . . , r −1}.\nTo inspect the SB-decompositions of Ar,s, let us study the S.P. partitions on\nthe set of its states. From the method for generating all S.P. partitions of an\nautomaton that is described in [5], we know that each nontrivial S.P. partition\ncan be obtained as a sum of some partitions πm\np,t, where πm\np,t denotes the minimal\n10"},{"page":11,"text":"S.P. partition such that it does not distinguish between states p and t, i.e., they\nbelong into the same block. Let us determine πm\np,t for various states p and t of\nAr,s.\nFirst, let us consider the case of πm\np,t such that p = qi,j, t = qi′,j′ and both\ninequalities i < i′ and j < j′ hold. Since qi,j ≡π qi′,j′, δ(qi,j, ai′−ibj′−j) = qi′,j′\nand δ(qi′,j′, ai′−ibj′−j) = q2i′−i,2j′−j (if 2i′ −i < r and 2j′ −j < s), as a\nconsequence of the substitution property of π, we obtain qi,j ≡π q2i′−i,2j′−j.\nBy applying this argument a finite number of times (keeping in mind the con-\nstruction of Ar,s), we obtain qi,j ≡π qr−1,s−1. Now let k ∈{i, . . ., r −1} and\nlet l ∈{i, . . . , s −1}.\nThen δ(qi,j, ak−ibl−j) = qk,l and δ(qi′,j′, ak−ibl−j) =\nqk+i′−i,l+j′−j (if such states exist), therefore qk,l ≡π qk+i′−i,l+j′−j. Again, we\ncan use the same argument to show that qk,l ≡π qr−1,s−1. Therefore, for this\ntype of π = πm\np,t, we have qk,l ≡π qk′,l′ for all k, l, k′, l′ such that i ≤k, k′ < r\nand j ≤l, l′ < s.\nNow let us consider the case of πm\np,t such that p = qi,j, t = qi′,j′ and it holds\ni > i′ and j < j′. Since qi,j ≡π qi′,j′, δ(qi,j, ar−1−ibs−1−j′) = qr−1,s−1−(j′−j)\nand δ(qi′,j′, ar−1−ibs−1−j′) = qr−1−(i−i′),s−1, as a consequence of the substitu-\ntion property of π, we have qr−1,s−1−(j′−j) ≡π qr−1−(i−i′),s−1. By exploiting\nthe substitution property again on this equivalence, using the words ai−i′−1,\nbj′−j−1 and bj′−j, we obtain qr−2,s−1 ≡π qr−1,s−1 ≡π qr−2,s−2. Therefore in\nthis case, no such πm\np,t partition can distinguish between states qr−2,s−1, qr−1,s−1\nand qr−2,s−2.\nThe last case to consider is the case of πm\np,t such that p = qi,j, t = qi′,j′ and\nit holds i = i′ (the case j = j′ is analogous). Without loss of generality, we\ncan assume that j < j′. Now, using the same arguments as in the first case, we\ncan show that qi,l ≡π qi,l′ for all l, l′ such that j ≤l, l′ < s. Therefore for each\ngiven k such that i ≤k < r, it holds that qk,l ≡π qk,l′ and all of the states not\nmentioned in this equivalence form separate blocks of πm\np,t.\nIt is easy to verify that one nontrivial ASB-decomposition of Ar,s is given\nby the S.P. partitions\nπ1\n=\n{{q0,0, . . . , q0,s−1}, {q1,0, . . . , q1,s−1}, . . . , {qr−1,0, . . . , qr−1,s−1}} and\nπ2\n=\n{{q0,0, . . . , qr−1,0}, {q0,1, . . . , qr−1,1}, . . . , {q0,s−1, . . . , qr−1,s−1}}\nNow we show that any other SB-decomposition of Ar,s is given by S.P. partitions\npreceding to π1 and π2 in the partial order ⪯and therefore is redundant.\nIndeed, notice that none of the πm\np,t partitions of the first and the second\ndiscussed type can distinguish between any of the states qr−2,s−1, qr−1,s−1 and\nqr−2,s−2, therefore no sum of them can, either. For the partitions of the third\ntype, it holds either qr−2,s−1 ≡π qr−1,s−1 or qr−1,s−1 ≡π qr−2,s−2, therefore it\nwill take two partitions to distinguish between these three states. Hence any\nnontrivial SB-decomposition is determined by two S.P. partitions, both of which\nmust be of the third type. But it is easy to see that for any partition π of this\ntype it holds either π ⪯π1 or π ⪯π2.\nDefinition 4.2. Let A = (K, Σ, δ, q0, F) be a deterministic finite automaton,\nlet K ∩{p0, p1, . . . , pk−1} = ∅and let c be a new symbol not included in Σ. We\nshall define a k-extension A′ of the automaton A by the following construction:\nA′ = (K ∪{p0, p1, . . . , pk−1}, Σ ∪{c}, δ′, p0, F), where the transition function δ′\n11"},{"page":12,"text":"is defined as follows:\n(∀q ∈K) (∀a ∈Σ);\nδ′(q, a)\n=\nδ(q, a)\n(∀q ∈K);\nδ′(q, c)\n=\nq\n(∀p ∈{p0, p1, . . . , pk−1}) (∀a ∈Σ);\nδ′(p, a)\n=\np\n(∀i ∈{0, 1, . . ., k −2});\nδ′(pi, c)\n=\npi+1\nδ′(pk−1, c)\n=\nq0.\nLemma 4.2. Let A be a DFA consisting of n states, all of which are reach-\nable. Let A′ be its k-extension. Then A has a nontrivial nonredundant SB-\ndecomposition (ASB-decomposition) consisting of automata having r and s states\niffA′ has a nontrivial nonredundant decomposition of the same type, consisting\nof automata having k + r and k + s states.\nProof. We will try to inspect S.P. partitions on the set of states of A′, using\nthe notation from Definition 4.2. Let us assume that π′ is an S.P. partition\non the set of states of A′ such that pi and pj are in the same block of π′;\ni, j ∈{0, 1, . . ., k −1}. As a consequence of the S.P. property, if i, j < k −1\nthen also pi+1 and pj+1 are in the same block of π′, because δ′(pi, c) = pi+1\nand δ′(pj, c) = pj+1. By applying this argument a finite number of times, we\ncan show that there exists some l ∈{0, 1, . . ., k −2} such that pl ≡π′ pk−1,\nand using the argument once more, we obtain pl+1 ≡π′ q0. However, it holds\nδ′(pl, a) = pl for all a ∈Σ, hence pl ≡π′ δ′(q0, w) for all w ∈Σ∗. Since all of the\nstates of A′ are reachable, we have pl ≡π′ q for all q ∈K. Thus such a partition\ncannot distinguish between the original states of the automaton A.\nNow let us suppose that π′ is an S.P. partition on the set of states of A′ such\nthat for some i ∈{0, 1, . . ., k −1}, pi ≡π′ q for some q in K. Then it also holds\nthat pi ≡π′ pi+1, because δ(pi, c) = pi+1 and δ(q, c) = q. But we have already\nshown that pi ≡π′ pi+1 implies that all of the states in K are equivalent modulo\nπ′, thus this S.P. partition cannot distinguish between the states of A, either.\n¿From these observations it follows that if π′ is any S.P. partition on the set\nof states of A′ such that the states of A are not all equivalent modulo π′, then\nπ′ must also contain k blocks, each of which contains only one state pi, where\ni ∈{0, 1, . . ., k −1}. Now we can prove the equivalence stated in the theorem.\nLet A have an SB-decomposition consisting of r and s states. Then there\nexist S.P. partitions π1 and π2 on the set of states of A having r and s blocks,\nsuch that π1 · π2 = 0.\nLet us now construct new partitions π′\n1 and π′\n2 on\nthe set of states of A′ by π′\n1 = π1 ∪{{p0}, {p1}, . . . , {pk−1}} and π′\n2 = π2 ∪\n{{p0}, {p1}, . . . , {pk−1}}. Obviously, π′\n1 and π′\n2 have substitution property, be-\ncause for the states in K this property is inherited from π1 and π2, and the new\nstates p0, p1, . . . , pk−1 cannot violate this property either, because each of these\nstates belongs to a separate block in π′\n1 and π′\n2, making the substitution prop-\nerty hold trivially. Neither do the new c-moves defined on the states from K\nviolate the substitution property. Finally, it holds that π′\n1 · π′\n2 = 0. To see this,\nnote that for a state q ∈K, it holds [q]π′\n1·π′\n2 = [q]π1·π2 = {q}, since π1 · π2 = 0.\nFor a state q ∈K′ −K, [q]π′\ni = {q} for i ∈{1, 2} thus [q]π′\n1·π′\n2 = {q}, too. Hence\neach state of A′ belongs to a separate block of π′\n1 · π′\n2, which implies π′\n1 · π′\n2 = 0.\nTherefore π′\n1 and π′\n2 induce an SB-decomposition of A′. It is also easy to see\nthat if π1 and π2 separate the final states of A, then also π′\n1 and π′\n2 separate the\nfinal states of A′, making the induced decomposition an ASB-decomposition.\n12"},{"page":13,"text":"On the other hand, let us now assume that A′ has an SB-decomposition and\nπ′\n1 and π′\n2 are the S.P. partitions on K′ that induce this decomposition, thus\nπ′\n1 · π′\n2 = 0. From the observations made in the beginning of this proof, we know\nthat any S.P. partition that can distinguish between the states in K in any way,\nmust contain each of the states p0, p1 . . . pk−1 in a separate block containing only\nthis state. As π′\n1 ·π′\n2 = 0, for all q1, q2 ∈K, at least one of these partitions must\ndistinguish between these states, i.e., [q1]π′\ni ̸= [q2]π′\ni. If one of the partitions\ndistinguished between all such pairs, it would imply that this partition must\ncontain a separate block for each one of the states in K′, thus becoming a trivial\npartition 0, resulting in a trivial decomposition. Therefore both π′\n1 and π′\n2 have\nto distinguish between some pair of states from K, which implies that they both\ncontain a separate block for each of the states p0, p1 . . . pk−1 containing no other\nstate. By removing these k blocks from π′\n1 and π′\n2, we obtain new partitions π1\nand π2 on the set K, such that π1 = π′\n1−{{p0}, {p1}, . . . , {pk−1}} and π2 = π′\n2−\n{{p0}, {p1}, . . . , {pk−1}}. These partitions preserve the substitution property,\nsince (∀a ∈Σ)(∀q ∈K): δ(q, a) ∈K and π′\n1 and π′\n2 were S.P. partitions. It also\nholds π1 ·π2 = 0, as for all q1, q2 ∈K, q1 ≡π1·π2 q2 implies q1 ≡π′\n1·π′\n2 q2 and that\nimplies q1 = q2. So π1 and π2 induce an SB-decomposition of A. As π′\n1 and π′\n2\nwere nontrivial, so are π1 and π2 and the obtained decomposition. It is again\neasy to see that if π′\n1 and π′\n2 separate the final states of A′, then also π1 and π2\nmust separate the final states of A.\nThe described relationship between the S.P. partitions on the set of states\nof A and the corresponding S.P. partitions on A′ also implies, that each de-\ncomposition of A is nonredundant iffthe corresponding decomposition of A′ is\nnonredundant, too.\nSince a k-extension of a minimal DFA is again a minimal DFA, we can\ncombine the lemmas to obtain the following theorem.\nTheorem 4.3. Let n ∈N be such that n = k + r.s, where r, s, k ∈N, r, s ≥2.\nThen there exists a minimal DFA A consisting of n states, such that it has only\none nontrivial nonredundant SB-decomposition (ASB-decomposition) up to the\norder of the automata in the decomposition, and this decomposition consists of\nautomata with k + r and k + s states.\nReferences\n[1] J. L. Balcazar, J. Diaz, J. Gabarro, Structural Complexity I., Springer-Verlag\nNew York, 1988\n[2] S. Even, A. L. Selman, Y. Yacobi, The Complexity of Promise Problems\nwith Applications to Public-Key Cryptography, Information and Control 61(2):\n159-173 (1984)\n[3] S. Yu, State Complexity: Recent Results and Open Problems, Fundamenta\nInformaticae 64: 471-480 (2005)\n[4] J. C. Birget, Intersection and Union of Regular Languages and State Com-\nplexity, Information Processing Letters 43: 185-190 (1992)\n[5] J. Hartmanis and R. E. Stearns, Algebraic Structure Theory of Sequential\nMachines, Prentice-Hall, 1966\n13"},{"page":14,"text":"[6] J. E. Hopcroft and J. D. Ullman, Introduction to Automata Theory, Lan-\nguages, and Computation, Addison-Wesley, 1979\n14"}]},"section_tree":[],"assets":{"figures":[],"tables":[],"images":[]},"math_expressions":{"inline":[],"block":[{"equation_id":"eq1","equation_number":null,"raw_text":"instead of looking for an automaton A such that L = L(A) we can look for a","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq2","equation_number":null,"raw_text":"(possibly simpler) automaton B such that L = L(B)∩L′. We can then say that","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq3","equation_number":null,"raw_text":"automaton A′ such that L′ = L(A′)) an advisor to B. In this case the advisor","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq4","equation_number":null,"raw_text":"Definition 2.1. A pair of DFAs (A1, A2), where A1 = (K1, Σ, δ1, q1, F1) and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq5","equation_number":null,"raw_text":"A2 = (K2, Σ, δ2, q2, F2), forms an acceptance-identifying decomposition (AI-","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq6","equation_number":null,"raw_text":"decomposition) of a DFA A = (K, Σ, δ, q0, F), if L(A) = L(A1) ∩L(A2). This","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq7","equation_number":null,"raw_text":"Definition 2.2. A pair of DFAs (A1, A2), where A1 = (K1, Σ, δ1, q1, F1) and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq8","equation_number":null,"raw_text":"A2 = (K2, Σ, δ2, q2, F2), forms a state-identifying decomposition (SI-decomposition)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq9","equation_number":null,"raw_text":"of a DFA A = (K, Σ, δ, q0, F), if there exists a mapping β : K1 × K2 →K, such","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq10","equation_number":null,"raw_text":"that it holds β(δ1(q1, w), δ2(q2, w)) = δ(q0, w) for all w ∈Σ∗. This decomposi-","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq11","equation_number":null,"raw_text":"Definition 2.3. A pair of DFAs (A1, A2), where A1 = (K1, Σ, δ1, q1, F1) and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq12","equation_number":null,"raw_text":"A2 = (K2, Σ, δ2, q2, F2), forms a weak acceptance-identifying decomposition","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq13","equation_number":null,"raw_text":"(wAI-decomposition) of a DFA A = (K, Σ, δ, q0, F), if there exists a relation","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq14","equation_number":null,"raw_text":"Definition 2.4. A DFA A′ = (K′, Σ, δ′, q′","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq15","equation_number":null,"raw_text":"behavior of a DFA A = (K, Σ, δ, q0, F) if there exists an injective mapping","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq16","equation_number":null,"raw_text":"(i) (∀a ∈Σ)(∀q ∈K); δ′(α(q), a) = α(δ(q, a)),","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq17","equation_number":null,"raw_text":"(ii) α(q0) = q′","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq18","equation_number":null,"raw_text":"Definition 2.5. The parallel connection of two DFA A1 = (K1, Σ, δ1, q1, F1)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq19","equation_number":null,"raw_text":"and A2 = (K2, Σ, δ2, q2, F2) is the DFA A = A1||A2 = (K1 × K2, Σ, δ, (q1, q2),","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq20","equation_number":null,"raw_text":"F1 × F2) such that δ((p1, p2), a) = (δ1(p1, a), δ2(p2, a)).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq21","equation_number":null,"raw_text":"A partition π on a set of states of a DFA A = (K, Σ, δ, q0, F) has substitution","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq22","equation_number":null,"raw_text":"a = a0, a1, a2, . . . , an = b, such that ai ≡π1 ai+1 ∨ai ≡π2 ai+1 for all","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq23","equation_number":null,"raw_text":"Definition 2.7. The partitions π1 = {R1, . . . , Rk} and π2 = {S1, . . . , Sl} on a","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq24","equation_number":null,"raw_text":"set of states of a DFA A = (K, Σ, δ, q0, F) are said to separate the final states","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq25","equation_number":null,"raw_text":"Rir) ∩(Sj1 ∪. . . ∪Sjs) = F.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq26","equation_number":null,"raw_text":"Theorem 3.1. A DFA A = (K, Σ, δ, q0, F) has a nontrivial SB-decomposition","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq27","equation_number":null,"raw_text":"such that π1 · π2 = 0. This decomposition is an ASB-decomposition if and only","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq28","equation_number":null,"raw_text":"Theorem 3.2. Let A = (K, Σ, δ, q0, F) be a deterministic finite automaton, let","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq29","equation_number":null,"raw_text":"Bk) ∩(C1 ∪. . . ∪Cl) = F. We shall construct two automata A1 and A2 having","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq30","equation_number":null,"raw_text":"nontrivial AI-decomposition of A. Let A1 = (π1, Σ, δ1, [q0]π1, {B1, . . . , Bk}) and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq31","equation_number":null,"raw_text":"A2 = (π2, Σ, δ2, [q0]π2, {C1, . . . , Cl}) be DFAs with δi defined by δi([q]πi, a) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq32","equation_number":null,"raw_text":"is an S.P. partition). We now need to prove that L(A) = L(A1) ∩L(A2).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq33","equation_number":null,"raw_text":"Since (B1 ∪. . . ∪Bk) ∩(C1 ∪. . . ∪Cl) = F we obtain","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq34","equation_number":null,"raw_text":"Theorem 3.3. Let A = (K, Σ, δ, q0, F) be a deterministic finite automaton,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq35","equation_number":null,"raw_text":"and π2 as follows: Ai = (πi, Σ, δi, [q0]πi, ∅), where δi([q]πi, a) = [δ(q, a)]πi and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq36","equation_number":null,"raw_text":"Theorem 3.4. Let A = (K, Σ, δ, q0, F) be a minimal DFA, let (A1, A2) be its","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq37","equation_number":null,"raw_text":"Proof. Since (A1, A2) is an AI-decomposition of A, L(A) = L(A1) ∩L(A2).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq38","equation_number":null,"raw_text":"the automaton A1||A2 such that L(A1||A2) = L(A). Since A is the minimal","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq39","equation_number":null,"raw_text":"0, w)) = δ(β(q′","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq40","equation_number":null,"raw_text":"Since A1||A2 is a parallel connection (i.e., K′ = K1 ×K2, q′","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq41","equation_number":null,"raw_text":"and Theorem 3.2. Now let (A1, A2) be an SB- and AI-decomposition of A =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq42","equation_number":null,"raw_text":"that δ(q0, w) = q. Then q ∈F ⇔w ∈L(A) ⇔w ∈L(A1) ∩L(A2) ⇔α(q) ∈","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq43","equation_number":null,"raw_text":"Definition 3.1. Let A1 = (K1, Σ, δ1, p1, F1) and A2 = (K2, Σ, δ2, p2, F2) be","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq44","equation_number":null,"raw_text":"a word w ∈Σ∗such that δ1(p1, w) = q and δ2(p2, w) = r.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq45","equation_number":null,"raw_text":"Theorem 3.6. Let A = (K, Σ, δ, q0, F) be a DFA and let (A1, A2) be its SB-","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq46","equation_number":null,"raw_text":"Proof. Let Ai = (Ki, Σ, δi, qi, Fi), i ∈{1, 2}. Since (A1, A2) is an SB-decomposition","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq47","equation_number":null,"raw_text":"α(q0) = (q1, q2) and (∀a ∈Σ)(∀p ∈K); α(δ(p, a)) = (δ1(p1, a), δ2(p2, a)), where","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq48","equation_number":null,"raw_text":"α(p) = (p1, p2). Let us define a new mapping β : K1 × K2 →K by","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq49","equation_number":null,"raw_text":"β(p1, p2) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq50","equation_number":null,"raw_text":"if ∃p ∈K, α(p) = (p1, p2)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq51","equation_number":null,"raw_text":"SI-decomposition, i.e., that (∀w ∈Σ∗); β(δ1(q1, w), δ2(q2, w)) = δ(q0, w). Since","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq52","equation_number":null,"raw_text":"α(q0) = (q1, q2) and all the pairs of states we encounter in the computation of","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq53","equation_number":null,"raw_text":"α: K →K1 × K2 by the equation α(q) = β−1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq54","equation_number":null,"raw_text":"language L = {a4kb4l|k ≥0, l ≥1} gives a counterexample. Inspecting its S.P.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq55","equation_number":null,"raw_text":"decomposed into minimal automata for languages L1 = {a4kbl|k ≥0, l ≥1} and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq56","equation_number":null,"raw_text":"L2 = {w|#b(w) = 4l; l ≥0}. According to Theorem 3.4, this AI-decomposition","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq57","equation_number":null,"raw_text":"mod 3 = 0 ∧#b(w)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq58","equation_number":null,"raw_text":"mod 5 = 0} ∪","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq59","equation_number":null,"raw_text":"mod 3 = 2 ∧#b(w)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq60","equation_number":null,"raw_text":"mod 5 = 4}","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq61","equation_number":null,"raw_text":"it forms a t-decomposition of A and n = k · l.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq62","equation_number":null,"raw_text":"Theorem 3.10. Let A = (K, Σ, δ, q0, F) be a DFA and let Amin be a min-","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq63","equation_number":null,"raw_text":"imal DFA such that L(A) = L(Amin).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq64","equation_number":null,"raw_text":"K such that it holds (∀w ∈Σ∗); α(δ1(q1, w), δ2(q2, w)) = δ(q0, w), where δi and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq65","equation_number":null,"raw_text":"β : K →Kmin such that (∀w ∈Σ∗); β(δ(q0, w)) = δmin(β(q0), w), where δmin is","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq66","equation_number":null,"raw_text":"since L(A) = L(Amin).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq67","equation_number":null,"raw_text":"Example 3.1. Let us consider the language L = {a2kb2l|k ≥0, l ≥1}. The","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq68","equation_number":null,"raw_text":"minimal automaton Amin = (K, ΣL, δ, a0, {a0, b0}) has its transition function","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq69","equation_number":null,"raw_text":"Now let us examine the automaton A′ = (K′, ΣL, δ′, a0, {a0, b0}) with the","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq70","equation_number":null,"raw_text":"L(A′) = L(Amin), but by inspecting the lattice of S.P. partitions of A′, we can","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq71","equation_number":null,"raw_text":"find the pair π1 = {{a0}, {a1}, {b0, b1}, {R0, R1}} and π2 = {{a0, a1, b0, R0}, {b1, R1}}","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq72","equation_number":null,"raw_text":"such that π1 · π2 = 0 and they separate the final states of A′. By Theorem 3.1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq73","equation_number":null,"raw_text":"Theorem 3.11. Let A = (K, Σ, δ, q0, F) be a deterministic finite automa-","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq74","equation_number":null,"raw_text":"ton and let Amin = (Kmin, Σ, δmin, qmin, Fmin) be the minimal DFA such that","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq75","equation_number":null,"raw_text":"L(A) = L(Amin). Let (A1, A2) be a nontrivial SB-decomposition of A consisting","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq76","equation_number":null,"raw_text":"Proof. Since Amin is the minimal DFA such that L(A) = L(Amin), there exists","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq77","equation_number":null,"raw_text":"a mapping f : K →Kmin such that (∀w ∈Σ∗); f(δ(q0, w)) = δmin(qmin, w).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq78","equation_number":null,"raw_text":"p ≡ρ q ⇔f(p) = f(q). Clearly, ρ is an S.P. partition.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq79","equation_number":null,"raw_text":"S.P. partitions π1 and π2 on the set of states of A such that π1 · π2 = 0. Let us","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq80","equation_number":null,"raw_text":"i| = |ρ + πi| ≤|πi|, therefore if","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq81","equation_number":null,"raw_text":"2 = 0, we can use them to","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq82","equation_number":null,"raw_text":"2 = 0. Let us assume that p′ and q′ are states","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq83","equation_number":null,"raw_text":"2 q′ and p, q are some states of A such that f(p) = p′","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq84","equation_number":null,"raw_text":"and f(q) = q′. Then p′ ≡π′","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq85","equation_number":null,"raw_text":"lattice of all S.P. partitions of A is distributive, we have (ρ + π1) · (ρ + π2) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq86","equation_number":null,"raw_text":"ρ + (π1 · π2) = ρ + 0 = ρ, therefore p ≡ρ q, which by definition of ρ implies that","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq87","equation_number":null,"raw_text":"f(p) = f(q), in other words p′ = q′. Hence π′","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq88","equation_number":null,"raw_text":"regular languages (e.g. L(n) = {ak|k ≥n −1} is wAI-undecomposable for each","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq89","equation_number":null,"raw_text":"each way (e.g. L(k,l) = {w ∈{a, b}∗|#a(w) mod k = 0 ∧#b(w) mod l = 0} has","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq90","equation_number":null,"raw_text":"Proof. Let us study the minimal automaton Ar,s = (K, Σ, δ, q0,0, F) defined","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq91","equation_number":null,"raw_text":"by K = {qi,j|i ∈{0, . . . , r −1}, j ∈{0, . . ., s −1}}, F = {qr−1,s−1} and the","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq92","equation_number":null,"raw_text":"p,t such that p = qi,j, t = qi′,j′ and both","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq93","equation_number":null,"raw_text":"inequalities i < i′ and j < j′ hold. Since qi,j ≡π qi′,j′, δ(qi,j, ai′−ibj′−j) = qi′,j′","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq94","equation_number":null,"raw_text":"and δ(qi′,j′, ai′−ibj′−j) = q2i′−i,2j′−j (if 2i′ −i < r and 2j′ −j < s), as a","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq95","equation_number":null,"raw_text":"Then δ(qi,j, ak−ibl−j) = qk,l and δ(qi′,j′, ak−ibl−j) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq96","equation_number":null,"raw_text":"type of π = πm","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq97","equation_number":null,"raw_text":"p,t such that p = qi,j, t = qi′,j′ and it holds","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq98","equation_number":null,"raw_text":"i > i′ and j < j′. Since qi,j ≡π qi′,j′, δ(qi,j, ar−1−ibs−1−j′) = qr−1,s−1−(j′−j)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq99","equation_number":null,"raw_text":"and δ(qi′,j′, ar−1−ibs−1−j′) = qr−1−(i−i′),s−1, as a consequence of the substitu-","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq100","equation_number":null,"raw_text":"p,t such that p = qi,j, t = qi′,j′ and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq101","equation_number":null,"raw_text":"it holds i = i′ (the case j = j′ is analogous). Without loss of generality, we","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq102","equation_number":null,"raw_text":"Definition 4.2. Let A = (K, Σ, δ, q0, F) be a deterministic finite automaton,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq103","equation_number":null,"raw_text":"let K ∩{p0, p1, . . . , pk−1} = ∅and let c be a new symbol not included in Σ. We","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq104","equation_number":null,"raw_text":"A′ = (K ∪{p0, p1, . . . , pk−1}, Σ ∪{c}, δ′, p0, F), where the transition function δ′","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq105","equation_number":null,"raw_text":"then also pi+1 and pj+1 are in the same block of π′, because δ′(pi, c) = pi+1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq106","equation_number":null,"raw_text":"and δ′(pj, c) = pj+1. By applying this argument a finite number of times, we","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq107","equation_number":null,"raw_text":"δ′(pl, a) = pl for all a ∈Σ, hence pl ≡π′ δ′(q0, w) for all w ∈Σ∗. Since all of the","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq108","equation_number":null,"raw_text":"that pi ≡π′ pi+1, because δ(pi, c) = pi+1 and δ(q, c) = q. But we have already","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq109","equation_number":null,"raw_text":"such that π1 · π2 = 0.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq110","equation_number":null,"raw_text":"1 = π1 ∪{{p0}, {p1}, . . . , {pk−1}} and π′","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq111","equation_number":null,"raw_text":"2 = 0. To see this,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq112","equation_number":null,"raw_text":"2 = [q]π1·π2 = {q}, since π1 · π2 = 0.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq113","equation_number":null,"raw_text":"i = {q} for i ∈{1, 2} thus [q]π′","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq114","equation_number":null,"raw_text":"2 = {q}, too. Hence","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq115","equation_number":null,"raw_text":"2 = 0. From the observations made in the beginning of this proof, we know","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq116","equation_number":null,"raw_text":"2 = 0, for all q1, q2 ∈K, at least one of these partitions must","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq117","equation_number":null,"raw_text":"i ̸= [q2]π′","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq118","equation_number":null,"raw_text":"and π2 on the set K, such that π1 = π′","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq119","equation_number":null,"raw_text":"1−{{p0}, {p1}, . . . , {pk−1}} and π2 = π′","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq120","equation_number":null,"raw_text":"holds π1 ·π2 = 0, as for all q1, q2 ∈K, q1 ≡π1·π2 q2 implies q1 ≡π′","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq121","equation_number":null,"raw_text":"implies q1 = q2. So π1 and π2 induce an SB-decomposition of A. As π′","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq122","equation_number":null,"raw_text":"Theorem 4.3. Let n ∈N be such that n = k + r.s, where r, s, k ∈N, r, s ≥2.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"}],"equation_refs":[]},"symbol_table":[],"references":{"in_text_markers":[],"bibliography":[]},"claims":[],"flow_graph":{"nodes":[],"edges":[]},"quality":{"text_char_count":39535,"parse_confidence":0.5,"equation_parse_rate_proxy":1.0,"citation_resolution_rate_proxy":0.0,"structure_coverage_proxy":0.0,"asset_coverage_proxy":0.0,"accepted_for_training":true}}