On the union of cylinders in 3-space Esther Ezra

On the union of cylinders On the union of cylinders in 3-spacein 3-space

EstherEsther Ezra Ezra Duke Duke

UniversityUniversity

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

Combinatorial problemWhat is the combinatorial complexity of the boundary of the union?(vertices/edges/faces of the arrangement A(K) of the cylinders that are not contained in the interior of any cylinder).

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(s(n) log n) [Efrat Katz. 1999].

Each of the angles

O(n1+) , for any >0 . r

r’

r’/r ,and 1.

r diam(C) ,

D C, < 1 is a constant.

rCD

depends linearly on 1/ .

DS-sequence of order s on n symbols. (s is a fixed constant). s(n) O(n) .

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) [Agarwalm 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

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 prisms, having cross sections of arbitrary sizes.Union complexity: O*(n2) .

Each of these bounds is nearly-optimal.

fat

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

Combinatorial problemWhat is the combinatorial complexity of the boundary of the union?

Trivial bound: O(n3).

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

Quadratic lower boundsQuadratic lower bounds

R

B

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

Each blue intersection line of a consecutive pair of cylinders

in B intersects all the red cylinders in R.

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.

h

h

h is the xy-plane.

““fatness” in the context of fatness” in the context of cylinderscylinders

Theorem:Let K’ K be a subset of K that captures most of the union vertices.There exists a direction d, such that K hd is fat, for any K K’, where hd is a plane perpendicular to d.

The 2D cross section of a cylinderK on hd is a fat ellipse.

If we sweep hd along K’, the 2D cross section is always fat.

hd

K

hd

d is the z-axis.

Envelopes in 3DEnvelopes in 3DInput:F = {F1, …, Fn} a collection of n bivariate 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 R2 .

The complexity envelopesThe complexity envelopes[Sharir 1994]The combinatorial 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 sandwich regionThe sandwich region[Agarwal etal. 1996, koltun sharir 2003]The combinatorial complexity of the sandwich regionenclosed between the lower envelope of n simple algebraic surfaces in 3-space and the upper envelope of another such collection is O*(n2) .

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

• Decompose space into prism cells .

• Partition the boundary of the cylindersinto canonical strips.

• Show that in each most of the union vertices v appear on the sandwich region enclosed between the lower envelope of the lower strips and the upper envelopeof the upper strips.

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

(1/r)-(1/r)-cutting:cutting:From cylinders to envelopes

K is a collection of n cylinders 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 elements of the input.

Constructing (1/r)-cuttings:

1. Project all the cylinders in K onto the xy-plane.

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

Each cylinder is projected to a

strip.

Constructing (1/r)-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)

C

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

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

Each prism subcell meets only n/r silhouette-lines of the cylinders in K .

The problem decompositionThe problem decompositionConstruct a (1/r)-cutting for F as above.Fix a prism-cell of .

Classify each cylinder K that meets as:

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

• small - otherwise.

HH’

w

The number of small cylinders in a single prism-cell

K

l1

l2

l1

l2

2r

The silhouette-lines of K do not meet .

The projection onto the xy-plane.

Claim:

A small cylinder K within must have a silhouette-line crossing .

w

The problem decomposition

Each prism-cell of meets

• At most n large cylinders.

• At most n/r small cylinders.

Next stage:Show that large cylinders behave as functions within .Process in recursion all the small cylinders.

Classification of the union vertices

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

• LLL – if all three cylinders that are incident to v are large in .

• LLS – if two of these cylinders are large and one is small in .

• LSS - if one of these cylinders is large and the other two are small in .

• SSS – if all these cylinders are small in .

Bounding the number of LLL-vertices

Theorem:The number of LLL-vertices in is O*(n2).

Proof sketch:Partition the boundary of the cylinders into M canonical strips .

A direction is good for a strip if:

• The angle between and the normal n to H (or H’) is small (in terms of M).

• Each line tangent to forms a large angle (in terms of M) with .

HH’

w

Large constant.

n

Bounding the number of LLL-vertices

A direction is good for a vertex v of the union,incident upon three strips 1, 2, 3,

if it is good for each of 1, 2, 3.

Lemma:Each vertex v of the union has at least one good direction , taken from a (small) set of overall O(1) directions.

1

2

3

v

n

Depends on M.

Bounding the number of LLL-vertices

Lemma:Let be a good direction for a

vertex v =12 3

of the union. Then:• Any line parallel to

intersects 1 at most once.• If we enter into the cylinder

K1 bounded by 1 in the -direction, we exit before leaving K1 .

HH’

n

w

vv’

Bounding the number of LLL-vertices

The strips behave as functions in the -direction inside .

Each LLL-vertex appears on the sandwich region enclosed between the upper envelope of the -upper strips and the lower envelope of the -lower strips.

Overall: O*(n2) .

The case of congruent cylinders

Since all cylinders have equal radii,all cylinders K meeting are either large or small within .

Each vertex v of the union that appears in is either LLL or SSS (no LLS, LSS).

The case of congruent cylinders• Construct a recursive (1/r)-cutting for K .

Number of cells in the cutting: O(r2) .Each cell meets at most

• n large cylibders of F . n/r small cylinders of F.

• Bound LLL-vertices in each before applying a new recursive step.

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

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

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

Number of (SSS) vertices on the

union boundary.

Cylinders with arbitrary radiiTheorem:The number of LLS- and LSS-vertices in is O*(n2).

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

• Bound LLL-, LLS-, LSS-vertices in each before applying a new recursive step.

• Bound SSS-vertices by brute-force at the bottom of the recursion.

The overall bound is: O*(n2) .

Thank youThank you

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

Union of “fat” tetrahedraUnion 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.

fat

thin

A cube can be decomposed into O(1) fat tetrahedra.