On Geometric Permutations Induced by Lines Transversal through a Fixed Point

Shakhar Smorodinsky Courant institute, NYU

Joint work with Boris Aronov

Please label me as a computational geometer and a combinatorial

geometer………

Geometric Permutations S - a set of disjoint convex bodies in Rd

A line transversal l of S induces a geometric permutation of S

l2

l11

2 3

l1: <1,2,3> l2: <2,3,1>

An example of S with 2n-2 geometric permutations

1

<2,3,…,n-2,1><3,..2,…,n-2,1>

2

3

n-2

Motivation?

YES!!!

Problem Statement

gd(S) = the number of geometric permutations of S

gd(n) = max|S|=n {gd(S)}

? < gd(n) < ?

Known Facts

g2(n) = 2n-2 (Edelsbrunner, Sharir 1990) gd(n) = (nd-1) (Katchalski, Lewis, Liu 1992) gd(n) = O(n2d-2) (Wenger 1990)

Special cases: n arbitrary balls in Rd have at most (nd-1) GP’s

(Smorodinsky, Mitchell, Sharir 1999) (nd-1) bound was extended to fat objects (Katz, Varadarajan 2001) n unit balls in Rd have at most O(1) GP’s (Zhou, Suri

2001)

4

What have we done?

A result and a damage!!! A result settling a specific general case!

Specific = all lines pass through a fixed point

General = arbitrary convex bodies We refute a conjecture of [Sharir, Smorodinsky 2003]

about the number of “neighbor pairs” and offer a “better” conjecture.

4

The Result

ĝd(S) = the number of geometric permutations of S

induced by lines passing through a fixed point

ĝd(n) = max|S|=n{ĝd(S)}

Thm: ĝd(n) = (nd-1)

4

ĝd(n) = (nd-1) (cont) Lemma: S = family of n convex bodies in Rd Two rays, r and r’ emanating from O and meet S

Then r and r’ must meet S in the same order!O

ĝd(n) = (nd-1) (cont)

r and r’ must meet S in the same order!For otherwise….

O

r r’

ĝd(n) = (nd-1) (cont)

l = oriented line transversal to S through O.

S-l = those intersected by l before the origin.

S+l = those intersected by l after the origin.

O

The lemma implies that two lines l and r through O induce the same GP iff

they induce the same (“before, after”) originpartition. That is:

l

(S-l , S+

l) = (S-r , S+

r )

ĝd(n) = (nd-1) (cont)

l = oriented line transversal to S through O.

S-l = those intersected by l before the origin.

S+l = those intersected by l after the origin.

The question is therefore:

How many such partitions (S-l , S+

l) exist?

body b S, take a hyperplane h separating it from the origin

ĝd(n) = (nd-1) (cont)

b O

h b1

O.

hUnit Sphere

Sd-1

B1 is crossed before O

Consider the arrangement of these great circles

A connected component C, corresponds to a set of line orientations with at most one

(“before, after”) partition.

A fixed permutation in C

C

Hence, #GP’s ≤ # faces which is O(nd-1).

The lower bound (nd-1) in

(Smorodinsky, Mitchell, Sharir 1999) is such that all lines pass through the origin!

The Damage!!!Many “Neighbors” can exist!

S- a set of convex bodies in Rd

Two bodies bi, bj in S are called neighbors

[Sharir, Smorodinsky 03]

If geometric permutation for which

bi, bj appear consecutive:

bi bj

Neighbors

1 2

3

n-2

Example

(2,3)

(2,4)

…..

(2,n-2)

Neighbors

No neighbors !!!

Neighbors Lemma [Sharir, Smorodinsky 03]

In Rd, if N is the set of neighbor pairs

of S, then gd(S)=O(|N|d-1).

Neighbors (cont)

Thm [Sharir, Smorodinsky] 03:

In the plane (d = 2) O(n) neighbors.

Conjectured: few neighbors in higher dimensions (d > 2)

Embarrassingly … we disprove it!!!

Many Neighbors (cont)

S1={s1,s2,…,sn/2}, S2 = {b1,b2,…,bn/2}

We realize the following GP’s for S1:

1: < sn/2,…, s3, s2 , O, s1 >

2: < sn/2,…, s3 , O, s1, s2 >…

i: < sn/2,…, si+1, O, s1, s2…si >

For any Such i, we can replace O with j:

j: < bj, …, b1, O, bj+1,…, bn/2>Hence: i, j {1,…,n/2} we realize

< sn/2,…, si+1, bj, …, b1, bj+1,…, bn/2, s1, s2…si >

Many Neighbors (cont)

Hence: i, j {1,…,n/2} we realize

< sn/2,…, si+1, bj, …, b1, bj+1,…, bn/2, s1, s2…si >

So i, j {1,…,n/2},

(si+1, bj) are neighbors

We can actually realize it:

1. With balls

2. All lines transversals pass through the origin

Compensation

Note that these neighbor pairs determines the GP.

“better” conjecture

Define neighbors as follows:

1. Be consecutive in at least “many” GP’s

2. “many”= some constant k > 1

“better conjecture”:

o(n2) such neighbors. If so, its good!!!

I really have to stop, the

neighborsare complaining!!!