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Almost tight bound for the union of fat tetrahedra in R 3 Esther Ezra Micha Sharir. Duke University. Tel-Aviv University. Arrangement of geometric objects. Input: S = {S 1 , …, S n } a collection of n simply geometric objects in d -space. The arrangement A( S ) is the
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Almost tight bound for the union of fat tetrahedra in R3EstherEzra Micha Sharir Duke University Tel-Aviv University
Arrangement of geometric objects Input: S = {S1, …, Sn} a collection of n simply geometric objects in d-space. The arrangementA(S) is the subdivision of space induced by S . The maximal number of vertices/edges/faces of A(S) is: (nd) Combinatorial complexity. Each object has a constant description complexity
Union of simply-shaped bodies:A substructure in arrangements Input: S = {S1, …, Sn} 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 inS? Trivial bound:O(nd) (tight!). Combinatorial complexity.
Previous results in 2D:Fat objects Each of the angles 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! depends linearly on 1/ . r’/r , and 1. r O(n1+), for any>0. r’ r diam(C) ,D C, < 1is a constant. C r D
Previous results in 3D:Fat Objects Congruent cubes narbitrarily aligned (nearly) congruent cubes. Union complexity: O*(n2)[Pach, Safruti, Sharir 2003] . Simple curved objects ncongruent inifnite cylinders. Union complexity: O*(n2)[Agarwal Sharir 2000]. n-round objects. Union complexity: O*(n2)[Aronov et al. 2006]. Each of these bounds is nearly-optimal. r diam(C) ,D C, < 1is a constant. C D r
Special case: Fat tetrahedra Input: T = {T1, …, Tn} a collection of n -fat tetrahedra in R3of arbitrary sized. A tetrahedron T is -fat if: All dihedral angles and solid angles are . Union complexity ? Trivial bound:O(n3). fat thin It is sufficient to bound the number of intersection vertices.
Our results: New union bounds Almost tight. • -fattetrahedra, of arbitrary sizes: O*(n2) . • arbitrary side-length cubes: O*(n2) . • -fattrihedral wedges: O*(n2) . • -fattriangular prisms, having cross sections of arbitrary sizes: O*(n2) . • Revisit union of fat trianlgles:O*(n) . A cube can be decomposed into O(1) fat tetrahedra. Follows easily by our analysis. Follows easily by our analysis.
The union of fat wedges [Pach, Safruti, Sharir 2003] The combinatorial complexity of the union of n-fatdihedral wedges:O*(n2). The bound depends linearly on 1/ . The dihedral angle. Thin dihedral wedges (almost half-planes) create a grid with Ω(n3) vertices.
Main idea: Reduce tetrahedra to dihedralwedges • Decompose space into cells. • Show that most of the cells meet at most two facets of the same tetrahedron. • Most of the union vertices aregenerated by intersections ofdihedral wedges. The tetrahedron is a dihedral wedge inside most of these cell. w v u Apply the bound O*(n2) of[Pach, Safruti, Sharir 2003].
(1/r)-cutting:From tetrahedra to wedges T is a collection of n-fat tetrahedra in R3. 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 mostn/r tetrahedra facets ofT .
Constructing (1/r)-cuttings: • Choose a random sample R of O(r log r) of the planes containing tetrahedra facets in T(r is a fixed parameter). • Form the arrangementA(R) of R:Each cell C of A(R) is a convex polyhedron.Overall complexity: O(r3 log3r). • Triangulate each cell C, and obtain a collection of O(r3 log3r) 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 forTas 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. Apply the nearly-quadratic bound of [Pach, Safruti, Sharir 2003]. At least one of these tetrahedra has three (or more) facets that meet .
Bounding the number of bad vertices meets all four facets of T. Fix a tetrahedron T T: A cell is called bad for T, if it meets at least three facets ofT. Goal: For each fixed tetrahedron T T, the number of bad cells is small. Lemma: There are onlyO*(r) bad cells for T.
Bad cells are scarce The trivial bound is O*(r2). The 2D cross-sections of all cells intersectingFis a 2D arrangement of lines. Overall number of cells: O*(r2) . F Our bound improves the trivial bound by roughly an order of magnitude.
The number of cells that meet two facets F1 F2 Two facets of T can meet Ω*(r2) cells. The construction is impossible for three facets of T !
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 forceat the bottom of the recursion. The overall bound is:O*(n2).
Union of “fat” tetrahedra Input: A set of nfat tetrahedra in R3of arbitrary sizes. Result: Union complexity:O(n2) Almost tight. Special case: Union of cubes of arbitrary sizes. fat thin A cube can be decomposed into O(1) fat tetrahedra.
-fatdihedral/trihedral wedges -fatdihedral 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 . Wis (,)-substantially fatif the sum of the angles of its three facets , and > 4/3. The solid angle.
-fattetrahedron A tetrahedron T is -fat if: • Each pair of its facets define an -fatdihedral wedge. • Each triple of its facets define an -fattrihedral wedge.
Classification of the intersection vertices Outer vertex: The intersection of an edge of a tetrahedron with a facet of another tetrahedron. Overall : O(n2) . Inner vertex: The intersection of three facets of three distinct tetrahedra. Overall : O(n3) . Reduce the problem to: How many inner vertices appear on the boundary of the union? v u
The union of fat wedges:A quadratic lower bound construction W R B Merge the wedges in R and in B so that they form a 2D-grid on 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(n2).
The union of fat trihedral wedges: An almost quadratic upper bound [Pach, Safruti, Sharir 2003] The union of n(,)-substantially fattrihedral wedges:O*(n2). The combinatorial complexity of the union of ncongruent arbitrarily aligned cubes is O*(n2). 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.
How to construct (1/r)-cuttings 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”. n/r The number of the non-sampled points is n/r, with high probability!
Constructing (1/r)-cuttings Use the hierarchical decomposition of Dobkin & Kirkpatrick • Choose a random sample R of O(r log r) of the planes containing tetrahedra facets in T(r is a fixed parameter). • Form the arrangementA(R) of R:Each cell C of A(R) is a convex polyhedron.Overall complexity: O(r3 log3r). • Triangulate each cell C, and obtain a collection of O(r3 log3r) simplices. Theorem [Clarkson & Shor] [Haussler & Welzl] : Each cell of is crossed by n/r tetrahedra facets of T, with high probability. C
Triangulating a cell: The hierarchical decomposition of Dobkin & Kirkpatrick Hierarchical representation of a convex polyhedron C(An informal description): • Construct a (large) independent setV1 of vertices of C=C1 . • Remove the vertices in V1 from C1:Fill each hole with simplicial subcells, and peel them off C1. • Obtain a new polyhedron C2 C1. • Apply steps 1—3 recursively.Bottom of recursion: The new polyhedron Ckis a simplex. C1=C C3 C2 k = O(log r) number of levels in the recursion.
The DK-hierarchical decomposition Claim: There exists a hierarchical representation for C that satisfies: • k = O(log r). • . Each line l that stabs C, crosses only O(log r) of its simplices. l
Properties of the overall decomposition The DK-decomposition properties imply: The overall number of cells of is O(r3 log3r). Each tetrahedron edge crosses at most O(r log2r) simplices of . Another crucial property to follow. There are O(r log r) planes in R. Due to the stabbing line property.