Searching for the center.

Don SheehyCMU Theory Lunch

October 8, 2008

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It’s a fine line between stupid and

clever.

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The Divide and Conquer Game

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The Divide and Conquer Game

How to win

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The Divide and Conquer Game

How to win

Pick a center point.

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The Divide and Conquer Game

How to win

Pick a center point.

Given a set S ! Rd, a center point p is a pointsuch that every closed halfspace with p on itsboundary contains at least n

d+1points of S.

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Some definitions you probably already know.

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!

pi!P

cipiLinear:

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!

pi!P

cipiLinear:

Nonnegative: ci ! 0

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!

pi!P

cipiLinear:

Nonnegative:

Affine:

ci ! 0

!ci = 1

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!

pi!P

cipiLinear:

Nonnegative:

Affine:

Convex: Affine and Nonnegative

ci ! 0

!ci = 1

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Radon => Helly => Center Points Exist.

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Radon’s TheoremIf P ! Rd has d+2 (or more) points then there is a partition of P into (U,U)

such that conv(U) " conv(U) is nonempty.

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Radon’s TheoremIf P ! Rd has d+2 (or more) points then there is a partition of P into (U,U)

such that conv(U) " conv(U) is nonempty.

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Radon’s TheoremIf P ! Rd has d+2 (or more) points then there is a partition of P into (U,U)

such that conv(U) " conv(U) is nonempty.

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Radon’s TheoremIf P ! Rd has d+2 (or more) points then there is a partition of P into (U,U)

such that conv(U) " conv(U) is nonempty.

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Radon’s TheoremIf P ! Rd has d+2 (or more) points then there is a partition of P into (U,U)

such that conv(U) " conv(U) is nonempty.

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Radon’s TheoremIf P ! Rd has d+2 (or more) points then there is a partition of P into (U,U)

such that conv(U) " conv(U) is nonempty.

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Radon’s TheoremIf P ! Rd has d+2 (or more) points then there is a partition of P into (U,U)

such that conv(U) " conv(U) is nonempty.

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Radon’s TheoremIf P ! Rd has d+2 (or more) points then there is a partition of P into (U,U)

such that conv(U) " conv(U) is nonempty.

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Radon’s TheoremIf P ! Rd has d+2 (or more) points then there is a partition of P into (U,U)

such that conv(U) " conv(U) is nonempty.

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Radon’s Theorem

d+2!

i=1

cipi = 0

d+2!

i=1

ci = 0

If P ! Rd has d+2 (or more) points then there is a partition of P into (U,U)such that conv(U) " conv(U) is nonempty.

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Radon’s Theorem

d+2!

i=1

cipi = 0

d+2!

i=1

ci = 0

I+

= {i : ci > 0}

I!

= {i : ci < 0}

!

i!I+

cipi =!

i!I!

(!ci)pi

If P ! Rd has d+2 (or more) points then there is a partition of P into (U,U)such that conv(U) " conv(U) is nonempty.

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Radon’s Theorem

d+2!

i=1

cipi = 0

d+2!

i=1

ci = 0

I+

= {i : ci > 0}

I!

= {i : ci < 0}

!

i!I+

cipi =!

i!I!

(!ci)pi

!

i!I+

ci =!

i!I!

(!ci)

If P ! Rd has d+2 (or more) points then there is a partition of P into (U,U)such that conv(U) " conv(U) is nonempty.

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Radon’s Theorem

d+2!

i=1

cipi = 0

d+2!

i=1

ci = 0

I+

= {i : ci > 0}

I!

= {i : ci < 0}

!

i!I+

cipi =!

i!I!

(!ci)pi

!

i!I+

ci =!

i!I!

(!ci)

If P ! Rd has d+2 (or more) points then there is a partition of P into (U,U)such that conv(U) " conv(U) is nonempty.

x =

!

i!I+

"

ci#

j!I+ cj

$

pi =

!

i!I!

"

!ci#

j!I!!cj

$

pi

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Helly’s TheoremGiven some convex sets in Rd such that every d + 1 sets

have common intersection, then the whole collection of setshas a common intersection.

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Helly’s TheoremGiven some convex sets in Rd such that every d + 1 sets

have common intersection, then the whole collection of setshas a common intersection.

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Helly’s TheoremGiven some convex sets in Rd such that every d + 1 sets

have common intersection, then the whole collection of setshas a common intersection.

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Proof Hint:Use Radon’s Theorem!

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Fun Exercise:Show that the Radon Point is in every set.

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More than d+2 sets?

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The Center Point Theorem

Consider the set of all minimal halfspaces containingat least dn

d+1+ 1 points.

Helly’s Theorem implies that all the halfspaces havea common intersection. The intersection is the setof center points.

Observe that every d + 1 have a common intersection.

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Tverberg’s Theorem

Let S be a set of at least (d + 1)(r ! 1) + 1 points in Rd.There exists a partition of S into r subsets X1, . . . , Xr

such that!r

i=1conv(Xi) "= #

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Tverberg’s Theorem

Let S be a set of at least (d + 1)(r ! 1) + 1 points in Rd.There exists a partition of S into r subsets X1, . . . , Xr

such that!r

i=1conv(Xi) "= #

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Tverberg’s Theorem

Let S be a set of at least (d + 1)(r ! 1) + 1 points in Rd.There exists a partition of S into r subsets X1, . . . , Xr

such that!r

i=1conv(Xi) "= #

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Tverberg’s Theorem

Let S be a set of at least (d + 1)(r ! 1) + 1 points in Rd.There exists a partition of S into r subsets X1, . . . , Xr

such that!r

i=1conv(Xi) "= #

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Tverberg’s Theorem

Let S be a set of at least (d + 1)(r ! 1) + 1 points in Rd.There exists a partition of S into r subsets X1, . . . , Xr

such that!r

i=1conv(Xi) "= #

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Tverberg’s Theorem

Let S be a set of at least (d + 1)(r ! 1) + 1 points in Rd.There exists a partition of S into r subsets X1, . . . , Xr

such that!r

i=1conv(Xi) "= #

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Tverberg’s Theorem

Let S be a set of at least (d + 1)(r ! 1) + 1 points in Rd.There exists a partition of S into r subsets X1, . . . , Xr

such that!r

i=1conv(Xi) "= #

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Tverberg’s Theorem

Let S be a set of at least (d + 1)(r ! 1) + 1 points in Rd.There exists a partition of S into r subsets X1, . . . , Xr

such that!r

i=1conv(Xi) "= #

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Tverberg’s Theorem

Let S be a set of at least (d + 1)(r ! 1) + 1 points in Rd.There exists a partition of S into r subsets X1, . . . , Xr

such that!r

i=1conv(Xi) "= #

Choose r = n/(d+1)It’s a center point!

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Proof via Helly’s Theorem

Proof via Tverberg’sTheorem

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Proof via Helly’s Theorem

Proof via Tverberg’sTheorem

coNP NP

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An Algorithm

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Approximating Center Points with Iterated Radon Points[ Clarkson, Eppstein, Miller, Sturtivant, Teng, 1993 ]

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Approximating Center Points with Iterated Radon Points[ Clarkson, Eppstein, Miller, Sturtivant, Teng, 1993 ]

1. Randomly sample points into sets of d+2.2. Compute the Radon point for each set. 3. Compute the Radon points of the Radon points 4. Continue until only one point remains.5. Return that point.

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Approximating Center Points with Iterated Radon Points[ Clarkson, Eppstein, Miller, Sturtivant, Teng, 1993 ]

1. Randomly sample points into sets of d+2.2. Compute the Radon point for each set. 3. Compute the Radon points of the Radon points 4. Continue until only one point remains.5. Return that point.

O!

n

d2

"

-center with high probability.

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Analysis looks like Helly-type proof.

Look at all projections to one dimensionat the same time.

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Let’s build an algorithm so that the analysis will look less like Helly and more like Tverberg.

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This almost works.

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Analysis: How good is the resulting center point?

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Analysis: How good is the resulting center point?

Say gn is the minimum guaranteed partition size on n points

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Analysis: How good is the resulting center point?

Say gn is the minimum guaranteed partition size on n points

Suppose for contradiction that gn < n

(d+1)2 .

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Analysis: How good is the resulting center point?

Say gn is the minimum guaranteed partition size on n points

Then, gn/2 < n2(d+1)2 and the corresponding partition

uses less than n2(d+1) points.

Suppose for contradiction that gn < n

(d+1)2 .

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Analysis: How good is the resulting center point?

Say gn is the minimum guaranteed partition size on n points

Then, gn/2 < n2(d+1)2 and the corresponding partition

uses less than n2(d+1) points.

So, with n points, we can construct d + 2 pointswith partitions of size gn/2.

Suppose for contradiction that gn < n

(d+1)2 .

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Analysis: How good is the resulting center point?

Say gn is the minimum guaranteed partition size on n points

Then, gn/2 < n2(d+1)2 and the corresponding partition

uses less than n2(d+1) points.

So, with n points, we can construct d + 2 pointswith partitions of size gn/2.

This means we can iterate the algorithm, andgn ! 2gn/2

Suppose for contradiction that gn < n

(d+1)2 .

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Analysis: How good is the resulting center point?

Say gn is the minimum guaranteed partition size on n points

Then, gn/2 < n2(d+1)2 and the corresponding partition

uses less than n2(d+1) points.

So, with n points, we can construct d + 2 pointswith partitions of size gn/2.

This means we can iterate the algorithm, andgn ! 2gn/2

Base case: gd+2 = 2.

Suppose for contradiction that gn < n

(d+1)2 .

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Analysis: How good is the resulting center point?

Say gn is the minimum guaranteed partition size on n points

Then, gn/2 < n2(d+1)2 and the corresponding partition

uses less than n2(d+1) points.

So, with n points, we can construct d + 2 pointswith partitions of size gn/2.

This means we can iterate the algorithm, andgn ! 2gn/2

Base case: gd+2 = 2. =! gn " 2log n

d+2 =n

d + 2

Suppose for contradiction that gn < n

(d+1)2 .

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Thank you.

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