Almost tight bound for the union of fat tetrahedra in R 3 Esther Ezra Micha Sharir Duke University Tel-Aviv University

Input: S = {S 1, …, S n } a collection of n simply geometric objects in d -space. The arrangement A(S) is the subdivision of space induced by S. The maximal number of vertices/edges/faces of A(S) is:  (n d ) Arrangement of geometric objects Combinatorial complexity. Each object has a constant description complexity

Input: S = {S 1, …, S n } a collection of n simply-shaped bodies in d -space of constant description complexity. The problem: What is the maximal number of vertices/edges/faces that form the boundary of the union of the bodies in S ? Trivial bound: O(n d ) (tight!). Union of simply-shaped bodies: A substructure in arrangements Combinatorial complexity.

Previous results in 2D: Fat objects n  -fat triangles. Number of holes in the union: O(n). Union complexity: O(n loglog n). [Matousek et al. 1994] Fat curved objects (of constant description complexity) n convex  -fat objects. Union complexity: O*(n) [Efrat Sharir. 2000]. n  -curved objects. Union complexity: O*(n) [Efrat Katz. 1999]. Union complexity is ~ “one order of magnitude” smaller than the arrangement complexity! Each of the angles   O(n 1+  ), for any  >0. r r’ r’/r  , and   1. r   diam(C), D  C,  < 1 is a constant. r C D depends linearly on 1/ .

Previous results in 3D: Fat Objects Congruent cubes n arbitrarily aligned (nearly) congruent cubes. Union complexity: O*(n 2 ) [Pach, Safruti, Sharir 2003]. Simple curved objects n congruent inifnite cylinders. Union complexity: O*(n 2 ) [Agarwal Sharir 2000]. n  -round objects. Union complexity: O*(n 2 ) [Aronov et al. 2006]. Each of these bounds is nearly-optimal. r C r   diam(C), D  C,  < 1 is a constant. D

Special case: Fat tetrahedra Input: T = {T 1, …, T n } a collection of n  -fat tetrahedra in R 3 of arbitrary sized. A tetrahedron T is  -fat if: All dihedral angles and solid angles are  . Union complexity ? Trivial bound: O(n 3 ). fat thin It is sufficient to bound the number of intersection vertices.  

Our results: New union bounds  -fat tetrahedra, of arbitrary sizes: O*(n 2 ). arbitrary side-length cubes: O*(n 2 ).  -fat trihedral wedges: O*(n 2 ).  -fat triangular prisms, having cross sections of arbitrary sizes: O*(n 2 ). Revisit union of fat trianlgles: O*(n). Almost tight. Follows easily by our analysis. A cube can be decomposed into O(1) fat tetrahedra.

The union of fat wedges [Pach, Safruti, Sharir 2003] The combinatorial complexity of the union of n  -fat dihedral wedges: O*(n 2 ). Thin dihedral wedges (almost half-planes) create a grid with Ω(n 3 ) vertices. The bound depends linearly on 1/ .  The dihedral angle.

Main idea: Reduce tetrahedra to dihedral wedges Decompose space into cells. Show that most of the cells meet at most two facets of the same tetrahedron. Most of the union vertices are generated by intersections of dihedral wedges.  u v w The tetrahedron is a dihedral wedge inside most of these cell. Apply the bound O*(n 2 ) of [Pach, Safruti, Sharir 2003].

(1/r)- cutting: (1/r)- cutting: From tetrahedra to wedges T is a collection of n  - fat tetrahedra in R 3. Use (1/r)- cutting in order to partition space. (1/r)- cutting: A useful divide & conquer paradigm. Fix a parameter 1  r  n. (1/r)- cutting: a subdivision of space into (openly disjoint) simplicial subcells , s.t., each cell  meets at most n/r tetrahedra facets of T. 

Constructing (1/r)- cuttings: 1.Choose a random sample R of O(r log r) of the planes containing tetrahedra facets in T ( r is a fixed parameter). 2.Form the arrangement A(R) of R : Each cell C of A(R) is a convex polyhedron. Overall complexity: O(r 3 log 3 r). 3.Triangulate each cell C, and obtain a collection  of O(r 3 log 3 r) simplices. Theorem [Clarkson & Shor] [Haussler & Welzl] : Each cell  of  is crossed by  n/r tetrahedra facets of T, with high probability. C

The problem decomposition Construct a (1/r)- cutting  for T as above. Fix a cell  of . Some of the tetrahedra in T may become half-spaces/  -fat dihedral wedges inside . Classify each vertex v of the union that appears in  as: Good - if all three tetrahedra that form v are half-spaces/  -fat dihedral wedges in . Bad - otherwise. At least one of these tetrahedra has three (or more) facets that meet . Apply the nearly-quadratic bound of [Pach, Safruti, Sharir 2003].

Bounding the number of bad vertices Fix a tetrahedron T  T : A cell  is called bad for T, if it meets at least three facets of T. Goal: For each fixed tetrahedron T  T, the number of bad cells is small. Lemma: There are only O*(r) bad cells for T.   meets all four facets of T.

Bad cells are scarce F The 2D cross-sections of all cells intersecting F is a 2D arrangement of lines. Overall number of cells: O*(r 2 ). Our bound improves the trivial bound by roughly an order of magnitude. The trivial bound is O*(r 2 ).

The number of cells that meet two facets Two facets of T can meet Ω*(r 2 ) cells. The construction is impossible for three facets of T ! F1F1 F2F2

The overall analysis Construct a recursive (1/r)- cutting  for T. Most of the vertices of the union become good at some recursive step. Bound the number of bad vertices by brute force at the bottom of the recursion. The overall bound is: O*(n 2 ).

Thank you

Union of “fat” tetrahedra Input: A set of n fat tetrahedra in R 3 of arbitrary sizes. Result: Union complexity:  O(n 2 ) Almost tight. Special case: Union of cubes of arbitrary sizes. fat thin A cube can be decomposed into O(1) fat tetrahedra.

 -fat dihedral/trihedral wedges  -fat dihedral wedge W : W is the intersection of two halfspaces. The dihedral angle  .  The dihedral angle.  -fat trihedral wedge W : W is the intersection of three halfspaces. The solid angle  . W is ( ,  )- substantially fat if the sum of the angles of its three facets  , and  > 4  /3.  The solid angle. 

 -fat tetrahedron A tetrahedron T is  -fat if: 1.Each pair of its facets define an  -fat dihedral wedge. 2.Each triple of its facets define an  -fat trihedral wedge.  

Classification of the intersection vertices Outer vertex: The intersection of an edge of a tetrahedron with a facet of another tetrahedron. Overall : O(n 2 ). Inner vertex: The intersection of three facets of three distinct tetrahedra. Overall : O(n 3 ). Reduce the problem to: How many inner vertices appear on the boundary of the union? u v

The union of fat wedges: A quadratic lower bound construction Merge the wedges in R and in B so that they form a 2D-grid on W. R B W The right facet of W “shaves” the edges of the wedges in R and in B. The number of vertices of the union is Ω(n 2 ).

The union of fat trihedral wedges: An almost quadratic upper bound [Pach, Safruti, Sharir 2003] The union of n ( ,  )- substantially fat trihedral wedges: O*(n 2 ). The combinatorial complexity of the union of n congruent arbitrarily aligned cubes is O*(n 2 ). Apply a reduction from cubes to wedges. Each cube intersects only O(1) cells of the grid.

More general union problems Union of arbitrary side-length cubes: Use the grid reduction? Does not work! Need to apply a more elaborate partition technique of space, so as to reduce cubes to wedges. Union of fat tetrahedra: The grid reduction does not work even when the tetrahedra are congruent!  Each tetrahedron induces at least one non-substantially fat trihedral wedge.

The 1-dim problem: We have a set of n points on the real line. Choose a random sample R of r log r points : With high probability, the points in R partition the real line into roughly “equal pieces”. How to construct (1/r)-cuttings The number of the non-sampled points is n/r, with high probability! n/r

Constructing (1/r)- cuttings 1.Choose a random sample R of O(r log r) of the planes containing tetrahedra facets in T ( r is a fixed parameter). 2.Form the arrangement A(R) of R : Each cell C of A(R) is a convex polyhedron. Overall complexity: O(r 3 log 3 r). 3.Triangulate each cell C, and obtain a collection  of O(r 3 log 3 r) simplices. Theorem [Clarkson & Shor] [Haussler & Welzl] : Each cell  of  is crossed by  n/r tetrahedra facets of T, with high probability. C Use the hierarchical decomposition of Dobkin & Kirkpatrick

Triangulating a cell: The hierarchical decomposition of Dobkin & Kirkpatrick Hierarchical representation of a convex polyhedron C (An informal description): 1.Construct a (large) independent set V 1 of vertices of C=C 1. 2.Remove the vertices in V 1 from C 1 : Fill each hole with simplicial subcells, and peel them off C 1. 3.Obtain a new polyhedron C 2  C 1. 4.Apply steps 1—3 recursively. Bottom of recursion: The new polyhedron C k is a simplex. C 1 =C C3C3 C2C2 k = O(log r) number of levels in the recursion.

Claim: There exists a hierarchical representation for C that satisfies: 1.k = O(log r). 2.. Each line l that stabs C, crosses only O(log r) of its simplices. The DK-hierarchical decomposition l

Properties of the overall decomposition  The DK-decomposition properties imply: The overall number of cells  of  is O(r 3 log 3 r). Each tetrahedron edge crosses at most O(r log 2 r) simplices  of . Another crucial property to follow. Due to the stabbing line property. There are O(r log r) planes in R.