Balancing location of points on a boundary of region

Balancing location of points on a boundary

of regionTakeshi TokuyamaTohoku University

Joint work with many collaborators

Prologue: balancing in real life

　　　 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 balance

Puzzle by Yoshio Okamoto

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:Any 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)

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

Minkowski sum⊕

⊕

⊕

⊕ ⊕

⊕ ⊕

⊕ ⊕

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) ⊕ ⊕ ⊕

　　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 mass

　 Proof

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 balance• Consider the second heaviest point and the third heaviest point, and move both to

the boundary• All three points are on the boundary !

• Locate three weights w(1) , w(2), and w(3) on the boundary such that their weighted barycenter is at pp

• 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)







p

q(1)

q(2,3)











(Use our Key Lemma)• Consider the second heaviest point and the

third heaviest point, and move both to the boundary














Exercise: Generalize it to general k weights case

　　 What happens in three 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 Q2. They are equidistant from p

• With a short thought, you see they form an equilateral triangle

Theorem 2. Tripodal location always exists (if Q is a polyhedron, a smooth manifold, or a PL manifold).

• 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 example

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 curve• Exercise: Find a peg position of the following shape

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 curve

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


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)):


n

p

f

a(t)

p

n

f

a(t)

p

n

f

a(t)

p


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)):


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 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 ETCWe should show that there exists t and θ 　 such that Sign (t, θ) = (0,0)

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 ETC

Obseration: Sign(0, θ ） = (+,+) (or (+,0) or (0,+) or (0,0) ) Sign (1, θ ）　＝ (-, -) (or (-,0), or (0,-), or (0,0))

　　 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.

　　 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.

　　 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 signature• We get contradiction (0,0) must exists.

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??

　　　 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 2• Recent result by Michael Dobbins: true if d is 2i3j

• Beautiful proof using characteristic class of vector bundle• Need mathematical tools such as homology theory

　　　 Balancing location on edges in 4-dim 　Set p = o (origin), and consider Q = P ∩ 　 (-P) • Q has a vertex v in S2(P) ∩ 　 S2(-P) (Sk(P) is the k-skeleton of P)• Caution: S1(P) ∩ 　 S3(-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 o

Future direction• Major conjectures remain unsolved• Existence of d-dimensional version of the tripodal location• Balancing location on 1-skeleton for a nonconvex 3-dim polytope

• Relation to Borsuk-Ulam theorem• Modern interpretation of Borsuk-Ulam theorem

• Euler class of vector bundle on d-dim sphere with Z2 action is nontrivial• Extension to fiber bundle of topological space with group action

• We need collaboration with researchers on algebraic topology

• More general location problems (higher motion etc)

Epilogue 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.

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!

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



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Balancing location of points on a boundary of region