Some Counterexamples for Compatible Triangulations
Cody Barnson, Dawn Chandler, Qiao Chen, Christina Chung, Andrew Coccimiglio, Sean La, Lily Li, Aïna Linn, Anna Lubiw, Clare Lyle, Shikha Mahajan, Gregory Mierzwinski, Simon Pratt, Yoon Su, Hongbo Zhang, Kevin Zhang
SSome Counterexamples for Compatible Triangulations
Cody Barnson Dawn Chandler Qiao Chen Christina Chung Andrew Coccimiglio Sean La Lily Li A¨ına Linn Anna Lubiw Clare Lyle Shikha Mahajan Gregory Mierzwinski Simon Pratt Yoon Su (Matthias) Yoo Hongbo Zhang Kevin Zhang December 16, 2016
Abstract
We consider the conjecture by Aichholzer, Aurenhammer, Hurtado, and Krasser that anytwo points sets with the same cardinality and the same size convex hull can be triangulatedin the “same” way, more precisely via compatible triangulations . We show counterexamples tovarious strengthened versions of this conjecture.
Let P and Q be two point sets, each with n points, and with h points on their convex hulls.Triangulations T P of P and T Q of Q are compatible if there is a bijection φ from P to Q such that abc is a clockwise-ordered triangle of T P if and only if φ ( a ) φ ( b ) φ ( c ) is a clockwise-ordered triangleof T Q .A main application of compatible triangulations is the problem of morphing. If we have com-patible triangulations of two point sets in the plane then the area bounded by the convex hull ofthe first set can be morphed to the area bounded by the convex hull of the second set. In the caseof very similar point sets, a linear motion of each triangle will work. For the general case, there aremore sophisticated planarity-preserving morphs [6, 2]. Department of Mathematics and Computer Science, University of Lethbridge, [email protected] School of Computing Science, Simon Fraser University, [email protected] School of Computing Science, Simon Fraser University, [email protected] Department of Computer Science, University of Toronto, [email protected] School of Computing Science, Simon Fraser University, [email protected] Department of Mathematics, Simon Fraser University, sean [email protected] Department of Mathematics and the School of Computing Science, Simon Fraser University, [email protected] Department of Computer Science, McGill University, [email protected] Cheriton School of Computer Science, University of Waterloo, [email protected] School of Computer Science, McGill University, [email protected] Cheriton School of Computer Science, University of Waterloo, [email protected] Department of Computer Science, Bishop’s University, [email protected] Cheriton School of Computer Science, University of Waterloo,
[email protected] Department of Computer Science, University of Manitoba, [email protected] Cheriton School of Computer Science, University of Waterloo, [email protected] Faculty of Applied Science, University of British Columbia, [email protected] a r X i v : . [ c s . C G ] D ec n 2003 Aichholzer, Aurenhammer, Hurtado, and Krasser [1] conjectured that compatible tri-angulations always exist: Conjecture 1. If P and Q are point sets in general position (i.e. with no 3 points collinear) withthe same cardinality and the same size convex hull then they have compatible triangulations. We prove that several strengthened forms of the conjecture are false. A triangulation of a finite set of points, P , in the plane is a maximal set of segments pq , p, q ∈ P such that no two segments ab and cd cross , i.e. intersect in a point that is not a common endpointof the segments. We assume that the points P do not all lie on one line. Then a triangulation is aplanar graph whose interior faces are all triangles [5].From the definition we immediately get: Claim 1.
Consider a set of points P . If a, b ∈ P have the property that the interior of segment ab is not intersected by any other segment pq with p, q ∈ P , then ab must be part of any triangulationof P . This claim implies that the convex hull segments are part of any triangulation. We can also usethe claim to show that certain other segments must be in any triangulation:
Claim 2.
Let p be a point of the convex hull of point set P . Suppose that when we remove p from the point set, the interior points q , . . . , q t become convex hull vertices. Then the segments pq i , i = 1 . . . t must be part of any triangulation of P .Proof. Each segment pq i satisfies the condition for Claim 1. Aichholzer et al. [1] gave a counterexample to show that Conjecture 1 fails if there are collinearpoints. Their example has n = 7 and is shown in Figure 1. In this section we give a smallercounterexample with n = 6. We also give an alternate counterexample with n = 7. Furthermore,we show that there is no counterexample with n ≤ P and Q . xyz ab c ab c Figure 3: T P , T Q , and T Q .Let P and Q be the left and right point sets of Figure 2, respectively. We claim that a compatibletriangulation between P and Q does not exist. To do this, first we will prove that the triangulations T P , T Q and T Q as shown in Figure 3 are the only triangulations for their point sets. Then, wewill show that T P is compatible with neither T Q nor T Q .First consider point set P . Observe that every edge of triangulation T P is forced by Claim 1.Now consider point set Q . Two edges from each point on the convex hull to a, b and c areforced by Claim 2. If we add an additional edge from a convex hull point to an interior point weget triangulation T Q , and its rotationally symmetric copies. If we do not add any additional edgesfrom convex hull points then triangulation T Q is the unique possibility.To show that P and Q are not compatible, we look at the degrees of the hull points of theirtriangulations. T P ’s hull points have degrees of 3, 5, and 5. T Q ’s have degrees 4, 4, and 4, and T Q ’s have degrees 4, 4, and 5. Therefore, the triangulation T P is compatible with neither T Q nor T Q . Since these are the only triangulations, we can conclude that no triangulation exists betweenthe point sets P and Q .We give another counterexample with n = 7 in Figure 4. In this subsection we prove that the above counterexample with n = 6 is the smallest possible. Inother words, we prove that any point sets with n ≤ Figure 4: Another counterexample to Conjecture 1 for the case of collinear points. Observe thatthe left hand triangulation is unique and point x has degree 6, but no point on the right can havedegree 6 in any triangulation.sets with n ≤ Theorem 1.
If two points sets of size n (possibly with collinear points) have 1 or 2 internal points,then they have a compatible triangulation. Aichholzer et al. [1] proved this for point sets with no collinearities. We give a proof that handlescollinearities. In fact they proved the stronger result that there is a compatible triangulation evenwhen the convex hull mapping is forced, i.e. when one point of the convex hull of the first point setis mapped to one point of the convex hull of the second point set. However, that stronger result isno longer true in the presence of collinearities as shown in Figure 5. Figure 5: With the indicated mapping of convex hull points, these point sets have no compatibletriangulation. Note that the triangulations are unique.
Proof of Theorem 1.
For one internal point, construct compatible triangulations by adding edgesfrom the internal point to all points on the convex hull.Now consider two points sets P and Q that each have two internal points. Following the proofof Aichholzer et al. [1] we will construct a line (cid:96) P through the two internal points of P and a line (cid:96) Q through the internal points of Q , and argue that we can map two points a P and b P of P thatlie on opposite sides of (cid:96) P , to two points a Q and b Q of Q that lie on opposite sides of (cid:96) Q . Then weconstruct compatible triangulations as shown in Figure 6.If P has a convex hull point c P on the line (cid:96) P (i.e. collinear with the two internal points) and Q also has a convex hull point c Q on the line (cid:96) Q then we map these points to each other, let a P and b P be the two neighbours of c P on the convex hull of P , and let a Q and b Q be the two neighbours(in the same clockwise order) of c Q on the convex hull of Q . Observe that a P and b P lie on strictly4 b Figure 6: When a and b are on opposite sides of the line through the two internal points, wetriangulate by joining a and b to the two internal points, then joining every other convex hull pointto the unique internal point it sees.opposite sides of (cid:96) P and similarly, a Q and b Q lies on strictly opposite sides of (cid:96) Q in Q . Thus wehave found the points we need in this case.In the remaining situations, one of P or Q has no convex hull points on the line through itsinterior points. Suppose without loss of generality that P has this property. Orient the lines (cid:96) P and (cid:96) Q vertically. Since the convex hull of P has at least 3 points, and none of them lie on (cid:96) P , oneside of (cid:96) P must have at least 2 convex hull points. Suppose this is the right-hand side. Let q bethe bottom point where (cid:96) Q intersects the convex hull of Q . Note that q may or may not be a pointof Q . See Figure 7.Starting from q and considering the clockwise-ordered convex hull of Q , let a Q be the first pointof Q strictly after q , and let b Q be the last point of Q strictly before q . If q is not in Q then a Q and b Q are adjacent on the convex hull, and otherwise, they are separated by one convex hull point.Now consider P . Let p be the bottom point where (cid:96) P intersects the convex hull of P . Note that p is not in P . Let a P be the first point of P after p in the clockwise-ordered convex hull. If q a and q b are adjacent on the convex hull of Q , let b P be the last point of P before p in the clockwise-orderedconvex hull (Figure 7(c), (d)), and otherwise, let b P be the second last point ((Figure 7(a), (b)).Because P has at least 2 convex hull points to the left of (cid:96) P , b P is to the left of (cid:96) P . Thus we havefound the points we need. Any compatible triangulations must map convex hull points to convex hull points, and in the sameclockwise order. There is still one free choice—point p on the convex hull of point set P can bemapped to any point of the convex hull of Q . Aichholzer, et al. make the stronger conjecture thatcompatible triangulations exist even when the mapping of p to a point of the convex hull of Q isfixed.We examined what happens when the mapping is specified for more points. Our example inFigure 8 shows that the conjecture fails if we specify the mapping of the convex hull and of oneinternal point. 5 P P Q qpa P a P b P b P P Q qpa P a P b P (a) (b)(c) (d) Figure 7: Finding the compatible pair a P and a Q . The convex hull edges must be present in any triangulation of a point set, i.e. they are “forced”.We examined what happens when we specify other edges that must be present in the triangulation.Note that two simple polygons on n points do not always have compatible triangulations [3], so ifwe specify n edges that must be in the triangulation then there may be no compatible triangluation.Our example in Figure 9 shows that even specifying one non-convex hull edge that must beincluded in the triangulation leads to a counterexample. Observe that for the point set on theleft, every convex hull point is forced to have edges to two internal points by Claim 2. However,the required edge on the right prevents the left-hand convex hull point from having edges to twointernal points. Another direction of research on compatible trianglations is to allow the addition of extra points,called
Steiner points .Although a proof of Conjecture 1 would imply that two point sets have compatible triangulationswith no Steiner points, there has been some partial progress on bounding the number of Steinerpoints needed. Danciger, et al. [4] showed that two Steiner points suffice if they may be placed faroutside the convex hull of the points. Aichholzer, et al. [1] showed that a linear number of Steinerpoints inside the convex hull always suffice. This is in contrast to the case of two polygons wherea quadratic number of Steiner points always suffice and are sometimes necessary [3].6 a b a Figure 8: When the mapping of points a and b is specified there is no compatible triangulationbecause the forced edges shown on the left cross on the right.Figure 9: If we require the internal edge shown on the right, then there is no compatible triangu-lation between these point sets.In this section we explore the situation, which arises in some applications, where the mappingof the points is given as part of the input. In this case, compatible triangulations might not exist,as shown in Figure 8. In order to get compatible triangulations we must add Steiner points.In Figure 10 we give an example of two mapped point sets with 6 points each, where compatibletriangulations require the addition of two Steiner points. A solution with two Steiner points is shownin Figure 11, and we argue below that one Steiner point is not sufficient.It would be interesting to find a polynomial time algorithm to test if two mapped point setshave compatible triangulations with no Steiner points, or to show that the problem is NP-complete. Claim 3.
The point sets shown in Figure 10 do not have compatible triangulations with just oneSteiner point.Proof.
Note that in compatible triangulations, each point must have the same neighbours in clock-wise order. Using this property, we can show that one Steiner point is not enough to compatiblytriangulate the two point sets with the given mapping.Claim 2 forces edges (3 , ,
5) in P and edges (2 , ,
5) in Q . Observe that edges (3 , ,
6) cross in P and edges (3 ,
6) and (2 ,
5) cross in Q . To obtain compatible triangulations ofthe point sets with a single Steiner point a , we must place a so that it eliminates the crossing edges.Hence a must be placed such that it breaks one of (3 ,
5) and (2 ,
6) and one of (3 ,
6) and (2 , a so that it breaks edge (3 ,
6) in P and edge (3 ,
5) in Q as shown inFigure 12. Note that in P , a must be placed to the left of the line through 1 and 6, and belowthe line through 2 and 6. Point a is restricted to the analogous region in Q . In P , by Claim 2, wemust have the edge (2 , Q we must have the edge (2 , P than in Q . Thus there can be no compatible triangulation of the point sets with the given mappingin this case. We use a similar argument for the case when a breaks edges (2 ,
5) and (2 , a so that it breaks edge (3 ,
6) in P and edge (2 ,
6) in Q . Claim 2 forcesedges ( a,
3) and ( a,
6) in P . To break edge (2 ,
6) in Q , a must be placed below the line through(3 , a comes before point 5 in the clockwise ordering of neighbours of 3 in Q . To obtainthe same ordering in P , we need a below the line (3 , a,
6) crosses edge (3 ,
5) as6 is above (3 ,
5) as shown in Figure 13. Therefore we still cannot compatibly triangulate the pointsets with the given mapping. Finally, the case where we place a to break edge (2 ,
5) in P and edge(3 ,
5) in Q is symmetric. Acknowledgements
This research was accomplished as part of a two day undergraduate research workshop at theUniversity of Waterloo in October 2016.
References [1] Oswin Aichholzer, Franz Aurenhammer, Ferran Hurtado, and Hannes Krasser. Towards com-patible triangulations.
Theoretical Computer Science , 296(1):3 – 13, 2003.8igure 11: The two triangulations (left and right) are compatible—this can be seen most easily bycomparing each one with the intermediate configuration (center).
12 345 12 34 5
P Qa a
Figure 12: Edges (2 ,
5) and (2 ,
6) appear in different order in the cyclic ordering of edges aroundpoint 2 in P and Q .[2] Soroush Alamdari, Patrizio Angelini, Fidel Barrera-Cruz, Timothy M Chan, GiordanoDa Lozzo, Giuseppe Di Battista, Fabrizio Frati, Penny Haxell, Anna Lubiw, Maurizio Pa-trignani, Vincenzo Roselli, Sahil Singla, and Bryan Wilkinson. How to morph planar graphdrawings. to appear in SIAM J. on Computing , 2016.[3] Boris Aronov, Raimund Seidel, and Diane Souvaine. On compatible triangulations of simplepolygons. Computational Geometry , 3(1):27–35, 1993.[4] Jeff Danciger, Satyan L Devadoss, and Don Sheehy. Compatible triangulations and point par-titions by series-triangular graphs.
Computational Geometry , 34(3):195–202, 2006.[5] Satyan L Devadoss and Joseph O’Rourke.
Discrete and Computational Geometry . PrincetonUniversity Press, 2011.[6] Craig Gotsman and Vitaly Surazhsky. Guaranteed intersection-free polygon morphing.
Com-puters & Graphics , 25(1):67–75, 2001. 9
P Q aa Figure 13: To obtain the same cyclic ordering of edges around point 3 in P and Q , we need Steinerpoint a below (3 ,
5) in P , causing ( a,
6) to cross (3 ,,