Download CS 372: Computational Geometry Lecture 14 Geometric

CS 372: Computational Geometry
Lecture 14
Geometric Approximation Algorithms
Antoine Vigneron
King Abdullah University of Science and Technology
December 5, 2012
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CS 372 Lecture 14
December 5, 2012
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1
Introduction
2
The diameter problem
3
A 2-approximation algorithm
4
Approximation scheme
5
Rounding to a grid
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Outline
Introduction to geometric approximation algorithms.
Example: Computing the diameter of a point-set.
Simple 2-approximation algorithm.
Two approaches for ε-approximation:
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Rounding.
Projection.
References:
Sariel Har Peled’s book.
Paper by T. Chan, Approximating the diameter, width, smallest
enclosing cylinder, and minimum-width annulus, Section 2.
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The Diameter Problem
P
Input: a set P of n points in Rd .
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The Diameter Problem
a∗
P
∆∗ = |a∗ b ∗ |
b∗
Input: a set P of n points in Rd .
Output: the maximum distance ∆∗ between any two points of P.
∆∗ is called the diameter of P.
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Brute Force Algorithm
Computing the distance between two points:
If a = (a1 , a2 , . . . , ad ) and b = (b1 , b2 , . . . , bd ), then
q
|ab| = (b1 − a1 )2 + (b2 − a2 )2 + · · · + (bd − ad )2 .
It takes O(d) time.
Here we assume that d is constant: d = O(1).
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We are in fixed dimension.
Then we can compute |ab| in O(1) time.
Computing the diameter:
Brute force: Check all pairs in P 2 and keep the maximum distance.
Brute force takes O(n2 ) time.
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Approximation Algorithms
Definition (c-factor approximation)
We say that ∆ is a c-factor approximation to ∆∗ if ∆ 6 ∆∗ 6 c∆.
Definition (c-approximation algorithm)
A c-approximation algorithm for a maximization problem is an algorithm
that computes a c-factor approximation of the optimum in polynomial
time.
We will give a 2-approximation algorithm for the diameter problem.
That is, we find ∆0 such that ∆0 6 ∆∗ 6 2∆0
O(n) time, very simple.
Best known exact algorithms are slower and complicated.
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2-Approximation Algorithm
P
a
Pick a point a ∈ P.
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2-Approximation Algorithm
P
b
a
Pick a point a ∈ P.
Find b such that |ab| is maximum.
∆0 = |ab|.
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Analysis
Pick any point a ∈ P.
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O(1) time.
Find b ∈ P such that |ab| is maximum.
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Just go through the list of points in P and keep the maximum distance
to a.
It takes O(n) time.
Conclusion: This algorithm runs in O(n) time.
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Proof of Correctness
We need to prove that ∆0 is a 2-factor approximation. of ∆∗ .
It means ∆0 6 ∆∗ 6 2∆0 .
Since (a, b) ∈ P 2 , clearly, ∆0 = |ab| 6 ∆∗ .
We also need to prove that ∆∗ 6 2∆0 .
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Let a∗ and b ∗ denote two points such that |a∗ b ∗ | = ∆∗ .
By definition of b we have |aa∗ | 6 |ab| and |ab ∗ | 6 |ab|.
By the triangle inequality
∆∗ = |a∗ b ∗ |
6 |a∗ a| + |ab ∗ |
6 |ab| + |ab|
= 2∆0 .
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Concluding Remark
The running time is optimal.
If we do not assume d = O(1):
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This algorithm still works.
But the running time has to be written O(dn).
It is still optimal as we need Θ(nd) time to read the input.
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(1 + ε)-Approximation Algorithms
We just found a 2-approximation algorithm.
We would like to obtain a better approximation.
Let ε > 0 be a real number.
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In this lecture, we assume ε < 1.
ε should be thought of as being small, say ε = 0.1 or ε = 0.01.
We want to design a (1 + ε)-approximation algorithm.
The result will be ∆ such that ∆ 6 ∆∗ 6 (1 + ε)∆.
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In other words, the relative error we allow is ε.
So ε = 0.01 means a 1% error.
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Analysis
How to analyze a (1 + ε)-approximation algorithm?
The running time is expressed as a function of n and ε using O(·)
notation.
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For instance, the last algorithm in this lecture runs in time.
3
O n + (1/ε) 2 (d−1)
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Since d = 0(1), it is polynomial in n and 1/ε.
F
It is an FPTAS (Fully Polynomial Time Approximation Scheme).
The running time is linear in n.
Problem: exponential in d.
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These algorithms are only interesting in low dimension d.
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Idea
Start with a set P of n points.
Transform it into a set P 0 such that:
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The cardinality |P 0 | is small.
The diameter ∆0 of P 0 is a good approximation of ∆.
Then find ∆0 by brute force
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Rounding to a Grid
P
0
0
Consider a regular grid over Rd .
The side length of the grid is 0 , to be specified later.
Intuition: we will choose 0 ' ∆∗ , it is the error we allow.
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Rounding to a Grid
P
Replace each point of P with the nearest grid point.
This operation is called rounding.
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Rounding to a Grid
P0
The grid points we obtain form the set P 0 .
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Rounding to a Grid
c0
∆0
d0
P0
Compute the diameter ∆0 of P 0 by brute force.
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Intuition
P 0 is a d-dimensional point set with diameter ∆0 .
The points are on a grid with side length 0 ; we will choose it such
that 0 ' ∆0 .
So in the worst case, there are about as many points in P 0 as in a
(1/) × (1/) · · · × (1/) grid in Rd .
There are O((1/)d ) such points.
We can compute ∆0 by brute force in time O((1/)2d ).
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How to Perform Rounding?
The grid points have coordinates (k1 0 , k2 0 , . . . kd 0 ) where each ki is
an integer.
Let p = (p1 , p2 , . . . pd ) ∈ P.
How can we find the closest grid point p 0 ?
We need to find the closest integer ki to pi /0 .
It is given by the formula
pi
1
ki = 0 +
.
2
p is rounded to p 0 = (k1 0 , k2 0 , . . . kd 0 ).
It takes O(d) = O(1) time.
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Rounding Error
Property
√
Let x ∈ Rd , and let x 0 be the closest grid point to x. Then |xx 0 | 6 0 d/2.
Proof: Apply previous slide formula.
Intuition:
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x is in a hypercube centered at x 0 with side length 0 .
√
0 d
x0
x
0
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0
√
This hypercube diagonal has length 0 d.
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Approximation Factor
Let a0 and b 0 be the closest grid points to a∗ and b ∗ , respectively.
√
√
Then from previous slide, |a0 a∗ | 6 0 d/2 and |b 0 b ∗ | 6 0 d/2.
By the triangle inequality,
∆∗ = |a∗ b ∗ |
6 |a∗ a0 | + |a0 b 0 | + |b 0 b ∗ |
√
6 |a0 b 0 | + 0 d
√
6 ∆0 + 0 d.
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Approximation Factor
Let c 0 and d 0 be two points of P 0 such that |c 0 d 0 | = ∆0 .
Let c and d be points of P that have been rounded to c 0 and d 0 ,
respectively.
√
√
Then |cc 0 | 6 0 d/2 and |dd 0 | 6 0 d/2.
So by the triangle inequality,
∆0 = |c 0 d 0 |
√
6 |cd| + 0 d.
Of course |cd| 6 ∆∗ .
It follows that
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√
∆0 6 ∆∗ + 0 d
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Approximation Factor
We obtained
√
√
∆0 − 0 d 6 ∆∗ 6 ∆0 + 0 d.
√
We choose ∆ = ∆0 − 0 d, so
√
∆ 6 ∆∗ 6 ∆ + 20 d.
√
We want to make sure that 20 d 6 ∆∗ /2.
√
Since ∆0 6 ∆∗ , it suffices that 20 d 6 ∆0 /2.
√
So we first compute ∆0 , and then we set 0 = ∆0 /(4 d).
It follows that ∆ 6 ∆∗ 6 ∆ + ∆∗ /2.
Problem: We wanted to show that ∆ 6 ∆∗ 6 ∆ + ∆.
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Technical Difficulty
Intuition: The relative error is less than , so it is OK.
Proof: We want to prove that ∆∗ 6 ∆ + ∆.
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We know that ∆∗ 6 ∆ + ∆∗ /2.
Then
1
∆
1 − /2
2
3
= 1+ +
+
+ . . . ∆.
2
4
8
∆∗ 6
I
Since < 1, we get
1
1 1 1
∆ 6 1+
+ + +
...
∆
2 4 8 16
∗
= (1 + )∆.
So ∆ is a (1 + )-factor approximation of ∆∗ .
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Cardinality of P 0
∆0
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Cardinality of P 0
r
√
r 6 ∆0 + 0 d/2
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Cardinality of P 0
r0
B
√
r 0 6 ∆0 + 0 d/2 + 0 /2
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Cardinality of P 0
All the points of P are in a sphere with radius ∆0 .
√
So all the points of P 0 are in a sphere with radius ∆0 + 0 d/2.
To each point p ∈ P 0 , we associate a ball b(p) centered at p with
radius 0 /2.
These balls
√ are disjoint and contained in a ball B with radius
∆0 + 0 ( d/2 + 1/2) 6 2∆0 .
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How Many Grid Points are There?
The volume of a ball with radius r in dimension d is Cd r d , where Cd
depends only on d.
So the number of balls b(p), p ∈ P 0 is at most
16√d d
Cd (2∆0 )d
= O((1/)d ).
=
Cd (0 /2)d
Each point p ∈ P 0 is in exactly one ball b(p).
So |P 0 | = O((1/)d ).
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Summary
Compute ∆0 in O(n) time.
√
Round all the points to a grid with side length 0 = ∆0 /(4 d).
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It takes O(n) time.
There are O((1/)d ) such points.
We denote by P 0 the set of rounded points.
Compute the diameter ∆0 of P 0 by brute force.
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It takes O((1/)2d ) time.
√
∆0 − 0 d is a (1 + )-factor approximation of the diameter ∆∗ of P.
Running time: O(n + (1/)2d ).
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Grid Cleaning: Example in R2
b̂
â
P0
â is the lowest point on a vertical line
b̂ is the highest point on a vertical line
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Grid Cleaning: Example in R2
b̂
â
P0
Idea: keep only the highest and the lowest point on each vertical line.
Then run the brute-force algorithm.
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Analysis in R2
Only O(1/) points remain after grid cleaning.
So the algorithm runs in O(n + (1/)2 ) time.
In higher dimension, same idea.
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Consider all the points that coincide in their first (d − 1) coordinates.
Keep only highest and lowest.
Then only O((1/)d−1 ) remain from P 0 .
So rounding + grid cleaning yields a running time O(n + (1/)2d−2 ).
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Projecting on Lines
We measure angles in radian.
That is, an angle is in [0, 2π].
Property
For any α,
1−
α2
6 cos α 6 1.
2
Idea: We get a relative error ε by choosing α to be roughly
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√
ε.
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Projecting on Lines
√
b
a
b0
`
a0
|a0 b 0 | 6 |ab| 6 (1 + ε)|a0 b 0 |
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Projecting on Lines
Assume that the angle between line ab and line ` is at most
a0
(resp.
b0 )
√
.
is the orthogonal projection of a (resp. b) into `.
Then
|a0 b 0 | 6 |ab|
1
1 − /2
2 8
0 0
+
+ ... .
= |a b | × 1 + +
2
4
8
6 |a0 b 0 | ×
Since we assume that < 1, we obtain
|a0 b 0 | 6 |ab| 6 |a0 b 0 |(1 + ).
In other words, |a0 b 0 | is a (1 + )–factor approximation of |ab|.
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Idea
P` obtained by projecting P onto a line `.
a∗
P
`
b∗
P`
Compute the diameter of P` .
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Can be done in O(n) time: Find maximum and minimum along `.
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Idea
√
If the angle between ` and a∗ b ∗ is less than , then the diameter of
P` is a (1 + )-factor approximation of ∆∗ .
√
How can we find a line that makes an angle with ab?
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Pick several lines.
√
√
In the plane: Pick direction k for each k ∈ [0, π/ ].
√
O( ) cones
√
with angular diameter Antoine Vigneron (KAUST)
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Algorithm in R2
√
For any integer k ∈ [0, π/ ], we denote by `k a line that makes
√
angle k with horizontal.
`k
√
k ε
Project P onto `k , obtaining P`k .
The maximum of the diameter of P`k over all k is a (1 + )-factor
approximation of the diameter or P.
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Algorithm in R2 : Analysis
Projecting onto a particular `k takes O(n) time.
Computing the diameter of P`k takes O(n) time.
√
There are O(1/ ) such lines.
√
Overall running time O(n/ ).
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Algorithm in R2 : Proof
Let θ be the angle of a∗ b ∗ with horizontal.
√
√
There exists k such that k 6 θ < (k + 1) .
√
The angle between a∗ b ∗ and `k is at most .
So the diameter of P`k is at least ∆∗ /(1 + ).
So the output of the algorithm is at least ∆∗ /(1 + ).
On the other hand, the algorithm only looks at distances between two
projected points, which are always smaller than ∆∗ .
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Generalization in Rd
d
S
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Generalization in Rd
Problem: Find a set of directions that approximates well the whole
set of directions.
Reformulation:
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Let S be the unit sphere in Rd .
Find D ⊂ S with small cardinality such that
√
∀x ∈ S there is a point d ∈ D such that |dx| 6 .
A point d ∈ D is associated with the line through√
the origin and d.
d handles a cone of direction with angular radius .
Such a set D with cardinality O((1/)(d−1)/2 ) can be computed
efficiently.
So the algorithm runs in time O(n(1/)(d−1)/2 ).
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Combining the two Techniques
Running times:
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Grid + cleaning: O(n + (1/)2d−2 ).
Projections: O(n(1/)(d−1)/2 ).
Approach:
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First round to P 0 and do grid cleaning.
We are left with O((1/)d−1 ) points.
Project on lines.
Overall running time: O(n + (1/)3(d−1)/2 )
Technical problem: the approximation factor is now about 1 + 2. (how
to solve it?)
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