how to cut pseudoparabolas into segments seminar on geometric incidences by: almog freizeit
TRANSCRIPT
How to Cut Pseudoparabolas into Segments
Seminar on Geometric Incidences
By: Almog Freizeit
A Reminder
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Székely’s method
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Our goal• We want to apply Székely’s method to circles with arbitrary
radii.
• The problem: the graph is not simple
• What can we do?
• We will make the Székely’s graph simple: Cutting into pseudo-segments.
• Each pair of pseudo-segments intersects at most once, and the resulting graph is guaranteed to be simple.
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Our goal
• Example:
Original P and C Cutting into pseudo-segments The Székely graph
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Our goal
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Hisao Tamaki and Takeshi Tokuyama, 1998
The bounds
are not tight
!!
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Terminologies
•Let Γ be an arrangement of pseudoparabolas. The arrangement subdivides the plane into faces. We use the terms cell, edge and vertex for two-, one- and zero-dimensional respectively.
•When two pseudoparabolas intersect twice, they form a closed curve, which we call a lens. We say a lens is a 1-lens if no curve crosses the lens.
•Observation: The cutting number of Γ is notless then the number of 1-lenses
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Lower bound
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Lower bound
• Very carefully, we countedthe number of incidences in this arrangement and succeeded to prove the desiredlower bound.
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Lower bound
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Lower bound
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Lower bound
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Lower bound
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Lower bound
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Upper bound
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Some notations about Hypergraphs
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Our Hypergraph• We define a hypergraph H(Γ)=(X,E):
• X: the set of edges of the arrangement Γ.
• E: each hyperedge is a set of nodes which its corresponding set of edges in the arrangement forms a lens.
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Computing a covering• A greedy algorithm for computing a covering is the following:
1.Find a node of maximum degree
2.Insert the node to the covering, and remove it and all hyperedges containing it.
3.If all hyperedges are covered, EXIT; Else GOTO 1.
Lovász showed that the greedy algorithm achieves a covering size at most logd(H)+1 times the size of the covering of H. We neither use nor prove this fact, yet we will use and prove a key inequality from his proof.
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Lovász’s Inequality
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Lovász’s Inequality
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Lovász’s Inequality
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So what we had so far?
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25)The graph is undirected(
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Upper envelope
• On the other hand, the upper envelope of A(C) has at most 5 edges, and the lower envelope of A(D) has at most 7 edges (board)
• Let's place those envelopes together on the plane
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32Extremal edges
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33Near 1-lens
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Upper bound
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What about circles?• We can obtain these bounds to an arrangement of arbitrary
circles as well:
• We are given an arrangement of n circles.
• Each pair of circles intersect at most twice, but a circle is not an x-monotone curve
• Let's cut each circle with its horizontal diameter, and divide it into an upper half-circle and a lower half-circle.
• Now we connect two vertical downward (resp. upward) rays to an upper (resp. lower) half-circle at its endpoints, and obtain an x-monotone curve separating the plane.
• It is easy to see that every pair of curves intersects at most twice
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What about circles?
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Overview
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Other results
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Terminologies
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Terminologies
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Bounding the number of lunes
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Let's define a graph
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Let's define a graph• Lemma: G is a planar
• Proof: we will show that the plane embedding of G defined before has no pair of crossing edges. This will be a special case of the following more general lemma:
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G is a planar
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G is a planar
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Case 1
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Case 1
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Case 1
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Case 2
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Case 3
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Bounding number of lunes
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Questions?
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Thank you!
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