Algebraic and Combinatorial Tools for State Complexity : Application to the Star-Xor Problem
JJ. Leroux and J.-F. Raskin (Eds.): Tenth International Symposiumon Games, Automata, Logics, and Formal Verification (GandALF’19).EPTCS 305, 2019, pp. 154–168, doi:10.4204 / EPTCS.305.11 c (cid:13)
Caron, Hamel-de le Court & LuqueThis work is licensed under theCreative Commons Attribution License.
Algebraic and Combinatorial Tools for State Complexity :Application to the Star-Xor Problem
Pascal Caron Edwin Hamel-de-le-courtJean-Gabriel Luque
LITIS, Université de Rouen,Avenue de l’Université,76801 Saint-Étienne du Rouvray Cedex,France {Pascal.Caron, Edwin.Hamel-de-le-court, Jean-Gabriel.Luque}@univ-rouen.fr
We investigate the state complexity of the star of symmetrical di ff erences using modifiersand monsters. A monster is an automaton in which every function from states to states isrepresented by at least one letter. A modifier is a set of functions allowing one to transforma set of automata into one automaton. These recent theoretical concepts allow one to findeasily the desired state complexity. We then exhibit a witness with a constant size alphabet. The state complexity of a rational language is the size of its minimal automaton and the statecomplexity of a rational operation is the maximal one of those languages obtained by applyingthis operation onto languages of fixed state complexities.The classical approach is to compute an upper bound and to provide a witness, that is aspecific example reaching the bound which is then the desired state complexity.Since the 70 s , the state complexity of numerous unary and binary operations has beencomputed. See, for example, [9, 11, 14, 15, 16, 19] for a survey of the subject. More recently, thestate complexity of combinations of operations has also been studied. In most cases the resultis not simply the mathematical composition of the individual complexities and studies lead tointeresting situations. Examples can be found in [7, 12, 17, 18].In some cases, the classical method has to be enhanced by two independent approaches.The first one consists in describing states by combinatorial objects. Thus the upper bound iscomputed using combinatorial tools. For instance, in [4], the states are represented by tableauxrepresenting boolean matrices and an upper bound for the catenation of symmetrical di ff erenceis given. These combinatorial objects will be used to compute an upper bound for the Kleenestar of symmetrical di ff erence. The second one is an algebraic method consisting in buildinga witness for a certain class of rational operations by searching in a set of automata with asmany transition functions as possible. This method has the advantage of being applied to alarge class of operations, but has the drawback of giving witnesses that have alphabets of non-constant size. Witnesses with small alphabets are indeed favoured in this area of research whenthey can be found, as evidenced by several studies ([5, 6]). This approach has been describedindependently by Caron et al. in [3] as the monster approach and by Davies in [8] as the OLPA(One Letter Per Action) approach but was implicitly present in older papers like [2, 10]. . Caron, E. Hamel-de le Court & J.-G. Luque ff erence. Furthermore, we improve the witness found by drastically reducingthe size of its alphabet to a constant size.The paper is organized as follows. Section 2 gives definitions and notations about automataand combinatorics. In Section 3, we recall the monster approach : we define modifiers , monsters ,and give some properties of these structures related to state complexity. In Section 5, the statecomplexity of star of symmetrical di ff erence is computed. Hence, in Section 6, we find witnessesfor this operation with an alphabet size of 17. The cardinality of a finite set E is denoted by E , the set of subsets of E is denoted by 2 E and the set of mappings of E into itself is denoted by E E . The symmetric di ff erence of two sets E and E is denoted by ⊕ and defined by E ⊕ E = ( E ∪ E ) \ ( E ∩ E ). For any positive integer n , letus denote { , . . . , n − } by (cid:126) n (cid:127) . denotes the identity mapping, the set of which depends oncontext. Let Σ denote a finite alphabet. A word w over Σ is a finite sequence of symbols of Σ . The length of w , denoted by | w | , is the number of occurrences of symbols of Σ in w . For a ∈ Σ , we denoteby | w | a the number of occurrences of a in w . The set of all finite words over Σ is denoted by Σ ∗ .A language is a subset of Σ ∗ .A complete and deterministic finite automaton (DFA) is a 5-tuple A = ( Σ , Q , i , F , δ ) where Σ is theinput alphabet, Q is a finite set of states, i ∈ Q is the initial state, F ⊆ Q is the set of final statesand δ is the transition function from Q × Σ to Q extended in a natural way from Q × Σ ∗ to Q .The cardinality of A is the cardinality of its set of states, i.e. A = Q . We will often use (cid:126) n (cid:127) forsome n ∈ N as the set of states for DFAs.Let A = ( Σ , Q , i , F , δ ) be a DFA. A word w ∈ Σ ∗ is recognized by the DFA A if δ ( i , w ) ∈ F . The language recognized by a DFA A is the set L( A ) of words recognized by A . Two DFAs are said tobe equivalent if they recognize the same language.For any word w , we denote by δ w the function q → δ ( q , w ). Two states q , q of D are equivalent if for any word w of Σ ∗ , δ ( q , w ) ∈ F if and only if δ ( q , w ) ∈ F . This equivalence relation is calledthe Nerode equivalence and is denoted by q ∼ Ner q . If two states are not equivalent, then theyare called distinguishable .A state q is accessible in a DFA if there exists a word w ∈ Σ ∗ such that q = δ ( i , w ). A DFAis minimal if there does not exist any equivalent DFA with less states and it is well knownthat for any DFA, there exists a unique minimal equivalent one ([13]). Such a minimal DFAcan be obtained from D by computing (cid:98) A / ∼ = ( Σ , Q / ∼ , [ i ] , F / ∼ , δ ∼ ) where (cid:98) A is the accessiblepart of A , and where, for any q ∈ Q , [ q ] is the ∼ -class of the state q and satisfies the property δ ∼ ([ q ] , a ) = [ δ ( q , a )], for any a ∈ Σ . The number of its states is denoted by Min ( A ). In a minimalDFA, any two distinct states are pairwise distinguishable.Let L be a regular language defined over an alphabet Σ . We denote by L ∗ { w = u · · · u n | u i ∈ L ∧ n ∈ N } .56 Algebraic Tools for State Complexity
The syntactic semigroup of L is the semigroup generated by the transition functions of allletters of the minimal DFA of L . A unary regular operation is a function from regular languages into regular languages of Σ .A k-ary regular operation over the alphabet Σ is a function from the set of k -tuples of regularlanguages of Σ into regular languages of Σ .The state complexity of a regular language L denoted by sc( L ) is the number of states of itsminimal DFA. This notion extends to regular operations: the state complexity of a unaryregular operation ⊗ is the function sc ⊗ such that, for all n ∈ N \ { (cid:48) } , sc ⊗ ( n ) is the maximum of allthe state complexities of ⊗ ( L ) when L is of state complexity n , i.e. sc ⊗ ( n ) = max { sc( ⊗ ( L )) | sc( L ) = n } .This can be generalized, and the state complexity of a k -ary operation ⊗ is the k -ary functionsc ⊗ such that, for all ( n , . . . , n k ) ∈ ( N ∗ ) k ,sc ⊗ ( n , . . . , n k ) = max { sc( ⊗ ( L , . . . , L k )) | for all i ∈ { , . . . , k } , sc( L i ) = n i } . (1)Then, a witness for ⊗ is a a way to assign to each ( n , . . . , n k ), assumed su ffi ciently big, ak-tuple of languages ( L , . . . , L k ) with sc( L i ) = n i , for all i ∈ { , . . . , k } , satisfying sc ⊗ ( n , . . . , n k ) = sc( ⊗ ( L , . . . , L k )). Let Σ and Γ be two alphabets. A morphism is a function φ from Σ ∗ to Γ ∗ such that, for all w , v ∈ Σ ∗ , φ ( wv ) = φ ( w ) φ ( v ). Notice that φ is completely defined by its value on letters.Let L be a regular language over alphabet Σ recognized by the DFA A = ( Σ , Q , i , F , δ ) and let φ be a morphism from Γ ∗ to Σ ∗ . Then, φ − ( L ) is the regular language recognized by the DFA B = ( Γ , Q , i , F , δ (cid:48) ) where, for all a ∈ Γ and q ∈ Q , δ (cid:48) ( q , a ) = δ ( q , φ ( a )). Therefore, note that we have Property 1
Let L be a regular language and φ be a morphism. We have sc( φ − ( L )) ≤ sc( L ) . We say that a morphism φ is 1 -uniform if the image by φ of any letter is a letter. In otherwords, a 1-uniform morphism is a (not necessarily injective) renaming of the letters and theonly complexity of the mapping stems from mapping a and b to the same image, i.e., φ ( a ) = φ ( b ). In [1], Brzozowski gives a series of properties that would make a language L n of state complexity n su ffi ciently complex to be a good candidate for constructing witnesses for numerous classicalrational operations. One of these properties is that the size of the syntactic semigroup is n n , which means that each transformation of the minimal DFA of L n can be associated toa transformation by some non-empty word. This upper bound is reached when the set oftransition functions of the DFA is exactly the set of transformations from state to state. We thusconsider the set of transformations of (cid:126) n (cid:127) as an alphabet where each letter is simply named bythe transition function it defines. This leads to the following definition : Definition 1 A -monster is an automaton Mon Fn = ( Σ , (cid:126) n (cid:127) , , F , δ ) defined by • the alphabet Σ = (cid:126) n (cid:127) (cid:126) n (cid:127) , . Caron, E. Hamel-de le Court & J.-G. Luque • the set of states (cid:126) n (cid:127) , • the initial state , • the set of final states F, • the transition function δ defined for any a ∈ Σ by δ ( q , a ) = a ( q ) .The language recognized by a -monster DFA is called a . Example 1
The -monster Mon { } is , [00] [11] , [10] [01] , [11][00] , [10] where, for all i , j ∈ { , } , the label [ i j ] denotes the transformation sending to i and to j, which isalso a letter in the DFA above. Let us notice that some families of 1-monster languages are witnesses for the Star andReverse operations ([3]). The following claim is easy to prove and captures a universality-likeproperty of 1-monster languages:
Property 2
Let L be any regular language recognized by a DFA A = ( Σ , (cid:126) n (cid:127) , , F , δ ) . The language L isthe preimage of L(Mon Fn ) by the -uniform morphism φ such that, for all a ∈ Σ , φ ( a ) = δ a , i.e.L = φ − (L(Mon Fn )) . (2)This is an important and handy property that we should keep in mind. We call it the restriction-renaming property.We can wonder whether we can extend the notions above to provide witnesses for k -aryoperators. In the unary case, the alphabet of a monster is the set of all possible transformationswe can apply on the states. In the same mindset, a k -monster DFA is a k -tuple of DFAs, andits construction must involve the set of k -tuples of transformations as an alphabet. Indeed, thealphabet of a k -ary monster has to encode all the transformations acting on each set of statesindependently one from the others. This leads to the following definition : Definition 2
A k-monster is a k-tuple of automata
Mon F ,..., F k n ,..., n k = ( M , . . . , M k ) where M j = ( Σ , (cid:126) n j (cid:127) , , F j , δ j ) for j ∈ { , k } is defined by • the common alphabet Σ = (cid:126) n (cid:127) (cid:126) n (cid:127) × . . . × (cid:126) n k (cid:127) (cid:126) n k (cid:127) , • the set of states (cid:126) n j (cid:127) , • the initial state , • the set of final states F j , • the transition function δ j defined for any ( a , . . . , a k ) ∈ Σ by δ j ( q , ( a , . . . , a k )) = a j ( q ) .A k-tuple of languages ( L , . . . , L k ) is called a monster k -language if there exists a k-monster ( M , . . . , M k ) such that ( L , . . . , L k ) = (L( M ) , . . . , L( M k )) . Algebraic Tools for State Complexity
Remark 1
When F j is di ff erent from ∅ and Q j , M j is minimal. Definition 2 allows us to extend the restriction-renaming property in a way that is still easy tocheck.
Property 3
Let ( L , . . . , L k ) be a k-tuple of regular languages over the same alphabet Σ . We assumethat each L j is recognized by the DFA A j = ( Σ , (cid:126) n j (cid:127) , , F j , δ j ) . Let Mon F ,..., F k n ,..., n k = ( M , . . . , M k ) . For allj ∈ { , . . . , k } , the language L j is the preimage of L( M j ) by the -uniform morphism φ such that, for alla ∈ Σ , φ ( a ) = ( δ a , . . . , δ ak ) , i.e. ( L , . . . , L k ) = ( φ − (L( M )) , . . . , φ − (L( M k ))) . (3)It has been shown that some families of 2-monsters are witnesses for binary boolean operationsand for the catenation operation [3]. Many papers concerning state complexity actually usemonsters as witnesses without naming them (e.g. [2]). Therefore, a natural question arises : canwe define a simple class of rational operations for which monsters are always witnesses ? Thisclass should ideally encompass some classical regular operations, in particular the operationsstudied in the papers cited above. In the next section, we define objects that allow us to answerthis question. We first describe a class of regular operations for which monsters are always witnesses in theunary case. Once again, the restriction-renaming property comes in handy and gives us theintuition we need. We call 1 -uniform any unary regular operation ⊗ that commutes with any1-uniform morphism, i.e. for every regular language L and every 1-uniform morphism φ , ⊗ ( φ − ( L )) = φ − ( ⊗ ( L )). For example, it is proven in [8] that the Kleene star and the reverse are1-uniform. Suppose now that ⊗ is a unary 1-uniform operation. Then, if L is a regular language, A = ( Σ , (cid:126) n (cid:127) , , F , δ ) its minimal DFA, and φ the 1-uniform morphism sending any letter of Σ intoits associated transition function in A , we have ⊗ ( L ) = ⊗ ( φ − (L(Mon Fn )) = φ − ( ⊗ (L(Mon Fn ))) . (4)It follows that sc( ⊗ ( L )) = sc( φ − ( ⊗ (L(Mon Fn )))) ≤ sc( ⊗ (L(Mon Fn ))) by Property 1. In addition,Remark 1 implies that L(Mon Fn ) has the same state complexity as L . Therefore, we have Theorem 1
Any -uniform operation admits a family of monster -languages as a witness. We now introduce the second central concept of our paper. In many cases, to compute statecomplexities, it is easier to describe regular operations as constructions on DFAs. We wouldtherefore like to find a class of operations on DFAs, that are naturally associated to 1-uniformoperations. Such an operation on DFAs needs to have some constraints that are described inthe following definitions.
Definition 3
The state configuration of a DFA A = ( Σ , Q , i , F , δ ) is the triplet ( Q , i , F ) . Definition 4 A is a unary operation on DFA m that produces a DFA such that : • For any DFA A, the alphabet of m ( A ) is the same as the alphabet of A. . Caron, E. Hamel-de le Court & J.-G. Luque • For any DFA A, the state configuration of m ( A ) depends only on the state configuration of theDFA A. • For any DFA A over the alphabet Σ , for any letter a ∈ Σ , the transition function of a in m ( A ) depends only on the state configuration of the DFA A and on the transition function of a in A. Example 2
The star modifier. For all DFA A = ( Σ , Q , i , F , δ ) , define Star ( A ) = ( Σ , Q , ∅ , { E | E ∩ F (cid:44) ∅} ∪ {∅} , δ ) , where δ is as follows : for all a ∈ Σ , δ a ( ∅ ) = (cid:40) { δ a ( i ) } if δ a ( i ) (cid:60) F { δ a ( i ) , i } otherwise and, for all E (cid:44) ∅ , δ a ( E ) = (cid:40) δ a ( E ) if δ a ( E ) ∩ F = ∅ δ a ( E ) ∪ { i } otherwise The modifier
Star describes the classical construction on DFA associated to the Star operationon languages, i.e. for all DFA A , L( A ) ∗ = L( Star ( A )). Example 3
If we apply the modifier
Star to the modifier
Mon { } described in Example 1, we obtain theDFA drawn in Figure 1. ∅ { }{ } { , } [01] , [00] [11] , [10][01] , [00] [11] , [10][10] , [00] [11] , [01][10] , [01] , [11][00] Figure 1:
Star (Mon { } ) From this, one deduces the action of the modifier
Star on any DFA with two states. For instance,applying
Star to DFA C (Figure 2) gives the DFA described in Figure 3. a b a , b Figure 2: The DFA C ∅ { }{ } { , } a ba b a , b a , b Figure 3:
Star ( C ) Remark that to apply
Star to C, we just take the subautomaton of
Star (Mon { } ) with letters beingexactly the transition functions of letters in C, and rename its letters by the letters of C of which they arethe transition functions. The transition labeled by b in Figure 2 is first assimilated to the transition [11] in Mon { } (see Example 1). Hence, the transition labeled by b in Star ( C ) is the same as the transitionlabeled by [11] in Star (Mon { } ) (Figure 1). Algebraic Tools for State Complexity
Theorem 2
A regular unary operation ⊗ is -uniform if and only if there exists a -modifier m suchthat for any regular language L and any DFA A recognizing L, ⊗ ( L ) = L( m ( A )) . Proof:
Let ⊗ be a 1-uniform unary operation. We define a 1-modifier m as follows. Forany DFA A = ( Σ , Q A , i A , F A , δ A ), we can rename its set of states so that A becomes the DFA D = ( Σ , (cid:126) n (cid:127) , , F , δ ). Let us denote by B = ( (cid:126) n (cid:127) (cid:126) n (cid:127) , Q (cid:48) , i (cid:48) , F (cid:48) , δ (cid:48) ) the minimal DFA of ⊗ (L(Mon Fn )).We set m ( A ) = ( Σ , Q (cid:48) , i (cid:48) , F (cid:48) , ˜ δ (cid:48) ), with ˜ δ (cid:48) ( q , a ) = δ (cid:48) ( q , δ a ). Notice that m is indeed a 1-modifier. First,( Q (cid:48) , i (cid:48) , F (cid:48) ) depends only on ( Q A , i A , F A ). Second, ˜ δ (cid:48) a depends only on δ a and on δ (cid:48) , which in turndepend only on ( Q A , i A , F A ) and δ aA .Furthermore, by construction, L( m ( A )) = φ − (L( B )), where φ is the 1-uniform morphismsuch that φ ( a ) = δ aD for all a ∈ Σ . Therefore, we have L( m ( A )) = φ − ( ⊗ (L(Mon Fn ))). And, since ⊗ is 1-uniform, we obtain L( m ( A )) = ⊗ ( φ − (L(Mon Fn ))) = ⊗ ( L ).Conversely, let ⊗ be a regular operation and let m be a 1-modifier such that for any regularlanguage L and any DFA A recognizing L , ⊗ ( L ) = L( m ( A )). We must prove that ⊗ is 1-uniform.Let Γ and Σ be two alphabets. Consider a 1-uniform morphism φ from Γ ∗ to Σ ∗ and a language L over Σ . Let A = ( Σ , Q , i , F , δ ) be any DFA recognizing L and let B = ( Γ , Q , i , F , ˜ δ ) the DFA suchthat ˜ δ a = δ φ ( a ) for any letter a ∈ Γ . We have L( B ) = φ − (L( A )).Let m ( A ) = ( Σ , Q , i , F , δ ) and m ( B ) = ( Γ , Q , i , F , δ ). Since the state configuration of A isthe same as the state configuration of B , we have ( Q , i , F ) = ( Q , i , F ). Furthermore, becausethe transition function of any letter a ∈ Γ in B is also the same as the transition function of φ ( a ) in A , we have δ a = δ φ ( a )1 . Hence, L( m ( B )) = φ − (L( m ( A ))), which implies that ⊗ (L( B )) = φ − ( ⊗ ( A )).Therefore, ⊗ ( φ − (L( A ))) = φ − ( ⊗ (L( A ))), as expected. (cid:3) We extend the previous theorems by generalizing the definitions to k -ary operations. Definition 5
A k-ary regular operation ⊗ is called -uniform if, for any k-tuple of rational languages ( L , . . . , L k ) , for any -uniform morphism φ , ⊗ ( φ − ( L ) , . . . , φ − ( L k )) = φ − ( ⊗ ( L , . . . , L k )) . Using the same arguments as in Theorem 1, we find
Theorem 3
Any k-ary -uniform operation admits a family of monster k-languages as a witness. Proof:
Suppose now that ⊗ is a k -ary 1-uniform operation. Then, if ( L , . . . , L k ) is a k -tuple ofregular languages over Σ , ( A , . . . , A k ) the k -tuple of DFAs such that each A j = ( Σ , Q j , i j , F j , δ j ) is theminimal DFA of L i , and φ the 1-uniform morphism such that, for all a ∈ Σ , φ ( a ) = ( δ a , . . . , δ ak ), and ifMon F ,..., F k n ,..., n k = ( M , . . . , M k ), then ⊗ ( L ) = ⊗ ( φ − (L( M )) , . . . , φ − (L( M k ))) = φ − ( ⊗ (L( M ) , . . . , L( M k ))).It follows that sc( ⊗ ( L )) = sc( φ − ( ⊗ (L( M ) , . . . , L( M k )))) ≤ sc( ⊗ (L( M ) , . . . , L( M k ))) by Property 1.In addition, each L( M j ) has the same state complexity as L j . (cid:3) Definition 6
A k-modifier is a k-ary operation on DFAs over the same alphabet that returns a DFA andsuch that : • The alphabet of m ( A , ..., A k ) is the same as the alphabet of each A j . • For any k-tuple of DFAs ( A , . . . , A k ) , the state configuration of m ( A , ..., A k ) depends only on thestate configurations of the DFAs A , . . . , A k . • For any k-tuple of DFAs ( A , . . . , A k ) where each DFA is over the alphabet Σ , for any letter a ∈ Σ ,the transition function of a in m ( A , . . . , A k ) depends only on the state configurations of the DFAsA , . . . , A k and on the transition functions of a in each of the DFAs A , ..., A k . . Caron, E. Hamel-de le Court & J.-G. Luque Example 4
For all DFAs A = ( Σ , Q , i , F , δ ) and B = ( Σ , Q , i , F , δ ) , define Xor ( A , B ) = ( Σ , Q × Q , ( i , i ) , ( F × ( Q \ F ) ∪ ( Q \ F ) × F ) , ( δ , δ ))The modifier Xor describes the classical construction associated to the operation Xor on couplesof languages, i.e for all DFAs A and B , L( A ) ⊕ L( B ) = L( Xor ( A , B )). Theorem 4
A regular k-ary operation ⊗ is -uniform if and only if there exists a k-modifier m such thatfor any k-tuple of regular languages ( L , . . . , L k ) and any k-tuple of DFAs ( A , . . . , A k ) such that each A j recognizes L j , we have ⊗ ( L , . . . , L k ) = L( m ( A , . . . , A k )) . The proof of Theorem 2 can be easily adapted to k -ary operations.The following proposition states the e ff ects of composition on modifiers and 1-uniformoperations and directly stems from Definitions 5 and 6. Proposition 1
Let ⊗ be a k -ary -uniform operation and ⊗ be a k -ary -uniform operation. The ( k + k ) -ary operation defined by ⊗ ( L , . . . , L k + k ) = ⊗ ( L , . . . , L l , ⊗ ( L l + , . . . , L l + k ) , L l + k , . . . , L k + k ) is -uniform. Furthermore, if m is a k -modifier associated with ⊗ and m is a k -modifier associatedwith ⊗ , the operation on ( k + k ) -tuples of DFAs defined by m ( A , . . . , A k + k ) = m ( A , . . . , A l , m ( A l + , . . . , A l + k ) , A l + k , . . . , A k + k ) (5) is a modifier associated to ⊗ . ff erence In this section, we compute the state complexity of the 2-ary regular operation L (cid:63) (cid:13) L = ( L ⊕ L ) ∗ . Examples 2 and 4 together with Proposition 1 show that (cid:63) (cid:13) is 1-uniform and that anassociated modifier can be defined by StX ( A , A ) = Star ( Xor ( A , A )). To be more precise, if A = ( Σ , Q , i , F , δ ) and A = ( Σ , Q , i , F , δ ), then StX ( A , A ) = ( Σ , Q × Q , ∅ , { E ∈ Q × Q | E ∩ F (cid:44) ∅} ∪ {∅} , δ )where F = ( F × Q ) ⊕ ( Q × F ) and, for all a ∈ Σ , δ a ( ∅ ) = (cid:40) { ( δ a ( i ) , δ a ( i )) } if ( δ a ( i ) , δ a ( i )) (cid:60) F { ( δ a ( i ) , δ a ( i )) , ( i , i ) } otherwiseand, for all E (cid:44) ∅ , δ a ( E ) = (cid:40) ( δ a , δ a )( E ) if ( δ a , δ a )( E ) ∩ F = ∅ ( δ a , δ a )( E ) ∪ { ( i , i ) } otherwise.Theorem 3 states that (cid:63) (cid:13) admits a family of 2-monsters as witness. For any positive integers n , n , let ( M , M ) = Mon { n − } , { } n , n . We are going to show that, for all ( n , n ) ∈ N ∗ , (L( M )) , L( M ))is indeed a witness for (cid:63) (cid:13) . This allows us to compute its state complexity. To be more precise,here is the outline of our proof. For any positive integers n , n , any F , F ⊆ (cid:126) n (cid:127) × (cid:126) n (cid:127) , let usdenote by M F , F the DFA StX (Mon F , F n , n ). We are going to minimize the DFA M { n − } , { } by firstcomputing its accessible states, and then, restricting it to its accessible states, by computingits Nerode equivalence. We will therefore have computed the minimal DFA equivalent toM { n − } , { } , and computing its size allows us to compute the state complexity of L(M { n − } , { } ).We then show that the state complexity of L(M { n − } , { } ) is the greatest out of all the statecomplexities of L(M F , F ), with ( F , F ) ⊆ (cid:126) n (cid:127) × (cid:126) n (cid:127) . Theorem 3 allows us to conclude that thestate complexity of L(M { n − } , { } ) is indeed sc (cid:63) (cid:13) ( n , n ).62 Algebraic Tools for State Complexity M { n − } , { } In order to understand more easily the next proofs, we associate elements of 2 (cid:126) n (cid:127) × (cid:126) n (cid:127) to booleanmatrices of size n × n . Such a matrice is called a tableau when crosses are put in place of 1s,and 0s are erased. We denote by the same letter the element of 2 (cid:126) n (cid:127) × (cid:126) n (cid:127) , the associated booleanmatrix, and the associated tableau. If T is an element of 2 (cid:126) n (cid:127) × (cid:126) n (cid:127) , we denote by T x , y the valueof the boolean matrix T at row x and column y . Therefore, the three following assertions meanthe same thing : a cross is at the coordinates ( x , y ) in T , T x , y =
1, ( x , y ) ∈ T .We say that a cross at coordinates ( x , y ) in an element of 2 (cid:126) n (cid:127) × (cid:126) n (cid:127) is in the final zone of M F , F if ( x , y ) ∈ ( F × (cid:126) n (cid:127) ) ⊕ ( (cid:126) n (cid:127) × F ). We remark that an element of 2 (cid:126) n (cid:127) × (cid:126) n (cid:127) is final in M F , F if andonly if it has a cross in the final zone of M F , F . We fix for the remainder of this section twopositive integers n and n . Lemma 1
The states of M { n − } , { } that are accessible are exactly the tableaux T of size n × n such that,if T has a cross in the final zone of M { n − } , { } , then T has cross at (0 , . Proof:
It is easy to see by the definition of the transition function of
StX that every tableau T with a cross in the final zone of M { n − } , { } and no cross at (0 ,
0) is not accessible.Let δ be the transition function of M { n − } , { } . If T is a tableau of size n × n , let nf T be thenumber of crosses of T which are not in the final zone of M { n − } , { } . Let us define an order < oncross matrices as T < T (cid:48) if and only if T < T (cid:48) or ( T = T (cid:48) and nf T < nf T (cid:48) ) .Let us prove every tableau T of size n × n such that, if T has a cross in the final ( { n − } , { } )-zone, then T has cross at (0 , < (the empty cross matrix is the initial state of M { n − } , { } , and so it is accessible).The only minimal cross matrix for non-empty matrices and the order < is the cross matrixwith only one cross at (0 , ∅ by reading the letter ( , ).Let us notice that each letter is a couple of functions of (cid:126) n (cid:127) (cid:126) n (cid:127) × (cid:126) n (cid:127) (cid:126) n (cid:127) . Now let us take across matrix T (cid:48) , and find a cross matrix T such that T < T (cid:48) , and T’ is accessible from T . Wedistinguish the cases : • T (cid:48) has no cross in the final zone, except maybe at (0 , – Case T (cid:48) n − , = . Let ( i , j ) be the index of a cross of T (cid:48) . Let ( f , g ) = ((0 , i ) , (0 , j )) where (0 , i )and (0 , j ) denote transpositions, and let T = ( f , g )( T (cid:48) ) where ( f , g )( T (cid:48) ) = { ( f ( i ) , g ( j )) | ( i , j ) ∈ T (cid:48) } . As ( f , g ) is a one-to-one transformation on (cid:126) n (cid:127) × (cid:126) n (cid:127) , as ( f , g )( T (cid:48) ) hasa cross at (0 ,
0) and as T (cid:48) does not have any crosses in the final zone, we have δ ( f , g ) ( T ) = ( f , g )( T ) = ( f , g )( f , g )( T (cid:48) ) = T (cid:48) . We also have T < T (cid:48) since T = T (cid:48) and nf T < nf T (cid:48) . – Case T (cid:48) n − , = . Let ( f , g ) = ((0 , n − , ) and let T = ( f , g )( T (cid:48) ). We have δ ( f , g ) ( T ) = T (cid:48) ,and T < T (cid:48) as nf T < nf T (cid:48) . • T (cid:48) has a cross in the final zone other than (0 , i , j ) be such a cross, and let ( f , g ) = ((0 , i ) , (0 , j )). Let T (cid:48)(cid:48) be the cross matrix obtainedfrom T (cid:48) by deleting the cross at (0 , T = ( f , g )( T (cid:48)(cid:48) ). As ( f , g ) is still one-to-one on (cid:126) n (cid:127) × (cid:126) n (cid:127) , we have T , = (( f , g )( T (cid:48)(cid:48) )) , = T (cid:48)(cid:48) i , j =
1, and (( f , g )( T )) = (( f , g )(( f , g )( T (cid:48)(cid:48) ))) = T (cid:48)(cid:48) .As T (cid:48)(cid:48) has a cross in the final zone, we therefore have δ ( f , g ) ( T ) = T (cid:48) and T < T (cid:48) as T < T (cid:48) . • The only cross of T (cid:48) which is in the final zone is (0 , . Caron, E. Hamel-de le Court & J.-G. Luque ×× ××× (cid:16)(cid:16) (cid:17) , (cid:17) ×× ××× Figure 4: The two tableaux T (cid:48) and T for case A . ×× ×× ((1 , , ) ×× ×× Figure 5: The two tableaux T (cid:48) and T for case ¬ A and T (cid:48) n − , = – Case A: there exists j such that T (cid:48) , j = . Let ( f , g ) = ( (cid:16) n − (cid:17) , ) and let T i , j = i , j ) = (0 , , T (cid:48) , j if i = n − ∧ j (cid:44) , T (cid:48) i , j otherwise.It is easy to check that δ ( f , g ) ( T ) = T (cid:48) , and T < T (cid:48) as nf T < nf T (cid:48) . – Case ¬ A and T (cid:48) n − , = . There exists ( i , j ) (cid:44) ( n − ,
0) such that i (cid:44) T (cid:48) i , j = f , g ) = (( i , n − , ) and let T = ( f , g )( T (cid:48) ). We have δ ( f , g ) ( T ) = T (cid:48) , and T < T (cid:48) as nf T < nf T (cid:48) . – Case ¬ A and T (cid:48) n − , = . Let ( f , g ) = (( n − , n − , ) and let T = ( f , g )( T (cid:48) ). We have δ ( f , g ) ( T ) = T (cid:48) , and T < T (cid:48) as nf T < nf T (cid:48) . (cid:3) For all ( F , F ) ⊆ (cid:126) n (cid:127) × (cid:126) n (cid:127) , let us now call (cid:98) M F , F the DFA M F , F restricted to states T such that,if T has a cross in the final zone of M F , F , then T has cross at (0 , StX . Remark 2
The accessible part of M F , F is included in (cid:98) M F , F . (cid:98) M { n − } , { } Definition 7
A tableau T in (cid:126) n (cid:127) × (cid:126) n (cid:127) is right-triangle free if ∀ x , x (cid:48) ∈ (cid:126) n (cid:127) such that x (cid:44) x (cid:48) and ∀ y , y (cid:48) ∈ (cid:126) n (cid:127) , such that y (cid:44) y (cid:48) , we have { ( x , y ) , ( x , y (cid:48) ) , ( x (cid:48) , y ) , ( x (cid:48) , y (cid:48) ) } ∩ T ) (cid:44) . Definition 8
If T and T (cid:48) are distinct tableaux, we define the transformation on tableaux → as T → T (cid:48) if T (cid:48) = T ∪ { ( i (cid:48) , j (cid:48) ) } , and there exists ( i , j ) such that { ( i , j ) , ( i (cid:48) , j ) , ( i , j (cid:48) ) } ⊆ T. The equivalence relation ∗ ↔ isdefined as the symmetric, reflexive and transitive closure of → . For any tableau T , we define Sat( T ) as the smallest tableau (relatively to inclusion) with noright-triangle containing T . The existence and the unicity of Sat( T ) are easy to check. It is therepresentative of the equivalence class of T . Two tableaux T and T (cid:48) are therefore equivalent ifSat( T ) = Sat( T (cid:48) ).64 Algebraic Tools for State Complexity ×××
Figure 6: A tableau with a right-triangle
Lemma 2
The tableau T in (cid:126) n (cid:127) × (cid:126) n (cid:127) is right-triangle free if and only if for all i , i (cid:48) ∈ (cid:126) n (cid:127) , the lines iand i (cid:48) are either the same (for all j ∈ (cid:126) n (cid:127) , T i , j = T i (cid:48) , j ), or disjoint (for all j ∈ (cid:126) n (cid:127) , T i , j = ∨ T i (cid:48) , j = ). Lemma 3
Let ( F , F ) ⊆ (cid:126) n (cid:127) × (cid:126) n (cid:127) , and let T and T (cid:48) be any two states of M F , F such that T → T (cid:48) .Then T is final if and only if T (cid:48) is final. Let us recall that the alphabet of M F , F is (cid:126) n (cid:127) (cid:126) n (cid:127) × (cid:126) n (cid:127) (cid:126) n (cid:127) . If ( f , g ) is such a letter and T = { ( x , y ) , . . . , ( x n , y n ) } is a tableau, then define ( f , g )( T ) as { ( f ( x ) , g ( y )) , . . . , ( f ( x n ) , g ( y n )) } . Lemma 4
Let ( F , F ) ⊆ (cid:126) n (cid:127) × (cid:126) n (cid:127) , and let T and T (cid:48) be any two states of (cid:98) M F , F such that T → T (cid:48) .Then, for any a ∈ (cid:126) n (cid:127) (cid:126) n (cid:127) × (cid:126) n (cid:127) (cid:126) n (cid:127) , δ a ( T ) → δ a ( T (cid:48) ) or δ a ( T ) = δ a ( T (cid:48) ) . Proposition 2
Let ( F , F ) ⊆ (cid:126) n (cid:127) × (cid:126) n (cid:127) , and let T , T (cid:48) be two states of M F , F . If T ∗ ↔ T (cid:48) , then T andT (cid:48) are not distinguishable. Proof:
From Lemma 4, it is easy to see by a simple induction that, for any word w if T → T (cid:48) then δ w ( T ) → δ w ( T (cid:48) ) or δ w ( T ) = δ w ( T (cid:48) ). From Lemma 3, if T → T (cid:48) , then T ∼ Ner T (cid:48) in the sense of theNerode equivalence. Thus, as ∗ ↔ is the symmetric and transitive closure of → , T ∗ ↔ T (cid:48) implies T ∼ Ner T (cid:48) . (cid:3) Lemma 5
All states of ( (cid:98) M { n − } , { } ) / ∗ ↔ are pairwise distinguishable. Proof:
Let δ be the transition function of ( (cid:98) M { n − } , { } ) / ∗ ↔ . Let T and T (cid:48) be the representativesof two states of ( (cid:98) M { n − } , { } ) / ∗ ↔ , such that T (cid:44) T (cid:48) . Let ( i , j ) be such that T i , j (cid:44) T (cid:48) i , j . Suppose, forexample that T i , j =
1. Take { i , . . . , i (cid:96) } = { α | T (cid:48) α, j = } and { j , . . . , j p } = { j } ∪ { β | T (cid:48) i ,β = } . We can seethat :1. By Lemma 2, lines i , . . . , i (cid:96) are the same, as they all have a cross on the column j . Columns { j , . . . , j p } are also the same, as they all have a cross on line i . It follows that, if ( i (cid:48) , j (cid:48) ) ∈ (cid:16) { i , . . . , i (cid:96) } × (cid:16) { , . . . , n − } \ { j , . . . , j p } (cid:17)(cid:17) ∪ (cid:16) ( { , . . . , n − } \ { i , . . . , i (cid:96) } ) × { j , . . . , j p } (cid:17) , then T (cid:48) i (cid:48) , j (cid:48) = j ∈ { j , . . . , j p } and i (cid:60) { i , . . . , i (cid:96) } .Let f ( i (cid:48) ) = (cid:40) n − i (cid:48) ∈ { i , . . . , i (cid:96) } , g ( j (cid:48) ) = (cid:40) j (cid:48) ∈ { j , . . . , j p } n − f ( i (cid:48) ) , g ( j (cid:48) )) is in the final zone of M { n − } , { } , then( i (cid:48) , j (cid:48) ) ∈ (cid:16) { i , . . . , i (cid:96) } × (cid:16) { , . . . , n − } \ { j , . . . , j p } (cid:17)(cid:17) ∪ (cid:16) ( { , . . . , n − } \ { i , . . . , i (cid:96) } ) × { j , . . . , j p } (cid:17) , and so the first point above gives us T (cid:48) i (cid:48) , j (cid:48) = . Caron, E. Hamel-de le Court & J.-G. Luque × ×× ⊗× × ji × ◦ ××× ××× ××××××× j j j i i i j Figure 7: An example of two tableaux T and T (cid:48) Therefore, δ ( f , g ) ( T (cid:48) ) has only at most two crosses, one in ( n − ,
0) and one in (0 , n − T i , j = δ ( f , g ) ( T ) , = δ ( f , g ) ( T ) is final. Thus, T and T (cid:48) are distinguishable. (cid:3) Proposition 2 and Lemma 5 give us that ( (cid:98) M { n − } , { } ) / ∗ ↔ is the minimal DFA equivalent to theDFA (cid:98) M { n − } , { } . The following corollary stems from this assertion combined with Lemma 1. Corollary 1 ( (cid:98) M { n − } , { } ) / ∗ ↔ is the minimal DFA equivalent to M { n − } , { } . M { n − } , { } The number of right-triangle free tableaux T of size (cid:126) n (cid:127) × (cid:126) n (cid:127) such that, if T has a cross in thefinal zone of M { n − } , { } , then T has cross at (0 ,
0) is exactly 2 α n − , n − + α (cid:48) n , n where α x , y is thenumber of right-triangle free tableaux of size x × y and α (cid:48) x , y the number of right-triangle freetableaux of size x × y having a cross in (0 , Lemma 6
The state complexity of
L(M { n − } , { } ) is α n − , n − + α (cid:48) n , n . Closed formulas for α ( x , y ) and α (cid:48) ( x , y ) are given in Corollary 20 and Proposition 22 of [4].In the next subsection, we prove that ( { n − } , { } ) is a couple of final states that maximizesthe size of the minimal DFA associated to any M F , F , with ( F , F ) ⊆ (cid:126) n (cid:127) × (cid:126) n (cid:127) . (cid:63) (cid:13) applied to monster -languages Let T be the set of right-triangle free tableaux of size n × n . For all ( F , F ) ⊆ (cid:126) n (cid:127) × (cid:126) n (cid:127) , let T F , F = (cid:98) M F , F ) / ∗ ↔ = { T ∈ T | T has a cross in the final zone implies T , = } . We show that :
Lemma 7
For any F × F ⊆ (cid:126) n (cid:127) × (cid:126) n (cid:127) such that F , F (cid:44) ∅ and F (cid:44) (cid:126) n (cid:127) ,F (cid:44) (cid:126) n (cid:127) , T F , F ≤T { n − } , { } . Therefore, by Remark 2, Proposition 2, and Corollary 1, for any F × F ⊆ (cid:126) n (cid:127) × (cid:126) n (cid:127) such that F , F (cid:44) ∅ and F (cid:44) (cid:126) n (cid:127) , F (cid:44) (cid:126) n (cid:127) , min (M F , F ) ≤ (cid:98) M F , F ) / ∗ ↔ ) = T F , F ≤ T { n − } , { } = (cid:98) M { n − } , { } ) / ∗ ↔ ) = min (M { n − } , { } ) . The cases where F = ∅ or F = ∅ or F = (cid:126) n (cid:127) or F = (cid:126) n (cid:127) are easy and proven by : Lemma 8
If F = ∅ or F = ∅ or F = (cid:126) n (cid:127) or F = (cid:126) n (cid:127) , then min (M F , F ) ≤ min (M { n − } , { } ) . Therefore, by Theorem 3 and Lemma 6,
Theorem 5
The state complexity of (cid:63) (cid:13) is α n − , n − + α (cid:48) n , n , i.e. for all n , n ∈ N ∗ , sc (cid:63) (cid:13) ( n , n ) = α n − , n − + α (cid:48) n , n . Algebraic Tools for State Complexity
We now prove that there is a finite-bounded-alphabet witness. Let n , n be two positive integersand let ( M , M ) = Mon { n − } , { } n , n . Recall that the letters of Mon { n − } , { } n , n are couples of mappingsand that is the identities bot in (cid:126) n (cid:127) and in (cid:126) n (cid:127) . Let B and B be the DFAs obtained byrestricting the letters of respectively M and M to the alphabet Σ (cid:48) = (cid:110) ((0 , . . . , n − , ) , ((1 , . . . , n − , ) , ( , (1 , . . . , n − , ((1 , . . . , n − , ) , (cid:0) ab (cid:1) ( , (1 , . . . , n − , ((0 , n − , ) , ( , (0 , n − , ((0 , , (0 , , ((0 , , ) , ( , (0 , , (( n − , n − , ) , ( (cid:16) (cid:17) , ) , ( , (cid:16) (cid:17) ) , ( (cid:16) n − n − (cid:17) , ) , ( , (cid:16) n − n − (cid:17) ) , ( (cid:16) n − (cid:17) , ) , ( , (cid:16) n − (cid:17) ) (cid:111) . Let B = StX ( B , B ), and (cid:98) B be the DFA obtained by restricting B to states T such that, if T has across in the final zone of M { n − } , { } , then T has cross at (0 , A = (cid:98) B / ∗ ↔ is obtained byrestricting the letters of (cid:98) M { n − } , { } to the alphabet Σ (cid:48) . We are going to show that A is minimal.Let us recall that all letters of Σ (cid:48) can be seen as a function acting on tableaux. Every word w of Σ (cid:48) acts on a tableau T by applying the composition of all letters of w to T : if w = a . . . a n ,define w ( T ) = a n ◦ . . . ◦ a ( T ). When it exists, we denote by w − the inverse function of a n ◦ . . . ◦ a .Let δ be the transition function of B . We first notice that w ( T ) is not necessarily equivalent to T (cid:48) = δ w ( T ) since (0 ,
0) is in T (cid:48) if T (cid:48) has a cross in the final zone. We denote by w [ i , j ] the subword a i · · · a j . By convention, if j < i, w [ i , j ] = ε . The proof of the following lemma is easy by induction. Lemma 9
Let w be a word of Σ (cid:48) , and T be a state of B. If, for any integer k < | w | , we have ( w [1 , k ]( T )) , = or w [1 , k ]( T ) has no cross in the final zone, then δ w ( T ) = w ( T ) . Lemma 10
All the states of (cid:98)
B are accessible.
Proof:
As the induction is the same as in Lemma 1, we only focus on cases of this previouslemma where the letters used are not in Σ (cid:48) . • If T (cid:48) has no cross in the final zone, according the previous remark, we have only to examinethe case where T (cid:48) n − , =
0. Let ( i , j ) be the index of a cross of T (cid:48) . Let w = (( , (0 , , . . . , n − , ) i ( , (1 , . . . , n − j − and let T = w − ( T (cid:48) ). We have T , = < k < | w | , for all( i , j ) (cid:44) ( n − ,
0) such that i = n − j =
0, ( w [1 , k ]( T )) i , j = ( w [ k , | w | − − ( T (cid:48) )) i , j =
0. Thus,by Lemma 9, δ w ( T ) = T (cid:48) . We also have T < T (cid:48) since T = T (cid:48) and nf T < nf T (cid:48) . • T (cid:48) has a cross in the final zone other than (0 , i , j ) be such a cross. We distinguishtwo cases. – If j =
0, we consider the word w = ((1 , . . . , n − , ) i . Let T (cid:48)(cid:48) be the tableau obtainedfrom w − ( T (cid:48) ) by deleting the cross at (0 ,
0) and let w = ((0 , , ) and T = w ( T (cid:48)(cid:48) ). It iseasy to see that T (cid:48) = w ( δ w ( T )). By Lemma 9, we have T (cid:48) = w ( δ w ( T )) = δ w w ( T ) in (cid:98) M { n − } , { } and T < T (cid:48) as T < T (cid:48) . – Otherwise define w = ((1 , . . . , n − , ) i − ( , (1 , . . . , n − j − . Let T (cid:48)(cid:48) be tableau ob-tained from w − ( T (cid:48) ) by deleting the cross at (0 , T (cid:48) = w ( T (cid:48)(cid:48) ) ∪ { (0 , } .Let w = ((0 , , (0 , T = w ( T (cid:48)(cid:48) ). Then we have δ w ( T ) = T (cid:48)(cid:48) ∪ { (0 , } or δ w ( T ) = T (cid:48)(cid:48) . ∗ If δ w ( T ) = T (cid:48)(cid:48) ∪ { (0 , } , then by Lemma 9, as w does not change the first line andthe first column we have δ w w ( T ) = δ w ( δ w ( T )) = w ( δ w ( T )) = w (( T (cid:48)(cid:48) ∪ { , } ) = w ( T (cid:48)(cid:48) ) ∪ { (0 , } = T (cid:48) . . Caron, E. Hamel-de le Court & J.-G. Luque ∗ If δ w ( T ) = T (cid:48)(cid:48) , then we set k = min { l < | w | | δ ( w [1 , l ] ( T (cid:48)(cid:48) ) , = } . For all integer l < k ,the tableau δ w [1 , l ] ( T (cid:48)(cid:48) ) has no cross in the final zone, and we can apply Lemma 9,and w ( w [1 , l ]( T )) = δ w w [1 , l ]( T ). Furthermore δ w w [1 , k ] ( T ) = ( w [1 , k ])( δ w ( T )) ∪{ (0 , } and as letters of w [ k + , | w | ] do not change the first line and the firstcolumn, we have δ w w ( T ) = w [ k + , | w | ]( w [1 , k ]( δ w ( T )) ∪ { (0 , } ) = w ( δ w ( T )) ∪{ (0 , } = w ( T (cid:48)(cid:48) ) ∪ { (0 , } = T (cid:48) . Moreover, as T < T (cid:48) , we have T < T (cid:48) . • The only cross of T (cid:48) which is in the final zone is (0 , j such that T (cid:48) , j = T (cid:48) n − , =
0. It follows that there exists ( i , j ) (cid:44) ( n − ,
0) such that i (cid:44) T (cid:48) i , j =
1. Let w = ((1 , . . . , n − , ) i and let T = w − ( T (cid:48) ). By Lemma 9, as for each proper prefix w (cid:48) of w ,( w (cid:48) ( T )) , =
1, we have δ w ( T ) = T (cid:48) in B , and T < T (cid:48) as nf T < nf T (cid:48) . (cid:3) Similarly, the following lemma is obtained by simulating with letters in Σ (cid:48) the transition func-tions used in Lemma 5. Lemma 11
All states of A are pairwise distinguishable.
Lemma 10 and 11 imply that A is minimal and that the following theorem holds. Theorem 6
The couple ( B , B ) is a witness for the operation (cid:63) (cid:13) . We have given the state complexity of the star of symmetrical di ff erence and have provided awitness with a constant alphabet size. We know that the bounded size of the alphabet that weexhibit is not optimal, but it simplifies the proof given. Moreover, proving the optimality of abound seems out of reach for now and would necessitate to introduce new tools.One of our future works will be to generalize the method used here to a whole well-definedclass of operations, in order to provide a witness with bounded alphabet size for all of them. Acknowlegedments
This work is partially supported by the projects MOUSTIC ( ERDF / GRR)and ARTIQ (ERDF / RIN)
References [1] Janusz A. Brzozowski (2013):
In Search of Most Complex Regular Languages . Intern. J. of Foundationsof Comp. Sc. http://dx.doi.org/10.1142/S0129054113400133 .[2] Janusz A. Brzozowski, Galina Jirásková, Bo Liu, Aayush Rajasekaran & Marek Szykula (2016):
On the State Complexity of the Shu ffl e of Regular Languages . In: Descriptional Complexity of FormalSystems - 18th IFIP WG 1.2 International Conference, DCFS 2016, Bucharest, Romania, July 5-8,2016. Proceedings , pp. 73–86, doi:10.1007 / New tools forstate complexity . CoRR abs / http://arxiv.org/abs/1807.00663 .[4] Pascal Caron, Jean-Gabriel Luque, Ludovic Mignot & Bruno Patrou (2016): State Complexity ofCatenation Combined with a Boolean Operation: A Unified Approach . Int. J. Found. Comput. Sci. / S0129054116500234.[5] Pascal Caron, Jean-Gabriel Luque & Bruno Patrou (2016):
State complexity of multiple catenation . CoRR abs / http://arxiv.org/abs/1607.04031 . Algebraic Tools for State Complexity [6] Pascal Caron, Jean-Gabriel Luque & Bruno Patrou (2017):
State complexity of catenation combined withboolean operations . CoRR abs / http://arxiv.org/abs/1707.03174 .[7] Bo Cui, Yuan Gao, Lila Kari & Sheng Yu (2011): State Complexity of Two Combined Operations:Catenation-Union and Catenation-Intersection . Int. J. Found. Comput. Sci. http://dx.doi.org/10.1142/S0129054111009045 .[8] Sylvie Davies (2018):
A General Approach to State Complexity of Operations: Formalization and Limita-tions . Developments in Language Theory , doi:10.1007 / State Complexity of Proportional Removals . Journal of Automata, Lan-guages and Combinatorics / jalc-2002-455.[10] Michael Domaratzki & Alexander Okhotin (2009): State complexity of power . Theoretical Com-puter Science / j.tcs.2009.02.025. Available at . Formal Languages and Appli-cations: A Collection of Papers in Honor of Sheng Yu.[11] Yuan Gao, Nelma Moreira, Rogério Reis & Sheng Yu (2017): A Survey on Operational State Complexity . Journal of Automata, Languages and Combinatorics / jalc-2016-251.[12] Yuan Gao, Kai Salomaa & Sheng Yu (2008): The State Complexity of Two Combined Operations: Star ofCatenation and Star of Reversal . Fundam. Inf. http://dl.acm.org/citation.cfm?id=1377804.1377812 .[13] J. E. Hopcroft & J. D. Ullman (1979):
Introduction to Automata Theory, Languages and Computation .Addison-Wesley, Reading, MA.[14] Jozef Jirásek, Galina Jirásková & Alexander Szabari (2005):
State complexity of concatenation andcomplementation . Int. J. Found. Comput. Sci. http://dx.doi.org/10.1142/S0129054105003133 .[15] Galina Jirásková (2005):
State complexity of some operations on binary regular languages . Theor. Comput.Sci. http://dx.doi.org/10.1016/j.tcs.2004.04.011 .[16] Galina Jirásková & Alexander Okhotin (2008):
State complexity of cyclic shift . ITA / ita:2007038.[17] Galina Jirásková & Alexander Okhotin (2011): On the State Complexity of Star of Union and Starof Intersection . Fundam. Inform. http://dx.doi.org/10.3233/FI-2011-502 .[18] Arto Salomaa, Kai Salomaa & Sheng Yu (2007):
State complexity of combined operations . Theor.Comput. Sci. http://dx.doi.org/10.1016/j.tcs.2007.04.015 .[19] Sheng Yu (2001):
State Complexity of Regular Languages . Journal of Automata, Languages andCombinatorics //