On the Union of Cylinders in Esther Ezra Duke University On the Union of Cylinders in 3 Esther Ezra Duke University.

On the Union of Cylinders On the Union of Cylinders in in 3

EstherEsther EzraEzra

Duke UniversityDuke University

S = {S1, …, Sn} a collection of n simply-shaped bodies in d-space.

The union of S consists of all region of d that are covered by at least one element of S .

Example: Union of triangles in the planeAll portions of the plane that are covered by the triangles.

Union of simply-shaped bodiesUnion of simply-shaped bodies

The union boundary

The union has two holes

The arrangement A(S) is the subdivision

of space induced by S .

The maximal number of

vertices/edges/faces of A(S) : (nd)

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!).

The Union as a substructure in The Union as a substructure in arrangementsarrangements

Combinatorial complexity.

Union is a substructure of

the arrangement

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/ log (1/) . [Pach, Tardos. 2002]

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] . Generalization: Arbitrary side-length cubesUnion complexity: O*(n2) [Ezra, Sharir 2007].

Simple curved objects n congruent infinite 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.

rC

r diam(C) , D C, < 1 is

a constant.

D

The case of infinite cylindersThe case of infinite cylinders

Input:K = {K1, …, Kn} a collection of n infinite cylinders in R3 of arbitrary radii.

What is the combinatorial complexity of the union?

Trivial bound: O(n3).

Conjectured by [agarwal, sharir 2000]:Upper bound: O(n2) (?)

It is crucial that the cylinders are infinite.Otherwise, the union complexity is (n3) .

The vertices of the unionThe vertices of the union

Each vertex v of the union is generated

by the intersection of a triple of cylinders.

Reduce the problem to:

How many (intersection) vertices

appear on the boundary of the union?

# edges + # faces = O(# vertices + n2)

v

Quadratic lower boundsQuadratic lower bounds

T

B

The number of vertices of the union is Ω(n2).

Each (top) intersection line of a consecutive pair of cylinders

in B intersects all the top cylinders in T.

B

T

Envelopes in Envelopes in dd-space-space

Input:

F = {F1, …, Fn} a collection of n

(d-1)-variate functions.

The lower envelope EF of F is the

pointwise minimum of these functions:

EF(x) = min{F F} F(x) , for x d-1 .

[Sharir 1994]

The complexity of the lower envelope

of n simple algebraic surfaces in

d-space is O*(nd-1) .

The lower envelope is monotone.

O*(n2) for d=3 .

The sandwich regionThe sandwich region

[Agarwal et al. 1996, koltun sharir 2003]The complexity of the sandwich region enclosed between the lower envelope of n simple algebraic surfaces in d-space and the upper envelope of another such collection is O*(nd-1) , for d 4.

For d=3, the complexity of the sandwich region is: O*(n2)

Main idea: Main idea: Reduce cylinders to envelopesReduce cylinders to envelopes

• Decompose space into vertical prism cells .

• Partition the boundary of the cylindersinto canonical strips.

• Show that in each cell most of the union vertices v appear on the sandwich region enclosed between two envelopes of the strips.

Apply the bound O*(n2) of [Agarwal et al. 1996].

From cylinders to envelopesFrom cylinders to envelopes

Use (1/r)-cuttings 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 elements of the input .

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 How to construct (1/r)-(1/r)-cuttingscuttings

The number of the non-sampled points is n/r,

with high probability!

n/r

Constructing Constructing (1/r)-(1/r)-cuttings:cuttings:

First step:

Construct a (1/r)-cutting in the xy-plane:

1. Project all the cylinders in K onto the xy-plane. Obtain a set of n strips.

Let L be the set of the bounding lines of the projections of K .

Each cylinder is projected to

a strip.

Constructing Constructing (1/r)-(1/r)-cuttings:cuttings:

2. Choose a random sample R of O(r log r) lines of L (r is a fixed parameter).

3. Form the planar arrangement A(R) of R: Each cell C of A(R) is a convex polygon.Overall complexity: O(r2 log2r).

4. Triangulate each cell C.Number of simplices: O(r2 log2r)

Theorem [Clarkson & Shor] [Haussler & Welzl] :Each simplicial cell is crossed by n/r lines of L, with high probability.

C

The cutting propertyThe cutting property

Second step:

5. Lift all the simplices in the z-direction into vertical prisms .Obtain a collection of O(r2 log2r) prisms.

Each prism cell meets only n/r

silhouette-lines of the cylinders in K .

The problem decompositionThe problem decompositionConstruct a (1/r)-cutting for K as above.Fix a prism-cell of .Classify each cylinder K that meets as:

• wide – if the radius of K satisfies: w/2, where w is the width of .

• narrow - otherwise.

Main goal:• Wide cylinders behave as functions within .• The number of narrow cylinders in is small ( n/r ):

A narrow cylinder in must have a silhouette-line crossing .

HH’

w

Classification of the union vertices

Each vertex v of the union that appears in is classified as:

• Good - if all three cylinders that are incident to v are wide in . v lies in the sandwich region enclosed between two envelopes.

• Bad - otherwise.

Framework:• Construct a recursive (1/r)-cutting for K .

• 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. O*(n2) .

Apply the nearly-quadratic bound of

[Agarwal et al. 1996].

Overall bound:O*(n2)

The recursion treeThe recursion tree

N = n, W = 0

N n/r, W n N n/r, W n N n/r, W n1

r22

= 3 Partitioning 3 into O*(r2) prisms: Some narrow become wide.

…

Dispose of overall O*(n2) good vertices.

Nn/r2, Wn/r Nn/r2, Wn/r Nn/r2, Wn/r ……

Dispose of overall O*(n2) good vertices.

N=O(1), Wn/r N=O(1), Wn/r N=O(1), Wn/r

… #good vertices = O*(n2)

#bad vertices = O*(n2)

The overall bound is: O*(n2) .# nodes in bottom of recurrence = O*(n2)

Thank YouThank You

Previous results in 3D:Previous results in 3D:Fat ObjectsFat Objects

Fat tetrahedran -fat tetrahedra of arbitrary sizes.Union complexity: O*(n2) [Ezra, Sharir 2007].

Special cases: n arbitrary side-length cubes.Union complexity: O*(n2) .

n -fat triangular (infinite) prisms, having cross sections of arbitrary sizes.Union complexity: O*(n2) .

n -fat triangles.A simpler proof that shows a bound of O*(n) .

Each of these bounds is nearly-optimal.

fat

Extend the notion of “fatness”Extend the notion of “fatness”A cylinder is not fat!

A wider definition for fatness:We can sweep K with a plane h whose 2D cross section with each K K is always fat.

The 2D cross section is a set of fat ellipses.

Lemma:Let K’ K be a subset of K that captures most of the union vertices.Then we can sweep K’ with a plane h whose 2D cross section with each K K’ is always fat.

h

h

h is the xy-plane.

Envelopes in Envelopes in dd-space-space

Input:

F = {F1, …, Fn} a collection of n (d-1)-variate functions.

The lower envelope EF of F is the pointwise minimum of these functions.

That is, EF is the graph of the following function:

EF(x) = min{F F} F(x) , for x d-1 .

The lower envelope is monotone.

The complexity of envelopesThe complexity of envelopes

[Sharir 1994]

The complexity of the lower envelope

of n simple algebraic surfaces in

d-space is O*(nd-1) .

For d=3, the complexity of

the lower envelope: O*(n2)

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 How to construct (1/r)-(1/r)-cuttingscuttings

The number of the non-sampled points is n/r,

with high probability!

n/r

The number of narrow cylinders in a single prism-cell

K

l1

l2

l1

l2

2

The silhouette-lines of K do not meet .

The projection onto the xy-plane.

Claim: A narrow cylinder K in must have a silhouette-line crossing .

w

Conclusion: By the cutting properties, meets only a small number of narrow cylinders.

Wide cylinders behave as functions Wide cylinders behave as functions inside inside

Main idea:Partition the boundary of the cylinders into canonical strips .

A direction is good for a strip if when entering from into the cylinder K bounded by in the -direction, we exit before leaving K.

Key lemma:Each strip of a wide cylinder has many good directions.

There is a common good direction ,for which all strips behave as functions.

The strips behave as functions in the

-direction inside .

Construct the envelopes in the

-direction.

From technical reasons.

Wide cylinders behave as functions Wide cylinders behave as functions inside inside

Main idea:Partition the boundary of the cylinders into canonical strips .

A direction is good for a strip if when entering from into the cylinder K bounded by in the -direction, we exit before leaving K.

is good for a vertex v of the union,if it is good for each of its three incident strips.

The strips behave as functions in the

-direction inside .

From technical reasons.

1

2

3

v

Wide cylinders behave as functions Wide cylinders behave as functions inside inside

Key lemma:

Each vertex v incident to three wide cylinders has many good directions.

There is a common good direction ,for which all strips behave as functions.

Construct the envelopes in the

-direction.

Wide cylinders behave as functions Wide cylinders behave as functions inside inside

Main idea:Partition the boundary of the cylinders into canonical strips .

A direction is good for a strip if when entering from into the cylinder K bounded by in the -direction, we exit before leaving K.

Key lemma:Each strip of a wide cylinder has many good directions.

There is a common good direction ,for which all strips behave as functions.

The strips behave as functions in the

-direction inside .

Construct the envelopes in the

-direction.

From technical reasons.

The overall analysisThe overall analysis

• Construct a recursive (1/r)-cutting for K .

• 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) .

The recursion treeThe recursion tree

N = n, W = 0

N n/r, W n N n/r, W n N n/r, W n1

r22

= 3 Partitioning 3 into O*(r2) prisms: Some narrow become wide.

…

Dispose of overall O*(n2) good vertices.

Nn/r2, Wn/r Nn/r2, Wn/r Nn/r2, Wn/r ……

Dispose of overall O*(n2) good vertices.

N=O(1), Wn/r N=O(1), Wn/r N=O(1), Wn/r

… #good vertices = O*(n2)

#bad vertices = O*(n2)

The overall bound is: O*(n2) .# nodes in bottom of recurrence = O*(n2)

The case of congruent cylinders:

A simple proofSimplifying the proof of [Agarwal, Sharir 2000]:Since all cylinders have equal radii,all cylinders K meeting are either wide or narrow within .

Each bad vertex v in must be incident to a triple of narrow cylinders!(v cannot be incident to both wide and narrow cylinders.)

Does not hold for cylinders with arbitrary radii

The case of congruent cylinders

• Construct a recursive (1/r)-cutting for K (r - a sufficiently large constant).Number of cells in the cutting: O(r2) .Each cell meets at most:

• n wide cylinders of K . n/r narrow cylinders of K.

• Bound # good vertices in each before applying a new recursive step.

1. Bound # bad vertices by brute-force at the bottom of the recursion.

U(n) = O*(n2) + O*(r2) U(n/r)

Solution: U(n) = O*(n2) .

Number of (bad) vertices on the

union boundary.