Title: Substructures in Geometric Arrangements Esther Ezra Advisor: Micha Sharir
1Substructures in Geometric ArrangementsEsther
Ezra Advisor Micha Sharir
2Problem IUnion of simply-shaped bodies
- 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.
3Problem II A single cell in an arrangement of
geometric objects
- Input
- S S1, , Sn a collection of n geometric
objects in - d-space of constant description complexity.
- A(S) The arrangement induced by S .
- The problem
- What is the maximal number of
- vertices/edges/faces that form the
- boundary of a single cell of A(S) ?
- Trivial bound O(nd) (NOT tight!).
A hole in the complement of the union.
4- Union of Geometric Objects
5Previous results in 2D
- The case of pseudodisks
- A collection of n compact connected sets, s.t.,
- the boundaries of any pair of sets intersect at
most twice. - Arise in the case of a convex translating robot R
- amid convex pairwise disjoint obstacles.
- Union complexity 6n-12 O(n)
- Kedem et al. 1986.
6Previous results in 2DFat objects
Each of the angles ? ?
- A set of 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) - A set of n convex ?-fat objects.
- Union complexity O(n) Efrat Sharir. 2000.
- A set of n ?-curved objects.
- Union complexity O(?s(n) log n) Efrat Katz.
1999.
r/r ? ? , and ? ?1.
r
O(n1?) , for any ?gt0 .
r
r ? ?? diam(C) , D ? C, ? lt 1 is a constant.
DS-sequence of order s on n symbols. (s is a
fixed constant). ?s(n) ? O(n) .
C
r
D
7Previous results in 3D
- Translational motion planning
- P is a set of k convex polyhedra with n facets
(overall) - that arise in the case of a convex translating
robot R - amid convex (pairwise disjoint) obstacles.
- Each P ? P is the Minkowski sums of (-R) and an
obstacle. - Union complexity O(nk log k) Aronov, Sharir
1997 . - The general problem
- P is any set of k convex polyhedra
- with n facets (overall).
- Union Complexity O(k3 nk log k) Aronov,
Sharir, Tagansky 1997 .
Cannot be applied when R is non-convex.
8Previous results in 3DFat Objects
- Congruent cubes
- A set of n arbitrarily aligned (nearly) congruent
cubes. - Union complexity O(n2) Pach, Safruti, Sharir
2003 . - Simple curved objects
- A set of n congruent inifnite cylinders.
- Union complexity O(n2) Agarwal Sharir 2000.
- A set of n ?-round objects.
- Union complexity O(n2) Aronov et al. 2006.
- Each of these bounds is nearly-optimal.
r ? ?? diam(C) , D ? C, ? lt 1 is a constant.
C
r
D
9Special case Fat tetrahedra
- Input
- T T1, , Tn a collection of n fat tetrahedra
in R3. - Combinatorial problem
- What is the combinatorial complexity of
- the boundary of the union?
- Trivial bound O(n3) .
fat
A cube can be decomposed into O(1) fat
tetrahedra.
thin
It is sufficient to bound the number of
intersection vertices.
10Classification 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
11?-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 ? gt 4?/3 .
?
The solid angle.
?
12?-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.
?
?
13The 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/? .
Thin dihedral wedges (almost half-planes) create
a grid with O(n3) vertices.
14The union of fat wedgesA quadratic lower bound
construction
R
W
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 O(n2).
15The 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.
16More 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.
17Our results
Almost tight.
- The combinatorial complexity of the union of n
?-fat tetrahedra, of arbitrary sizes, is O(n2) . - The combinatorial complexity of the union of
polyhedra, of arbitrary sizes, that can be
decomposed or covered by n fat tetrahedra, is
O(n2) . - The combinatorial complexity of the union of n
?-fat trihedral wedges is O(n2) . - The combinatorial complexity of the union of n
?-fat triangular prisms, having cross sections
of arbitrary sizes is O(n2) .
Special case Union of cubes of arbitrary sizes.
Follows easily by our analysis.
Follows easily by our analysis.
18From 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 is a subdivision of
- space into openly disjoint simplicial
- subcells ?, s.t., each cell ? meets at
- most n/r tetrahedra facets of T .
?
19Constructing (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 REach 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
20Triangulating 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 CC1 . - Remove the vertices in V1 from C1Fill 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.
C1C
C3
C2
k O(log r) number of levels in the recursion.
21The 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
22Properties 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.
23The problem decomposition
- Construct an edge-sensitive (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 (inner) vertex v of the union that
appears in ? as - Good - if all three tetrahedra that are incident
to 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 ? .
24Bounding 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.
?
Immediate when the cells meet an edge of T
(stabbing line property).
25Goal 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.
26The frequency of bad cells
F1
F2
Two facets of T can meet O(r2) cells.
The construction is impossible for three facets
of T !
27The overall analysis
- Construct a recursive edge-sensitive
(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) .
28SpecializationUnion 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 ?.
29The 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-tetrahedra in ?. - WT if one of these triangles is a W-triangle
and the other is a T-trianglein ?. - TT - if both of these triangles are T-triangles
in ?.
?
w
v
u
30Bounding the number of intersection vertices of
the union
- WW vertices
- Easy reduce to the case of
- ?-fat wedges.
- WT, TT vertices
- More involved.
- In particularWT verticesObtain a
nearly-linear bound.
Apply the linear bound of Efrat, Rote, Sharir
1994.
Use a non-trivial variant of the construction.
31T-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.
32The 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 before applying a new
recursive step. - Bound TT vertices by brute-force at the bottom of
the recursion. - U(n) O(n) O(r2)?U(n/r2)
- Solution U(n) O(n) .
Number of vertices on the union boundary.
33- A single cell in geometric arrangements
34A single cell in 2D arrangements
- Arrangement of lines
- Complexity of arrangement O(n2) .
- Complexity of a single cell O(n) .
- Arrangement of segments/triangles
- Complexity of arrangement O(n2) .
- Complexity of a single cell O(n) ?
- Actual bound ?(n?(n)) Guibas et al. 1989
C
The inverse Ackermann function.
35Previous results
Some constant.
- 2D arrangements
- n Jordan arcs, each pair of arcs intersect in at
most s point. - Single cell complexity O(?s2(n))? O(n) Guibas
et al. 1989. - k convex polygons with n edges in total.
- Single cell complexity T(n?(k)) Aronov Sharir
1997. -
- 3D arrangements
- A set T of n triangles.
- The complexity of all non-convex cells in A(T)
O(n7/3 log n) - Aronov Sharir 1990.
- A set S of n low degree algebraic surface
patches. - Single cell complexity O(n2) Halperin Sharir
1995.
Segments s1.
Many components
Curved simply-shaped regions
36Previous results
- d-dimensional arrangements
- A set of n (d-1)-simplices.
- Single cell complexity O(nd-1log n) Aronov
Sharir 1994. - A set S of n low degree algebraic surface
patches. - Generalizing the bound of Halperin Sharir
1995. - Single cell complexity O(nd-1) Basu 2003.
- All bounds are almost tight in the worst case.
- The case of k convex polyhedra with n facets in
R3 - Use Aronov Sharir 1994 O(n2 log n) .
- This bound does not depend on k.
- Conjecture Aronov, Sharir, Tagansky 1997
- The actual upper bound is close to O(nk) .
37Lower bounds for the unbounded cell in 3D
O(nk) vertices
O(k2) vertices
Can be modified to O(nk?(k)) vertices
38Our result
- The combinatorial complexity of a single cell of
A(? ) is O(nk1?) ,? ? gt 0 . - We use a variant of the technique of Halperin
Sharir 1995 . - We present a deterministic algorithm that
constructs a single cell in O(nk1? log3 n) time,
? ? gt 0 .
The bound depends on the number k of polyhedra
Crucial The input regions are of constant
description complexity
39Classification of the intersection vertices
- Outer vertex The intersection of an edge of a
polyhedron with a facet of another
polyhedron.Overall number O(nk) . - Inner vertexThe intersection of three facets of
three distinct polyhedra.Overall number O(nk2)
. - Reduce the problem to
- How many inner vertices appear
- on the boundary of a single cell?
u
40Analysis Exposed convex chains
Classify each vertex v by How long can we freely
go from v when alternating out-of/into the
unbounded cell.
Not meeting any polyhedra
?
1 step
?
After the removal of P 4 steps
41Analysis Continue
- We trace this way Exposed convex chains.
- Number of steps length of the chain
- V(j)(? ) the number of inner vertices of the
unbounded cell of A(? ) with ? j steps.
5 steps
?
?
V(0)(? ) bounds the overall number of inner
vertices of the unbounded cell.
42The overall complexity of exposed chains
- Exposed chains of length ? 4
- Use recurrence V(j)(? ) ? V(j1)(? )
- Exposed chains of length 4 or 5
- Lemma
- The number of vertices on exposed chains of
length ? 5 is O(nk) . - The number of vertices on exposed closed chains
(of length 4) is O(nk) .
Multiply by O(k?).
?
This is the only interesting case.
43Solving the recurrence
- V(j)(? ) O(nk1?) ,? ? gt 0, 0 ? j ? 4
- The combinatorial complexity of a single cell of
A(? ) is O(nk1?) ,? ? gt 0 .
44 45Exposed chains of length ? 5
?
?
?M ? P_3
?M ? P
?
?
MF_1 ? P_2
46Motivation Translational motion planning
The workspace.
- Input
- Robot R , a set A A1, , Ak of k disjoint
obstacles. - The robot and the obstacles are
- (not necessarily convex) polyhedra in 3-space.
- The free space
- The set of all legal placements of R,
- while translating R in 3-space.
collision .
R does not intersect any of the obstacles in A.
47The configuration space
reference point.
- The robot R is mapped to a point.
- Each obstacle Ai is mapped to the set
- Pi (x,y,z) R(x,y,z) ? Ai ? ?
Ai?(-R(0,0,0)) -
- A point p in Pi corresponds to an illegal
placement of R and vice versa.
The forbidden placements of R .
The Minkowski sum.
The expanded obstacle.
48The free space
49Union of convex polyhedra
- Input
- P P1, , Pk a collection of k convex
polyhedra in 3-space with n facets in total (k
n). - Combinatorial problem
- What is the maximal number of vertices/edges/faces
that form the boundary of the union of the
polyhedra in P ?Trivial bound O(n3) (tight!).
It is sufficient to bound the number of
intersection vertices.
50Union of convex polyhedra
- General goal
- Obtaining (natural) special cases where the union
has subcubic/nearly-quadratic complexity. - Computational problem
- An algorithm that constructs the union?
-
- Not efficient when the complexity of the whole
union is high (cubic).
51RestrictionA single component of the free space
- A single component of
- The subset of all placements reachable from a
given initial free placement of R via a
collision-free motion.
52Restatement A single component in the
complement of the union
- Input
- ? P1, , Pk a collection of k convex
polyhedra in 3-space with n facets in total. - The problem
- What is the maximal number of vertices/edges/face
s that formthe boundary of a single component
of ?
Minkowski sum of a convex obstacle with a convex
part of R.
It is sufficient to bound the number of
intersection vertice.
A single cell of A(? ).
53Single (bounded) cell in 2D