Balancing location of points on a boundary of region - PowerPoint Presentation

Slide1

Balancing location of points on a boundary of region

Takeshi Tokuyama

Tohoku University

Joint work with many collaboratorsSlide2

Prologue: balancing in real lifeSlide3

　　　Center of mass

and barycenter

Given a set of weighted points q(

i

) with weight w(i)

i=1,2,..,k such that

their center of mass is at the barycenter point

・

Balancing location of points: Given a target point p and weight set w(

i

) , find a set of points q(1),q(2),..,q(k) on a given curve (or surface) C such that their weighted barycenter is at p.

Basic problem (in engineering and in daily life): Adjusting the center of mass by giving some weights on a specified curve

Japanese weight measure: move one weight to balancePuzzle by Yoshio Okamoto

 Slide4

A very special case: Antipodal pair

For any bounded region Q with boundary C and a target point p in Q, there are two points (antipodal pair) on C such that their midpoint becomes p

Balancing location of two points with a same weight.

How to prove it??? (easy if shown, but difficult to find by yourself)

Famous related theorem:

Borsuk-Ulam

theorem:

A

ny continuous function F from d-dimensional sphere to d-dimensional real space has a point x such that F(x)= F(-x)

On the earth, there

are

opposite points

with

the same temperature and the same height.

Many applications (Monograph of

Matousek

)Slide5

　　Another special case: Motion of robot arm

Consider

a robot arm with k links of lengths w(1), w(2)..,w(k) with total length 1. If one end is fixed at the origin,

the other end can reach every point in the unit ball??

(we do not mind crossing of links)

Equivalent problem:

A balancing location of k weighted points on a sphere C such that the barycenter becomes the center exists?

Another equivalent problem:

w(1)S ⊕w(2)S⊕

…⊕w(k)S = B , where B is the unit ball and

w(k)S is the sphere with radius w(k)　

and

⊕

is the Minkowski sum X ⊕　Y =

Exercise 1: Prove the equivalence of above three problemsExercise 2: Show

what is 2/3 C ⊕　1/3 C for a unit circle C

Exercise 3: Give a sufficient and necessary condition for weights.

 Slide6

Minkowski

sum

⊕

⊕

⊕Slide7

⊕

⊕

⊕

⊕

⊕

⊕Slide8

Balancing location on a boundary of region

Definition: A weight set W = w(1), w(2),..,w(k) is monopolistic if its largest weight is larger than half of the total weight.

Theorem 1. If the weight set is not monopolistic, for any closed region Q and a target point p in Q, there exists a balancing location of k weighted points on the boundary curve

of Q.

Theorem 1 is written as: If W is not monopolistic, then

w(1)

⊕

w(2)

⊕

..⊕ w(k)  Slide9

　　Ｋｅｙ　ｌｅｍｍａ

If we have a heavier weight on the boundary and the other in the interior of Q, then we can move both to the boundary keeping the center of massSlide10

ProofSlide11

Algorithm for non-monopolistic weight set (k=3)

Take any line L through the target point p

Place the heaviest weight

w

(1) at the nearest intersection q(1) of L and the boundary The other two points w(2) and w(3) are located at the same interior point q(2,3) such that they are balancing (i.e. center of mass is at p)

This is possible since weights are not monopolistic

Consider the heaviest point and the second heaviest point, and move both to the boundary (using the lemma) keeping the balanceConsider the second heaviest point and the third heaviest point, and move both to the boundary

All three points are on the boundary !Slide12

Locate three weights

w(1)

,

w(2)

,

and

w(3)

on the boundary such that their weighted barycenter is at p

pSlide13

Take any line L through the target point p

Place the heaviest weight w(1) at the nearest intersection q(1) of L and the boundary

The other two points w(2) and w(3) are located at the same interior point q(2,3) such that they are balancing

Consider the heaviest point and the second heaviest point, and move both to the boundary keeping the balance

Consider the second heaviest point and the third heaviest point, and move both to the boundary

All three points are on the boundary !

p

L

q

(1)Slide14

Take any line L through the target point p

Place the heaviest weight w(1) at the nearest intersection q(1) of L and the boundary

The other two points w(2) and w(3) are located at the same interior point q(2,3) such that they are balancing

Consider the heaviest point and the second heaviest point, and move both to the boundary keeping the balance

Consider the second heaviest point and the third heaviest point, and move both to the boundary

All three points are on the boundary !

p

q

(1)

q(2,3)Slide15

Take any line L through the target point p

Place the heaviest weight w(1) at the nearest intersection q(1) of L and the boundary

The other two points w(2) and w(3) are located at the same interior point q(2,3) such that they are balancing

Consider the heaviest point and the second heaviest point, and move both to the boundary keeping the balance

Consider the second heaviest point and the third heaviest point, and move both to the boundary

All three points are on the boundary !Slide16

Take any line L through the target point p

Place the heaviest weight w(1) at the nearest intersection q(1) of L and the boundary

The other two points w(2) and w(3) are located at the same interior point q(2,3) such that they are balancing

Consider the heaviest point and the second heaviest point, and move both to the boundary keeping the balance

(Use our Key Lemma)

Consider the second heaviest point and the third heaviest point, and move both to the boundary

All three points are on the boundary !Slide17

Take any line L through the target point p

Place the heaviest weight w(1) at the nearest intersection q(1) of L and the boundary

The other two points w(2) and w(3) are located at the same interior point q(2,3) such that they are balancing

Consider the heaviest point and the second heaviest point, and move both to the boundary keeping the balance

Consider the second heaviest point and the third heaviest point, and move both to the boundary

All three points are on the boundary !Slide18

Take any line L through the target point p

Place the heaviest weight w(1) at the nearest intersection q(1) of L and the boundary

The other two points w(2) and w(3) are located at the same interior point q(2,3) such that they are balancing

Consider the heaviest point and the second heaviest point, and move both to the boundary keeping the balance

Consider the second heaviest point and the third heaviest point, and move both to the boundary

All three points are on the boundary !

Exercise: Generalize it to general k weights caseSlide19

What happens in thre

e dimensions?

In real life it is more stable to use three weights than two weights

Given a three dimensional region Q, can we find balancing location on

for a target point p and a non-monopolistic weight set?

Yes, since we can find a balancing location in a two-dimension slice of

Give more constraint! Tripodal

location: Can we find three points on

such that

1. Their center of mass is at a given point p of Q

2. They are equidistant from pWith a short thought, you see they form an equilateral triangleTheorem 2. Tripodal location always exists (if Q is a polyhedron, a smooth manifold, or a PL manifold).  Slide20

Impossible in 2D in general

Exercise: Give a counter example

In 3D, possible!

Equilateral triangle is “special”.

No analogue for other triangle shape than the equilateral triangle

Exercise: Give a counter exampleSlide21

famous problem

Toeplitz

’ Square Peg Problem (1911)

Given a closed curve in the plane, is it always possible to find four points on the curve forming vertices of a square?Yes, if the curve is smooth!, Unknown for a general curveExercise: Find a peg position of the following shapeSlide22

famous problem

Toeplitz

’ Square Peg Problem (1911)

Given a closed curve in the plane, is it always possible to find four points on the curve forming vertices of a square?Yes, if the curve is smooth!, Unknown for a general curveSlide23

Existence of

tripodal

location

Consider the nearest and farthest points n and f from p.

Consider a path

A= { a(t): 0 < t <1} from n to f ,such that n = a(0) and t=a(1). We fix “base

hyperplane H(t)” through a(t) and p which is continuous in t. (This is called “vector field” in mathematics)

For each t, we define base triangle T(t, 0) = ( a(t), b(t

）, c(t)): Equilateral triangle on H(t) locating a vertex a(t) and center at p

Slide24

Existence of

tripodal

location

Consider the nearest and farthest points n and f from p.

Consider a path

A

= { a(t): 0 < t <1} from n to f ,such that n = a(0) and f=a(1). We fix “base

hyperplane

H(t)” through a(t) and p which is continuous in t. (This is called “vector field” in mathematics)

For each t, we define base triangle T(t, 0) = ( a(t), b(t

）

, c(t)): Equilateral triangle on H(t) locating a vertex a(t) and center at p

n

p

f

a

(t)

p

n

f

a

(t)

p

n

f

a

(t)

pSlide25

Existence of

tripodal

location

Consider the nearest and farthest points n and f from p.

Consider a path

A= { a(t): 0 < t <1} from n to f ,such that n = a(0) and f=a(1). We fix “base

hyperplane H(t)” through a(t) and p which is continuous in t. (This is called “vector field” in mathematics)

For each t, we define base triangle T(t, 0) = (

a(t), b(t）

, c(t)):

Equilateral triangle on H(t) locating a vertex a(t) and center at p

Slide26

Existence of tripodal location

Consider the nearest and farthest points n and f from p.

Consider a path A= { a(t): 0 < t <1} from n to f on the

boundary,such

that n = a(0) and t=a(1). We fix “base

hyperplane

H(t)” through a(t), which is continuous in t.For each t, we define base triangle T(t, 0) = ( a(t), b(t

）

, c(t)): Equilateral triangle on H(t) locating a vertex a(t) and center at pT(t, θ）：

Triangle obtained by rotating T(t, 0) by θ about the line a(t)p

Sign(t, θ）　= (+,+) if both b(t) and c(t) are outside of Q

= (-,

+) if b(t) is inside Q and c(t)is outside = (0,+ ) if b(t) is on the boundary and c(t) is outside ETCWe should show that there exists t and θ　such that Sign (t, θ) = (0,0) Slide27

Existence of tripodal

location

Consider the nearest and farthest points n and f from p.

Consider a path G= { a(t): 0 < t <1} from n to f ,such that n = a(0) and f=a(1). We fix “base

hyperplane

H(t)” through a(t) and p , which is continuous in t.

For each t, we define base triangle T(t, 0) = ( a(t), b(t）

, c(t)): Equilateral triangle on H(t) locating a vertex a(t) and center at p

T(t, θ）：

Triangle obtained by rotating T(t, 0) by θ about the line a(t)p

Sign(t, θ）　= (+,+) if both b(t)

and

c(t)

are outside of Q = (-, +) if b(t) is inside Q and c(t)is outside = (0,+ ) if b(t) is on the boundary and c(t) is outside ETCObseration: Sign(0, θ） = (+,+) (or (+,0) or (0,+) or (0,0) ) Sign (1, θ）　＝(-, -) (or (-,0), or (0,-), or (0,0))Slide28

Transition of signature

(+,+) cannot change to (-,-) directly if t (or θ

）

continuously changes(+,+) 

(+,0)  (+,-)

 (0,-)

 (-,-)(+,+)  (00)  (-,-)

Form a walk a graph G from (+,+) to (-,-) if we fix θ

　and move t from 0 to 1

A walk on a circle C if (0,0) does not appear.Slide29

Transition of signature

Form a walk a graph G from (++) to (--) if we fix θ

and move t from 0 to 1

A walk on a circle C if (0,0) does not appear.

Key observation: If we change θ, the parity of the number of the red edges on the walk is unchanged.Slide30

Transition of signature

Key observation: If we change θ, the parity of the number of the red edges on the walk is unchanged.

Each walk use

red

and

green

odd times in total

Consider T(t, θ）　

and T(t, -θ

）,

then

they have opposite signatureWe get contradiction  (0,0) must exists.Slide31

OPEN PROBLEM

Is it true that we have balancing location of four points such that they form a equilateral tetrahedron ????

How about in d-dimensions??Slide32

　　　Can we find balancing location on edges?

In two dimensional space, boundary of a polygon consists of edges, and we can find antipodal pair on edges.

In higher dimensional case, boundary and edge-skeleton (union of edges) are different.

Conjecture: Given a convex polytope P in d-dimensional space and a target point p in P, can we locate d points on the edge-skeleton such that their center of mass is p?

Conjecture is true if d is a power of 2Recent result by Michael Dobbins: true if d is 2i

3jBeautiful proof using characteristic class of vector bundle

Need mathematical tools such as homology theorySlide33

B

alancing location on edges in 4-dim

Set p = o (origin), and consider Q = P

∩　

(-P)

Q has a vertex v in S

2

(P)

∩　

S

2

(-P) (

S

k

(P) is the k-skeleton of P)

Caution: S

1

(P

)

∩　

S

3

(-

P

) may be empty.

This means P has antipodal pair each point on 2-dim faces

Each point 2-dim face has antipodal pair on edges

Center of mass of these four points is oSlide34

Future direction

M

ajor conjectures remain

unsolved Existence of d-dimensional version of the tripodal location

Balancing location on 1-skeleton for a nonconvex 3-dim polytopeRelation to

Borsuk-Ulam theoremModern interpretation of

Borsuk-Ulam theoremEuler

class of vector bundle on d-dim sphere with Z2 action is nontrivial

Extension to fiber bundle of topological space with group actionWe need collaboration with researchers

on algebraic topologyMore general location problems (higher motion

etc

)Slide35

E

pilogue

Importance of International

Collaboration

Collaborators:

Luis Barba (ULB,

Bellguim and Carleton, Canada), Jean Lou de

Carufel (Carleton), Otfried

Cheong(KAIST, Korea), Michael Dobbins(POSTECH, Korea), Rudolf Fleischer (Fukdan, China and

Gutech, Oman), Akitoshi

Kawamura(U. Tokyo, Japan), Matias Korman

(

Katalonia

Polytech, Spain), Yoshio Okamoto (UEC, Japan), Janos Pach (EPFL Swiss, and Reny Inst, Hungary), Yan Tang (Fudan), Sander Verdonschot(Carleton), Tianhao Wang (Fudan) 12 collaborators from 10 countries I thank to organizers of international workshops, which were essential to accomplish this work.Slide36
Slide37
Slide38

Huge database of unfolding: How to handle?

http://www.al.ics.saitama-u.ac.jp/horiyama/research/unfolding/catalog.new/index.ja.html

Concluding remark: We need to be clever to handle geometric computation with help of power of mathematics!