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On Geometric Permutations Induced by Lines Transversal through a Fixed Point

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.

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On Geometric Permutations Induced by Lines Transversal through a Fixed Point

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  1. 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………

  2. 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. 3 2 n-2 1 An example of S with2n-2 geometric permutations <2,3,…,n-2,1> <3,..2,…,n-2,1>

  4. Motivation? YES!!!

  5. Problem Statement gd(S) = the number of geometric permutations of S gd(n) = max|S|=n {gd(S)} ? < gd(n) < ?

  6. 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

  7. 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

  8. 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

  9. 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!

  10. r’ r O ĝd(n) = (nd-1) (cont) r and r’ must meet S in the same order! For otherwise….

  11. 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 )

  12. 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 bS, take a hyperplane h separating it from the origin

  13. Unit Sphere Sd-1 h b1 . O B1 is crossed before O ĝd(n) = (nd-1) (cont) h b O

  14. 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).

  15. The lower bound (nd-1)in (Smorodinsky, Mitchell, Sharir 1999)is such that all lines pass through the origin!

  16. 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:

  17. 3 2 n-2 1 Neighbors Example (2,3) (2,4) ….. (2,n-2)

  18. Neighbors No neighbors !!!

  19. Neighbors Lemma [Sharir, Smorodinsky 03] In Rd, if Nis the set of neighbor pairs of S, then gd(S)=O(|N|d-1).

  20. 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!!!

  21. 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>

  22. 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

  23. 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!!!

  24. I really have to stop, the neighbors are complaining!!!

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