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 ∃n0 ∀n≥n0 ∀x∈M,xn = xn+1. Here, n0 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 pnqrn = q, ∀n > 0 If n ≥ N, then pn+1qrn = pnqrn = q Hence p(pnqrn) = 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

• peration . 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 ηI 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) = (ηF(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 = {x1, x2, ...., xk}, h−1(X) is the union of the sets h−1(x1), h−1(x2), ...... h−1(xk). 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 Ea = ¯ φ.a.¯ φ defines a language La contained in L.Thus, we could focus our attention on writing star-free expressions to cover words in L\ ∪a∈A La.

• 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 ∈ La.

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(w1).a.y.b.h(w2) ∈ I for each w1, w2 ∈ 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 La 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

Proof (Contd)

• Thus, we have picked an arbitrary element w of h−1(xM\F(x)) and

shown that there is a star-free regular expression involving h−1(y) for some y with F(y) strictly containing F(x) that describes a language containing w and which is contained in h−1(xM\F(x)) ∪ h−1(F(x)).

• Since M and Σ are finite sets, it follows that we can write a star-free

regular expression involving some elements y1, y2, ...yk of M, where F(yi) strictly contains F(yi) for each i, that describes a language containing h−1(xM\F(x)) and which is contained in h−1(xM\F(x)) ∪ h−1(F(x)).

• Following Lemma 4, induction hypothesis and the fact that boolean
• perations can be used in the star-free expressions, we can get a star-free

expression for h−1(xM\F(x)). We can use the similar proof for h−1(Mx\F(x)). Hence, the lemma statement is proved.

G Sreedurga and Amishi Singh (IISc) Schutzenberg’s Theorem November 30, 2018 16 / 18

Conclusion

Let X ⊆ M such that preimage of X under the homomorphism h gives the language L recognized by the monoid M. From Lemma 6, we proved that for each x ∈ X , h−1(x) can be expressed as a star-free expression involving h−1(y) and other languages definable via smaller aperiodic

• monoids. By applying induction, h−1(x) can thus be expressed as a

star-free expression. Union of all such h−1(xi)’s will also be a star-free

• expression. Thus, h−1(X) i.e L can be recognized via star-free expression.

G Sreedurga and Amishi Singh (IISc) Schutzenberg’s Theorem November 30, 2018 17 / 18

The End

G Sreedurga and Amishi Singh (IISc) Schutzenberg’s Theorem November 30, 2018 18 / 18