Universit�at KonstanzMinimum Strictly Convex Quadrangulationsof Convex Polygons

Matthias M�uller-HannemannKarsten WeiheKonstanzer Schriften in Mathematik und InformatikNr. 13, Juli 1996ISSN 1430{3558

c Fakult�at f�ur Mathematik und InformatikUniversit�at KonstanzPostfach 5560/D 188, 78434 Konstanz, GermanyEmail: preprints@informatik.uni{konstanz.deWWW: http://www.informatik.uni{konstanz.de/Schriften

Minimum Strictly Convex Quadrangulations of Convex PolygonsMatthias M�uller{Hannemann1 Karsten Weihe213/1996Universit�at KonstanzFakult�at f�ur Mathematik und InformatikAbstractWe present a linear{time algorithm that decom-poses a convex polygon conformally into a min-imum number of strictly convex quadrilaterals.Moreover, we characterize the polygons that canbe decomposed without additional vertices insidethe polygon, and we present a linear{time algo-rithm for such decompositions, too. As an ap-plication, we consider the problem of construct-ing a minimum conformal re�nement of a mesh inthe three{dimensional space, which approximatesthe surface of a workpiece. The latter problemhas resulted from a cooperation with an engineer-ing company that sells CAD packages. We haveproved that this problem is strongly NP{hard,and we present a linear-time algorithm with a con-stant approximation ratio of 4.1 IntroductionIn this paper, we introduce a new approach tothe decomposition of convex polygons and, as anapplication, to the conformal re�nement of �niteelement meshes into strictly convex quadrilater-als.Much work has been done on decompositions1Technische Universit�at Berlin, Sekr. MA 6{1, Str.d.17.Juni 136, 10623 Berlin, Germany, mhannema@math.tu{berlin.de, http://winnie.math.tu{berlin.de/~mhannema.This author was supported by the special program \E�-cient Algorithms for Discrete Problems and Their Applica-tions" of the Deutsche Forschungsgemeinschaft under grantMo 446/2-2.2Universit�at Konstanz, Fakult�at f�ur Mathematik undInformatik, Postfach 5560/D188, 78434 Konstanz, Ger-many, karsten.weihe@uni{konstanz.de, http://www.infor-matik.uni{konstanz.de/~weihe.

into triangles; see [4] and [1] for surveys. How-ever, there is much less work on quadrilaterals,although quadrilaterals are more appropriate incertain situations. Ref. [5] states one possiblereason for that: \good quadrilateral meshes areharder to generate than good triangular meshes."See [12] for a survey of the work on related topics.Conformal decompositions of polygons. Apolygon is a closed sequence of straight lines(edges) in the plane or, more generally, ofsmooth curves in a smooth surface in the three{dimensional space. A polygon is simple if it doesnot cross itself, and convex if the internal angle ateach vertex is at most �. A vertex is a corner ifits internal angle is strictly less than �. A polygonis strictly convex if all its vertices are corners.Throughout this paper, a conformal decomposi-tion of a simple polygon is a decomposition of itsenclosed area into strictly convex quadrilateralssuch that the following holds (see Fig. 1):1. The quadrilaterals are openly disjoint.2. Each edge of the polygon is an edge of exactlyone quadrilateral. In particular, additionalvertices on the boundary of the polygon arenot allowed (nonetheless in the interior).3. If two quadrilaterals share more than one cor-ner, they share exactly one edge as a whole.There is a conformal decomposition for a poly-gon P if and only if the number of vertices of Pis even (cf. Lemma 2.2). This fact will be provedonly for convex polygons, because this is our focusin this work, but the proof can be easily extendedto general simple polygons.We call a conformal decomposition of a convexpolygon P perfect if it has no internal vertices. In1

Figure 2: a mesh modeling a pump. This meshhas been constructed by a German car company. Figure 3: The conformal re�nement produced bythe heuristic in [7]. Note that the decompositionof an original polygon is usually not conformal.Figure 1: A convex polygon with 7 corners and16 vertices and a conformal decomposition with7 additional, internal vertices.other words, all internal edges are chords of P . IfP admits a perfect conformal decomposition, wecall P perfect, too. In Sect. 2, we will prove auseful characterization of perfect polygons (Lem-mas 2.6 and 2.8). Based on this result, we willpresent a linear{time algorithm that tests whethera given polygon P is perfect and, if so, constructsa perfect decomposition.In Sect. 3, we will present a linear{time algo-rithm that constructs a conformal decompositionof a convex polygon with a minimum number ofquadrilaterals. This problem is closely relatedto perfect decompositions, since for perfect poly-gons, the perfect decompositions are exactly theminimum decompositions (cf. Corollary 3.3).To our knowledge, neither problem has beendealt with so far. Nonetheless, a few related prob-lems have found some attention. Ref. [2] giveslower and upper bounds on the number of quadri-laterals in a conformal decomposition of simple,not necessarily convex polygons (even with holes),

but in contrast they also allow additional verticesto be placed on the boundary. Refs. [9, 10] investi-gate perfect decompositions of (star-shaped) rec-tilinear polygons into non{strictly convex quadri-laterals, and [6] considers perfect decompositionsof non-convex polygons but even allows overlap-ping internal edges. See [12] for a systematic sur-vey.Conformal mesh re�nement. Section 4 isdevoted to an application, which has motivatedthe research on decompositions of quadrilaterals:Conformal re�nements of �nite{element meshesare important for the computer aided design ofmachines, vehicles, and many other kinds of tech-nical devices. Such a mesh approximates the sur-face of a workpiece; see Figs. 2 and 3. This pro-cedure prepares an interactively generated modelof the workpiece for numerical computations.A mesh is a complex of openly disjoint, con-vex polygons. In a conformal re�nement of amesh, each polygon is decomposed into strictlyconvex quadrilaterals, but not necessarily confor-mally. Instead, the common decomposition of allpolygons ful�lls the �rst and third conditions ofconformality; see Fig. 3. The solution in Fig. 3is constructed by a heuristic, which is based onbidirected network ows [7].In [7] and all other work we know of, the num-ber of quadrilaterals is not considered explicitly,although this number is crucial for the run timeof all later steps of the CAD process. For this2

reason, the problem we consider in this paper isthe following:Find a conformal re�nement with a minimumtotal number of quadrilaterals (minimum confor-mal mesh re�nement problem).This problem de�nition might not re ect allpractical aspects, but allows better insights intothe mathematics of the problem. In Sect. 4 wediscuss the problem further and present the fol-lowing results:� The minimum conformal mesh re�nementproblem is NP{hard.� There is a linear-time approximation algo-rithm which yields an approximation ratio of4.This algorithm relies heavily on the results aboutdecompositions of polygons in Sects. 2 and 3.2 Perfect PolygonsIn this section, we sketch the proof of the followingtheorem.Theorem 2.1 There is a linear{time algorithmwhich tests whether or not a polygon is perfectand, if so, constructs a perfect conformal decom-position.To prove Theorem 2.1, we will need some gen-eral notation. An interval on a polygon P is apath on its boundary. An interval is trivial if itconsists of exactly one edge of P . A segment Sis an interval between two successive corners ofP . Let e1 and e2 be two di�erent edges of P .Then I[e1; e2] denotes the interval counterclock-wise from e1 to e2, including neither e1 nor e2.A conformal decomposition of P is usually iden-ti�ed with the planar, embedded graph G =(V;E) whose outer face is P and whose inter-nal faces are the quadrilaterals. For such a graphG = (V;E), the dual graph Gd = (V d; Ed) is de-�ned as usual [8]: V d is the set of all faces of G,and two vertices in V d are adjacent if and onlyif they are incident to a common edge of G. LetG� = (V �; E�) be the graph arising from Gd byremoving the vertex corresponding to the outerface of G. See Fig. 4. A planar graph whose ver-tices all lie on a common face is called outerplanar.

Hence, a perfect decomposition corresponds to anouterplanar graph.Lemma 2.2 [5] A convex polygon P admits aconformal decomposition if and only if the numberof vertices of P is even.For completeness, we sketch an illustrativeproof.Proof: The if part follows from the constructionin Fig. 5.Fig. 4 demonstrates an obvious fact: that thedegree of a face v� 2 E� has in G� the same parityas the number of edges of P incident to v�. Nowthe only{if part follows from the fact that the sumof all degrees in an undirected graph G� is alwayseven. 2Assumption 2.3 Throughout Sects. 2 and 3, weassume that the number of vertices of P is even.Note that G� is always connected. The fol-lowing observation is well-known for outerplanargraphs and is also obvious from Fig. 4.Observation 2.4 G is a perfect decomposition ifand only if G� is a tree.Each degree{one vertex v� of G� points to atrivial segment of P (see Fig. 4). In the quadrilat-eral corresponding to v�, this is the edge oppositeto the edge crossed by the unique incident dualedge. We will sometimes identify such a vertexwith this trivial segment. Since each tree has atleast two leaves, we obtain:Corollary 2.5 Every perfect polygon contains atleast two trivial segments.The length L(I) of an interval I is the number ofits edges. Moreover, K(I) denotes the maximumsize of a choice of strictly convex internal verticesof I such that no two of them are neighbored on P .We often denote (L� 2K)(I) := L(I)� 2 �K(I).Note that (L� 2K)(I) is always nonnegative.Roughly speaking, Lemmas 2.6 and 2.8 showthat, if a polygon is perfect, it even admits a per-fect decomposition of a very restricted shape: Thereduced dual graph G� is either a path, or it hasa single branching vertex, which has degree three.3

eFigure 4: The polygon P (solid) and the confor-mal decomposition G (dashed) of Fig. 1, and thegraph G� (solid). The unique degree{one vertexof G� points to e. Figure 5: proof of Lemma 2.2. First the con-vex polygon is reduced to a strictly convex core(thick solid line), then the core can be re�ned ina straightforward manner. Obviously, this proce-dure does not minimize the number of quadrilat-erals in general.In the former case, it is called a path decomposi-tion, and in the latter case, a K1;3{decomposition,because then G� is a subdivision of the completebipartite graph K1;3 on 1 + 3 vertices.Lemma 2.6 Let P be a polygon with exactly twotrivial segments e1 and e2, and let I1 := I[e1; e2]and I2 := I[e2; e1]. W.l.o.g. we have L(I1) �L(I2). Then P is perfectly decomposable if andonly if (L � 2K)(I1) � L(I2). In this case, itadmits a path decomposition.Proof: As each tree with two leaves is a path, anyperfect decomposition is a path decomposition.When G� is a path, each vertex of G� correspondsto a quadrilateral that shares exactly two edgeswith I1[I2. If these two edges are on the same in-terval I1 or I2, they are incident and enclose a cor-ner of P . (Otherwise, they would enclose a thirdleaf of G�.) Therefore, we have L(I1) � L(I2) �2 � K(I1) and L(I2) � L(I1) � 2 � K(I2). Obvi-ously, these two inequalities are also su�cient forperfectness. The latter inequality is always truebecause of L(I1) � L(I2) � (L� 2K)(I2). Hence,the former inequality determines perfect decom-posability, which proves Lemma 2.6. 2In Lemma 2.8, we assume the following sce-nario.Scenario 2.7 Let P be a polygon with at leastthree trivial segments. Let e1, e2, and e3 bethree trivial segments such that the counterclock-wise order around P is e1 � e2 � e3 � e1. Let

I1 := I[e1; e2]; I2 := I[e2; e3]; I3 := I[e3; e1], andw.l.o.g. L(I1) � L(I2) and L(I1) � L(I3). As-sume that L(I1) is minimum subject to all theseconditions.Fig. 6 shows an example for this scenario.Lemma 2.8 In Scenario 2:7, P is perfectly de-composable if and only if (L� 2K)(I1) � L(I2) +L(I3) + 1. In this case, there is a perfect decom-position that is either a path decomposition withleaves e1 and e2, or it is a K1;3{decompositionwith leaves e1, e2, and e3.Proof: We consider four di�erent cases sepa-rately.Case I : (L� 2K)(I1) = L(I2) + L(I3) + 1.Then L(I1) � L(I2)+L(I3)+1 = L(I2[e3[I3),and the existence of a path G� with leaves e1 ande2 follows analogously to Lemma 2.6.Case II : L(I2) = 0 or L(I3) = 0.W.l.o.g., let L(I2) = 0. By Assumption 2.3,L(I1) + L(I2) + L(I3) is odd. Hence, L(I1) andL(I3) have di�erent parities, and the overall as-sumption L(I1) � L(I3) can be strengthened toL(I1) � L(I3)+1 = L(I2[ e3[ I3). Again the ex-istence of a path G� with leaves e1 and e2 followsanalogously to Lemma 2.6.Case III : (L � 2K)(I1) < L(I2) + L(I3) + 1 andL(I2) > 0 and L(I3) > 0.4

This is the case shown in Fig. 6. We �rst de-�ne three values b1; b2, and b3. Recall the over-all assumption L(I1) � L(I2) and L(I1) � L(I3).In particular, this yields L(I1) � jL(I2) � L(I3)j.Together with the condition de�ning Case III, weobtain that the interval I :=h(L�2K)(I1); L(I1)i\hjL(I2)�L(I3)j; L(I2)+L(I3)iis not empty. If jIj = 1, the unique element is ei-ther (L�2K)(I1) or L(I1), because L(I2); L(I3) >0 implies j[jL(I2) � L(I3)j; L(I2) + L(I3)]j > 1.In this case, we de�ne b1 as this unique elementof I, and hence b1 + L(I2) + L(I3) is odd. Onthe other hand, if jIj > 1, we choose b1 2 I sothat b1 + L(I2) + L(I3) is odd again. Next de�neb2 := L(I2) � 1 and b3 := L(I3) and consider thefollowing linear equation system:0@ 0 1 1 b11 0 1 b21 1 0 b3 1AThe unique solution isx1 = (b2 + b3 � b1)=2;x2 = (b1 + b3 � b2)=2;x3 = (b1 + b2 � b3)=2:By de�nition of the bi, b1 + b2 + b3 is even.Hence, the xi are integral. Next we show that thexi are nonnegative. Because of b1 2 I, we haveb1 � L(I2)+L(I3), and since b1 and L(I2)+L(I3)have di�erent parities, we even have b1 � b2 + b3,which means x1 � 0. Moreover, b1 2 I immedi-ately yields b1 � jL(I2)�L(I3)j, and by the sameparity argument, we even have b1 � jb2�b3j. Thisimplies x2 � 0 and x3 � 0.All these facts together su�ce, because we canconstruct a subdivision of K1;3 from the valuesxi (cf. Fig. 6): x1 is the number of quadrilateralssharing one edge with I2 and one with I3. Anal-ogously, x2 quadrilaterals share an edge with I1and I3, and x3 with I1 and I2.There are as many as 12 (L(I1) � (x2 + x3))quadrilaterals that share two edges with I1. Notethat the number of quadrilaterals that share twoedges with I1 is geometrically feasible, becausewe obtain 12 (L(I1) � (x2 + x3)) � K(I1) from(L� 2K)(I1) � b1 = x2 + x3.

I1

I

1

3

3

ee2

e

1 23

4

6

7

8

5

9 10 11

I2

Figure 6: an example for Scenario 2.7. Here wehave L(I1) = 11, (L � 2K)(I1) = 7, L(I2) = 6,L(I3) = 4, and I = [7; 11] \ [2; 10] = [7; 10].For this decomposition, we have chosen b1 = 7,b2 = 5, and b3 = 4. This yields x1 = 1 (inducingquadr. no. 6), x2 = 3 (nos. 1,2,3), and x3 = 4(nos. 7,9,10,11).As b2 = L(I2)�1, there is one quadrilateral thatshares one edge with I2 and no edge with P n I2.This is the branching vertex of the subdivision ofK1;3 (no. 5 in Fig. 6).Case IV : (L� 2K)(I1) > L(I2) + L(I3) + 1.First we show that I1 contains no trivial seg-ment. To see this, assume I1 does contain a trivialsegment e4. Let I4 := I[e1; e4] and I5 := I[e4; e2].Clearly, we have L(I4) < L(I1) and L(I5) < L(I1).In Case IV, we also have L(I2 [ e3 [ I3) < L(I1).However, this contradicts the speci�c choice offe1; e2; e3g in the scenario of Lemma 2.8, becausefe1; e4; e2g would be a strictly better choice.Now we know that I1 contains no trivial seg-ment. Suppose that nonetheless P admits a per-fect decomposition G = (V;E) of P . Then allleaves of G� belong to e2 [ I2 [ e3 [ I3 [ e1. Inother words, the quadrilaterals incident to I1 formone single path p of G�.Let Q1 denote the quadrilaterals sharing oneedge with I1 and one or more edges with I2 [ e3[I3, let Q2 be the quadrilaterals sharing exactlyone edge with I1 and none with I2 [ e3 [ I3, andlet Q3 be the quadrilaterals sharing at least oneedge with I2 [ e3 [ I3 and none with I1. SinceI1 contains no trivial segments, no quadrilateralshares more than two edges with I1. Note thatat most K(I1) quadrilaterals share exactly twoedges with I1. Therefore, we have jQ1j + jQ2j �(L� 2K)(I1).5

Next note that each quadrilateral in Q2 is abranching vertex of the tree G�. The branch thatdoes not belong to p contains at least one leaf, andthis leaf belongs to Q3. This implies jQ3j � jQ2j.Clearly, we have jQ1j + jQ3j � L(I2) + L(I3) +1. Altogether, we obtain the contradiction (L �2K)(I1) � jQ1j + jQ2j � jQ1j + jQ3j � L(I2) +L(I3) + 1. 2Now we are going to introduce the linear{timealgorithm for perfect convex polygons. Recallfrom Corollary 2.5 that polygons with less thantwo trivial segments are not perfect. Lemma 2.6immediately translates into a linear{time algo-rithm for polygons with exactly two trivial seg-ments.So consider the case that P has more thantwo trivial segments. Clearly, the only problemis to �nd three trivial segments e1, e2 and e3as described in Scenario 2.7, because the proofof Lemma 2.8, Cases I{III, immediately yields alinear{time algorithm for the rest.To �nd suitable trivial segments e1, e2, e3, weapply the following strategy. Let n denote thenumber of edges of P . We number all edges coun-terclockwise 0 : : : n� 1, starting anywhere. In thefollowing, all additions are performed modulo n.In particular, a subtraction x�y denotes the coun-terclockwise distance from y to x around P .For a real number x 2 [0; n], `(x) 2 f0; : : : ; n�1g denotes the �rst value i in the sequence(bxc; bxc � 1; bxc � 2; : : :) such that the edgeno. i is a trivial segment. Analogously, r(x) 2f0; : : : ; n � 1g denotes the �rst value i in the se-quence (dxe; dxe + 1; dxe + 2; : : :) such that theedge no. i is a trivial segment. Clearly, all thesevalues can be computed in linear time during onesingle pass around P .For an i 2 f0; : : : ; n� 1g that is the number ofa trivial segment, let M(i) denote the minimumpossible value of maxfi0 � i; i00 � i0; i � i00g suchthat i0 and i00 are numbers of trivial segments andi < i0 < i00 < i is the cyclic order modulo n. Itsu�ces to computeM(i) for each i and to take theoverall minimum. For real numbers x; y 2 [0; n]let A(x; y) := 8<: x+y2 ; x � y;x+y2 + n2 ; x > y:

Intuitively, A(x; y) is the middle of the interval ofP that goes counterclockwise from x to y. Thefollowing lemma �lls in the remaining gap.Lemma 2.9 For i 2 f0; : : : ; n � 1g, M(i) is at-tained by the following choices of i0 and i00.1. If all trivial segments except i belong to [i +1; bi+ 13nc], we choose i00 = `(i+ 13n), and i0is an arbitrary trivial segment di�erent fromi and i00.2. Analogously, if all trivial segments except ibelong to [di� 13ne; i�1], we choose i0 = r(i�13n) and i00 arbitrarily.3. Otherwise, one of the following choices is ap-propriate:i0 i00`(i+ 13n) `(A(i0; i)) or r(A(i0; i))`(A(i; i00)) or r(A(i; i00)) r(i� 13n)r(i+ 13n) `(i� 13n)Proof: In the �rst case of Lemma 2.9, we havei0�i � n=3, i00�i0 � n=3, and i�i00 � 2n=3 for anychoice of i0 and i00. Therefore, i00 is best chosencounterclockwise \farthest away" from i, whichmeans i00 = `(i + 13n). Clearly, i0 may be chosenarbitrarily. The second case is mirror{symmetric.In the third case, we apply a further case dis-tinction. For i0 2 [i + 1; bi + 13nc], we havei0 � i � maxfi00 � i0; i � i00g, because i0 � i � 13nand all three di�erences sum up to n. Hence, thebest choice for i0 is i0 = `(i + 13n). To minimizemaxfi00 � i0; i � i00g, we clearly choose i00 as closeto A(i0; i) as possible, that is, i00 = `(A(i0; i)) ori00 = r(A(i0; i)). The case i00 2 [di � 13ne; i � 1] ismirror{symmetric.It remains the case i0; i00 2 [i+ 13n; i� 13n] (notethat i� 13n = i+ 23n). Then we have i00� i0 � 13n,i0 � i � 13n, and i� i00 � 13n, which means that i0and i00 should be chosen \farthest away" from eachother, that is, i0 = r(i + 13n) and i00 = `(i � 13n).26

3 Non{Perfect PolygonsRemember Assumption 2.3. For a conformal de-composition G = (V;E) of polygon P let q(G) de-note the number of internal, quadrangular faces.In this section, we will sketch the proof of thefollowing result.Theorem 3.1 There is a linear{time algorithmthat constructs, for a convex polygon, a conformaldecomposition G that minimizes q(G).Let i(G) denote the number of internal vertices,that is, the members of V that do not lie on P . Inparticular, a decomposition is perfect if and onlyif i(G) = 0. The following lemma relates i(G) toq(G).Lemma 3.2 For a �xed polygon P , q(G) growsstrictly monotonously with i(G). Therefore, tominimize q(G) it su�ces to minimize i(G).Proof: Euler's formula says jV j + (q(G) + 1) =jEj+2. Let denote (V1; E1) = P , and let E2 := EnE1 denote the set of all internal edges of G. Thenwe have jV1j = jE1j, and since all internal facesof G are quadrilaterals, we also have 4 � q(G) =jE1j+2 � jE2j. Together, this yields q(G) = i(G)+jE1j=2� 1. 2Corollary 3.3 If existing, a perfect decomposi-tion G of a polygon P minimizes q(G).Therefore, the algorithm from Sect. 2 solves theproblem for perfect polygons. For non{perfectpolygons, we abstract from all geometrical aspectsand restrict our attention to purely structural as-pects. More precisely, in Lemma 3.4 we �rst char-acterize the planar graphs induced by conformaldecompositions.Lemma 3.4 Let P be a convex polygon and G =(V;E) a biconnected planar graph together with adesignated face F such that1. F is isomorphic to P ;2. all other faces of G are quadrilaterals;3. the following vertices have degree at least 3:all internal vertices and all vertices on F thatdo not correspond to corners of P ;

wv

(a)

F’

1v

v

vv4

3

2

(b)Figure 7: The schematic situations prevented bythe fourth and �fth condition in Lemma 3.4, re-spectively.4. if F shares two vertices v; w 2 G with an-other face F 0 such that v and w correspondto vertices on the same segment of P , thenv and w are adjacent in G (see Fig. 7(a) forthe situations prevented by this condition);5. when two non{adjacent vertices, v1 and v3,on a cycle (v1�v2�v3�v4�v1) of length 4 areremoved, each connected component of G nfv1; v3g contains v2 or v4 (see Fig. 7(b) forthe situations prevented by this condition).Then (and only then) there is a planar embeddingof G such that the outer face is mapped onto Pand all internal faces are strictly convex.Proof: Omitted (see Appendix A). 2It is easy to see that all these preconditions arenecessary. Our proof of su�ciency is constructiveand yields a linear{time algorithm for a geomet-rically feasible embedding of such a graph G (alldetails omitted for brevity). Hence, from now onwe may restrict our attention to the structural as-pects of G. It is also easy to see from Fig. 7 thatthe last two conditions of Lemma 3.4 are ful�lledfor each optimal decomposition. Thus, we mayeven restrict our attention to the �rst three con-ditions. These conditions will be trivially ful�lledby our construction.Lemma 3.5 below is the basis of the linear{timealgorithm for non{perfect polygons. To state thislemma, we need some more terminology.We will denote a polygon by the counterclock-wise sequence of the lengths of its segments.For example, (1; 1; 1; 1) denotes the strictly con-vex quadrilateral, (1; 1; 2) = (1; 2; 1) = (2; 1; 1)the quadrilateral degenerated to a triangle, and7

(d)(c) (e)

(a) (b)

Figure 8: The �rst class of cut components inLemma 3.5(2). The solid lines belong to P ,and the dashed lines are internal edges. ByLemma 3.4, only the structure of the graph mat-ters; the concrete lengths and angles are only ex-emplary. Figure 9: The �ve smallest cut components of thesecond class. The de�nition of the whole (in�nite)class might be obvious.(4; 1; 2; 3; 2; 2; 2) = (1; 2; 3; 2; 2; 2; 4) = � � � thepolygon in Fig. 1. By Lemma 3.4, this notationencodes all information that we need for our pur-pose.Recall the de�nition of G� = (V �; E�) from thebeginning of Sect. 2. Let G = (V;E) be an undi-rected planar, embedded graph. An area com-ponent of G is a subgraph G0 induced by a con-nected component of G�. More precisely, G0 con-sists of all vertices and edges incident to the poly-gons that correspond to this component of G�.An area decomposition of G is a collection of areacomponents such that the inducing componentsof G� partition all vertices in V �. Intuitively, thismeans that the internal faces of G are partitionedand covered by closed, but openly disjoint, con-nected areas.Lemma 3.5 For P 62 f(2; 1; 1); (3; 3; 3; 3);(4; 3; 3); (4; 2; 2)g, there is a conformal decompo-sition G with minimum q(G) such that there is anarea decomposition of G with the following prop-erties:1. The area decomposition consists of at mostfour area components.2. All area components except one are isomor-phic to one of the components depicted inFigs. 8 and 9. These area components arehenceforth called the cut components.3. The remaining area component is outerpla-nar. This area component is henceforth calledthe core component.4. No two cut components share an edge.

5. All cut components except at most one are oftype (c), (d), or (e) in Fig. 8.6. If a cut component of type (a) or (b) in Fig. 8or a cut component in Fig. 9 occurs, the corecomponent admits a path decomposition.7. If a cut component of the type in Fig. 9 oc-curs, the area decomposition contains at mosttwo cut components; if a cut component oftype (a) or (b) in Fig. 8 occurs, this is theonly cut component.Proof: Omitted (see Appendix B). 2In the following, we will only sketch the linear{time algorithm that performs this reduction. SeeFigs. 10 and 11 for examples of optimal solutionsconstructed by the algorithm.By Lemma 3.4, constructing an outerplanar de-composition of an area component amounts toapplying the algorithm from Sect. 2. Hence, byLemma 3.2, it su�ces to �nd a collection of cutcomponents so that the remainder is a suitablecore component and the number of internal ver-tices introduced by these cut components is min-imized.We need some more terminology. Let G0 bea partial conformal decomposition of polygon P .This means that the outer face of G0 is isomorphicto P , and that all internal faces except one arequadrilaterals. (The exceptional face is yet to bere�ned.) Let F 0 denote this face. We call a vertexof F 0 relatively convex if it is either a corner of P ,or it has degree at least three in G0. A relativesegment of F 0 is an interval of F 0 between twosuccessive relative corners. By Lemma 3.4, we can8

Figure 10: An optimal decomposition with onecut component of the type in Fig. 8(c) and oneof the type in Fig. 8(d). The dashed lines are thechords of the (outerplanar) core component. Inthis example, the core component is decomposedaccording to a dual subdivision of K1;3.Figure 11: An optimal decomposition with onecut component of the type in Fig. 8(d) and oneof the type in Fig. 9. By Lemma 3.5(6), the corecomponent admits a path decomposition.extend G0 to a full conformal decomposition of Pby decomposing F 0 and thereby treating relativelyconvex vertices as corners and every other vertexof F 0 as a vertex having internal angle �.Note that all cut components except those inFig. 8(a) and (b) introduce trivial relative seg-ments for the core component. The idea is toconstruct a collection of cut components such thatthe induced trivial relative segments allow a pathdecomposition or a K1;3{decomposition.Consequently, the algorithm consists of twostages: The �rst stage tries to �nd such a collec-tion so that the core component admits a K1;3{decomposition; in the second stage, the core com-ponent shall instead be a path decomposition. Ifsucceeding, each stage determines a solution witha minimum number of vertices. By Lemmas 2.6,2.8, and 3.5(3), at least one stage will succeed,and if both stages succeed, the better result is theoverall optimum.Remember the de�nition of r(�) and `(�) forLemma 2.9. For a real number x 2 [0; n], we nextintroduce the values `j(x) and rj(x) for j = 0; 1; 2.This is a generalization of the values `(x) andr(x) introduced in Sect. 2. Roughly speaking,`j(x) (rj(x), resp.) denotes the clockwise (coun-terclockwise) distance from x to the next possibil-ity to create a trivial relative segment spending jadditional internal vertices. In particular, we have`0(x) = `(x) and r0(x) = r(x).More formally, `1(x) is the �rst value i in thesequence (bxc; bxc � 1; bxc � 2; : : :) such that theedges no. i; : : : ; i � 3 may belong to a cut com-

ponent of type (c) in Fig. 8. In other words, thevertices between the edges i and i � 1 and be-tween the edges i � 2 and i � 3 are corners of P .Analogously, `2(x) is the �rst value i in this se-quence such that the edges no. i; : : : ; i � 3 maybelong to a cut component of type (d), or edgesno. i; : : : ; i � 4 by a component of type (e). Thismeans that either the vertex between the edgesi� 1 and i� 2 is a corner or the vertices betweenthe edges i and i� 1 and the edges i� 3 and i� 4are corners. The de�nitions of r1(x) and r2(x)are mirror{symmetric. Clearly, we can computeall these values in linear time during one singlepass around P .Next we explain the �rst stage of the algorithmin detail, that is, �nding a minimum conformaldecomposition such that the core component ad-mits a K1;3{solution. See Fig. 10 for an examplewhere this stage succeeds and yields the overalloptimum. Like in Sect. 2, we try to �nd opti-mal i0 and i00 for each trivial segment i. However,here we do the same also for each possible triv-ial relative segment that may result from cuttingo� an area component of type (c),(d) or (e) inFig. 8. For each such trivial segment or trivialrelative segment e, we apply 9 variants on theprocedure in Lemma 2.9: one for each combina-tion j1; j2 2 f0; 1; 2g. For j1 = j2 = 0, this isexactly the algorithm in Lemma 2.9. The othervariants di�er (up to minor details) from this onein that `0 and r0 are replaced by `j1 and rj1 fori0, and by `j2 and rj2 for i00. The best result of allthese variants is the optimum for e, and the best9

result for all e is the overall optimum.It remains to explain the second stage of thealgorithm in detail, that is, �nding a minimumconformal decomposition such that the core com-ponent admits a path solution. See Fig. 11. Be-cause of Lemma 3.5(7), all possibilities to applytype (a) or (b) in Fig. 8 may be tested separately.On the other hand, all possible reductions thatneed only components of types (c), (d) or (e) inFig. 8 may be examined analogously to the �rststage of the algorithm.Lemma 3.5(5) and (7) say that we need at mostone cut component of the type in Fig. 9, and thatthere is at most one further cut component, whichis of type (c), (d), or (e) in Fig. 9. In other words,we may assume that one leaf of the decomposi-tion according to Lemma 2.6 is either an originaltrivial segment or is created by type (c), (d) or (e)in Fig. 8. Let i be the index of this (relatively)trivial segment. Then it is easy to see that, in anoptimal solution, a component of Fig. 9 can onlybe applied to a corner incident to the segment towhich the index i + n=2 belongs. Therefore, atmost two possibilities of this type must be exam-ined after �xing the �rst trivial segment.This concludes the sketch of the algorithm.4 Conformal Mesh Re�nementsAs an application for the decomposition of simplepolygons, we consider in this section the problemof constructing a minimum conformal re�nementof a mesh in the three{dimensional space. Thisproblem has resulted from a cooperation with anengineering company that sells CAD packages.Throughout this section, we identify a meshwith the undirected graph G = (V;E) that is in-duced by all corners of these polygons. In otherwords, each polygon is a \hole" of G, and sincea side of a polygon may consist of more than oneedge of G (e.g., in Figs. 2 and 12), these polygonsof G are not strictly convex, but still convex.Note that there may be holes of G that donot belong to the approximation of the surface,namely holes of the surface itself. (These holesmay not be convex.) Note further that thegraph G of a mesh need not be planar; for exam-ple, a mesh approximating the surface of a torushas genus one. Even more, a mesh may contain

folding edges, that is, edges incident to more thantwo polygons (Fig. 12). We call a mesh homoge-neous if it does not contain folding edges.Figure 12: a small mesh with three homogeneouscomponents and �ve folding edges.Most work on conformal re�nements in the lit-erature is based on the template model, becausethis is easier to approach. In that model, thepossibilities of decomposing a single polygon arerestricted to a few classes of templates. For exam-ple, the most important template for quadrilateralpolygons is an (m � n){grid, where m and n arevariable.Ref. [11] introduces an integer programming ap-proach, which relies on very strong template re-strictions. In [7], we reported on a cooperationwith an engineering company, which sells CADpackages. In this cooperation, we developed andimplemented an algorithm for mesh quadrangula-tion, which is based on a more exible templatemodel and applies network ow techniques. Thisalgorithm performs very well in practice, but it isa pure heuristic.In this paper now, we attack the problem froma theoretical point of view and introduce a newapproach that does not rely on template restric-tions.Unfortunately, the minimum conformal meshre�nement problem is highly intractable:Theorem 4.1 The minimum conformal mesh re-�nement problem is strongly NP{hard even forhomogeneous meshes.Proof: by a reduction from Exact Cover by 3{Sets [3]; all details are omitted. 2However, for the asymptotic complexity of thenumerical analysis it is not necessary to �nd theexact minimum of quadrilaterals, but to achievea constant approximation ratio.10

The following lemma shows that there is an al-gorithm that runs in linear time and achieves asmall constant approximation ratio.Theorem 4.2 There is a linear{time algorithmthat constructs a conformal re�nement of a meshsuch that the number of quadrilaterals is at mostfour times the optimum.Proof (sketch): Let G = (V;E) denote themesh. We subdivide each edge in E exactly once.Obviously, each polygon is even afterwards, andwe may apply the algorithm from Sect. 3 sepa-rately to each polygon. This already completesthe description of the algorithm.To prove the approximation ratio, we need somefurther terminology. For a convex polygon P withan even number of vertices, min(P ) denotes theminimum number of quadrilaterals required byany conformal decomposition of P .For an arbitrary convex polygon P with edgeset EP , a mapping X : EP ! N0 is called feasi-ble if Pe2EP X(e) has the same parity as EP . Inparticular, if jEP j is even, X � 0 is possible, too.The polygon PX is constructed from P by sub-dividing each edge e 2 EP exactly X(e) times.Hence, X feasible means that PX admits a con-formal decomposition. Moreover, Min(P ) denotesthe minimum number of quadrilaterals in any con-formal decomposition of any polygon PX ,Min(P ) := minfq(G)jG conformal decompositionof PX ; X : EP ! N0 feasibleg :Let P be a polygon of G, let X : EP ! N0 beuniformly equal to 1, X � 1. The heart of theproof (all details omitted; see Appendix D) is toshow that min(PX) � 2�Min(P )+jEP j�2 for thisspecial class of weightingsX. Because of this fact,it su�ces to prove jEP j � 2�Min(P )+2. However,this follows from the proof of Lemma 3.2: LetY : EP ! N0 such that Min(P ) = min(PY ), andlet i(Y ) denote the number of internal vertices ofan optimal conformal decomposition of PY . Thenwe have min(PY ) = i(Y ) + jEP j=2 � 1, and theclaim follows. 2References[1] M. Bern and D. Eppstein. Mesh generation andoptimal triangulation. In D.-Z. Du and F. Hwang,

editors, Computing in Euclidean Geometry, pages23{90. World Scienti�c, 1992.[2] H. Everett, W. Lenhart, M. Overmars, T. Sher-mer, and J. Urrutia. Strictly convex quadrilat-eralizations of polygons. In Proceedings FourthCanadian Conference on Computational Geome-try, pages 77{83, 1992.[3] M. R. Garey and D. S. Johnson. Computers andIntractability: A Guide to the Theory of NP{Completeness. W H Freeman & Co Ltd, 1979.[4] K. Ho-Le. Finite element mesh generation meth-ods: a review and classi�cation. Computer{AidedDesign, 20:27{38, 1988.[5] B. Joe. Quadrilateral mesh generation in polygo-nal regions. Computer{Aided Design, 27:209{222,1995.[6] A. Lubiw. Decomposing polygonal regions intoconvex quadrilaterals. In Proceedings of the ACMSymposium on Computational Geometry, pages97{106, 1985.[7] R. H. M�ohring, M. M�uller-Hannemann, andK. Weihe. Using network ows for surfacemodelling. In Proceedings of the 6th AnnualACM-SIAM Symposium on Discrete Algorithms,SODA'95, pages 350{359, 1995.[8] T. Nishizeki and N. Chiba. Planar Graphs: The-ory and Algorithms, volume 32 of Annals of Dis-crete Mathematics. North{Holland, 1988.[9] J. R. Sack. An O(n logn) algorithm for decom-posing simple rectilinear polygons into convexquadrilaterals. In Proceedings 20th Conference onCommunications, Control, and Computing, pages64{74, 1982.[10] J. R. Sack and G. Toussaint. A linear{time al-gorithm for decomposing rectilinear star{shapedpolygons into convex quadrilaterals. In Proceed-ings 19th Conference on Communications, Con-trol, and Computing, pages 21{30, 1981.[11] T. Tam and C. Armstrong. Finite element meshcontrol by integer programming. Int. J. Numer.Methods Eng., 1993.[12] G. Toussaint. Quadrangulations of planar sets. InProceedings 4th International Workshop on Algo-rithms and Datastructures, WADS'95, Kingston,Canada, August 16-18, 1995, pages 218{227,1995.11

AppendixA Proof of Lemma 3.4Remark A.1 For convenience, the proof is for-mulated only for the case of polygons in the plane;the extension to polygons in other smooth sur-faces is straightforward. The proof disregards fur-ther interesting aspects such as internal angles ofembedded quadrilaterals. However, the decompo-sitions constructed in Sects. 2 and 3 are of a veryrestricted shape, and further criteria may be ful-�lled to some extent, too.Proof (sketch): The idea of the proof is to useinduction on the number of internal faces jF j of G.For technical reasons, however, we use a slightlydi�erent induction hypothesis. Let us relax thedegree constraints of Item 3 in the following way:3' Every internal vertex of G has degree 3.4' If F shares two or more vertices with anotherface F 0, these vertices form a common inter-val of both F and F 0.Hence we allow non-corners of P to have degree2.Claim A.2 Let P and G and the Items 1,2, and5 as in Lemma 3.4, but replace Item 3 by 3' andItem 4 by 4'.Then there is a planar embedding of G such thatthe outer face is embedded like P and all internalfaces are strictly convex except those which areincident to non-corners of degree 2. Nonetheless,the latter faces are still convex.Clearly, it su�ces to prove this claim, as Item 3implies that all faces of G can be embeddedstrictly convex.The case jF j = 1 is trivial. Hence, assume jF j >1. The conditions in Lemma 3.4 imply that thereis a path P which connects two vertices v and won the boundary of P such that all edges of p areinternal edges. In particular, we assume that the

vertices v; w and the path p are chosen such thatp is shortest subject to this condition.As v and w belong to di�erent segments of Pand P is convex, we can embed p on a straightline between v and w (such that the vertices ofp are equally spaced, say). Regard p as orientedfrom v to w.Now consider the polygons P` = p [ IP (w; v),Pr = p [ IP (v; w) (IP (�; �) is the counterclockwiseinterval with respect to P ), and the subgraphs G`and Gr which are the restrictions of P and G tothe left hand and right hand side of p, respectively.We want to apply the induction hypothesis tothe convex polygon P` and G`, and to Pr andGr. However, to this end we have to modify p toensure all preconditions. Clearly, G` and Gr ful�llItems 1 and 2 of the preconditions, and each hasfewer internal faces than G. It is also easy to seethat the property stated in Item 5 carries over tothem.At this point it becomes clear why we used therelaxed version of Item 3: We cannot be sure thateach internal vertex of p has degree 3 or more inG` and Gr.Suppose Item 4' is not valid anymore for P` orPr, say for P`. Then there is a face f with twonon-adjacent vertices v1; v2 on p. Denote by v3; v4the two other vertices of this face, and let v3 be theone which is enclosed by the cycle formed by theedges (v4; v1); (v4; v2) and the interval between v1and v2 on p in the embedding of G. The distanceof v1 and v2 on p must be equal to 2, as otherwisep would not be a shortest path between v andw. Let v5 be the vertex between v1 and v2 on p.Then remove from p the edges (v1; v5); (v5; v2) andreplace them by (v1; v3); (v3; v2) to yield a newpath p0 between v and w. Assume that we havechosen p0 instead of p in the beginning, and iteratethis modi�cation step if necessary. Observe thatItem 5 now guarantees that the change from p top0 cannot cause a (new) violation of Item 4' in theother subgraph. (In fact, this is the place wherewe need Item 5.)Hence, we �nally �nd P` and Pr to which the in-12

duction hypothesis applies. Thus, both polygonsadmit the desired strictly convex embeddings withthe exception of only those faces incident internalvertices of p which have degree 2 in G` or Gr.But now we use the fact that for any strictlyconvex polygon there is an � > 0 such that we canshift any of its vertices from its current positioninto an arbitrary direction within some disk of ra-dius �, and the \shifted" polygon remains strictlyconvex. Hence, if necessary, we can readjust in-ternal vertices of p (and keep all other verticesstationary) such that all faces of G are simultane-ously strictly convex. 2B Proof of Lemma 3.5In this section, we give a detailed proof ofLemma 3.5. As the proof is quite long and in-volved and includes several extensive case dis-tinctions, the proof is divided into three parts(Sects. B.1{B.3).In Sect. B.1, we consider internal quadrilater-als, that is, quadrilaterals that share no edge withthe boundary of the polygon (though possibly ver-tices). Roughly speaking, we prove that everyeven polygon admits an optimal conformal decom-position such that an internal quadrilateral ap-pears only in a con�guration as shown in Fig. 13(except for the polygon (2; 1; 1), to be precise).As a by{product (Lemma B.3), we characterizethe unique optimal conformal decompositions forpolygons of the type (2k; 1; 1), k 2 N.Then, in Sect. B.2, we show that we may alsorestrict our attention to optimal conformal de-compositions where internal vertices appear onlyin very speci�c situations. Based on this charac-terization, we prove that there is an optimal con-formal decomposition with an area decompositionsuch that all but at most one area components be-long to one of the types depicted in Figs. 8 and 9,the cut components.Finally, in Sect. B.3, we prove that there isan optimal conformal decomposition such that anarea decomposition contains only a few of thesecut components. From this we conclude correct-ness of Lemma 3.5 and are done.Throughout Sect. B, we need some further ter-minology. For two polygons P = (s1; :::; sk) and

Q = (t1; :::; t`) with k > `, we write P � Q ifthere are i0 = 0 < i1 < � � � < i` = k such thattj = ijXi=ij�1+1 sifor j = 1; : : : ; `. In other words, Q may be con-structed from P by \stretching" some strictly con-vex internal angles to �. Obviously, \�" is a par-tial order. We write P � Q if P � Q or P = Q.B.1 Internal QuadrilateralsLemma B.1 is auxiliary (nonetheless insightful),Lemma B.2 shows that internal quadrilaterals ap-pear only in con�gurations as depicted in Fig. 13,and Lemma B.3 characterizes the optimal confor-mal decompositions of polygons (2k; 1; 1).Lemma B.1 No two internal quadrilaterals areadjacent in an optimal conformal decompositionif at least one of them is incident to an internalvertex.Proof: Let Z be be an inclusion{maximal con-nected component of internal quadrilaterals (con-nectivity with respect to the reduced dual graphG�, see Sect. 2). Suppose for a contradiction thatZ is incident to an internal vertex and consists ofmore than one quadrilateral.Let C be the cycle separating Z from therest of the plane, and let e1 = (v0; v1); e2 =(v1; v2); : : : ; ek = (vk�1; v0) be the edges of C(in this counterclockwise cyclic order). Moreover,let V1; : : : ; Vk be the quadrilaterals incident toe1; :::; ek outside Z. (Observe that these \outer"quadrilaterals are not necessarily pairwise di�er-ent.)We show in the following that we can alwaysmodify the given conformal decomposition suchthat L(C) shrinks by at least two unless L(C) = 4.This su�ces for an induction on L(C).Case I: No two quadrilaterals Vi are identical.Then for all vi , i = 0; : : : ; k�1, there is an edgeleaving Z at vi. Thus Z itself may be treated likea polygon that is everywhere strictly convex (andthus admits a perfect conformal decomposition).Case I(a): Vi is adjacent to Vi+1 for all i =0; : : : ; k � 1 (with indices taken modulo k).13

For i = 0; : : : ; k � 1 let e0i be the edge oppositeto ei on Vi. Then e01 � e02 � � � � � e0k � e01 is theoriginal polygon P itself. Hence, there must bean index `, such that the common vertex of e0̀and e0̀+1 is a corner. Now identify the verticesv`�1 and v`+1, and delete vertex v` and all itsincident edges. This lets L(C) shrink by two, butZ remains perfect.Case I(b): At least two consecutive quadrilater-als are not adjacent.Then there is i 2 f0; : : : ; k�1g such that Vi andVi+1 are not adjacent and vi�1 is an internal ver-tex. To see this, distinguish two cases: If no twoconsecutive quadrilaterals are adjacent, vi�1 maybe an arbitrary internal vertex (whose existenceis assumed in Lemma B.1). Otherwise, there isvi�1 such that Vi�1 and Vi are adjacent and Viand Vi+1 are not. Clearly, when Vi�1 and Vi areadjacent, vi�1 is internal.Now we know that there is i 2 f0; : : : ; k � 1gsuch that Vi and Vi+1 are not adjacent and vi�1is an internal vertex. Then the vertices vi�1 andvi+1 can be identi�ed with each other. As a result,L(C) shrinks by two, but Z still allows a perfectconformal decomposition.Case II: There is an index i with Vi = Vi+1.Case II(a): Vi 6= Vi�1 and Vi+1 6= Vi+2.The quadrilateral Vi is incident to the edgesei and ei+1. Denote by e0i and e0i+1 the oppositesides of ei and ei+1 in Vi. Since Z is inclusion{maximal, Vi is not an internal quadrilateral, soe0i or e0i+1 (say e0i+1) belongs to the boundary ofP . The vertex vi is strictly convex relative to theunion of Z and Vi. Thus, we may remove ei andei+1 and replace them by an edge e = (vi�1; vi+2),thereby creating a new quadrilateral V with edgese; ei�1; e0i; e0i+1. The modi�ed C has been shrunkby two edges, and outside C nothing changes.Case II(b): Vi = Vi+1 = Vi+2.In this case, e = (vi�1; vi+2) is a boundary edgeof P , and vi�1 and vi+2 are both strictly convexrelative to the union of Z and Vi. Delete the ver-tices vi and vi+1 with all their incident edges, andinsert a new edge e0 = (vi�2; vi+3). This edge e0is feasible, as neither vi�2 nor vi+3 can belongto the same segment as e. By that, Vi is re-

VV

V VV1

2 34Figure 13: Lemma B.2 means that an internalquadrilateral V occurs only in a con�guration likethis, where the four horizontal edges below areconsecutive edges of the same segment of the poly-gon.placed by the quadrilateral consisting of the fouredges e; ei�1; e0; ei+4. This time, L(C) shrinks by4 units. (Note that L(C) � 8 before this modi�-cation.) 2Lemma B.2 Let P 6= (2; 1; 1) be a convex poly-gon, and let G be an optimal conformal decompo-sition of P with a minimum number of internalquadrilaterals. Assume that G contains an in-ternal quadrilateral V that is incident to at leastone internal vertex, and let V be adjacent to thequadrilaterals V1; V2; V3; V4 in cyclic order. Then(for an appropriate choice of V1) Vi and Vi+1 areadjacent for i = 1; 2; 3, and all four quadrilater-als are incident to consecutive edges of the samesegment of P (see Fig. 13).Proof: By Lemma B.1 we may assume that no Viis an internal quadrilateral. Let ei = (wi; wi+1) bethe edge incident to both V and Vi for i = 1; : : : ; 4(with w5 = w1). Furthermore, denote by e0i theopposite edge of ei in Vi.Case I: degree(wi) = 3 for i = 1; : : : ; 4.As no Vi is an internal quadrilateral, all edgese0i lie on the boundary of P . By assumption,P 6= (2; 1; 1), thus P = (1; 1; 1; 1), and this casecontradicts optimality.Case II: At least three vertices, say w1; w2; w3,have degree � 4.Then w4 can be assumed to be the internal vertexof V , and therefore be identi�ed with w2. Thisgives a feasible conformal decomposition withfewer internal vertices.Case III: Exactly two opposite vertices, say w1and w3, have degree � 4.14

Then the identi�cation of w2 and w4 gives an im-proved feasible conformal decomposition.Case IV: Two adjacent vertices, say w1 and w2,have degree � 4, and w3; w4 have degree = 3.Case IV(a): degree(w2) > 4. (The casedegree(w1) > 4 is mirror-symmetric.)The solution can always be improved, see Fig.14.Drawn is the \worst case", i.e. the case withthe smallest number of strictly convex vertices.Hence, the following solution works for all othercases, too.First identify w2 and w4, then delete w3 andall its incident edges. Delete furthermore the �rsttwo edges of the cyclic adjacency list of w2 be-tween (w2; w3) and (w2; w1). Introduce a newedge (w03; z), and complete the construction in theobvious way.Case IV(b): degree(w1) = degree(w2) = 4.First suppose, that there is a quadrilateral V5neighbored to V1 and to V2, and let z the ver-tex incident to both V1 and V5. If z is strictlyconvex relative to the subpolygon Q1 formed bythe quadrilaterals V; V1; :::; V5 then we can �nd asolution for Q1 which needs only 2 internal ver-tices. If z is not strictly convex, but z is an inter-nal vertex, then there is a third quadrilateral V6adjacent to z. The subpolygon Q2 = Q1 [V6 cur-rently uses 4 internal vertices. However, the twopossibilities with fewest segments Q2 = (5; 2; 1)or Q2 = (4; 3; 1) can be solved with less internalvertices. Now consider the case, that z is on theboundary of P and not strictly convex. If w1 isalso on the boundary, the possibilityQ1 = (4; 2; 2)has an optimal solution without internal quadri-lateral, and otherwise Q1 � (4; 2; 2) needs lessinternal vertices than the current solution. If w1is an internal vertex, we have the same subcasesas with Q2.It remains the case, where V5 does not exist.In other words, w2 lies on the boundary of P . Bysymmetry, we may also assume, that w1 lies on theboundary. But then, we can �nd an alternativesolution for the quadrilaterals V; V1; :::; V4 whichuses the same number of internal vertices, but nointernal quadrilateral.

Case V: degree(w2) = degree(w3) =degree(w4) = 3, degree(w1) � 4.Case V(a): The common vertex of e01 and e02 isstrictly convex relative to Q3 = V [V1[V2[V3[V4.Then Q3 � (3; 1; 1; 1), and so needs at most twointernal vertices.Case V(b): The common vertex of e02 and e03 isstrictly convex relative to Q3.In this case, Q3 � (2; 2; 1; 1), and so needs at mostone internal vertex.Case V(c): All 4 quadrilaterals V1; :::; V4 are in-cident to the same segment of P .This �nal case was stated in the lemma. 2Lemma B.3 The optimal conformal decomposi-tion of the polygon (2k; 1; 1) is unique for allk � 1. The polygon (2; 1; 1) needs four internalvertices, and for k � 2, the polygon (2k; 1; 1) needsk + 1 internal vertices.Proof: The proof in done by induction on k. De-note by a; b; c the strictly convex vertices of thepolygon (2k; 1; 1), and let v1; v2; :::; v2k�1 be theremaining vertices on the base segment betweena and b.In any solution, each edge of the base segmentis incident to one quadrilateral, and all of themhave to be pairwise di�erent. This gives an im-mediate lower bound of 2k quadrilaterals for anyconformal decomposition. For k � 2, the given so-lution needs exactly 2k+1 quadrilaterals. Hence,a solution cannot be optimal if it uses 2 quadri-laterals not incident to the base segment in thiscase.Case k � 2:Easy to see.Case k > 2:Let G be an optimal conformal decomposition.If degree(a) = degree(b) = 2 and degree(c) =3, then there is a vertex v incident to c andv1; v2k�1. This means that the induced subpoly-gon (v; v1; :::; v2k�1; v) has the shape of a (2k �2; 1; 1), and we are done by induction.Suppose degree(a) > 2. This implies the exis-tence of a quadrilateral V1 = (a; va; vc; c) not inci-dent to the base segment. The internal vertex vamust have degree � 3, and therefore be incidentto an edge (va; vd) (with a 6= vd 6= vc). If vd canbe chosen as an internal vertex, then we may also15

w1 w1V w2w3w03z

w03zw4

Figure 14: Case IV(a) in Lemma B.2.k = 1 k = 2 k > 2Figure 15: Illustration of optimal conformal decompositions of (2k; 1; 1)assume that vd; va and vc belong to a further inter-nal quadrilateral V2. If vd belongs to the boundaryof P , then this implies a triangle-shaped subpoly-gon (2k0; 1; 1) with a; va; vd as strictly convex ver-tices. By induction hypothesis (note that vd 6= b),this subpolygon needs at least one internal quadri-lateral. Thus, the case degree(a) > 2 is alwayssuboptimal, and by symmetry this holds also fordegree(b) > 2. But if degree(a) = degree(b) = 2,we have necessarily degree(c) � 3.It remains to show that degree(c) > 3 is sub-optimal. Any solution which includes an edge(c; vi) cannot be optimal, as such an edge eitherinduces two odd polygons (which is infeasible) orinduces two triangle shaped polygons for whichthe induction hypothesis applies and so is subop-timal. If degree(c) > 4, the solution has two inter-nal quadrilaterals which are adjacent. By LemmaB.1, this is not optimal. Thus, the only interest-ing case is degree(c) = 4. Let (c; d) and (c; e) bethe two edges starting at c and going into the in-ternal of P . Clearly both edges are incident to aninternal quadrilateral V . Denote by f the fourthvertex of V . Then f must be an internal vertex

and degree(f) = 3, as otherwise d and e could beidenti�ed. In addition, there are edges (v1; d) and(v2k�1; e), because degree(a) = degree(b) = 2.Furthermore, we cannot have degree(d) � 4 anddegree(e) � 4, as otherwise f and c could beidenti�ed. Suppose therefore that degree(d) = 3.But this implies the existence of edges (v2; f) and(v3; e), as no internal quadrilateral can be adja-cent to V in an optimal conformal decomposition.But now v3; e; v2k�1 induce a triangle-shaped sub-polygon for which the induction hypothesis can beapplied. This observation �nishes the inductionstep. 2B.2 Internal ComponentsFor the set of operations illustrated in Fig. 8 weuse the following terminology.Terminology B.4 Operation (a) in Fig. 8 iscalled a prolongation as it enlarges the length ofsome interval. Operation (b) has (possibly) a neg-ative e�ect on the reduced length (L � 2K) andis therefore said to be a 1{reduction. Those op-erations which create relatively trivial segments16

vv

v

3

2

1Figure 16: Lemma B.6 says that, if two incidentedges (v1; v2); (v2; v3) between internal verticesare not incident to the same internal quadrilat-eral, then they form a constellation like this.are called cuts, operation (c) is a 1{cut, whereasoperations (d) and (e) are 2{cuts. Finally, thecuts described in Fig. 9 are called (2k; 1; 1){cuts.A (2k; 1; 1){cut requires a segment S of lengthL(S) � 2k, which is henceforth called the basesegment of this cut.Observation B.5 Let G be a feasible conformaldecomposition of the polygon P . If v is an internalvertex of G and connected to vertices v1 and v2 onthe same segment of P , then the edges fv; v1g andfv; v2g are not incident to a common face.Fig. 13 shows that v1 and v2 may be on the samesegment if fv; v1g and fv; v2g are not incident toa common face.The following three lemmas, Lemma B.8, B.9,and B.6, restrict the ways internal vertices mayoccur in optimal conformal decompositions.But �rst, we need some more terminology. Fora conformal decomposition G of polygon P let Gidenote the subgraph induced by all internal ver-tices. We call an internal vertex isolated if it is notincident to another internal vertex. Analogously,an edge of Gi is isolated if it is not incident toany other edge of Gi. Lemma B.6 characterizesthe connected components of Gi that are not iso-lated vertices or edges. Afterwards, this result isused to characterize the area components of G.Lemma B.6 For any polygon P 62 f(3; 3; 3; 3);(4; 3; 3); (4; 2; 2)g, there is an optimal conformaldecomposition G such that for any two incidentedges (v1; v2); (v2; v3) in Gi one of the followingholds:� Either at least one of (v1; v2) and (v2; v3) isincident to an internal quadrilateral,

� or the con�guration in Fig. 16 occurs: Allfour incident quadrilaterals are pairwise dif-ferent, and the four edges opposite to (v1; v2)and (v2; v3) belong to the same segment of P .Proof: Let G be an optimal conformal decom-position of P . For internal components which areisolated vertices or single edges nothing is to show.We treat all larger internal components by an ex-tensive case distinction. Throughout this proof wemay assume that internal quadrilaterals of G (ifany) are of the form described in the statement ofLemma B.2. Thus let v1; v2; v3 be internal verticesof G which are connected by the edges (v1; v2)and (v2; v3). If one of the quadrilaterals incidentto (v1; v2) or to (v2; v3) is an internal quadrilat-eral, the lemma holds. Hence, in the following weassume the converse.Case I: degree(v2) = 3.This means that two quadrilaterals, say V1 andV3, coincide. But then V1 is an internal quadri-lateral, a contradiction.Case II: degree(v1) = degree(v3) = 3,degree(v2) � 4.Consider the bounding cycle C of all quadrilat-erals which are incident to v1; v2; v3.Case II(a): degree(v2) = 4.C = (a� b� c� w0 � f � e� d� w � a), withvertices named as in Fig. 17. We may assume,that the vertices a; b; :::; f all lie on the boundaryof P , as otherwise there is an internal quadrilat-eral as stated in the lemma. Because of Obser-vation B.5, the edges (a; b) and (d; e), as well as(b; c) and (e; f) lie on di�erent segments of P . IfC � (4; 3; 1), there is a better solution for C withonly two internal vertices. Hence, none of the ver-tices a; d; c; f is strictly convex relative to C. Butthen P � (2; 2; 2; 2) which has also an improvedsolution, or we have the exception P = (4; 2; 2).Case II(b): degree(v2) = 5.Similar to case 2a), take the cycle C = (a �b1 � w00 � b2 � c � w0 � f � e � d � w � a). Seeagain Fig. 17. Either we have the exception P =(4; 3; 3), or some additional vertex is strictly con-vex relative to C. But then the four possibilities17

w

a b c

w’

d e f

v vv

vv

23

13

1 v2

w

a

w’’

b c

w’

d e f

vv

2

13v

b1

2Figure 17: Characterization of internal components | Cases II(a) - II(c) in Lemma B.6w w’ w w’

Figure 18: Characterization of internal components | Case II(d) in Lemma B.6 before and aftermodi�cationC � (3; 3; 2; 2), C � (3; 3; 3; 1), C � (4; 3; 2; 1),or C � (4; 3; 1; 2) all allow solutions with at mosttwo internal vertices.Case II(c): degree(v2) = 6, and exactly twoedges in the cyclic adjacency list of v2 are betweenthe edges (v1; v2) and (v3; v2).We may have the exception P = (3; 3; 3; 3), orC � (3; 3; 3; 2; 1), and needs at most 2 internalvertices.Case II(d): degree(v2) = k + ` � 6, with k � 3,and exactly k edges in the cyclic adjacency list ofv2 are between the edges (v1; v2) and (v3; v2).Let V5 (V6) be the third quadrilateral incidentto v1 (v3), and call w (w0) the vertex incident toV5 (V6) but not adjacent to v1 (v3).The third and �fth vertex on the intervalI(w;w0) on C must be strictly convex relative to C(they exist by the condition de�ning Case II(d).Hence we can apply a 1{cut based between thethird and �fth vertex. In addition we apply a 2{cut at w0. By that, C can be reduced to a polygonof the form (1; 1; 1; 2; :::; 2; 1; 1; 1; 2; :::; 2), which isclearly a perfect polygon with a path solution.Moreover, there is a path solution such that noinserted edge has both endvertices on I(w;w0) or

on I(w0; w). This means that all edges are feasible(they do not lie on the same segment) by Obser-vation B.5, applied to v1 and resp. to v3. Wehave to show that we do not create a new compo-nent of internal vertices by one of these new edges.Suppose we do. Then such an edge must be inci-dent to a vertex which has not been adjacent tov1; v2; v3 before. But this would imply an inter-nal quadrilateral in G with two vertices of degreelarger than 3, in contradiction to our assumptionthat G ful�lls Lemma B.2.Finally, the component of v1; v2; v3 has beensplit by the new decomposition of C. This�nishes case 2).In all remaining cases we may assume thatdegree(v3) � 4. Consider Fig. 19.The edges (a1; a2) and (b1; b2) lie on the bound-ary of P . To see this, observe that if (a1; a2)does not lie on the boundary of P then thequadrilateral V formed by (a1; v3; v2; a2) is aninternal quadrilateral with degree(v2) � 4 anddegree(a1) � 4 which contradicts the situation ofLemma B.2. By symmetry, the same holds for(b1; b2).If (a1; a2) and (b1; b2) do not lie on the samesegment of P we may replace the edge (v2; v3) by18

a3 a2 a1b3 b2 b1v3v2v1

Figure 19: Characterization of internal components | Case III in Lemma B.6v

v

v

vw

a b b b ba

w1

3

2 1

1

a2 1 21 2

22

3

Figure 20: Characterization of internal components | Case IV in Lemma B.6: before and after themodi�cation.(b1; a2). By that modi�cation, the internal ver-tices v2 and v3 are no longer connected, and nonew component is produced, as both b1 and a2lie on the boundary of P . Hence from now on wewill assume that both edges (a1; a2) and (b1; b2)belong to the same segment of P .Case III: degree(v2) = 4 and degree(v3) � 4.If furthermore degree(v1) � 4, then with thesame argument as before the edges (a2; a3) and(b2; b3) also may be assumed to belong to the samesegment of P . But that means that V1; V2; V3; V4all are incident to the same segment of P , asclaimed. Otherwise, degree(v1) = 3. If eitherb3 or a3 are strictly convex relative to C, we mayimprove our current solution by deleting v1 fromG with all its incident edges and by insertion ofthe edge (a2; b3) or (b2; a3). If neither b3 or a3are strictly convex, then both b2 and a2 must bestrictly convex, as each polygon has at least three

corners. Hence, we may delete both v1 and v2,and insert edges (a3; v3) and (b3; v3), yielding animproved conformal decomposition.Case IV: degree(v2) > 4 and degree(v3) � 4.Suppose that k > 1 vertices in the cyclic adja-cency list of v2 are between v1 and v3. Call themw1; w2; :::; wk(= a2). As neither V1 nor V2 are in-ternal quadrilaterals, w1 is a boundary vertex ofP .Suppose that the vertices w1 and b1 do not lieon the same segment of P . Then the followingmodi�cation is feasible, see Fig. 20. \Shift" the(2k; 1; 1) formed by the interval I(a1; b1) and theedges (b1; v3) and (v3; a1) by one edge in directionto a2. Replace the edges (wi; v2) by (wi; v3) fori = 2; :::; k, and insert edges (w1; v3) and (w1; b1).By that, we have again a feasible conformal de-composition, and v2 and v3 are no longer in thesame component of internal vertices. Note that19

by the insertion of new edges, no vertices becomemember of the same component which have notbeen before.It remains to show that we may assume thatw1 and b1 do not belong to a common segment.First, consider the subcase that the vertex w1 lieson the same segment of P as the edge (b1; b2), andthereby induces a (2k; 1; 1) in the current solution,namely formed by the edges v2; b2 and v2; w1 andthe interval I(w1; b2) of P . But the current solu-tion looks di�erent from the unique optimal solu-tion of such polygons we have shown in LemmaB.3. By the same argument it would also be sub-optimal if w1 lies on the same segment of P as theedge (b1; b2), and thereby induces a (2k; 1; 1) inthe current solution, in this case formed by v2; a2,v2; w1 and the interval I(a2; w1) of P . 2Corollary B.7 There is a conformal decomposi-tion G of the polygon P with minimum q(G) suchthat a connected component of Gi which containsmore than two vertices induces a subgraph in Gwhich is isomorphic to the optimal decompositionof some (2k; 1; 1) in G, and one of the trivial seg-ments of the (2k; 1; 1) is an edge of P .Proof: By Lemma B.6, there is an optimal de-composition such that for each connected com-ponent of Gi with more than two vertices anytwo incident edges form a situation as depictedin Fig. 13 or 16. Denote by G0 the area compo-nent of G which is induced by all quadrilateralsincident to a vertex of some connected componentof Gi with more than two vertices. (If no such ex-ists, nothing has to be proved.) By Lemma B.6,this G0 contains a (2k; 1; 1). Let PT be the largest(2k; 1; 1) (by inclusion) contained in G0.The lemma claims that we may assume thatone of the trivial segments of PT lies on P . Tosee this, suppose that both trivial segments areinternal edges. Then there is a quadrilateral Qincident to one of these two trivial segments butnot contained in PT . Now we have two subcases.Either the union of PT = (2k; 1; 1) and Q formsa (2k + 1; 1; 1; 1) or it forms a (2k; 1; 1; 1; 1) rel-ative to P . It is easy to see that the latter casecontradicts the optimality of the current decom-position. In the �rst case, however, we can modifyour current decomposition by exchanging Q and

PT within the (2k + 1; 1; 1; 1). By that, eitherone trivial segment of PT immediately becomesa boundary edge of P or we can continue in thesame way. If we choose always the same trivialsegment in these modi�cations steps, then thisprocedure clearly stops after a �nite number ofsteps. 2Lemma B.8 There is an optimal decompositionsuch that degree(v) = 3 for all isolated internalvertices, unless P = (2; 2; : : : ; 2) (i.e., an arbi-trary number of segments, each of length 2). Thisdecomposition has also the properties of Corol-lary B.7.Proof: Suppose we have a situation where anisolated internal vertex v has degree(v) � 4.Let C be the surrounding cycle of the set of allquadrilaterals incident to v. If C 6= (2; 2; : : : ; 2)we have C � (2; 2; : : : ; 2), because of Obser-vation B.5. Thus, C � (1; 1; 2; 2; : : : ; 2) con-tains two trivial segments. Let v1; v2; : : : ; vk bethe vertices adjacent to v in G in cyclic order.Assume that v1 is the vertex which is incidentto the two trivial segments of C. Let v0 bethe vertex on C between vk and v1. Replacethe edges (v; v1); (v; v2); : : : ; (v; vk�3) by edges(v0; v2); (v0; v3); : : : ; (v0; vk�3). This modi�cationstill yields a conformal decomposition, but nowwe have degree(v) = 3. Observe that v0 cannotbe an internal vertex ofG, as otherwiseG containsthe internal quadrilateral (v; vk; v0; v1), in contra-diction to Lemma B.2, Case II. This means thatour modi�cation cannot create a new violation ofthe claim with v0 instead of v, and does not changethe properties of other internal vertices. 2Note that the exception P = (2; 2; : : : ; 2) doesnot contradict Lemma 3.5, as its (unique) optimaldecomposition can be interpreted as a single 1{cutplus a perfect core polygon.Lemma B.9 There is an optimal decompositionsuch that for each component of Gi which is a sin-gle edge, say e = (v1; v2), we have degree(v1) =degree(v2) = 3, unless P = (3; 3; 2; 2; : : : ; 2).This decomposition has also the properties ofCorollary B.7 and Lemma B.9.20

Proof: Take an arbitrary optimal decompositionof P and suppose e = (v1; v2) is a connected com-ponent of Gi. Let C be the surrounding cycleformed by all quadrilaterals incident to v1 and v2.We may assume that degree(v1) � degree(v2).Case I: degree(v1) > 3 and degree(v2) > 3.Delete the edge e and replace it by some edge(w;w0) which makes all faces to quadrilaterals. ByObservation B.5, this modi�cation is feasible, andv1 and v2 are now isolated.Case II: degree(v1) > 3 and degree(v2) = 3, andsome vertex w incident to v2 is strictly convexrelative to C.Then the current solution cannot be optimal, aswe may replace v2 and all its incident edges by asingle edge.Case III: degree(v1) > 3 and degree(v2) = 3,and some vertex w incident to v1 is strictly con-vex relative to C.In this case, replace the edge (w; v1) by a new one,say (w0; w00), such that G still consists of quadri-laterals. By that modi�cation, the degree of v1decreases by one, and we can continue by induc-tion. Observe that neither w0 nor w00 can be aninternal vertex of G, as otherwise G contains aninternal quadrilateral, in contradiction to LemmaB.2, Case II.Case IV: degree(v1) > 3 and degree(v2) = 3,and no vertex incident to v1 or v2 is strictly convexrelative to C.Then P = (3; 3; 2; 2; : : : ; 2).Case V: degree(v1) = 3 and degree(v2) = 3.Then nothing is to show. 2Note that the exception P = (3; 3; 2; 2; : : : ; 2)does not contradict Lemma 3.5 either; its (unique)optimal decomposition can be regarded as a 2{cutplus a perfect core polygon.B.3 Combinations of Internal Compo-nentsNow we are in a position to restrict our attentionessentially to the area components described inLemma 3.5.

Lemma B.10 For any polygon P 62 f(3; 3; 3; 3);(2; 1; 1); (4; 3; 3); (4; 2; 2)g, there is a conformaldecomposition G with minimum q(G) which hasan area decomposition such that there is at mostone outerplanar area component and all other areacomponents are isomorphic to one of the cut com-ponents in Figs. 8 or 9. Furthermore, no two cutcomponents share an edge in that area decompo-sition.Proof: By Lemma B.6, there is an optimal de-composition that each connected component of Gifalls into one of the following categories:1. it is an isolated vertex,2. it is an isolated edge, or3. any two incident edges form a situation asdepicted in Fig. 13 or 16.By Lemmas B.8 and B.9, the vertices which fallinto the �rst two categories can be assumed tohave degree three. By Corollary B.7, all largerinterior components induce a (2k; 1; 1) with onetrivial segment belonging to P .Given such a decomposition, we have to �nd anarea decomposition as claimed in the lemma.If there is no interior component, the decom-position is perfect and we have just one single,outerplanar area component, namely G itself.So let us assume that G has interior compo-nents. We give an algorithmic proof which buildsthe area decomposition by induction on the num-ber of interior components. Start with an area de-composition which has only one area componentG0 := G.If there is an interior component of more thantwo vertices, all quadrilaterals incident to thiscomponent together form a cut component of thetype in Fig. 9, as we have seen above. Hence, wecan take this area component away from G0 andcontinue with one interior component less.Next consider the case of an isolated internalvertex v. As v has degree three, it is incident toexactly three quadrilaterals which form a cycle Cof length six. Note that C contains an intervalof length four which belongs to P , as otherwisethe cycle C can be re�ned without internal vertexwhich contradicts optimality. If two of the three21

vl0 vltvr0 vrt vltvr0 vrt vlt+lvl2Figure 21: (p; q){shift - here: a (2; 0){shiftquadrilaterals of C are incident to two edges lyingon P they form a cut component of type (c) inFig. 8. If only one quadrilateral is incident to twoedges of P , we take this as a cut component oftype (b). At least one quadrilateral must havethe latter property, and so we can also reduce G0in this case.Finally, consider an isolated internal edge e =(v1; v2). As both v1; v2 have degree three, they areincident to exactly four quadrilaterals which alsoform a cycle C of length six. Exactly three con-secutive edges of C belong to the same segmentof P as otherwise the current decomposition ofC cannot be optimal. If two more edges of C be-long to P , we either form a cut component of type(d) or (e) in Fig. 8. Otherwise, the quadrilateralincident to the second edge of the three which be-long to the same segment of P is taken as an areacomponent of type (a) in Fig. 8.Note that the area component G0 always re-mains connected if we take away one of the cutcomponents in the above procedure. Note furtherthat if we de�ne the area components as we did,no two cut components can share an edge. 2With these lemmata at hand it is not too di�-cult to prove the remaining statements of Lemma3.5. The proof technique is as follows: We supposefor contradiction that there is no optimal confor-mal decomposition such that G has the form ac-cording to Lemma 3.5. Since we know so far thatthere is an optimal conformal decomposition withan area decomposition into cut components plusat most one single core component, this meansthat we have applied too many cut operations,or equivalently, G has too many internal compo-nents. Hence, we have to show that there is an-other optimal conformal decomposition G0 whichuses at least one internal component less, andthen we are done by induction. The other pos-

sibility is to derive a contradiction to optimality.This is achieved in a case analysis for the di�erentkinds of operations. Roughly speaking, the ideais to \unite" two internal components to a singlelarger one (with not more internal vertices in to-tal) and to rebuild the conformal decompositionafterwards.The latter involves the \shifting" of a wholebunch of quadrilaterals ofG along some dual path.Let us �rst de�ne what we mean by a (p; q){shiftapplied to some shortest dual path in a conformaldecomposition, see Fig. 21 for an intuitive under-standing.Let G� be the dual internal graph of some feasi-ble conformal decomposition, and V �1 � V �2 � :::�V �s�1 � V �s some shortest path in G� between V �1and V �s . Corresponding to a dual path, there aretwo primal paths pl; pr, incident to these quadri-laterals on the dual path's \left-hand" or \right-hand" side. A vertex of pl or pr is covered if it isan endvertex of an edge going from pl to pr, anduncovered otherwise.The shift consists of three steps:1. Contract all those quadrilaterals which areincident to an uncovered vertex. That is,identify each uncovered vertex with its twoneighbors on pl or pr. As a result, we getpaths ~pl and ~pr of the same length t.2. Let vl0; :::; vlt (vr0; :::; vrt ) be the vertices of ~pl(~pr). Assume further, that there is a path ofp (primal) edges starting at vlt, whose verticesvlt+1; :::; vlt+p we want to cover. Similarly,there is a path of r edges, starting at vr0, withvertices vr�1; :::; vr�q . In this step, the edgesei = (vli; vri ) are replaced (i.e. \shifted") byedges ~ei = (vli+p; vri�q).3. Reverse the contraction of the �rst step.22

V �s�1V �q V �s V �p V �s�1 V �s�1V �s V �sV �p V �pV �q V �q

Figure 22: How to avoid (2k; 1; 1)-cuts in non path-solutionsThe whole operation is called a (p; q){shift.Lemma B.11 If a (2k; 1; 1){cut appears in someoptimal conformal decomposition, it is a leaf in apath-solution.Proof: Suppose we have an optimal conformaldecomposition G of some polygon P which in-volves some (2k; 1; 1){cut.Let G�t be the graph obtained from the dual in-ternal graph G� where all the vertices belongingto faces of the (2k; 1; 1){cut are identi�ed to Vt.We will show that there is no path V �t � V �2 �::: � V �s�1 � V �s in G�t , such that there are twoother neighbors V �p ; V �q of V �s which form an inde-pendent set with V �s�1. Having proved that, thelemma is immediate, since a K1;3{like solution (orthe (2k; 1; 1){cut being no leaf) obviously impliessuch a vertex V �s in its internal dual graph.Hence, assume �rst that there is such a pathstarting at V �t , and take a shortest one. We wantto show that there is a conformal decompositionwhich uses one internal vertex less than the cur-rent solution, which is a contradiction. We haveto distinguish two cases: k = 2 and k � 3. (Ob-serve that a cut with k = 1 will never appear inan optimal solution, unless P = (2; 1; 1).)

Case k � 3:The idea is to replace the (2k; 1; 1){cut be a(2k � 2;1; 1){cut. By Lemma B.3, this saves exactly oneinternal vertex. The cut edge, however, changesits position and the last two vertices of the basesegment of the (2k; 1; 1){cut are not covered any-more. Take the base segment as the \left-hand-side" of the dual path. To cover them, we applya (2; 0){shift to the path V �s�1� :::�V �2 (the pathdirected in this way), if this path is not empty.This then would give us two uncovered verticesincident to the former V �s�1.Case k = 2:If k = 2, the (4; 1; 1){cut is replaced by a 2{cut.From the two trivial segments produced by the 2{cut, we select the one incident to the base segmentof the (4; 1; 1){cut as the cut edge. Again, we takethe base segment as the \left-hand-side" of thedual path, but this time we apply a (1; 1){shift tothe path V �s�1 � :::� V �2 .In any case, we are now able to use the existenceof V �s and its independent neighbors V �p ; V �q tocover all vertices without introducing new internalvertices. The di�erent cases (with the exceptionof symmetric ones) are illustrated in Fig. 22. 223

Lemma B.12 (Restriction of cuts, reductions,and prolongations)For P 62 f(2; 1; 1); (3; 3; 3; 3); (4; 3; 3); (4; 2; 2)g,there is an optimal conformal decomposition ofone of the following types:a) the core polygon P 0 only admits a path solu-tion, but not a K1;3{like solution.a1) If a leaf is cut, all applied operationsyield cut edges, i.e. no 1-reduction orprolongation occurs. Furthermore, atmost one (2k; 1; 1){cut is used.a2) If no leaf is cut, we may use a single 1{reduction, or a single prolongation, butnot more.b) the core polygon P 0 admits a K1;3{like solu-tion. Cuts are only used to create the corre-sponding trivial segments, if necessary, andno other operation is applied elsewhere.Proof: By Lemma B.10, there is an optimal con-formal decomposition G of P which has an areadecomposition into cut components and one per-fect core component.Suppose we have such an optimal conformal de-composition G of P , as derived in the algorithmicproof of Lemma B.10 but G has not the form ofone of the above types as claimed in the lemma.Let P 0 be the surrounding cycle of the core compo-nent in the given area decomposition, henceforthcalled the core polygon.Case a): The core polygon P 0 only admits apath-solution (but no K1;3{like solution).Let s1 and s2 be the trivial segments which arethe leaves in the path-solution for P 0. If s1 (s2)is not an edge of P , let G1 (G2) be the cut com-ponent in the area decomposition incident to s1(s2). Denote by P 00 the polygon which remains ifwe delete all quadrilaterals of G1 and G2 from G.Denote by R1 and R2 the intervals of P 00 betweens1 and s2. Assume further L(R1) � L(R2). If thisrelation is ful�lled with equality, no further oper-ation is necessary at all. Clearly, a prolongationhas been applied to R2, and a 1{reduction only toR1.Suppose that a 1{cut or a 2{cut has been ap-plied which does not yield a leaf for P 0. We

claim that this contradicts our assumption thatP 0 does not admit a K1;3{like solution). To seethis, start with the area decomposition as de�nedin Lemma B.10. As all internal vertices incidentto 1{cuts and 2{cuts have degree three, there isexactly one quadrilateral incident to each suchcut which belongs to the conformal decomposi-tion GP 0 of the core polygon P 0. If we removeall these quadrilaterals from GP 0 , the remaininggraph is still outerplanar, and so admits eithera path-solution or a K1;3{like solution. In anycase, if we add to such a solution the quadrilater-als which we just removed, this decomposition ofP 0 cannot be a path.Case a1): At least one leaf is a cut edge, andsome other operation has been applied which doesnot yield the second leaf of the path.Since a 1{cut or a 2{cut would allow a K1;3{like solution, the type of this further operationcan only be a 1{reduction or a prolongation.The e�ect of a prolongation is to enlarge L(R2)by two units. As we get the same e�ect if wereduce (L � 2K)(R1) by two units, the idea isto use a \larger" cut instead of a prolongation.Hence, if we currently use a 1{cut, it will be re-placed by a 2{cut, a 2{cut will be replaced by a(4; 1; 1){cut, and a (2k; 1; 1){cut can be replacedby a (2k+2; 1; 1){cut. Note that by the de�nitionof a (2k; 1; 1){cut, the base segment has to be asegment of the original polygon. Thus it mightbe impossible to apply a (2k; 1; 1){cut if the cor-responding segment is too short. But in thesesituations we can always use the strictly convexvertex which makes the base segment too short,and apply there a cheaper cut. We so still yieldthe desired e�ect on (L�2K)(R1). In fact, a caseanalysis shows that the existence of a prolonga-tion always leads to a contradiction to optimality.A 1{reduction has the e�ect to shorten(L � 2K)(R1) by two units (or is useless). Butthe same idea as for prolongations works in thiscase: we replace the current cut in the samemanner as above, and delete the 1{reductionwhich is closest on R1 to the leaf. By that, weonly change the place on R1 where to shortenits length, but not (L � 2K)(R1) itself. Acomplete decomposition can be rebuilt using anappropriate shift.24

So far we have ruled out the use of 1{reductionsor prolongations. It remains the case, that at bothleaves a (2k; 1; 1){cut has been applied.Hence, suppose that an optimal conformal de-composition consists of some (2k1; 1; 1){cut andsome (2k2; 1; 1){cut, with k1 � k2, and that thereis a unique dual path V �1 � : : : � V �s of s quadri-laterals between the two cut edges. It is easy tosee that we may assume s � 1 (otherwise the de-composition cannot be optimal).Let v1 and v2 be the two corners at whichthe cuts have been applied, and I1 = I[v1; v2],I2 = I[v2; v1] the corresponding intervals in P . Ifone base segment lies on I1 and the other on I2,the current solution cannot be optimal, as we mayreplace both (2ki; 1; 1){cuts by 2{cuts at the samecorners v1 and v2 and rebuild a feasible decompo-sition by applying a (2; 2){shift to V �1 � : : :� V �s .Hence, we may assume that both base segmentsbelong to I1.If the base segment, say S1, of the (2k1; 1; 1){cut has length L(S1) � 2k1+3, we may replace the(2k1; 1; 1){cut by a (2k1 +2; 1; 1){cut. If the basesegment S1 has only length L(S1) � 2k2 + 2, westill use a (2k1; 1; 1){cut, but at the other cornerof S1.In any case, we replace the other (2k2; 1; 1){cutby a 2{cut, if k2 = 2, or by a (2k2 � 2; 1; 1){cut, otherwise, and can rebuild a conformal de-composition by an appropriate shift. Clearly, wecan continue in the same way, until the second(2k2; 1; 1){cut vanishes.This �nishes the case of two (2k; 1; 1){cuts asleaves of a path-solution.Case a2): If no leaf is a cut edge, but a prolonga-tion and one other operation (i.e., another prolon-gation or a 1{reduction) has been applied, theseuse together at least 3 internal vertices. Hence,we may also use a (4; 1; 1){cut at the former leafto yield the same e�ect.It remains the case that no leaf is a cut edge,but more than one 1{reduction has been used.The idea is then to replace two 1{reductions by a2{cut which is based at the strictly convex vertexof one of the 1{reductions. Let us choose two 1{reductions where the interval between them on R1does not contain another 1{reduction. Apply the

2{cut instead of one of them and take as a cut edgethe trivial segment closer on R1 to the other 1{reduction. Observe that we may assume becauseof Lemma B.8 that the internal vertices belongingto the 1{reductions all have degree 3, and that oneof the three edges of an internal vertex is incidentto R2. In particular, this observation applies tothe chosen second 1{reduction. The correspond-ing vertex incident to R2, say v3, must be strictlyconvex relative to the subpolygon formed by allquadrilaterals of the path solution between thetwo 1{reductions. The two other vertices lying onR1, say v1; v2, have to be covered when we deletethe second 1{reduction from the current solution.Both can be covered by the �rst neighbor of v3in direction to the new 2{cut. The rest of thesubpolygon is then perfect. Note that we can dothis modi�cation in such a way that for no inter-nal vertex which belongs to some other operationthe degree changes. Hence, we can continue byinduction.Case b): The current core polygon P 0 has aK1;3{like solution, and some operation is applied, whichdoes not create one of the leaves.We want to show that such a situation contra-dicts optimality. Denote by s1; s2; s3 the trivialsegments which form the leaves of the current so-lution, and R1 the interval between s1 and s2, R2the interval between s2 and s3, R3 the intervalbetween s3 and s1, the intervals Ri taken with re-spect to the core polygon P 0. Assume further,that the quadrilateral VC corresponding to thedual vertex of degree 3 in the core polygon P 0has an edge (v1; v2) incident to R1.The applied operation cannot be a (2k; 1; 1){cut, as we have seen in Lemma B.11. If thereis a 2{cut with an e�ective reduction of 4 units,we may instead apply a 1{reduction at the samestrictly convex vertex which has only an e�ectivereduction of 2 units. All other types of reductionsand cuts have an e�ective reduction of 2 units.Hence, if we undo one such operation or do thereplacement to the 1{reduction in the other case,we get exactly 2 uncovered vertices.Suppose �rst, that the two uncovered verticeslie on R1. We may assume that the two uncoveredvertices are directly neighbored to v1; v2 by usingan appropriate (k; 0){shift, but delay the last re-25

s1s3VC s1

s3s1s3

s1s3VC

s2 s2s2s2

Figure 23: Sketch: How to avoid cuts and reductions in K1;3-like solutionsversing step of the shift. Using the quadrilateraladjacent to VC on the path leading to s3, the cen-ter VC itself and the two uncovered vertices, wecan �nd a perfect conformal decomposition of theinduced (3; 1; 1; 1; 1; 1), see a sketch in Fig. 23.(Note that this transformation changes the K1;3{like solution to a path solution if the path from VCto s3 contains only a single quadrilateral.) Finally,we rebuild those parts which have been shrunkenaway in the shift.Suppose now, that the two uncovered verticeslie on R2 (or on R3, the symmetric case). Againwe may assume that the two uncovered verticesare directly neighbored to VC by using an appro-priate (k; 0){shift without the last reversing step.In this case we take only VC and the two uncov-ered vertices which together induce a (2; 1; 1; 1; 1).Clearly, the latter polygon is perfect. The laststep is again the rebuilding of shrunken quadri-laterals.It remains the case that a prolongation has beenapplied. But observe that the e�ect of a prolon-gation is just the opposite one of an reduction orcut. Thus by reversing the above transformationsfor cuts and reductions, we easily get improvedsolutions. 2

C Proof of Theorem 4.1Proof: Exact Cover by 3-Sets is the followingproblem. Given a ground set U = fu1; :::; u3mgand a family S1; S2; :::; Sn of subsets of U , each ofcardinality three, is there a subfamily ofm subsetsthat covers U ? This problem is well-known to bestrongly NP{complete [3]. The reduction fromExact Cover by 3-Sets to the minimum conformalmesh re�nement problem is as follows.For each subset Si we create a polygon as in Fig.24. For simplicity we will refer to it as polygonSi. Edge elSi with l 2 f1; 2; 3g is to correspondwith the l-th element of Si. If an edge elSi issubdivided this will mean that subset Si coversthe corresponding element.f1

iS

3Sei

eixS

1Sie 2Se

i

f f2 3

Figure 24: Polygon for a subset Si.Suppose that the edges f1; f2; f3 cannot be sub-divided in any optimal solution. This can be as-26

sumed by the following construction. Add a ho-mogeneous bunch of conforming quadrilaterals tosuch an edge e such that the shortest path fromit to the mesh boundary has a length of at leastk, where k is some constant to be speci�ed later.Each bunch needs at most O(k2) quadrilaterals.By parity, the subdivision of fi by one unit wouldimply at least k additional quadrilaterals in anyfeasible solution. If no edge of Si is subdivided atall, then we clearly need exactly 3 quadrilaterals.If one subdivides each of the edges elSi and xSiexactly once, then we need exactly 7 quadrilater-als. If only one edge elSi is subdivided once, thenxSi must be subdivided by parity reasons, and weneed 5 quadrilaterals. If exactly two edges elSi aresubdivided once, one needs exactly 6 quadrilater-als.Hence, in an exact cover one needs exactly7m + 3(n �m) = 3n + 4m quadrilaterals for allsubsets Si together. If an exact cover exists, itis not worth to cover three times only one ele-ment, as this needs 15 quadrilaterals comparedto 7 + 2 � 3 = 13. Furthermore, it is more ex-pensive to cover once only one element and oncetwo elements (and once no element), as this needs5+6+3 = 14 quadrilaterals. Finally, it is more ex-pensive to cover three times two elements, as thisneeds 18 quadrilaterals compared with 17 quadri-laterals in an exact cover. If no exact cover existsone needs more than 3n + 4m quadrilaterals forthe re�nement of all subsets Si.Let ki denote the total number of appearancesof ui in the subsets Sj. De�ne for each element uia polygon as in Fig. 25.ui

h(1+k )ui ih1u i

e1u ie

y

k u

uxui

i i

i

Figure 25: Polygon for an element ui.The parity of such an \element" poly-gon ui is odd. Suppose that none of itsedges h1ui ; :::; hkiui ; h(1+ki)ui ; xiui ; yiui is subdi-vided. This can be achieved by the same trick

with additional quadrilaterals as above. Then aconformal re�nement of ui requires that an oddnumber of additional points is placed on the edgese1ui ; :::; ekiui . In particular, the optimum re�ne-ment for ui is achieved if we add exactly one ad-ditional point on one of these edges. In this case,ui becomes perfect and needs exactly 1+ki quadri-laterals.Now we establish the relationship between el-ements ui and subsets Sj . If ui is an element ofSj, we connect an edge elSj with a correspondingedge erui by a chain of q conforming quadrilater-als (q is some small constant) as in Fig. 26 whichwe call channels.erui lSe

i

Figure 26: \Channel" between element ui andsubset Si before and after blowing up. Thepointed line denotes the split if this channel isused.Clearly, all these channels can easily be embed-ded into the three-dimensional space such thatthey do not cross each other. The purpose of theseconnections is to make sure that each element uimay be covered by any subset Sj it belongs to.All channels have the same length such that itmakes no di�erence which one is chosen to coveran element.We also want that if erui is subdivided oncethen also elSj is subdivided once in any optimalsolution. Hence, it must be cheaper to split eachquadrilateral on the channel into two quadrilater-als than to use any other feasible re�nement. Tothis end, we \blow up" each channel: we take fourcopies of each channel and glue them together asin Fig. 26. As all these quadrilaterals are conform-ing, a subdivision by one unit of the original edgeerui implies by parity arguments an odd subdivi-sion of either the original edge elSj as intended orof one of the copies of erui or elSj . In the �rstcase, the original path is split into two chains. To27

rule out the latter case, we have to make such asubdivision expensive. This can be done by thesame trick with additional bunches of conform-ing quadrilaterals as above. If the constant k ischosen larger than q, then it is easy to see thatany optimal solution uses exactly one channel perelement ui. From the above discussion it is alsoclear that an exact cover is cheaper than any othercover of U .Observe that no folding edge occurs in our con-struction. Hence, the theorem remains true in thespecial case of homogeneous meshes. 2D Detailed Proof of Theo-rem 4.2Proof: It remains to showmin(PX) � 2 �Min(P ) + jEP j � 2 (1)for X � 1. Note that all segments of PX haveeven lengths. Hence, PX is di�erent from allpolygons listed in Lemma 3.5 except possibly for(4; 2; 2). Ineq. (1) is easy to check by hand forPX = (4; 2; 2), which means P = (2; 1; 1). There-fore, in the rest of this proof we assume that PXis none of the polygons listed in Lemma 3.5.Let i(X) denote the number of internal verticesin an optimal conformal decomposition of PX .The proof of Lemma 3.2 yields that this valueis the same for all optimal conformal decomposi-tions, namelymin(PX) = i(X) + jEPX j2 � 1= i(X) + jEP j � 1 : (2)Hence, it su�ces to showi(X) � 2 �Min(P )� 1 : (3)The proof of Lemma 3.2 also yieldsMin(P ) � jEP j2 � 1 : (4)Therefore, Ineq. (1) is shown ifi(X) � jEP j � 3 : (5)

Since PX is none of the polygons listed inLemma 3.5, there is an optimal decompositionfor PX that can be partitioned into one outerpla-nar area component G0 and cut components Gjwith j = 1; : : : ; `, ` � 3. For j = 0; : : : ; `, letEj denote the edges of PX belonging to Gj. Forj = 1; : : : ; `, let i(j;X) denote the number ofinternal vertices introduced by the cut compo-nent Gi. For a cut component Gj of type (b)or (c) in Fig. 8, this means i(j;X) = 1; for theother types in Fig. 8 this means i(j;X) = 2; andfor a cut component of the type in Fig. 9 with2k + 1 edges this means i(j;X) = k + 1. Clearly,jEP j = X̀j=0 jEj j ; (6)and i(X) = X̀j=1 i(j;X) : (7)Obviously, we have jEj j � i(j;X) � 2 for eachj = 1; : : : ; k. Hence, if the optimal conformaldecomposition of PX contains two or more cutcomponents, Ineq. (5) is immediate. Otherwise,the core component contains a leaf that is a triv-ial segment of PX . This means jE0j � 1, andIneq. (5) is immediate again. 2

28