1. Almost tight bound for the union of fat tetrahedra in R3 Esther Ezra Micha Sharir
Duke University
Tel-Aviv University

2. 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/faces
of A(S) is: (nd)
Each object has a constant description complexity
Combinatorial complexity.

3. 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 in S ?
Trivial bound: O(nd) (tight!).
Combinatorial complexity.

4. 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(s(n) log n) [Efrat Katz. 1999].
depends linearly on 1/ .
r’/r   ,
and  1. r
O(n1+) , for any >0 . r’
r   diam(C) , D  C,  < 1 is a constant.
DS-sequence of order s on n symbols. (s is a fixed constant). s(n)  O(n) . C r D

5. Previous results in 3D: Fat Objects
Congruent cubes
n 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].
Union complexity is ~ “one order of magnitude” smaller than the arrangement complexity!
Each of these bounds is nearly-optimal.
r   diam(C) , D  C,  < 1 is a constant. C D r

6. Special case: Fat 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.

7. Our results: New union bounds
Almost tight.
-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) .
A cube can be decomposed into O(1) fat tetrahedra.
Follows easily by our analysis.
Follows easily by our analysis.

8. 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

9. The bound depends linearly on 1/ .
The union of fat wedges
[Pach, Safruti, Sharir 2003]
The combinatorial complexity of the union
of n -fat dihedral 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.

10. 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.
The right facet of W “shaves” the edges of the wedges in R and in B.
The number of inner vertices of the union is Ω(n2).

11. 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.
The tetrahedron is a dihedral wedge inside most of these cell.  w v u
Apply the bound O*(n2) of [Pach, Safruti, Sharir 2003].

12. (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 most n/r
tetrahedra facets of T . 

13. 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!

14. 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/r
tetrahedra facets of T , with high probability. C

15. k = O(log r) number of levels in the recursion.
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.

16. The DK-hierarchical decomposition
Claim:
There exists a hierarchical representation for C that satisfies:
k = O(log r) . Stabbing line property: Each line l that stabs C, crosses only O(log r) of its simplices. l
# simplices in the triangulation is O(|C|) .

17. Properties of the overall decomposition 
Follows by
The DK-decomposition properties imply:
The overall number of cells  of  is O(r3 log3r) .
Each tetrahedron edge crosses at most O(r log2r)  O(r) simplices  of  .
Another crucial property to follow.
Due to the stabbing line property.
Relatively small number.

18. 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.
Apply the nearly-quadratic bound of [Pach, Safruti, Sharir 2003].
At least one of these tetrahedra has three (or more) facets that meet  .

19. 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 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. 

20. Our bound improves the trivial bound by roughly an order of magnitude.
Bad cells are scarce
The trivial bound is O*(r2).
The 2D cross-sections of all cells intersecting F is a
2D arrangement of lines.
Overall number of cells: O*(r2) . F
Our bound improves the trivial bound by roughly an order of magnitude.

21. 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 !

22. The overall bound is: O*(n2) .
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) .

23. Specialization: Union of fat triangles
Input:
T = {T1, …, Tn} a collection of n -fat triangles in the plane.
Construct a (1/r)-cutting  for T.
Number of cells in the cutting: O(r2) .
Each cell  meets at most n/r triangles edges of T .
Fix a cell  of .
Classify each triangle T  T as:
W-triangle in  , if  meets at most two edges of T.
T-triangle in  , otherwise.
T behaves as a half-plane or a wedge inside .
T behaves as a triangle inside .

24. The classification of the vertices of the union
Each vertex v of the union that appears in  is classified as:
WW – if the two triangles that are incident to v are W-triangles in .
WT– if one of these triangles is a W-triangle and the other is a T-triangle in .
TT - if both of these triangles are T-triangles in .  w v u

25. Bounding the number of intersection vertices of the union
WW vertices
Easy – reduce to the case of
-fat wedges.
WT, TT vertices
More involved.
Apply the linear bound of [Efrat, Rote, Sharir 1994].

26. T-triangles are scarce
Only a single triangle  can meet all three edges of T.  T
On average, each cell  meets at most n/r2 triangle edges of T-triangles.

27. Use a non-trivial variant of the construction.
The recursive scheme
Construct a (1/r)-cutting  for T. Number of cells in the cutting: O(r2) . Each cell  meets at most
n/r W-triangles edges of T .
 n/r2 T-triangles edges of T.
Bound WW and WT vertices in each  before applying a new recursive step.
Bound TT vertices by brute-force at the bottom of the recursion.
#WT vertices = O*(n) .
Use a non-trivial variant of the construction.

28. The recursive relationship
U(n) = O*(n) + O(r2)U(n/r2)
Solution: U(n) = O*(n) .
Number of TT vertices on the union boundary.

29. Thank you

30. Union 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.

31. -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. 

32. -fat tetrahedron A 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.  

33. 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

34. 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.
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).

35. 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.

36. 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.

37. 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!

38. 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/r
tetrahedra facets of T , with high probability.
Use the hierarchical decomposition of Dobkin & Kirkpatrick C

39. k = O(log r) number of levels in the recursion.
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.

40. 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

41. 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.