aa r X i v : . [ c s . C G ] J u l NONOBTUSE TRIANGULATIONS OF PSLGS
CHRISTOPHER J. BISHOP
Abstract.
We show that any planar straight line graph with n vertices has a con-forming triangulation by O ( n . ) nonobtuse triangles (all angles ≤ ◦ ), answeringthe question of whether any polynomial bound exists. A nonobtuse triangulationis Delaunay, so this result also improves a previous O ( n ) bound of Eldesbrunnerand Tan for conforming Delaunay triangulations of PSLGs. In the special case thatthe PSLG is the triangulation of a simple polygon, we will show that only O ( n )triangles are needed, improving an O ( n ) bound of Bern and Eppstein. We alsoshow that for any ǫ >
0, every PSLG has a conforming triangulation with O ( n /ǫ )elements and with all angles bounded above by 90 ◦ + ǫ . This improves a result ofS. Mitchell when ǫ = π = 67 . ◦ and Tan when ǫ = π = 42 ◦ . Date : March 2011; revised Nov 2014; re-revised Jan 2016.1991
Mathematics Subject Classification.
Primary: 68U05 Secondary: 52B55, 68Q25 .
Key words and phrases. nonobtuse triangulation, acute triangulation, conforming triangulation,PSLG, Delaunay triangulation, Gabriel condition, nearest-neighbor learning, quadrilateral mesh,Voronoi diagram, thick-thin decomposition, polynomial complexity, propagation paths, return re-gions, spirals.The author is partially supported by NSF Grant DMS 13-05233. Introduction
A planar straight line graph Γ (or PSLG from now on) is the disjoint union of afinite number (possibly none) of non-intersecting open line segments (called the edgesof Γ) together with a disjoint finite point set (the vertices of Γ) that includes all theendpoints of the line segments, but may include other points as well.If V is a finite point set in the plane, a triangulation of V is a PSLG with vertexset V and a maximal set of edges. If Γ is a PSLG with vertex set V , then a conformingtriangulation for Γ is a triangulation of a point set V ′ that contains V and such thatthe union of the vertices and edges of the triangulation covers Γ. We allow V ′ to bestrictly larger than V ; in this case V ′ \ V are called the Steiner points . We wantto build conforming triangulations for Γ that have small complexity (the number oftriangles used) and good geometry (the shape of each triangle; no angles too large ortoo small), but these two goals are often in conflict. In this paper, we are interestedin finding the best angle bounds on the triangles that allow us to polynomially boundthe number of triangles needed in terms of n , the number of vertices of Γ.If we triangulate a 1 × r rectangle into a fixed number of elements, it is easy to checkthat some angles must tend to zero as r → ∞ , so there is no uniform, strictly positivelower angle bound possible, if we want the number of triangles to be bounded onlyin terms of the number of vertices of the given PSLG. Since the angles of a trianglesum to 180 ◦ , if we had an upper bound of 90 ◦ − ǫ on the angles of a triangulation,then we also have a 2 ǫ lower angle bound. Therefore, no upper bound strictly lessthan 90 ◦ is possible. Thus nonobtuse triangulation (all angles ≤ ◦ ) is the bestwe can hope for.In 1960 Burago and Zalgaller [15] showed that any polyhedral surface has an acutetriangulation (all angles < ◦ ), but without giving a bound on the number oftriangles needed. This was used as a technical lemma in their proof of a polyhedralversion of the Nash embedding theorem. In 1984 Gerver [25] used the Riemannmapping theorem to show that if a polygon’s angles all exceed 36 ◦ , then there existsa dissection of it into triangles with maximum angle 72 ◦ (in a dissection, adjacenttriangles need not meet along an entire edge). In 1988 Baker, Grosse and Rafferty [2]again proved that any polygon has a nonobtuse triangulation, and their constructionalso gives a lower angle bound. As noted above, in this case no complexity bound in CHRISTOPHER J. BISHOP terms of n alone is possible, although there is a sharp bound in terms of integratingthe local feature size over the polygon. For details, see [7], [40] or the survey [18].A linear bound for nonobtuse triangulation of point sets was given by Bern, Epp-stein and Gilbert [7], and Bern and Eppstein [6] gave a quadratic bound for simplepolygons with holes (this is a polygonal region where every boundary component isa simple closed curve or an isolated point). Bern, Dobkin and Eppstein [4] improvedthis to O ( n . ) for convex domains. Bern, S. Mitchell and Ruppert [9] gave a O ( n )algorithm for nonobtuse triangulation of simple polygons with holes in 1994 and theirconstruction uses only right triangles. We shall make use of their result in this paper.These and related results are discussed in the surveys [5] and [10]. Other papers thatdeal with algorithms for finding nonobtuse and acute triangulations include [20], [33],[35], and [37]. Giving a polynomial upper bound for the complexity of nonobtuse tri-angulation of PSLGs has remained open (e.g., see Problem 3 of [5]). We give such abound by proving: Theorem 1.1.
Every PSLG with n vertices has a O ( n . ) conforming nonobtusetriangulation. Maehara [36] showed that any nonobtuse triangulation using N triangles can berefined to an acute triangulation (all angles < ◦ ) with O ( N ) elements. A differentproof was given by Yuan [55]. In our proof of Theorem 1.1 the triangulation willconsist of all right triangles, but the arguments of Maehara or Yuan then show thetheorem also holds with an acute triangulation, at the cost of a larger constant in the O ( n . ). As noted above, simple examples give a quadratic lower bound for PSLGs(see [6]), so a gap remains between our upper bound and the worst known example.However, this gap can be eliminated in some special cases, e.g., Theorem 1.2.
A triangulation of a simple n -gon has a O ( n ) nonobtuse refinement. This improves a O ( n ) bound given by Bern and Eppstein [6]. We can also approachthe quadratic lower bound if we consider “almost nonobtuse” triangulations: Theorem 1.3.
Suppose θ > . Every PSLG with n vertices has a conforming trian-gulation with O ( n /θ ) elements and all angles ≤ ◦ + θ . This improves a result of S. Mitchell [38] with upper angle bound π = 157 . ◦ anda result of Tan [47] with π = 132 ◦ . ONOBTUSE TRIANGULATIONS OF PSLGS 3
A triangulation is called
Delaunay if whenever two triangles share an edge e , thetwo angles opposite e sum to 180 ◦ or less. If all the triangles are nonobtuse, then thisis certainly the case, so Theorem 1.1 immediate implies Corollary 1.4.
Every PSLG with n vertices has a O ( n . ) conforming Delaunaytriangulation. This improves a 1993 result of Edelsbrunner and Tan [19] that any PSLG has aconforming Delaunay triangulation of size O ( n ). Conforming Delaunay triangula-tions for Γ are also called Delaunay refinements of Γ. There are numerous papersdiscussing Delaunay refinements including [18], [21], [39], [40], [41] and [45]. Theargument in this paper does not seem to give a better estimate for Delaunay triangu-lations than for nonobtuse triangulations, nor does the proof appear to simplify in theDelaunay case. Finding an improvement (either for the estimate or the argument) inthe Delaunay case would be extremely interesting.An alternative formulation of the Delaunay condition is that every edge in the tri-angulation is the chord of an open disk that contains no vertices of the triangulation.We say the triangulation is Gabriel if every edge is the diameter of such disk. It iseasy to check that a nonobtuse triangulation must be Gabriel, so we also obtain astronger version of the previous corollary:
Corollary 1.5.
Every PSLG with n vertices has a O ( n . ) conforming Gabriel tri-angulation. Given a finite planar point set V , and a point v ∈ V , the Voronoi cell corre-sponding to v is the open set of points that are strictly closer to v than to any otherpoint of V . The union of the boundaries of the all the Voronoi cells is called the Voronoi diagram of V . In [42], it is shown that given a nonobtuse triangulationwith N elements, one can find a set of O ( N ) points so that the Voronoi diagram ofthe point set covers all the edges of the triangulation. Thus we obtain Corollary 1.6.
For every PSLG Γ with n vertices, there is a point set S of size O ( n . ) whose Voronoi diagram covers Γ . The authors of [42] were interested in a type of machine learning called “nearestneighbor learning”. Given Γ and S as in the corollary, and any point z in the plane, CHRISTOPHER J. BISHOP we can decide which complementary component of Γ z belongs to by finding theelement w of S that is closest to z ; z and w must belong to the same complementarycomponent of Γ. Thus the corollary says that a partition of the plane by a PSLGof size n can be “learned” from a point set of size O ( n . ). This answers a questionfrom [42] asking if a polynomial number of points always suffices.Acute and nonobtuse triangulations arise in a variety of other contexts. In recre-ational mathematics one asks for the smallest triangulation of a given object intoacute or nonobtuse pieces. For example, a square can obviously be meshed withtwo right triangles, but less obvious is the fact that it can be acutely triangulatedwith eight elements but not seven; see [16]. For further results of this type see [23],[24], [26], [27], [28], [29], [30], [31], [43], [56], [57], [58], the 2002 survey [60] and the2010 survey [59]. There is less known in higher dimensions, but recent work hasshown there is an acute triangulation of R , but no acute triangulation of R n , n ≥ n of triangulation elements; for general triangulations his estimate is exponen-tial in n , but for nonobtuse triangulations it is only linear in n . Other examples wherenonobtuse or Delaunay triangulations give simpler or faster methods include: discretemaximum principles [12], [17], [53]; Stieltjes matrices in finite element methods [13],[46]; convenient description of the dual graph [8]; the Hamilton-Jacobi equation [3];the fast marching method [44]; the tent pitcher algorithm for meshing space-time [1],[48], [49].The ideas in this paper are used in a companion paper [11] to obtain conformingquadrilateral meshes for PSLGs that have optimal angle bounds and optimal worstcase complexity. The precise statement from this paper used in [11] is Lemma 13.1;this follows from a slight modification of the proof of Theorem 1.3. The result ob-tained in [11] says that every PSLG has an O ( n ) conforming quadrilateral meshwith all angles ≤ ◦ and all new angles ≥ ◦ . The angle bounds and quadraticcomplexity bound are both sharp. ONOBTUSE TRIANGULATIONS OF PSLGS 5
Many thanks to Joe Mitchell and Estie Arkin for numerous conversations aboutcomputational geometry in general and the results of this paper in particular. Alsothanks to two anonymous referees for many helpful comments and suggestions on twoearlier versions of the paper; their efforts greatly improved the presentation in thisversion.In Section 2 we recall a theorem of Bern, Mitchell and Ruppert [9] that connectsnonobtuse triangulation to finding Gabriel edges, and we sketch the proof of thisresult in Section 3. In Section 4 we use their theorem to give a simple proof ofTheorem 1.2. In Sections 5-8 we discuss propagation paths, dissections and returnregions; these are used in the proofs of both Theorems 1.1 and 1.3. Sections 9–12 givethe proof of Theorem 1.3 and Section 13 summarizes facts from the proof that areused in the sequel paper [11]. Section 14 gives an overview of the proof of Theorem1.1 and Sections 15–21 provide the details.2.
The theorem of Bern, Mitchell and Ruppert
Given a point set V and two points v, w ∈ V , the segment vw is called a Delaunayedge if it is the chord of some open disk that contains no points of V and is called a Gabriel edge if it is the diameter of such a disk (see [22]). We will call a PSLG Γwith vertex set V and edge set E Gabriel if every edge in E is Gabriel for V . Givena PSLG which is not Gabriel, can we always add extra vertices to the edges, makinga new PSLG that is Gabriel? We are particularly interested in the case when P = T is a triangle. See Figure 1. Figure 1.
A triangle that is not Gabriel and one way to add pointsso that it becomes Gabriel.The connection between Gabriel triangles and nonobtuse triangulation is given bythe following result of Bern, Mitchell and Ruppert [9]. Suppose T is a triangle with CHRISTOPHER J. BISHOP three vertices V = { a, b, c } and suppose V is a finite subset of the edges of T . Then T \ ( V ∪ V ) is a finite union of segments and we let V ′ denote the midpoints of thesesegments. Theorem 2.1.
Suppose we add N points V to the edges of a triangle T , so that thetriangle becomes Gabriel. Assume further that no point of the interior of T is in morethat two of the Gabriel disks. Then the interior of T has a nonobtuse triangulationconsisting of O ( N ) right triangles and the triangulation vertices on the boundary of T are exactly the vertices of T and the points in V and V ′ . This follows from the proof in [9], but this precise statement does not appear there,so in the next section we will briefly describe now to deduce Theorem 2.1 from thearguments in [9].We say that one triangulation T is a refinement of another triangulation T , ifeach triangle in T is a union of triangles in T . If Γ is a PSLG that is a triangulationand we add enough points to the edges of Γ to make every triangle Gabriel, then theresulting nonobtuse refinements of each triangle agree along any common edges (theset of boundary vertices is e ∩ ( V ∪ V ′ ) for both triangles with edge e ). Thus we get: Corollary 2.2.
Suppose that Γ is a planar triangulation with n elements, and thatwe can add N vertices to the edges of Γ so that every triangle becomes Gabriel. Then Γ has a refinement consisting of O ( n + N ) right triangles. It is fairly easy to see that we can always add a finite number of vertices andmake each triangle Gabriel. Thus nonobtuse refinement of a triangulation is alwayspossible, but the difficulty is to bound N in terms of n . Since any PSLG with n vertices can be triangulated using O ( n ) triangles (and the same vertex set), Theorem1.1 is reduced to Theorem 2.3.
Given any triangulation with n elements we can add O ( n . ) verticesto the edges so that every triangle becomes Gabriel. We will prove this for general planar triangulations later in the paper. We startwith the simpler case of triangulations of simple polygons (Theorem 1.2) in Section4. The key feature of this special case is that the elements of such a triangulation
ONOBTUSE TRIANGULATIONS OF PSLGS 7 form a tree under adjacency (sharing an edge); this fails for general triangulations,and dealing with this failure is the main goal of this paper.In Theorem 2.3 we only need to check that each triangle becomes Gabriel, not thatthe point set is Gabriel for the whole triangulation. The difference is that in thefirst case if we take a disk D with an edge e as its diameter, we only have to makesure that D does not contain any vertices belonging to the two triangles that have e on their boundary. In the second case, we would have to check that D does notcontain any vertices at all. This apparently stronger condition follows from the weaker“triangles-only” condition, but we don’t need this fact for the proof of Theorem 1.1.3. Sketch of the proof of Theorem 2.1
As noted above, Theorem 2.1 is due to Bern, Mitchell and Ruppert in [9], but theprecise result is not given in that paper. For the convenience of the reader, we brieflysketch how our statement is deduced using the argument in [9].The basic idea in [9] is to pack the interior of a polygon P (which in our case is justa triangle T ) with disks until the remaining region is a union of pieces that are eachbounded by three or four circular arcs, or a segment lying on the polygon’s boundary.These are called the remainder regions.Each remainder region R is associated to a simple polygon R + , called the aug-mented region of R as follows. Each circular arc in the boundary of R lies on somecircle,and we add to R the sector of the circle subtended by this arc. Doing this foreach boundary arc of R gives the polygon R + . See Figure 2. Figure 2.
The light gray is a 3-sided remainder region R and thedark gray is the corresponding augmented region R + . CHRISTOPHER J. BISHOP
The augmented regions decompose the original polygon into simple polygons andthe authors of [9] show how each augmented region can be meshed with right triangles.Moreover, the mesh of R + only has vertices at the vertices of R + (the centers of thecircles), the endpoints of the boundary arcs (the tangent points between disks) or onthe straight line segments that lie on the boundary of P = T .We modify their construction first placing the Gabriel disks along the edge ofthe triangle, as shown in Figure 3. The disk packing construction of [9] will onlybe applied to the part of the triangle outside the Gabriel disks, hence none of theremainder regions that are formed will have straight line boundary segments on theboundary of T . Figure 3.
The white points are the set V that makes the triangle T Gabriel. The Gabriel disks may intersect in pairs, but we assume thatno three of them intersect.Whenever two Gabriel disks overlap, we place a small disk D near each of theintersection points of their boundaries. The disk D is tangent to both overlappingdisks and its interior is disjoint from all the Gabriel disks. See Figures 4 and 5 fortwo situations where this can occur: the boundaries of the Gabriel disks (which wecall the Gabriel circles) either have two intersections inside T or they have just oneintersection inside T .Figure 4 shows what happens when the boundaries intersect at two points. Weplace two disks near the intersections points (the dashed disks in the figure), and weform a quadrilateral by connecting the centers of the four circles. This quadrilateralis then meshed with 16 right triangles as shown in the figure. The overall structureis shown on the left, and an enlargement is shown on the right. The black point on ONOBTUSE TRIANGULATIONS OF PSLGS 9 the right where six triangles meet is the common intersection point of three lines:the line L through the two intersection points of the Gabriel circles, and the twotangent lines l , l between the dashed disk and the the Gabriel disks. It is proven in[9] that these three lines meet at a single point, as shown. The center of the dashedcircle in Figure 4 is not necessarily on L (although it appears this way in the figure).Figure 5 shows the case when two Gabriel circles have one intersection inside T (andthe other is at vertex of T ). The proof that all the triangles are right is fairly evident(see [9]). Figure 4.
Gabriel disks whose boundaries intersect twice inside T .We add tangent disks as shown (dashed) and triangulate as shown. Thepicture on the right shows more detail near one of the added disks. Thepoint where six triangles meet is the intersection of the line throughthe intersection points of the Gabriel circles and the two tangent linesbetween the added disk and the Gabriel disks.From this point the proof follows [9]. Lemma 1 of the paper shows that we canadd disjoint disks until all the remainder regions have three or four sides and thatthe number of disks added is comparable to the number of Gabriel disks we startedwith. See Figure 6.Because of the Gabriel disks, none of the remainder regions have straight lineboundary arcs, so all the augmented regions are meshed by right triangles whosevertices are either interior to T or lie in the set V or in the set V ′ (the centers of theGabriel disks) and every such point is used. This gives Theorem 2.1. Figure 5.
Figure 5 is analogous to Figure 4, but shows that casewhen two Gabriel circles have a single intersection inside T (the otheris at a vertex of T ). In this case, the mesh is a sub-picture of theprevious case and only uses 10 right triangles. Figure 6.
After adding the disks (dashed) tangent to the intersectingGabriel disks, we pack the remaining region with disjoint disks (gray)until only circular arc regions with three or four sides remain. Thisstep and the rest of the proof follow the proof in [9] exactly.
ONOBTUSE TRIANGULATIONS OF PSLGS 11 Proof of Theorem 1.2
A PSLG γ is a simple polygon if it is a simple closed Jordan curve. Suppose γ isa simply polygon and { T k } N is a triangulation of γ with no Steiner points. The unionof the edges and vertices of the triangulation is the PSLG Γ in Theorem 1.2. Foreach triangle T k , let C k be the inscribed circle and let T ′ k ⊂ T k be the triangle withvertices at the three points where C k is tangent to T k . Note that the arcs betweenpoints of C k all have angle measure < π , so T ′ k must be acute. These vertices of T ′ k will sometimes be called the cusp points of T k . See Figures 7 and 8. T T k k C k Figure 7.
The definition of C k and T ′ k . Figure 8.
A triangulation { T k } of a simple polygon, the inscribedcircles (dashed) and triangles { T ′ k } (shaded).Also note that T k \ T ′ k consists of three isosceles triangles, each with its base asone side of T ′ k and its opposite vertex a vertex of T k . Foliate each isosceles trianglewith segments that are parallel to its base (foliate simply means to write a regionas a disjoint union of curves). We call these P-segments (since they are “parallel” to the base and they will also allow us to “propagate” certain points through thetriangulation).Given a vertex v of some T ′ k , this point is either on γ (the boundary of the trian-gulation) or it is on the side of some other triangulation element T j , j = k . In thefirst case do nothing. In the second case, either v is also a vertex of T ′ j or it is not.In the first case, again do nothing. In the second case, build a polygonal path whosefirst edge is the P -segment in T j that has v as one endpoint. The other endpoint isa point w on a different side of T j . If w is on γ , or is a vertex of some T ′ i , i = j , thenend the polygonal arc at w . Otherwise, w is on the side of some third triangle T i andwe can add the P -segment in T i that has one endpoint at w . See Figures 9 and 10. a bdc Figure 9.
A cusp point (vertex of a shaded triangle) can either be on γ (the boundary of the triangulation), can be a vertex of another shadedtriangle, or neither. In the last case we can “propagate” the vertexusing a P -segment and continue the process until one the first twoconditions holds. Vertices a and b represent the first two possibilitiesand vertices c and d both represent the third case; the dashed linesshow how these points propagate until they hit the boundary.We continue in this way, adding P -segments to our polygonal path until we eitherreach a point on the boundary of the triangulation or hit a point that is the vertexof some triangle T ′ m . We call the path formed by adjoining P -segments a P -path .Since our triangles come from a triangulation of a simple polygon, they form a treeunder edge-adjacency and so the P -paths starting at the three vertices of T ′ k mustcross distinct triangles and hence can use at most n − ONOBTUSE TRIANGULATIONS OF PSLGS 13
Figure 10.
Here is the same triangulation as in Figure 9 with allcusp points propagated until they terminate. This refines the triangula-tion into acute triangles (shaded) and isosceles triangles and trapezoids(white). We claim the new vertices make all the original trianglesGabriel.every P -path must terminate and all the P -paths formed by starting at all verticesof all the { T ′ k } can create at most n ( n −
1) new vertices altogether.
Lemma 4.1.
The set U of vertices created by these P -paths crossing edges of { T k } makes every triangle Gabriel.Proof. Note that U contains every vertex of every T ′ k and we also include in U the allvertices of all the T k ’s. To prove the lemma, consider a segment e that is a connectedcomponent of T k \ U (so e is a diameter of one of our Gabriel disks). Since U containsthe vertices of T ′ k , e lies on a non-base side of one of the three isosceles triangles inside T k . If we reflect e over the symmetry axis of this isosceles triangle we get an edge e ′ on the other non-base side. Moreover, e ′ is also one of the edges created by adding U to the triangulation, and the disks D , D ′ with diameters e and e ′ respectively, arereflections of each other through the symmetry line of the isosceles triangle. Thusthe boundaries of D and D ′ intersect, if at all, on the line of symmetry, so D ∩ e ′ ⊂ e ′ and D ′ ∩ e ⊂ e . In particular, the open disk D does not contain the endpoints of e ′ and vice versa. This implies D can’t contain any of the points of U that lie on thesame side of T k as e ′ . Clearly D does not contain any points of U on the side of T k containing e ′ . Finally, D does not contain any points of U that are on the third sideof T k because D is contained in the disk D ′′ , centered at the vertex of T k where the sides containing e and e ′ intersect and passing through two vertices of T ′ k , and thisdisk does not hit the third side of T k . See Figure 11. (cid:3) D DT k ee D DT k Figure 11.
The points where propagation paths cross triangle edgesdefine Gabriel edges in each triangle. Note that at most two Gabrieldisks can intersect when the cusp points are included in set of endpoints(the set V in Theorem 2.1).This argument also shows that two Gabriel disks can only intersect if they lie ondifferent non-base sides of one of the three isosceles sub-triangles. Thus no three diskscan intersect and the condition in Theorem 2.1 is automatically satisfied wheneverthe set V contains the vertices of T ′ k for every k .Thus for each triangle T k , the points U ∩ T k makes T k Gabriel. By Theorem 2.1, T K has a nonobtuse triangulation with only these boundary vertices and using O ( U ∩ T k )) triangles. These triangulations fit together to form a nonobtuse refinement ofthe original triangulation of size O ( U ) + n ) = O ( n ), which proves Theorem 1.2.We can make a slight improvement to the algorithm above. As we propagated eachvertex, we could have stopped whenever the path encountered any isosceles trianglewith angle ≥ ◦ . In this case, the Gabriel condition will be satisfied no matter howwe add points to a non-base sides of the isosceles triangle, since the correspondingdisks don’t intersect the other non-base side of the triangle. See Figure 12. In somecases this might lead to a smaller nonobtuse triangulation. This observation will alsobe used later in the proof of Theorem 1.1, when it will be convenient to assume weare dealing only with isosceles triangles that are acute. ONOBTUSE TRIANGULATIONS OF PSLGS 15
Figure 12. P -paths can be stopped when they hit an isosceles tri-angle with angle ≥ ◦ since the corresponding Gabriel disks can’t hitthe other sides of the triangle.5. Dissections and quadrilateral propagation
We now start to prepare for the proofs of Theorems 1.1 and 1.3. The definitionsand results in this and the next three sections will be used in both proofs.Suppose Ω is a domain in the plane (an open connected set). We say Ω has apolygonal dissection if there are a finite number of simple polygons (called the pieces of the dissection) whose interiors are disjoint and contained in Ω and so thatthe union of their closures covers the closure of Ω. A mesh is a dissection whereany two of the closed polygonal pieces are either (1) disjoint or (2) intersect in apoint that is a vertex for both pieces or (3) intersect in a line segment that is anedge for both pieces. See Figure 13 for an example. A dissection is also called a non-conforming mesh . A vertex of one dissection piece that lies on the interior ofan edge for another piece is called a non-conforming vertex . If there are no suchvertices, then the dissection is actually a mesh.Given any convex quadrilateral with vertices a, b, c, d (say in counterclockwise or-der), there is a unique affine map from [ a, b ] to [ c, d ] that takes a to d and b to c .A propagation segment in the quadrilateral is a segment connecting a point in [ a, b ]to its affine image point in [ c, d ] (or connecting a point in [ b, c ] to its affine image in[ d, a ] under the analogous map for that pair of sides). See Figure 14. In a triangle A, B, C with marked vertex, say A , propagation paths either connect points on [ A, B ]to their linear images on [
A, C ] or they connect any point on [
B, C ] to the single point
Figure 13.
On the left is a polygon dissected into quadrilateralsand triangles and on the right is the standard propagation of the non-conforming vertices until the propagation paths leave the polygon. Ifall the paths terminate, this gives a mesh, as described in the text. A (this is what we would get if we think of the triangle as a degenerate quadrilateral A, B, C, D with A = D , i.e., one side of length zero).Given θ >
0, we say a quadrilateral is θ -nice if all the angles are within θ of 90 ◦ .In this paper we will always assume θ < ◦ so the quadrilateral is convex. We say atriangle with a marked vertex is θ -nice if the two unmarked vertices have angles thatare within θ of 90 ◦ . ab c d ACB Figure 14.
Standard propagation segments for a quadrilateral anda triangle with a marked vertex (A). In both cases, “ θ -niceness” ispreserved by cutting a piece into sub-pieces by such segments. Lemma 5.1.
Suppose θ < ◦ and that Q is a θ -nice quadrilateral. If Q is sub-dividedby a propagation line, then each of the resulting sub-quadrilaterals is also θ -nice.Proof. Set a t = (1 − t ) a + tb and c t = (1 − t ) d + tc . Let I t = [ a t , c t ] be the segmentconnecting these points and let θ ( t ) be the angle formed by the segments [ a, b ] and I t . ONOBTUSE TRIANGULATIONS OF PSLGS 17
It suffices to show this function is monotone in t . If it were not monotone, then therewould be two distinct values of s, t ∈ [0 ,
1] where I s and I t were parallel. Becauseboth endpoints move linearly in t , this implies I r is parallel to I s for all s ≤ r ≤ t .Because θ ( t ) is analytic in t , this means it is constant on [0 , θ is eitherstrictly monotone or is constant; in either case it is monotone, as desired. (cid:3) Similarly (but more obviously), when a θ -nice triangle is cut by such a propaga-tion segment (of either type) the resulting pieces are θ -nice quadrilaterals or θ -nicetriangles. Lemma 5.2.
Suppose Ω has a dissection into θ -nice pieces (triangles and quadrilat-erals). Suppose that every non-conforming vertex can be propagated so that it reachesthe boundary of Ω or hits another vertex after a finite number of steps. Then theresulting paths cut the θ -nice dissection pieces into θ -nice triangles and quadrilateralsthat mesh Ω . The proof is evident since when we are finished, there are no vertices that arein the interior of any edge of any piece. See Figure 13. What is not so clear iswhether, in general, the propagation paths have to end; in the proof of Theorem 1.2every propagation path did end within a fixed number of steps, but in general, thepaths may never terminate (see Figure 15) or may only terminate only after a hugenumber of steps. Later in this paper we discuss two ways of “bending” the standardpropagation paths so that they terminate within a certain number of steps, and sothat the pieces formed satisfy certain geometric conditions.6.
Isosceles dissections
Next we discuss a special type of polygonal dissection. An isosceles triangle isa triangle T with a marked vertex v so that the two sides adjacent to v have equallength. An equilateral triangle can be considered as isosceles in three ways, but weassume that if such triangles occur, a vertex is specified.The side opposite v is called the base of T and the other two sides are the non-base sides of T . The angle of an isosceles triangle will always refer to the interiorangle at the vertex opposite the base edge. A P -segment is a segment in T withendpoints on the non-base sides that is parallel to the base. This is a special case of Figure 15.
The upper lefts shows an isosceles dissection of a region.The horizontal segment is S = [0 ,
1] and the six vertices shown on thissegment are (left to right) 0 , α , , α, α ,
1. A path started at a point x ∈ S will visit x − α mod 1 after propagating once through the upperhalf-plane and once through the lower half-plane. The upper right andlower left pictures show a path after 2 visits to the lower half-planeand 50 visits for α = 1 / √
2. If α is irrational, then the propagationpath becomes dense in the dissected region. When α is rational, thepropagation paths either connect non-conforming vertices or are loops;the connecting paths give a mesh but there is no uniform bound onthe number of elements. The connecting paths for α = . P -loops.the propagation segments for marked triangles in the previous section. We require theinterior of the segment to be in the interior of T , so the base itself is not a P -segment.We say a triangle is θ -nice if all its angles are bounded above by 90 ◦ + θ .An isosceles trapezoid is a quadrilateral that has a line of symmetry that bisectsopposite sides. This is equivalent to saying that there is at least one pair of parallel ONOBTUSE TRIANGULATIONS OF PSLGS 19 sides (called the base sides) that have the same perpendicular bisector and the otherpair of sides (the non-base sides) have the same length as each other. We allowrectangles, but in this case we specify a pair of opposite sides as the base sides. The angle of an isosceles trapezoid is the angle made by the lines that contain the twonon-base sides; we take this to be zero if these sides are parallel (when the trapezoidis a rectangle). The vertex of the trapezoid is the point where these same linesintersect (in the case when they are not parallel; otherwise we say the vertex is at ∞ ). See Figure 16. We say a quadrilateral is θ -nice if all its interior angles arebetween 90 ◦ − θ and 90 ◦ + θ (inclusive). This is the same as saying the angle of thetrapezoid is ≤ θ . wv θ Figure 16.
An isosceles trapezoid. The base sides are vertical inthis picture. The vertex is the point v and the width w is the lengthof the non-base sides. The angle of the trapezoid is θ .As with isosceles triangles, we can define P -segments in an isosceles trapezoid assegments in the trapezoid that are parallel to the base sides (again, these correspondto propagation segments for quadrilaterals). A P -path is a simple polygonal arcformed by adjoining P -segments end-to-end.An isosceles piece is either an isosceles triangle or an isosceles trapezoid. We willuse this term when it is unimportant which type of shape it is. When referring to abase side of an isosceles piece we mean either base side for a trapezoid and the baseside or the opposite vertex for a triangle; the vertex is considered as a segment oflength 0, so when we refer to the length of the smaller base side of an isosceles piece,we mean zero if the piece is a triangle.A Q -segment in an isosceles triangle is a segment joining a point of the baseto the vertex opposite the base. The two non-base sides do count as Q -segments,and we shall also call these the Q -sides of the isosceles triangle. A Q -segment for an isosceles trapezoid is a propagation segment that connects the base sides of thetrapezoids. As with triangles, we count the non-base sides as Q -segments and callthese the Q -sides of the trapezoid. The width of the piece is the length of a Q -side(both Q sides have the same length). It might be more natural to define the width asthe distance between the base sides, but the definition as given will simplify matterswhen we later join isosceles pieces to form tubes.Suppose that Ω is a domain in the plane (an open connected set). As mightbe expected, an isosceles dissection of Ω is a finite collection of disjoint, openisosceles triangles and trapezoids contained in Ω, so that the union of their closurescovers all of Ω. However, we also require that when two pieces have sides with non-trivial intersection, these sides are both Q -sides. We do this so that in an isoscelesdissection, a P -path can always be continued unless the path hits a vertex of thedissection, or hits the boundary of the dissected region. A θ -isosceles dissection isan isosceles dissection where every piece is θ -nice.For example, Figure 8 in Section 4 shows a triangulated polygon. Let Ω be the partof interior of the polygon with the (closed) shaded triangles removed; the remainingwhite region it is a union of isosceles triangles that only meet along non-base sides.Thus Ω has an isosceles dissection; note that it is not a mesh since the isoscelestriangles do not always meet along full edges. When we remove the P -paths generatedby propagating the vertices of the central triangles, we obtain a mesh into isoscelestriangles and trapezoids (as required by Lemma 5.2). See Figure 10. We call this an isosceles mesh .Figure 15, upper left, shows an isosceles dissection of a region using 18 triangles.In that example, if α is irrational, then the P -paths never hit the boundary of theregion and can continue forever without terminating. For α rational the P -pathsstarting at non-conforming vertices terminate at other non-conforming vertices andthese paths create an isosceles mesh. However, the number of mesh elements dependson the choice of α and may be arbitrarily large.In Figure 17, we show an isosceles dissection using only trapezoids, and an isoscelesmesh generated by propagating non-conforming vertices along P -paths.A chain in a dissection is a maximal collection of distinct pieces T , . . . , T k so thatfor j = 1 , . . . k − T j and T j +1 share a Q -side (the sides are identical, not just ONOBTUSE TRIANGULATIONS OF PSLGS 21
Figure 17.
A domain W dissected by isosceles trapezoids and a meshgenerated by propagating non-conforming vertices along P -paths. The P -sides of the trapezoids are drawn thicker.overlapping). If a piece in the dissection does not share Q -side with any other piece,we consider it as a chain of length one. For example, the dissection in Figure 17 haschains of length 2,4, 5 and 7 and four chains of length 1. The Q -ends of a chainare the Q -side of T not shared with T , and the Q -side of T k not shared with T k − .When T and T k also share a Q -side, then the chain forms a closed loop. We willcall this a closed chain (this case is not of much interest to us since no propagationpaths will ever occur inside such a closed chain.)Suppose T is an isosceles piece. Given θ >
0, a θ -segment is a segment in T withone endpoint on each non-base side, and so that the segment is within angle θ ofbeing parallel to the base. We allow one endpoint of a θ -segment to be a vertex of T ,but the interior of the segment must be contained in the interior of T , so we don’tconsider base sides of T to be θ -segments. A θ -path is a polygonal arc made up of θ -segments joined end-to-end. We shall sometimes refer to this as a θ -bent path .Note that if T is an θ -nice isosceles piece that is cut by a P -segment, then it is cutinto two θ -nice isosceles pieces. If it is cut by a θ -segment, then we get two pieces(triangles or quadrilaterals) that are 2 θ -nice (but not isosceles unless θ = 0). SeeFigure 18.We say that a finite set of points on the Q -sides of an isosceles piece make thatpiece Gabriel if the following holds. Each Q side is split into several segments bythese points and we require that the open disks with these segments as diameters donot contain any of the added points or corners of the piece. See Figure 19. < θ < θ Figure 18. A θ -segment (dashed) makes angle at most θ with P -segments (solid). A θ -path is made up out of θ -segments. Figure 19.
On the left the points make this piece Gabriel, on theright they do not. 7.
Tubes
Two P -paths γ , γ are parallel if each point of γ can be connected to a pointof γ by a Q -segment (equivalently, the paths cross the same sequence of isoscelespieces, in the same order). A tube in Ω is the union of two parallel P -paths γ , γ and all the Q segments that connect the first to the second. The P -paths γ , γ arecalled the P -sides or P -boundaries of the tube. See Figure 20.Suppose the endpoints of γ and γ are { x , y } and { x , y } respectively and that[ x , x ] and [ y , y ] are Q -segments. These segments are called the ends or Q -ends of the tube. These may or may not be disjoint segments. The two ends of a tubehave the same length, and this common length of each of the two Q -ends is calledthe width of the tube (a tube can be thought of as a union of isosceles pieces, allof the same width, joined end-to-end along their Q -sides). The points { x , y , x , y } are the corners of the tube (although in some cases, these need not be four distinct ONOBTUSE TRIANGULATIONS OF PSLGS 23 points in the plane, e.g. pure spirals that we will discuss later).
Opposite corners of a tube mean either the pair { x , y } or the pair { x , y } . A maximal width tube is the union of all P -paths parallel to a given one. If a tube is maximal width, theneach of the P -path boundaries contains segments that are bases for at least one piecethat the tube crosses (otherwise we could widen the tube). See Figure 20. y y γ x x γγ γ Figure 20.
Given a P -path γ that returns to the same Q -segment,there is an associated widest return region consisting of all parallelpaths. The P -boundary of this tube consists of the two curves γ , γ ;by maximality, each must contain a base edge of a dissection piece(highlighted with darker edges in the figure). The points x , y formone pair of opposite corners; x , y the other pair. We are interested injoining opposite corners by a θ -path crossing the tube (dashed curveconnecting x to y ).We say a path strictly crosses a tube if it is contained in the tube and has oneendpoint on each Q -end. We say a path crosses a tube if it contains a sub-path thatstrictly crosses the tube. The P -paths that strictly cross a tube can be parameterizedas γ t with t ∈ [0 ,
1] where γ and γ are the P -sides of the tube as discussed aboveand γ t has endpoints x t = (1 − t ) x + tx and y t = (1 − t ) y + ty . Moreover ℓ ( γ t ) = (1 − t ) ℓ ( γ ) + tℓ ( γ ) , where ℓ ( γ ) denotes the length of a path γ . This formula is obvious for tubes thathave a single isosceles piece, and it follows in general since a sum of affine functionsis affine. The path with t = 1 / center path of the tube. Note that ℓ ( γ / ) = 12 ( ℓ ( γ ) + ℓ ( γ )) . (7.1)The length ℓ ( T ) of a tube T is the minimum length of the two P -sides, i.e., ℓ ( T ) = min( ℓ ( γ ) , ℓ ( γ )) . It is possible for a tube to have length zero, e.g., when all the pieces are triangleswith a common vertex. See the left side of Figure 21.The length of an isosceles piece is the length of its shorter base edge (zero fortriangles). If a tube T is made up of isosceles pieces { T k } then it is possible to haveboth ℓ ( T ) > ℓ ( T k ) = 0 for all k . See the right side of Figure 21. We definethe minimal-length of a tube to be˜ ℓ ( T ) = X ℓ ( T k ) , i.e., we sum over the minimal base length for each piece of the tube, whereas ℓ ( T ) isdefined by summing over all segments in one P -boundary of the tube or all segmentsin the other. Clearly ˜ ℓ ( T ) ≤ ℓ ( T ). Figure 21.
On the left is a tube of length zero. On the right is atube with positive length, but zero minimal-length.As noted above, it is possible to have both ˜ ℓ ( T ) = 0 and ℓ ( T ) >
0. However, if wesplit a tube into two parallel tubes using the center path then this cannot happen foreither sub-tube:
Lemma 7.1.
Let Q be a tube and Q , Q the parallel sub-tubes obtained by splitting Q by its center path γ . Then ℓ ( Q ) ≤ ℓ ( Q ) and ℓ ( Q ) ≤ ℓ ( Q ) . ONOBTUSE TRIANGULATIONS OF PSLGS 25
Proof.
Let γ be the common P -boundary of Q and Q , and γ the common P -boundary of Q and Q . See Figure 22. By Equation (7.1), the center path γ of Q has length between the lengths of γ and γ . Thus ℓ ( Q ) = min( ℓ ( γ ) , ℓ ( γ )) ≤ min( ℓ ( γ ) , ℓ ( γ )) = ℓ ( Q ) . This is the first inequality in the lemma. QQ γγ γ Q Figure 22.
The tube Q is split into two parallel tubes Q , Q , by itsmid-path γ .To prove the second inequality, suppose L = ℓ ( Q ) is the length of the tube Q and γ is the P -boundary shared by Q and Q . The length of γ is the sum of the lengthsof its segments and we group this sum into two parts, depending on whether or notthe segments are the longer or shorter base sides of the corresponding isosceles piecesin Q (if the piece has equal length bases, its makes no difference in which sub-sum weplace the segment). Call the two sums L and L where these give the sums over theshorter edges and longer edges respectively. By definition L = L + L . If L ≥ L ,then the short sides of Q (and hence the short sides of Q ) add up to at least L . Soin this case ˜ ℓ ( Q ) ≥ L ≥ L , as desired. If L < L , then L > L and hence thecorresponding mid-segments of the pieces add up to L (since the mid-segment haslength as least half the longer base side). But these mid-segments are the shorter basesides of pieces in Q , so again ˜ ℓ ( Q ) ≥ L , as desired. This proves the lemma. (cid:3) The P -segments give an identification between the non-base sides of an isoscelespiece that preserves length. The displacement of a θ -segment [ a, b ] is | p − b | where[ a, p ] is a P -segment. It is easy to check that this is unchanged if we reverse the roles of a and b . Similarly, the two ends of a tube are identified by an isometry inducedby the parallel P -paths defining the tube. The displacement of a path that strictlycrosses the tube is | b − p | where a, b are the endpoints of the path and the P -pathstarting at a hits the other end at p . The most important estimates in the remainderof the paper involve how much displacement a path can have, given that it satisfiescertain limitations on its “bending” across each isosceles piece. For Theorem 1.3, thebending is bounded by a fixed angle θ , and for Theorem 1.1 the amount of bendingdepends on the piece and is determined by the Gabriel condition.8. Return regions
In this section we introduce a collection of regions, one of which must be hit by any P -path that is sufficiently long (in terms of the number of pieces it crosses). We willclassify the regions into four types, and bound the total number of regions needed toform such an unavoidable collection.Suppose, as above, that Ω has an isosceles dissection. A return path is a P -paththat begins and ends on the same Q -side of some piece of the dissection, and thatintersects any Q segment at most three times. Figure 23 shows four ways that thiscan happen: C-curve: both ends of γ hit the same side of S and γ ∪ S separates the endpointsof S from ∞ , S-curve: γ starts and ends on different sides of S , crossing S exactly once inbetween, and this crossing point separates the endpoints of γ on S , G1-curve: γ starts and ends on different sides of S , crossing S exactly once,and this crossing point does not separate the endpoints of γ on S , G2-curve: γ starts and ends on different sides of S with no crossings S .When we refer to a G-curve, we can mean either a G1 or a G2-curve. Lemma 8.1.
Suppose n is the number of isosceles pieces in an isosceles dissection.Every P -path γ with n + 1 segments contains a sub-path that is a return path of oneof the four types described above.Proof. A P -path γ with 2 n + 1 segments must cross some dissection piece three timesby the pigeon hole principle. Therefore γ crosses a non-base side S of such a piece at ONOBTUSE TRIANGULATIONS OF PSLGS 27
Figure 23.
A C-curve, S-curve and the two types of G-curve. Eachis named for the letter it vaguely resembles.least three times. By passing to a sub-path, if necessary, we can also assume γ doesnot hit any Q -side more than three times. Suppose that γ does not contain a C-curveor a G2-curve as a subpath. Then the sub-path between its first and second visit to S starts and ends on the same side of S and the same for the sub-path between itssecond and third visit (but now it starts and ends on the other side of S ). Thus thesubpath formed between the first and third visits is either a S-curve or a G1-curve.Thus one of the four types of curve must occur as a subpath. (cid:3) A tube consisting of parallel return paths will be called a return tube and becalled a C-tube, S-tube, G1-tube or G2-tube depending on the type of curves itcontains (clearly all parallel curves must be of the same type). We want to show thatthe length of a return tube cannot be too small compared to its width. We do this byconsidering the different types of tubes one at a time. We call a return tube a simpletube if the two ends have disjoint interiors (they may share a corner); otherwise theregion is called a spiral . A C-tube or S-tube must be a simple tube; a G-tube canbe either be a simple tube or a spiral. As the name suggests, simple tubes are easierto understand and we start with this case. See Figure 24.
Lemma 8.2.
The length L of a C-tube is at least twice its width w .Proof. The ends of a C-tube are disjoint intervals on the same Q -segment S , so thelength of S is at least 2 w . But both P -sides of the tube cover S when projected x yy x Figure 24.
A G-region can form a single tube with disjoint ends (asimple G-tube) or the two ends can overlap (a spiral).orthogonally onto the line containing S , so both P -sides have length ≥ | S | ≥ w .The length of the tube is the minimum of these two path lengths, so is also ≥ w . (cid:3) Lemma 8.3.
The length L of a S-tube is at least twice its width w .Proof. Split the S-tube into four sub-tubes as follows. Each P -path in the tube iscut by a point where it crosses S and this cuts the tube into two sub-tubes that alsohave width w , say U , U , which meet end-to-end. Each of these are split into twothinner tubes by the central P -path γ / , giving four sub-tubes called U i , U o , U i , U o ,e.g., U i is the inner part of U and its endpoints on S separate the endpoints of the outer part U o on S . See Figure 25. Note that the length of the original S-tube isat least the minimum of the lengths of the two outer tubes (since they each have a P -boundary contained in the P -boundary of the S-tube). w w U o1 U i2 U o2 U i1 Figure 25.
The outer tubes of a S-tube are shaded.
ONOBTUSE TRIANGULATIONS OF PSLGS 29
However, the endpoints of the outer P -boundary of an outer part are separated byat least distance 2 w , and each P -boundary of S contains the outer P -boundary ofone of its outer parts. Thus both P -boundaries have length at least 2 w . (cid:3) Lemma 8.4.
The length L of a simple G-tube is at least its width.Proof. Suppose I and J are the ends of the spiral. If we project either P -side of thetube orthogonally onto S , then it covers either I or J , so both sides of the tube arelonger than the tube is wide. (cid:3) Putting together the last three lemmas we get
Corollary 8.5.
The length L of a simple return tube is at least its width w . If thetube is not a G tube, then L ≥ w . The more interesting and difficult return regions are the spirals: G-tubes wherethe Q -ends [ x , x ] and [ y , y ] overlap but are not identical.Suppose S is a spiral return tube and the corners are ordered on S as x < y A pure spiral with N turns and width w has length at least N w .Proof. Without loss of generality we may scale the spiral so the width w = 1. Use thesegment [ x , y ] to cut the entire spiral into N simple G-tubes. The first has lengthat least w , because both P -boundaries project orthogonally onto one of the Q -ends.In general, both the parts of the P -boundary paths of the j th sub-tube have length y x zx y Figure 26. A spiral can always be divided into a simple tube(shaded) and a pure spiral. The pure spiral can be thought of as manyparallel simple G-tubes that don’t cross S , or as one very long tubethat crosses S multiple times. The pure spiral here has five windings.at least 2 j − 1. To see this, consider the curves in each half-plane defined by theline L through x , y , and project orthogonally onto L . The part in one half-planeprojects to a segment of length at least j − j . See Figure 27. Hence the j th tube has length at least 2 j − 1. Summing1 + 3 + 5 + · · · + (2 N − 1) = N gives the result. (cid:3) jj+1 Figure 27. Estimating the length of a G-curve.The argument proving Lemma 8.6 will be used again in Section 18. Next we slightlyrefine Lemma 8.1 Lemma 8.7. Suppose n is the number of isosceles pieces in an isosceles dissection.Every P -path γ with n + 1 segments contains a sub-path that begins and ends on the Q -end of a chain and is a return path of one of the four types described above (S, C,G1, G2). ONOBTUSE TRIANGULATIONS OF PSLGS 31 Proof. Apply Lemma 8.1 to the initial path of 2 n + 1 steps, to get a return path ofone of the four types. If the path already begins and ends on the Q -end of a chainthere is nothing to do. If it begins and ends at an interior Q -segment of the chainthen by deleting or extending the paths as shown in Figure 28 we can obtain a paththat begins and ends on the Q -end of the chain. The number of steps added is lessthan n , so the new path must be a sub-path of γ . (cid:3) C SG1 G2 Figure 28. A return path ending inside a chain can easily be mod-ified to start and finish on the Q -end of the chain (the extensions areshown as dashed lines and follow P -paths). Black dots are the originalendpoints of the path an the white dots are the modified endpoints.We say that a return region is standard if is of one of the types (C, S, G1, G2)discussed above and if it begins and ends on segments that are the Q -ends of somechains (possibly the same chain or two different chains). The following is one of thekey estimates of this paper. Lemma 8.8. If Ω has an isosceles dissection into n pieces with M chains, then thereare O ( M ) standard return regions with disjoint interiors so that any P -path withmore than n + 1 segments must hit one of the regions.Proof. Each chain in the dissection has two P -boundaries. Each of these P -boundariesmay or may not be part of the P -boundary of a standard return region. If it is, thenassociate to the P -boundary of the chain the maximal width standard return region that contains the P -boundary of the chain in its own boundary. Note that at most2 M return regions can be selected in this way, since there are M chains and each hastwo P -boundaries.We claim that any P -path γ in the dissection with 5 n + 1 steps contains a sub-path that crosses one of the selected return regions. Let γ ′ be the path obtained bydeleting n steps from each end of γ . By Lemma 8.7, γ ′ contains a sub-path γ ′′ thatis a return path of one of the four standard types and which begins and terminateson the Q -end of a chain. Thus the set of paths parallel to γ ′′ forms a standard returntube T of maximal width. If T is one of the chosen return regions, then we are donesince γ ′′ ⊂ γ crosses T .On the other hand, suppose T is not one of the chosen regions. Since T is maximal,it contains the P -boundary of some chain C within its own P -boundary. Since T wasnot chosen, there must be another return region T ′ , at least as wide as T , that waschosen and T ∩ C ⊂ T ′ ∩ C . Thus every path crossing T hits T ′ . Thus γ ′′ hits T ′ . Ifwe add n steps to both ends of γ ′′ , the new, longer path must now cross T ′ , but itis still a sub-path of γ . Thus γ crosses some return region in the chosen collection.This proves the claim.The collection of maximal width return regions defined above may overlap. To getdisjointness, we order the chosen regions R , . . . , R m , m = O ( M ), from widest to nar-rowest and label the first region “protected” and label the remainder “unprotected”.At each stage we look at the first unprotected region R k in the current list and see ifthere are any P -paths strictly crossing it that intersect a protected region (anythingearlier in the list). If there is no such path, then label R k protected, and move to thenext region.If there is a P -path γ strictly crossing R k that hits a protected region R j , thenremove from R k any P -paths that hit R j . Since R j is at least as wide as R k , removingthese paths gives a connected return region R ′ k ⊂ R k (possibly empty). Now re-sortthe list by width. R ′ k either stays where it is or moves later in the list; the protectedregions all stay where they are. Since two regions will never overlap after the first timethey were compared, this process stops after at most m steps, and gives a collectionof return regions with disjoint interiors. Moreover, once a region is protected, it isnever modified again and is part of the final collection. ONOBTUSE TRIANGULATIONS OF PSLGS 33 Finally, we have check that every long enough P -path hits one of the disjointregions. If γ is any path with 5 n + 1 segments then it crossed some region in theoriginal list. Suppose R γ is the first region on the original sorted list that is crossedby γ . If a part of R γ containing γ is deleted in the construction, then γ must hita protected return region. Thus γ hits a return region in the final collection, asdesired. (cid:3) This lemma is used in the proofs of both Theorems 1.1 and 1.3. In both cases wewill reduce to meshing a region Ω that has an isosceles dissection. We will propagatethe non-conforming vertices until they either hit the boundary of Ω or hit the Q -endof a return region. By Lemma 8.8, one of these two options must occur with O ( n )steps. The proofs of the two theorems differ mostly in how we construct Ω and howwe mesh inside the return regions.9. Proof of Theorem 1.3: reduction to a meshing lemma This is the first of four sections that construct the almost nonobtuse triangulationin Theorem 1.3. Unlike the proofs of Theorems 1.1 and 1.2, the proof of Theorem 1.3does not make use of Theorem 2.1 to create the triangulation; we shall construct thetriangulation directly. However, we will use the result of Bern, Mitchell and Ruppert[9] that any simply n -gon has a nonobtuse triangulation with O ( n ) triangles. We willalso make use of return regions and bending paths, both ideas we shall use again inthe proof of Theorem 1.1.As in the proof of Theorem 1.2, we start with a PSLG that is a triangulation. Inthat earlier proof we divided each triangle T into a central triangle and three isoscelestriangles. Here we will replace the single central triangle by a simple polygon thatapproximates a triangle with circular edges.Given a triangle T with vertices v , v , v , let C be the inscribed circle and let z , z , z be the three points where this circle is tangent to the triangle (numbered sothat z k lies on the side of T opposite v k ). Any pair of the z ’s are equidistant from oneof the vertices of T and hence are connected by a circular arc centered at this vertex.This defines a central region bounded by three circular arcs, each pair of arcs tangentwhere they meet (see the shaded area on the left of Figure 29). We will replace eachof these circular arcs by a polygonal path inscribed in the arc. For example, let γ be a polygonal arc inscribed in the circular arc connecting z and z . If γ consistsof m equal length segments, then the angle subtended from v by these segments isless than π/m , (since γ subtends at most angle π ). If we then connect the verticesof γ to v we obtain a chain of isosceles triangles all with angle ≤ θ = π/m . See theright side of Figure 29. Taking the union of these isosceles triangles over all T in theoriginal triangulation gives the region Ω, that clearly has a θ -nice isosceles dissectionwith O ( n ) chains and O ( n/θ ) pieces. v v z z γ zv Figure 29. We define a simple polygon by inscribing polygon arcson circular arcs as shown above. If we use m evenly points on each arcthen remaining region clearly has a θ -isosceles dissection for θ = π/m . Lemma 9.1. Suppose Ω is a region that has a θ -nice isosceles dissection (both trian-gles and quadrilaterals are allowed). Assume that the dissection has O ( n ) chains and O ( n/θ ) pieces. Then there is a mesh of Ω using O ( n /θ ) θ -nice quadrilaterals andtriangles. Moreover, each dissection piece T contains at most O ( n/θ ) mesh elementsand every mesh element Q contained in T is bounded by at most two sub-segments ofthe Q -sides of T (possibly points) and at most two θ -paths in T (possibly the vertexof T , if T is a triangle). We will prove this in the next three sections. We can deduce Theorem 1.3 fromLemma 9.1 as follows. By a result of Bern, Mitchell and Ruppert, each central polygonhas a nonobtuse triangulation with at most O (1 /θ ) triangles. This triangulation mayplace extra vertices on the edges of the central polygon, but not more than O (1 /θ )vertices in total. Each such edge e is the base of one of the isosceles triangles T in ONOBTUSE TRIANGULATIONS OF PSLGS 35 the dissection, and we connect the extra vertex on e to the opposite vertex of T bya Q -segment S . See Figure 30. Figure 30. Each isosceles triangles T in the dissection of Ω is meshedby at most O ( n/θ ) quadrilaterals and triangles (here a chain of fourtriangles is shown). Connecting “extra” vertices (white dots) on thebase of T to the opposite vertex thus creates at most O ( n/θ ) extramesh pieces per dissection triangle.This creates a new θ -nice quadrilateral or triangle for each θ -path that S crosses.Since there are at most O ( n/θ ) such paths per piece, each extra vertex on e createsat most O ( n/θ ) new elements of the mesh. Since there are O ( n ) central polygonsand each has at most O (1 /θ ) extra boundary points coming from its nonobtusetriangulation, at most O ( n /θ ) extra pieces are created overall.As the final step, we add diagonals to the quadrilateral pieces of the mesh, gettinga triangulation. Since all the quadrilaterals are θ -nice, the resulting triangles havemaximum angle 90 ◦ + θ , which proves Theorem 1.3.Our application of Lemma 9.1 to Theorem 1.3 only needs to apply to dissectionsconsisting entirely of triangles, but has been stated for more general isosceles dissec-tions which may use both triangles and quadrilaterals. The extra generality does notlengthen the proof at all, but it is useful for the application to optimal quad-meshinggiven in [11]. That application involves an isosceles dissection that uses only trape-zoid pieces; the precise variant of Lemma 9.1 that is needed in that paper will bestated and proved in Section 13. Proof of Lemma 9.1: outside the return regions We continue with the proof of Theorem 1.3, by starting the proof of Lemma 9.1.In this section we will mesh the part of Ω that is outside the return regions.Let { R k } N be the disjoint return regions for Ω given by Lemma 8.8. Since thereare O ( n ) chains there are O ( n ) return regions. For each triangle T k , and each of thethree vertices of T ′ k on its boundary, construct the P -path starting at this point andcontinued until it hits another cusp point, leaves Ω or enters a return region. Lemma5.2 says these paths cut the isosceles pieces of the dissection into isosceles pieces thatform a mesh. By Lemma 8.8, each P -path we generate terminates within O ( n/θ )steps and there are less than 3 n of these paths (at most three per triangle), so a totalof O ( n /θ ) mesh pieces are created outside the return regions. Moreover, each suchpath crosses a single dissection piece at most O (1) times. Thus each dissection piececan be crossed at most O ( n ) times but such paths.Next we place O (1 /θ ) evenly spaced points on both Q -sides of each return region(the reason for this will be explained in the next section). Each of these points ispropagated by P -paths outside the return region it belongs to, until it runs into theboundary of Ω or hits the Q -side of some return region (possibly the same one theystarted from). As above, this generates a θ -nice mesh outside the return regions.There are O ( n ) return regions and O (1 /θ ) points per region to be propagated. Eachpath continues for at most O ( n/θ ) steps, so at most O ( n /θ ) mesh elements arecreated in total. Moreover, each dissection piece is crossed at most O (1) times byeach path, so is crossed O ( n/θ ) times in total by such paths.11. Proof of Lemma 9.1: the simple tubes Next, we have mesh inside the return regions. In this section, we deal with thereturn regions that are simple tubes and in the next section we deal with spirals.For the first time we will use θ -paths rather than P -paths (recall that a θ -path ismade up of segments that are within θ of parallel to the base of the isosceles piece;when we cut a θ -nice piece by a θ -path we get two 2 θ -nice pieces). We need thefollowing lemma. Lemma 11.1. Suppose Q is a tube whose width w is at most sin θ times its minimal-length ˜ ℓ . Then opposite corners of Q can be connected by a θ -path inside the tube. ONOBTUSE TRIANGULATIONS OF PSLGS 37 Proof. Suppose T is a θ -nice isosceles piece and x, y are the endpoints of a P -segment S = [ x, y ] crossing T . Then any point z on the same side as y and within distancesin( θ ) | x − y | can joined to x by a θ -path. See Figure 31. Thus if { T k } is an enumerationof the pieces making up the tube and ℓ k is the minimal base length of the k th piece,then we can create a θ -path that crosses the tube and whose endpoints are displacedby P k ℓ k sin θ = ˜ ℓ sin θ with respect to a P -path. This proves the lemma. (cid:3) y θ qpxw Figure 31. Clearly | p − y | ≥ | q − y | = | x − y | sin θ ≥ ˜ ℓ sin θ where q is the closest point to y on the line making angle θ with the segment xy . Corollary 11.2. If R is return region that is a C-tube, S-tube or simple G-tube, thenwe can cut R into O (1 /θ ) parallel sub-tubes and connect opposite corners of the eachtube by a θ -path contained in that tube.Proof. Suppose the return region R has length L and width w . Choose an even integer M ≥ / sin θ and split the return region R into the disjoint union of M thinner tubes { T j } of width w/M . By Lemmas 8.2, 8.3 and 8.4 each of these new tubes has lengththat is at least w and width equal to w/M . Thus each has length that is at least M times as long as its width.Since M ≥ 2, each of our thin tubes is half of a thicker tube that is still inside thegiven tube R . By Lemma 7.1˜ ℓ ( T j ) ≥ ℓ ( T j ) ≥ ℓ ( R ) ≥ w. On the other hand, the width of T j is w/M . Hence the minimal-length of each tube T j is more than M/ / sin θ times its width, so the previous Lemma 11.1 appliesto T j , as desired. (cid:3) We can now continue with the proof of Lemma 9.1. We then cut the return regioninto O (1 /θ ) parallel tubes as described above, and divide each tube by a θ -pathconnecting opposite corners. This meshes each tube using 2 θ -nice pieces. Any P -path hitting a Q -end of one of these tubes is then propagated to a corner on theopposite end of the tube by standard quadrilateral propagation paths. This gives a2 θ -nice mesh of the tube that is consistent with all the meshes created outside thetube.Since there were at most O ( n/θ ) P -paths that might terminate and each returnregion has at most O ( n/θ ) isosceles pieces, at most O ( n /θ ) 2 θ -nice triangles andquadrilaterals are created inside all the return regions. Moreover, each dissectionpiece is crossed by at most O ( n/θ ) paths (there are O (1 /θ ) paths per return regionand O ( n ) return regions), so it contains at most this many mesh pieces.12. Proof of Lemma 9.1: the spirals This is the final section in the proof of Theorem 1.3. Here we prove Lemma 9.1inside the spiral return regions.Since any spiral can be divided into a simple G-tube and a pure spiral, and we cantreat the simple tube as above, it suffices to deal with the pure spirals. Let N be thewinding number of the spiral; we may assume N ≥ 2, since otherwise the spiral canbe treated as a tube and can be triangulated as in the previous section. Let p be thenumber of isosceles pieces that are hit by the spiral (this is the number of steps ittake to complete one winding of the spiral).The spiral can be divided into N tubes joined end-to-end, each starting and endingon the same Q -edge of some isosceles pieces. We divide the first and last of thesetubes into O (1 /θ ) parallel thin tubes. Then any P -path that enters the tube fromeither end can be θ -bent so that it terminates at the corner of one of the thin tubesafter winding once around the spiral.If N = O (1 /θ ), then we simply propagate all the interior corners of the thin tubesat one end of the spiral around the spiral until they run into the corners of the thintubes at the other end. This generates O ( N · p · θ − ) = ( pθ − ) new vertices. SeeFigure 32. ONOBTUSE TRIANGULATIONS OF PSLGS 39 Figure 32. Cut the inner and outer tubes into O (1 /θ ) parallel nar-row tubes and θ -bend all entering P -paths so they terminate insidethese narrow tubes. The corners of the narrow inner tubes are thenpropagated around the N turns of the spiral until they hit the cornersof the narrow outer tubes. This creates a 2 θ -nice mesh of the spiralusing O ( nN/θ ) pieces. The figure shows p = 9 and N = 6.If N & θ − ≥ / sin( θ ) then after O ( θ − ) spirals, we can create a θ curve that isa closed loop and we let the paths generated by the interior corners of the inner thinpart hit this closed loop. We create another θ -bent closed loop at radius N − O ( pθ − ) verticesare used. See Figure 33.The propagation paths cut the spiral into 2 θ -nice triangles and quadrilaterals.Moreover, as in the case of simple tubes, it is easy to check that each dissection pieceis crossed by at most 1 /θ paths in the construction of each spiral. Since there are O ( n ) spirals, this means there are at most O ( n/θ ) such crossings of a dissection piecein total. The propagation paths that enter each spiral cross each dissection piece atmost once, and there are O ( n/θ ) such paths in total, hence O ( n/θ ) such crossings ofeach piece.This completes the proof of Lemma 9.1, and hence the proof of Theorem 1.3.13. A lemma for quadrilateral meshing We now restate our conclusions in a form that is useful for proving the theorem onoptimal quad-meshing in [11]. Readers interested only in Theorem 1.1 may skip thissection. Figure 33. If the winding number is much larger than θ , then after O (1 /θ ) windings we can create a closed θ -bent loop (the dashed curve).We then propagate the corners of the narrow inner tunes until they hita vertex of this closed loop. We can also create a θ -bent loop onewinding in from the outer tube and use it similarly to propagate thecorners of the narrow outer tubes until they hit a vertex of this loop. Theorem 13.1. Suppose that W is a polygonal domain with an isosceles trapezoiddissection with n pieces. Suppose also that ◦ ≤ θ ≤ ◦ and that every dissectionpiece is θ -nice. Finally, suppose the number of chains in the dissection is M . Thenwe can remove O ( M/θ ) θ -nice quadrilaterals of uniformly bounded eccentricity from W so that the remaining region W ′ has a θ -nice quadrilateral mesh with O ( nM/θ ) elements. At most O ( M/θ ) new vertices are created on the Q -boundary of W ′ . Atmost O ( M ) vertices are created on the P -boundary of W ′ , and no more than O (1) vertices are placed in any single P -side of any dissection piece of W ′ . For this quad-mesh, any boundary point on a Q -side of W ′ propagates to another boundary pointafter crossing at most O ( n ) quadrilaterals.Proof. The proof is exactly the same as the argument of the last few sections, exceptfor some slight modifications inside the return regions.For each return region we place O (1 /θ ) equally space points along the two Q -sides of the region and propagate these outside the return regions until they hit theboundary of Ω or hit the Q -side of some return region. There are O ( M/θ ) suchpaths and they generate at most O ( M n/θ ) quadrilaterals and O ( M/θ ) endpoints on Q -sides of ∂ Ω. ONOBTUSE TRIANGULATIONS OF PSLGS 41 First consider return regions that are simple tubes. As before, split each suchregion into O (1 /θ ) parallel sub-tubes and so that in each sub-tube we can connectopposite corners by a θ -path. Now, however, we remove a small quadrilaterals ata pair of opposite corners of the tube. These quadrilaterals have one edge on a P -boundary of the tube, one edge on a Q -end of the tube, one vertex in the interior ofthe tube and the two edges adjacent to this vertex are chosen to lie a P -segment anda Q -segment. See Figure 34. Figure 34. We place quadrilaterals (shaded) at opposite corners of atube, and connect the internal corners by a θ -path. Every path enteringtube either immediately hits the shaded quadrilateral at that end, orpropagates to hit the shaded quadrilateral at the other end. We alsohave to propagate the interior corner of the shaded quadrilateral alonga Q -path. This gives a mesh of every tube by 2 θ -nice quadrilaterals.We then connect the interior corner of each of the two quadrilaterals by a θ -curve.This requires less displacement than connecting the corners, so it is clearly possibleto do this (to make it easier to see, we could always increase the number of tubesand decrease their width by a fixed factor). We have freedom in choosing the sizeof the quadrilaterals, and so we can arrange for all the quadrilaterals chosen in thesame dissection trapezoid to have sides along the same Q -segment. Thus when we Q -propagate the corners of the quadrilaterals, only two extra points will be createdon the P -side of the dissection piece containing such a quadrilateral.If we apply quadrilateral propagation to each P -path entering the tube from eitherend, it crosses the tube and hits a Q -side of the removed quadrilateral at the other end of the tube. See Figure 34. This gives a 2 θ -nice quadrilateral mesh inside themodified tubes.Inside the spirals we do a similar thing. In the previous proof, paths inside spiralswere terminated by bending them in a sub-tube of the spiral until they hit a corneron the opposite side of the tube from where they entered, in order to form a loop.so the same construction works. Outside the return regions, the P -paths convert the θ -nice dissection into a θ -nice quadrilatal mesh (previously the only triangles createdby the P -paths were in triangular pieces of the dissection, which we now assume don’texist). See Figures 35 and 36. (cid:3) Figure 35. For spirals with ≫ θ − windings, we can make a θ -pathloop in the j th spiral when j & /θ (solid thick curve). Then placea quadrilateral as shown with one edge on the loop; one corner is θ -propagated around the spiral once to hit the center of a side of thesame quadrilateral (thin dashed curve). The boundary of the spiral is θ -bent to hit the other corner (thick dashed curve).14. Overview of the proof of Theorem 1.1 The remainder of the paper gives the proof of Theorem 1.1. In this section wegive the overall strategy of the proof and we will provide the details in the followingsections.The proof combines ideas already seen in the proofs of Theorems 1.2 and 1.3 butrequires a different displacement estimate in tubes and a more intricate construction ONOBTUSE TRIANGULATIONS OF PSLGS 43 Figure 36. The quadrilateral construction near the outside of a largespiral. This is similar to the construction in Figure 35, but we can do itin the sub-tube adjacent to the outermost one. The two constructionsgive a 2 θ -nice mesh of the entire spiral (minus the two quadrilaterals).in the spirals. As explained in Section 2, it suffices to prove Theorem 2.3: assumeΓ is a triangulation and show we can place O ( n . ) points along the edges so thateach triangle becomes Gabriel. As in the proof of Theorem 1.2 we start taking thedissected domain Ω to be the original triangles { T k } with the central triangles { T ′ k } removed (recall the vertices of T k are the three points where the inscribed circletouches the triangle T k ). We do not use the “approximate circular-arc triangles” thatwere used in the proof of Theorem 1.3.For each triangle T k , remove the closed triangle T ′ k as in Section 4. As before, T k \ T ′ k is a union of three isosceles triangles. Keep the isosceles triangles with angle < ◦ ; as explained at the end of Section 4, isosceles triangles with angles ≥ ◦ canbe ignored because adding any set of points to the Q -edges will make the triangleGabriel. The remaining region Ω thus has an isosceles dissection by O ( n ) acutetriangles. We construct return regions for Ω just as before.Each vertex of each T ′ k are propagated by P -paths until then leave Ω or hit the Q -side of a return region. This creates O ( n ) crossing points on Γ.If a return region has k isosceles pieces then we will place O ( √ k ) even spaced pointsin each Q -end of the region and propagate these until they leave Ω or hit a returnregion. Since k = O ( n ), this creates at most O ( n . ) new points. If different return regions had to use distinct isosceles pieces this estimate would be O ( n ) instead. Im-proving the exponent in Theorem 1.1 seems to be entirely a matter of understandingthe behavior of distinct return regions that share isosceles pieces.Why do we split the Q -ends of the return regions into O ( √ k ) pieces? When we bendthe P -paths inside the return regions, we must verify that the Gabriel condition issatisfied by the points that we generate. This is a more restrictive condition than the θ -bending of the earlier proof, so paths can be bent less and hence take a more stepsto terminate. The difference is illustrated in Figure 37. The left side shows the rangeof options for a θ -segment crossing a single rectangle; the allowable displacement isroughly θ | a − b | . The center and right pictures of Figure 37 show the restrictions ona Gabriel path. Note that there are two such restrictions: the exit point b must bebetween the Gabriel disks tangent at a and the entrance point a must be between thedisks tangent at b . This restricts b to an interval of length approximately | a − b | /w ,where w is the width of the piece. This estimate will be made more precise in thenext section; the main point is that it shrinks quadratically with | a − b | whereasthe estimate for θ -paths decreased linearly. Thus the proof of Theorem 1.1 requireslonger, narrower tubes than the proof of Theorem 1.3.To illustrate the idea, consider a simple case: a square divided into k thin parallelrectangles. See Figure 38. A Gabriel path crossing the square takes k steps, eachwith displacement ≃ /k , so the total displacement is ≃ /k . At first glance, thisseems to say we should cut the square into O ( k ) parallel tubes; then we could get allentering paths to terminate before hitting the far side of tube. This works, but leadsto the estimate O ( n ) in Theorem 1.1.We can do better. Cut the square into √ k tubes instead. Now the tangent diskshave diameter k − / and the cusp regions where we choose our next point have height ≃ √ k ( √ k / k ) = k − / . Thus a Gabriel path takes k steps, each with displacement ≃ k − / , for a total displacement ≃ k − / , which is the approximate width of thetube. Thus using only O ( √ k ) tubes, we can bend Gabriel paths enough to hit thefar corner of the tube (and thus terminate).We shall prove in the next two sections this holds for any return regions that aresimple tubes, not just squares with rectangular pieces. Each return region that is aC-tube, S-tube or simple G-tube will be split into O ( √ k ) narrow parallel tubes and ONOBTUSE TRIANGULATIONS OF PSLGS 45 a b a b a b Figure 37. A θ -bent path can reach any point defined by a conewith angle 2 θ , but a Gabriel bent path can only reach points definedby the cusp between two tangent disks. Moreover, this is a two partcondition: the exit point must be the cusp defined by the entrancepoint and the entrance point must be in the cusp defined by the exitpoint. Figure 38. A Gabriel-bent path must stay outside certain pairs oftangent disks. When we cut a unit square into k × O (1 /k ). If we cut thesquare into O ( √ k ) horizontal tubes, paths in each tube take k stepswith displacement of k − / and hence total displacement of 1 / √ k . Sincethis is also the width of the tube, we can Gabriel-bend a path to hitthe side of a tube before leaving it.the entering propagation paths will be Gabriel bent until they a far corner of thetube; here k is the number of isosceles pieces forming the tube. We also place O ( √ k ) narrow parallel sub-tubes at the two ends of spiral returnregions, i.e., we subdivide the innermost and outermost windings of the spiral. Aswith simple tubes, all paths entering the spiral can be bent within these narrow tubesto terminate within O ( k ) steps. But then we have to propagate both the externaland internal corners of the narrow tubes. The external corners propagate outside thespiral until they terminate just as described for the corners for narrow tubes in theprevious paragraph.The most difficult part of the proof of Theorem 1.1 is dealing with the √ k internalcorners that propagate through the spiral; since we have no bound for the number N of windings of the spiral in terms of n , this could produce arbitrarily many newvertices. Thus propagation paths of the internal corners must be bent to terminateearlier. Consider the case of paths that start near the inner end of the spiral (theouter part is handled in the same way, but is easier, since the windings of the spiralare longer). We consider what happens for very large spirals (where the number ofwindings N is bigger than the number k of isosceles pieces in the spiral; for smallervalues of N we truncate the construction at the appropriate stage.)We first bend the propagation paths so that adjacent paths merge, and then mergeadjacent merged paths, and continue until all the propagation paths generated by the O ( √ k ) internal corners have merged into a single path. This occurs around winding k / . This path is then propagated as a P -path out to winding k / . See Figure 39.At this stage we have enough freedom to bend the curve to hit itself, forming aclosed loop that wraps once around the spiral. This is similar to what we did in theproof of Theorem 1.3, but in this case, in order for this closed loop to be Gabriel, theremust be another (larger) closed loop parallel to it. This did not occur in Theorem1.3. This constraint requires us to construct a sequence of parallel closed loops in thespiral between windings k / and k . The closed loops gradually can become fartherand farther apart; only O ( √ k ) loops are used in all. At winding k , there is no needfor a “next” loop and the sequence of closed loops ends. The part of the spiralbeyond winding k is an “empty” region until we reach a closed loop coming from theanalogous construction in the outer half of the spiral.In the remainder of the paper we give the details of the argument sketched above. ONOBTUSE TRIANGULATIONS OF PSLGS 47 Figure 39. This illustrates the stages in the spiral construction.First (light gray), we divide the tube into thinner tubes and enteringpaths are bent to hit the sides of these. Next (white), the thin tubesare bent and collapsed in pairs; in this figure four tubes are mergedinto one after two windings. In the third stage (gray), the single tubeis propagated until we can bend it to intersect itself. Next (white)comes a sequence of closed loops that gradually grow further apart.Finally we reach the empty region (gray), where no paths propagate.This figure gives a rough idea of the construction, but scales have beendrastically compressed to make all the stages visible in the same picture.15. Gabriel bending in isosceles pieces This section contains the main estimate used in proof of Theorem 1.1.Suppose a and b are endpoints of a P -path in an isosceles piece T . If we keep a fixed, how far we can move b and still have the Gabriel condition hold? Moreprecisely, can we find an ǫ > Q -side as b that arewithin distance ǫ of b can be connected to a by a Gabriel segment? If this holds wesay that the allowable displacement for the piece is at least ǫ . Lemma 15.1. Suppose T is an isosceles piece of width w . Suppose [ a, b ] is a P -segment crossing T and R is the distance of a from the vertex of the piece ( R = ∞ ifthe piece is a rectangle). Then if c is a point on the same Q -side as b and is withindistance ǫ = | a − b | w , R ) , (15.1) of b , then [ a, c ] is a Gabriel segment crossing T . In particular, the allowable displace-ment is at least ǫ .Proof. First suppose the isosceles piece is a rectangle ( R = ∞ ). Consider disjointsub-segments of the Q -side containing a that have a as a common endpoint andconsider the disks D , D with these segments as diameters. See Figure 40. Assumethese disks have radii r and s . The diameters of these disks are disjoint segmentsthat both lie on the same non-base side of an isosceles piece, so their length adds upto be less than the width of the piece, i.e., 2 r + 2 s ≤ w . Thus max(2 r, s ) ≤ w . sr abc d wD D Figure 40. The segment [ a, b ] is a P -segment for a rectangular piece.A simple estimate shows | c − b | ≥ | a − b | /r and | d − b | ≥ | a − b | /s .Suppose [ c, d ] is the Q -segment containing b that is disjoint from these disks (again,see Figure 40). We want to estimate | c − b | and | d − b | from below. Such a lowerbound gives the desired lower bound on the allowable displacement.If the disk D is too small, i.e., r < | a − b | , then the disk D does not hit the Q -side containing b and the Gabriel condition is automatically satisfied. Thus wemay assume r ≥ | a − b | . Then by the Pythagorean theorem | c − b | = r − p r − | a − b | , ONOBTUSE TRIANGULATIONS OF PSLGS 49 or (using 1 − √ − y ≥ y/ , r | c − b | = 1 − r − | a − b | r ≥ | a − b | r so | c − b | ≥ | a − b | / r . The calculation for the other disk is identical, so the twodisks omit all points within distance ǫ = | a − b | min( 1 r , s ) ≥ | a − b | w . of b . Since 1 /R = 0 in this case, this implies (15.1).Next we consider what happens when the piece is not a rectangle. To be concrete,we assume one Q -side lies on the real axis, the vertex of the piece is at − R and the P -path connects a = 0 to b in the upper half-plane. Suppose the piece has angle θ . Some elementary trigonometry shows that the disk D does not hit the Q -sidecontaining b if (see Figure 41) r < ( R − r ) sin θ. r s θ abc dR R−rR+s Figure 41. If r, s are small enough then the Gabriel disks for one Q -side don’t hit the other Q -side.Since | a − b | = 2 R sin θ/ 2, this is equivalent to r < R sin θ θ = | a − b | sin θ θ/ θ ) . (15.2)By the double angle formula, for 0 ≤ θ ≤ π/ θ = 2 sin θ cos θ , sosin θ θ/ θ ) = cos θ/ θ ) ≥ cos π/ π/ 2) = 12 √ . (15.3) Hence (15.2) holds if r < | a − b | / √ 2. If this condition holds, then the point c is acorner of the piece T , so the estimate holds trivially to the left of b . Therefore we mayassume r ≥ | a − b | / √ w ≥ | a − b | / √ r ≤ w/ D does not hit the opposite Q -side if s < | a − b | sin θ sin( θ/ − sin θ ) = 2 | a − b | cos θ/ − sin θ . See Figure 41. The trigonometric function on the far right is increasing for θ ∈ [0 , π/ θ = 0. Hence D doesnot hit the opposite Q -side if s < | a − b | . In this case d is a corner of the isoscelespiece and the lemma holds trivially to the right of b . Therefore we may assume s ≥ | a − b | in what follows.Now suppose we have an isosceles piece with angle θ > 0. We normalize the pictureas in Figure 42 with one Q -side along the real axis, the vertex of the piece at − R .The other Q -side is labeled L . We consider a P -path with one endpoint at the originand the other endpoint (labeled b in the figure) on L in the upper half-plane. Wealso consider disks D , D , D centered at points − R , − r , s on the real line that aretangent at the origin. We let [ c, d ] be the segment of L \ ( D ∪ D ) that contains b .See Figure 42. D D D R θ r sb =0 ac d L Figure 42. We want a lower bound on | b − c | and | b − d | in termsof | a − b | , r , s and R . We prove this by applying the transformation z → /z to this picture, to get the picture in Figure 43. ONOBTUSE TRIANGULATIONS OF PSLGS 51 s r bL c dD D D > 1/(4r) > 1/(2s) 12R 0 Figure 43. This is the inversion of Figure 42. The point a = 0maps to a ′ = ∞ and the circles through 0 map to vertical lines, whosedistance apart is easy to compute. The line L maps to a circle L ′ .Now apply the map z → /z . The origin is mapped to infinity and the circlescentered on the real line passing through 0 now map to vertical lines. (The map z → /z is a linear fractional transformation that maps 0 to ∞ , so circles through 0map to circles through ∞ , i.e., lines.) See Figure 43. The boundary of D maps to L = { x = − / r } , the boundary of D maps to L = { x = 1 / s } and the boundaryof D maps to L = { x = 1 / R } . Since R ≥ r , we have the distance between L and L is at least 1 / r .The line L is distance ρ = R sin θ from the origin, whereas | a − b | = 2 R sin( θ/ , so by a similar calculation to (15.3), we get ρ = | a − b | sin θ θ/ ≥ | a − b |√ , since we assume 0 ≤ θ ≤ π/ 2. This means that on the line L , the derivative of 1 /z is bounded above by 2 / | a − b | . Therefore the image of the segment [ c, b ] has length at most 2 | c − d | / | a − b | . Thus | c ′ − b ′ | ≤ | c − d || a − b | . However, the image of this segment is a circular arc that connects the lines L and L and hence has length at least | c ′ − b ′ | ≥ | r − R | ≥ r since r ≤ R/ 2. Combining these inequalities gives2 | c − b || a − b | ≥ r . Since r ≤ w/ r ≤ R/ | c − b | ≥ | a − b | r ≥ max( | a − b | w , | a − b | R ) . Similar calculations show 2 | d − b || a − b | ≥ | b ′ − d ′ | , and | b ′ − d ′ | ≥ R + 12 s ≥ R + 1 w from which we deduce | d − b | ≥ | a − b | R + 1 w ) ≥ | a − b | R , w ) . (cid:3) Gabriel bending in tubes Next we apply the Cauchy-Schwarz inequality to our displacement estimate forpieces, to get a displacement estimate for tubes: Lemma 16.1. Suppose T is a tube of width w , minimal-length L = ˜ ℓ ( T ) and consistsof p isosceles pieces. Let a, b be points on opposite ends of T that are connected by a P -path in T . Then a can be connected by a Gabriel path to any point on the oppositeend of T that is within distance d = L / (4 pw ) of b . ONOBTUSE TRIANGULATIONS OF PSLGS 53 Proof. For the j th piece in the tube, let ℓ j be the length of the piece (the shorterof its two base lengths, zero for triangles). Then L = P j ℓ j by definition. By theCauchy-Schwarz inequality L = ( X j ℓ j ) ≤ ( X j ) · ( X j ℓ j ) = p X j ℓ j . The allowable displacement of each piece is at least ℓ j / w , so the total allowabledisplacement is at least X j ℓ j w ≥ L pw . (cid:3) Corollary 16.2. If T is a tube with minimal-length L , width w composed of p piecesand w ≤ L/ √ p , then opposite corners of the tube can be connected by a Gabriel pathinside the tube.Proof. By assumption we have L ≥ w √ p so the allowable displacement is at least(2 w √ p ) wp ≥ w, where w is the width of the tube. Thus opposite corners can be connected. (cid:3) Corollary 16.3. Suppose T is a C-tube, S-tube or simple G-tube composed of p isosceles pieces and we cut T into M parallel, equal width sub-tubes with M ≥ √ p .Then the opposite corners of each sub-tube can be connected by a Gabriel path in thatsub-tube.Proof. Since M ≥ 2, each sub-tube is half of a wider tube inside T and hence haslength that is at most four times its minimal length by Lemma 7.1. Moreover, thelength of each sub-tube is bounded below by the length of T . By Corollary 8.5 everysimple return tube has length that is at least its width w and hence its minimal-lengthis at least w/ 4. Thus each sub-tube has minimal-length at least M/ ≥ √ p timeslonger than its width. The conclusion then follows from Corollary 16.2. (cid:3) Gabriel bending in spirals Finally we have to consider bending in pure spirals. This is the final, but mostcomplicated, step in the proof of Theorem 1.1, and we break the construction intoseveral steps described in different sections. Lemma 17.1. Suppose S is a pure spiral made up of at most k isosceles pieces.Then we can mesh the interior of the spiral using at most O ( k . ) quadrilaterals andtriangles so that the added vertices make every isosceles piece in the spiral Gabriel.Also, every path entering the spiral can be Gabriel bent to terminate within one wind-ing. The bound on the number of points added is independent of N , the number ofwindings of the spiral. Without loss of generality we will assume the bound k on the number of isoscelespieces is a power of 4, k = 4 K = 2 K and that it is at least 16 times larger thanthat actual number of isosceles pieces in the spiral. We do this so that we can applyCorollary 16.3 with the value M = √ k instead of M = 16 √ k . This will make notationslightly easier and does not affect the statement of the lemma.For simplicity, we rescale so that the entrance and exit segments of the spiral havelength 1, i.e., the width of the tubes in the spiral is w = 1. The spiral is a topologicalannulus, with one bounded and one unbounded complementary component. Thetwo ends of the spiral corresponding to these two regions will be called the inner-end and outer-end respectively. The spiral is a union of N simple G-tubes joinedend-to-end. We number these consecutively starting at the inner-end and denotethem T , . . . , T N . We will call T and T N the innermost and outermost tubes (orwindings) respectively.If the number of windings satisfies 1 < N ≤ 100 then we can treat the spiral likea simple tube. We divide both T and T N into √ k = 2 K parallel sub-tubes just aswe did for simple G-tubes. We call these the narrow tubes . Then every P -pathentering the innermost tube either hits a corner of one of the narrow tubes, or itenters one of the narrower sub-tubes. In the latter case, it can be Gabriel bent tohit a far corner of this narrow sub-tube. Similarly for paths entering the outermosttube. The √ k corners of the narrow tubes are propagated as P -paths through the ONOBTUSE TRIANGULATIONS OF PSLGS 55 spiral until they hit the corners of the narrow tubes at the other end. This createsat most O ( k . ) vertices in the spiral. See Figure 44. Figure 44. For small spirals (less than 100 windings) we divide thetubes into √ k narrow sub-tubes. Every entering path can be Gabrielbent to terminate within one winding. Here we show N = 2.For larger N there are four different types of construction we will use, dividedby certain “phase transitions” in our ability to bend curves as N increases. Thetransitions occur at N ≃ k / , k / and k , as described in the next few sections.18. 100 < N ≤ k / : Dyadic merging of paths We now consider the case 100 < N ≤ k / . As above, we subdivide the innermostand outermost tubes into √ k = 2 K narrow, parallel sub-tubes, we terminate allentering paths within these narrow sub-tubes and we propagate the 2 K corners ofthe narrow sub-tubes through the spiral. However, if we propagate each corner of anarrow tube as a separate path all the way through the spiral we will generate toomany new vertices. The solution is to merge paths as they propagate through thespiral.The general idea is to take pairs of adjacent paths and bend them towards eachother until they meet; this reduces the number of paths from √ k to √ k/ 2. Wethen define new adjacent pairs and bend these towards each other, until they merge,creating √ k/ as pairs in the previous stage, we need a total displacement that is twice as long asbefore. Moreover, since the tubes are twice as wide, the allowable displacement ishalf as large per piece. Since the formula for displacement in Lemma 16.1 involves afactor of L , the length only needs to double to merge the next set of tubes.However, this does not mean we need twice as many windings. Since the windingsof the spiral become longer as we move outwards, fewer windings are needed to achievea given length, so the number of pieces crossed grows, but does not double, i.e., itgrows at a geometric rate that is strictly less than 2. This will allow us to obtain thedesired bound. We do the same construction starting at both ends of the spiral, sothat when paths meet at the center of the spiral, the surviving paths from both endsmatch up. Next we give the details about how the merging and matching processworks. See Figure 45.If N is even write N = 2 M ; otherwise write N = 2 M + 1. We will give the detailsfor the merging process in the inner-half of the spiral, i.e., T ∪ · · · ∪ T M . The sameprocedure is used in the outer half T N − M ∪ . . . T N ; these tubes are all longer than thecorresponding tubes in the inner half of the spiral, and so the allowable displacementswill be larger; thus any displacement that can be achieved in the inner half can also(more easily) be attained in the outer half. Thus we can assume the merged pathsin the inner half terminate at the same points as the merged paths in the outer halfwhen N is even and that they can be joined by P -paths in T M +1 when N is odd.Let λ = 2 / < λ j be the integer part of λ j . Let S j = T λ j +1 ∪ · · · ∪ T λ j +1 ,i.e., this is a sub-tube of the spiral that goes from winding λ j to winding λ j +1 . Bythe argument proving Lemma 8.6 this part of the tube has length at least λ j +1 X i = λ j +1 (2 i − ≥ λ j +1 − ( λ j + 1) ≥ ( λ j +1 − − ( λ j + 1) = [ λ j +2 − λ j +1 + 1 − λ j − λ j − λ j [ λ − λ − j +1 − − λ − j ]An explicit calculation shows that for j = 5 λ − λ − j +1 − − λ − j ≈ . ≥ , and since this function is increasing in j , we deduce that S j has length ≥ λ j if j ≥ λ = 2 / ≈ . ≥ ONOBTUSE TRIANGULATIONS OF PSLGS 57 Figure 45. Between windings 100 and 16 k / we merge the pathsgenerated by the internal corners of the narrow tubes at the entrancesto the spiral. Here we show four tubes merging into two tube in the firstwinding and the remaining two tubes merging into one in the secondwinding. (In general though, it will take more windings to merge widertubes.)Suppose S j is divided into 2 m parallel sub-tubes of width w = 2 − m . By Lemma16.1 a path entering one of these sub-tubes can be Gabriel bent within the tube tohit either of the far corners of the tube if w ≤ L pw . In this case L ≥ λ j , p = kλ j (each piece is repeated in the tube kλ j times; once perspiral), and w = 2 − m , so this inequality becomes2 − m ≤ ( λ j ) λ j k − m and this occurs iff 4 k − m ≤ λ j . Since 2 j = λ j and k = 2 K , this is equivalent to2 − m ≤ j − / √ k = 2 j − − K . So for j ≥ S j into 2 K − j +1 parallel sub-tubes. The tube S j shares anend with S j − . The corners of the narrow sub-tubes of S j on this end form a subsetof the corners of the narrow sub-tubes of S j − on the shared end. The remainingcorners are mid-points of one end of the narrow tubes in S j and can be propagated by Gabriel paths through the narrow sub-tubes of S j so that they hit far corners ofthese sub-tubes, merging another set of paths.The total number of vertices created in S j is therefore O ( λ j · K − j · k ) (the numberof windings of the tube, times the number of paths propagating through the tube,times the number of pieces crossed in each winding). The inner half of the spiralcontains tubes S , . . . , S m as long as λ m = 2 m/ < M , or m ≤ log M . Summingover all j ∈ [5 , m ] gives the upper bound O ( log M X j =5 λ j K − j K ) = O (2 K ∞ X j =5 j − j ) = O (2 K ) = O ( k . ) . Therefore if N = O ( k / ), we can terminate every path entering the spiral using only O ( k . ) new vertices in the spiral.19. 16 k / < N ≤ k / : a single path If 16 k / < N ≤ k / we duplicate the construction of the previous section upto and including S K +5 and after this point we simply let the single remaining pathpropagate as a P -path. Since the path winds around the spiral at most √ k times,at most O ( k . ) new vertices are created between 2 k / and k / , and the path in theinner half eventually hits the analogous path from the outer half of the spiral.20. 8 k / < N ≤ k : multiple closed curves When j = 8 √ k we hit a new transition point: S j now has length 8 √ k = 2 K +3 ,hence minimal length ≥ √ k . Hence Lemma 15.1 says that if j ≥ √ k , then theallowable displacement in S j is at least˜ ℓ ( S j )4 k = (2 j ) k ≥ , i.e., the allowable displacement is larger than the width of the tube, so we can connectopposite corners of S j by a Gabriel path (Gabriel assuming the tangent disks havediameter 1, the width of the tube).The idea is that for j ≥ √ k we cut the spiral by closed curves γ j in S j . See Figure46. If we choose one such curve from every S j for √ k ≤ j ≤ k then we generate ≃ k curves, each with k vertices, giving a total of k new vertices. This will eventuallylead to a O ( n ) estimate in Theorem 1.1 (still a polynomial bound, but larger than we ONOBTUSE TRIANGULATIONS OF PSLGS 59 want). Instead, we will select a subsequence of tubes { S j p } to contain closed curves.We want to choose at most O ( √ k ) indices between √ k and k and we want γ j p to beGabriel with respect to annular region bounded between γ j p − and γ j p +1 . The lengthof the Q -segments connecting these two curves is exactly δ p ≡ j p +1 − j p − and theminimal-length of S j p is at least j p , so by Lemma 16.1, (taking L = j p / p = k , w = δ p ) the Gabriel condition can be met if( j p / kδ p ≥ , or equivalently δ p ≤ j p k Figure 46. Between windings k / and k we use closed Gabriel loopsthat start with unit spacing near k / and eventually reach spacing ≃ k .We shall see in the next section that beyond radius ≃ k , no more loopsare needed.Suppose we have an increasing sequence of integers { j i } Pi =1 with(1) j = √ k = 2 K ,(2) j P = k = 2 K ,(3) P = O ( √ k ),(4) j i +1 − j i ≤ k j i .The fourth condition implies | j i +1 − j i − | ≤ | j i +1 − j i | + | j i − j i − | ≤ k ( j i + j i − ) ≤ k j i , since j i − ≤ j i . So if we can find such a set of integers, we will have constructed thedesired closed loops.We divide the interval [ √ k, k ] = [2 K , K ] into geometrically increasing blocks[2 K , K +1 ] , . . . , [2 K + q , K + q +1 ] , . . . , [2 K − , K ] . In the q th block we can take our sequence to be separated by gaps of size k (2 q √ k ) =2 q − and hence 2 K +7 − q evenly spaced integers suffice to cross this block with the de-sired spacing. Summing over all q ’s shows that at most K X q =0 K +7 − q ≤ · K = 128 √ k, integers j p are needed. Thus at most O ( k . ) new vertices are created in this phaseof the construction. 21. N > k : the empty region In the previous section we created Gabriel loops in the spiral, assuming there wereother loops nearby to prevent the Gabriel disks from getting too large; thus we neededa sequence of larger and larger loops. However, once j ≥ k , we no longer need tolimit the size of the Gabriel disks and this sequence of loops can end. More precisely, Lemma 21.1. Suppose the spiral has k pieces and the spiral has more than k/ windings. Then each sub-tube T j with j ≥ k/ is crossed by a closed loop so that thevertices of the loop make each isosceles piece Gabriel.Proof. As above, we assume T j has width 1. Lemma 15.1 implies that the allowabledisplacement for an isosceles piece is at least | a − b | / R , where R is the distance tothe vertex of the piece. We also have | a − b | = 2 R sin( θ/ 2) where θ ∈ [0 , π ) is theangle of the piece. Since sin( θ/ ≥ √ θ/π , the allowable displacement for such apiece is at least R sin ( θ/ ≥ R π θ . If we only bend the path for pieces where the vertex direction points towards thecenter of the spiral, the sum of the angles is at least 2 π , so by the Cauchy-Schwarzinequality, the sum of the angles squared is at least (2 π ) /k . Hence the total allowabledisplacement in the tube T j is at least 8 R/k times the minimal-length of T j . In T j ONOBTUSE TRIANGULATIONS OF PSLGS 61 every piece with vertex towards the center of the spiral has R ≥ m − ≥ m/ ≤ π/ 2, there are at least 4 of them) so the allowabledisplacement is at least 4 m/k ≥ 1, the width of T j . (cid:3) Up to this point, we have divided the spiral into inner and outer halves, but at thisstage this is not necessary. The outermost tube contains a Gabriel curve that connectsopposite corners, forming a loop and every entering path can be Gabriel propagatedto hit this same corner. 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