Quantitative Steinitzs Theorems with Applications to

Multingered Grasping

David Kirkpatrick Bhubaneswar Mishra and CheeKeng Yap

Courant Institute of Mathematical Sciences

New York University

New York NY

Abstract

We prove the following quantitative form of a classical theorem of Steinitz Let m

b e suciently large If the convex hull of a subset S of Euclidean dspace contains

a unit ball centered on the origin then there is a subset of S with at most m p oints

whose convex hull contains a solid ball also centered on the origin and having residual

radius

2

d1

d

d

m

The case m d was rst considered by Barany Katchalski and Pach We

also show an upp er b ound on the achievable radius the residual radius must b e less

than

2

d1

d

m

These results have applications in the problem of computing the socalled closure

grasps by an mngered rob ot hand The ab ove quantitative form of Steinitzs the

orem gives a notion of eciency for closure grasps The theorem also gives rise to

some new problems in computational We present some ecient algorithms

for these problems esp ecially in the two dimensional case

Septemb er

1

Supp orted by NSF Grants DCR and DCR ONR Grant NJ NSF

Grant Sub contract CMU Chee Yap is supp orted in part by the German Research Foundation

DFG and by the ESPRIT II Basic Research Actions Program of the EC under contract No pro ject

ALCOM

2

Department of Computer Science University of British Columbia Vancouver Canada

Section

Intro duction

d

Caratheo dorys theorem states that given a subset S of Euclidean dspace E any p oint

inside its convex hull is also inside the convex hull of some subset of S with at most d p oints

d

Steinitzs theorem states that given a subset S E any p oint in the interior of its

convex hull is also in the interior of the convex hull of some subset of S with at most d p oints

Barany Katchalski and Pach showed that the following quantitative version of Steinitzs

theorem holds

Theorem Quantitative Steinitzs Theorem

d d

For any positive d there is a constant r r d d such that given any set S E of points

in dspace whose convex hul l contains the unit bal l centered at the origin o there is a subset

X S with at most d points whose convex hul l contains a bal l centered at o with radius r d

bdc

In fact notes that r d ced d for some constant c In this pap er we generalize

this quantitative Steinitzs theorem and study various algorithmic questions related to it

d

We intro duce the following terminology For any set S E let the residual bal l of S refer

to the largest closed ball BS centered at the origin o such that the interior of BS is either

fully outside or fully contained inside the convex hull of S The residual radius of S denoted

r S is the signed radius of this residual ball where the sign is zero if BS is a p oint otherwise

the sign is p ositive or negative dep ending resp ectively on whether BS lies inside or outside

the convex hull Let r m S or r m S if d is clear from the context denote the largest

d

residual radius of a subset X of S with at most m p oints Let r m denote the minimum value

d

d

of r m S as S ranges over all subsets S E with r S Hence for m d the result

d

d

of Barany Katchalski and Pach shows that d r m Here we derive tighter upp er

d

and lower b ounds for r m Note that the notation r d in the Quantitative Steinitzs theorem

d

ab ove is simply the case of r d

d

Application in Rob otics Our interest in these theorems comes from the study of rob ot

hand grasps We are interested in hands with m frictionless p ointngers A grasp in this

mo del consists of m p oints on the b oundary of the b o dy that we want to grasp To actually

grasp the b o dy we must then sp ecify forces to b e applied at these m grasp p oints A desirable

notion of grasping is that of a closure grasp see for example Intuitively a closure

grasp has the ability to resp ond to any external force or torque by applying appropriate forces

at the grasp p oints The quantitative Steinitz theorem gives us a measure of the eciency of

such closure grasps Roughly sp eaking the eciency of the grasp is given by the ratio of largest

external forcetorque that can b e resisted by applying at most unit forces at each of the grasp

p oints so a ratio of one corresp onds to the most ecient grasp The quantity r m in the

d

quantitative Steinitzs theorems gives this eciency directly

Computational problems These theorems naturally lead to new problems in computa

tional geometry For instance given a nite set S of p oints and a numb er m we want to nd

an msubset of S that achieves the residual radius r m S We will present algorithms for such

problems Here our strongest results are in two dimensions

The rest of this pap er is organized as follows In section we explain more precisely the

connection b etween quantitative Steinitzs theorem and grasping Sections and prove the

generalized Steinitzs theorems Sections and present algorithms for computational versions

of the generalized Steinitzs theorems We conclude in section

Application to Multingered Positive Grasps

We refer to for a general survey of the eld and to for the theory of rob ot grasping

as used in this work Consider an idealized dextrous hand consisting of m indep endently

movable forcesensing ngers These ngers can only contact ob jects at their tips and can thus

b e represented as p oints in threedimensional space The goal is to grasp a closed b ounded

and connected rigid ob ject K A nger can only apply a force on the ob ject K at the p oint of

contact with K We assume that the p oints of contact are nonsingular ie the surface of K

has a unique surface normal at each such p oint and frictionless and hence the force can only

b e applied along the surfacenormal at the p oint of contact directed inward into the ob ject K

An interesting task for such a hand is that of grasp selection for a given ob ject K by a grasp

we mean a set of m p oints on the b oundary of K We also describ e such grasps in our mo del

as positive b ecause the ngers can only push into the b o dy but not pull at the b o dy as might

happ en if we p ostulate sticky ngers

The ob ject K to b e grasp ed is assumed to have a piecewise smo oth b oundary Assume

that the grasp p oints are to b e chosen from a given subset S of the nonsingular p oints of

For example S may consist of all nonsingular p oints of by denition the surface normals at

nonsingular p oints are uniquely dened Or again S may b e a set of nitely many preselected

p oints For any p oint r in S let nr denote the unit surface normal directed inward at r

Dene the function mapping S into the sixdimensional forcetorque space as follows

S R

r nr r nr

where here denotes the vector crosspro duct of dimensional vectors Essentially maps r

to the p oint r in the forcetorque space that represents the eects of applying a unit force at

r in the direction nr

If X S is a set of mp oints we call X an mnger closure grasp if the interior of the convex

hull of X fr r X g contains the origin o It is shown in that for some m an

mnger closure grasp of K exists if and only if

o interiorconvexhull S

Moreover if is not a surface of revolution the ab ove condition is always satised with S equal

to the set of nonsingular p oints of For p olyhedral ob jects Mishra Schwartz and Sharir

also gave an algorithm to nd a nger closure grasp in linear time However in the absence

of any measure of go o dness for closures grasps the synthesized grasp may not b e very robust

The motivation for our work is to quantify the go o dness of closure grasps and to synthesize

provably go o d closure grasps

One criterion for go o dness is the eciency of a grasp which measures the amount of

external force and torque that can b e resisted by applying at most a unit of force at each grasp

p oint This is precisely the value r m S To see this note that if we cho ose m p oints in

Section

S with residual radius r then any forcetorque vector v whose Euclidean norm is at most

r can b e written as a convex combination of the m chosen p oints So if v is any external

forcetorque that is applied to the b o dy K and v lies in the residual ball of radius r we can

counter this external forcetorque by applying suitable forces of magnitude at most at the

grasp p oints such that these forces sum to v hence we maintain the b o dy in equilibriu m

These concepts can b e sp ecialized to the case where K is a planar b o dy in which case the

forcetorque space is dimensional The numb er of ngers for closure grasps can b e reduced

to in this case

Quantitative Steinitzs Theorem in Dimensions

Let S E b e a subset of the Euclidean plane and P its convex hull Without loss of generality

we may assume that P has at least four vertices Also it is assumed that a unit disc B centered

ab out the origin o is contained inside P In general let B denote the closed disc of radius

centered at the origin Our goal is to develop techniques given S and m for cho osing a set X

of at most m p oints from S so that the residual radius of X is maximized

Lemma Given S as above for any integer k we can nd a subset X of at most k

points of S such that the convex hul l of X contains B with cos

k

proof

Take k equally spaced rays from o making sure that one of them pass through a of P

the convex hull of S Let these rays intersect the unit circle centered at o at the p oints v

v For each ray if it intersects a vertex of P then we cho ose that vertex and if it intersects

k

an edge we cho ose the two vertices of that edge Thus we cho ose at most k p oints of S

forming the subset X S Clearly the convex hull of X contains the p oints v v and

k

hence it contains the B with cos k

We now show that this b ound is asymptotically tight It is imp ortant to note that this b ound

is achieved by cho osing vertices of the convex hull of S

Theorem For al l m we have

r m

m m

proof

The upp er b ound on r m comes from the previous lemma which shows that r m

cos m and from the fact that cos x x

For the lower b ound we let S b e the vertices of a regular m gon that just contains the

unit disc B Then the omission of any p oint of S gives a residual disc of radius

cos

m

cos

m

Thus r m is upp er b ounded by this radius

4 2

2 4

m m

r m

2

2

m

m m m

m

The sp ecial case where m is of particular interest We now give some sp ecial arguments

for this Starting with S as ab ove the preceding lemma shows how to cho ose at most p oints

of S whose convex hull contains the disc B B It is not hard to argue that one of

cos

these p oints has the prop erty that its two neighb oring p oints span an angle of at most

ab out the origin o and hence if we delete this p oint the remaining four p oints has a residual

radius of at least cos We can do b etter with the following argument

Theorem

cos

sin r

cos

proof

This upp er b ound is achieved by the regular p entagon which is the sp ecial case m of the

pro of in the previous theorem showing that for S given by the vertices of a regular m gon

cos r m S cos

m m

As b efore let P denote the convex hull of S For the lower b ound x to b e any angle

b etween and For any vertex v of P we dene its forbidden zone which consists of two

disjoint cones each spanning an angle of at the origin o and such that the two bisectors of

these cones together with the ray ov are equally spaced at apart See gure

A vertex v of P is bad if there is another vertex v of P that lies in its forbidden zone First

if v is bad then we can cho ose three other vertices v v v as follows Let v b e a vertex of P

that lies in the forbidden zone of v Let R b e the ray originating from o and bisecting the larger

of the two angles dened by the two rays from o through v and v resp ectively Let v v b e

the endp oints of the edge of P that R intersects It is not hard to see that the

v v v v contains a circle of radius at least

sin

Now supp ose that P has no bad vertices Assume that v is vertically ab ove o and the two

forbidden cones C C of v are b ounded by the rays oR oR and oR oR resp ectively where

the R s are p oints on the unit circle Since v is not bad each ray in C intersects a common

i

edge of P say v v and similarly each ray in C intersects a common edge of P say v v See

gure

First supp ose that the angle \v ov we always measure angles clo ckwise from v to v in

this notation is at most Then it is easy to see that the residual radius of v v v v

is at least sin

Section

v

o

Figure Forbidden zone of v is shaded

Hence assume that the angle \v ov Without loss of generality assume that

\R ov \v oR Then we have

\R ov \R ov

It is easy to see that Thus the distance from o to the line through v R is at least sin

the distance from o to the line through v v resp ectively v R is at least sin The

distance from o to the line through R R is at least We conclude that the residual radius of

v v v v which is at least the residual radius of v v R R is at least

min sin sin

to maximize this expression This where We cho ose ie

proves our lower b ound

Quantitative Steinitzs Theorems in Higher Dimensions

We now consider the ddimensional case for d The techniques are slightly weaker than the

dimensional case

v

v

o

v

R

R

R R

v

v

Figure Case of no bad vertices

Lower Bound

d+3

d

2

We rst give a lower b ound for r m for suciently large m in particular for all m d

d

Thus m is chosen to b e large enough to guarantee that

d

m

k

d

p

de takes integral values greater than d

d d

Lemma For any set S E whose convex hul l contains the unit bal l B centered at the

origin o we can nd a set X S of at most m points with residual radius

2

d1

d+3

d

d

2

r X d for al l m d

m

proof

Let k b e dened as a function of d and m as b efore It suces to show that

d d

r X

k k

in the given range for k

Section

Henceforth P will stand for the convex hull of S Let C b e the ddimensional cub e whose

faces are normal to the appropriate co ordinate axes of sidelength and containing the unit

d

ball B On each of C we place a k k k d times grid so the grid p oints have

co ordinates that are integer multiples of and two adjacent grid p oints are apart Note

k k

d

that there are fewer than dk md grid cub es on the union of the d faces of C Through

d

each grid p oint p we pass a ray R from the origin Let R intersect the unit sphere S at xR

For each such ray R we cho ose at most d vertices of P the convex hull of S as follows If the

ray passes through an iface of P we cho ose i vertices of P whose convex span intersects

that ray and is contained in that iface Thus the set X of chosen vertices has at most m p oints

The convex hull of X contains the set X of all p oints of the form xR where R is a ray passing

through the grid p oint

Let R b e any ray originating from o and supp ose it intersects some face of C at a p oint a

where a lies inside a grid cub e S Consider the oab where b is any other p oint on the

b oundary of S

sin \oab

sin \aob jabj

jobj

p

d

k

Cho ose to b e

p

d

arcsin

k

Let q b e any p oint at distance cos from the origin We show that q lies in the convex

hull of X Let R b e the ray from o through q and supp ose R intersects the grid cub e S

Let K b e the cone b ounded by the set of rays originating from o that makes an angle of with

R Hence each ray that passes through a vertex of S is contained in K There is a unique

d

hyp erplane H containing K S Note that q R H Let

T fxR R passes through a vertex of S g

and

T fR H R passes through a vertex of S g

By denition T X Note that each p oint in T lies on the side of H not containing the

origin This means that the convex hull of X contains the set T But the convex span of the

set T contains the p oint q R H This proves r X r X cos

cos sin

X

sin sin

i

sin

i

sin sin sin

sin

sin

since sin

d

k

This proves the lower b ound lemma

Upp er Bound

In this subsection we derive an upp er b ound for r m For this purp ose we let S b e all the

d

p oints on the unit sphere and then b ound the largest radius of a ball contained in the convex

hull of m p oints on the unit sphere The convex hull of any such m p oints forms a p olytop e

The pro of relies on the facts that any long edges of this p olytop e b ound the radius of the

contained ball and since the p olytop e has only m vertices it must have some long edges

The detailed calculations provide an appropriate numerical b ound

d

Lemma Let S E be the set of al l points on the surface of the ddimensional unit bal l

d

centered at the origin o Thus the convex hul l of S contains the unit bal l B centered at the

origin o Then any set X S of at most m points has a residual radius

2

d1

d

d

r X for al l m d

m

proof

The pro of pro ceeds in two steps We rst show that for all m and for all

2

d1

tan d

A

cos r X max

m

Then by an appropriate choice of the parameter we obtain the claimed b ound

d

Let X b e a set of m p oints in E all lying on the surface of a unit ball and P

ConvexHull X Let P b e the p olyhedron obtained from P by triangulating the nonsimplici al

facets of P Let pq b e an edge of the p olyhedron P Then

\poq

r X cos

Thus if

max \poq

0

pq edge of P

is the maximum of all such angles then

r X cos

If then

r X cos

Section

Henceforth we assume that Let t stand for tan thus t

Let p X b e any p oint and dene its truncated cone K as follows

p

K fx \xop and x p g

p

Now if q is an arbitrary p oint on the surface of the unit ball then the oq b elongs to

K for each vertex p of some simplicial of P As each such facet has d vertices

p

the collection of truncated cones cover each p oint in the unit ball at least d times Thus we see

that

m Volume K d Volumeunit ball

p

Let V r stand for the volume of a ddimensional ball of radius r

d

d

V r V r

d d

Thus the volume of the ddimensional unit ball is given by

Z

V sin sin d V

d d

Z

d

sin d V

d

K dV

d

where K d is dened by the last equation The volume of each K is given by

p

Z

V r tan dr Volume K

d p

Z

d

r dr V tan

d

V tan

d

d

Substituting the volumes into the preceding inequality we get

d

tan V

d

m dK dV

d

d

Hence

1

d1

d K d

cd m t tan tan

m

where cd m is dened in the last equation Using the inequality cd m t we get

cos cd m cd m t cd m

cd m

Hence

1

t

2

t cd m cd m cos cos

and

t

cd m cos

Finally we get

1

2

t t

cos cd m cd m

Hence

2

d1

d K d t

r X

m

Note that eg page

Z

d

sin d K d

k

if d k even

k

k

if d k o dd

k

d1

2

Here k stands for k k k terminating in or dep ending on

whether k is o dd or even Thus

2

d1

tan d

r X

m

The stated b ound follows with appropriate choice of the parameter as shown b elow

d

Let m d then

2

d1

d

m

Cho ose the parameter and observe that

2

d1

d

cos

m

Since tan

2 2

d1 d1

tan d d

m m

If we cho ose in the preceding pro of we can show that for all m

2

d1

d

r X

m

Section

If m d then

2

d1

d

r X

m

On the other hand if m d we get the result since cos and tan

Summarizing lemmas and

d+3

d

2

Theorem For al l m d

2 2

d1 d1

d d

r m d

d

m m

Computational Problems in the Plane

Finding m Vertices of a Convex Hull

The quantitative Steinitzs theorem p oses several interesting and new problems in computational

geometry We b egin with the simplest version of such problems given a convex ngon P whose

interior contains the origin nd four vertices of P whose residual radius is maximum In this

case we are able to give an elegant and simple linear time metho d Without loss of generality

we assume that n and the interior angles at each vertex of P is less than

Theorem There is a linear time algorithm for nding a set Q of four vertices of a convex

ngon P such that Q has the maximum residual radius r Q r P

We use the following general notations Assume that u s i m m are p oints

i

distinct from the origin Let ou denote the ray from origin o through u The notation

i i

u u u

m

says that the rays ou s are distinct and the ray ou is encountered b efore ou when sweeping

i i i

a ray originating from o counterclo ckwise from ou to ou We extend this notation to the case

m

where the u s are not necessarily distinct but we still require that the rays ou and ou b e

i m

distinct For instance we may write

u u u or u u u

For any p oint u on the b oundary of the p olygon P let the successor succ u of u denote the

vertex immediately following u when we traverse the b oundary of P clo ckwise If u is a vertex

of P we insist that succu is the next vertex of P

Our algorithm is simple to describ e its correctness is slightly harder to see Supp ose that

we have four vertices u u u u of P such that there are at least three distinct vertices among

them and

u u u u

The notation here makes sense since at most one of the inequalities is nonstrict Let

Q u u u u denote the p olygon formed by these vertices so Q is a triangle or a quadrilateral

Our goal is to rep eatedly cho ose one of these four p oints say u for some i to

i

advance ie set u to succu in the hop e of attaining a larger residual radius The criteria

i i

for cho osing the vertex to advance dep ends on the following two cases Remark Here all

arithmetic on subscripts are mo dulo

Q is a triangle Supp ose for some i u and u are coincident that is u u

i i i i

Then we advance u the forward vertex

i

Q is a quadrilateral An edge u u of the quadrilateral is limiting if the residual circle of Q

i i

touches that edge We then advance u the backward vertex where u u is any such

i i i

limiting edge

We make some observations

In case Q is a triangle advancing u can in turn make u and u coincident causing u

i i i i

to b e advanced in the next iteration However there cannot b e more than three consecutive

iterations in which Q is a triangle Note that a triangle Q can have nonp ositive residual

radius

In case Q is a quadrilateral and the residual radius r Q is nonp ositive the limiting edge

and hence u is uniquely determined After advancing u provided u u remains limiting

i i i i

the radius r Q will increase This same vertex is rep eatedly chosen at least until the rst

time t the edge u u is no longer limiting Observe that there are two p ossibilities at

i i

time t a r Q b ecomes p ositive b r Q remains nonp ositive In the latter case the

edge u u b ecomes limiting and we next start to advance u

i i i

To complete the description of this algorithm we must initialize the four p oints and give the

termination condition

The Algorithm Initially we pick any four consecutive vertices of the p olygon to

serve as Q u u u u We record the initial p osition of u Then we iterate the

basic step of picking and advancing an u up dating if necessary the largest value of

i

r Q encountered so far The algorithm halts when u returns to its initial p osition

and outputs the largest value of r Q recorded

It is clear that the algorithm makes at most n iterations when it halts

For the next lemma we need some notations Supp ose a a a a are the vertices of P that

achieves the maximum residual radius r r P Without loss of generality we may assume

that all four vertices are distinct and

a a a a

This partitions the vertices of the convex p olygon into four nonempty sections named

W W W W where

W a a W a a W a a W a a

Section

W

a

a

W

W

o

a

a

W

Figure Q and four sections of the p olygon P

See gure The notation a a refers to the consecutive subsequence of vertices from a

counterclo ckwise to but not including a Also let Q a a a a

The expression at time t u is advanced from a vertex a means that at time t u is at

j j

vertex a and at time t it is at succ a

Lemma Suppose at time t some u j is advanced from a of Q Hence

j i

at instant t u is in section W If r is not yet attained by the algorithm by time t then

j i

u is not in W at time t

j i

proof

Without loss of generality assume that i j in the statement of the lemma That is at

time t u is advanced from vertex a By way of contradiction supp ose u is in W at time t

There are two cases

Case Supp ose Q is a triangle at time t Let t t b e the last instant when Q was a

quadrilateral By a previous observation t t Note that at time t for some

u is advanced so that u and u b ecame coincident Thus u and u are adjacent at

time t If r Q at time t then the origin O is on the side of the line u u opp osite to the

other vertices of P which is imp ossible This shows that r Q Since Q is a quadrilateral

we only advance u u b ecause u u is limiting But u u determines a radius greater than

r which leads to a contradiction

Case Supp ose Q is a quadrilateral at time t Then u u is limiting at time t If r Q were

p ositive we deduce that r Q is at least r which is a contradiction If r Q were nonp ositive

then in order that u u b e limiting the origin must lie to the left of the line directed from u to

u This forces the origin to lie outside Q again leading to a contradiction

We are now ready to show

Lemma The algorithm is correct

proof

Supp ose the algorithm halts when u returns to its original p osition b Without loss of generality

assume that b lies in the section W a a Let t b e the instant when u is advanced from

a into W If r Q has already achieved the maximum value of r b efore time t then we are

done Otherwise we obtain a contradiction as follows By the previous lemma u do es not lie

in W at time t

Claim u do es not lie in W W at time t for if u lies in W then u would b e forced to

b e in W as well this is a contradiction So it remains to show u do es not lie in W If it do es

then b oth u and u lie in W Let t t b e the last time that u do es not lie in W such

an instant is welldened So u was advanced from a at time t Now an application of the

previous lemma again shows that r Q would have attained the maximum value r b efore time

t which is a contradiction This proves the claim

We can rep eat the argument of this claim to show that u do es not lie in W W W at

time t Hence u lies in W Thus b oth u and u lies in W Again let t t b e the last

time that u do es not lie in W Then an application of the ab ove lemma to u at time t yields

the contradiction

We easily extend the ab ove metho d to nding the b est m vertices of the p olygon P

Now we need O log m p er iteration using a priority queue to nd the limiting edge of the

current mgon and the numb er of iterations is at most mn This yields the following theorem

Theorem For any m and n m there is an O nm log m time algorithm which on

any input convex ngon P computes the value of r m P

Finding m Points in the General Case

The ab ove section considers algorithms to compute r m S for the sp ecial cases where S is the

set of vertices of a convex p olygon In general S is an arbitrary set of p oints in the plane and

supp ose P is the subset of S consisting of all the vertices of the convex hull of S We note that

r m S is in general larger than r m P As an example let P b e the vertices of a regular

p entagon and S contains in addition to P for each edge of the p entagon a p oint in the interior

of P but lying very close to the midp oint of that edge Then r S r P

On the other hand r m P is a reasonably go o d lower b ound to r m S This follows from

our general constructions in section where the asymptotically tight lower b ound for r m S

is obtained by cho osing p oints on the convex hull of S Nevertheless we may want to nd the

exact value of r m S This subsection gives such an algorithm Again we b egin with the case

m

Theorem There is an O n log n algorithm to nd a set X of four points in a set S of n

points such that the residual radius of X is maximized

Let S b e a set of p oints with p ositive residual radius Let the p oints of S b e assumed to b e

arranged by their counterclo ckwise angular order as in the previous subsection We want to

nd four p oints in S with the largest residual radius For any pair u v of distinct p oints let

Section

w

x

u v

u v

C u v

u v

Figure A quadrilateral w uv x containing the circle C u v

C u v denote the circle centered at the origin which has the line uv as tangent and let radu v

b e the radius of C u v If for some two p oints w and x of S w u v and u v x C u v

is contained in the quadrilateral w uv x then the residual radius of w uv x is equal to radu v

Note that it is p ossible that w x In the remaining p ortion of this subsection we show how

for a given pair u and v such a choice of w and x if they exist can b e made in logarithmic

time thus providing an O n log ntime algorithm for the problem

First we need some notations Let C b e a circle and let u and v b e p oints of S outside

C For a given p oint u let u denote the p oint diametrically opp osite to u with resp ect to the

origin o We say u and v are mutual ly C visible if the line segment connecting u and v do es

not intersect the interior of C Also we say v is covered by u relative to C if v is mutually

C visible with u and b elongs to the smallest cone with its ap ex at u and containing C See

gure We say a p oint is relevant for C if it is outside C and not covered by any other p oints

of S We omit references to the circle C if it is apparent from the context

We have the following observations

Lemma

Let w uv x be a convex quadrilateral containing C u v Consider two points w and x such

that w covers w and x covers x Then one of the w uv x or x uv w also

contains C u v

Let w uv x be a convex quadrilateral containing C u v Consider two relevant points w

and x satisfying the fol lowing conditions u w w u v x x v w and u are

mutual ly C u v visible and x and v are mutual ly C u v visible Then either w uv x or

x uv w is a convex quadrilateral and contains C u v

w

u

v

o

Figure u covers v but not w w covers neither u nor v

Let C C be two concentric circles with C being the larger of the two If u is not relevant

to C then u is also not relevant to C

Let u b e a p oint outside C and W S b e the set of p oints w w relevant to C such

that

u w w u

Let w b e a p oint in W with the largest index such that

j

for all i j u and w are mutually C visible

i

Then w is the rightmost element in the set W mutually visible with u Supp ose there is

j

another element w W k j that is mutually visible with u Then k j and w is

k j

not mutually visible with u in this case w would b e covered by w thus contradicting the

j k

hyp othesis that all p oints of W are relevant We say w is the rightmost C partner of u denoted

j

RP u Similarly we may dene the leftmost C partner of u LP u We conclude that RP u

and LP v can b e computed whenever they exist in logarithmic time if we have a balanced

search structure that contains only the p oints relevant to C u v sorted by their angular order

Now using the preceding lemma we observe that if for some w and x the quadrilateral

w uv x contains the circle C u v then so do es one of the quadrilaterals RP u uv LP v and

LP v uv RP u Thus it suces to check that C u v is tangent to uv at some p oint in

the segment uv and RP u and LP v are mutually C u v visible

The basic idea is to put all vertices u and also all unordered pairs fu v g of p oints in S into

a single priority queue We use the Euclidean distance b etween u and the origin o as priority of

u and the value of radu v as priority of fu v g We may omit all fu v gs with radu v

Section

and b egin the pro cessing of the queue by successively extracting items with the smallest priority

Note that we could assume that the rst item extracted is a pair fu v g

For the rst pair fu v g extracted from the queue we need to initialize a data structure to

store the p oints of S relevant to C u v according to their angular order We may break the

circular ordering into a linear ordering at some arbitrary breakp oint We store these p oints as

a linear ordering in the leaves of a balanced binary tree T

In the general step supp ose we extract from the priority queue either a pair fu v g or vertex

u There are two cases

Case A pair fu v g is extracted and C u v touches uv at some p oint z in the segment

uv Then assuming u z v we use the search tree T to determine the rightmost

C u v partner w of u and the leftmost C u v partner x of v and check if w and x are

mutually C u v visible If so we have found a larger residual radius

Case a A pair fu v g is extracted and C u v touches uv at some p oint outside the

segment uv Assume that of the two p oints u v the p oint u is the closer to o Then we

note that for subsequent larger radii circles C the vertex v covers u relative to C Hence

we can delete the element u from the data structure T b Similarly if u is extracted we

can delete u from T

Clearly each op eration takes O log n time Since there are O n elements in the queue the

overall complexity is O n log n Note that at any instant when a pair fu v g is b eing considered

only relevant p oints of S are left in T More precisely any p oint in the interior of C u v or

covered by some other p oint would have already b een deleted A p oint in the interior of C u v

is at a smaller distance from o than radu v and if v covers u then radu v radu v u v

is touched by C u v outside the segment u v and u is closer to o than v Thus it is clear

that the algorithm is correct

Again the metho d generalizes to nding any numb er m of p oints that has the b est residual

radius This yields the following theorem

Theorem For any m and n m there is an O n m log n time algorithm which on

any input set S of n points in the plane computes the value of r m S

We note that a faster algorithm is p ossible if we are willing to settle for a go o d to within

a factor approximation of r m S for all m Sp ecically we can

determine if r m S is or a xed value r in time O nm log n In O n log n time we can

determine the set of p oints that are relevant for a circle of radius r For each relevant p oint u in

O m log n time we can determine if there exist m other additional p oints such that the

set containing u together with these p oints has a residual radius of r or larger To b egin with

we cho ose the b est set of four vertices on the convex hull of S and call its residual radius r

Using an O n log n time convex hull algorithm and the algorithm of the previous subsection

we guarantee that this step takes no more than O n log n time Thus

r

r m S

Using k O log comparisons we can p erform a binary search to improve the approxima

tion to

r

k

r m S

Thus the resulting algorithm computes a go o d approximation with a relative error of in time

O nm log n log

Computational Problems in Higher Dimensions

In this section we study the following algorithmic problem

Given a set S of n p oints in ddimensional Euclidean space whose residual radius

r S is p ositive nd a subset X S of at most m p oints such that the following

inequality holds

2

d1

d r X

r m d

d

r S m

d+3

d

2

Here m and n are assumed to b e suciently large ie n m d

We shall not discuss the more general optimization problem of nding a subset of m p oints

that maximizes the preceding ratio for two reasons rstly for large m the approximate solution

provides a reasonably go o d answer secondly any hop e for nding such a set in time p olynomial

in b oth d and n seems rather dim While an investigation of this optimization problem is called

for we simply leave it as an op en problem

Returning to the stated problem we see that this problem can b e solved by essentially

following the ideas outlined in lemma We rst cho ose a set X of at most md p oints on

the surface of the unit ball such that the residual radius of X is no smaller thanr m We can

d

then determine a set X S of at most m p oints such that for some r S the convex

min

hull of X contains the set of p oints

X q q X

min min

Thus

r X r X r m r S r m

min min d d

The p oints of X are chosen as follows Let C b e the ddimensional cub e comprising the

p oints x x with jx j for i d On each face of C we place a k k k

d i

d times grid with k taking the value

1

(d1)

m

d

Let

o n

d

op S p is a grid p oint X

Section

d

Thus jX j dk md For each q X we determine an appropriate set S S of at

q

most d p oints such that

oq ConvexHull S ConvexHull S

q

thus for some

q

q ConvexHull S

q q

Let X b e

X S

q

0

q X

0

Evidently r S with taking the value min

q min min q X

Note that jX j m and

X ConvexHull X

min

This demonstrates the correctness of the algorithm since we know that the residual radius of

X is b ounded from b elow by r m see the pro of of lemma

d

In order to complete the algorithm we show how to eciently compute the set S for any

q

p oint q using the following linear programming formulation Let S fp p p g Without

n

loss of generality we assume that the p oints of S are in general position ie at most d points of S

may lie on any d dimensional hyperplane If not the original p oints of S may b e p erturb ed

using generic p erturbation metho ds see for example the following discussions still apply

th

mutatis mutandis Dene the d n matrix A whose j column consists of the co ordinates of

the p oint p Corresp onding to the p oint q dene a column dvector b The linear programming

j

problem LP is given as follows

Given A d n matrix A and a column dvector b

Solve

minimize

sub ject to Ax b

T

e x

x

T T T

where x x x e and are column nvectors

n

Let x b e an optimal solution of LP Then is the maximum value of such that

n

X

q x p

i

i

i

P

n

and x with x

i

i i

Now consider the following dual of the LP which will b e referred to as DLP

maximize y

d

sub ject to a y a y y

d d d

a y a y y

d d d

a y a y y

n dn d d

b y b y

d d

2

d

This problem can b e solved in O n time by using ClarksonDyers improvement on Megiddos

multidimensional search technique Let us now see how to recover the solution to the

original problem

Clearly b oth LP and DLP have optimal solutions Let an optimal solution for DLP b e

y y y y

d d

Let I fng b e the set of all the indices j such that

q

a y a y a y y

j j dj

d d

T

where a a a By the Complementary Slackness Theorem see this implies

j j dj

then i I By virtue of our nondegeneracy hyp othesis that for all i n if x

q

i

ab out the p oints of S we see that jI j d We now claim that S fp j I g can serve as a

q q j q

desired solution Clearly S S has at most d p oints and

q

oq ConvexHullS ConvexHull S

q

Note that even if the original set had b een p erturb ed by a suciently small amount the set

S chosen from the unp erturb ed set S still provides the desired solution

q

To summarize

d d

Theorem For n m d we can nd a set X of at most m points from an input

set S of n points such that

2

d1

d r X

r m d

d

r S m

2

d

in time O mn

Final Remarks

d

It is natural to seek improved forms of Steinitzs theorem for certain subsets S E In other

d

words if k is any numb er b etween d and d we want to characterize those subsets S E

whose residual radius is p ositive and are such that S contains a subset X of at most k p oints

where X has a p ositive residual radius For instance in the plane

Lemma Let S E be any set with positive residual radius Then there is a subset of three

points in S with positive residual radius if and only if S is not contained in two lines through

the origin

We omit the easy pro of It would b e interesting to develop an appropriate quantitative forms

of this lemma We see that an obvious quantitative version for this lemma fails That is there

do es not exist a constant with the following prop erty

Supp ose the residual radius of S E is at least one and S do es not lie in two lines

through the origin Then there exists three p oints in S whose residual radius is at

least

Section

To see this consider the set S fA B C D E g where A B C D

L and E L for L L suciently large Then no subset of S with three

p oints has residual radius at least

Yet another area of research that calls for further investigation arises from the observation

that the torque and force dimensions are really noncomparable We want a notion of grasp

eciency that can take this into account A related issue is that the current approach dep ends

on the origin of the reference frame in which the torques are measureed Is there an origin

indep endent approach to eciency and other metrics of a grasp

Acknowledgment

We are pleased to acknowledge some helpful discussions with Professor S Rao Kosara ju of Johns

Hopkins University and Professor Richard Pollack of the Courant Institute We also thank an

anonymous referee who p ointed out a calculation error in the lemma

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