240 likes | 318 Views
Explore the fascinating world of geometric permutations induced by transversal lines passing through a fixed point. Discover how these permutations can be maximized and delve into known facts and special cases in computational and combinatorial geometry. Refute a conjecture, settle specific cases, and offer a new conjecture. Learn about the impact of neighbor pairs and delve into the intricacies of geometric permutations in various scenarios. Find out about the consequences of neighbor pairs and challenge existing theories in higher dimensions. Uncover the complexity of arranging convex bodies and lines to maximize geometric permutations.
E N D
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: <2,3,1> l1: <1,2,3> l2 2 3 1 l1
3 2 n-2 1 An example of S with2n-2 geometric permutations <2,3,…,n-2,1> <3,..2,…,n-2,1>
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: • narbitraryballs in Rd have at most (nd-1)GP’s (Smorodinsky, Mitchell, Sharir 1999) • (nd-1)boundwasextended to fat objects (Katz, Varadarajan 2001) • nunit balls in Rdhave at mostO(1)GP’s (Zhou, Suri 2001) 4
A result and a damage!!! A result settling a specificgeneral 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. What have we done? 4
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) The Result 4
Lemma: S = family of n convex bodies in Rd Two rays, r and r’ emanating from O and meet S O ĝd(n) = (nd-1) (cont) Then r and r’ must meet S in the same order!
r’ r O ĝd(n) = (nd-1) (cont) r and r’ must meet S in the same order! For otherwise….
l = oriented line transversal to S through O. S-l = those intersected bylbefore the origin. S+l =those intersected bylafter the origin. ĝd(n) = (nd-1) (cont) l O The lemma implies that two lines landrthrough O induce the same GP iff they induce the same (“before, after”) origin partition. That is: (S-l ,S+l) = (S-r ,S+r )
l = oriented line transversal to S through O. S-l = those intersected bylbefore the origin. S+l =those intersected bylafter the origin. ĝd(n) = (nd-1) (cont) The question is therefore: How many such partitions (S-l ,S+l) exist? body bS, take a hyperplane h separating it from the origin
Unit Sphere Sd-1 h b1 . O B1 is crossed before O ĝd(n) = (nd-1) (cont) h b 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!
bi bj 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, bjappear consecutive:
3 2 n-2 1 Neighbors Example (2,3) (2,4) ….. (2,n-2)
Neighbors No neighbors !!!
Neighbors Lemma [Sharir, Smorodinsky 03] In Rd, if Nis 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 replaceO 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: • With balls • All lines transversals pass through the origin
Compensation Note that these neighbor pairs determines the GP. “better” conjecture Define neighbors as follows: • Be consecutive in at least “many” GP’s • “many”= some constant k > 1 “better conjecture”: o(n2) such neighbors. If so, its good!!!
I really have to stop, the neighbors are complaining!!!