Mathematical morphology and poset geometry

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PII. S0161171201006718 http://ijmms.hindawi.com © Hindawi Publishing Corp.

MATHEMATICAL MORPHOLOGY AND POSET GEOMETRY

ALAIN BRETTO and ENZO MARIA LI MARZI (Received 16 February 2001)

Abstract.The aim of this paper is to characterize morphological convex geometries (resp., antimatroids). We deﬁne these two structures by using closure operators, and kernel operators. We show that these convex geometries are equivalent to poset geometries. 2000 Mathematics Subject Classiﬁcation. 37F20, 06A07.

1. Introduction. Convex geometries are particular closure operators. These objects link the theory of convex sets to combinatorial theory. More precisely poset geometries are basic structures, because any convex geometry can be generated from them [6].

Mathematical morphology, introduced by Serra [7] is a very important tool in im-age processing and pattern recognition. The framework of mathematical morphology consists in erosions and dilations (resp., Galois functions) which result in morpholog-ical closures and kernels. These particular operators are well adapted to algorithmic computation [5].

Theorem 4.2proves that poset geometries and morphological geometries deﬁned below are equivalent. The convexity is very important in mathematical morphology [5]. So this article tries to demonstrate the relation between combinatorial convexity, mathematical morphology, and image processing.

2. Basic concepts. In this section, we introduce the deﬁnitions that will be needed. We will usexfor{x}, andP(S)will be the power set ofS. Throughout (untilSection 6) we will suppose that all sets are ﬁnite.

2.1. Convex geometries. LetSbe a set, consider the familyᏯof subsets ofSwith the following properties:

∅ ∈Ꮿ, S∈Ꮿ, (2.1)

A, B∈Ꮿ implies A∩B∈Ꮿ. (2.2)

This family gives rise to the closure operator

φ(X)= ∩{A∈Ꮿ, X⊆A}. (2.3)

Conversely, every closure operator deﬁnes a family Ꮿ with the properties (2.1) and (2.2). Elements ofᏯ, or elements deﬁned byφwill be called convex.

The couple(S, φ)is aconvex geometryifφveriﬁes the anti-exchange axiom

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In the same way [4]

Ifφ(X)≠S,∃p∈S\φ(X), φ(X∪p)=φ(X)∪p. (2.5)

2.1.1. Poset geometries. LetP be a partially ordered set andXbe a subset ofP, we deﬁne

Dp(X)= {y∈P , y≤xfor somex∈X}, (2.6)

(P , Dp)is a convex geometry called poset geometry. Convex geometries give rise to poset geometries which are easily characterized by the following result.

Theorem2.1. The convex geometry(S, φ)arises from the poset geometry on a poset

Pif and only if

φ(A∪B)=φ(A)∪φ(B) ∀A, B⊆S. (2.7)

Proof. See [4].

2.2. Antimatroid. On the setSwe consider the familyᏲof subsets ofS. The couple(S;Ᏺ)is anantimatroidifᏲveriﬁes the following axioms:

(i) ∅ ∈Ᏺ,Ᏺis closed under union.

(ii) ForX∈Ᏺ, X≠∅,there exists anx∈Xsuch thatX\x∈Ᏺ.

2.3. Morphological operators

2.3.1. Erosion. Anerosionis deﬁned by an operator

ε:P (S) →P (S) (2.8)

with the following properties:

ε(S)=S, A, B∈P (S), ε(A∩B)=ε(A)∩ε(B). (2.9)

2.3.2. Dilation. Adilationis deﬁned by an operator

δ:P (S)→P (S) (2.10)

with the following properties:

δ(∅)= ∅, A, B∈P (S), δ(A∪B)=δ(A)∪δ(B). (2.11)

Erosion and dilation are linked by themorphological duality.

Theorem_2.2. _Let_S_{be a set, a monotone mapping}_δ_{is a dilation if and only if there}

exists another operatorεsuch that

δ(X)⊆Y⇐⇒X⊆ε(Y ). (2.12)

The operatorεis unique and monotone and given by

ε(X)= ∪B, B∈P (S), δ(B)⊆X, (2.13)

we have

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Proof. _{See [7].}

Remark_2.3. _{Morphological duality in [5] is called}_adjuction_and _{Galois function}

in [1].

The operatorφ=ε◦δdeﬁnes a closure calledmorphological closureandφ∗=δ◦ε deﬁnes a kernel calledmorphological kernel.

2.4. Structuring function and dilation. LetSbe a set, we deﬁne astructuring func-tionunderP(S)as any mappingδfromSintoP(S). Letδbe a structuring function, considerX∈P(S)its image underδ, deﬁned by

δ(X)=

x∈X

δ(x). (2.15)

Thusδ deﬁnes a dilation fromP(S)intoP(S). Conversely, any dilation induces a structuring function by (2.15) symbolized again byδ.

3. Convex geometries, antimatroid, and duality. In the literature antimatroids are deﬁned as (i) and (ii) above, and rarely with kernel operators.

On the setSwe consider a familyᏲof subsets ofSverifying

∅ ∈Ᏺ, S∈Ᏺ, (3.1)

A, B∈Ᏺ implies A∪B∈Ᏺ. (3.2)

This family gives rise to the kernel operator

φ∗(X)= ∪{A∈Ᏺ, A⊆X}. (3.3)

Conversely, every kernel operator results in the familyᏲwith the properties (3.1) and (3.2).

Lemma_3.1. _{The pair}₍_S_{, φ}∗₎_{is an antimatroid if}_φ∗_{veriﬁes the following axiom:}

Forφ∗(X)≠∅,∃p∈φ∗(X); φ∗(X\p)=φ∗(X)\p. (3.4)

Proof. It is suﬃcient to verify that (ii) is equivalent to (3.4). We suppose (ii) is true.

Sinceφ∗(X\p)⊆X\p sop∈φ∗(X\p). From monotonicity, X\p⊂X implies φ∗(X\p)⊂φ∗(X). Soφ∗(X\p)⊆φ∗(X)\p.

Conversely,φ∗(X)\p⊆X\pimpliesφ∗(φ∗(X)\p)⊆φ∗(X\p). From (ii),φ∗(X)\ p⊆φ∗(X\p), consequently,φ∗(X)\p=φ∗(X\p).

The converse is obvious.

We have deﬁned convex geometries (resp., antimatroids) thanks to closure opera-tors (resp., kernel operaopera-tors). These concepts are dual. Indeed, in Boolean lattices, the complementationX→Xc_{induces a duality that produces a correspondence between} any closureφfromP(S)intoP(S), and its dualφ∗, as deﬁned by

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withφ∗ as a kernel. So suppose that (S;φ) is a convex geometry; sinceφis closed under intersection,φ∗is closed under union, andφveriﬁes (2.5). We have

φ∗(X)\p=φcXc\p=φcXc∩pc=φXc∪pc

=φXc_∪_pc₌_φ_X_∩_pccc₌_φ_(X_\_p)cc

=φ∗(X\p).

(3.6)

So we have a new proof of the well-known result due to Bjorner [8].

Theorem3.2. The couple (S, φ)is a convex geometry if and only if (S, φ∗)is an

antimatroid.

4. Poset geometries and mathematical morphology. The following theorem gives a characterization of convex geometries in the particular case of morphological clo-sure operators (these convex geometries will be called morphological geometries). We show that these morphological geometries are equivalent to poset geometries.

We say that(S;δ)is separated in a primary sense, ifδveriﬁes the following two properties:

(a) For any family(xi)i∈I of elements ofS and for any elementx ∈S verifying

δ(x)⊆i∈Iδ(xi), there existsj∈Isuch thatδ(x)⊆δ(xj). (b)δ(x)=δ(y)is equivalent tox=yfor anyx, y∈S.

We will callmorphological geometry(resp.,morphological poset) any convex geom-etry (resp., poset geomgeom-etry)(S;φ)such thatφ=ε◦δ.

For any dilation we can canonically associate a binary relation deﬁned byxRy is equivalent tox∈δ(y)forx, y∈S. Another interesting binary relation is given by xRy is equivalent to δ(x)⊆δ(y), for x, y∈S. We are now going to clarify the relations betweenRandR.

Lemma_4.1. _Let_δ_{be a dilation on}_S_,_R_{its binary relation canonically associated with}

it. The following two assertions are equivalent: (1)Ris reﬂexive and transitive.

(2)xRyis equivalent toδ(x)⊆δ(y).

Proof. _{The proof that (2) implies (1) is trivial.}

SupposingR reﬂexive and transitive. Consequently, δ(x)⊆ δ(y) and x ∈δ(x) imply thatx∈δ(y)which is equivalent toxRy. Soδ(x)⊆δ(y)impliesxRy. Sup-posingxRy. For anyz∈δ(x), we havezRy by transitivity, that impliesz∈δ(y), consequentlyδ(x)⊆δ(y).

Theorem_4.2. _Let_S_{be a set and}_φ=ε◦_δ_{, a morphological closure. The following}

three assertions are equivalent:

(i) δseparatesSin a primary sense. (ii) (S;φ)is a morphological poset. (iii) (S;φ)is a poset geometry.

Proof. Suppose thatδseparatesSa primary sense. The relationxRyis equivalent

toδ(x)⊆δ(y)is an order relation. Letφ(X)=ε◦δ(X), it is easy to verify that

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We have two cases.

•There existsx∈Xsuch thatδ(x)⊆δ(y), which is equivalent toxRy.

• There exists a family(xi), i∈ {1,2, . . . , n}(n >1) of elements of X such that

δ(y)⊆δ(x1)∪δ(x2)∪···∪δ(xn), but from our hypothesisδseparatesS a primary sense, consequentlyδ(y)⊆δ(x1)orδ(y)⊆δ(x2)or···orδ(y)⊆δ(xn), soyRx1 oryRx2or···oryRxn, which leads to

φ(X)=y∈S, δ(y)⊆δ(X)= {y∈S, yRxforx∈X}. (4.2)

This set is an ideal ofS. So any closed set fromφgives rise to an ideal ofS. Under the hypothesis (i),Ris an order relation, it is easy to show that any ideal ofS gives rise to a closed set. The couple (S;φ) is a poset geometry.

Suppose that (S;φ) is a morphological poset. We haveu, v ∈φ(X), v ≠u and u∈φ(X∪v)involvesv∈φ(X∪u), we havev∈ {y, δ(y)⊆δ(X)∪δ(u)}butu∈ {y, δ(y)⊆δ(X)∪δ(v)}. So the conditionδ(u)=δ(v)is equivalent tou=v, which means that the axiom (b) is veriﬁed.

Moreover, we haveφ(X)= {y∈S, δ(y)⊆δ(X)} = {y∈S, δ(y)⊆x∈Xδ(x)} which implies that for anyy∈S, δ(y)⊆δ(X)there existsx∈Xsuch thatδ(y)⊆ δ(x). ConsequentlyδseparatesS in a primary sense.

If(S, φ)is a morphological poset, a fortiori it is a poset geometry.

Suppose now that (S, φ) is a poset geometry. We haveφ(X)= {y∈S, y ≤xfor x∈X}. FromLemma 4.1, we can associate a dilation to this relationδ, soφ(X)= {y∈S, y≤xforx∈X} = {y∈S, δ(y)⊆δ(x)forx∈X} = {y∈S, δ(y)⊆δ(X)} which is a morphological poset.

5. Poset geometry, mathematical morphology, and topology. We are now going to extend this result to the case whereSis an inﬁnite set. Before we will give some deﬁnitions and remarks.

A topological space is anAlexandroﬀ spaceif the intersection of any family of open sets is open (resp., the union of any family of closed sets is closed) [3].

A topological spaceSis aT0-spacewhenever, forxandy are distinct elements of Sifx∈φ(y)theny∉φ(x).

LetS be an inﬁnite set, we now rewrite the axioms of closure operators [2]. The family of subsetsᏯofS, with the following properties:

(i) ∅ ∈Ꮿ, S∈Ꮿ,

(ii) Ꮿis preserved under intersection, (iii) Ꮿis preserved by nested union,

is equivalent to give a closure operatorφwith the following condition calledalgebraic condition:

∀x∈φ(X)there exists a ﬁnite setF include inXsuch thatx∈φ(F ). (5.1)

Families satisfying (iii) are the inductive systems. This axiom is equivalent to the al-gebraic condition [2].

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Theorem_5.1. _Let_S_{be an inﬁnite set and}_φ=_ε◦_δ_{, a morphological closure. The}

following four assertions are equivalent: (i) δseparatesSin a primary sense. (ii) (S;φ)is a morphological poset. (iii) (S;φ)is a poset geometry. (iv) (S;φ)is aT0-Alexandroﬀ space.

Proof. _{It is easy to adapt the proof of} _{Theorem 4.2} _{to prove the equivalences}

between (ii), (iii), and (iv). So we will just prove the equivalence between (i) and (ii). Suppose that (S;φ) is a T0-Alexandroﬀ space, x, y∈S, x≠y withx, y ∉φ(X), andx∈φ(X∪ {y}), implies thatx∈φ(y), but (S;φ) is aT0-space, so we havey∉ φ(X∪x). Consequently,

∀x, y∈φ(X), x≠y, x∈φ(X∪y) implies y∉φ(X∪x). (5.2)

The anti-exchange axiom is veriﬁed.

Suppose thatx∈φ(X), (S;φ) being an Alexandroﬀ space there existsy∈Xsuch thatx∈φ(y). Consequently, for allx∈φ(X), there existsF⊆X,Fbeing a ﬁnite set such thatx∈φ(F ). So (S;φ) is a poset geometry.

Suppose now that (S;φ) is a poset geometry. For allx∈φ(X), there existsF ⊆X, F, being a ﬁnite set such thatx∈φ(F ). So for allx∈φ(X)implies thatx∈φ(y)for somey∈X. It is easy to deduce that(S;φ)is an Alexandroﬀ space.

Moreover,φveriﬁes the anti-exchange axiom, from this axiom it is easy to deduce that(S;φ)is aT0-space.

Corollary_5.2. _Let_S_{be an inﬁnite space and let}_φ_{be a closure operator.The couple}

(S;φ)is a convex geometry if and only if for allX⊆S(S\φ(X);ψ)is aT0-Alexandroﬀ space whereψ(A)=y∈Aφ(φ(X)∪y)∩S\φ(X).

Proof. _{Suppose that}₍_S;_φ)_{is a convex geometry.}

We will just prove thatψ(ψ(A))=ψ(A). The other axioms can be easily shown.

ψψ(A)=

y∈A





z∈φ(φ(X)∪y)∩S\φ(X)

φφ(X)∪y∩S\φ(X)  

⊆

y∈A

φφ(X)∪φφ(X)∪y∩S\φ(X)∩S\φ(X)

=

y∈A

φφφ(X)∪y∩S\φ(X)

=

y∈A

φφ(X)∪y∩S\φ(X)=ψ(A).

(5.3)

Soψ(ψ(A))=ψ(A).

Letx,ybe two distinct elements ofS\φ(X)by application of anti-exchange axiom, we have: ifx∈ψ(y)thenx∉ψ(x). Hence(S\φ(X);ψ)is aT0-Alexandroﬀ space.

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FromTheorem 3.2andCorollary 5.2we have the following known result [4].

Corollary_5.3. _{The couple}₍_S;_φ)_{is a convex geometry if and only if the relation}

x∈φ(φ(X)∪y)∩S\φ(X)is a partial order onS\φ(X), for allX⊆S.

6. Concluding remarks. We can deﬁne a partial order on the set ofT0-Alexandroﬀ spaces (denoted byT0-Alex(CG(S))on a convex geometry(S, φ)by

S\φX1, ψ1)≤S\φX2, ψ2⇐⇒S\φX1⊆S\φX2, (6.1)

∀x∈S\φX1, ψ1(x)⊆ψ2(x). (6.2)

The setT0-Alex(CG(S))contains a maximal element. Indeed, let(S\φ(Xi), ψi)i∈Ibe a chain, it is suﬃcient to prove that(_i∈IS\φ(Xi);ψ)(withψ(A)=i∈I[y∈Aφ(φ(Xi)

∪y)∩(S\φ(Xi))i∈I]) is aT0-Alexandroﬀ and to apply Zorn’s lemma. This remark leads us to the following problem.

Problem_6.1. _{Can we classify any convex geometry}₍_S_{, φ)}_{from the maximal}

ele-ments ofT0-Alex(CG(S))?

Acknowledgement. This work was supported by Ministero dell’Università e della

Ricerca Scientiﬁcae Tecnologica (MURST): National Project “Geometric Structures, Combinatorics, their Applications” (Italy), and by the Combinatorial and Image Pro-cessing Group (France).

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Alain Bretto: University of Saint-Etienne, LIGIV, Site G.I.A.T Industries3, Rue Javelin Pagnon, BP505,42007Saint-Etienne Cedex1, France

E-mail address:[email protected]

Enzo Maria Li Marzi: Department of Mathematics, University of Messina, Contrada Papardo, Salita Sperone31 98166Sant’Agata, Messina, Italy