On Geometric Permutations Induced by Lines Transversal through a Fixed Point
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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………
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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>
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An example of S with 2n-2 geometric permutations
1
<2,3,…,n-2,1><3,..2,…,n-2,1>
2
3
n-2
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Motivation?
YES!!!
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Problem Statement
gd(S) = the number of geometric permutations of S
gd(n) = max|S|=n {gd(S)}
? < gd(n) < ?
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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)
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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.
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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)
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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
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ĝd(n) = (nd-1) (cont)
r and r’ must meet S in the same order!For otherwise….
O
r r’
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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 )
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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
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ĝd(n) = (nd-1) (cont)
b O
h b1
O.
hUnit Sphere
Sd-1
B1 is crossed before O
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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).
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The lower bound (nd-1) in
(Smorodinsky, Mitchell, Sharir 1999) is such that all lines pass through the origin!
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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
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Neighbors
1 2
3
n-2
Example
(2,3)
(2,4)
…..
(2,n-2)
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Neighbors
No neighbors !!!
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Neighbors Lemma [Sharir, Smorodinsky 03]
In Rd, if N is the set of neighbor pairs
of S, then gd(S)=O(|N|d-1).
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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!!!
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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 >
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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
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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!!!
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I really have to stop, the
neighborsare complaining!!!