Lecture 6: Smallest enclosing circles and more · Introduction Smallest enclosing circle algorithm Randomized incremental construction Smallest enclosing circles and more Computational

IntroductionSmallest enclosing circle algorithm

Randomized incremental construction

Smallest enclosing circles and more

Computational Geometry

Lecture 6: Smallest enclosing circles and more

Computational Geometry Lecture 6: Smallest enclosing circles and more



Facility locationProperties of the smallest enclosing circle

Facility location

Given a set of houses and farmsin an isolated area. Can we placea helicopter ambulance post sothat each house and farm can bereached within 15 minutes?

Where should we place anantenna so that a number oflocations have maximumreception?





Facility location in geometric terms

Given a set of points in the plane.Is there any point that is within acertain distance of these points?

Where do we place a point thatminimizes the maximum distanceto a set of points?





Facility location in geometric terms

Given a set of points in theplane, compute the smallestenclosing circle





Smallest enclosing circle

Observation: It must pass throughsome points, or else it cannot besmallest

Take any circle that encloses thepoints, and reduce its radiusuntil it contains a point p

Move center towards p whilereducing the radius further, untilthe circle contains anotherpoint q






Move center on the bisector of pand q towards their midpoint,until:(i) the circle contains a thirdpoint, or(ii) the center reaches themidpoint of p and q






Question: Does the “algorithm” ofthe previous slide work?






Observe: A smallest enclosing circlehas (at least) three points on itsboundary, or only two in which casethey are diametrally opposite

Question: What is the extraproperty when there are three pointson the boundary?




Randomized incremental constructionA more restricted problemA yet more restricted problemEfficiency analysis


Construction by randomized incremental construction

incremental construction: Add points one by one andmaintain the solution so far

randomized: Use a random order to add the points





Adding a point

Let p1, . . . ,pn be the points in random order

Let Ci be the smallest enclosing circle for p1, . . . ,pi

Suppose we know Ci−1 and we want to add pi

If pi is inside Ci−1, then Ci = Ci−1

If pi is outside Ci−1, then Ci will have pi on its boundary





Adding a point

Ci−1

pi

Ci−1

pi





Adding a point

Question: Suppose we remembered not only Ci−1, but alsothe two or three points defining it. It looks like if pi is outsideCi−1, the new circle Ci is defined by pi and some points thatdefined Ci−1. Why is this false?





Adding a point





Adding a point

How do we find the smallestenclosing circle of p1 . . . ,pi−1 with pi

on the boundary?

We study the new(!) geometricproblem of computing the smallestenclosing circle with a given point pon its boundary

p





Smallest enclosing circle with point

Given a set P of points and onespecial point p, determine thesmallest enclosing circle of P thatmust have p on the boundary

Question: How do we solve it? p






Construction by randomized incremental construction

incremental construction: Add points one by one andmaintain the solution so far

randomized: Use a random order to add the points





Adding a point

Let p1, . . . ,pi−1 be the points in random order

Let C′j be the smallest enclosing circle for p1, . . . ,pj (j≤ i−1)and with p on the boundary

Suppose we know C′j−1 and we want to add pj

If pj is inside C′j−1, then C′j = C′j−1

If pj is outside C′j−1, then C′j will have pj on its boundary(and also p of course!)





Adding a point

C ′j−1

pj

C ′j−1

pj

p p





Adding a point

How do we find the smallestenclosing circle of p1 . . . ,pj−1 with pand pj on the boundary?

We study the new(!) geometricproblem of computing the smallestenclosing circle with two given pointson its boundary

pq





Smallest enclosing circle with two points

Given a set P of points and twospecial points p and q, determine thesmallest enclosing circle of P thatmust have p and q on the boundary

Question: How do we solve it? pq





Two points known

p

q q

p





Two points known

p

q q

p





Algorithm: two points known

Assume w.lo.g. that p and q lie on a vertical line. Let ` be theline through p and q and let `′ be their bisector

For all points left of `, find the one that, together with p andq, defines a circle whose center is leftmost → pl

For all points right of `, find the one that, together with pand q, defines a circle whose center is rightmost → pr

Decide if C(p,q,pl) or C(p,q,pr) or C(p,q) is the smallestenclosing circle





Two points known

p

q q

p p

qq

p

pl pr

C(p, q, pr)

C(p, q, pl)





Analysis: two points known

Smallest enclosing circle for n pointswith two points already known takesO(n) time, worst case

pq





Algorithm: one point known

Use a random order for p1, . . . ,pn; start withC1 = C(p,p1)for j← 2 to n doIf pj in or on Cj−1 then Cj = Cj−1; otherwise, solvesmallest enclosing circle for p1, . . . ,pj−1 with two pointsknown (p and pj)

C ′j−1

pj

C ′j−1

pj

p p





Analysis: one point known

If only one point is known, we used randomized incrementalconstruction, so we need an expected time analysis

C ′j−1

pj

C ′j−1

pj

p p






Backwards analysis: Consider the situation after adding pj,so we have computed Cj

p Cjp Cj






The probability that the j-th addition was expensive is thesame as the probability that the smallest enclosing circlechanges (decreases in size) if we remove a random point fromthe j points

p Cjp Cj






This probability is 2/j in the left situation and 1/j in the rightsituation

p Cjp Cj






The expected time for the j-th addition of a point is

j−2j·Θ(1)+

2j·Θ(j) = O(1)

or

j−1j·Θ(1)+

1j·Θ(j) = O(1)

The expected running time of the algorithm for n points is:

Θ(n)+n

∑j=2

Θ(1) = Θ(n)






Smallest enclosing circle for n pointswith one point already known takesΘ(n) time, expected

p





Algorithm: smallest enclosing circle

Use a random order for p1, . . . ,pn; start withC2 = C(p1,p2)for i← 3 to n doIf pi in or on Ci−1 then Ci = Ci−1; otherwise, solvesmallest enclosing circle for p1, . . . ,pi−1 with one pointknown (pi)

Ci−1

pi

Ci−1

pi





Analysis: smallest enclosing circle

For smallest enclosing circle, we used randomized incrementalconstruction, so we need an expected time analysis

Ci−1

pi

Ci−1

pi






Backwards analysis: Consider the situation after adding pi,so we have computed Ci

Ci Ci






The probability that the i-th addition was expensive is thesame as the probability that the smallest enclosing circlechanges (decreases in size) if we remove a random point fromthe i points

Ci Ci






This probability is 3/i in the left situation and 2/i in the rightsituation

Ci Ci






The expected time for the i-th addition of a point is

i−3i·Θ(1)+

3i·Θ(i) = O(1)

or

i−2i·Θ(1)+

2i·Θ(i) = O(1)

The expected running time of the algorithm for n points is:

Θ(n)+n

∑i=3

Θ(1) = Θ(n)





Result: smallest enclosing circle

Theorem The smallest enclosingcircle for n points in plane can becomputed in O(n) expected time




ConditionsDiameter and closest pairWidthMore examples

When does it work?

Randomized incremental construction algorithms of this sort(compute an ‘optimal’ thing) work if:

The test whether the next input object violates thecurrent optimum must be possible and fast

If the next input object violates the current optimum,finding the new optimum must be an easier problem thanthe general problem

The thing must already be defined by O(1) of the inputobjects

Ultimately: the analysis must work out





Diameter, closest pair

Diameter: Given a set of n points inthe plane, compute the two pointsfurthest apart

Closest pair: Given a set of n pointsin the plane, compute the two pointsclosest together





Width

Width: Given a set of n points in theplane, compute the smallest distancebetween two parallel lines thatcontain the points (narrowest strip)





Rotating callipers

The width can be computed using the rotating callipersalgorithm

Compute the convex hull

Find the highest and lowest point on it; they define twohorizontal lines that enclose the points

Rotate the lines together while proceeding along theconvex hull





Rotating callipers





Rotating callipers





Rotating callipers





Rotating callipers





Rotating callipers





Rotating callipers





Width

Property: The width is alwaysdetermined by three points of the set

Theorem: The rotating callipersalgorithm determines the width (andthe diameter) in O(n logn) time





Width by RIC?

Property: The width is alwaysdetermined by three points of the set

We can maintain the two linesdefining the width to have a fast testfor violation





Adding a point

Question: How about adding apoint? If the new point lies insidethe narrowest strip we are fine, butwhat if it lies outside?





Adding a point





Adding a point





Width

A good reason to be very suspiciousof randomized incrementalconstruction as a working approachis non-uniqueness of a solution





Minimum bounding box

Question: Can we compute theminimum axis-parallel bounding boxby randomized incrementalconstruction?





Minimum bounding box

Yes, in O(n) expected time

. . . but a normal incrementalalgorithm does it in O(n) worst casetime





Lowest point in circles

Problem 1: Given n disks in theplane, can we compute the lowestpoint in their common intersectionefficiently by randomized incrementalconstruction?

Problem 2: Given n disks in theplane, can we compute the lowestpoint in their union efficiently byrandomized incrementalconstruction?





Red-blue separation

Problem: Given a set of n red andblue points in the plane, can wedecide efficiently if they have aseparating line?





One-guardable polygons

Problem: Given a simple polygonwith n vertices, can we decideefficiently if one guard is enough?





One-guardable polygons

It can easily happen that aproblem is an instance of linearprogramming

Then don’t devise a newalgorithm, just explain how totransform it, and show that it iscorrect (that your problem isreally solved that way)


Documents

Lecture 6: Smallest enclosing circles and more · Introduction Smallest enclosing circle algorithm Randomized incremental construction Smallest enclosing circles and more Computational