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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 Triangles Esther Ezra

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Joint Work with Boris Aronov and Micha Sharir

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

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?

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

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

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

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

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|) .

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.

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.

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

[Matousek et al. 1994]

Expand to obtain fat triangles

(n2) holes

O(n) holes

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.

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!

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

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

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).

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.

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

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.

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

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

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

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

Thank You