Cut Locus Realizations on Convex Polyhedra
CCut Locus Realizations on Convex Polyhedra
Joseph O’Rourke and Costin VˆılcuFebruary 23, 2021
Abstract
We prove that every positively-weighted tree T can be realized asthe cut locus C ( x ) of a point x on a convex polyhedron P , with T weights matching C ( x ) lengths. If T has n leaves, P has (in general) n + 1 vertices. We show there are in fact a continuum of polyhedra P each realizing T for some x ∈ P . Three main tools in the proof areproperties of the star unfolding of P , Alexandrov’s gluing theorem,and a cut-locus partition lemma. The construction of P from T issurprisingly simple. There is a long tradition of reversing, in some sense, the construction of agraph G from a geometric set. The geometric set may be a point set, apolygon, or a polyhedron, and the graph G could be the Voronoi Diagram,the straight skeleton, or the cut locus, respectively. Reversing would startwith, say, the straight skeleton, and reconstruct a polygon with that skeleton.Here we start with the cut locus and construct polyhedra P on which the cutlocus is realized for a point x ∈ P . (The cut locus is defined in Section 2.1below.)The literature has primarily examined three models for the graph G , oftenspecialized (as here) to trees T :(1) Unweighted tree : The combinatorial structure of T , without furtherinformation. 1 a r X i v : . [ c s . C G ] F e b Length tree : T with positive edge weights representing Euclidean lengths,and with given circular order of the edges incident to each node of T .Called “ribbon trees” in [CDLR14], and “ordered trees” in [BGP + Geometric tree : Given by a drawing, i.e., coordinates of nodes, deter-mining lengths and angles.Our main result is this:
Theorem 1.
Given a length tree T of n leaves, we can construct a continuumof star-unfoldings of convex polyhedra P of n +1 vertices, each of which, whenfolded, realizes T as the cut locus C ( x ) for a point x ∈ P . Each star-unfoldingcan be constructed in O ( n ) time. Thus, every length tree is isometric to a cut locus on a convex polyhedron.
The computer science literature is extensive, and we cite just a few results: • Every unweighted tree can be realized as the Voronoi diagram of a setof points in convex position [LM03]. • Every length tree can be realized as the furthest-point Voronoi diagramof a set of points [BGP + • Every unweighted tree can be realized as the straight skeleton of aconvex polygon, and conditions for length-tree realization are known[CDLR14] [ABH +
15] [BGP + O ( n ) or O ( n log n )for trees of n nodes. Although all these results can be viewed as variationson realizing Voronoi diagrams, and a cut locus is a subgraph of a Voronoidiagram, it appears that prior work does not imply our results.Our inspiration derives from two results in the convexity literature: • Every length graph can be realized as a cut locus on a Riemanniansurface [IV15]. The result is non-constructive. • Every unweighted tree can be realized as a cut locus on a doubly coveredconvex polygon, and length trees can be realized on such polygons whenseveral conditions are satisfied [IV04].2
Background
In this section we describe the tools needed to prove our main theorem,drawing heavily on our [OV20].
The cut locus C ( x ) of a point x on (the surface of) a convex polyhedron P is the closure of the set of points to which there are more than one shortestpath from x . This concept goes back to Poincar´e [Poi05], and has beenstudied algorithmically since [SS86] (under the name “ridge tree”). Somebasic properties and terminology: • C ( x ) is a tree whose endpoints are vertices of P , and all vertices of P are in C ( x ). • Points interior to C ( x ) of tree-degree 3 or more we will call ramificationpoints . • The edges of C ( x ) are geodesic segments on P , geodesic shortest pathsbetween their endpoints [AAOS97]. The star unfolding S P ( x ) of P with respect to x is formed by cutting ashortest path from x to every vertex of P [AO92] [AAOS97]. This unfolds toa simple non-overlapping planar polygon S = S P ( x ) of 2 n vertices: n images x i of x , and n images of the vertices v i of P . The connection between the cutlocus and the star unfolding is that the image of C ( x ) in S is the restrictionto S of the Voronoi diagram of the images of x [AO92]. See Fig. 1. We rely on Alexandrov’s celebrated “Gluing” Theorem [Ale05, p.100].
Theorem AGT.
If the boundaries of planar polygons are glued together (byidentifying portions of the same length) such that(1) The perimeters of all polygons are matched (no gaps, no overlaps).(2) The resulting surface is a topological sphere.
23 658 B TKL R
11 2 23 3658 744
Tx KFL R (a) (b) F Figure 1: (a) Cut segments to the 8 vertices of a cube from a point x on thetop face. T, F, R, K, L, B = Top, Front, Right, Back, Left, Bottom. (b) Thestar-unfolding from x . The cut locus C ( x ) (red) is the Voronoi diagram ofthe 8 images of x (green). Two pairs of fundamental triangles are shaded./4
3) At most π surface angle is glued at any point.Then the result is isometric to a convex polyhedron P , possibly degenerated toa doubly-covered convex polygon. Moreover, P is unique up to rigid motionand reflection. The proof of this theorem is nonconstructive, and there remains no effec-tive procedure for constructing the polyhedron guaranteed to exist by thistheorem.
The following lemma is one key to our proof. See Fig. 1(b).
Lemma 1 (Fundamental Triangles [INV12]) . For any point x ∈ P , P canbe partitioned into flat triangles whose bases are edges of C ( x ) , and whoselateral edges are geodesic segments from x to the ramification points or leavesof C ( x ) . Moreover, those triangles are isometric to plane triangles, congruentby pairs. The overall form of our proof of Theorem 1 is to create a star unfolding S by pasting together x i -apexed fundamental triangles straddling each edgeof T , and then applying Alexandrov’s theorem to conclude that the foldingof S yields a convex polyhedron P which realizes C ( x ). The last tool we need is a generalization of lemmas in [INV12]. On a poly-hedron P , connect a point x to a point y ∈ C ( x ) by two geodesic segments γ, γ (cid:48) . This partitions P into two “half-surface” digons H and H . If wenow zip each digon separately closed by joining γ and γ (cid:48) , AGT leads to twoconvex polyhedra P and P . The lemma says that the cut locus on P is the“join” of the cut loci on P i . See Fig. 2. Lemma 2.
Under the above circumstances, the cut locus C ( x, P ) of x on P is the join of the cut loci on P i : C ( x, P ) = C ( x, P ) (cid:116) y C ( x, P ) , where (cid:116) y joins the two cut loci at y . And starting instead from P and P , the naturalconverse holds as well. roof. ( Sketch . See [OV20] for a formal proof.) All geodesic segments start-ing at x into H i remain in H i , because geodesic segments do not branch.Therefore, H has no influence on C ( x, P ) and H has no influence on C ( x, P ).Figure 2: Geodesic segments γ and γ (cid:48) (purple) connect x = x = x to y = y = y . P folds to a tetrahedron, and P to an 8-vertex polyhedron, with x and y vertices in each. P and P are cut open along geodesic segments from x i to y i and glued together to form P . Based on the cube unfolding in Fig. 1(b). If T has a node of degree-2, then its incident edges may be merged and theiredge-lengths summed. So henceforth we assume T has no nodes of degree-2.6e start with T a star-tree: one central node u of degree- m with edges tonodes u , u , . . . , u m .A cone in the plane is the unbounded region between two rays from itsapex. Set λ > L to be longer than L , the length of the longest edge of T .We realize T within a cone of apex angle α , with 0 < α ≤ π . See Fig. 3.Identify points x and x m +1 on the cone boundary, with | ux i | = λ . Insidethe cone, place x images x , . . . , x m , with each | ux i | = λ . Finally place u i sothat uu i bisects ∠ ( ux i , ux i +1 ), i = 1 , . . . , m . Chose u i so that | uu i | matches T ’s edge weights. Finally, connect( u, x , u , x , . . . , x m , u m , x m +1 )into a simple polygon. u=yx x x x u u u γλ γʹ Figure 3: α = 120 ◦ , m = 3, T edge lengths (2 , ,
1) (red), λ = 4. In theinduction proof of Lemma 3, ¯ P ( T ) (yellow) is joined to ¯ P ( T ) (blue).Before we proceed with the proof, we emphasize that there are severalfree choices in this construction, illustrated in Fig. 4: • The angle α at the root is arbitrary. • The angular distribution of the ux i segments is arbitrary. • λ > L needs to be “sufficiently large” in a sense we will quantify, butotherwise is arbitrary. Note that we allow α > π ; α = 2 π represents the whole plane. If α = 2 π , x = x m +1 and u is interior to the polygon.
7n general we will distribute x i equi-angularly. Choosing α = 2 π results in apolyhedron of n + 1 vertices; for α < π , u is an additional vertex.We call the described star- T construction a triangle packing . Lemma 3.
A triangle packing of star- T , for sufficiently large λ , is the star-unfolding of a polyhedron P with respect to a point x such that T = C ( x ) .Proof. The proof is by induction on the number m of edges of T , which isthe degree of the root node u . If T is a single edge e = uu , then foldingthe twin triangles by creasing e and joining x and x , the two images of x ,leads to a doubly covered triangle with x at the corner opposite e . (See theyellow triangles in Fig. 3.) Clearly e = C ( x ) is the restriction of the Voronoidiagram of x and x , the bisector of x and x . Concerning λ , we need thatthe angle θ x incident to x is at most 2 π . In this base case, θ + θ ≤ π isimmediate, so λ is sufficiently large.Now let n >
1, and partition T into T ∪ u T with root degrees m and m respectively, where ∪ u indicates joining the trees at the root u . We willuse ¯ P ( T ) for the planar triangle packing for T , and P ( T ) for the foldedpolyhedron.We first address λ . In order to apply AGT, we need that θ x , the sum ofthe angles at the tips of the triangles—the images of x —is at most 2 π .Let λ be the larger of λ and λ for ¯ P ( T ) and ¯ P ( T ) respectively, andstretch the smaller λ i so that they both share the same λ . Form ¯ P ( T ) byjoining the two now-compatible triangle packings. Fixing α and the sectorangles, it is clear that the angle at each triangle tip decreases monotonicallyas λ increases. So increase λ as needed so that θ x ≤ π . Somewhat abusingnotation, call these possibly enlarged packings ¯ P ( T ), ¯ P ( T ) and ¯ P ( T ).Now we aim to show that ¯ P ( T ) is the star-unfolding of a polyhedronwith T = C ( x ). Certainly ¯ P ( T ) folds to a polyhedron P , because (a) byconstruction the edges incident to u , . . . , u k match in length, and (b) wehave ensured that θ x ≤ π . So Alexandrov’s theorem applies. Now identify γ and γ (cid:48) on P from x to y = u separating the surface into pieces correspondingto ¯ P ( T ) and ¯ P ( T ), which fold to P and P respectively. (Refer again toFig. 3.) By the induction hypothesis, T = C ( x, P ) and T = C ( x, P ).Applying Lemma 2, we have C ( x, P ) = C ( x, P ) (cid:116) y C ( x, P ), where the twocut loci are joined at y = u . And so indeed T ∪ u T = T = C ( x ).In Section 4.1 we will calculate the needed λ explicitly.8 a) (b)(c) Figure 4: Star- T with edge lengths (3 , , , , α = 270 ◦ , equiangular x i , λ = 6. (b) α = 180 ◦ , random x i , λ = 5. (c) α = 360 ◦ , equiangular x i , λ = 6. 9 General Length-Trees T We now generalize the above to arbitrary length-trees T , using the examplein Fig. 5 as illustration.
22 ½1 2 1 1 1111 11 u x x x x x x x x u u u Figure 5: Realization of a length-tree T of height 3, using α = 2 π and λ = L = 5. This polygon folds to a polyhedron of 9 vertices: the 8 leaves of T , and x .Given T , select any node u to serve as the root. Fix any α , and choose λ to exceed the length L of the longest path from u to a leaf in T . Now createa triangle packing for T as follows.First create a triangle packing for u and its immediate children u , . . . , u m ,just as previously described. With λ > L , the external angle at u i is10trictly less than π , i.e., it forms a “ V -shape” there. Call this a cup , c i =( x i , u i , x i +1 ) with α i external angle at u i . Let u i have children u i , u i , . . . , u il .So u i , ( u i , . . . , u il ) is a star-tree. Fill in the cup c i by inserting a trianglepacking for this sub-star-tree, with apex at u i , angle α i , and λ -length | u i x i | ,the distance from u i to the tips of the cup.After filling the c i cups for all the u i at level-2 of T , repeat the processwith level-3 of T , and so on. Throughout the construction, the locations for x i remain fixed after their initial placement. And with sufficiently large λ ,all the cups form V -shapes.Note that the triangles incident to an internal node u i of T (neither theroot nor a leaf) leave no gaps: they cover the 2 π surrounding u i . Lemma 4.
A triangle packing for any T , as just described, for sufficientlylarge λ , is the star-unfolding of a polyhedron P with respect to a point x suchthat T = C ( x ) .Proof. The proof is by induction, and parallels the proof of Lemma 3 closely.Consequently, we only sketch the proof.The base of the induction is a star-graph, settled by Lemma 3. Let T be an arbitrary length-tree, and partition T into two smaller trees T and T sharing the root u , so T = T ∪ u T . Select an α for T and α i for T i , i = 1 , α = α + α .By the induction hypothesis, T i can be realized in cups of angle α i . More-over, ¯ P ( T i ) folds to P i and T i = C ( x, P i ). Stretch λ i as needed to allow ¯ P ( T )to share λ with ¯ P ( T ) at u , and stretch again so that θ x ≤ π .Form ¯ P ( T ) by adjoining ¯ P ( T ) and ¯ P ( T ) at u , with cup apex α . Fold¯ P ( T ) to P by AGT. Apply Lemma 2 to conclude C ( x, P ) = C ( x, P ) (cid:116) y C ( x, P ), where y = u . And so T ∪ u T = T = C ( x ).Note that all ramification points of C ( x ) are flat on P , with 2 π incidentsurface angle. If θ x is strictly less than 2 π , then the source x is a vertex on P . If at the root, α < π , then in addition u is a vertex on P . So P has n , n + 1, or n + 2 vertices.Lemmas 3 and 4, together with Lemma 5 (below) prove Theorem 1. Theconstruction of the triangle packing can be achieved in O ( n ) time: we aregiven the cyclic ordering of the edges incident to each node, so sorting is notnecessary, and the level-by-level packing construction is proportional to thethe number of edges. 11 .1 Total angle at x We derive here a sufficient condition on the value of the parameter λ for theroot cup to guarantee that θ x ≤ π . Lemma 5. If T has m edges, and L is the longest path from the root in T then θ x ≤ π when λ ≥ L (cid:104) (cid:16) πm (cid:17)(cid:105) . Figure 6: (a) θ i ≤ φ i . (b) Increasing λ increases d . L is the length of thelongest path in T , D the target bound for λ ≥ L + D . Proof.
First we establish notation, illustrated in Fig. 6(a). As before, u is theroot of T , and let u i be a leaf of T , with edge uu i shared by twin triangles,one of which is (cid:52) i . Let (cid:96) i = | uu i | and d i the distance from the tip of (cid:52) i to u i , and θ i the angle at that tip, an image of x . Because the fundamentaltriangles come in pairs, and there are m edges, we have that the total angleat x satisfies θ x = 2 (cid:80) θ i .Consider now a right triangle (cid:52) (cid:48) i having the same base as (cid:52) i and height d i ,and denote by φ i its angle opposite to the base uu i . Then φ i = arctan (cid:18) (cid:96) i d i (cid:19) and θ i ≤ φ i ; see again Fig. 6.Because arctan is an increasing function, we can obtain an upper boundby replacing (cid:96) i with the longest edge length (cid:96) , and replacing d i by the shortest12f the xu j diagonals, call it d : (cid:96) = max i (cid:96) i , d = min i d i . So φ i ≤ arctan (cid:18) (cid:96)d (cid:19) ,and therefore θ x = 2 (cid:88) θ i ≤ (cid:88) φ i ≤ m arctan (cid:18) (cid:96)d (cid:19) . The expression 2 m arctan (cid:18) (cid:96)d (cid:19) decreases as d increases. For it to evaluateto at most 2 π , we must have d ≥ (cid:96) cot (cid:16) πm (cid:17) . The longest path L from root to leaf is at least as long as the longest edge, L ≥ (cid:96) , so this bound will more than suffice: d ≥ L cot (cid:16) πm (cid:17) = D .
Now we show that if λ is long enough, then d ≥ D . Consider Fig. 6(b), where L is the longest path from u to a leaf u i . Let the external angle at u i be α i .By definition, there is some index j such that d = d j . The triangle inequalitydirectly implies λ ≤ (cid:96) j + d j ≤ L + d .With L and α i fixed, increasing λ increases d . If we substitute the neededlower bound D for d in the expression (see Fig. 6(b)), then λ ≥ L + D forces d ≥ D . Explicitly λ ≥ L (cid:104) (cid:16) πm (cid:17)(cid:105) suffices to guarantee that θ x ≤ π .The bound—approximately L (1 + m/π )—is far from tight. For example, inFig. 5, m = 13 and L = 5 leads to λ ≥
26, but λ = 5 (illustrated) leads to θ x ≈ ◦ , and so easily suffices. One further example is shown in Fig. 7. It is the star unfolding of a poly-hedron of 49 vertices, whose resemblance to a fractal suggests there mightat a deeper connection. We will only mention that fractals play a role inthe folding of specific convex polyhedra in [Ueh20], and fractal cut loci on13igure 7: Regular degree-3 tree, random edge lengths, α = 2 π , n = 48, θ x ≈ ◦ . 14n k -differentiable Riemannian and Finslerian spheres are shown in [IS16] toexist for any 2 ≤ k < ∞ .A natural question is whether geometric trees—drawings embedded in theplane, and so providing angles between adjacent edges—can be realized as cutloci on convex polyhedra. Certainly not all geometric trees are realizable, forthere are constraints on the angles: around a ramification point, no angle canexceed π , and the angles must sum to ≤ π . And the sum of the curvaturesat the u i and at x must be 4 π to satisfy the Gauss-Bonnet theorem.We leave it as an open problem to characterize those geometric trees thatare realizable as the cut locus on a convex polyhedron. References [AAOS97] Pankaj K. Agarwal, Boris Aronov, Joseph O’Rourke, and Cather-ine A. Schevon. Star unfolding of a polytope with applications.
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