Lecture 2: Radon’s theorem
Nabil H. Mustafa
Dept. of Computer Science, LUMS.
http://russell.lums.edu.pk/~nabil
Convex-hull of a point set
Given a set P = {p1 , . . . , pn } of n points in Rd , the convex-hull of P,
denoted conv (P), is the intersection of all convex sets containing P.
Convex-hull of a point set
Given a set P = {p1 , . . . , pn } of n points in Rd , the convex-hull of P,
denoted conv (P), is the intersection of all convex sets containing P.
Convex-hull of a point set
Given a set P = {p1 , . . . , pn } of n points in Rd , the convex-hull of P,
denoted conv (P), is the intersection of all convex sets containing P.
Convex-hull of a point set
Given a set P = {p1 , . . . , pn } of n points in Rd , the convex-hull of P,
denoted conv (P), is the intersection of all convex sets containing P.
Equivalently, a point p ∈P
conv (P) iff there exist nonnegative
real
P
numbers t1 , . . . , tn with ni=1 ti = 1 such that p = ni=1 ti · xi .
Caratheodory’s theorem
Claim: Given a set P of points in R2 , if q ∈ conv (P), then q also lies in a
triangle defined by three points of P.
Caratheodory’s theorem
Claim: Given a set P of points in R2 , if q ∈ conv (P), then q also lies in a
triangle defined by three points of P.
Caratheodory’s theorem
Claim: Given a set P of points in R2 , if q ∈ conv (P), then q also lies in a
triangle defined by three points of P.
Caratheodory’s theorem
Claim: Given a set P of points in R2 , if q ∈ conv (P), then q also lies in a
triangle defined by three points of P.
Caratheodory’s theorem
Claim: Given a set P of points in R2 , if q ∈ conv (P), then q also lies in a
triangle defined by three points of P.
Caratheodory’s theorem
Given a set P of points in Rd , if q ∈ conv (P), then there exists a set
P 0 ⊆ P, |P 0 | ≤ (d + 1), such that q ∈ conv (P 0 ).
A partitioning problem
Given a set P of points in R2 , divide P into two ‘overlapping’ groups.
A partitioning problem
Given a set P of points in R2 , divide P into two ‘overlapping’ groups.
A partitioning problem
Given a set P of points in R2 , divide P into two ‘overlapping’ groups.
A partitioning problem
Given a set P of points in R2 , divide P into two ‘overlapping’ groups.
A partitioning problem
Given a set P of points in R2 , divide P into two ‘overlapping’ groups.
A partitioning problem
Given a set P of points in R2 , divide P into two ‘overlapping’ groups.
A partitioning problem
Given a set P of points in R2 , divide P into two ‘overlapping’ groups.
A partitioning problem
Given a set P of points in R2 , divide P into two ‘overlapping’ groups.
A partitioning problem
Given a set P of points in R2 , divide P into two ‘overlapping’ groups.
A partitioning problem
Given a set P of points in R2 , divide P into two ‘overlapping’ groups.
A partitioning problem
Given a set P of points in R2 , divide P into two ‘overlapping’ groups.
Question: What is the smallest number of ‘dividable’ points in R2 ?
A partitioning problem
Given a set P of points in R2 , divide P into two ‘overlapping’ groups.
Question: What is the smallest number of ‘dividable’ points in R2 ? 4?
A partitioning problem
Given a set P of points in R2 , divide P into two ‘overlapping’ groups.
Question: What is the smallest number of ‘dividable’ points in R2 ? 4?
Question: What is the smallest number of ‘dividable’ points in Rd ?
Radon’s Theorem
Radon’s theorem
Given any set P of d + 2 points in Rd , one can partition P into two sets
P1 , P2 , such that conv (P1 ) ∩ conv (P2 ) 6= ∅.
Radon’s Theorem
Radon’s theorem
Given any set P of d + 2 points in Rd , one can partition P into two sets
P1 , P2 , such that conv (P1 ) ∩ conv (P2 ) 6= ∅.
Note: Easy to see that d + 2 is the best possible bound.
Radon’s Theorem
Radon’s theorem
Given any set P of d + 2 points in Rd , one can partition P into two sets
P1 , P2 , such that conv (P1 ) ∩ conv (P2 ) 6= ∅.
Note: Easy to see that d + 2 is the best possible bound.
Proof in R
Radon’s Theorem
Radon’s theorem
Given any set P of d + 2 points in Rd , one can partition P into two sets
P1 , P2 , such that conv (P1 ) ∩ conv (P2 ) 6= ∅.
Note: Easy to see that d + 2 is the best possible bound.
Proof in R
Radon’s Theorem
Radon’s theorem
Given any set P of d + 2 points in Rd , one can partition P into two sets
P1 , P2 , such that conv (P1 ) ∩ conv (P2 ) 6= ∅.
Note: Easy to see that d + 2 is the best possible bound.
Proof in R
Radon’s Theorem
Proof in R2
Radon’s Theorem
Proof in R2
Radon’s Theorem
Proof in R2
Radon’s Theorem
Proof in R2
Radon’s Theorem
Proof in R2
Radon’s Theorem
Proof in R2
Radon’s Theorem
Proof in R2
Proof of Radon’s theorem
Claim: Given a set P = {p1 , . . . , pd+2 } of d + 2 points in Rd , one can find
real numbers α1 , . . . , αd+2 , not all of them zero, such that
X
X
αi · pi = 0
and
αi = 0
i
i
Proof of Radon’s theorem
Claim: Given a set P = {p1 , . . . , pd+2 } of d + 2 points in Rd , one can find
real numbers α1 , . . . , αd+2 , not all of them zero, such that
X
X
αi · pi = 0
and
αi = 0
i
i
Consider the d + 1 points p2 − p1 , p3 − p1 , . . . , pd+2 − p1 .
Proof of Radon’s theorem
Claim: Given a set P = {p1 , . . . , pd+2 } of d + 2 points in Rd , one can find
real numbers α1 , . . . , αd+2 , not all of them zero, such that
X
X
αi · pi = 0
and
αi = 0
i
i
Consider the d + 1 points p2 − p1 , p3 − p1 , . . . , pd+2 − p1 .
They are linearly dependent, so there must exist β3 , . . . , βd+2 such that:
(p2 − p1 ) =
d+2
X
i=3
βi · (pi − p1 )
Proof of Radon’s theorem
Claim: Given a set P = {p1 , . . . , pd+2 } of d + 2 points in Rd , one can find
real numbers α1 , . . . , αd+2 , not all of them zero, such that
X
X
αi · pi = 0
and
αi = 0
i
i
Consider the d + 1 points p2 − p1 , p3 − p1 , . . . , pd+2 − p1 .
They are linearly dependent, so there must exist β3 , . . . , βd+2 such that:
(p2 − p1 ) =
d+2
X
βi · (pi − p1 )
i=3
Rearranging the terms, we have
(β3 + · · · + βd+2 − 1)p1 + p2 + (−β3 )p3 + · · · + (−βd+2 )pd+2 = 0
Proof of Radon’s theorem
Claim: Given a set P = {p1 , . . . , pd+2 } of d + 2 points in Rd , one can find
real numbers α1 , . . . , αd+2 , not all of them zero, such that
X
X
αi · pi = 0
and
αi = 0
i
i
Consider the d + 1 points p2 − p1 , p3 − p1 , . . . , pd+2 − p1 .
They are linearly dependent, so there must exist β3 , . . . , βd+2 such that:
(p2 − p1 ) =
d+2
X
βi · (pi − p1 )
i=3
Rearranging the terms, we have
(β3 + · · · + βd+2 − 1)p1 + p2 + (−β3 )p3 + · · · + (−βd+2 )pd+2 = 0
Set α1 = β3 + · · · + βd+2 − 1, α2 = 1, αi = −βi , and we’re done.
Proof of Radon’s theorem
Given P = {p1 , . . . , pd+2 } of d + 2 points in Rd , let the real numbers
α1 , . . . , αd+2 , not all of them zero, be such that
X
X
α i · pi = 0
and
αi = 0
i
i
Proof of Radon’s theorem
Given P = {p1 , . . . , pd+2 } of d + 2 points in Rd , let the real numbers
α1 , . . . , αd+2 , not all of them zero, be such that
X
X
α i · pi = 0
and
αi = 0
i
i
Set P1 = {i | αi > 0} and P2 = {i | αi < 0}. Both sets non-empty.
Proof of Radon’s theorem
Given P = {p1 , . . . , pd+2 } of d + 2 points in Rd , let the real numbers
α1 , . . . , αd+2 , not all of them zero, be such that
X
X
α i · pi = 0
and
αi = 0
i
i
Set P1 = {i | αi > 0} and P2 = {i | αi < 0}. Both sets non-empty.
Set A1 = {pi : i ∈ P1 } and A2 = {pi : i ∈ P2 }.
Proof of Radon’s theorem
Given P = {p1 , . . . , pd+2 } of d + 2 points in Rd , let the real numbers
α1 , . . . , αd+2 , not all of them zero, be such that
X
X
α i · pi = 0
and
αi = 0
i
i
Set P1 = {i | αi > 0} and P2 = {i | αi < 0}. Both sets non-empty.
Set A1 = {pi : i ∈ P1 } and A2 = {pi : i ∈ P2 }.
Set S =
P
i∈P1
αi . Note that S = −
P
i∈P2
αi .
Proof of Radon’s theorem
Given P = {p1 , . . . , pd+2 } of d + 2 points in Rd , let the real numbers
α1 , . . . , αd+2 , not all of them zero, be such that
X
X
α i · pi = 0
and
αi = 0
i
i
Set P1 = {i | αi > 0} and P2 = {i | αi < 0}. Both sets non-empty.
Set A1 = {pi : i ∈ P1 } and A2 = {pi : i ∈ P2 }.
Set S =
P
i∈P1
The point x =
αi . Note that S = −
αi
i∈P1 S pi
P
P
i∈P2
αi .
lies in conv (A1 ).
Proof of Radon’s theorem
Given P = {p1 , . . . , pd+2 } of d + 2 points in Rd , let the real numbers
α1 , . . . , αd+2 , not all of them zero, be such that
X
X
α i · pi = 0
and
αi = 0
i
i
Set P1 = {i | αi > 0} and P2 = {i | αi < 0}. Both sets non-empty.
Set A1 = {pi : i ∈ P1 } and A2 = {pi : i ∈ P2 }.
Set S =
P
i∈P1
The point x =
Since
P
i∈P1
αi . Note that S = −
αi
i∈P1 S pi
P
α i pi +
P
i∈P2
P
i∈P2
αi .
lies in conv (A1 ).
αi pi = 0, we have x =
P
i∈P2
−αi
S pi
Proof of Radon’s theorem
Given P = {p1 , . . . , pd+2 } of d + 2 points in Rd , let the real numbers
α1 , . . . , αd+2 , not all of them zero, be such that
X
X
α i · pi = 0
and
αi = 0
i
i
Set P1 = {i | αi > 0} and P2 = {i | αi < 0}. Both sets non-empty.
Set A1 = {pi : i ∈ P1 } and A2 = {pi : i ∈ P2 }.
Set S =
P
i∈P1
The point x =
Since
P
i∈P1
αi . Note that S = −
αi
i∈P1 S pi
P
α i pi +
P
i∈P2
P
i∈P2
αi .
lies in conv (A1 ).
αi pi = 0, we have x =
Therefore x also lies in conv (A2 ).
P
i∈P2
−αi
S pi
A more intuitive proof of Radon’s theorem in R3
A more intuitive proof of Radon’s theorem in R3
We are given 5 points in R3
A more intuitive proof of Radon’s theorem in R3
First, find a dividing plane containing three points of P
A more intuitive proof of Radon’s theorem in R3
We can construct the Radon partition directly now
A more intuitive proof of Radon’s theorem in R3
We can construct the Radon partition directly now
Problems?
A more intuitive proof of Radon’s theorem in R3
A more intuitive proof of Radon’s theorem in R3
A more intuitive proof of Radon’s theorem in R3
A more intuitive proof of Radon’s theorem in R3
A more intuitive proof of Radon’s theorem in R3
A more intuitive proof of Radon’s theorem in R3
A more intuitive proof of Radon’s theorem in R3
A more intuitive proof of Radon’s theorem in R3
A more intuitive proof of Radon’s theorem in R3
Geometric proof of Radon’s theorem in Rd
Given a set P of d + 2 points in Rd , one can always find a plane h
containing P 0 ⊂ P, |P 0 | = d, and where the remaining two points of
P \ P 0 lie on different sides of h.
Geometric proof of Radon’s theorem in Rd
Given a set P of d + 2 points in Rd , one can always find a plane h
containing P 0 ⊂ P, |P 0 | = d, and where the remaining two points of
P \ P 0 lie on different sides of h. Why?
Geometric proof of Radon’s theorem in Rd
Given a set P of d + 2 points in Rd , one can always find a plane h
containing P 0 ⊂ P, |P 0 | = d, and where the remaining two points of
P \ P 0 lie on different sides of h. Why?
Let P 00 = P \ P 0 , and q be the point conv (P 00 ) ∩ h.
Geometric proof of Radon’s theorem in Rd
Given a set P of d + 2 points in Rd , one can always find a plane h
containing P 0 ⊂ P, |P 0 | = d, and where the remaining two points of
P \ P 0 lie on different sides of h. Why?
Let P 00 = P \ P 0 , and q be the point conv (P 00 ) ∩ h.
The set P 0 ∪ {q} lies on h, and has size d + 1.
Geometric proof of Radon’s theorem in Rd
Given a set P of d + 2 points in Rd , one can always find a plane h
containing P 0 ⊂ P, |P 0 | = d, and where the remaining two points of
P \ P 0 lie on different sides of h. Why?
Let P 00 = P \ P 0 , and q be the point conv (P 00 ) ∩ h.
The set P 0 ∪ {q} lies on h, and has size d + 1.
By induction in Rd−1 , there exists a Radon partition P10 , P20 of the set
P 0 ∪ {q}.
Geometric proof of Radon’s theorem in Rd
Given a set P of d + 2 points in Rd , one can always find a plane h
containing P 0 ⊂ P, |P 0 | = d, and where the remaining two points of
P \ P 0 lie on different sides of h. Why?
Let P 00 = P \ P 0 , and q be the point conv (P 00 ) ∩ h.
The set P 0 ∪ {q} lies on h, and has size d + 1.
By induction in Rd−1 , there exists a Radon partition P10 , P20 of the set
P 0 ∪ {q}.
Say q belongs to P10 . As q ∈ conv (P 00 ), we have
conv (P10 ) ⊆ conv (P10 − {q} + P 00 )
Geometric proof of Radon’s theorem in Rd
Given a set P of d + 2 points in Rd , one can always find a plane h
containing P 0 ⊂ P, |P 0 | = d, and where the remaining two points of
P \ P 0 lie on different sides of h. Why?
Let P 00 = P \ P 0 , and q be the point conv (P 00 ) ∩ h.
The set P 0 ∪ {q} lies on h, and has size d + 1.
By induction in Rd−1 , there exists a Radon partition P10 , P20 of the set
P 0 ∪ {q}.
Say q belongs to P10 . As q ∈ conv (P 00 ), we have
conv (P10 ) ⊆ conv (P10 − {q} + P 00 )
Therefore, the sets P10 − {q} + P 00 and P20 are a Radon partition of P.
So far...
Centerpoint theorem
Given any set P of n points in Rd , there exists a point q such that any
closed halfspace containing q contains at least n/(d + 1) points of P.
So far...
Centerpoint theorem
Given any set P of n points in Rd , there exists a point q such that any
closed halfspace containing q contains at least n/(d + 1) points of P.
Radon’s theorem
Given any set P of d + 2 points in Rd , one can partition P into two sets
P1 and P2 such that conv (P1 ) ∩ conv (P2 ) 6= ∅.
So far...
Centerpoint theorem
Given any set P of n points in Rd , there exists a point q such that any
closed halfspace containing q contains at least n/(d + 1) points of P.
Radon’s theorem
Given any set P of d + 2 points in Rd , one can partition P into two sets
P1 and P2 such that conv (P1 ) ∩ conv (P2 ) 6= ∅.
Is there a similarity between these two theorems?
So far...
Centerpoint theorem
Given any set P of n points in Rd , there exists a point q such that any
closed halfspace containing q contains at least n/(d + 1) points of P.
Radon’s theorem
Given any set P of d + 2 points in Rd , one can partition P into two sets
P1 and P2 such that conv (P1 ) ∩ conv (P2 ) 6= ∅.
Is there a similarity between these two theorems?
It turns out that these two theorems are special cases of a more
general theorem, called Tverberg’s Theorem.
Tverberg’s Theorem
Tverberg’s theorem was proved by H. Tverberg in 1966:
Tverberg’s theorem
d
Given a set P of (r − 1)(d +
T1) + 1 points in R , one can partition P into
r sets P1 , . . . , Pr such that i conv (Pi ) 6= ∅.
Tverberg’s Theorem
Tverberg’s theorem was proved by H. Tverberg in 1966:
Tverberg’s theorem
d
Given a set P of (r − 1)(d +
T1) + 1 points in R , one can partition P into
r sets P1 , . . . , Pr such that i conv (Pi ) 6= ∅.
r = 4, n = 10
Tverberg’s Theorem
Tverberg’s theorem was proved by H. Tverberg in 1966:
Tverberg’s theorem
d
Given a set P of (r − 1)(d +
T1) + 1 points in R , one can partition P into
r sets P1 , . . . , Pr such that i conv (Pi ) 6= ∅.
r = 4, n = 10
Tverberg’s Theorem
Tverberg’s theorem was proved by H. Tverberg in 1966:
Tverberg’s theorem
d
Given a set P of (r − 1)(d +
T1) + 1 points in R , one can partition P into
r sets P1 , . . . , Pr such that i conv (Pi ) 6= ∅.
r = 4, n = 10
Relations
Tverberg’s theorem
d
Given a set P of (r − 1)(d +
T1) + 1 points in R , one can partition P into
r sets P1 , . . . , Pr such that i conv (Pi ) 6= ∅.
Relations
Tverberg’s theorem
d
Given a set P of (r − 1)(d +
T1) + 1 points in R , one can partition P into
r sets P1 , . . . , Pr such that i conv (Pi ) 6= ∅.
Radon’s theorem: Apply Tverberg’s theorem with r = 2.
Relations
Tverberg’s theorem
d
Given a set P of (r − 1)(d +
T1) + 1 points in R , one can partition P into
r sets P1 , . . . , Pr such that i conv (Pi ) 6= ∅.
Radon’s theorem: Apply Tverberg’s theorem with r = 2.
Centerpoint theorem: Apply Tverberg’s theorem with r = dn/(d + 1)e.
Relations
Tverberg’s theorem
d
Given a set P of (r − 1)(d +
T1) + 1 points in R , one can partition P into
r sets P1 , . . . , Pr such that i conv (Pi ) 6= ∅.
Radon’s theorem: Apply Tverberg’s theorem with r = 2.
Centerpoint theorem: Apply Tverberg’s theorem with r = dn/(d + 1)e.
Possible, as (r − 1)(d + 1) + 1 = (dn/(d + 1)e − 1)(d + 1) + 1 ≤ n
Relations
Tverberg’s theorem
d
Given a set P of (r − 1)(d +
T1) + 1 points in R , one can partition P into
r sets P1 , . . . , Pr such that i conv (Pi ) 6= ∅.
Radon’s theorem: Apply Tverberg’s theorem with r = 2.
Centerpoint theorem: Apply Tverberg’s theorem with r = dn/(d + 1)e.
Possible, as (r − 1)(d + 1) + 1 = (dn/(d + 1)e − 1)(d + 1) + 1 ≤ n
T
Let q ∈ i conv (Pi ) be a common point.
Relations
Tverberg’s theorem
d
Given a set P of (r − 1)(d +
T1) + 1 points in R , one can partition P into
r sets P1 , . . . , Pr such that i conv (Pi ) 6= ∅.
Radon’s theorem: Apply Tverberg’s theorem with r = 2.
Centerpoint theorem: Apply Tverberg’s theorem with r = dn/(d + 1)e.
Possible, as (r − 1)(d + 1) + 1 = (dn/(d + 1)e − 1)(d + 1) + 1 ≤ n
T
Let q ∈ i conv (Pi ) be a common point.
Any closed halfspace containing q must contain at least one point
from each Pi
Relations
Tverberg’s theorem
d
Given a set P of (r − 1)(d +
T1) + 1 points in R , one can partition P into
r sets P1 , . . . , Pr such that i conv (Pi ) 6= ∅.
Radon’s theorem: Apply Tverberg’s theorem with r = 2.
Centerpoint theorem: Apply Tverberg’s theorem with r = dn/(d + 1)e.
Possible, as (r − 1)(d + 1) + 1 = (dn/(d + 1)e − 1)(d + 1) + 1 ≤ n
T
Let q ∈ i conv (Pi ) be a common point.
Any closed halfspace containing q must contain at least one point
from each Pi
Therefore any closed halfspace containing q contains at least r points.
A partial Tverberg’s theorem
d
Claim: Given a set P of n points in
T R , one can always partition P into
n/d(d + 1) sets P1 , . . . such that i conv (Pi ) 6= ∅
A partial Tverberg’s theorem
d
Claim: Given a set P of n points in
T R , one can always partition P into
n/d(d + 1) sets P1 , . . . such that i conv (Pi ) 6= ∅
Exercise
QUESTIONS?