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A talk on centerpoint presented at the Carnegie Mellon Theory Lunch
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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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