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Operations Research Letters 32 (2004) 304–308

OperationsResearchLetters

www.elsevier.com/locate/dsw

The vertex-adjacency dual of a triangulated irregular networkhas a Hamiltonian cycle

John J. Bartholdi III∗, Paul GoldsmanSchool of Industrial and Systems Engineering, Georgia Institute of Technology, 765 Ferst Drive, Atlanta, GA 30332-0205, USA

Received 27 June 2003; received in revised form 11 November 2003; accepted 11 November 2003

Abstract

Triangulated irregular networks (TINs) are common representations of surfaces in computational graphics. We de4ne thedual of a TIN in a special way, based on vertex-adjacency, and show that its Hamiltonian cycle always exists and can befound e5ciently. This result has applications in transmission of large graphics datasets.c© 2003 Elsevier B.V. All rights reserved.

Keywords: Hamiltonian cycle; Triangulated irregular network; Triangle mesh; Triangulation

1. Introduction

Following Alfeld [1]: A triangulated irregular net-work (TIN, or triangle mesh, or triangulation) T isa collection of n triangles satisfying the following:

(1) The interiors of the triangles are pairwise dis-joint;

(2) Each edge of a triangle in T is either a commonedge of two triangles in T or else it is on theboundary of the union of all the triangles;

(3) The union of all the triangles is homeomorphicto a square.

Alfeld observes that

The 4rst requirement says that triangles donot overlap. The second requirement rules out

∗ Corresponding author.E-mail address: john.bartholdi@isye.gatech.edu

(J.J. Bartholdi III).

combinations of triangles where one has a vertexin the interior of an edge of another triangle, andthe strange sounding last requirement rules outholes, pinchpoints (where just 2 triangles meet ina single point) and disjoint sets of triangles.

TINs are commonly used to represent surfaces suchas terrains and 3D computer graphics models (see forexample [6]).

A dual graph of a TIN can be constructed byde4ning a vertex (node) for each triangle, and anedge for each pair of triangles that share an edge.A TIN is said to be Hamiltonian if this dual has aHamiltonian cycle (or path). Thus a Hamiltonian cycleorders the triangular cells of a TIN in a way that pre-serves edge-adjacency. The idea of 4nding sequencesof edge-adjacent triangles has been used to compresslarge polyhedral surface models [3,5,7–9]. The prob-lem arises because each vertex in a TIN may belongto multiple triangles and so a naive method of trans-mitting or processing a TIN, such as by sending thetriangles individually, would contain much duplicated

0167-6377/$ - see front matter c© 2003 Elsevier B.V. All rights reserved.doi:10.1016/j.orl.2003.11.005

J.J. Bartholdi III, P. Goldsman /Operations Research Letters 32 (2004) 304–308 305

data. If the TIN can be disassembled into a Hamil-tonian path of triangles—that is, if the dual graph isHamiltonian—then the TIN can be transmitted veryeconomically, by a strategy of sending the singleremaining vertex of the next triangle along the path.In this way, each point would be processed once.

Unfortunately, it is NP-Complete to determinewhether or not a TIN is Hamiltonian [2]. Conse-quently, this method of compression is generallyimplemented by trying to 4nd long sequences ofedge-adjacent triangles (segments of paths in the dualgraph). Evans et al. [4] discuss some typical greedyheuristics to 4nd long paths in the dual in hopes ofachieving greater compression. However, increasingcomputation risks oGsetting any gains in compression.

We show that a weaker version of the dual guaran-tees the existence of a Hamiltonian circuit. We sug-gest considering the vertex-adjacency dual of a TIN,which is constructed by de4ning a vertex for each tri-angle, and an edge for each pair of triangles that share avertex or an edge. A triangle has more vertex-adjacentneighbors than it does edge-adjacent neighbors; weshow that this extra connectivity is su5cient to guar-antee the existence of a Hamiltonian circuit. More-over, such a circuit can be constructed in linear time,so that it seems to oGer a practical basis for data com-pression.

Fig. 1 illustrates the three approaches to trans-mitting a TIN that we have mentioned: (i) naiveapproach, in which each triangle is sent individually,resulting in the transmission of 15 vertices; (ii) tri-angle strip approach, in which the TIN is dividedinto two edge-adjacent triangle strips, and 9 verticesare transmitted; (iii) vertex-adjacency approach, inwhich 8 vertices are transmitted.

1

2

3 1

2

3 1

2

3

13

14

15

7

8

9

1110

12

54

6

4

5

76

89

4

5

6 7

8

(i) (ii) (iii)

Fig. 1. Three approaches to transmitting TIN data. (i) as individual triangles, (ii) by edge-adjacency, (iii) by vertex-adjacency.

2. Constructing a Hamiltonian cycle

We show how to construct a Hamiltonian cyclethrough the vertex-adjacency dual of a TIN T . Thecycle of the nodes in the dual corresponds to a cyclicordering of the triangles in the TIN. For convenience,where there can be no confusion we will speak inter-changeably of cycles of either nodes (in the dual) ortriangles (in the TIN).

Each triangle t has two neighbors in a cycle: apredecessor pred(t) and a successor succ(t). Neigh-boring triangles of T sharing a common edge (twovertices) will be called 2-adj; neighboring trianglessharing exactly one common vertex will be called1-adj.

The cycle that we create has the following specialproperty that is useful in avoiding an awkward casein the construction of the cycle.

Property 1. No triangle is 1-adj, at the same vertex,to both of its neighbors in the cycle.

Theorem 1. The vertex-adjacency dual of a TIN isHamiltonian.

Proof (Constructive):

1. Initialization.Assume TIN T has n triangles. The theorem is triv-ially true for n=1 or n=2, so assume n ≥ 3. Chooseany triangle a, a second triangle b that is 2-adj toa, and a third triangle c that is 2-adj to either aor b. Assume without loss of generality that c is2-adj to b. Let C signify the subset of T composedof triangles a, b, and c. C is itself a TIN, and sincetriangle a is clearly 1-adj or 2-adj to triangle c,

306 J.J. Bartholdi III, P. Goldsman /Operations Research Letters 32 (2004) 304–308

st

Fig. 2. Case I: t is 1-adj (at a vertex on e) to at least one neighbor.

a Hamiltonian cycle of the vertex-adjacency dual ofC is simply: a; b; c; a. Property 1 is satis4ed, sinceeach triangle is 2-adj to at least one other triangle.The procedure continues by adding triangles to C,one at a time, and updating the cycle of its dualat each step, until C includes all of T . After eachupdate, we must verify that the new cycle is stillHamiltonian (by checking adjacencies around thechange). We also verify that Property 1 remainssatis4ed by showing, for each triangle involvedin the update, that either: its neighbors are un-changed; it has a 2-adj neighbor; it is 1-adj toboth neighbors at diGerent vertices; or, it has ex-changed one 1-adj neighbor for a diGerent 1-adjneighbor at the same vertex. The straightforwarddetails of all these checks are deferred to theAppendix.

2. IterationIf C includes all the triangles in T then the proce-dure is done. Otherwise, 4nd a triangle s, not in C,that shares edge e with some triangle t in C. Adds to C, updating the cycle according to the follow-ing cases, which are partitioned according to t’srelationship with its neighbors:Case I: t is 1-adj, on e, to (at least) one of itsneighbors (Fig. 2). Assume, without loss of gener-ality, that t is 1-adj, on e, to pred(t).Solution: Add s to the cycle as follows:pred(t); s; t;succ(t). (Note that pred(t) now actually becomespred(s), but for clarity we continue to refer to ver-tices by their initial labels, here and elsewhere.)The curve in Fig. 2 is shown to originate at thecenter of pred(t), and terminate at the center of t.This is meant to indicate that we do not restricthow pred(t) is adjacent to pred(pred(t)), or howt is adjacent to succ(t)—because all (legal) pos-sibilities are covered by the solution shown. Forexample, t and succ(t) could be 2-adj at either oft’s open edges, or they could be 1-adj at any of t’s

IIa st

IIb Otherwise,

st

Fig. 3. Case II: t is 2-adj to both neighbors. Subcase IIa:pred(pred(t)) is 1-adj to pred(t) on e, and succ(succ(t)) is 1-adjto succ(t) on e. Subcase IIb: Otherwise.

vertices (aside from the vertex at which t is 1-adjto pred(t)).Case II: t is 2-adj to both neighbors (Fig. 3). Notethat in the 4gure some curves begin (end) at avertex, indicating that the 4rst triangle is adjacent toits predecessor (successor) at the vertex indicated.When a curve begins (ends) at a triangle center, thepreceding triangle may be adjacent by any of itsedges or vertices, as long as the adjacency is legal.◦ IIa: pred(pred(t)) is 1-adj to pred(t) on e, andsucc(succ(t)) is 1-adj to succ(t) on e.Solution: pred (pred (t)); s; t; pred(t); succ(t);succ(succ(t)).

◦ IIb: Otherwise (pred(pred(t)) is not 1-adj topred(t) on e, or succ(succ(t)) is not 1-adj tosucc(t) on e, or both are not). Assume, withoutloss of generality, that pred(pred(t)) is not 1-adjto pred(t) on e.Solution: pred(t); s; t; succ(t).

Case III: t is 2-adj to one neighbor, and 1-adj, o<e, to the other neighbor (Fig. 4). Assume, withoutloss of generality, that t is 2-adj to pred(t), and1-adj, oG e, to succ(t).◦ IIIa: pred(t) is 1-adj to pred(pred(t)), on e.

Solution: pred(pred(t)); s; t; pred(t); succ(t).◦ IIIb: Otherwise (pred(pred(t)) is not 1-adj topred(t) on e).Solution: pred(t); s; t; succ(t).

The procedure consumes one triangle with eachiteration, and so terminates after n− 3 steps. To ver-ify that the cases comprise the complete set of pos-sibilities, observe that: Each triangle t in the current

J.J. Bartholdi III, P. Goldsman /Operations Research Letters 32 (2004) 304–308 307

IIIa st

IIIb Otherwise,

st

Fig. 4. Case III: t is 2-adj to one neighbor, and 1-adj (at t’s vertexopposite e) to the other neighbor. Subcase IIIa: pred(t) is 1-adj(on e) to pred(pred(t)). Subcase IIIb: Otherwise.

cycle (at each step) is either 1-adj or 2-adj to each ofits two neighbors pred(t) and succ(t). Case I coversthe case where t is 1-adj to both neighbors (becauseat least one of the common vertices must be on e, byProperty 1); Case II covers the case where t is 2-adjto both neighbors; Case III covers the case where tis 2-adj to one neighbor and 1-adj to the other (ex-cept where the 1-adjacency is on e, which is coveredin Case I). Cases II and III, are each partitioned bya pair of mutually-exclusive subcases. The remainingpossibility, in which both of t’s neighbors are 1-adj atthe same vertex, is forbidden by Property 1 and neveroccurs.

3. Conclusions

By accepting vertex-adjacency, rather than insistingon edge-adjacency, we have shown that, for any TIN,one can construct, in linear time, a single sequenceof adjacent triangles that encompasses the entire TIN.This allows an alternative strategy for compression.

Recent eGorts have focused on 4nding a small num-ber of path segments that span all triangles of the TIN.Each path segment can then be sent e5ciently—butthere is extra work to 4nd the path segments and extrawork to shift from one path segment to the next. Our

method quickly 4nds a single path, but it may not com-press as well as the ideal one based on edge-adjacency(which may not exist and can be hard to 4nd in anycase).

Potential duplication arises in our method whenevertwo neighboring triangles (in the cycle) share just onevertex, so that we must send two new vertices ratherthan one to specify the second triangle. This neverrequires more than two points per triangle, which isworse than the (perhaps unachievable) lower boundof one point per triangle, but better than sending eachtriangle separately (3 points per triangle). More to thepoint, we would not expect vertex-adjacent triangles todominate; informal tests suggest that there will likelybe runs of edge-adjacent triangles interspersed withoccasional vertex-adjacent triangles.

Comparison of compression methods based onedge-adjacency and methods based on vertex-adjacency is an empirical question that will beexplored in subsequent work.

Acknowledgements

We appreciate the support of the O5ce ofNaval Research through Grant #N00014-89-J-1571,the National Science Foundation through Grant#DMI-9908313, The Logistics Institute at GeorgiaTech and its sister organization, The Logistics Insti-tute Asia Paci4c in Singapore. We are also gratefulfor the useful comments of an anonymous referee.

References

[1] P. Alfeld, Triangulations, http://www.math.utah.edu/∼alfeld/MDS/triangulation.html, March 1999.

[2] E.M. Arkin, Martin Held, J.S.B. Mitchell, S.S. Skiena,Hamiltonian triangulations for fast rendering, in: Visualization’95: Proceedings, October 29–November 3, Atlanta, GA, 1995.

[3] J.J. Bartholdi, III, P. Goldsman, Multiresolution indexing oftriangulated irregular networks, IEEE Trans. Visual. Comp.Graph. to appear.

[4] F. Evans, S. Skiena, A. Varshney, Optimizing triangle stripsfor fast rendering, in: Proceedings, IEEE Visualization 1996,San Francisco, CA, 1996, pp. 319–326.

[5] M. Isenburg, Triangle strip compression, in: Proceedings ofGraphics Interface 2000, Montreal, Quebec, Canada, May2000, pp. 197–204.

308 J.J. Bartholdi III, P. Goldsman /Operations Research Letters 32 (2004) 304–308

[6] R. Laurini, D. Thompson, Fundamentals of Spatial InformationSystems, Academic Press, Ltd., San Diego, CA, 1992.

[7] J. Rossignac, Edgebreaker: connectivity compression fortriangle meshes, IEEE Trans. Visual. Comp. Graph. 5 (1)(1999) pp. 47–61.

[8] G. Taubin, J. Rossignac, Geometric compression throughtopological surgery, ACM Trans. Graph. 17 (2) (1998) 84–115.

[9] C. Touma, C. Gotsman, Triangle mesh compression, in:Proceedings of Graphics Interface, Vancouver, BC, 1998, pp.26–34.