Schutzenbergs Theorem of Aperiodic monoids and Star-free languages - PowerPoint PPT Presentation

Schutzenberg’s Theorem of Aperiodic monoids and Star-free languages G Sreedurga and Amishi Singh Indian Institute of Science November 30, 2018 G Sreedurga and Amishi Singh (IISc) Schutzenberg’s Theorem November 30, 2018 1 / 18

Schutzenberg’s Theorem Statement Let φ : Σ ∗ → M be a monoid morphism, for an aperiodic monoid M . Then any language reocgnizable by M can be described via star-free regular expressions. G Sreedurga and Amishi Singh (IISc) Schutzenberg’s Theorem November 30, 2018 2 / 18

Definitions Aperiodic Monoid A monoid M is said to be aperiodic iff ∃ n 0 ∀ n ≥ n 0 ∀ x ∈ M , x n = x n +1 . Here, n 0 is called the index of the monoid. Star-free Languages The class of star-free languages is the smallest class SF(Σ) of languages L ∈ Σ ∗ such that: (i) φ, { ǫ } ∈ SF (Σ) and { a } ∈ SF (Σ) ∀ a ∈ Σ (ii) If X , Y ∈ SF (Σ) , then X ∪ Y , XY and ¯ X ∈ SF (Σ) G Sreedurga and Amishi Singh (IISc) Schutzenberg’s Theorem November 30, 2018 3 / 18

The Simplification Rule Lemma 1 Let M be an aperiodic monoid and let p , q , r ∈ M . If pqr = q , then pq = q = qr . Let N be the index of M . If pqr = q , then p n qr n = q , ∀ n > 0 If n ≥ N , then p n +1 qr n = p n qr n = q Hence p ( p n qr n ) = pq = q . Similarly, q = qr . G Sreedurga and Amishi Singh (IISc) Schutzenberg’s Theorem November 30, 2018 4 / 18

Definitions Ideal We say that a subset I of M is an ideal if IM ⊆ I and MI ⊆ I . Thus, an ideal is a subset that is closed w.r.t. multiplication (on both sides) by the elements of the monoid. Forbidding ideal With each element x of a monoid M , we can associate an interesting ideal F ( x ), called the forbidding ideal of x . F ( x ) = { y | ∀ p , q pyq � = x } It consists of all the elements that cannot divide or cannot generate x via multiplication. F ( x ) is clearly an ideal. G Sreedurga and Amishi Singh (IISc) Schutzenberg’s Theorem November 30, 2018 5 / 18

Definition Let M be an monoid and let I be an ideal of M . Then, there is a natural monoid M \ I whose elements are M – I ∪ { i } and whose multiplication operation . is defined as follows: • x.i = i.x = i.i =i • x.y = i, if x.y ∈ I • x.y is the same as x.y in M otherwise. • M \ I is a monoid and there exists a morphism η I from M to M \ I which is identity on the elements of M – I and maps every element of I to i . • If a language L is recognized by a morphism h as h − 1 (I) in M then the same is recognised by η o h to M \ I as the pre-image of { i } . We shall simply write h to denote the composed map η o h to M \ I . G Sreedurga and Amishi Singh (IISc) Schutzenberg’s Theorem November 30, 2018 6 / 18

Lemma 2 Let M be an finite aperiodic monoid, I be an ideal and let either I ⊆ X or I ∩ X = φ . Then, any language L recognized as the preimage of X is recognized via the monoid M \ I . In particular, if I has atleast 2 elements then L is recognized by a smaller aperiodic monoid. Lemma 3 Let h be a morphism from Σ ∗ to M. Then, h − 1 ( x ) = ( η F ( x ) o h ) − 1 ( x ). Thus, if F(x) has at least 2 elements then the language recognized as h − 1 ( x ) can be recoginized using a smaller monoid. G Sreedurga and Amishi Singh (IISc) Schutzenberg’s Theorem November 30, 2018 7 / 18

Outline We proceed by induction on the size of the monoid M . If M is the trivial monoid, then the only languages recognised via M are φ and Σ ∗ and clearly both are star-free languages. For the induction step, consider any language L recognized via some monoid M . For any X = { x 1 , x 2 , ...., x k } , h − 1 ( X ) is the union of the sets h − 1 ( x 1 ), h − 1 ( x 2 ), ...... h − 1 ( x k ). Now, it suffices to show that h − 1 ( x ) can be expressed as a star-free expression involving languages definable using aperiodic monoids smaller than M . If F ( x ) has at least two elements, we can directly use Lemma 3. G Sreedurga and Amishi Singh (IISc) Schutzenberg’s Theorem November 30, 2018 8 / 18

Outline (Contd) Else, we show that there is a set Y ⊆ M such that F ( y ) contains more elements than F ( x ) for each y ∈ Y and further h − 1 ( x ) can be expressed as a star-free expression involving h − 1 ( y ) and other languages definable using small aperiodic monoids. Then, by definition of a natural monoid and Lemma 2, we can say that h − 1 ( y ) is recognizable via a smaller monoid M \ F ( y ) for each y ∈ Y . This would show that h − 1 ( x ) can be described via a star-free expression that only involves languages definable via smaller aperiodic monoids and we can apply the induction hypothesis to conclude that h − 1 ( x ) can be described via a star-free expression. G Sreedurga and Amishi Singh (IISc) Schutzenberg’s Theorem November 30, 2018 9 / 18

Lemma 4 If L = h − 1 (I) for some ideal I in M then L can be expressed using star-free expressions involving languages which are recognized by smaller aperiodic monoids. Proof : Case 1 (I = φ ) : If I = φ then L = φ . So the regular expression for L is φ (star- free). Case 2 ( | I | > 0) : • Let A = { a | h ( a ) ∈ I } . ∀ a ∈ A , the expression E a = ¯ φ . a .¯ φ defines a language L a contained in L .Thus, we could focus our attention on writing star-free expressions to cover words in L \ ∪ a ∈ A L a . • For any word w in L , Consider a minimal substring of u of w such that u ∈ L . If u = ǫ then M = L thus L = Σ ∗ If u = a for some a ∈ Σ then w ∈ L a . G Sreedurga and Amishi Singh (IISc) Schutzenberg’s Theorem November 30, 2018 10 / 18

Proof (Contd) • Consider the case when u = avb for some v ∈ Σ ∗ and a , b ∈ Σ . Let h ( v ) = y . • Since I is an ideal, h ( w 1 ) . a . y . b . h ( w 2 ) ∈ I for each w 1 , w 2 ∈ Σ ∗ . Thus, φ. a . h − 1 ( y ) . b . ¯ ¯ φ ⊆ L . • Next we show that F ( y ) has at least two elements, thus establishing that h − 1 ( y ) can be accepted via a smaller monoid ( M \ F ( y )). Since , y / ∈ I and I is an ideal, I ⊆ F ( y ). Since , I is nonempty, F ( y ) is also non empty. • Claim : There exists at least one other element in F ( y ). Proof : Consider h ( a ) . y such that h ( a ) . y / ∈ F ( y ) then there must be p , q such that y = p . h ( a ) . y . q . y = p . h ( a ) . y ⇒ y . h ( b ) = p . h ( a ) . y . h ( b ) Since h ( a ) . y . h ( b ) ∈ I , p . h ( a ) . y . h ( b ) ∈ I . But this contradicts the minimality of u. Thus, h ( a ) . y ∈ F ( y ) \ I . Thus F(y) has at least two elements. G Sreedurga and Amishi Singh (IISc) Schutzenberg’s Theorem November 30, 2018 11 / 18

Proof (Contd) • The monoid M and the alphabet Σ are finite, therefor there are finitely many choices for triples of the form ( a , y , b ). • Thus, we can describle all of L \ ∪ a ∈ A L a as a finite union of languages of the form ¯ φ. a . h − 1 ( y ) . b . ¯ φ , with | F ( y ) | > 1. G Sreedurga and Amishi Singh (IISc) Schutzenberg’s Theorem November 30, 2018 12 / 18

Lemma 5 x = ( xM ∩ Mx ) \ F ( x ) Proof: • y ∈ ( xM ∩ Mx ) \ F ( x ) ⇒ ∃ p , q , r , s such that y = px , y = xq and x = rys . By Lemma 1, y = xq = rysq ⇒ y = ry and y = px = prys ⇒ y = ys . Thus, y = rys = x . G Sreedurga and Amishi Singh (IISc) Schutzenberg’s Theorem November 30, 2018 13 / 18

Lemma 6 Let x ∈ M , x � = e , then there is a subset Y ⊆ M such that ∀ y ∈ Y , F ( y ) strictly contains F ( x ) and h − 1 ( x ) can be expressed as a star-free expression involving h − 1 ( y ) , y ∈ Y and other languages definable via smaller aperiodic monoids. Proof: • From Lemma 5, x = ( xM ∩ Mx ) \ F ( x ). h − 1 (( xM ∩ Mx ) \ F ( x )) = h − 1 ( xM \ F ( x )) ∩ h − 1 ( Mx \ F ( x )) and h − 1 ( xM \ F ( x )) = h − 1 ( xM ) \ h − 1 ( F ( x )). • Let w ∈ h − 1 ( xM \ F ( x )). Let u be the shortest prefix of w such that h ( u ) ∈ ( xM \ F ( x )). If u = ǫ , then e = xd for some d ∈ M . By Lemma 1, this implies x = e . Contradiction. So, u � = ǫ . • u = va for some v such that h ( v ) = y / ∈ ( xM \ F ( x )). Claim: h − 1 ( y ) . a . ¯ φ ⊆ h − 1 ( xM ) Proof: h ( vaw ′ ) = h ( va ) . h ( w ′ ) = xm . d = xm ′ . Thus it is in xM . Thus, h − 1 ( y ) . a . ¯ φ \ h − 1 ( F ( x )) is a subset of h − 1 ( xM \ F ( x )) containing w . G Sreedurga and Amishi Singh (IISc) Schutzenberg’s Theorem November 30, 2018 14 / 18

Proof (Contd) • y ∈ F ( x ) ⇒ yh ( a ) ∈ F ( x ) ⇒ h ( u ) ∈ F ( x ) ⇒ Contradiction. This implies y / ∈ F ( x ). So, F ( x ) ⊆ F ( y ). • Let yh ( a ) / ∈ F ( y ) ⇒ ∃ p , q ∈ M such that y = pyh ( a ) q . By Lemma 1, y = yh ( a ) q ⇒ y = xdq (Since yh ( a ) = h ( u ) ∈ xM ) ⇒ y ∈ xM . We showed y / ∈ F ( x ). So, y ∈ xM \ F ( x ). This contradicts the minimality of u . So, yh ( a ) ∈ F ( y ). F ( y ) has more elements than F ( x ). G Sreedurga and Amishi Singh (IISc) Schutzenberg’s Theorem November 30, 2018 15 / 18