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On the Union of Fat Triangles Esther Ezra PowerPoint Presentation

On the Union of Fat Triangles Esther Ezra

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On the Union of Fat TrianglesEstherEzra

Joint Work with Boris Aronov and Micha Sharir

Union of simply-shaped planar regions

S = {S1, …, Sn} a collection of nsimply-shaped

planar regions.

The union of Sconsists of all points in 2 that

are covered by at least one element of S.

Example:

Union of triangles in the plane

All portions of the plane that are

covered by the triangles.

The union has two holes

The union boundary

The union complexity

Combinatorial complexity of the union:

The maximal number of vertices and edges that form

the boundary of the union of the regions inS.

Problem: Bound this complexity.

Trivial bound:O(n2)

It is sufficient to bound #vertices:

Each pair of regions meets in O(1) points.

Bound is tight?

Previous results: Fat objects

Piecewise linear objects:

n -fat wedges.

Union complexity: O(n) . [Efrat et al. 1993]

n -fat triangles.

Number of holes in the union: O(n) .

Union complexity: O(n log log n) . [Matousek et al. 1994]

Worst case lower bound: (n (n) ) .

Bound depends on

Angle at apex

Bound depends on

Each of the angles

Previous results: Fat objects

Fat curved objects (simply-shaped)

n convex -fat objects.

Union complexity: O*(n)[Efrat, Sharir 2000]

Locally -fat objects:

Union complexity: O(n log n) [De Berg 2010]

r

r’

r/r’ ,

and 1.

O(n1+), for any >0.

O

area(D O) area(D)

0 < 1

D

Previous results: Fat objects

Summary:

Union complexity is ~ “one order of magnitude” smaller than the worse-case (n2) bound!

Back to -fat triangles:

The “flagship” problem: simple and natural.

Previous results:

Number of holes in the union: O(n) .

Union complexity: O(n log log n) .

Worst case lower bound: (n (n) )

Each of the angles

Bounds depends on

Our results

Union complexity of n -fat triangles :

O(n 2(n) log* n) .

Follow-up study [with Mark De Berg]:

Union complexity of n-fat triangles :

O(n log* n) .

Union complexity of n locally -fat objects:

n 2 O(log* n ) .

Bound depends on

O

D

Consequences to geometric set-cover

Input:

P - points in the plane,

T – a set of-fat triangles.

A set-cover for (P,T) is a subset

S T that covers all the points in P.

Goal: find smallest set-cover.

Problem is NP-hard to solve.

Approximation factor (for the abstract setting):

O(log |P|) .

Set-cover Vs. Union

Theorem [Aronov Ezra, Sharir 2009] [Clarkson Varadarajan 2007]:

The complexity of the union is O(n (n))

Approximation factor, achieved in polynomial time: O(log (OPT)) .

Fat triangles:

Previous bound on union complexity: O(n log log n)

Approximation factor : O(log log log(OPT)).

New bound on union complexity: O(n 2(n) log* n)

New approximation factor : O(log log*(OPT)) .

() is a slowly growing function.

OPT = size of the smallest set-cover.

Union complexity: Analysis

Main idea: Reduce triangles to wedges

- Decompose plane into trapezoidal cells .
- Show that in each cell most of the triangles appearas -fat wedges.
- Process the remaining (survival) triangles in recursion.At the bottom of the recursion we are left with only a fewof them.

Apply the O(n) bound on their union.

Ingredient I: #holes in the union

Number of holes in the union is only O(n) .

[Matousek et al. 1994]

Expand to obtain fat triangles

(n2) holes

O(n) holes

Ingredient II: The Combination Lemma

We have two sets of triangles

(not necessarily fat):

T1 – Blue collection

T2 – Red collection

U(T1 ) , U(T2 )are given.

What is the joint complexity

U(T1 T2)?

Boundary complexity of the union: # vertices.

Ingredient II: The Combination Lemma

Combination Lemma [Edelsbrunner etal. 1990]:

T1, T2 two sets with total of n triangles.

Then:

U(T1 T2) U(T1 ) + U(T2) + O( n + H(T1 T2) )

Remarks:

- The lemma is fairly general, and does not assume fatness .
- The coefficients of U(T1 ) , U(T2) are precisely 1 .

# holes in the joint union.

Crucial!

Curse and recurs

T – a set of n-fat triangles .

Derive an O(n log n) bound for T :

Use a divide and conquer scheme:

- Partition T into two equal-size subsets T1, T2 , |T1| = |T2| = n/2.
- Use Combination Lemma + linear number of holes:U(T1 T2) U(T1 ) + U(T2) + O(n)
f(n) 2 f(n/2) + O(n)

Solution: f(n) = O(n log n)

Bottom of recursion:

n=1.

maximum union complexity ofn-fat triangles

Our recurrence

Goal: Obtain a recurrence of the form:

f(n) (n / log n) f( log n) + O(n 2 (n) )

Solution:Use induction…

f(n) = A n 2 (n) log* n, A > 0 constant.

Strategy:Use a D&C mechanism in order to partition the plane.

maximum union complexity ofn-fat triangles

f(n) (n / log n) f( log n) + O(n 2 (n) )

Lemma:

Fix a parameter r := n / log n .

There is a subdivision of the plane into ~ r

openly disjoint trapezoidal subcells , s.t.,

each meets at mostn/r = log n

triangles edges .

Main property: Number of cells is small! (due to fatness)

Easy to obtain O(r 2).

f(n) (n / log n) f( log n) + O(n 2 (n) )

Goals:

- Reduce triangles to wedges in each cell , and apply the linear bound on their complexity.
- Bound union complexity of triangles recursively(show that the number of survival triangles is small).
- Combine union complexity of wedges and triangles by the Combination Lemma and the fact that #holes is linear.

How to obtain the subdivision?Reduce to semi-canonical triangles

Reduce the problem to

semi-canonical triangles:

“almost isosceles” right triangles .

Theorem [Matousek et al. 1994]:

Let T be a set of n semi-canonical

triangles, all of which lie on a

common horizontal line. Then:

U(T) = O( n 2(n) )

Hypotenuse orientation is 1351 degrees

( ) is the inverse Ackermann function

A tree partitioning for T

At most n/2 triangles above l

Use an interval tree on the

y-projections of the

triangles in T.

Obtain at most k = O(log n)

subfamilies T1, …, Tk , T = i=1k Ti ,

For each subfamily Tiwe have:

U(Ti ) O( ni 2(n) ) , ni = |Ti | .

T2

l

T1

T2

At most n/2 triangles below l

The “lower” and the “upper” blue sets are pairwise disjoint.

The key for reducing the number of cells

Build decomposition only with the exterior portions

Lemma:

Overall number of segments:O( n2(n) ) .

Overall number of intersections: O( n2(n) log n).[ much smaller than O(n2) ]

The key to reducing the number of cells

Number of intersections among the full triangles

Wrapping up: the actual recurrence

Actual choose of r :n2(n) / log n

We have a decomposition into O(r) cells ,

each of which meets n 2(n)/ r (boundary of)

triangles.

Number of “wedge triangles” in :

O(n 2(n)/ r) = O(log n) .

Their union complexity in : O(log n) .

Their overall union complexity, over all : O(n 2(n) ) .

Wedge-triangles

Wrapping up : the actual recurrence

Number of “real” triangles in :

O(n 2(n)/ r) = O(log n) .

BUT there is only a single cell where

a triangle can survive!

So on average, there are only

O(log n / 2(n) ) such triangles.

Their union complexity in : f(log n / 2(n) ) .

Real triangle

Wrapping up : the actual recurrence

By the Combination Lemma,

total union complexity in :

O(log n) + f(log n / 2(n) ) + O(log n)

Summing over all :

f(n) (n 2(n)/ log n) f( log n / 2(n)) + O(n 2 (n) )

Solution: f(n) = A n 2 (n) log* n, A > 0 constant.

Wedge triangles

Real triangles

# holes

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