Almost tight bound for the union of fat tetrahedra in R 3 Esther Ezra Micha Sharir Duke University...
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Transcript of Almost tight bound for the union of fat tetrahedra in R 3 Esther Ezra Micha Sharir Duke University...
Almost tight bound for the Almost tight bound for the union of fat tetrahedra in Runion of fat tetrahedra in R33
EstherEsther Ezra Micha SharirEzra Micha Sharir
Duke Duke UniversityUniversity
Tel-Aviv Tel-Aviv UniversityUniversity
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)
Arrangement of geometric objectsArrangement of geometric objects
Combinatorial complexity.
Each object has a constant description
complexity
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/facesthat form the boundary of the union of the bodies in S ?
Trivial bound: O(nd) (tight!).
Union of simply-shaped bodies:Union of simply-shaped bodies: A A substructure in arrangementssubstructure in arrangements
Combinatorial complexity.
Previous results in 2D:Previous results in 2D:Fat objectsFat 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(n1+) , for any >0 .r
r’
r’/r ,and 1.
r diam(C) ,
D C, < 1 is a constant.
rC
D
depends linearly on 1/ .
Previous results in 3D:Previous results in 3D:Fat ObjectsFat Objects
Congruent 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.
rC
r diam(C) , D C, < 1 is
a constant.
D
Special case: Special case: FatFat tetrahedra tetrahedra
Input:
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).
fat
thin
It is sufficient to bound the number of
intersection vertices.
Our results: New union boundsOur 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 wedgesThe 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 Ω(n3) vertices.
The bound depends
linearly on 1/ .
The dihedral angle.
Main idea: Reduce tetrahedra to Main idea: Reduce tetrahedra to dihedraldihedral wedgeswedges
• 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.
u
v
w
The tetrahedron is a dihedral wedge
inside most of these cell.
Apply the bound O*(n2) of [Pach, Safruti, Sharir 2003].
(1/r)-(1/r)-cutting: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 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(r3 log3r).
3. 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 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
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 facets
Two facets of T can meet Ω*(r2) cells.
The construction is impossible for three facets
of T !
F1
F2
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) .
Thank youThank you
Union of “fat” tetrahedraUnion of “fat” tetrahedra
Input:
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.
fat
thin
A cube can be decomposed into O(1) fat tetrahedra.
-fat-fat dihedral/trihedral wedges 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-fat tetrahedron 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 Classification of the intersection verticesvertices
Outer 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?
u v
The union of fat wedges:The union of fat wedges:A quadratic lower bound constructionA quadratic lower bound construction
Merge the wedges in R and in B so that they form a 2D-grid on W.
RB
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: The union of fat trihedral wedges: An almost quadratic upper boundAn 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 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)-cuttings1. 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(r3 log3r).
3. 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
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):
• 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 1—3 recursively.Bottom of recursion: The new polyhedron Ck is a simplex.
C1=C C3 C2
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
|)(|||1
COCk
ii
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.