Automaton (Semi)groups (Basic Concepts)
aa r X i v : . [ m a t h . G R ] J a n Automaton (Semi)groups (Basic Concepts)
J¯anis Buls, L¯ıga Uˇzule, Aigars Valainis
Department of Mathematics, University of Latvia, Zel¸l¸u iela 25,R¯ıga, LV-1002 Latvia, [email protected]; [email protected]; [email protected]
Abstract
In this papper, we give an introduction to basic concepts of automatonsemigroups. While we must note that this paper does not contain newresults, it is focused on extended introduction in the subject and detailedexamples.
Keywords automaton semigroups
1. Preliminaries
Let A be a finite non-empty set and A ∗ the free monoid generated by A . The set A is also called an alphabet , its elements are called letters and those of A ∗ are called finite words . The identity element of A ∗ is called an empty word and denoted by λ . We set A + = A ∗ \{ λ } .A word w ∈ A + can be written uniquely as a sequence of lettersas w = w w . . . w l , with w i ∈ A , 1 ≤ i ≤ l , l >
0. The integer l iscalled the length of w and denoted by | w | . The length of λ is 0. Weset w = λ and ∀ i ∈ N w i +1 = w i w . The word w ′ ∈ A ∗ is a factor (or subword ) of w ∈ A ∗ if thereexists u, v ∈ A ∗ such that w = uw ′ v . The words u and v are called,respectively, a prefix and a suffix . A pair ( u, v ) is called an occurrence of w ′ in w . A factor w ′ is called proper if w = w ′ . We denote,respectively, by F( w ), Pref( w ) and Suff( w ) the sets of w factors,prefixes and suffixes.An (indexed) infinite word x on the alphabet A is any total map-ping x : N → A . We shall set for any i ≥ x i = x ( i ) and write x = ( x i ) = x x . . . x n . . . . The set of all the infinite words over A is denoted by A ω .The word w ′ ∈ A ∗ is a factor of x ∈ A ω if there exists u ∈ A ∗ , y ∈ A ω such that x = uw ′ y . The words u and y are called,respectively, a prefix and a suffix . We denote, respectively, by F( x ),1 ✛ h Q, A, B i u u . . . u n v v . . . v n
1. Figure: An abstract Mealy machine.Pref( x ) and Suff( x ) the sets of x factors, prefixes and suffixes. Forany 0 ≤ m ≤ n , x [ m, n ] denotes a factor x m x m +1 . . . x n . The word x [ m, n ] is called an occurrence of w ′ in x if w ′ = x [ m, n ]. The suffix x n x n +1 . . . x n + i . . . is denoted by x [ n, ∞ ).If v ∈ A + , then we denote by v ω the infinite word v ω = vv . . . v . . . . The concatenation of u = u u . . . u k ∈ A ∗ and x ∈ A ω is theinfinite word ux = u u . . . u k x x . . . x n . . . For denoting concatenation we sometimes use symbol x is called ultimately periodic if there exists words u ∈ A ∗ , v ∈ A + such that x = uv ω . In this case, | u | and | v | are called, respectively,an anti-period and a period .We use notation 0 , n to denote set { , , ..., n } .
2. Serial composition of Mealy machines
A 3-sorted algebra V = h Q, A, B, ◦ , ∗i is calleda Mealy machine if Q, A, B are finite, nonempty sets, the mapping Q × A ◦ −→ Q is a total function and the mapping Q × A ∗ −→ B is atotal surjective function. The, set Q is called state set , sets A, B are called input and outputalphabet , respectively. The mappings ◦ and ∗ may be extended to Q × A ∗ by defining q ◦ λ = q, q ◦ ( ua ) = ( q ◦ u ) ◦ a,q ∗ λ = λ, q ∗ ( ua ) = ( q ∗ u ) q ◦ u ) ∗ a ) , for each q ∈ Q , ( u, a ) ∈ A ∗ × A . See 1. fig. for interpretation ofMealy machine as a word transducer. Henceforth, we shall omitparentheses if there is no danger of confusion. So, for example, wewill write q ◦ u ∗ a instead of ( q ◦ u ) ∗ a. Similarly, we will write q ◦ q ′ ∗ a instead of q ◦ ( q ′ ∗ a ) where q ′ ∈ Q .2 .2. Definition. A 3-sorted algebra V = h Q, A, B, q , ◦ , ∗i iscalled an initial Mealy machine if h Q, A, B, ◦ , ∗i is a Mealy machineand q ∈ Q . Let ( q, x, y ) ∈ Q × A ω × B ω . We write y = q ∗ x if ∀ n ∈ N y [0 , n ] = q ∗ x [0 , n ] and say machine V transforms x to y . We say initialmachine V transforms x to y if q = q . We refer to words x and y as machines input and output , respectively. Look at 2. fig. for examples of machines V and V . There we have V : a = (0 , , a = (0 , , a = (1 , V : q ◦ a = q + a (mod2) , q ∗ a = q · a (mod2).We might refer to operations ◦ and ∗ as machine transition and output functions , respectively. From now on we would use notation h Q, A, B i to denote Mealy machine without specifying operations ◦ and ∗ . Similarly we would use notation h Q, A, B ; q i to denote initialMealy machine. Henceforth we would use terms machine and initialmachine to refer to, respectively, Mealy machine and initial Mealymachine.Suppose that we are given two initial machines V = h Q, A, B ; q , ◦ , ∗i and V ′ = h Q ′ , A ′ , B ′ ; q ′ , ´ ◦ , ´ ∗i . Schematicallyit is shown in 3.a. fig.We want to connect output of machine V to input of machine V ′ (shown in 3.b. fig.). Clearly, in this situation, we have v = v ′ .Suppose that B ⊆ A ′ , then for machine V ′ input we always canuse word v = q ∗ u . Therefore word w is correctly defined as w ↽ q ′ ´ ∗ ( q ∗ u ) . Our goal is to create initial machine V ′′ = h Q ′′ , A, B ′ ; q ′′ , ¨ ◦ , ¨ ∗i (shown3.c. fig.) such as ∀ u ∈ A ∗ q ′′ ¨ ∗ u = q ′ ´ ∗ ( q ∗ u ) . The new machine V ′′ is called a serial or cascade composition (or connection ) of machines V and V ′ . To denote serial composition ✍✌✎☞ V V ❄ ✍✌✎☞ ❄ ✍✌✎☞ ❄ ✍✌✎☞ ❄⑦ ❂⑥ ❃ a a /
11 0 / a a /
00 0 / a / / / a /
2. Figure: Machines V and V .3 ✛ ✛ ✛ ✛ uvv ′ w ′ v VV ′ b ✛ ✛ ✛ uvw VV ′ c ✛ ✛ uw V ′′
3. Figure: Serial composition.we will use notation V ❀ V ′ . Formally, lets define a class V ❀ V ′ ↽ {h Q ′′ , A, B ′ ; q ′′ , ¨ ◦ , ¨ ∗i | ∀ u ∈ A ∗ q ′′ ¨ ∗ u = q ′ ´ ∗ ( q ∗ u ) } . Assume that • Q ′′ ↽ Q ′ × Q ; • q ′′ ↽ ( q ′ , q ); • ( q ′ , q )¨ ◦ a ↽ ( q ′ ´ ◦ q ∗ a, q ◦ a ); • ( q ′ , q )¨ ∗ a ↽ q ′ ´ ∗ q ∗ a . For ∀ u ∈ A ∗ ( q ′ , q )¨ ◦ u = ( q ′ ´ ◦ q ∗ u, q ◦ u );( q ′ , q )¨ ∗ u = q ′ ´ ∗ q ∗ u. ✷ Proof of lemma is inductive. Let v = ua , where u ∈ A ∗ and4 ∈ A , then( q ′ , q )¨ ◦ v = ( q ′ , q )¨ ◦ ua = ( q ′ , q )¨ ◦ u ¨ ◦ a = ( q ′ ´ ◦ q ∗ u, q ◦ u )¨ ◦ a = (cid:0) ( q ′ ´ ◦ q ∗ u )´ ◦ (( q ◦ u ) ∗ a ) , ( q ◦ u ) ◦ a (cid:1) = (cid:0) q ′ ´ ◦ ( q ∗ u )´ ◦ ( q ◦ u ∗ a ) , q ◦ u ◦ a (cid:1) = (cid:0) q ′ ´ ◦ (( q ∗ u ) q ◦ u ∗ a )) , q ◦ ua (cid:1) = ( q ′ ´ ◦ q ∗ ua, q ◦ v ) = ( q ′ ´ ◦ q ∗ v, q ◦ v );( q ′ , q )¨ ∗ v = ( q ′ , q )¨ ∗ ua = (cid:0) ( q ′ , q )¨ ∗ u (cid:1) q ′ , q )¨ ◦ u ¨ ∗ a = ( q ′ ´ ∗ q ∗ u ) q ′ ´ ◦ q ∗ u, q ◦ u )¨ ∗ a = (cid:0) q ′ ´ ∗ q ∗ u (cid:1) (cid:0) ( q ′ ´ ◦ q ∗ u )´ ∗ (( q ◦ u ) ∗ a ) (cid:1) = (cid:0) q ′ ´ ∗ ( q ∗ u ) (cid:1) (cid:0) ( q ′ ´ ◦ ( q ∗ u ))´ ∗ ( q ◦ u ∗ a ) (cid:1) = q ′ ´ ∗ (cid:0) ( q ∗ u ) q ◦ u ∗ a ) (cid:1) = q ′ ´ ∗ ( q ∗ ua ) = q ′ ´ ∗ q ∗ v. V ′′ ∈ V ❀ V ′ .
3. Sequential functions
A total mapping f : A ∗ → B ∗ is called a se-quential function if (i) ∀ u ∈ A ∗ | u | = | f ( u ) | ;(ii) u ∈ Pref( v ) ⇒ f ( u ) ∈ Pref( f ( v )) . For all sequential functions, we have that if u ∈ Pref( v ) ∩ Pref( w ) , then f ( u ) ∈ Pref( f ( v )) ∩ Pref( f ( w )) . It states that if words u and v have matching prefixes of length k , then words f ( u ) and f ( v ) have matching prefixes of length k . ✷ Suppose that u ∈ Pref( v ) ∩ Pref( w ), then accordingly withthe definition of sequential function f ( u ) ∈ Pref( f ( v )) and f ( u ) ∈ Pref( f ( w )). Let f : A ∗ → B ∗ be a sequential function and u ∈ A ∗ , then f u ( v ) define a prefix of mapping f ( uv ) with length | v | .The mapping f u is called a quotient of sequential function f . f ( uv ) = f ( u ) f u ( v ) . uw ∈ Pref( uv ) ⇔ w ∈ Pref( v ) . ⇒ If uw ∈ Pref( uv ), then there exists word w ′ such as uww ′ = uv . Hence ww ′ = v and therefore w ∈ Pref( v ). ⇐ If w ∈ Pref( v ), then exists word w ′ such as ww ′ = v . Hence uww ′ = uv and therefore uw ∈ Pref( uv ). The quotient f u is a sequential function. ✷ Accordingly with the definition of quotient | v | = | f u ( v ) | . Inthe same time, if w ∈ Pref( v ), then [by lemma3.5.] uw ∈ Pref( uv ),therefore f ( u ) f u ( w ) = f ( uw ) ∈ Pref( f ( uv )) . It holds that Pref( f ( uv )) = Pref( f ( u ) f u ( v )), therefore f ( u ) f u ( w ) ∈ Pref( f ( u ) f u ( v )). Hence [3.5. Lemma] f u ( w ) ∈ Pref( f u ( v )). If mapping f : A ∗ → B ∗ is a sequential function,then f u ( vw ) = f u ( v ) f uv ( w ) . ✷ Suppose that w ∈ A ∗ , then we have f ( uvw ) = f ( u ) f u ( vw ) ,f ( uvw ) = f ( uv ) f uv ( w ) = f ( u ) f u ( v ) f uv ( w ) . Hence f u ( vw ) = f u ( v ) f uv ( w ) . Let mapping f : A ∗ → B ∗ be a sequentialfunction. If f u = f u ′ , then ∀ v ∈ A ∗ f uv = f u ′ v . ✷ Let w ∈ A ∗ , then f u ( v ) f uv ( w ) = f u ( vw ) = f u ′ ( vw ) = f u ′ ( v ) f u ′ v ( w ) . We have if | f u ( v ) | = | f u ′ ( v ) | , then f uv ( w ) = f u ′ v ( w ). Hence f uv = f u ′ v . Let mappings f : A ∗ → B ∗ and g : B ∗ → C ∗ be sequential functions, then ∀ u ∈ A ∗ ( g ◦ f ) u = g f ( u ) ◦ f u . ✷ We have g ( f ( ux )) = g ( f ( u ) f u ( x )) = g ( f ( u )) g f ( u ) ( f u ( x )) . Hence ( g ◦ f ) u ( x ) = g f ( u ) ( f u ( x )) . . Restricted sequential functions Let mapping f : A ∗ → B ∗ be a sequential func-tion. The function f defines set Q f ↽ { f u | u ∈ A ∗ } , where f u is a quotient of f . The function f is called a restrictedsequential function (or sequential function on restricted domain) ifthe set Q f is finite. For each function f : A ∗ → B ∗ there exists ainitial machine V = h Q f , A, B ; q i such as ∀ v ∈ A ∗ f ( v ) = q ∗ v. We might say that output function of machine V is equal withfunction f . ✷ Let Q f = { q , q , . . . , q k } , where element q is a set containingquotient f λ . Suppose that f u ∈ q , a ∈ A , f ua ∈ q ′ and f u ( a ) = b ,then q ◦ a ↽ q ′ , q ∗ a ↽ b. (i) f u ∈ q ⇒ f uv ∈ q ◦ v. ▽ If v = λ , then f uλ = f u ∈ q = q ◦ λ .Further proof is inductive, given that f uv ∈ q ◦ v . Hence accord-ingly with the definition of the transition function ◦ of machine f uva ∈ ( q ◦ v ) ◦ a = q ◦ va. △ (ii) f u ∈ q ⇒ q ∗ v = f u ( v ) . ▽ If v = λ , then q ∗ λ = λ = f u ( λ ).Further proof is inductive, given that q ∗ v = f u ( v ). Since f uv ∈ q ◦ v , then accordingly with the definition of output function ∗ ofmachine f uv ( a ) = q ◦ v ∗ a . Hence q ∗ va = q ∗ v q ◦ v ∗ a = f u ( v ) f uv ( a ) = f u ( va ) . △ We note that f λ = f hence f ( v ) = f ( λ ) f λ ( v ) = f λ ( v ). Accord-ingly with the definition of element q we have f λ ∈ q hence f ( v ) = f λ ( v ) = q ∗ v. Let P A be a set, elements of which are all possible restrictedsequential functions f : A ∗ → A ∗ . Serial composition of two Mealymachines is a Mealy machine, thereby composition f ( g ( u )) of tworestricted sequential functions f : A ∗ → A ∗ , g : A ∗ → A ∗
7s restricted sequential function. Thereby we have proved that P A isa semigroup. The operation of semigroup is composition of restrictedsequential functions.In particular, when A = 0 , ∗ we would use simpler notation P for denoting semigroup P , .
5. Group AS h G, ◦i be a monoid. An element x ∈ G has inverse (is invertible )if ∃ y ∈ G ( y ◦ x = λ = x ◦ y );where λ is neutral element of monoid G . Element y is called dual (or inverse ) of element x and is usually denoted as x − . A monoid G , each element of whom is invert-ible is called a group. The set AS ↽ { f ∈ P | f is bijection } is a group in which the group operation is composition of restrictedsequential functions. ✷ Let f ∈ AS , then there exists a Mealy machine V = h Q, , , , q , ◦ , ∗i and for ∀ u ∈ , ∗ f ( u ) = q ∗ u .We henceforth use fallowing notation: ¯0 ↽ ↽
0. Let V ′ = h Q, , , , q , ´ ◦ , ´ ∗i be a new Mealy machine, where q ´ ∗ a ↽ q ∗ a,q ´ ◦ a ↽ ( q ◦ a, if q ∗ a = a ; q ◦ ¯ a, if q ∗ a = ¯ a. The Mealy machine V ′ define restricted sequential function g ( u ) ↽ q ´ ∗ u .We need to prove that g = f − , i.e., that g is inverse function of f . (i) Suppose that Q ′ ↽ { q | ∃ u ∈ , ∗ q = q ◦ u } . Lets prove that ∀ q ∈ Q ′ ∀ a ∈ , q ∗ ( q ∗ a ) = a. (1)Let q = q ◦ u and thus f ( ua ) = q ∗ ua = q ∗ u q ◦ u ∗ a = q ∗ u q ∗ a,f ( u ¯ a ) = q ∗ u ¯ a = q ∗ u q ◦ u ∗ ¯ a = q ∗ u q ∗ ¯ a. Given that f is a bijection, we have that q ∗ a = q ∗ ¯ a. (2)8 If q ∗ a = a , then q ∗ ( q ∗ a ) = q ∗ a = a . • If q ∗ a = ¯ a , then by (2) we have that q ∗ ¯ a = a . Hence q ∗ ( q ∗ a ) = q ∗ ¯ a = a .This proves that f g ( a ) = a .(ii) Suppose that q ∈ Q ′ and a ∈ , • If q ∗ a = a , then q ´ ◦ a = q ◦ a . Hence q ´ ◦ ( q ∗ a ) = q ´ ◦ a = q ◦ a. • If q ∗ a = ¯ a , then by (2) q ∗ ¯ a = a . Hence q ´ ◦ ( q ∗ a ) = q ´ ◦ ¯ a = q ◦ ( q ∗ ¯ a ) = q ◦ a. This proves that q ´ ◦ ( q ∗ a ) = q ◦ a. (3)(iii) Suppose that q = q ◦ u and f q : 0 , ∗ → , ∗ : v q ∗ v, then f ( uv ) = q ∗ uv = q ∗ u q ◦ u ∗ v = f ( u ) f q ( v ). Suppose that w ∈ , ∗ and we have that f is a surjection therefore there exists u such as f ( u ) = f ( u ) w. Hence u = u u , where | u | = | f ( u ) | and | u | = | w | . Then thereexists u such as f ( u u ) = f ( u ) w . Then u = u because f isan injection. This proves f ( u ) w = f ( uu ) = f ( u ) f q ( u ). Hence w = f q ( u ), and f q is a surjection.If f q ( u ) = w , then f ( uu ) = f ( u ) w . Function f is an injection,thereby u = u . This proves that function f q is an injection.Finally, we conclude that f q is a bijection for each q ∈ Q ′ .(iv) Suppose that f ′ q is a restricted sequential function defined byMealy machine V ′ such as f ′ q : 0 , ∗ → , ∗ : u q ´ ∗ u. In section (i) we proved (equation (7)), i.e., f q f ′ q ( a ) = a , if a ∈ , q ∈ Q ′ and all words u of length n holds f q f ′ q ( u ) = u .Suppose that a ∈ ,
1, then f q f ′ q ( au ) = q ∗ q ´ ∗ au = q ∗ ( q ´ ∗ a q ´ ◦ a ´ ∗ u ) = q ∗ ( q ∗ a q ´ ◦ a ´ ∗ u )= q ∗ q ∗ a q ◦ ( q ∗ a ) ∗ ( q ´ ◦ a ´ ∗ u )= a q ´ ◦ a ∗ ( q ´ ◦ a ´ ∗ u ) = a f q ´ ◦ a f ′ q ´ ◦ a ( u ) = au.
9e note that q ´ ◦ a = q ◦ ( q ∗ a ) ∈ Q ′ . This concludes inductive partof the proof. We note that f ( u ) = f q ( u ) and f − ( u ) = f ′ q ( u ).(v) Let f and g be restricted sequential functions from the set AS , Then there exists Mealy machines V = h Q, , , , q , ◦ , ∗i , V ′ = h Q ′ , , , , q ′ , ´ ◦ , ´ ∗i such that for ∀ u ∈ , ∗ f ( u ) = q ∗ u ∧ g ( u ) = q ′ ´ ∗ u The serial composition V ❀ V ′ is machine, which realizes composi-tion of functions f and g . In other words, if ˘ V = h ˘ Q, , , ,
1; ˘ q , ˘ ◦ , ˘ ∗i ∈ V ❀ V ′ , then ∀ u ∈ , ∗ gf ( u ) = ˘ q ˘ ∗ u.V ✍✌✎☞ ❄ q / I : 0 , ∗ → , ∗ : u u is a restrictedsequential function. The machine implementing identity functiongiven as V (shown in 4.fig.).(vii) We note that compositions of functions is associative. Fi-nally, we have proven (in (iv) – (vi)) that AS is a group.Suppose that X ⊆ G , then h X i ↽ \ X ⊆ H ≤ G H, i.e., we consider intersection of all subgroups of group G containingset X . Notification . In this situation similarly as in case of the semi-groups we use notation h X i . The group h X i is called subgroup generated byset X . If h X i = G , then set X is called the set of generators of group G . Elements of set X are called generators . If X = { x , x , . . . , x n } ,then the following notification is used h x , x , . . . , x n i ↽ h X i . In this case (set X — finite), group G is finitely generated. .4. Definition. A group G is called cyclic group if ∃ g ∈ G ∀ x ∈ G ∃ n ∈ Z x = g n . In this case an element g is called a generator of cyclic group G . Let h G, ⊙i be group. The cardinality of set G ,denoted as | G | , is called the order of group G . A group G is called a finite group if its cardinality | G | is a naturalnumber, denoted | G | < ℵ . In opposite case, a group G is called an infinite group and denotated as | G | ≥ ℵ . Let a be an element of group G , then subgroups h a i order is called the order of element a . We use notification o ( a ) to denote order of element, i.e., o ( a ) ↽ |h a i| .
6. Elements of graph theory
A 2-sorted algebra h V, E, s, t i is called pseudo-graph if: (i) V is nonempty set; (ii) s, t are total mappings E s → V, E t → V , respectively. The elements of the set V are called vertices ,the elements of the set E are called arcs . If sets V and E are finite then the pseudograph G is called finite. In this case to denote a directed arc l we usenotation ( s ( l ) , t ( l )). The mapping s is called an initial vertex , whilethe mapping t is called a terminal vertex . We say that an arc l is incident out of the vertex s ( l ) and incident in to the vertex t ( l ). A pseudograph G ′ = h V ′ , E ′ , s ′ , s ′ i is called asub-pseudograph of pseudograph G = h V, E, s, t i if : (i) V ′ ⊆ V ; (ii) E ′ ⊆ n l ∈ E | s ( l ) ∈ V ′ ∧ t ( l ) ∈ V ′ o ; (iii) s ′ = s | E ′ ∧ t ′ = t | E ′ . a pseudograph h V, E, s, t i is called an orientedgraph, if : (i) ∀ l ∈ E s ( l ) = t ( l )(ii) ∀ l l ∈ E ( s ( l ) = s ( l ) ∧ t ( l ) = t ( l ) ⇒ l = l ) . This means that oriented graph can be given by two sets V , E ⊆ V .Usually in this case notation G ( V, E ) is used.
An oriented graph is called a graph (non ori-ented graph), if ∀ l ∈ E, ∃ l ∈ E : ( s ( l ) = t ( l ) ∧ s ( l ) = t ( l )) . In this case the pair of symmetrical arcs l , l is identified by the set { s ( l ) , s ( l ) } . This pair is called an edge connecting vertexes s ( l ) and s ( l ) .
11o denote edge { s ( l ) , s ( l ) } we use notification [ s ( l ) , s ( l )]. A n + 1 tuple c = ( v , l , v , l , ..., l n , v n +1 ) iscalled a walk in pseudograph h V, E, s, t i if (i) ∀ i ∈ , n + 1 , v i ∈ V ; (ii) ∀ j ∈ , n, l j ∈ E ; (iii) ∀ j ∈ , n, s ( l j ) = v j ∧ t ( l j ) = v j +1 . We say that a walk c = ( v , l , v , l , ..., l n , v n +1 ) of length startsat the vertex v and ends at the vertex v n +1 , connecting vertex v i with v j , i < j, i, j ∈ , n + 1. From the definition of graph, for eachwalk c = ( v , l , v , l , ..., l n , v n +1 ) there exists a reverse walk c ′ = ( v n +1 , l n , v n , l n − , ..., l , v ) , therefore we may say that the walks c connects vertices v i and v j , i, j ∈ , n + 1.Let m = ( m , m , . . . , m k ) be a walk, here the integer k is called extended length of walk m . A vertex v occurs in a walk m , if there ∃ j v = m j . An arc l occurs in a walk m if ∃ i l = m i . An edge e = { s ( l ) , s ( l ) } occurs in a walk if ∃ i l = m i ∨ l = m i . The walk m ′ = ( m ′ , m ′ , . . . , m ′ l ) is a part of walk m = ( m , m , . . . , m k ) if ∃ j l ≤ k − j ∧ ∀ i ∈ , lm ′ i = m j + i . A walk is called • closed, if it starts and ends at the same vertex; • trail, if it is without repeated arcs (and edges); • cycle, if it is nonempty closed trail; • simple trail (or a path), if it is without repeated vertices. If there exists a walk connecting vertices v and v , then there exists a trail connecting v and v . ✷ The proof is inductive by induction on extended length k ofwalk m = ( m , m , . . . , m k ). If the extended length of walk is 1 thenit is a trail.If m is not a trail, then there exists an edge l such that l = m i = m i + j , where 1 < i < i + j < k . Then for vertices m i − = m i + j − , m i +1 = m i + j +1 . The extended length of walks( m , . . . , m i − , l, m i + j +1 , m i + j +2 , . . . , m k )is less than k , therefore accordingly to the inductions hypothesisthere exist a trail connecting v and v . If there exists two different trails connectingvertices v and v , then there exist a cycle. Suppose that ( v , α , α , . . . , α k , v ) and ( v , β , β , . . . , β n , v )are two different trails connecting vertices v and v . Suppose that α = β , α = β , . . . , α s = β s , α s +1 = β s +1 , then α s is a vertex and( α s , α s +1 , . . . , α k , v ) , ( β s , β s +1 , . . . , β n , v )are two different trails connecting vertex α s = β s and vertex v .Suppose that α t is the first element of sequence α s +1 , α s +2 , . . . , α k , matching with some element β τ in { β s +1 , β s +2 , . . . , β n } . Then( α s , α s +1 , . . . , α t = β τ , β τ − , . . . , β s +1 , β s = α s )is a cycle, otherwise( α s , α s +1 , . . . , α k , v , β n , . . . , β s +1 , β s )is a cycle. An edge [ v , v ] of graph is called to be incidentto vertex v , and also to vertex v . Such vertices v and v are calledneighboring vertices. Thus vertex v is a neighbor of vertex v andvice versa. The number of edges incident to a vertex v is called thedegree of vertex. In this case we use notification deg ( v ). A vertex v is called isolatedvertex , if deg ( v ) = 0. A vertex v is called a leaf if deg ( v ) = 1. Let G ( V, E ) be a finite graph, then X v ∈ V deg ( v ) = 2 | E | ✷ As each edge is incident to two vertices, the sum of all degreesof vertices is twice the number of edges 2 | E | . If G ( V, E ) is a graph in which the degree ofeach vertex v is deg ( v ) ≥ , then for sets V, E holds | E | ≥ | V | . ✷ | E | = P v ∈ V deg ( v ) ≥ | V | . Hence | E | ≥ | V | . A graph is called connected if for each twovertices v , v there exist a walk m starting at v and ending at v .Otherwise graph is called disconnected. An edge [ a, b ] of connectedgraph G(V,E) is called a bridge if graph G ( V, E \ { [ a, b ] } ) is discon-nected. A connected graph G ( V , E ) is called connected component of G ( V, E ) if it is maximal connected subgraph. .13. Lemma. If | E | = | V | − , then the graph G ( V, E ) is dis-connected. ✷ Proof done by induction on n = | V | . If n = 2, then n − G ( V, E ) be a graph with n + 1 vertices and n − v with degree 1 or 0. If vertex v is isolated then graph G ( V, E ) is disconnected. Otherwise, if v isa leaf, we exclude vertex v and edge incident to it from the graph.The obtained graph G ′ has n vertices and n − G ′ is not connected. Therefore thereexist two vertices v and v for whom there is no trail connectingthem. If such trail κ would exists in the graph G ( V, E ) then vertex v would occur in it. We have that v = v = v hence walk κ can’t be atrail because vertex v is a leaf. Hence graph G ( V, E ) is disconnected.
A graph P is called an underlying graph oforiented graph G , if its obtained by replacing all arcs of graph G with edges. Replacement operation involves assigning to each arc in the originalgraph an arc symmetrical to it.
A connected graph without cycles is called atree. An oriented graph is called a oriented tree, if its underlyinggraph is a tree. An oriented tree is called a rooted oriented tree, ifthere exist a vertex σ such that for every other vertex v there exista walk connecting σ and v . Such vertex σ is called a root. The fallowing properties of graph T are equiv-alent:1. T = T ( V, E ) is a tree;2. T is connected and each edge is a bridge;3. T is connected and contains n − edge;4. T is without cycles and contains n − edge;5. Each two vertices of T are connected only one trail;6. T is without cycles, but connecting any two non neighboringvertices with a new edge would create a cycle. ✷ . ⇒ . The graph T is connected by definition, by the ex-clusion of an edge we cannot maintain connected graph, otherwise,there should be a cycle.2 . ⇒ . The proof is done by an induction on n . Let excludeone edge, say, [ v , v ]. We obtain a new graph T ( V, E ), where14 ↽ E \ { [ v , v ] } . Let define a subset of vertex set of the graph T ( V, E ): V ↽ { v | there exist a walk connecting vertex v and v } ,V ↽ { v | there exist a walk connecting vertex v and v } . (i) The graph T ( V, E ) is disconnected, therefore V ∩ V = ∅ .Otherwise there would exist a vertex v ∈ V ∩ V , implying existenceof walks µ , µ connecting vertex v to v and, respectively, vertex v to v . Therefore there would exist a walk m = ( m , m , . . . , m s )connecting vertices v v and v .Let w , w be two arbitrary chosen vertices of the graph T ( V, E ).As the graph T is connected, there exist a walk connecting w and w . Therefore (by proposition 6.7) there exist a trail κ =( κ , κ , . . . , κ ν ) connecting w and w . If the edge [ v , v ] is notoccurring in the trail κ , then κ is a trail in graph T ( V, E ) connect-ing vertices w and w . If trail κ contains edge [ v , v ], then existsindex i such as [ v , v ] = κ i . Hence for the part of trail we have( κ i − , κ i , κ i +1 ) = ( v , [ v , v ] , v ) , or ( κ i − , κ i , κ i +1 ) = ( v , [ v , v ] , v ) . In first case( κ , . . . , κ i − , m , m , . . . , m s − , κ i +1 , κ i +2 , . . . , κ ν )is a walk in graph T ( V, E ) connecting vertices w and w . In secondcase ( κ , . . . , κ i − , m s − , m s − , . . . , m , κ i +1 , κ i +2 , . . . , κ ν )is walk in graph T ( V, E ) connecting vertices w and w . Thereforegraph T ( V, E ) is connected. Thus we have arrived at a contradic-tion!(ii) Lets prove that V ∪ V = V . Suppose that v ∈ V ∪ V , theethere exist a trail α = ( α , α , . . . , α t ) , in graph T connecting vertices v and v . If edge [ v , v ] occurs intrail α , then v ∈ V . If edge [ v , v ] occurs in trail α , then there existindex i such that α i = [ v , v ]. Hence for the part of trail( α i − , α i , α i +1 ) = ( v , [ v , v ] , v ) , or ( α i − , α i , α i +1 ) = ( v , [ v , v ] , v ) .
15n first case ( α , α , . . . , α i − ) is a trail in graph T ( V, E ) connecting v and v , therefore v ∈ V . In second case ( α , α , . . . , α i − ) is a trailin T ( V, E ) connecting v and v , therefore v ∈ V .Hence graph T ( V, E ) contains two connected components T ( V , E ) and T ( V , E ), E = E | V × V , E = E | V × V . As eachedge in graph T is a bridge, then in graphs T ( V , E ) and T ( V , E )all edges are bridges.According to the induction hypothesis: | E | = | V |− | E | = | V | −
1. As a result we have | E | = | E | + | E | + 1 = | V | − | V | − | V | − . . ⇒ . If graph would contain a cycle then excluding some edgebelonging to this cycle would not make graph disconnected. Thegraph T obtained in this way would be connected and would contain n − n vertices. This is in contradiction by lemma 6.13.4 . ⇒ . From given is clear that no two vertices are connectedwith more than one trail, otherwise there would exist a cycle. Weneed to prove that graph is connected. Suppose that graph is dis-connected, and contains multiple connected components. Each con-nected component is without a cycles, therefore (by definition 6.15)they are trees. Hence 3. condition is true, i.e., each connected com-ponent G ( V i , E i ) has property | E i | = | V i | −
1. Hence, if number ofconnected component are k : k X i =1 | E i | = k X i =1 | V i | − k,n − | E | = k X i =1 | E i | = k X i =1 | V i | − k = | V | − k = n − k. Therefore k = 1, i.e., we have only one connected component.5 . ⇒ . If there exist a cycle in graph T , then at least two distinctvertices could be connected with two different trails. This leads tocontradiction and therefore graph T is without cycles.Suppose that vertices v , v aren’t neighbors, then exists a trail( v , κ , . . . , κ k , v ) , connecting v and v . Let add a edge [ v , v ], then( v , κ , . . . , κ k , v , [ v , v ] , v )is a cycle.6 . ⇒ . Suppose that v , v aren’t neighbors, but adding edge[ v , v ] create a cycle. Therefore [ v , v ] occurs in a cycle. Let sup-pose that this new cycle is( v , [ v , v ] , v , κ , . . . , κ k , v ) , v , κ , . . . , κ k , v ) is a walk connecting v and v . Therefore T is a connected graph, additionally T is without cycles. Hence,accordingly to the definition of tree, T is a tree. The integer deg − ( v ) ↽ |{ l | t ( l ) = v }| is calledthe indegree of vertex v of pseudograph h V, E, s, t i . The integer deg + ( v ) ↽ |{ l | s ( l ) = v }| is called the outdegree of vertex v of pseu-dograph h V, E, s, t i . A pseudograph is called out– p –regular , if outdegree of all vertices is p . Pseudograph G is out–regular , if there exists such integer p , that G is out– p –regular pseudograph. A rooted oriented tree is calledregular ( p –regular), if it is out– p –regular. P v ∈ V deg + ( v ) = | E | . ✷ Each arch starts with a vertex.
A graph G ( V, E ) is called vertex infinite, ifits vertex set V is infinite. If p > , then each out– p –regular rooted treeis infinite. ✷ If infinite rooted tree G ( V, E ) is out– p –regular, then p | V | = X v ∈ V deg + ( v ) S . = | E | . If p >
0, then | V | ≤ p | V | = | E | , but for all finite trees holds | E | = | V | −
1. This is a contradiction!
The indegree for root of an out– p –regular rootedtree is deg − ( σ ) = 0 . ✷ Suppose that σ is the root of a out– p –regular tree and deg − ( σ ) >
0, then there exist a vertex v and an arch l such as s ( l ) = v and t ( v ) = σ . As σ is the root, then there exist a walk( σ, c , c , . . . , c n , v ) , connecting σ with v . Hence( σ, c , c , . . . , c n , v, l, σ )is a cycle. A contradiction! If there exist a trail connecting vertices v and v , then there exist a simple trail (path) connecting v and v . Suppose that c = ( c , c , . . . , c n ) is a trail connecting v and v , i.e., c = v and c n = v . If v = v , then trail c = ( v ) is simple.Let v = v . The proof is done by induction on length of trail c .If c length is 3, then it is a simple trail. Suppose that n > c is not a simple trail, then there exist vertex c i such that c i = c i + j ,where 1 ≤ i < i + j ≤ n. We can not have both 1 = i and i + j = n in the some time because c = v = v = c n . Trail( c , c , . . . , c i , c i + j +1 , . . . , c n − , c n )connects v and v . Length of the trail is less than n, therefore bythe induction hypothesis exists a simple trail connecting v and v .If i + j = n , then trail ( c , c , . . . , c i ) connects v and v . The lengthof this trail also is less than n, therefore by the induction hypothesisexists simple trail connecting v and v . If v is a vertex of a out– p –regular rooted treeand v is not a root then deg − ( v ) = 1 . ✷ Let G be a p –regular tree and P an underlying graph of G .Suppose that deg − ( v ) >
1, then exists at least 2 vertices v and v such that ( v , v ) and ( v , v ) are arcs. Suppose that σ is the root,then exists simple trails( σ, α , α , . . . , α k , v ) and ( σ, β , β , . . . , β n , v ) . (i) Is the walk c ↽ ( σ, α , α , . . . , α k , v , ( v , v ) , v ) a simple trail?If we assume contrary, then there exists a vertex α i such as α i = v . Hence, if orientation of arcs is removed),( α i , α i +1 , . . . , α k , [ v , v ] , v )is a cycle in underlying graph P . This is a contradiction, since P is a tree. Therefore c is a simple trail. Similarly is provable that c ↽ ( σ, β , β , . . . , β n , v , ( v , v ) , v ) is a simple trail.(ii) If the orientation of arcs is removed, then c and c are 2distinct trails in the underlying graph P connecting vertices σ and v . This is a contradiction, since P is a tree. For each vertex v of a p –regular tree thereexist only one simple trail connecting root σ with vertex v . ✷ Let c v be the shortest trail connecting root σ with vertex v ,i.e., if c is a trail connecting σ with v , then trail c v is not longer than c . As deg − ( σ ) = 0, then only trail connecting σ with σ is of length0. 18urther proof is inductive, given that for each vertex v such as c v ≤ l exists only one simple trail connecting root σ with v . Supposethat w is vertex such as trail c w = ( σ, α , α , . . . , α n , α n +1 , w ) is oflength l + 1.Here c ↽ ( σ, α , α , . . . , α n ) is a trail of length l , α n +1 = ( α n , w )is an arc and α n is a vertex. As deg − ( w ) = 1, then for each trail( σ, β , . . . , β k , β k +1 , w ) connecting σ with w , arch β k +1 = ( α n , w )and vertex β k = α n .Therefore ( σ, β , . . . , β k ) is trail connecting σ with α n , i.e.,( σ, β , . . . , β k − , β k ) = ( σ, β , . . . , β k − , α n ) . The length of trail c is l , therefore length of c α n is not larger than l . From induction hypothesis, there exist only one trail connecting σ with α n . Then c α n = c = ( σ, α , α , . . . , α n )= ( σ, β , . . . , β k − , α n )= ( σ, β , . . . , β k − , β k ) . Hence( σ, β , . . . , β k − , β k , β k +1 , w ) = ( σ, α , α , . . . , α n , β k +1 , w )= ( σ, α , α , . . . , α n , ( α n , w ) , w )= ( σ, α , α , . . . , α n , α n +1 , w )= c w . Finally, we conclude that unity of trail is proved.
The integer l is called level of vertex v if l islength of trail connecting root σ with vertex v . For each vertex v of a p –regular tree existsonly one level. Suppose that G ( V, E ) is a p –regular tree with root σ . Let l ( v )denote length of trail connecting root σ with vertex v . L ( n ) ↽ { v | l ( v ) = n } . Elements of set L ( n ) are called n –th level vertices. From collorary6.26 set L ( n ) is defined uniquely and correctly. | L ( n ) | = p n for each p –regular tree G ( V, E ) . ✷ As G ( V, E ) is p –regular, then | L (1) | = p . Further proof isinductive , given that | L ( n ) | = p n . Suppose that v , v ∈ L ( n ) and( v , u ) , ( v , u ) are arcs of tree G ( V, E ), then u = u . Otherwise,19here would exist 2 distinct trails connecting root to vertex u . Thisis in contradiction with proposition 6.24.Suppose G ( V, E ) is a p –regular, therefore for each vertex v oflevel n the outdegree deg + ( v ) = p . Hence | L ( n + 1) | = | L ( n ) | p = p n p = p n +1 . Pair of mappings ( f : V → V , f : E → E ) is called a homomorphism of pseudographs h V , E , s , t i h V , E , s , t i , if for each arch l of pseudograph h V , E , s , t i holds: (i) f ( s l )) = s ( f ( l )) , (ii) f ( t ( l )) = s ( f ( l )) . If mappings f and f are bijections, then homomorphism is calledan isomorphism . If h V , E , s , t i = h V , E , s , t i , then homomor-phism is called an endomorphism . If additionally f , f are bijec-tions, then endomorphism is called an automorphism . A 3-sorted algebra h V, E, A, s, t, i i is called alabeled pseudograph if : (i) h V, E, s, t i is a pseudograph, (ii) i is a total mapping E i → A . Image i ( l ) ∈ A is called a label of arc l ∈ E . ✫✪✬✩✫✪✬✩ ✫✪✬✩✫✪✬✩ ✲✲❄ ❄ s ( l ) t ( l ) s ( f ( l )) t ( f ( l )) i ( l ) li ( f ( l )) f ( l ) f f
5. Figure: Isomorphism of labeled pseudographs.
Labeled pseudographs h V , E , s , t , i i , h V , E , s , t , i i re isomorph labeled pseudographs, if there exist bijections f : V → V , f : E → E , such that for each arch l of a labeled pseudograph h V , E , s , t , i i holds: (i) f ( s ( l )) = s ( f ( l )) , (ii) f ( t ( l )) = s ( f ( l )) , (iii) i ( l ) = i ( f ( l )) . Look for illustration in 5. fig.If pseudographs h V , E , s , t i , h V , E , s , t i are oriented graphs G ( V , E ) , G ( V , V ), then mapping f : V → V uniquely definesmapping f : E → E : f : E → E : ( u, v ) ( f ( u ) , f ( v )) . Therefore for the definition of homomorphism of oriented graphsonly one mapping is needed.
A mapping f : V → V is called a homo-morphism of (oriented) graphs G ( V , E ) , G ( V , E ) , if for each arc ( u, v ) ∈ E there exist ( f ( u ) , f ( v )) ∈ V . If G ( V , E ) = G ( V , E ), then a homomorphism is called an endo-morphism of oriented graph G ( V , E ).The situation is different with (oriented) graph isomorphism.Condition stating that f : V → V is a bijection is not enough(look 6. fig.). Mapping f : u v , u v , u v is a bijection of vertex sets of graphs G , G , therefore an homomor-phism, but it does not define the bijection of sets of arcs. A homomorphism f of (oriented) graphs G , G is called an isomorphism if f is a bijection of the vertex sets and f − is a graph homomorphism of G , G . If G = G , then isomorphism is called an automorphismu of graph G . Let G ( V , E ) , G ( V , E ) be oriented rootedtrees. Then a homomorphism f of oriented graphs G ( V , E ) , G ( V , E ) is called a homomorphism of oriented rooted trees G ( V , E ) , G ( V , E ) , if f ( σ ) = σ , where σ , σ are the roots oftrees G ( V , E ) , G ( V , E ) , respectively. The homomorphism of oriented rooted trees f is called an isomor-phism of oriented rooted trees, if f is an isomorphism of oriented21 ✉✉ ✲ u u u G ✉ ✉✉ ✲✁✁✁✁✁✁✕ v v v G
6. Figure: Nonisomorphic graphs.graphs G ( V , E ) , G ( V , E ). If G ( V , E ) = G ( V , E ), then homo-morphism of oriented rooted trees f is called an endomorphism oforiented rooted trees. If endomorphism f is an isomorphism, thenendomorphism f is called an automorphism of oriented rooted tree. If mappings f, g are automorphisms of ori-ented rooted tree G ( V, E ) , then f g is an automorphism of tree G ( V, E ) . ✷ Suppose that f, g are automorphisms of oriented rooted tree G ( V, E ) and s is the root of G ( V, E ). Then we have sf g = sg = s .If ( u, v ) is an arc of tree G ( V, E ), then ( uf, vf ) also is an arc of tree G ( V, E ) (by definitions 6.31 and 6.33 ) because f is an endomor-phism. As ( uf, vf ) is an arc of tree G ( V, E ), then ( uf g, vf g ) also isan arc of tree G ( V, E ) because g also is an endomorphism. Therefore f g is endomorphism of tree G ( V, E ).As f, g are automorphisms of tree G ( V, E ), then f − , g − areendomorphisms of the same tree (by definition 6.33). As f − , g − are endomorphisms, then g − f − is an endomorphism. Therefore( f g ) − is an endomorphism, because, as f, g are bijections, we have( f g ) − = g − f − .Finally, we have proven that f g is an automorphism of orientedrooted tree G ( V, E ).By Aut( G ( V, E )) we denote set of all automorphisms of orientedrooted tree G ( V, E ). h Aut( G ( V, E )) , ⊗i is a group.Here ⊗ : Aut( G ( V, E )) × Aut( G ( V, E )) → Aut G ( V, E ) is the composition of mappings. . Machine semigroups We would use notation M = h Q, A, ◦ , ∗i to denote Mealy machine M = h Q, A, A, ◦ , ∗i . In this section the input and output alphabetsof all Mealy machines are one and the same, additionally, we dropthe requirement that function Q × A ∗ → A must be surjective.For each q ∈ Q function Q × A ∗ → A defines a restricted sequentialfunction ¯ q : A ∗ → A ∗ : u q ∗ u. Fallowing definition is inductive. Let x ∈ Q + , then with xq wedenote function xq : A ∗ → A ∗ : u ( u ¯ x )¯ q ↽ ¯ q (¯ x ( u )) . Let M = h Q, A, ◦ , ∗i be a Mealy machine, thensemigroup generated by set { ¯ q | q ∈ Q } is called a machine semigroup M (automaton semigroup, semigroup of M ). We use notification h M i + to denote the semigroup of machine M . h M i + = { ¯ x | x ∈ Q + } Why we are interested in machine semigroups?The definition of machine semigroup implies that it is a convenientway for represent particular semigroup. In other words if we aregiven a Mealy machine M = h Q, A, ◦ , ∗i , then its machine semigroupis given as well.Another question arises:Is there a particular benefit from this representation? How informa-tive is particulat representation?It turns out that even a very small size Mealy machine givesinfinite semigroups. ✍✌✎☞ V V ❄ ✍✌✎☞ ❄ ✍✌✎☞✍✌✎☞ ❄✛ ✛ p q p q / / /
10 01 / /
17. Figure: |h V i + | = ∞ , |h V i + | = 4First we prove that semigroup h V i + V is infinite.We choose infinite word (01) ω ∈ , ω , it fallows that23 ¯ q ((01) ω ) = 1 (10) ω and qp ((01) ω ) = ¯ p (1 (10) ω ) = 0 (01) ω Further proof is inductive given( qp ) n − q ((01) ω ) = 1 n (10) ω and ( qp ) n ((01) ω ) = 0 n (01) ω . Hence ( qp ) n q ((01) ω ) = ¯ q (( qp ) n ((01) ω )) = ¯ q (0 n (01) ω )= 1 n (10) ω = 1 n +1) (10) ω and ( qp ) n +1 ((01) ω ) = ¯ p ( q ( qp ) n ((01) ω ))= ¯ p (1 n +1) (10) ω ) = 0 n +1) (01) ω . We have inductively proven that elements of semigroup h V i + ¯ q, qp, qpq, . . . , q ( pq ) n , ( qp ) n +1 , . . . are unique. Thereby semigroup h V i + is infinite.Now lets consider semigroup of machine V . Suppose that a ∈ , x ∈ , ω , then • ¯ p ( x ) = ¯ x, p ( x ) = ¯ p (¯ p ( x )) = ¯ p (¯ x ) = x = I ( x ) • ¯ q ( ax ) = a ¯ x, q ( ax ) = ¯ q (¯ q ( ax )) = ¯ q ( a ¯ x ) = x = I ( x ) • qp ( ax ) = ¯ p (¯ q ( ax )) = ¯ p (¯ ax ) = a ¯ x, pq ( ax ) = ¯ q (¯ p ( ax )) = q (¯ a ¯ x ) = ¯ ax Therefore we have pq = qp , t.i., ¯ p ¯ q = ¯ q ¯ p .Hence I ¯ p ¯ q pq I I ¯ p ¯ q pq ¯ p ¯ p I pq ¯ q ¯ q ¯ q pq I ¯ ppq pq ¯ q ¯ p I It fallows that h V i + is the Klain 4–group Z × Z . Lets choose an alphabet A = { a , a , . . . , a p } .Lets define a p –regular rooted tree ¯ K ( A ) as: • the words A ∗ are vertices of the tree; • the empty word λ is root of the tree; • the set { ( u, ua ) | u ∈ A ∗ ∧ a ∈ A } is the set of arcs.From construction we have that deg − ( λ ) = 0, ∀ u ∈ A + deg − ( u ) =1 and ∀ u ∈ A ∗ deg + ( u ) = p . The underlying graph of graph ¯ K ( A ) is a tree. υ υ υ υ υ υ λ ✍✌✎☞ ✁✁✁☛ ❆❆❆❯ ✍✌✎☞ ✁✁✁☛ ❆❆❆❯ ✍✌✎☞ ✁✁✁☛ ❆❆❆❯ ✍✌✎☞ ✁✁✁☛ ❆❆❆❯ ✍✌✎☞ ✁✁✁☛ ❆❆❆❯ ✍✌✎☞ ✁✁✁☛ ❆❆❆❯ ✍✌✎☞ ✁✁✁☛ ❆❆❆❯ ✍✌✎☞ ✁✁✁☛ ❆❆❆❯ ✍✌✎☞ ✁✁✁☛ ❆❆❆❯ ✍✌✎☞ ✁✁✁☛ ❆❆❆❯ ✍✌✎☞ ✁✁✁☛ ❆❆❆❯ ✍✌✎☞ ✁✁✁☛ ❆❆❆❯ ✍✌✎☞ ✡✡✡✡✡✢ ❏❏❏❏❏❫ ✍✌✎☞ ✡✡✡✡✡✢ ❏❏❏❏❏❫ K K (0 ,
8. Figure: Labeled rooted trees K and K (0 , ✷ For the underlying graph of ¯ K ( A ) the set of edges is { [ u, ua ] | u ∈ A ∗ ∧ a ∈ A } . Lets exclude an arbitrary edge [ u, ua ]. Let provethat for vertex w , which reachable from vertex ua , holds that ua ∈ Pref( w ). If length of walk is zero, then proposition holds. Supposethat proposition holds for all walks of length n starting at vertex ua . Lets choose an arbitrary walk( ua, l , v , l , v , . . . , l n , v n , l n +1 , v n +1 )of length n + 1, then ( ua, l , v , l , v , . . . , l n , v n ) is walk of length n ,and accordingly to the induction hypothesis ua ∈ Pref( v n ).(i) If v n = ua , then l n +1 = [ ua, uab ] for some b ∈ A because edge[ u, ua ] is excluded. Hence v n +1 = uab and ua ∈ Pref( uab ) = Pref( v n +1 ) . (ii) If v n = ua , then accordingly with induction hypothesis v n = uavb , where v ∈ A ∗ and b ∈ A . Thus p + 1 edge is incident to vertex v n , i.e.,[ uavb, uav ] , [ uavb, uavba ] , [ uavb, uavba ] , . . . , [ uavb, uava p ] . Therefore we have that v n +1 ∈ { uav, uavba , uavba , . . . , uava p } . Hence ua ∈ Pref( v n +1 ). This concludes inductive part of the proof.We have proven that vertex u is not reachable. This means thateach edge [ u, ua ] is a bridge, therefore (proposition 6.16) the under-lying graph is a tree.As the underlying graph of ¯ K ( A ) is a tree, deg − ( λ ) = 0, ∀ u ∈ A + deg − ( u ) = 1 and ∀ u ∈ A ∗ deg + ( u ) = p , thus we have proven,that ¯ K ( A ) is a p –regular rooted tree. In case when | A | = 2, the tree¯ K ( A ) is called a rooted binary tree .25 .6. Proposition. All rooted p –regular trees are isomorphic. ✷ Suppose that G ( V, E ) is a p –regular rooted tree with root σ .Lets prove that G ( V, E ) is isomorphic to tree ¯ K (0 , p − L ( n ) is set of n –th level vertices of graph G ( V, E ).Lets inductively define bijection f : V → , p − ∗ . f ( σ ) ↽ λ .Suppose that f already is defined for all elements of set S nk =0 L ( k ), v ∈ L ( n ) and f ( v ) = u ∈ , p − n , then deg + ( v ) = p . Suppose that( v, v ) , ( v, v ) , . . . , ( v, v p − )are arcs of tree G ( V, E ), then f ( v ) ↽ u , f ( v ) ↽ u , . . . , f ( v p − ) ↽ u ( p − . From the definition fallows that f and f − are bijections. We needto make sure that both mappings are homomorphisms. Suppose that( v, w ) ∈ E , then there exist n such as v ∈ L ( n ) and w ∈ L ( n + 1).According to the definition of f ∃ u ∈ , p − n , f ( v ) = u ∧ ∃ a ∈ , p − , f ( w ) = ua. Therefore( f ( v ) , f ( w )) = ( u, ua ) ∈ U ↽ { ( u, ua ) | u ∈ , p − ∗ ∧ a ∈ , p − } . We note that U is the set of arcs of tree ¯ K (0 , p − u, ua ) ∈ U , then exists n and v ∈ L ( n ) such as f ( v ) = u . According to the definition of f there exist w ∈ L ( n + 1)such as ( v, w ) ∈ E and f ( w ) = ua . Hence( f − ( u ) , f − ( ua )) = ( v, w ) ∈ E. As result with the precision up to isomorphism there exist onlyone p –regular rooted tree. For this reason, unless required otherwise,we choose tree ¯ K (0 , p −
1) as p –regular rooted tree.Let σ a → v a → · · · a n → v n (4)be a walk in labeled tree K (0 ,
1) (look 8. fig.), then v = a , v = a a , . . . , v n = a a . . . a n . Hence if we are interested only in those walks starting at root σ ,we can restrict our attention to ¯ K (0 , v n uniquely definespreviously mentioned walk (4).We will restrict our attention only on the endomorphisms of tree¯ K ( A ). With End( A ∗ ) we denote the set of all endomorphisms of tree¯ K ( A ). With Aut( A ∗ ) we denote the set of all automorphisms of tree¯ K ( A ). 26 .7. Lemma. If f ∈ End( A ∗ ) , then ∀ u ∈ A ∗ | f ( u ) | = | u | . ✷ We have f ( λ ) = λ because λ is a root. The set or arcs of tree¯ K ( A ) is U = { ( u, ua ) | u ∈ A ∗ ∧ a ∈ A } . As f is an endomorphism, then ∀ a ∈ A f ( λ, f ( a )) ∈ U . Hence ∃ b ∈ A ( f ( λ ) , f ( a )) = ( λ, f ( a )) = ( λ, λb ) = ( λ, b ) . Therefore f ( a ) = b and | f ( a ) | = | b | = 1 = | a | .Further proof is inductive, given u ∈ A n , then | f ( u ) | = | u | = n .Suppose that w ∈ A n +1 , then ∃ u ∈ A n ∃ a ∈ A w = ua . Hence( u, w ) = ( u, ua ) ∈ U . As f is an endomorphism, then ( f ( u ) , f ( w )) =( f ( u ) , f ( ua )) ∈ U . Hence | f ( ua ) | = | f ( u ) | + 1. Therefore | f ( w ) | = | f ( ua ) | = | f ( u ) | + 1 = | u | + 1 = | ua | = | w | . h End( A ∗ ) , ·i is a monoid. Here End( A ∗ ) × End( A ∗ ) · → End( A ∗ ) is a composition of mappings. ✷ The identity mapping I : A ∗ → A ∗ : u u is an endomorphismof tree ¯ K ( A ) and serves as neutral element. We need to prove, thatthe composition of endomorphisms is an endomorphism.Suppose that f, g ∈ End( A ∗ ) and ( u, v ) is an arc of tree ¯ K ( A ),then ( f ( u ) , f ( v )) is an arc of tree ¯ K ( A ), therefore ( g ( f ( u )) , g ( f ( v )))is an arc of tree ¯ K ( A ). Finally, we have that g ( f ( λ )) = g ( λ ) = λ .Hence g · f is an endomorphism. A mapping f : A ∗ → A ∗ is a sequential func-tion, if and only if f ∈ End( A ∗ ) . ✷ ⇒ Suppose that f : A ∗ → A ∗ is sequential function, then f ( λ ) = λ because accordingly to the definition of sequential function | f ( λ ) | = | λ | = 0. Suppose that ( u, ua ) is arc of tree ¯ K ( A ), then f ( u ) ∈ Pref( f ( ua )). Therefore | u | = | f ( u ) | and | ua | = | f ( ua ) | .Hence there exist some b ∈ A such as f ( ua ) = f ( u ) b . It means that( f ( u ) , f ( ua )) = ( f ( u ) , f ( u ) b ) is arc of tree ¯ K ( A ). ⇐ Suppose that f ∈ End( A ∗ ), then (Lemma 7.7) ∀ u ∈ A ∗ | f ( u ) | = | u | . Suppose that u ∈ A ∗ and a ∈ A , then ( u, ua ) is arc of tree ¯ K ( A ),therefore ( f ( u ) , f ( ua )) is an arc of tree ¯ K ( A ). Hence there exist some b ∈ A such as f ( ua ) = f ( u ) b . Therefore f ( u ) ∈ Pref( f ( ua )).Further proof is inductive, given u ∈ Pref( v ) ⇒ f ( u ) ∈ Pref( f ( v ))only if | v | − | u | ≤ n . 27uppose that u ∈ Pref( v ) and | v | − | u | = n + 1, then exists w ∈ A ∗ and a ∈ A such as v = uwa and | w | = n . As ( uw, uwa )is an arc of tree ¯ K ( A ), then f ( uw ) ∈ Pref( f ( uwa )). Accordinglywith the induction hypothesis f ( u ) ∈ Pref( f ( uw )), therefore f ( u ) ∈ Pref( f ( uwa )) = Pref( f ( v )). If mapping f : A ∗ → A ∗ is a restricted se-quential function, then f ∈ End( A ∗ ) . Suppose that q ∈ Q , then mapping ¯ q : A ∗ → A ∗ sometimes iscalled q action on tree ¯ K ( A ). The mapping φ : Q + → End( A ∗ ) : x ¯ x is ahomomorphism of semigroups. ✷ Suppose that u ∈ A ∗ and x, y ∈ Q + , then uφ ( xy ) = uxy =( u ¯ x )¯ y and ( uφ ( x )) φ ( y ) = ( u ¯ x )¯ y . Hence φ ( xy ) = φ ( x ) φ ( y )Suppose that u ∈ A ∗ and x ∈ Q + , then we would use notation ux ↽ uφ ( x ). For denoting semigroup Im( φ ) we would use notationΣ( M ), showing to which machine correspond given semigroup . A semigroup P is called a machine semigroup(automaton semigroup) if there exist a Mealy machine M such that P ∼ = Σ( M ) . As Ker( φ ) = { ( x, y ) | φ ( x ) = φ ( y ) } is a congruence, then Q + / Keris semigroup and Q + / Ker( φ ) ∼ = Im( φ ) = Σ( M ). Accordingly to theKer( φ ) definition x and y belong to the same coset, if φ ( x ) = φ ( y ).Hence ∀ u ∈ A ∗ ux = uφ ( x ) = uφ ( y ) = uy. Therefore ∀ α ∈ A ω αx = αy , where αx ↽ lim n →∞ α n , α n is aprefix of ω –word α .
8. Machine groups
Machine M = h Q, A, ◦ , ∗i is called to be in-vertible if for ∀ q ∈ Q mapping ¯ q | A is a bijection. If M = h Q, A, ◦ , ∗i is an invertible machine,then ∀ q ∈ Q ¯ q : A ∗ → A ∗ is a bijection. ✷ (i) Accordingly to the definition ¯ q | A is a bijection. We needto prove that ∀ n ∈ N ¯ q | A n is a bijection. This in turn would provethat ¯ q : A ∗ → A ∗ is a bijection.(ii) The proof is done by induction on n . Let w, w ′ ∈ A n +1 , thenexists words v, v ′ ∈ A n and a, a ′ ∈ A such that w ¯ q = q ∗ w = q ∗ va = q ∗ v q ◦ v ∗ a,w ′ ¯ q = q ∗ w ′ = q ∗ v ′ a ′ = q ∗ v ′ q ◦ v ∗ a ′ . ∀ q ∈ Q ¯ q | A n is a bijection and v = v ′ , then q ∗ v = q ∗ v ′ ,and therefore w ¯ q = w ′ ¯ q .Suppose that v = v ′ and a = a ′ . As for ∀ q ∈ Q mapping ¯ q | A n is a bijection, then q ◦ v | A is a bijection. Hence q ◦ v ∗ a = q ◦ v ∗ a ′ ,and therefore w ¯ q = w ′ ¯ q . Hence ¯ q | A n +1 is an injection.(iii) Suppose that w ∈ A n +1 , then there exist words v ∈ A n and b ∈ A such that w = vb . Suppose that ∀ q ∈ Q ¯ q | A n is a bijection,then there exist u ∈ A n such that v = u ¯ q = q ∗ u .If ∀ q ∈ Q ¯ q | A n is a bijection, then q ◦ u | A is a bijection, thereforeexists a ∈ A such that q ◦ u ∗ a = b . Hence ua ¯ q = q ∗ ua = q ∗ u q ◦ u ∗ a = vb = w. Thus ¯ q | A n +1 is a surjection. If M = h Q, A, ◦ , ∗i is an invertible machine,then ∀ q ∈ Q ∀ v ∈ A ∗ ∃ ! u ∈ A ∗ q ∗ u = v . ✷ ¯ q : A ∗ → A ∗ is bijection.Lets define for each invertible machine M = h Q, A, ◦ , ∗i Q − ↽ { q − | q ∈ Q } ;¯ q − : A ∗ → A ∗ : v vq − , where vq − ↽ u , if q ∗ u = v . ∀ q ∈ Q, ∀ u ∈ A ∗ , uqq − = uq − q = u . If q ∈ Q , then mapping ¯ q ∈ End( A ∗ ) for eachinvertible machine M = h Q, A, ◦ , ∗i . ✷ The arcs of tree K ( A ) are ( u, ua ), where u ∈ A ∗ and a ∈ A .Mapping ¯ q ∈ End( A ∗ ) if ( uq − , uaq − ) is an arc of tree K ( A ). Aswe are interested in endomorphisms of rooted oriented tree K ( A ),then we need to prove that λq − = λ . Last identity arises from thedefinition of mapping ¯ q , i.e., λq − = u ′ , if q ∗ u ′ = λ . The only word u ′ ∈ A ∗ , with property q ∗ u ′ = λ , is u ′ = λ . Therefore λq − = λ .Suppose that uq − = v and uaq − = w . According with defi-nition of ¯ q fallows that, q ∗ v = u and q ∗ w = ua . Suppose that w = v ′ b , v ′ ∈ A ∗ and b ∈ A , then ua = q ∗ w = q ∗ v ′ b = q ∗ v ′ q ◦ v ′ ∗ b. Hence q ∗ v ′ = u = q ∗ v . As ¯ q is a bijection, then v = v ′ . This meansthat uaq − = w = v ′ b = vb . Hence ( uq − , uaq − ) = ( v, w ) = ( v, vb ),where ( v, vb ) is an arc in tree K ( A ).For each q − ∈ Q − mapping ¯ q ∈ End( A ∗ ). We further woulduse notation q − ↽ ¯ q .Further definition is inductive. Suppose that x ∈ ( Q ∪ Q − ) + ,then with xq we would denote function xq : A ∗ → A ∗ : u ( u ¯ x )¯ q ↽ ¯ q (¯ x ( u )) , q ∈ Q . Contrary if q − ∈ Q − , then with xq − we denote function xq − : A ∗ → A ∗ : u ( u ¯ x )¯ q ↽ ¯ q (¯ x ( u )) . If M = h Q, A, ◦ , ∗i is an invertible machine, thenmapping φ : ( Q ∪ Q − ) + → End( A ∗ ) : x ¯ x is a homomorphism of semigroups. ✷ Suppose that u ∈ A ∗ and x, y ∈ ( Q ∪ Q − ) + , then uφ ( xy ) = uxy = ( u ¯ x )¯ y and ( uφ ( x )) φ ( y ) = ( u ¯ x )¯ y . Therefore φ ( xy ) = φ ( x ) φ ( y ). If M = h Q, A, ◦ , ∗i is invertible machine, thenhomomorphism φ : ( Q ∪ Q − ) + → End( A ∗ ) : x ¯ x is called the natural homomorphism. If M = h Q, A, ◦ , ∗i is invertible machine, then ∀ x ∈ ( Q ∪ Q − ) + mapping ¯ x is automorphism of ordered rooted tree ¯ K ( A ) . ✷ Suppose that q ∈ Q . We have already proved (by lemma 8.5),that ¯ q ∈ End( A ∗ ). By corollary 8.4 we have that ¯ q ¯ q = ¯ q ¯ q = I , i.e.,compositions of those mappings is the identity mapping. Therefore¯ q is the inverse mapping of ¯ q and an endomorphism. Hence ¯ q, ¯ q ∈ Aut(A ∗ ).Further proof is inductive. Let x ∈ ( Q ∪ Q − ) + , | x | = n and¯ x ∈ Aut( A ∗ ). If q ∈ Q ∪ Q − , then ¯ q ∈ Aut(A ∗ ). Hence xq = ¯ x ¯ q ∈ Aut(A ∗ ) (Proposition 6.34). Let M = h Q, A, ◦ , ∗i be an invertible ma-chine, then image Im φ of natural homomorphism φ : ( Q ∪ Q − ) + → End( A ∗ ) : x ¯ x is a subgroup of group Aut( A ∗ ) . ✷ Accordingly to corollary 8.4 we have I ∈ Im φ . If f, g ∈ Im φ ,then exists x, y ∈ ( Q ∪ Q − ) + such that xφ = f and yφ = g . Hence(by lemma 8.6) f g = ( xφ )( yφ ) = xyφ ∈ Im φ .If | x | = 1, then x ∈ Q ∪ Q − . Then we have x = q or x = q − forsome q ∈ Q .(i) Suppose that x = q , then q − φ ∈ Im φ and f ( q − φ ) = ( xφ )( q − φ ) = ( qφ )( q − φ ) = ( qq − φ ) = qq − S . = I . Therefore f − = q − φ ∈ Im φ . 30ii) Suppose that x = q − , then qφ ∈ Im φ and f qφ = ( xφ )( qφ ) = ( q − φ )( qφ ) = ( q − q ) φ = q − q S . = I . Therefore f − = qφ ∈ Im φ .(iii) Suppose that g ∈ Im φ, g = zφ , z ∈ ( Q ∪ Q − ) k and k ≤ n ,then exists z ′ ∈ ( Q ∪ Q − ) k such that g − = z ′ φ ∈ Im φ .Further proof is inductive. Let f ∈ Im φ and there exist x ′ ∈ ( Q ∪ Q − ) n +1 such that x ′ φ = f . If so, then here exists x ∈ ( Q ∪ Q − ) n and q ∈ Q ∪ Q − such that x ′ = xq . Accordingly to theinduction hypothesis here exists y ′ ∈ ( Q ∪ Q − ) n and q ′ ∈ Q ∪ Q − such that ( xφ ) − = y ′ φ ∈ Im φ and ( qφ ) − = q ′ φ ∈ Im φ . Hence q ′ y ′ ∈ ( Q ∪ Q − ) n +1 and ( q ′ y ′ ) φ ∈ Im φ , and we have also f (( q ′ y ′ ) φ ) = ( x ′ φ )(( q ′ y ′ ) φ ) = (( xq ) φ )(( q ′ y ′ ) φ ) = ( xqq ′ y ′ ) φ = ( xφ )(( qq ′ ) φ ) y ′ φ = ( xφ )(( qφ )( q ′ φ )) y ′ φ = ( xφ )(( qφ )( qφ ) − ) y ′ φ = ( xφ ) I y ′ φ = ( xφ ) y ′ φ = ( xφ )( xφ ) − = I . Therefore f − = ( q ′ y ′ ) φ ∈ Im φ . This concludes the inductive proof.For denoting a group Im φ we would use notation Γ( M ) showingthe corresponding machine . A group G is called a machine group (au-tomaton group), if there exist an invertible Mealy machine M suchthat G ∼ = Γ( M ) . The set AS A ↽ { f ∈ P A | f is bijection } is a group in which group operation is composition of restricted se-quential functions. ✷ Suppose that f ∈ AS A , then exists Mealy machine V = h Q, A, A ; q , ◦ , ∗i , such that ∀ u ∈ A ∗ f ( u ) = q ∗ u .Lets define a new Mealy machine V ′ = h Q, A, A ; q , ´ ◦ , ´ ∗i , where q ´ ∗ a ↽ b, ja q ∗ b = a, (5) q ´ ◦ a ↽ q ◦ b, ja q ∗ b = a. (6)The new Mealy machine defines restricted sequential function g ( u ) ↽ q ´ ∗ u .We need to prove that g = f − , i.e., g is the inverse function of f . 31i) Suppose that Q ′ ↽ { q | ∃ u ∈ A ∗ q = q ◦ u } . Lets prove ∀ q ∈ Q ′ ∀ a ∈ A q ∗ ( q ´ ∗ a ) = a. (7)Suppose that q = q ◦ u , then f ( ua ) = q ∗ ua = q ∗ u q ◦ u ∗ a = q ∗ u q ∗ a. Accordingly to the definition of q ´ ∗ a if q ∗ b = a , then q ´ ∗ a = b . As f is a bijection, therefore exists unique b ∈ A . Hence q ∗ ( q ´ ∗ a ) = q ∗ b = a. (8)We have proven, that f g ( a ) = a .Accordingly to q ´ ◦ a we have that q ◦ ( q ´ ∗ a ) = q ◦ b = q ´ ◦ a. (9)Further proof is inductive, given ∀ q ′ ∈ Q ′ ∀ v ∈ A n q ′ ∗ ( q ′ ´ ∗ v ) = v. (10)Suppose that w ∈ A n +1 , then there exist a ∈ A and v ∈ A n suchthat w = av . Hence q ∗ ( q ´ ∗ w ) = q ∗ ( q ´ ∗ av ) = q ∗ ( q ´ ∗ a q ´ ◦ a ´ ∗ v )= q ∗ ( q ´ ∗ a ) q ◦ ( q ´ ∗ a ) ∗ ( q ´ ◦ a ´ ∗ v )= (8) a q ◦ ( q ´ ∗ a ) ∗ ( q ´ ◦ a ´ ∗ v )= (9) a q ´ ◦ a ∗ ( q ´ ◦ a ´ ∗ v )= (10) a v = w. This concludes the inductive part of the proof.In a particular case we have f g ( w ) = q ∗ ( q ´ ∗ w ) = w . Therefore g = f − .(ii) Suppose that f and g are two restricted sequential functionsof the set AS A , then there exist two Mealy machines V = h Q, A, A ; q , ◦ , ∗i , V ′ = h Q ′ , A, A ; q ′ , ´ ◦ , ´ ∗i such that ∀ u ∈ A ∗ ( f ( u ) = q ∗ u ∧ g ( u ) = q ′ ´ ∗ u ) . The serial composition of these machines V ❀ V ′ is a machineimplementing the composition of functions f and g . To be moreprecise, if ˘ V = h ˘ Q, A, A ; ˘ q , ˘ ◦ , ˘ ∗i ∈ V ❀ V ′ , then ∀ u ∈ , ∗ gf ( u ) = ˘ q ˘ ∗ u. (iii) The identity function I : A ∗ → A ∗ : u u is restrictedsequential function. Machine implementing identity function givenas V a (shown in 8.fig.).(iv) We note that the composition of functions is associative.Finally, we have proven that AS A is a group.32 a ✍✌✎☞ ❄ q a a · · · a m /a a · · · a m
9. Figure: Implementation of identity function. If M = h Q, A, ◦ , ∗i is an invertible machine,then Γ( M ) is a subgroup of group AS A . If | A | = | B | , then AS A ∼ = AS B . ✷ If | A | = | B | , then exists bijection ϕ : A → B . Suppose that f ∈ AS A , then exists Mealy machine V = h Q, A, A ; q , ◦ , ∗i , ka ∀ u ∈ A ∗ f ( u ) = q ∗ u . Lets define a new Mealy machine V ′ = h Q, B, B ; q , ´ ◦ , ´ ∗i , where q ´ ◦ ϕ ( a ) ↽ q ◦ a,q ´ ∗ ϕ ( a ) ↽ ϕ ( q ∗ a ) . Machine V is invertible therefore ¯ q | A : A → A : a q ∗ a is abijection. A composition of bijections is a bijection therefore˜ q | B ↽ ϕ (¯ q | A ) ϕ − : B → B : b ϕ ( q ∗ ϕ − ( b )) = q ´ ∗ ϕϕ − ( b ) = q ´ ∗ b is a bijection, where ∀ v ∈ B ∗ ˜ q ( v ) ↽ q ´ ∗ v . Hence V ′ is a invertiblemachine. This whole construction describes mapping ψ : AS A → AS B : f ψ ( f ) , where ψ ( f )( v ) = q ´ ∗ v is a restricted sequential function. In presentcase f = ¯ q and ψ ( f ) = ˜ q .Suppose that g ∈ AS A , then there exist a Mealy machine W = h Q ′ , A, A ; t , ⊙ , ⊛ i , such that ∀ u ∈ A ∗ g ( u ) = t ⊛ u . Let define a new Mealy machine W ′ ↽ h Q ′ , A, A ; t , ´ ⊙ , ´ ⊛ i , where q ´ ⊙ ϕ ( a ) ↽ q ⊙ a,q ´ ⊛ ϕ ( a ) ↽ ϕ ( q ⊛ a ) . From previous construction we have that ˜ t = ψ (¯ t ) = ψ ( g ).The serial composition V ❀ W of machines V, W is a machineimplementing the composition of functions f and g . Formally, V W ↽ h Q ′ × Q, A, A ; ( t , q ) , ˙ ◦ , ˙ ∗i , t, q ) ˙ ◦ a ↽ ( t ⊙ q ∗ a, q ◦ a );( t, q ) ˙ ∗ a ↽ t ⊛ q ∗ a, then ∀ u ∈ A ∗ gf ( u ) = t ⊛ q ∗ u . By previously described construc-tion machine V ′ W ↽ h Q ′ × Q, B, B ; ( t , q ) , ¨ ◦ , ¨ ∗i , where ( t, q )¨ ◦ ϕ ( a ) ↽ ( t, q ) ˙ ◦ a, ( t, q )¨ ∗ ϕ ( a ) ↽ ϕ (( t, q )¨ ∗ a ) = ϕ ( t ⊛ q ∗ a ) , implementing mapping ψ ( gf ), i.e., ∀ v ∈ B ∗ ψ ( gf )( v ) = ( t , q )¨ ∗ v = ϕ ( t ⊛ q ∗ ϕ − ( v )) . The mapping ϕ is inductively extended on set A ∗ , by a condition ϕ ( ua ) ↽ ϕ ( u ) ϕ ( a ) . ( ψ ( g ) ψ ( f ))( v ) = ˜ t (˜ q ( v )) = ˜ t ( ϕ ( q ∗ ϕ − ( v )))= ϕ ( t ⊛ ϕ − ϕ ( q ∗ ϕ − ( v )))= ϕ ( t ⊛ ( q ∗ ϕ − ( v ))) = ψ ( gf )( v ) . Therefore mapping ψ : AS A → AS B is a group homomorphism.Suppose that f = g , then ∃ u ∈ A ∗ f ( u ) = g ( u ), i.e., q ∗ u = ¯ q ( u ) = f ( u ) = g ( u ) = ¯ t ( u ) = t ⊛ u. Hence ψ ( f )( ϕ ( u )) = ˜ q ( ϕ ( u )) = q ´ ∗ ϕ ( u ) = ϕ ( q ∗ u ) = ϕ ( t ⊛ u ) = t ´ ⊛ ϕ ( u ) = ˜ t ( ϕ ( u )) = ψ ( g )( ϕ ( u )) . Therefore mapping ψ is an injection.Suppose that h ∈ AS B , then exists Mealy machineˇ M = h R, B, B ; r , ˇ ◦ , ˇ ∗i such that ∀ v ∈ B ∗ h ( v ) = r ˇ ∗ v . We definenew machine ˆ M ↽ h R, A, A ; r , ˆ ◦ , ˆ ∗i , where r ˆ ◦ ϕ − ( b ) ↽ r ˇ ◦ b,r ˆ ∗ ϕ − ( b ) ↽ ϕ − ( r ˇ ∗ b ) . M we have ∀ u ∈ A ∗ ¯ r ( u ) ↽ r ˆ ∗ u . By previously describedconstruction we have that machine ˆ M is invertible, therefore ¯ r ∈ AS A .We apply previously described construction to machine ˆ M andcreate machine ˘ M = h R, B, B ; r , ˘ ◦ , ˘ ∗i , where r ˘ ◦ ϕ ( a ) ↽ r ˆ ◦ a = r ˆ ◦ ϕ − ϕ ( a ) = r ˇ ◦ ϕ ( a ) ,r ˘ ∗ ϕ ( a ) ↽ ϕ ( r ˆ ∗ a ) = ϕ ( r ˆ ∗ ϕ − ϕ ( a )) = ϕϕ − ( r ˇ ∗ ϕ ( a )) = r ˇ ∗ ϕ ( a ) . Hence ψ (¯ r ) = h . Therefore ψ is a surjection.Finally, we have that mapping ψ is an isomorphism of groups AS A , AS B , therefore AS A ∼ = AS B . ✍✌✎☞ V V ′ ❄ ✍✌✎☞ ❄ ✍✌✎☞ ❄ ✍✌✎☞ ❄⑦ ❂⑥ ❃ q q q q / / / / / / / / V .Machine V (10.fig.) is invertible because ¯ q is a bijection. Ma-chine V ′ (10.fig.) is constructed according to expressions (5), (6): q ´ ∗ , jo q ∗ ,q ´ ◦ q ◦ q , jo q ∗ ,q ´ ∗ , jo q ∗ ,q ´ ◦ q ◦ q , jo q ∗ ,q ´ ∗ , jo q ∗ ,q ´ ◦ q ◦ q , jo q ∗ ,q ´ ∗ , jo q ∗ ,q ´ ◦ q ◦ q , jo q ∗ . Mapping ˜ q ( u ) ↽ q ´ ∗ u is inverse of mapping ¯ q ( u ) ↽ q ∗ u . Let ˙ V = h ˙ Q, ˙ A, ˙ B ; ˙ ◦ , ˙ ∗i , ¨ V = h ¨ Q, ¨ A, ¨ B ; ¨ ◦ , ¨ ∗i be Mealy machines. A total map µ = ( µ , µ , µ ) : ( ˙ Q, ˙ A, ˙ B ) −→ ( ¨ Q, ¨ A, ¨ B )35 s called a homomorphism µ : ˙ V −→ ¨ V if µ ( q ˙ ◦ a ) = µ ( q )¨ ◦ µ ( a ) ∧ µ ( q ˙ ∗ a ) = µ ( q )¨ ∗ µ ( a ) for ∀ ( q, a ) ∈ Q × A . Lets consider machines V u V ′ (10.fig.) and consider mapping µ = ( µ , µ , µ ) : V → V ′ , where µ : q q , q q ; µ : 0 , µ : 0 , .µ ( q ◦
0) = µ ( q ) = q = q ´ ◦ µ ( q )´ ◦ µ (0) ,µ ( q ∗
0) = µ (1) = 1 = q ´ ∗ µ ( q )´ ∗ µ (0) ,µ ( q ◦
1) = µ ( q ) = q = q ´ ◦ µ ( q )´ ◦ µ (1) ,µ ( q ∗
1) = µ (0) = 1 = q ´ ∗ µ ( q )´ ∗ µ (1) ,µ ( q ◦
0) = µ ( q ) = q = q ´ ◦ µ ( q )´ ◦ µ (0) ,µ ( q ∗
0) = µ (0) = 1 = q ´ ∗ µ ( q )´ ∗ µ (0) ,µ ( q ◦
1) = µ ( q ) = q = q ´ ◦ µ ( q )´ ◦ µ (1) ,µ ( q ∗
1) = µ (1) = 1 = q ´ ∗ µ ( q )´ ∗ µ (1) . Let ˙ V = h ˙ Q, ˙ A, ˙ B ; ˙ ◦ , ˙ ∗i be a Mealy machine,and let there exist state ˙ q ∈ ˙ Q and letter of input alphabet ˙ a ∈ ˙ A such that ˙ q ˙ ◦ ˙ a = ˙ q . If so, then for each Mealy machine h Q, A, B ; ◦ , ∗i there exist a morphism of machines µ : V → ˙ V . ✷ Suppose that µ = ( µ , µ , µ ) : Q × A × B → ˙ Q × ˙ A × ˙ B is amapping defined by conditions ∀ q ∈ Q µ ( q ) = ˙ q, ∀ a ∈ A µ ( a ) = ˙ a, ∀ b ∈ B µ ( b ) = ˙ q ˙ ∗ ˙ a, then µ ( q ◦ a ) = ˙ q = ˙ q ˙ ◦ ˙ a = µ ( q ) ˙ ◦ µ ( a ) ,µ ( b ) = ˙ q ˙ ∗ ˙ a = µ ( q ) ˙ ∗ µ ( a ) . This shows that µ : V → ˙ V is a homomorphism of machines.In general case this result is not true even for invertible machines.Moreover, as shown in next example, we can choose invertible ma-chines V = h Q, A, A ; q , ◦ , ∗i , V ′ = h Q ′ , A ′ , A ′ ; q ′ , ´ ◦ , ´ ∗i , q ´ ∗ u is the inverse of mapping q ∗ u , but theredoesn’t exist nether homomorphism V → V ′ nor homomorphism V ′ → V . ✍✌✎☞ V V ′ ✍✌✎☞ ✍✌✎☞✍✌✎☞ ✍✎ ❲ ❲✗✾③❄✛(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)✒ ✍✌✎☞✍✌✎☞ ✍✌✎☞ ✍✎ ❲ ❲✗✾③❄✛(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)✒ ✍✌✎☞✍✌✎☞ ✍✌✎☞ q q q q q q ′ q ′ q ′ q ′ q ′ / / / / / / / / / / / / / / / / / / / / V . Suppose that µ = ( µ , µ , µ ) : V → V ′ is a homomorphism.(i) If µ ( q ) = q ′ , then we have cases: µ : 0 , , , , . (i a ) µ : 0 , µ ( q ) = µ ( q ◦
1) = µ ( q )´ ◦ µ (1) = q ′ ´ ◦ q ′ ,µ (1) = µ ( q ∗
0) = µ ( q )´ ∗ µ (0) = q ′ ´ ∗ ,µ (1) = µ ( q ∗
1) = µ ( q )´ ∗ µ (1) = q ′ ´ ∗ . Hence 0 = µ (1) = 1. A contradiction!(i b ) µ : 0 , µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
1) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ . Hence q ′ = µ ( q ) = q ′ . A contradiction!37i c ) µ : 0 , µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
1) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ . Hence q ′ = µ ( q ) = q ′ . A contradiction!(i d ) µ : 0 , µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
1) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ . Hence q ′ = µ ( q ) = q ′ . A contradiction!(ii) Suppose that µ ( q ) = q ′ .(ii a ) µ : 0 , µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
1) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ . Hence q ′ = µ ( q ) = q ′ . A contradiction!(ii b ) µ : 0 , µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
1) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ . Hence q ′ = µ ( q ) = q ′ . A contradiction!(ii c ) µ : 0 , µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
1) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ . Hence q ′ = µ ( q ) = q ′ . A contradiction!(ii d ) µ : 0 , µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ ,µ (0) = µ ( q ∗
1) = q ′ ´ ∗ ,µ (0) = µ ( q ∗
0) = q ′ ´ ∗ . Hence 0 = µ (0) = 1. A contradiction!(iii) Suppose that µ ( q ) = q ′ .38iii a ) µ : 0 , µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
1) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ . Hence q ′ = µ ( q ) = q ′ . A contradiction!(iii b ) µ : 0 , µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
1) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ . Hence q ′ = µ ( q ) = q ′ . A contradiction!(iii c ) µ : 0 , µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
1) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ . Hence q ′ = µ ( q ) = q ′ . A contradiction!(iii d ) µ : 0 , µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
1) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ . Hence q ′ = µ ( q ) = q ′ . A contradiction!(iv) Suppose that µ ( q ) = q ′ .(iv a ) µ : 0 , µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
1) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ . Hence q ′ = µ ( q ) = q ′ . A contradiction!(iv b ) µ : 0 , µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
1) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ . Hence q ′ = µ ( q ) = q ′ . A contradiction!(iv c ) µ : 0 , µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ ,µ (0) = µ ( q ∗
1) = q ′ ´ ∗ ,µ (0) = µ ( q ∗
0) = q ′ ´ ∗ . µ (0) = 0. A contradiction!(iv d ) µ : 0 , µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
1) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ . Hence q ′ = µ ( q ) = q ′ . A contradiction!(v) Suppose that µ ( q ) = q ′ .(v a ) µ : 0 , µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
1) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ . Hence q ′ = µ ( q ) = q ′ . A contradiction!(v b ) µ : 0 , µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
1) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ . Hence q ′ = µ ( q ) = q ′ . A contradiction!(v c ) µ : 0 , µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
1) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ . Hence q ′ = µ ( q ) = q ′ . A contradiction!(v d ) µ : 0 , µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
1) = q ′ ´ ◦ q ′ ,µ ( q ) = µ ( q ◦
0) = q ′ ´ ◦ q ′ . Hence q ′ = µ ( q ) = q ′ . A contradiction!In all cases we have arrived at contradictions. This proves thatthere doesn’t exist a homomorphism µ = ( µ , µ , µ ) : V → V ′ .Suppose that µ = ( µ , µ , µ ) : V ′ → V is a homomorphism.(i) If µ ( q ′ ) = q , then we have cases: µ : 0 , , , , . a ) µ : 0 , µ ( q ′ ) = µ ( q ′ ´ ◦
1) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q . Hence q = µ ( q ′ ) = q . A contradiction!(i b ) µ : 0 , µ ( q ′ ) = µ ( q ′ ´ ◦
1) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q . Hence q = µ ( q ′ ) = q . A contradiction!(i c ) µ : 0 , µ ( q ′ ) = µ ( q ′ ´ ◦
1) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q . Hence q = µ ( q ′ ) = q . A contradiction!(i d ) µ : 0 , µ ( q ′ ) = µ ( q ′ ´ ◦
1) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q . Hence q = µ ( q ′ ) = q . A contradiction!(ii) Suppose that µ ( q ′ ) = q .(ii a ) µ : 0 , µ ( q ′ ) = µ ( q ′ ´ ◦
1) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q . Hence q = µ ( q ′ ) = q . A contradiction!(ii b ) µ : 0 , µ ( q ′ ) = µ ( q ′ ´ ◦
1) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q . Hence q = µ ( q ′ ) = q . A contradiction!(ii c ) µ : 0 , µ ( q ′ ) = µ ( q ′ ´ ◦
1) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q . q = µ ( q ′ ) = q . A contradiction!(ii d ) µ : 0 , µ ( q ′ ) = µ ( q ′ ´ ◦
1) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q . Hence q = µ ( q ′ ) = q . A contradiction!(iii) Suppose that µ ( q ′ ) = q .(iii a ) µ : 0 , µ ( q ′ ) = µ ( q ′ ´ ◦
1) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q . Hence q = µ ( q ′ ) = q . A contradiction!(iii b ) µ : 0 , µ ( q ′ ) = µ ( q ′ ´ ◦
1) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q . Hence q = µ ( q ′ ) = q . A contradiction!(iii c ) µ : 0 , µ ( q ′ ) = µ ( q ′ ´ ◦
1) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q . Hence q = µ ( q ′ ) = q . A contradiction!(iii d ) µ : 0 , µ ( q ′ ) = µ ( q ′ ´ ◦
1) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q . (iv) Suppose that µ ( q ′ ) = q .(iv a ) µ : 0 , µ ( q ′ ) = µ ( q ′ ´ ◦
1) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q . Hence q = µ ( q ′ ) = q . A contradiction!42iv b ) µ : 0 , µ ( q ′ ) = µ ( q ′ ´ ◦
1) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q . Hence q = µ ( q ′ ) = q . A contradiction!(iv c ) µ : 0 , µ ( q ′ ) = µ ( q ′ ´ ◦
1) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q . Hence q = µ ( q ′ ) = q . A contradiction!(iv d ) µ : 0 , µ ( q ′ ) = µ ( q ′ ´ ◦
1) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q . Hence q = µ ( q ′ ) = q . A contradiction!(v) Suppose that µ ( q ′ ) = q .(v a ) µ : 0 , µ ( q ′ ) = µ ( q ′ ´ ◦
1) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q . Hence q = µ ( q ′ ) = q . A contradiction!(v b ) µ : 0 , µ ( q ′ ) = µ ( q ′ ´ ◦
1) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q . Therefore q = µ ( q ′ ) = q . A contradiction!(v c ) µ : 0 , µ ( q ′ ) = µ ( q ′ ´ ◦
1) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q . Therefore q = µ ( q ′ ) = q . A contradiction!(v d ) µ : 0 , µ ( q ′ ) = µ ( q ′ ´ ◦
1) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q ,µ ( q ′ ) = µ ( q ′ ´ ◦
0) = q ◦ q . q = µ ( q ′ ) = q . A contradiction!In all cases we have arrived at contradictions. This proves thatthere doesn’t exist a homomorphism µ = ( µ , µ , µ ) : V ′ → V . We say that machine ′ V simulates machine V if there exist mappings Q h −→ ′ Q, A h −→ ′ A, ′ B h −→ B such that the diagram Q × A ∗ ∗ −→ B ∗ h ↓ ↓ h ↑ h ′ Q × ′ A ∗ ∗ −→ ′ B ∗ commutes. That is if ∀ ( q, u ) ∈ Q × A ∗ q ∗ u = h ( h ( q ) ∗ h ( u )) . As it turns out (consider next example), we can choose such in-vertible machines V = h Q, A, A ; q , ◦ , ∗i , V ′ = h Q ′ , A ′ , A ′ ; q , ´ ◦ , ´ ∗i , that ´ q ´ ∗ u is the inverse mapping of q ∗ u , but machine V ′ doesn’tsimulate machine V , and V doesn’t simulate V ′ . Suppose that machine V ′ simulates machine V (10. fig.), i.e.,exists mappings h , h , h , that q ∗ u = h ( h ( q )´ ∗ h ( u ))for ∀ u ∈ , ∗ and for all states q of machine V . Therefore0 = q ∗ h ( h ( q ))´ ∗ h (1)) , q ∗ h ( h ( q )´ ∗ h (0)) . This shows that mappings h and h are bijections. Therefore existsinverse mapping (a bijection) h of mapping h . Hence h ( q ∗ u ) = h ( h ( h ( q )´ ∗ h ( u ))) = h ( q )´ ∗ h ( u ) . We also have that h (1) = h ( q ∗
0) = h ( q )´ ∗ h (0) ,h (0) = h ( q ∗
0) = h ( q )´ ∗ h (0) . As h is bijection, then we have that h also is bijection. Hence allmappings h, h , h are bijections.44i) Suppose that h : q q , q q , then h (10) = h ( q ∗
00) = h ( q )´ ∗ h (00)) = q ´ ∗ h (00) . As h is bijection, then h (0) = 1. Mapping h also is a bijection,therefore h (1) = 0. Hence h (10) = h ( q ∗
11) = h ( q )´ ∗ h (11) = q ´ ∗
00 = 00 . Contradiction because h is a bijection.(ii) Suppose that h : q q , q q , then h (10) = h ( q ∗
00) = h ( q )´ ∗ h (00)) = q ´ ∗ h (00) . As h is a bijection, then h (0) = 1. Mapping h also is a bijection,therefore h (1) = 0. Hence h (10) = h ( q ∗
11) = h ( q )´ ∗ h (11) = q ´ ∗
00 = 11 . This is a contradiction because h is a bijection.We have obtained contradictions in both cases, consequentiallymachine V ′ is not capable of simulating machine V .Suppose that machine V is simulating machine V ′ (look at 10.fig.), i.e., there exist mappings h , h , h such that q ´ ∗ u = h ( h ( q ) ∗ h ( u ))for all u ∈ , ∗ and for all states q of machine V ′ . Therefore0 = q ´ ∗ h ( h ( q )) ∗ h (1)) , q ´ ∗ h ( h ( q ) ∗ h (0)) . This shows that h and h are bijections. Hence there exist a inversemapping h of mapping h , which also is a bijection. Hence h ( q ´ ∗ u ) = h ( h ( h ( q ) ∗ h ( u ))) = h ( q ) ∗ h ( u ) . We also have that h (1) = h ( q ´ ∗
0) = h ( q ) ∗ h (0) ,h (0) = h ( q ´ ∗
0) = h ( q ) ∗ h (0) . Therefore h is a bijection, consequentially we have, that h is abijection. Hence all mappings h, h , h are bijections.(i) Suppose that h : q q , q q , then h (01) = h ( q ´ ∗
11) = h ( q ) ∗ h (11) = q ∗ h (11) . As h is a bijection, then h (1) = 0. Hence h (10) = h ( q ´ ∗
11) = h ( q ) ∗ h (11) = q ∗
00 = 00 . h is a bijection.(ii) Suppose that h : q q , q q , then h (10) = h ( q ´ ∗
11) = h ( q ) ∗ h (11)) = q ∗ h (11) . As h is a bijection, then h (1) = 0. The mapping h also is abijection, therefore h (1) = 0. Hence h (01) = h ( q ´ ∗
11) = h ( q ) ∗ h (11) = q ∗
00 = 00 . This is a contradiction because h is a bijection.Thus we have obtained contradictions in both cases, consequen-tially the machine V is not capable of simulating V ′ . References [1]
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