cs 372: computational geometry lecture 4 lower bounds for ... · element distinctness theorem in...

CS 372: Computational GeometryLecture 4

Lower Bounds for Computational Geometry Problems

Antoine Vigneron

King Abdullah University of Science and Technology

September 20, 2012

Antoine Vigneron (KAUST) CS 372 Lecture 4 September 20, 2012 1 / 28

1 Introduction

2 Algebraic computation trees

3 Topological lower bound

4 Discussion

5 3SUM-hardness


OutlineHardness results for some geometric problems.

Examples:

Line segment intersection detection.

Diameter of a point-set.

Degeneracy testing.

Two techniques:

Topological lower bound.

Reduction from 3-SUM.

References:

Textbook by Preparata and Shamos.

Dave Mount’s lecture notes, Lecture 26.

Ben-Or’s paper.

Gajentaan and Overmars paper.


http://www.cs.umd.edu/~mount/754

http://www.cs.huji.ac.il/~benor/AlgCompTrees83.pdf

http://dx.doi.org/10.1016/0925-7721(95)00022-2

Introduction

In the algorithms course (CS 260), you saw that:

Theorem (Lower bound for sorting)

Any comparison-based sorting algorithm makes Ω(n log n) comparisons inthe worst case.

Proof (sketch):

Model the algorithm as a binary decision tree.

Each internal node is a comparison, branching to its two children.

There are n! possible outcomes, i.e. permutation of the input.

Hence there are at least n! leaves.

So the tree has height Ω(log(n!)) = Ω(n log n).


Introduction

In Lecture 2, we showed that it yields the same Ω(n log n) lower bound forcomputing a convex hull.

Proof: After mapping a set of numbers to a parabola, the numbersappear in sorted order along the convex hull.

So the argument holds because a convex hull algorithm outputs apoint sequence.

This argument does not work if the output has constant size.Examples:

Intersection detection. (Output: a boolean.)

Diameter. (Output: a real number.)

We will give a different technique that yields an Ω(n log n) lower bound forthese two problems, and others.


Algebraic Computation Trees

fv1 := x3 − x1Node v1:

fv1 ≥ 0?Node v3:

fv2 ≥ 0? fv2 ≥ 0?

≥ <


v4 v5

≥ <

fv6 := x2 − x3 YES fv8 := x4 − x1YES

≥ <

fv6 ≥ 0? fv8 ≥ 0?≥

YES NO

<

YES NO

≥<

Example (1)

Given (x1, x2, x3, x4) ∈ R4

such that x1 6 x2 andx3 6 x4, this AlgebraicComputation Tree (ACT)decides whether[x1, x2] ∩ [x3, x4] 6= ∅.




fv3 := fv1× fv2Node v3:


NOYES


fv6 := fv4× fv5Node v6:


fv7 := fv3− fv6Node v7:

fv7 > 0?Node v8:

Example (CCW predicate)

Given (x1, x2, x3, x4, x5, x6) ∈ R6, thisACT decides whether the triangle((x1, x2), (x3, x4), (x5, x6)) iscounterclockwise.


Algebraic Computation TreesFormal definition from Ben-Or’s paper:

Definition (Algebraic computation tree)

An algebraic computation tree with input (x1, . . . , xn) ∈ Rn is a binary treeT with a function that assigns:

to any vertex v with exactly one child an operational instruction ofthe form

fv := f 1v f 2

v or fv := c f 1v or fv :=

√f 1v

where f iv = fvi for an ancestor vi of v , or f i

v ∈ x1, . . . , xn, ∈ +,−,×, /, and c ∈ R is a constant.

to any vertex v with two children (branching vertex) a test instructionof the form

f 1v > 0 or f 1

v > 0 or f 1v = 0.

where f 1v is fv1 for an ancestor v1 of v , or f 1

v ∈ x1, . . . , xn.to any leaf an output YES or NO.


http://www.cs.huji.ac.il/~benor/AlgCompTrees83.pdf


Informally:

Given an input x = (x1, . . . , xn) ∈ Rn, the program traverses a pathP(x) from the root to a leaf of T .

Along the path, it applies operations +,−, /,×,√· to input numbersxi or intermediate results obtained at previous nodes along P(x).

It may also branch using a test >,>,=.

At the leaf, it outputs YES or NO.

Definition

Let T be an algebraic computation tree with input x ∈ Rn. Let W ⊂ Rn

be the set of points x ∈ Rn such that T outputs YES. We say that Tdecides W .


Topological Lower Bound

Theorem (Ben-Or, 1983)

Any algebraic computation tree that decides a set W ⊂ Rn has heightΩ(log(#W )− n), where #W is the number of connected components ofW .

Interpretation:

The height of an ACT is its worst-case running time.

A program can often be unfolded onto an ACT.I Then its worst-case running time is at least the height of the ACT.

But some operations cannot be simulated by an ACT in O(1) time.Examples:

I The floor function.I Bitwise operations on integers (AND, OR, XOR).I Random number generation.


Element Distinctness

Problem (Element Distinctness)

Determine whether the elements of a list of numbers are distinct. That is,given (x1, . . . , xn) ∈ Rn, determine whether xi 6= xj for all i 6= j .

Let W + denote the set of positive instances of Element Distinctness:

W + = (x1, . . . , xn) ∈ Rn | ∀i 6= j , xi 6= xj.

Note that an instance of Element Distinctness is modeled as a singlepoint in Rn. Hence W + is a subset of Rn.

How many connected components are there in W +?



Lemma

The set W + of positive instances of Element Distinctness has exactly n!connected components.

Proof.

Done in class.



Theorem

In the algebraic computation tree model, the complexity of elementdistinctness is Θ(n log n).

Proof.

By Ben-Or’s theorem, since W + has n! connected components, any ACTsolving the element distinctness problem has height Ω(log(n!)− n), whichis Ω(n log n).

Conversely, there is an ACT with depth O(n log n) that solves elementdistinctness: First it sorts the input, then checks consecutive values. Forinstance, mergesort can be unfolded into a Θ(n log n)-depth ACT.



Let W− denote the set of negative instances of Element Distinctness:

W− = Rn \W +.

How many connected components are there in W−?


Application: Line Segment Intersection Detection

Theorem

Any ACT that solves the line segment intersection detection problem hasheight Ω(n log n). Hence, the complexity of line segment intersectiondetection is Θ(n log n) in the ACT model.

Proof.

Suppose T is an ACT that solves the line segment intersection detectionproblem. Let (x1, . . . , xn) ∈ Rn. We construct an ACT T ′ by plugging(x1, 0, x1, 1, x2, 0, x2, 1, . . . , xn, 0, xn, 1) to the input of T . That is, T ′

detects intersection between the segments [(xi , 0), (xi , 1)]. So T ′ solvesthe element distinctness problem. Thus height(T ′) = Ω(n log n). But byconstruction, height(T ′) = height(T ) + 4n. Thereforeheight(T ) = Ω(n log n).


Set Disjointness

Problem (Set Disjointness)

Given two sequences (a1, . . . , an) ⊂ R and (b1, . . . , bm) ⊂ R such thatm 6 n, determine whether there exists a pair i , j such that ai = bj .

Theorem

Any ACT that solves the set disjointness problem has height Ω(n log n).

Proof?

Done in class.

Note: The numbers a1, . . . , an are not necessarily distinct, so a directreduction to Element Distinctness fails.


Diameter

P

diam(P )

Problem (Diameter of a Point-Set)

The diameter diam(P) of a set P = p1, . . . , pn of n points is themaximum distance between any two points:

diam(P) = maxi ,j

d(pi , pj).


Maximum Gap

Problem (maximum gap)

Given a set of n (unsorted) numbers, the maximum gap problem is to findthe largest gap between two consecutive numbers in sorted order.

Example:

INPUT: 0.5, 2, 5, 4.5, -1

OUTPUT: 2.5

−1 0.5 2 4.5 5

0Maxgap


Optimization Algorithms

The ACT model deals with decision problems: problems with aYES-NO answer.

We will abuse notation and use ACTs for optimization problems:I The 2D-diameter problem.I Maximum gap.

In fact, our lower bound argument will be for the associated decisionproblems: For some δ ∈ R,

I Is the diameter at least δ?I Is the maximum gap at most δ?

An optimization problem is at least as hard as the correspondingdecision problem, because once you have the optimal value, you cancompare it with δ in O(1) time.

So a lower bound on the decision problem implies a lower bound onthe optimization problem.


Diameter

Ox

y

y = bjx

y = aix

pi

p′j

1

1

Given an instance A,B of setdisjointness, we map A to theright side and B to the leftside of the unit circle,respectively.


Diameter

Ox

y

pi

p′j

1

1

The diameter is 2 iff twopoints pi , p

′j are diametrically

opposed, that is ai = bj .


Diameter

Theorem

In the ACT model, the complexity of the 2D-diameter problem isΩ(n log n).

Proof.

An instance of set disjointness is mapped to the unit circle as follows:

Each ai is mapped to the point pi = (xi , yi ) such that yi = aixi ,xi > 0, and x2

i + y2i = 1.

Each bj is mapped to the point p′j = (x ′j , y′j ) such that yj = bjxj ,

xj < 0, and x2j + y2

j = 1.

Then the diameter of p1, . . . , pn, p′1, . . . , p

′n is 2 iff pi = −p′j for some

i , j , which means that ai = bj and hence A ∩ B 6= ∅.

Remark: it is crucial in this argument that the coordinates of pi and p′i canbe computed in constant time by an ACT.


Maximum Gap

Theorem

In the ACT model, the maximum gap problem has complexity Θ(n log n).

Proof?

Problem: there is a simple O(n)-time algorithm.

This linear-time algorithm uses the floor-function, hence it cannot betranslated into an ACT.


Discussion

It would seem that the ACT model is not powerful enough, as the lowerbound for maxgap can be broken using the floor function.

However, it is not reasonable to allow the use of the floor function:It has been proved that the floor function together with +,−,×, /allows to solve NP-hard problems in polynomial time.(Schonhage, On the power of random access machines, ICALP 79.)

In fact, the ACT model is very powerful:

It allows exact computation in constant time per arithmetic operation.

Even with high degree polynomials, or square roots.


Discussion

Ben-Or’s theorem often gives Ω(n log n) lower bounds for geometricproblems.

Optimal for several fundamental problems. (Intersection detection,2D-diameter...)

But it often gives the same bound in cases where no near-linear timealgorithm is known.

I Example: 4D diameter, Hopcroft’s problem.I Does not always give Ω(n log n): it gives Ω(n2) for knapsack. But this

problem is NP-hard.

Another approach: show that your problem is harder than somewell-known problem, for which no near-linear time algorithm is known.

3-SUM. Best algorithm runs in O(n2). (See following slides.)

Hopcroft’s problem. Best algorithm runs in roughly O(n4/3).


3SUM

Problem (3SUM)

Given a set S of n real numbers, is there a triple a, b, c ∈ S such thata + b + c = 0?

3SUM can be solved in O(n2) time. (How?)

There is an Ω(n log n) lower bound in the ACT model. (How?)

Despite a lot of effort, no better bound is known with usual models ofcomputation.

Hence, if we can argue that a problem is harder than 3SUM, it meansthat an o(n2)-time algorithm is currently out of reach.


3SUM-Hardness

Definition (3SUM-hard)

A problem is 3SUM-hard if an o(n2)-time algorithm for this problemimplies an o(n2)-time algorithm for 3SUM.

We give an example in the next slide. Other examples can be found in thepaper by Gajentaan and Overmars, for instance.


http://dx.doi.org/10.1016/0925-7721(95)00022-2

3SUM-Hardness

Problem (2D degeneracy-testing)

Given a set P of n points in the plane, the 2D degeneracy-testing problemis to decide whether there are three collinear points in P.

Theorem

2D degeneracy-testing is 3SUM-hard.

Proof.

Based on the following fact:For any three distinct numbers a, b, c ,a + b + c = 0 iff the points (a, a3), (b, b3), (c , c3) are collinear.


cs 372: computational geometry lecture 4 lower bounds for ... · element distinctness theorem in...

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