Problem session { Ferran homage€¦ · Problem session { Ferran homage This is an informal...
Transcript of Problem session { Ferran homage€¦ · Problem session { Ferran homage This is an informal...
Problem session – Ferran homage
This is an informal recollection of the slides used at EGC, during the problem session devotedto honor Ferran. Some problems are classic, and solved or partially solved, some other arestill open. All of them are to be seen as a small sample of the well-known ability of Ferranin asking interesting questions, and proposing nice open problems.
• Oriol Serra: Midpoints problem
• Prosenjit Bose: The last open problem Ferran asked me
• Bernardo Abrego: Crossings in proximity graphs
• Jean Cardinal: Alternating cell paths in bichromatic arrangements
• David Orden (on behalf of the Madrid group): Art gallery problems (variants)
• Alberto Marquez: Heterochromatic triangulations
• Oswin Aichholzer: Blocking Delaunay triangulations
• David Rappaport: Coins problems
• Alfredo Garcıa: Open problems on compatible graphs
• Joseph Mitchell: Reflexity of point sets
Midpoints ProblemP a set of points in the plane in general positionM(P) set of midpoints of the
(n2
)segments determined by the points.
Question (Ferran): Determine µ(n) = min{|M(P)| : |P| = n}.
O. Serra (UPC) Midpoints problem Barcelona 2015 1 / 1
Midpoints ProblemP a set of points in the plane in general positionM(P) set of midpoints of the
(n2
)segments determined by the points.
Question (Ferran): Determine µ(n) = min{|M(P)| : |P| = n}.
M(P) =1
2|A+ A|
Hurtado, Urrutia [??] µ(n) = (nlog2 3).
Pach (2003) µ(n) ≤ nec log n and lim supn→∞ µ(n)/n =∞(Behrend construction+Freiman Theorem: general position→ sets with no3–term arithmetic progressions.)
Sanders (2010) µ′(n) ≥ n(log n)1/3
log log n
Problem 8.3.(Sanders 2010) Find a constant an absolute constant c > 2/3 suchthat
µ(n) ≥ n logc n.
O. Serra (UPC) Midpoints problem Barcelona 2015 1 / 1
Introduction Results
The last open problem Ferran asked me
Prosenjit Bose
School of Computer ScienceCarleton University
Canada
The last open problem Ferran asked me
Introduction Results
Recent Area of Interest
Biplane Graph
A straight-line embedding of a graph on a point set is biplaneprovided that the graph is the disjoint union of two plane graphs.
The last open problem Ferran asked me
Introduction Results
Open Problem
Question
Given a set of points in the plane, can we build a biplane graphwhose spanning ratio is better than any plane graph?
The last open problem Ferran asked me
Introduction Results
Introduction
Geometric t-spanner
The last open problem Ferran asked me
Introduction Results
Introduction
Geometric t-spanner
The last open problem Ferran asked me
Introduction Results
Introduction
Geometric t-spanner
The last open problem Ferran asked me
Introduction Results
Introduction
Geometric t-spanner
Given a set P of points in the plane, a graph G is a t-spanner witht ≥ 1 provided that
∀x , y ∈ P, dG (x , y) ≤ t · |xy |
The last open problem Ferran asked me
Introduction Results
Introduction
Current Best Result
Xia (2012) showed that the spanning ratio of the Delaunaytriangulation is at most 1.998? Can we do better for biplanegraphs?
The last open problem Ferran asked me
Introduction Results
Biplane graph with lower spanning ratio
Two rotated copies of the empty equilateral triangle Delaunaytriangulation has better spanning ratio.
Spanning ratio of half-θ6-graph is√3 · cosα+ sinα,
where α is angle between uw and closest bisector
The last open problem Ferran asked me
Introduction Results
Biplane graph with lower spanning ratio
Two rotated copies of the empty equilateral triangle Delaunaytriangulation has better spanning ratio.
Spanning ratio of half-θ6-graph is√3 · cosα+ sinα,
where α is angle between uw and closest bisector
The last open problem Ferran asked me
Introduction Results
Rotating the Cones
Two rotated copies of the empty equilateral triangle Delaunaytriangulation has better spanning ratio.
Spanning ratio of half-θ6-graph is√3 · cosα+ sinα,
where α is angle between uw and closest bisector
The last open problem Ferran asked me
Introduction Results
Rotating the Cones
Two rotated copies of the empty equilateral triangle Delaunaytriangulation has better spanning ratio.
Spanning ratio of half-θ6-graph is√3 · cosα+ sinα,
where α is angle between uw and closest bisector
The last open problem Ferran asked me
Introduction Results
Rotating the Cones
Two rotated copies of the empty equilateral triangle Delaunaytriangulation has better spanning ratio.
Spanning ratio of half-θ6-graph is√3 · cosα+ sinα,
where α is angle between uw and closest bisector
The last open problem Ferran asked me
Introduction Results
Rotating the Cones
Two rotated copies of the empty equilateral triangle Delaunaytriangulation has better spanning ratio.
Spanning ratio of half-θ6-graph is√3 · cosα+ sinα,
where α is angle between uw and closest bisectorRotating by π/6 gives spanning ratio of roughly 1.932
The last open problem Ferran asked me
(2008) What is the list number of crossings in a geometric proximity graph of n points in the plane?
k-nearest neighbor graph
each point is adjacent to its k nearest neighbors
Problem. If k=9 or 10, what is the least number of crossings in the k-nearest neighbor graph of an n-element point set?
Crossings of proximity graphs
Crossings of proximity graphs
Cell-Paths
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Alternating Cell-Paths in Bichromatic Arrangements
Ferran, Sevilla, february 2013
Analogy: long plane alternating paths in red-blue points in convexposition (famous poison problem – cf. Delia’s talk this morning)
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Questions
• Maximum value p(n) such that every bichromatic linearrangement in the plane has an alternating cell-path oflength at least p(n)
• Maximum value f (n) such that every line arrangement in theplane has a cell-path of length at least f (n)
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Results
Two main papers:
1. Aichholzer, C., Hackl, Hurtado, Korman, Pilz, Silveira,Uehara, Vogtenhuber, Welzl (Sevilla ComPoSe Workshop,CCCG’13)
2. Hoffmann, Kleist, Miltzow (to appear in MFCS’15)
Results:
• f (n) ∼ n2/3.
• p(n) ≥ n.
• p(n) ≤ 2n − 1 – only one red line!
• Some “near-balanced” bichromatic arrangements do not havealternating paths longer than 2.8n
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Improved Upper Bound
• Hoffmann et al.: For arbitrary k, there exists an arrangementof 3k blue lines and 2k red lines with no alternating pathlonger than 14k
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Art Gallery Problem (variants)The guards are watched by another guards (summer of 1992)
P polygon
G set of guards
Conditions on Vis(P, G)
Art Gallery Problem (variants)The guards are watched by another guards (summer of 1992)
Vis(P,G) is connected COOPERATIVE GUARDSVis(P,G) no isolated vertices GUARDED GUARDSConsecutive vertices CONSECUTIVE GUARDS…………………………………………..
90’s Hernández, García00’s Michael, Pinciu, Zylinski, Kosowski, Malafiejski
Orthogonal polygons, grids, coloring proofs, etc.
Heterochromatic triangulations
Heterochromatic triangulations
Heterochromatic triangulations
Given two point sets (red and blue), does it admit a heterochromatictriangulation?
Heterochromatic triangulations
Given two point sets (red and blue), does it admit a heterochromatictriangulation?
Minimum number of edges to guarantee a heterochromatic triangulation
Algorithm?
Blocking Delaunay triangulations
Given is a set B of n blue points in general position in the plane.Add a set R of red points, such that the Delaunay triangulation ofB ∪R does not contain an edge between two blue points. We saythat the set R blocks the Delaunay edges of B. What is theminimal cardinality of R?
There are sets which require n points to be blocked [1], and it canbe shown that |R| ≤ (3/2)n is sufficient [2].
Close the gap!Conjecture [2]: n points are always necessary and sufficient.
Algorithmic question: How fast can a minimal blocking set R becomputed?
This problem was stated in [1] and also presented by Ferran duringthe open problem session in Dagstuhl 2011.
[1] B. Aronov, M. Dulieu, and F. Hurtado. Witness (delaunay) graphs. CG:TA 44(6-7):329-344, 2011.[2] O. Aichholzer, R. Fabila-Monroy, T. Hackl, M. van Kreveld, A. Pilz, P. Ramos, and B. Vogtenhuber. BlockingDelaunay Triangulations. CG:TA 46(2):154-159, 2013.
Problem Statement
• Given a two sets of n non-overlapping unit discs (coins) determine the distance between them.
Earth Movers Distance
• The Earth movers distance is the amount of work (moves) needed to transform one set to another.
• Work is measured by the number of straight line collision free translations of a coin (without lifting it) along a line.
3n/2 moves are sometimes necessary to move congruentcoins from one configuration to another.
At least two moves are needed to get the firstcoin in place.
At least two moves are needed to get the firstcoin in place.
At least two moves are needed to get the firstcoin in place.
One more for the second coin.
And so on.
Summary
2. A. Dumitrescu and M. Jiang, On reconfiguration of disks in the plane and related problems. Comput. Geom.. 2013, 191-202.
1. M. Abellanas, S. Bereg, F. Hurtado, A.G. Olaverri, D. Rappaport, and J. Tejel, Moving coins. Comput. Geom.. 2006, 35-48.
References
Some open problems on
Compatible Graphs
Alfredo García and Javier Tejel
Universidad de Zaragoza, Spain
XVI EGC, Barcelona, July 1-3, 2015
Compatible graphs
Given a set S of n points on the plane, two plane geometric graphs on S, G1 = (S, E1)
and G2 = (S, E2), are compatible if G1 U G2 = (S, E1 U E2) is also a plane geometric
graph.
They are disjoint if E1 ∩ E2 is empty.
Compatible graphs
Given a set S of n points on the plane, two plane geometric graphs on S, G1 = (S, E1)
and G2 = (S, E2), are compatible if G1 U G2 = (S, E1 U E2) is also a plane geometric
graph.
They are disjoint if E1 ∩ E2 is empty.
Compatible matchings
Conjecture: Given a perfect matching M on S, |S| = 2n and n even, is there another
perfect matching M’, disjoint with M ?
Compatible matchings
Conjecture: Given a perfect matching M on S, |S| = 2n and n even, is there another
perfect matching M’, disjoint with M ?
Solved in “Disjoint Compatible Geometric Matchings”, by M. Ishaque, D. Souvaine,
C.D. Tóth, Discrete and Computational Geometry, 49, 89-131,2013.
Compatible spanning trees
Given a plane geometric spanning tree T1, is there a compatible tree T2 disjoint with T1 ?
dS(T1) = Min. number of common edges between T1 and any other compatible tree T2
What is the value of dn= Min dS(T1) ? (|S| = n, T1 spanning tree)
Compatible spanning trees
Given a plane geometric spanning tree T1, is there a compatible tree T2 disjoint with T1 ?
dS(T1) = Min. number of common edges between T1 and any other compatible tree T2
What is the value of dn= Min dS(T1) ? (|S| = n, T1 spanning tree)
Compatible spanning trees
Given a plane geometric spanning tree T1, is there a compatible tree T2 disjoint with T1 ?
dS(T1) = Min. number of common edges between T1 and any other compatible tree T2
What is the value of dn= Min dS(T1) ? (|S| = n, T1 spanning tree)
Compatible spanning trees
4
3
5
2 nd
nn
Conjecture: Given a plane geometric
spanning tree T1, there is another
compatible tree T2 having in common
with T1 at most n/5 + O(1) edges.
Compatible graphs: path and matching
common edges 4
n
20
9n
Compatible tree and matching
common edges 4
n
20
9n
Conjecture: Given a plane Hamiltonian path P1 there exists a compatible perfect
matching M1 having in common with P1 at most n/4 edges.
Reflexivity of Point Sets• Given a set S of n points in the plane, the
reflexivity, r(S), is the minimum number of reflex vertices in a polygonalization of S
• Compute r(S) exactly or approximately?• Let R(n) = max r(S), for |S|=n• What is R(n)?– [Arkin et al]: floor(n/4) ≤ R(n) ≤ ceil(n/2)– [Ackerman,Aichholzer,Keszegh’08]: R(n) ≤ (5/12)n+o(1)
• Reflexivity of imprecise point sets?
Conjecture: floor(n/4) ≤ R(n) ≤ ceil(n/2)Conjecture: R(n) ≤ R(n+1)
r(S) is a measure of “convexity”
7 reflex 3 reflex