Almost tight bound for Almost tight bound for the union of fat the union of fat tetrahedra in R tetrahedra in R 3 3 Esther Esther Ezra Micha Ezra Micha Sharir Sharir Duke Duke Universi Universi ty ty Tel-Aviv Tel-Aviv Universi Universi ty ty
Jan 15, 2016
Almost tight bound for the union of fat tetrahedra in R3
Esther Ezra Micha Sharir
Duke UniversityTel-Aviv University
Arrangement of geometric objects Input:S = {S1, , Sn} 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/facesof A(S) is: (nd)
Combinatorial complexity.Each object has a constant description complexity
Union of simply-shaped bodies: A substructure in arrangementsInput: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/facesthat form the boundary of the union of the bodies in S ?
Trivial bound: O(nd) (tight!).
Combinatorial complexity.
Previous results in 2D:Fat objectsn -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(n1+) , for any >0 .rrr/r ,and 1.r diam(C) , D C, < 1 is a constant. rCDdepends linearly on 1/ .
Previous results in 3D:Fat ObjectsCongruent cubesn arbitrarily aligned (nearly) congruent cubes. Union complexity: O*(n2) [Pach, Safruti, Sharir 2003] .
Simple curved objects n congruent 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.rCr diam(C) , D C, < 1 is a constant. D
Special case: Fat tetrahedraInput:T = {T1, , Tn} a collection of n -fat tetrahedra in R3 of arbitrary sized.
A tetrahedron T is -fat if:All dihedral angles and solid angles are .
Union complexity ?Trivial bound: O(n3).
fatthinIt is sufficient to bound the number of intersection vertices.
Our results: New union bounds
-fat tetrahedra, of arbitrary sizes: O*(n2) .
arbitrary side-length cubes: O*(n2) .
-fat trihedral wedges: O*(n2) .
-fat triangular prisms, having cross sections of arbitrary sizes: O*(n2) .
Revisit union of fat trianlgles: O*(n) .
Almost tight.Follows easily by our analysis.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*(n2) .
Thin dihedral wedges (almost half-planes) create a grid with (n3) 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.
uvwThe tetrahedron is a dihedral wedge inside most of these cell.Apply the bound O*(n2) of [Pach, Safruti, Sharir 2003].
(1/r)-cutting:From tetrahedra to wedgesT 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 most n/r tetrahedra facets of T .
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 arrangement A(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/rtetrahedra facets of T , with high probability.C
The problem decompositionConstruct 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 verticesFix 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 FThe 2D cross-sections of all cells intersecting F is a2D arrangement of lines.Overall number of cells: O*(r2) . Our bound improves the trivial bound by roughly an order of magnitude. The trivial bound is O*(r2).
The number of cells that meet two facetsTwo facets of T can meet *(r2) cells.The construction is impossible for three facets of T !F1F2
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*(n2) .
Thank you
Union of fat tetrahedraInput:A set of n fat tetrahedra in R3 of arbitrary sizes.
Result:Union complexity:O(n2)Almost tight.
Special case:Union of cubes of arbitrary sizes.
fatthinA 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 tetrahedronA tetrahedron T is -fat if:Each pair of its facets define an -fat dihedral wedge.Each triple of its facets define an -fat trihedral wedge.
Classification of the intersection verticesOuter vertex: The intersection of an edge of atetrahedron with a facet ofanother tetrahedron.Overall : O(n2) .
Inner vertex:The intersection of three facets ofthree distinct tetrahedra.Overall : O(n3) .
Reduce the problem to:How many inner vertices appearon the boundary of the union?uv
The union of fat wedges:A quadratic lower bound constructionMerge the wedges in R and in B so that they form a 2D-grid on W.RBWThe right facet of W shaves the edges of the wedges in R and in B. The number of vertices of the union is (n2).
The union of fat trihedral wedges: An almost quadratic upper bound[Pach, Safruti, Sharir 2003]The union of n (,)-substantially fat trihedral wedges: O*(n2).
The combinatorial complexity of the union of n congruent 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 problemsUnion 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. The number of the non-sampled points is n/r, with high probability!n/r
Constructing (1/r)-cuttingsChoose a random sample R of O(r log r) of the planes containing tetrahedra facets in T (r is a fixed parameter).
Form the arrangement A(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/rtetrahedra facets of T , with high probability.CUse the hierarchical decomposition of Dobkin & Kirkpatrick
Triangulating a cell: The hierarchical decomposition of Dobkin & KirkpatrickHierarchical representation of a convex polyhedron C (An informal description):Construct a (large) independent set V1 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 13 recursively. Bottom of recursion: The new polyhedron Ck is a simplex.C1=CC3C2k = O(log r) number of levels in the recursion.
The DK-hierarchical decompositionClaim: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.Due to the stabbing line property.There are O(r log r) planes in R.