size functions and morphological transformations

Acta Applicandae Mathematicae 49: 85–104, 1997. 85c© 1997 Kluwer Academic Publishers. Printed in the Netherlands.

Size Functions and Morphological Transformations

PATRIZIO FROSINI1 ? and CLAUDIA LANDI21Dipartimento di Matematica dell’Universita di Bologna, Piazza Porta San Donato, 5, I-40127Bologna, Italy, e-mail: [email protected] di Matematica dell’Universita di Pisa, Via F. Buonarroti, 2, I-56127 Pisa, Italy,e-mail:[email protected]

(Received: 2 October 1995; in final form: 13 August 1996)

Abstract. In this paper changes of size functions under morphological transformations are studied.Some inequalities concerning size functions of a subset of the Euclidean plane, its dilation by anopen disk and its skeleton are proved. Such inequalities prove the stability of size functions withrespect to these morphological transformations and give a new approach for computation in SizeTheory.

Mathematics Subject Classifications (1991): Primary: 51M16, 68U10; Secondary: 55P99.

Key words: Skeleton, dilation, size function, mathematical morphology.

1. Introduction

Mathematicians are becoming more and more involved in the task of giving awell-structured mathematical answer to the problems of Shape Recognition andImage Analysis (see, e.g., [2] and [14]). Therefore, many efforts have been madein order to find solutions to the mathematical problem of comparing ‘shapes’ oftopological spaces. Size Theory and Mathematical Morphology are two differentapproaches in this direction. Our aim is to investigate the interaction betweenthese two mathematical tools used in Image Analysis.

In previous papers [4–8], the intuitive concept of shape of an object wasformalized by means of integer functions of two real variables, called size func-tions, defined through ‘measuring functions’. The idea underlying the conceptof size functions is that of setting metric obstructions to the classical notion ofhomotopy. Thus size functions convey information on both topological and met-ric properties of the set describing the viewed shape. This approach has turnedout to be useful for quite a lot of applications (see, e.g., [3, 17–19, 21–23]).

The main objective of Mathematical Morphology [16] is to reveal the structureof the objects by transforming the sets which model them. Some morphologicaltransformations, such as dilation, can be used to simplify the shape of the set or

? This work was performed within the Subproject ‘Geometria delle Varieta Differenziabili’ ofthe MURST (Italy) and partially supported by ELSAG-Bailey (Italy) and ASI (Italy).

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86 P. FROSINI AND C. LANDI

to emphasize its spatial distribution; others, such as the skeleton, can be used toemphasize a feature of a set associated with its connectivity.

In carrying out this study we have considered subsets of the Euclidean plane,since this is the most common and useful case in Image Analysis. The mainresults we have obtained in this paper are some inequalities involving size func-tions relative to these sets, to the sets obtained from these ones by some morpho-logical operations, namely dilation by an open disk and skeleton (Theorems 4.1and 5.1) and to their boundaries (in case of dilation; see Theorem 4.2).

In Section 2 we shall give some preliminary definitions about size functions,dilation and skeleton. Some useful results about size functions will be given inSection 3. The core of this paper is contained in Sections 4 and 5, where weshall treat extensively the problem of getting estimates of size functions of a settransformed by dilation by an open disk and skeleton, respectively, and of itsboundary (in case of dilation).

2. The Basic Definitions

We begin by recalling some preliminary definitions about size functions. Through-out the rest of the paper X will denote a subset of the Euclidean plane E2 endowedwith the induced topology.

We shall call any continuous function ϕ: X → R a measuring function.

DEFINITION 2.1. Let us consider a measuring function ϕ. For every y ∈ R,we can define an equivalence relation on X and all its subsets in the followingway: two points Q and R in X are said to be 〈ϕ 6 y〉-homotopic if either theyare the same point or a continuous map H: [0, 1] → X exists such that thefollowing properties hold: H(0) = Q, H(1) = R and ϕ

(H(t)

)6 y for every

t ∈ [0, 1]. H will be called a 〈ϕ 6 y〉-homotopy between Q and R. If Q and Rare 〈ϕ 6 y〉-homotopic we shall write Q ∼=ϕ6y R.

A 〈ϕ 6 y〉-homotopy is actually just a path in X where, for no point of thepath, ϕ takes a value grater than y.

DEFINITION 2.2. Let x and y be two real numbers. Let X〈ϕ 6 x〉 denote theset of all the points in X at which the measuring function ϕ takes a value lessthan or equal to x. We shall denote by `(X ,ϕ)(x, y) the finite or infinite number ofequivalence classes into which X〈ϕ 6 x〉 is divided by the equivalence relation∼=ϕ6y . Such a function `(X ,ϕ): R2 → N ∪ {+∞} will be called the size functionassociated to the pair (X , ϕ).

Remark 2.1. We shall denote by ‘∞’ any possible infinite cardinality. Wheny < x each point P such that y < ϕ(P ) 6 x identifies a different equivalenceclass. Thus if X〈ϕ 6 x〉 \ X〈ϕ 6 y〉 contains an infinite number of points, weget `(X ,ϕ)(x, y) = +∞.

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SIZE FUNCTIONS AND MORPHOLOGICAL TRANSFORMATIONS 87

Figure 1. Examples of size functions.

In general the presence of many discontinuities of the size function `(X ,ϕ)

reveals the complexity of shape of the underlying set X with respect to thechosen measuring function. In order to illustrate the capability of size functionsas shape descriptors, we show two examples of size functions in Figure 1. Theyare computed with respect to the two sets in the Euclidean plane displayedin Figure 1, by using in both cases as measuring function the distance from thecentre of mass of the set. Observe that the presence of small triangular regions onthe diagonal is due to the irregularity of the boundary of the considered set. Thedomains of the two size functions are divided in regions: the number displayedin each region denotes the value taken by the size function in that region. Weemphasize that since size functions are integer-valued, their discontinuities aregiven by integer jumps. Anyway, different measuring functions could point outdifferent properties of the shape of our sets.

Let us now recall a number of notions about the morphological transformationswe are concerned with, namely dilation by an open disk and skeletonization.

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Figure 2. Example of dilation.

Henceforth, we shall denote with B(P, λ) the open disk with centre at P andradius λ, and with Bλ the open disk centred at the origin with radius λ.

DEFINITION 2.3. Given the set X , the dilated set Y of X by Bλ is defined asthe locus of points P for which B(P, λ) has nonempty intersection with X . Weshall call this transformation dilation and denote it by the symbol ⊕. We have

X ⊕ Bλ = {P ∈ E2 | B(P, λ) ∩ X 6= ∅}.

Remark 2.2. By noting that X⊕Bλ = {P+Q: P ∈ X , Q ∈ Bλ}, we recognizethe classical Minkowski addition ([12]).

In the following any dilation is meant to be a dilation by an open disk.For λ great enough, dilation by Bλ removes from a given figure small holes

and joins parts that are separated by a narrow background portion; moreoversmall asperities in the contour are smoothed out. Thus essential parts of thefigure are enhanced at expense of small details (see Figure 2).

The function that takes a set to its skeleton is a morphological transformationthat not only provides a simplified version of the set, like dilation does, but it isalso useful to isolate connected components. Moreover, under suitable conditions,the skeleton preserves the homotopy type. Now, we shall introduce formally theconcept of skeleton.

In the following part of this Section we suppose that X is a non-empty openset.

DEFINITION 2.4. We say that the disk B(P, ρ), ρ > 0, is a maximal open diskin X if it is included in X and there do not exist a Q ∈ X and a ρ′ > 0 withB(Q, ρ′) properly containing B(P, ρ) and included in X , too.

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Figure 3. Example of skeletonization.

Remark 2.3. In case the open set X is bounded, each open ball B ⊆ X iscontained in at least one maximal open disk in X , since we can consider amaximal chain (Vi)i∈I with respect to the order given by the relation ⊆ in theset of the open balls containing B and included in X : the set

⋃i∈I Vi is the

wanted maximal open disk. On the other hand, in case it is unbounded, the setX may contain no maximal open disks (for example, if X is the complement ofa closed convex set).

DEFINITION 2.5. The skeleton of X , denoted by S(X ), is the set of centres ofthe maximal open disks contained in X .

As an example of skeleton see Figure 3.Historically, this approach to the notion of skeleton comes from Motzkin’s

characterization of closed convex sets in terms of nearest-point properties ([13]).A sufficient condition for a point P to belong to the skeleton of a set X is knownas Blum’s prairie fire model ([1]):

PROPOSITION 2.1. Assume P ∈ X 6= E2 and ρ = d(P,X c). Suppose that theclosed disk B(P, ρ) meets the boundary of X at least at two distinct points P0

and P1. Then P belongs to S(X ).Proof. The boundary of any open disk B(Q, ρ′) containing B(P, ρ) and con-

tained in X must be tangent to ∂B(P, ρ) at both the points P0 and P1. Thisimplies Q = P and hence B(P, ρ) must be maximal; that is to say P ∈ S(X )(see Figure 4). 2

The distance from X c associated with each point of S(X ) is usually called thequench function.

Another simple proof of this statement can be obtained by considering theconcepts of downstream and multiple point (see [10]).

Note that in general the skeleton is not necessarily a closed set; indeed, if weassume that S(X ) is closed, we can prove that a (deformation) retraction from Xto S(X ) exists: this is a key fact to compute the size function of S(X ) by usingthe size function of X (see Theorem 5.1). Therefore it is interesting to know

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Figure 4.

for which X ’s the skeleton is a closed set. J. A. Riley succeeded in proving thefollowing result in the Euclidean plane ([15]):

THEOREM 2.1. If ∂X is a locally finite union of curves Ci which meet, if at all,at their end points, and which possess continuously differentiable curvatures anda locally finite set of critical points for the curvature, then the skeleton S(X ) isa closed set.

We refer to [10] (Corollary 1 and 2) for other results about the closure of theskeleton.

3. Some Basic Results about Size Functions

A question that may arise is whether size functions can take the value +∞ alsofor x < y: in such a case a size function might not convey relevant information.We shall prove that under reasonable conditions on the setX , `(X ,ϕ)(x, y) < +∞for every x < y.

THEOREM 3.1. Let X be a non-empty compact, locally arcwise connected sub-set of E2 and let ϕ: X → R be a measuring function. Then, for every pair(x, y) ∈ R2 with x < y, we have `(X ,ϕ)(x, y) < +∞.

The proof of this theorem is a direct consequence of the following lemma.

LEMMA 3.1. Assume X ⊆ E2 nonempty, compact and locally arcwise con-nected. Let ϕ: X → R be a measuring function. Assume x < y. Then there

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exists a µ > 0 such that the conditions P1 ∈ X〈ϕ 6 x〉, d(P1, P2) < µ implyP1∼=ϕ6y P2.Proof. Since X is compact and since ϕ is continuous, the function ϕ is

uniformly continuous in X ; therefore, for every ε > 0, there exists a real numberδ > 0 such that |ϕ(P1) − ϕ(P2)| < ε when d(P1, P2) < δ. Moreover, sinceX is locally arcwise connected, any point P in X has an arcwise connectedneighbourhood basis. Thus, for every P ∈ X , the ball with centre P and radiusδ/2 contains an open neighbourhood UP of P which is arcwise connected. Thesets UP , when P sweeps X , form a covering for X . Since X is compact, wecan consider the Lebesgue number for the covering {UP }P∈X . Denote it byµ. Now consider two points P1, P2 ∈ X such that d(P1, P2) < µ; there mustnecessarily exist a point P ∈ X such that UP contains P1 and P2. Thus, P1 andP2 are connected by a path γ entirely contained in UP . Since UP ⊂ B(P, δ/2),each point in γ([0, 1]) has a distance from P less than δ/2. Hence, each pointin γ([0, 1]) has a distance from P1 less than δ. This implies that if P1 belongsto X〈ϕ 6 x〉, the path γ is a 〈ϕ 6 x + ε〉-homotopy between P1 and P2. Bychoosing ε small enough, we have that x+ ε < y and therefore P1

∼=ϕ6y P2. 2

Let us now give a proof of the above Theorem 3.1.

Proof. Suppose it were `(X ,ϕ)(x, y) = +∞. This assumption implies thatin X〈ϕ 6 x〉 there exists an infinite number of points which are pairwise not〈ϕ 6 y〉-homotopic; in fact X〈ϕ 6 x〉

/∼=ϕ6y must necessarily contain an infinitenumber of classes. This fact, together with the compactness of X , implies thatan infinite sequence of such points exists, converging in X〈ϕ 6 x〉. Hence, wewould have a violation of the previous lemma by which any two points closeenough in such a sequence must necessarily be 〈ϕ 6 y〉-homotopic. 2

It is easy to construct counterexamples to prove that the above assumptionscannot be weakened.

Another interesting question about size functions is how they change whenthe sets they describe change. By definition of size functions one expects thatto ‘small’ variations in the shape correspond still ‘small’ variations in the sizefunction. The following theorem formalizes this intuition.

THEOREM 3.2. Let X and Y be two nonempty subsets of the Euclidean planeand let ϕ: X → R, ψ: Y → R be two measuring functions. Assume there is ahomeomorphism f from X to Y verifying the following property: sup{|ϕ(P ) −ψ(f(P )

)|: P ∈ X} 6 ε. Then the following inequalities hold:

`(X ,ϕ)(x− ε, y + ε) 6 `(Y ,ψ)(x, y) 6 `(X ,ϕ)(x+ ε, y − ε)

for every x, y in R.

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Proof. Let us first prove that `(X ,ϕ)(x−ε, y+ε) 6 `(Y ,ψ)(x, y). In order to doso we shall construct an injective map G from X〈ϕ 6 x− ε〉

/∼=ϕ6y+ε to Y〈ψ 6x〉/∼=ψ6y . This fact will imply that the cardinality of X〈ϕ 6 x− ε〉

/∼=ϕ6y+ε isless than or equal to the cardinality of Y〈ψ 6 x〉

/∼=ψ6y ; by definition of sizefunction the considered inequality will follow immediately.

We construct the function G as follows: for each equivalence class E inX〈ϕ 6 x− ε〉

/∼=ϕ6y+ε , one point P ∈ E is arbitrarily fixed and G(E) is definedas the equivalence class of f(P ) in Y〈ψ 6 x〉

/∼=ψ6y .We shall write E = [P ]. Observe that f(P ) ∈ Y〈ψ 6 x〉 since ϕ(P ) 6 x− ε

and sup{|ϕ(P ) − ψ(f(P )

)|: P ∈ X} 6 ε. Thus f maps X〈ϕ 6 x − ε〉 into

Y〈ψ 6 x〉.Let us prove that G is injective. Consider E1 = [P1], E2 = [P2] belonging

to X〈ϕ 6 x − ε〉/∼=ϕ6y+ε . If G(E1) = G(E2) then either f(P1) = f(P2) or

there exists a 〈ψ 6 y〉-homotopy H from f(P1) to f(P2). In the first case, sincef is injective, it follows P1 = P2 that is to say E1 = E2. In the second case,let us consider the function K: [0, 1] → X defined by setting K = f−1 ◦ H;K is a 〈ϕ 6 y + ε〉-homotopy from P1 to P2 because f is a homeomorphismand sup{|ϕ(P ) − ψ

(f(P )

)|: P ∈ X} 6 ε. In other words f−1 transforms a

〈ψ 6 y〉-homotopy into a 〈ψ 6 y + ε〉-homotopy. Hence we have proved that inevery case the hypothesis G(E1) = G(E2) implies E1 = E2, i.e. G is an injectivefunction.

The inequality `(Y ,ψ)(x, y) 6 `(X ,ϕ)(x + ε, y − ε) follows from the previousone by inverting X with Y , ϕ with ψ, f with f−1 and by replacing x and y byx+ ε and y − ε, respectively. 2

We are now ready to discuss the main subject of this paper: the study of howa size function `(X ,ϕ) changes when X is replaced by its dilation, the boundaryof its dilation and its skeleton, respectively. We shall examine these cases in thenext two sections.

4. Size Functions and Dilation

As stressed before, dilations are very useful because they allow to simplify theshape of the set under study. Unfortunately, this fact implies that, in general, itis not trivial to restore information about the size function of the original setX . However, we have proved that under suitable conditions for X , it is alwayspossible to evaluate the size function of the original set by using the values ofthe size function associated to the dilated set.

THEOREM 4.1. Let X be a nonempty closed proper subset of E2 and set d =d(X , S(X c)

)in case S(X c) 6= ∅, d = +∞ otherwise. Consider a positive

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Figure 5.

number r < d. Let ϕ: X ⊕ Br → R be a continuous function and ω: R→ R itscontinuity modulus. Assume ω(r) < +∞. Then the following inequalities hold:

`(X ,ϕ|X )

(x, y + ω(r)

)6 `(X⊕Br ,ϕ)(x, y)

for every x, y in R,

`(X⊕Br ,ϕ)(x, y) 6 `(X ,ϕ|X )

(x+ ω(r), y

)for every x, y in R with x+ ω(r) 6 y.

Remark 4.1. Note that when d = 0 this statement is ‘empty’. However it isworth pointing out that in such a case it is not possible to formulate an analogousresult. As an example consider the set X = {(u, v) ∈ R2: v > |1/u|} displayedin Figure 5.

In this case d(X , S(X c)) = 0. For every choice of r ∈ R+, X ⊕Br is arcwiseconnected. On the other hand X is not. Hence our inequalities clearly do nothold. In order to verify it, take the ordinate of the point as measuring function.Then for x sufficiently large and x < y we have `(X ,ϕ|X )

(x, y + ω(r)

)= 2 and

`(X⊕Br ,ϕ)(x, y) = 1.

Proof. Consider the function f : X ⊕ Br → X that takes each point P ofX ⊕ Br to the nearest point in X . Such a point is unique since if it were notso, Proposition 2.1 would imply P ∈ S(X c), against the hypothesis r < d. Thefunction f is continuous. In fact, consider a sequence (Pi) in X ⊕ Br. By thetriangular inequality, for every P and P ′ in E2 we have |d(P,X )− d(P ′,X )| 6d(P,P ′). Assume that the sequence (Pi) converges to P and take any ε > 0.There is some n ∈ N such that for i > n we have d(P,Pi) < ε and hence

|d(P, f(P )) − d(Pi, f(Pi))| = |d(P,X ) − d(Pi,X )| 6 d(P,Pi) < ε,

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so that d(P, f(Pi)) < d(P, f(P )) + 2ε. Thus the sequence (f(Pi)) for i > n isincluded in the compact set B(P, r + 2ε), and so if it did not converge to f(P ),there would be a subsequence converging to a limit point Q 6= f(P ); clearlyQ ∈ X since X is closed. There is thus some i > n with d(f(Pi), Q) < ε, sothat d(P,Q) 6 d(P, f(Pi)) + d(f(Pi), Q) < d(P, f(P )) + 3ε; as ε was arbitrary,we get d(P,Q) 6 d(P, f(P )), contradicting the definition of f .

To prove the first inequality, consider the function

G: X〈ϕ|X 6 x〉/∼=ϕ|X6y+ω(r) → X ⊕Br〈ϕ 6 x〉

/∼=ϕ6y

so defined: for each class E ∈ X〈ϕ|X 6 x〉/∼=ϕ|X6y+ω(r) , we choose an arbi-

trary point P ∈ E and set G(E) equal to the class of P in X⊕Br〈ϕ 6 x〉/∼=ϕ6y .

G is injective; to prove this fact observe that G([P1]) = G([P2]) impliesthat P1

∼=ϕ6y P2; hence either P1 = P2, and therefore [P1] = [P2] in

X〈ϕ|X 6 x〉/∼=ϕ|X6y+ω(r) , or there exists a continuous path H: [0, 1]→ X⊕Br

with H(0) = P1 , H(1) = P2 , ϕ(H(τ)

)6 y ∀τ ∈ [0, 1]; in this latter case

consider the continuous path f ◦ H: [0, 1] → X , where f is the function pre-viously defined; it is f

(H(0)

)= f(P1) = P1, f

(H(1)

)= f(P2) = P2 and

ϕ|X(f(H(τ))

)6 y + ω(r) because for every τ in [0, 1] we have ϕ

(H(τ)

)6 y

and d(H(τ), f(H(τ))

)< r (since H(τ) ∈ X ⊕ Br and by the definition of

f ). Hence, in either case [P1] = [P2] in X〈ϕ|X 6 x〉/∼=ϕ|X6y+ω(r) and G is

injective. This proves the first inequality in the thesis.As for the second inequality, consider the function

K: X ⊕ Br〈ϕ 6 x〉/ ∼=ϕ6y→ X〈ϕ|X 6 x+ ω(r)〉/∼=ϕ|X6y

so defined: for each class F ∈ X⊕Br〈ϕ 6 x〉/∼=ϕ6y we choose an arbitrary point

Q ∈ F and set K(F) equal to the class of f(Q) in X〈ϕ|X 6 x+ω(r)〉/∼=ϕ|X6y .

It is easy to see thatK is well defined: from Q ∈ X⊕Br and the definition of fit follows that d

(Q, f(Q)

)< r; this fact, together with the inequality ϕ(Q) 6 x,

implies that ϕ|X(f(Q)

)6 x+ ω(r).

Also K is injective: K([Q1]) = K([Q2]) implies [f(Q1)] = [f(Q2)] in

X〈ϕ|X 6 x + ω(r)〉/∼=ϕ|X6y , that is to say either f(Q1) = f(Q2) or a

continuous path H: [0, 1] → X exists, with H(0) = f(Q1), H(1) = f(Q2),ϕ|X

(H(τ)

)6 y.

In the first case we have Q1, Q2 ∈ B(f(Q1), r

)⊆ X ⊕Br; consider the path

γ: [0, 1] → X ⊕ Br such that γ(τ) = (1 − 2τ)Q1 + 2τf(Q1) if τ ∈ [0, 1/2],γ(τ) = (2 − 2τ)f(Q1) + (2τ − 1)Q2 if τ ∈ [1/2, 1]; its image is the union ofsegments Q1, f(Q1) and f(Q1), Q2; the distance between any point in γ([0, 1])and the set {Q1, Q2} is less than r; hence the condition ϕ(Q1), ϕ(Q2) 6 x impliesϕ(γ(τ)) 6 x+ ω(r), ∀τ ∈ [0, 1]. Therefore, when x+ ω(r) 6 y, [Q1] = [Q2] inX ⊕ Br〈ϕ 6 x〉

/∼=ϕ6y .

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Figure 6. Size functions of a set and of its dilation.

In the case f(Q1) 6= f(Q2) consider a continuous path from Q1 to Q2 whoseimage is the union of segment Q1, f(Q1) with H([0, 1]), and with segmentf(Q2), Q2; define this path in X ⊕ Br as the path product γ1 ∗ γ2 ∗ γ3 whereγ1(τ) = (1 − τ)Q1 + τf(Q1), γ2(τ) = H(τ), γ3(τ) = (1 − τ)f(Q2) + τQ2

with τ ∈ [0, 1]; for any point in γ1 ∗ γ2 ∗ γ3([0, 1]) the measuring function takesa value less than y, because ϕ|X (H(τ)) 6 y, ϕ

(γ1(τ)

)6 x + ω(r) 6 y and

ϕ(γ3(τ)

)6 x + ω(r) 6 y. Thus, it has been proved that anyway [Q1] = [Q2]

in X ⊕ Br〈ϕ 6 x〉/∼=ϕ6y , that is to say K is injective. This proves the second

inequality in the assertion. 2

As an example of the robustness of size functions under dilation, in Figure 6the size functions associated to the sets displayed in Figure 2 are given. Theconsidered measuring function is ϕ(x, y) = xy + 5y.

Now, we shall prove another useful result, concerning size functions of theboundary of a dilated set.

Let us first prove a lemma.

LEMMA 4.1. Assume X is a nonempty subset of the Euclidean plane. ThenP ∈ ∂(X ⊕ Br) if and only if d(P,X ) = r.

Proof. By definition of dilation, Q ∈ X ⊕Br if and only if B(Q, r)∩X 6= ∅,that is to say d(Q,X ) < r. Let P ∈ ∂(X ⊕Br); this means that for every ε > 0,B(P, ε) intersects both X ⊕ Br and its complement, in other words there existQ,R with d(P,Q) < ε, d(P,R) < ε, d(Q,X ) < r and d(R,X ) > r. We deducethen that

r − ε < d(R,X )− d(P,R) 6 d(P,X ) 6 d(Q,X ) + d(P,Q) < r + ε,

which implies that d(P,X ) = r. Conversely, if d(P,X ) = r, then P ∈ (X⊕Br)c,and for every ε > 0, there is some Q ∈ X such that d(P,Q) < r + ε; there is

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then a point R on the line PQ such that d(P,R) < ε and d(R,Q) < r so thatR ∈ X ⊕ Br. Therefore P ∈ X ⊕ Br, and as P ∈ (X ⊕ Br)c, this means thatP ∈ ∂(X ⊕ Br). 2

PROPOSITION 4.1. Let X be a nonempty bounded open subset of the Euclideanplane, and set M = ∂X . Let d1 be the distance between M and S(X ) and d2

the distance betweenM and S(X c) in case it is nonempty, d2 = +∞ otherwise.Suppose d1 > 0 and d2 > 0. Consider the set obtained by dilating X by an opendisk with radius r < d2. Let M′ denote its boundary: M′ = ∂(X ⊕ Br). Thenthere exists a homeomorphism f from M′ to M such that d

(P, f(P )

)= r for

every P ∈M′.Proof. Consider the function f : M′ →M such that ∀Q ∈ M′ f(Q) is the

nearest point to Q in M (obviously this is also the nearest one in X ). FromLemma 4.1 and the fact that r < d2 it follows that Q 6∈ S(X c). So, fromProposition 2.1, such a point f(Q) is unique and therefore f is well defined. fturns out to be a homeomorfism.

In order to prove this assertion we will first show that every point P inM lieson the boundary of an open disk BP contained in X as well as on the boundaryof an open disk B′P contained in X c.

In fact consider a sequence (Pi) of points in X converging in X to P ∈M.Since X is open, for every Pi we can take an open disk contained in X andcentred at Pi. X bounded implies that such a disk is contained in a maximal opendisk Bi in X : let Ci be its centre and ρi its radius. Obviously c = inf{ρi} > d1 >0. Consider the sequences (Ci) and (ρi): since X is bounded we can extract asubsequence (Pij ) from (Pi) such that the induced subsequences (Cij ) and (ρij )

are both converging (in X and in R, respectively). Thus it is sufficient to take BPequal to the disk with centre at C = limj→∞Cij and radius ρ = limj→∞ ρij(>0).

As for B′P we can take a sequence (P ′i ) of points in X c converging to P .Such a sequence exists because P ∈ ∂X . Since X c is open, for every P ′i wecan take a small open disk (with radius at least smaller than r) contained in X cwith centre at P ′i . Such a disk is contained either in a maximal open disk in X cwhose radius we shall denote by ρ′i, or in a halfspace included in X c. In the firstcase we set ρ′i = min{ρ′i, r}, in the second case we set ρ′i = r. In any case thedisk containing P ′i is contained in a disk B′i with radius ρ′i 6 r and centre atC ′i contained in X c. Obviously, condition d2 > 0 implies that c′ = inf{ρ′i} > 0.Consider the sequences (C ′i) and (ρ′i): since limi→∞ P ′i = P and ρ′i 6 r forevery i, we have that d(C ′i, P ) 6 2r for each index i large enough. Hence wecan extract a subsequence (P ′ij ) from (P ′i ) such that the induced subsequences(C ′ij ) and (ρ′ij ) are both converging (in the plane and in R, respectively). Thusit is sufficient to take B′P equal to the disk with centre at C ′ = limj→∞C ′ij andradius ρ′ = limj→∞ ρ′ij (r > ρ

′ > 0).

ACAP1252.tex; 8/10/1997; 15:07; v.7; p.12


So, we have proved the existence of the wanted open disks BP and B′P .Let us now prove that f is surjective; for any point P ∈M consider the unit

vector N(P ) = (C ′ − P )/|C ′ − P | and the point Q = P + rN(P ). Obviouslyd(Q,X ) 6 r. We have that B(Q, r)∩X = ∅: in fact if B(Q, r) contained a pointof X , in X c there would exist a maximal open disk B∗ with B′P ⊆ B∗ ⊂ B(Q, r)(we recall that B′P ⊆ B(Q, r), since r > ρ′), and hence it should be r > d2,against our hypothesis. So d(Q,X ) > r. Therefore d(Q,X ) = d(Q,P ) = r and,by Lemma 4.1, Q is in M′ and f(Q) = P .

Moreover f is injective; assume that f(Q1) = f(Q2) = P ; then, by Lem-ma 4.1, both B(Q1, r) and B(Q2, r) contain P on their boundary and are con-tained in X c, so that such disks cannot intersect BP . These conditions imply thatit must be (Q1 − P ) = (Q2 − P ). Thus Q1 = Q2.

By the same argument as that used in Theorem 4.1 one sees that f is contin-uous.

Now, the continuity of f−1 follows immediately from the fact that f is acontinuous function with inverse defined on a compact set. 2

The property stated in Proposition 4.1 is also known as the parallel normaltransport (see [20]).

THEOREM 4.2. In the same assumptions as in the above proposition, let ϕ: E2 →R be a continuous function and ω: R → R be its continuity modulus. Assumeω(r) < +∞. Then for every pair (x, y) of real numbers it holds:

`(M,ϕ|M)

(x− ω(r), y + ω(r)

)6 `(M′,ϕ|M′)(x, y)

6 `(M,ϕ|M)

(x+ ω(r), y − ω(r)

).

Proof. The above inequalities follow directly by setting ε = ω(r) and apply-ing Theorem 3.2 to the homeomorphism f−1 defined in Proposition 4.1. 2

5. Size Functions and the Skeleton

Since the datum of an open set X is equivalent to that of its skeleton togetherwith the distance from X c associated with each point of S(X ), it is natural tolook for a correlation between the size function of S(X ) and the size functionof X . Indeed, such a correlation does exist and we shall prove it in this section(Theorem 5.1). First of all we need a preliminary result.

In the following, X will denote a bounded nonempty open set of the Euclideanplane.

ACAP1252.tex; 8/10/1997; 15:07; v.7; p.13


Figure 7. Retraction of a set to its skeleton.

PROPOSITION 5.1. Suppose that S(X ) is closed. Then there exists a retractionf : X → S(X ). Furthermore, we have

d(P, f(P )

)< max

P ′∈S(X )d(P ′,X c)

for every P ∈ X .Proof. For every point P ∈ X , let δ(P ) be its distance from the complement

of X , X c. Since X c is closed, δ(P ) is greater than zero. Consider the diskB(P, δ(P )

). We claim that B

(P, δ(P )

)is contained in a unique maximal open

disk in X (the existence of such a disk is trivial; see Remark 2.3). In fact, supposethat B1, B2 are two such maximal disks. Then, by recalling the definition of δ(P ),∂Bi is tangent to ∂B

(P, δ(P )

)at a point Qi ∈ ∂X for i = 1, 2 (possibly at other

points of ∂X , too). In case Q1 = Q2, the maximal disks B1, B2 are tangenteach to the other and, because of their maximality, they must coincide. In caseQ1 6= Q2 Proposition 2.1 implies that B

(P, δ(P )

)itself is a maximal disk in X

and, hence, B1 = B2 = B(P, δ(P )

). In both cases we have B1 = B2: so our

claim is proved. Now, consider the function f : X → S(X ) that takes each pointP ∈ X to the centre of the unique maximal disk in X containing B

(P, δ(P )

)(see Figure 7).

By definition of f , for every P in S(X ) we have f(P ) = P ; therefore f issurjective and f|S(X ) is the identity function.

ACAP1252.tex; 8/10/1997; 15:07; v.7; p.14


Finally, to prove that f is continuous, consider any sequence of points in X ,(Pi), converging to a point P ∈ X : since S(X ) is compact, it will be sufficient toprove that in case

(f(Pi)

)is converging then limi→+∞ f(Pi) = f(P ). So, sup-

pose(f(Pi)

)is converging and, for each i, choose a point QPi in ∂B

(Pi, δ(Pi)

)∩

∂X . Since X is bounded we can suppose that also (QPi) is converging in X (pos-sibly by extracting a subsequence from (Pi)). Set Q=def limi→+∞QPi . Since thefunction δ, distance from a closed set, is continuous, Q is in ∂B

(P, δ(P )

). On

the other hand Q is in ∂X because it is the limit of a sequence of points in ∂X ,which is a closed set. Now, recall that S(X ) is closed and hence observe thatlimi→+∞ f(Pi) ∈ S(X ), that is to say

Bα =def B(

limi→+∞

f(Pi), δ(

limi→+∞

f(Pi)))

is a maximal open disk in X . Let us consider also the maximal open disksBβ =def B

(f(P ), δ

(f(P )

))and Bi =def B

(f(Pi), δ

(f(Pi)

)), for every index i.

Because of its definition, Bi cannot contain the point QPi ∈ X c. Anyway, it mustcontain the disk B

(Pi, δ(Pi)

), whose boundary contains QPi : so ∂Bi is tangent to

∂B(Pi, δ(Pi)

)at QPi . By computing the limit for i→ +∞ we obtain that ∂Bα

is tangent to ∂B(P, δ(P )

)at Q. Furthermore, the definition of Bβ implies that it

cannot contain the point Q ∈ X c. Anyway, it must contain the disk B(P, δ(P )

),

whose boundary contains Q: so ∂Bβ must be tangent to ∂B(P, δ(P )

)at the point

Q. In conclusion, we have that ∂Bα and ∂Bβ are tangent to each other at Q.Thus from the maximality of Bα and Bβ it follows Bα = Bβ. Therefore theequality limi→+∞ f(Pi) = f(P ) holds and the function f is continuous.

The inequality d(P, f(P )

)< maxP ′∈S(X ) d(P ′,X c) for every P ∈ X is an

immediate consequence of the definition of f , since

f(P ) ∈ S(X ) and d(P, f(P )

)< d

(f(P ),X c

). 2

Moreover, it is easy to see that f is a strong deformation retraction of X . Aboutthe previous result, see also [16, p. 380].

Observe that the condition that S(X ) be closed is necessary in order that thefunction f defined in the above proposition is continuous.

As a counterexample, consider the set

X =def

+∞⋃i=1

B(Pi, ri) ∪ B((0, 0), 2

),

where |Pi| = 1, limi→+∞ Pi = (0,−1) and both the points Pi and the radiiri > 1 have been chosen in such a way that B(Pi, ri)∩ B(Pj , rj) ⊆ B

((0, 0), 2

)for every i and j, i 6= j (see Figure 8).

It is S(X ) =⋃+∞i=1 {O + λ(Pi − O): λ ∈ [0, 1]} (the skeleton is displayed in

Figure 8). Each point Pi belongs to S(X ), but the limit point (0,−1) does not,

ACAP1252.tex; 8/10/1997; 15:07; v.7; p.15


Figure 8. Example of a set which does not admit a retraction onto its skeleton.

because the maximum disk centred at (0,−1) and contained in X is containedin the disk with centre in (0, 0) and radius 2, also contained in X .

One can easily see that in this case the function f defined in the proof ofProposition 5.1 is not a continuous function. In fact we have that f(Pi) = Pi forevery i and limi→+∞ f(Pi) = (0,−1) 6= f

((0,−1)

)= (0, 0).

Moreover, for this choice of X there doesn’t exist any retraction to the skele-ton. Indeed if there existed such a retraction σ it should take any compact neigh-bourhood K of S(X ) contained in X onto S(X ). But σ(K) = S(X ) impliesS(X ) compact contradicting the fact that S(X ) is not a closed set.

However, according to Theorem 2.1, a large class of subsets of the Euclideanplane exists, whose skeleton is closed.

Now, we can give the main result in this Section.

THEOREM 5.1. Suppose that S(X ) is closed. Let ϕ: X → R be a continu-ous function whose modulus of continuity is the function ω. Finally setρ=def maxP∈S(X ) d(P,X c). Assume ω(ρ) < +∞. Then the following inequali-ties hold:

`(S(X ),ϕ|S(X))

(x, y + ω(ρ)

)6 `(X ,ϕ)(x, y)

for every x, y in R,

`(X ,ϕ)(x, y) 6 `(S(X ),ϕ|S(X))

(x+ ω(ρ), y

)for every x, y in R with x+ ω(ρ) 6 y.

Proof. Since S(X ) is a closed set contained in the bounded set X , S(X )itself is compact. Thus ρ is well defined.

ACAP1252.tex; 8/10/1997; 15:07; v.7; p.16


To prove the first inequality, consider the function

G: S(X )〈ϕ|S(X ) 6 x〉/ ∼=ϕ|S(X)6y+ω(ρ)→ X〈ϕ 6 x〉/ ∼=ϕ6y

so defined: for each class E ∈ S(X )〈ϕ|S(X ) 6 x〉/ ∼=ϕ|S(X)6y+ω(ρ) we choose anarbitrary point P ∈ E and set G(E) equal to the class of P in X〈ϕ 6 x〉

/∼=ϕ6y .G is injective; to prove this fact observe that G([P1]) = G([P2]) implies that

P1∼=ϕ6y P2; hence either P1 = P2, and therefore [P1] = [P2] in S(X )〈ϕ|S(X ) 6

x〉/ ∼=ϕ|S(X)6y+ω(ρ), or there exists a continuous path H: [0, 1]→ X with H(0) =

P1, H(1) = P2, ϕ(H(τ)

)6 y ∀τ ∈ [0, 1]; in this latter case consider the

continuous path f ◦ H: [0, 1] → S(X ), where f is defined in Proposition 5.1;it is f

(H(0)

)= f(P1) = P1, f

(H(1)

)= f(P2) = P2 and ϕ|S(X )

(f(H(τ))

)6

y + ω(ρ) because for every τ in [0, 1], d(H(τ), f(H(τ))

)< ρ. Hence, anyway

[P1] = [P2] in S(X )〈ϕ|S(X ) 6 x〉/ ∼=ϕ|S(X)6y+ω(ρ) and G is injective. Thisproves the first inequality in the thesis.

As for the second inequality, consider the function

K: X〈ϕ 6 x〉/∼=ϕ6y → S(X )〈ϕ|S(X ) 6 x+ ω(ρ)〉/ ∼=ϕ|S(X)6y

so defined: for each class F ∈ X〈ϕ 6 x〉/∼=ϕ6y we choose an arbitrary point

Q ∈ F and set K(F) equal to the class of f(Q) in S(X )〈ϕ|S(X ) 6 x +ω(ρ)〉/ ∼=ϕ|S(X)6y.

Now, let us see thatK is well defined; the inequality d(Q, f(Q)

)< ρ, together

with the fact that ϕ(Q) 6 x, implies that ϕ|S(X )

(f(Q)

)6 x+ ω(ρ).

Also K is injective; in fact K([Q1]) = K([Q2]) implies [f(Q1)] = [f(Q2)] inS(X )〈ϕ|S(X ) 6 x+ω(ρ)〉/ ∼=ϕ|S(X)6y ., that is to say either f(Q1) = f(Q2) or acontinuous path H: [0, 1] → S(X ) exists, with H(0) = f(Q1), H(1) = f(Q2),ϕ|S(X )

(H(τ)

)6 y ∀τ ∈ [0, 1].

In the former case Q1 and Q2 belong to the same maximal disk in X withcentre f(Q1); consider the path γ: [0, 1] → X such that γ(τ) = (1 − 2τ)Q1 +2τf(Q1) if τ ∈ [0, 1/2], γ(τ) = (2− 2τ)f(Q1) + (2τ − 1)Q2 if τ ∈ [1/2, 1]; itsimage is the union of segments Q1, f(Q1) and f(Q1), Q2; the distance betweenany point in γ([0, 1]) and the set {Q1, Q2} is less than ρ; hence the conditionsϕ(Q1), ϕ(Q2) 6 x imply ϕ(γ(τ)) 6 x + ω(ρ), ∀τ ∈ [0, 1]. Therefore, whenx+ ω(ρ) 6 y, [Q1] = [Q2] in X〈ϕ 6 x〉

/∼=ϕ6y .In the case f(Q1) 6= f(Q2), consider a continuous path from Q1 to Q2 whose

image is the union of segment Q1, f(Q1) with H([0, 1]), and with segmentf(Q2), Q2; define this path as the path product γ1 ∗ γ2 ∗ γ3 where γ1(τ) =(1−τ)Q1 +τf(Q1), γ2(τ) = H(τ), γ3(τ) = (1−τ)f(Q2)+τQ2 with τ ∈ [0, 1];for any point in γ1∗γ2∗γ3([0, 1]) the measuring function takes a value less than y,because ϕ|S(X )(H(τ)) 6 y, ϕ

(γ1(τ)

)6 x+ω(ρ) 6 y and ϕ

(γ3(τ)

)6 x+ω(ρ) 6

y. Thus, it has been proved that anyway [Q1] = [Q2] in X〈ϕ 6 x〉/∼=ϕ6y , that

is to say K is injective. This proves the second inequality in the assertion. 2

ACAP1252.tex; 8/10/1997; 15:07; v.7; p.17


Figure 9. Example of the change of size functions under skeletonization.

In Figure 9 we show the size functions associated to the sets represented inFigure 3, by using as measuring function the abscissa of the point. According tothe previous theorem size functions appear to be stable under skeletonization.

Remark 5.1. It must be noticed that for x+ω(ρ) < y all the previous inequal-ities are not trivial because the considered size functions can take only finite val-ues. Indeed the assumptions that X is a bounded open set of the Euclidean planeand S(X ) is closed, together with the fact that x+ω(ρ) < y, imply that every con-dition in Theorem 3.1 is satisfied in order that `(S(X ),ϕ|S(X))

(x+ω(ρ), y

)< +∞.

To see this, observe that since S(X ) is a bounded and closed set, it is compact;moreover by Proposition 5.1 it is a retract of X ; on the other hand X is an openset in the Euclidean plane and therefore it is locally arcwise connected; since ifa space X is locally arcwise connected then so is every retract A of X (see [9,p. 28]), S(X ) is locally arcwise connected.

Remark 5.2. Consider two nonempty open homeomorphic subsetsA and B ofthe Euclidean plane and suppose that a homeomorphism f between A and B anda continuous function ϕ: E2 → R exist with sup{|ϕ(P )−ϕ(f(P ))|: P ∈ A} 6 ε.By Theorem 3.2, this means that the two sets are similar from the size functionviewpoint. It is interesting to point out that Theorem 3.2 and Theorem 5.1 leaddirectly to estimates involving the size functions associated to the skeletons of Aand B. On the other hand, it must be noticed that a direct proof of these estimateswould be quite a labourious task because the skeletons of two homeomorphicsets can be non-homeomorphic. Incidentally, these facts underline the usefulnessand the robustness of the size function approach in Image Analysis.

ACAP1252.tex; 8/10/1997; 15:07; v.7; p.18


6. Concluding Remarks

The inequalities proved in this paper enlighten the robustness of size functionsunder morphological transformations and give new methods to compute sizefunctions of subsets of the Euclidean plane. From the applicative viewpoint thisapproach allows to evaluate the size function of a set by studying the correspond-ing size functions of its dilation and its skeleton.

Acknowledgements

The authors wish to thank M. Ferri and M. Serena for their helpful advice andthe anonymous referees for their suggestions which allowed to simplify manyproofs. Moreover the authors thank the referees for pointing out that the mainidea of Proposition 5.1 can be found in [11].

Anyway, the authors are the sole responsible for any mistakes.

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size functions and morphological transformations

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