On Corners of Objects Built from Parallelepiped Bricks Mirela Damian¤ Joseph O'Rourkey Abstract always have corners: the zonohedra, and objects built from orthogonal bricks, i.e., rectangular boxes. We investigate a question initiated in the work of Sib- Zonohedra are a particular class of genus-0 objects. ley and Wagon, who proved that 3 colors su±ce to color Second, we found a genus-3 object with no cor- any collection of 2D parallelograms glued edge-to-edge. ners [DO03a], and conjectured that every genus-0 Their proof relied on the existence of an \elbow" par- object built from bricks has a corner. allelogram. We explore the existence of analogous \cor- In this paper we re¯ne the 2D results and show ner" parallelepipeds in 3D objects, which would lead to that they fail to extend to topological balls in 3D. 4-coloring. Our results are threefold. First, we re¯ne We deviate from a concentration on genus and in- the 2D proof to render information on the number and stead posit two properties (called S and I=E) such location of the 2D elbows. Second, we extend the 2D re- that, if an object possesses these properties, then sults to 3D for objects satisfying two properties. Third, it must have corners. Finally, we prove that topo- we exhibit a genus-0 object (a topological ball) that sat- logical balls have the ¯rst property (S), but not the is¯es one but not both of our properties, and fails the second (I/E). 3D extension theorem, establishing that this theorem is, in a sense, tight. 2 An Object with No Corners: The ZZ-Object 1 Introduction The genus-3 example from [DO03a] serves as a counterexample to many hypotheses, and will be Sibley and Wagon [SW00] proved that any collec- important to illustrate our de¯nitions. The over- tion of parallelograms glued whole-edge to whole- all design is shown in Fig. 1a. It consists of two edge must have at least one elbow: a parallelo- Z-shaped paths connecting four cubes. Each of the gram with two edges incident to one of its ver- long connectors has no corner when split lengthwise tices exposed in the sense that neither is glued to into four bricks. Similarly, the four cubes have no another parallelogram. Elbows have at most two corners when split into eight cubes. However, as neighbors, which enabled them to prove that such is evident from (a), it is self-intersecting. The self- tilings are 3-colorable. The analogous question in e 3D is [Wag02, DO03b]: Must every object built from parallelepipeds (henceforth, bricks) have at least one corner, a brick with three faces incident to one of its vertices, exposed? The bricks must be properly joined: each pair is either disjoint, or in- tersects either in a single point, a single whole edge of each, or a single whole face of each. (a) (b) Just as in 2D, the existence of corner bricks would Figure 1: (a) A self-intersecting object with no lead to 4-colorability of the \brick graph." corners (after re¯nement) (b) A genus 3 non-self- But Robertson, Schweitzer, and Wagon [RSW02] intersecting object built from 14 bricks with no cor- found a polyhedron with no corner. Their ex- ners (after re¯nement). ample, a \buttressed octahedron," has genus 13. Since then, there have been two developments con- intersection can be removed by zig-zagging one of cerning corners of 3D brick objects. First, it was the Zs. The resulting object is shown in Fig. 1b. shown in [GO03] that two classes of such objects 3 2D Brick Objects Revisited ¤Dept. Comput. Sci., Villanova Univ., Villanova, PA Sibley and Wagon [SW00] proved that every 2D ob- 19085, USA. [email protected]. ject built from parallelograms has an elbow. In this yDept. Comput. Sci., Smith College, Northampton, MA 01063, USA. [email protected]. Supported by NSF section we re¯ne their theorem both quantitatively Distinguished Teaching Scholars award DUE-0123154. and qualitatively. First, we need some de¯nitions. 1 A pseudoline L[e] is a longest sequence of adja- the angle needed to turn vi ¡ vi¡1 to vi+1 ¡ vi; ¿i is cent bricks, all of whose shared edges are parallel to ¼ minus the internal angle at vi. e; it has two \sides" of edges, the top and bottom. A Let the parallelogram contributing edge ei = collapse of a pseudoline L[e] is a chain of edges ob- (vi; vi+1) to P have angles ®i at vi and so ¼ ¡ ®i at tained by shrinking all the edges of L[e] parallel to e vi+1. If O has no corners, then every vertex of P to length zero. A monotone chord is a boundary-to- must have at least two incident parallelograms of O; boundary chain of edges that is strictly monotone for just one incident parallelogram would be a cor- with respect to some direction. Intuitively, a mono- ner. So ¿i = ¼¡[®i+(¼¡®i¡1)+¯i] = ®i¡1¡®i¡¯i, tone chord z is the result of collapsing a pseudoline where ¯i ¸ 0 is the angular contribution of other L[e]. parallelograms incident to vi. We know that for any A path of bricks crosses a pseudoline L if it con- simple polygon, Pi ¿i = 2¼. In forming this sum tains a brick of L. A pseudoline L is separating (S) via the expression in terms of the ®i's, it is clear that if there is no path of bricks connecting a point on the parallelogram sharing ei has a net zero contri- its top to a point on its bottom, without crossing L; bution, because its terms in ¿i and in ¿i+1 cancel L is called interior (I) if none of its side edges are out. Thus each parallelogram sharing an edge of exposed; exterior (E) if all of its edges to one side P contributes 0 to the turn angle sum. We thus are exposed; and I/E if either I or E. Finally, L is have Pi ¿i = ¡ Pi ¯i · 0. This contradicts the an SI/E-pseudoline, if it is both S and I/E. Similar fact that the sum must equal 2¼, contradicting our de¯nitions hold for monotone chords. Fig. 2a shows assumption that there are no elbows. examples of monotone chords and pseudolines that We have now reproved the Sibley-Wagon theo- are interior, exterior, and separating. rem. But we have more, for each elbow vertex can contribute only strictly less than ¼ to the turn angle (e.g., a sharp parallelogram vertex with nearly 0 in- L2[ e2] ternal angle). So, for Pi ¿i to reach 2¼, we need at L [ e ] c e1 1 1 least three elbow vertices. If these three elbow ver- e2 k1 e3 [ e ] tices are vertices of distinct elbows, we are ¯nished. k L3 3 3 Only a one-parallelogram object can have three el- (a) (b) bow vertices on one elbow, and here the lemma is Figure 2: (a) L1[e1] and its collapse k1 are interior clearly true. So assume there is an elbow with two and non-separating; L2[e2] is interior and separat- elbow vertices on P . Then together those two ver- ing; L3[e3] and its collapse k3 are exterior (b) A 2D tices contribute exactly ¼ to the turn, still leaving object with three elbows. a gap to reach 2¼. So the claim of the lemma is established. ¤ An object O is SI=E-collapsible if it is either empty, or contains one pseudoline L that satis¯es That this result is best possible is established by property SI=E, and a collapse of L reduces O to an Fig. 2b. The second extension of Sibley-Wagon is SI=E-collapsible object. qualitative, yielding information about where the elbows are: one to each side of every monotone 3.1 Re¯nement of the Sibley-Wagon Theorem chord. We can prove this using turn angles, with- We will distinguish between an elbow parallelogram, out other assumptions. We choose instead to prove and an elbow vertex, the vertex of the elbow with in- a weaker version which extends almost directly to cident exposed edges. The ¯rst extension of Sibley- 3D. Wagon is quantitative: Lemma 2 A collapse of a pseudoline is a mono- Lemma 1 Every 2D object built from parallelo- tone chord. grams has at least three elbow vertices. Lemma 3 A monotone chord can cross a pseudo- Proof. Any object O built from parallelograms has line at most once. an outer boundary P that is a simple polygon. The Lemma 4 If O0 is an object obtained by collapsing edges of P are all edges of parallelograms that are a pseudoline in O, then a monotone chord k in O on the exterior face (the unbounded component of reduces to a monotone chord k0 in O0. R2 n O). Note that P is a simple polygon regardless of the genus of O. Theorem 5 For any SI=E-collapsible object O and Orient the boundary of P counterclockwise, and any SI=E-monotone chord z of O, there is an elbow de¯ne the turn angle ¿i at a vertex vi of P to be vertex strictly to each nonempty side of z. 2 Proof. The proof is by induction on the number of If L is interior, then k is interior, with none of its pseudolines. The base case is an object with one vertices exposed. Refer to Fig. 4. This implies that pseudoline L. By Lem.