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An Invitation to Discrete Geometry
Nabil H. Mustafa
Dept. of Computer Science, LUMS.
http://russell.lums.edu.pk/~nabil
Two statistical measures
Let P = {a1, . . . , an} be a set of n integers.
Mean: Average of all numbers in P.
Mean(P) =
∑ni=1 ai
n.
Median: The middle smallest number of P.
Median(P) = x if and only if |{i | ai ≤ x}| = n/2.
Two statistical measures
Let P = {a1, . . . , an} be a set of n integers.
Mean: Average of all numbers in P.
Mean(P) =
∑ni=1 ai
n.
Median: The middle smallest number of P.
Median(P) = x if and only if |{i | ai ≤ x}| = n/2.
Two statistical measures
Let P = {a1, . . . , an} be a set of n integers.
Mean: Average of all numbers in P.
Mean(P) =
∑ni=1 ai
n.
Median: The middle smallest number of P.
Median(P) = x if and only if |{i | ai ≤ x}| = n/2.
Two statistical measures
Let P = {a1, . . . , an} be a set of n integers.
Mean: Average of all numbers in P.
Mean(P) =
∑ni=1 ai
n.
Median: The middle smallest number of P.
Median(P) = x if and only if |{i | ai ≤ x}| = n/2.
Two statistical measures
Let P = {a1, . . . , an} be a set of n integers.
Mean: Average of all numbers in P.
Mean(P) =
∑ni=1 ai
n.
Median: The middle smallest number of P.
Median(P) = x if and only if |{i | ai ≤ x}| = n/2.
Two statistical measures
Let P = {a1, . . . , an} be a set of n integers.
Mean: Average of all numbers in P.
Mean(P) =
∑ni=1 ai
n.
Median: The middle smallest number of P.
Median(P) = x if and only if |{i | ai ≤ x}| = n/2.
Two statistical measures
Let P = {a1, . . . , an} be a set of n integers.
Mean: Average of all numbers in P.
Mean(P) =
∑ni=1 ai
n.
Median: The middle smallest number of P.
Median(P) = x if and only if |{i | ai ≤ x}| = n/2.
Two statistical measures
Let P = {a1, . . . , an} be a set of n integers.
Mean: Average of all numbers in P.
Mean(P) =
∑ni=1 ai
n.
Median: The middle smallest number of P.
Median(P) = x if and only if |{i | ai ≤ x}| = n/2.
Two statistical measures
Let P = {a1, . . . , an} be a set of n integers.
Mean: Average of all numbers in P.
Mean(P) =
∑ni=1 ai
n.
Median: The middle smallest number of P.
Median(P) = x if and only if |{i | ai ≤ x}| = n/2.
Which is the better measure?
Both mean and median have advantages and disadvantages.
The main disadvantage of mean is that it is sensitive to actual values,so gets distorted by even a little corrupt data.
Median robust to noise; a few erroneously large numbers don’tchange its value too much.
Which is the better measure?
Both mean and median have advantages and disadvantages.
The main disadvantage of mean is that it is sensitive to actual values,so gets distorted by even a little corrupt data.
Median robust to noise; a few erroneously large numbers don’tchange its value too much.
Which is the better measure?
Both mean and median have advantages and disadvantages.
The main disadvantage of mean is that it is sensitive to actual values,so gets distorted by even a little corrupt data.
Median robust to noise; a few erroneously large numbers don’tchange its value too much.
Which is the better measure?
Both mean and median have advantages and disadvantages.
The main disadvantage of mean is that it is sensitive to actual values,so gets distorted by even a little corrupt data.
Median robust to noise; a few erroneously large numbers don’tchange its value too much.
Which is the better measure?
Both mean and median have advantages and disadvantages.
The main disadvantage of mean is that it is sensitive to actual values,so gets distorted by even a little corrupt data.
Median robust to noise; a few erroneously large numbers don’tchange its value too much.
Which is the better measure?
Both mean and median have advantages and disadvantages.
The main disadvantage of mean is that it is sensitive to actual values,so gets distorted by even a little corrupt data.
Median robust to noise; a few erroneously large numbers don’tchange its value too much.
Which is the better measure?
Both mean and median have advantages and disadvantages.
The main disadvantage of mean is that it is sensitive to actual values,so gets distorted by even a little corrupt data.
Median robust to noise; a few erroneously large numbers don’tchange its value too much.
Which is the better measure?
Both mean and median have advantages and disadvantages.
The main disadvantage of mean is that it is sensitive to actual values,so gets distorted by even a little corrupt data.
Median robust to noise; a few erroneously large numbers don’tchange its value too much.
Which is the better measure?
Both mean and median have advantages and disadvantages.
The main disadvantage of mean is that it is sensitive to actual values,so gets distorted by even a little corrupt data.
Median robust to noise; a few erroneously large numbers don’tchange its value too much.
Generalizing the mean to R2
Let P = {p1, . . . , pn} be a set of n points in R2.
It is straightforward to generalize the mean to higher dimensions.
Centroid
The Centroid of P is defined as the point:
Centroid(P) =
(∑ni=1 pi (x)
n,
∑ni=1 pi (y)
n
).
Problem: again, distorted by noise.
Generalizing the mean to R2
Let P = {p1, . . . , pn} be a set of n points in R2.
It is straightforward to generalize the mean to higher dimensions.
Centroid
The Centroid of P is defined as the point:
Centroid(P) =
(∑ni=1 pi (x)
n,
∑ni=1 pi (y)
n
).
Problem: again, distorted by noise.
Generalizing the mean to R2
Let P = {p1, . . . , pn} be a set of n points in R2.
It is straightforward to generalize the mean to higher dimensions.
Centroid
The Centroid of P is defined as the point:
Centroid(P) =
(∑ni=1 pi (x)
n,
∑ni=1 pi (y)
n
).
Problem: again, distorted by noise.
Generalizing the mean to R2
Let P = {p1, . . . , pn} be a set of n points in R2.
It is straightforward to generalize the mean to higher dimensions.
Centroid
The Centroid of P is defined as the point:
Centroid(P) =
(∑ni=1 pi (x)
n,
∑ni=1 pi (y)
n
).
Problem: again, distorted by noise.
Generalizing the mean to R2
Let P = {p1, . . . , pn} be a set of n points in R2.
It is straightforward to generalize the mean to higher dimensions.
Centroid
The Centroid of P is defined as the point:
Centroid(P) =
(∑ni=1 pi (x)
n,
∑ni=1 pi (y)
n
).
Problem: again, distorted by noise.
Generalizing the mean to R2
Let P = {p1, . . . , pn} be a set of n points in R2.
It is straightforward to generalize the mean to higher dimensions.
Centroid
The Centroid of P is defined as the point:
Centroid(P) =
(∑ni=1 pi (x)
n,
∑ni=1 pi (y)
n
).
Problem: again, distorted by noise.
Generalizing the median to R2
It is less straightforward to generalize the median.
Intuitively, want the median to be at the ‘center’ of P.
Generalizing the median to R2
It is less straightforward to generalize the median.
Intuitively, want the median to be at the ‘center’ of P.
Generalizing the median to R2
It is less straightforward to generalize the median.
Intuitively, want the median to be at the ‘center’ of P.
Generalizing the median to R2
It is less straightforward to generalize the median.
Intuitively, want the median to be at the ‘center’ of P.
A first try
Let l be a vertical line that divides P into two equal parts.
Let h be a horizontal line that divides P into two equal parts.
Let x be the intersection point of h and l .
Claim: x should behave like a median in R2.
A first try
Let l be a vertical line that divides P into two equal parts.
Let h be a horizontal line that divides P into two equal parts.
Let x be the intersection point of h and l .
Claim: x should behave like a median in R2.
A first try
Let l be a vertical line that divides P into two equal parts.
Let h be a horizontal line that divides P into two equal parts.
Let x be the intersection point of h and l .
Claim: x should behave like a median in R2.
A first try
Let l be a vertical line that divides P into two equal parts.
Let h be a horizontal line that divides P into two equal parts.
Let x be the intersection point of h and l .
Claim: x should behave like a median in R2.
A first try
Let l be a vertical line that divides P into two equal parts.
Let h be a horizontal line that divides P into two equal parts.
Let x be the intersection point of h and l .
Claim: x should behave like a median in R2.
A first try
Let l be a vertical line that divides P into two equal parts.
Let h be a horizontal line that divides P into two equal parts.
Let x be the intersection point of h and l .
Claim: x should behave like a median in R2.
A first try
x
Let l be a vertical line that divides P into two equal parts.
Let h be a horizontal line that divides P into two equal parts.
Let x be the intersection point of h and l .
Claim: x should behave like a median in R2.
A first try
x
Let l be a vertical line that divides P into two equal parts.
Let h be a horizontal line that divides P into two equal parts.
Let x be the intersection point of h and l .
Claim: x should behave like a median in R2.
Not so fast
It looks like a very reasonable claim
... or is it?
Not so fast
It looks like a very reasonable claim ... or is it?
Not so fast
It looks like a very reasonable claim ... or is it?
x
Not so fast
It looks like a very reasonable claim ... or is it?
x
What we really want
Here is a definition that better captures what we really want out of theanalog of the median in R2:
Median in R2
A point x ∈ R2 is called a median of a set P of n points iff each linethrough x contains at least n/2 points of P on both its sides.
What we really want
Here is a definition that better captures what we really want out of theanalog of the median in R2:
Median in R2
A point x ∈ R2 is called a median of a set P of n points iff each linethrough x contains at least n/2 points of P on both its sides.
What we really want
Here is a definition that better captures what we really want out of theanalog of the median in R2:
Median in R2
A point x ∈ R2 is called a median of a set P of n points iff each linethrough x contains at least n/2 points of P on both its sides.
What we really want
Here is a definition that better captures what we really want out of theanalog of the median in R2:
Median in R2
A point x ∈ R2 is called a median of a set P of n points iff each linethrough x contains at least n/2 points of P on both its sides.
What we really want
Here is a definition that better captures what we really want out of theanalog of the median in R2:
Median in R2
A point x ∈ R2 is called a median of a set P of n points iff each linethrough x contains at least n/2 points of P on both its sides.
What we really want
Here is a definition that better captures what we really want out of theanalog of the median in R2:
Median in R2
A point x ∈ R2 is called a median of a set P of n points iff each linethrough x contains at least n/2 points of P on both its sides.
A problem
The definition looks great, but can one always find such a point?
No such point exists for this pointset!
A problem
The definition looks great, but can one always find such a point?
No such point exists for this pointset!
A problem
The definition looks great, but can one always find such a point?
No such point exists for this pointset!
A problem
The definition looks great, but can one always find such a point?
No such point exists for this pointset!
Not all is lost
As previous example shows, n/2 is not possible.
But notice: n/3 is possible.
Not all is lost
As previous example shows, n/2 is not possible.
But notice: n/3 is possible.
Not all is lost
As previous example shows, n/2 is not possible.
But notice: n/3 is possible.
Not all is lost
As previous example shows, n/2 is not possible.
But notice: n/3 is possible.
Not all is lost
As previous example shows, n/2 is not possible.
But notice: n/3 is possible.
The centerpoint theorem
No matter what P is given, does there always exist a point x such thatany line through x contains at least n/3 points of P on both sides?
Yes
This point is called the centerpoint.
Centerpoint theorem
For any set P of n points in R2, a centerpoint exists.
Our next goal is to prove the centerpoint theorem.
The centerpoint theorem
No matter what P is given, does there always exist a point x such thatany line through x contains at least n/3 points of P on both sides?
Yes
This point is called the centerpoint.
Centerpoint theorem
For any set P of n points in R2, a centerpoint exists.
Our next goal is to prove the centerpoint theorem.
The centerpoint theorem
No matter what P is given, does there always exist a point x such thatany line through x contains at least n/3 points of P on both sides?
Yes
This point is called the centerpoint.
Centerpoint theorem
For any set P of n points in R2, a centerpoint exists.
Our next goal is to prove the centerpoint theorem.
The centerpoint theorem
No matter what P is given, does there always exist a point x such thatany line through x contains at least n/3 points of P on both sides?
Yes
This point is called the centerpoint.
Centerpoint theorem
For any set P of n points in R2, a centerpoint exists.
Our next goal is to prove the centerpoint theorem.
The centerpoint theorem
No matter what P is given, does there always exist a point x such thatany line through x contains at least n/3 points of P on both sides?
Yes
This point is called the centerpoint.
Centerpoint theorem
For any set P of n points in R2, a centerpoint exists.
Our next goal is to prove the centerpoint theorem.
An equivalence
The key to solving our problem is to connect it to the following, seeminglyunrelated, problem.
Convex Sets
An equivalence
The key to solving our problem is to connect it to the following, seeminglyunrelated, problem.
Convex Sets
An equivalence
The key to solving our problem is to connect it to the following, seeminglyunrelated, problem.
Convex Sets
An equivalence
The key to solving our problem is to connect it to the following, seeminglyunrelated, problem.
Convex Sets
An equivalence
The key to solving our problem is to connect it to the following, seeminglyunrelated, problem.
Convex Sets
An equivalence
The key to solving our problem is to connect it to the following, seeminglyunrelated, problem.
Convex Sets
An equivalence
The key to solving our problem is to connect it to the following, seeminglyunrelated, problem.
Convex Sets
An equivalence
The key to solving our problem is to connect it to the following, seeminglyunrelated, problem.
Convex Sets
An equivalence
The key to solving our problem is to connect it to the following, seeminglyunrelated, problem.
Convex Sets
An equivalence
The key to solving our problem is to connect it to the following, seeminglyunrelated, problem.
Consider the set C of all convex polygons that contain greater than 2n/3points of P.
Claim: Let x be a point that lies in all the polygons of C. Then x is acenterpoint.
An equivalence
The key to solving our problem is to connect it to the following, seeminglyunrelated, problem.
Consider the set C of all convex polygons that contain greater than 2n/3points of P.
Claim: Let x be a point that lies in all the polygons of C. Then x is acenterpoint.
An equivalence
The key to solving our problem is to connect it to the following, seeminglyunrelated, problem.
Consider the set C of all convex polygons that contain greater than 2n/3points of P.
n = 12
Claim: Let x be a point that lies in all the polygons of C. Then x is acenterpoint.
An equivalence
The key to solving our problem is to connect it to the following, seeminglyunrelated, problem.
Consider the set C of all convex polygons that contain greater than 2n/3points of P.
n = 12
Claim: Let x be a point that lies in all the polygons of C. Then x is acenterpoint.
An equivalence
The key to solving our problem is to connect it to the following, seeminglyunrelated, problem.
Consider the set C of all convex polygons that contain greater than 2n/3points of P.
n = 12
Claim: Let x be a point that lies in all the polygons of C. Then x is acenterpoint.
An equivalence
The key to solving our problem is to connect it to the following, seeminglyunrelated, problem.
Consider the set C of all convex polygons that contain greater than 2n/3points of P.
n = 12
Claim: Let x be a point that lies in all the polygons of C. Then x is acenterpoint.
An equivalence
The key to solving our problem is to connect it to the following, seeminglyunrelated, problem.
Consider the set C of all convex polygons that contain greater than 2n/3points of P.
n = 12
Claim: Let x be a point that lies in all the polygons of C. Then x is acenterpoint.
An equivalenceClaim: Suppose x lies in all the polygons of C. Then x is a centerpoint.
An equivalenceClaim: Suppose x lies in all the polygons of C. Then x is a centerpoint.
An equivalenceClaim: Suppose x lies in all the polygons of C. Then x is a centerpoint.
x
An equivalenceClaim: Suppose x lies in all the polygons of C. Then x is a centerpoint.
x
For contradiction, assume x is not a centerpoint
An equivalenceClaim: Suppose x lies in all the polygons of C. Then x is a centerpoint.
x
For contradiction, assume x is not a centerpoint
Then there is a line l through x that has < n3 on one side.
An equivalenceClaim: Suppose x lies in all the polygons of C. Then x is a centerpoint.
x
l
For contradiction, assume x is not a centerpoint
Then there is a line l through x that has < n3 on one side.
An equivalenceClaim: Suppose x lies in all the polygons of C. Then x is a centerpoint.
x
l
For contradiction, assume x is not a centerpoint
Then there is a line l through x that has < n3 on one side.
Then the other side has > 2n3 points.
An equivalenceClaim: Suppose x lies in all the polygons of C. Then x is a centerpoint.
x
l
For contradiction, assume x is not a centerpoint
Then there is a line l through x that has < n3 on one side.
Then the other side has > 2n3 points.
But then a polygon in C not containing x
An equivalenceClaim: Suppose x lies in all the polygons of C. Then x is a centerpoint.
x
l
For contradiction, assume x is not a centerpoint
Then there is a line l through x that has < n3 on one side.
Then the other side has > 2n3 points.
But then a polygon in C not containing x
An equivalenceClaim: Suppose x lies in all the polygons of C. Then x is a centerpoint.
For contradiction, assume x is not a centerpoint
x
Then there is a line l through x that has < n3 on one side.
l
Then the other side has > 2n3 points.
But then a polygon in C not containing x
A contradiction, as x was in all polygons of C
Existence of the required point
Claim: There exists a point x lying in all convex polygons containinggreater than 2n/3 points of P.
Pick the pair of polygons in C whose lowest common intersectionpoint is the highest of all pairs.
Existence of the required point
Claim: There exists a point x lying in all convex polygons containinggreater than 2n/3 points of P.
Pick the pair of polygons in C whose lowest common intersectionpoint is the highest of all pairs.
Existence of the required point
Claim: There exists a point x lying in all convex polygons containinggreater than 2n/3 points of P.
Pick the pair of polygons in C whose lowest common intersectionpoint is the highest of all pairs.
Existence of the required point
Claim: There exists a point x lying in all convex polygons containinggreater than 2n/3 points of P.
Pick the pair of polygons in C whose lowest common intersectionpoint is the highest of all pairs.
> 2n3
Existence of the required point
Claim: There exists a point x lying in all convex polygons containinggreater than 2n/3 points of P.
Pick the pair of polygons in C whose lowest common intersectionpoint is the highest of all pairs.
> 2n3
> 2n3
Existence of the required point
Claim: There exists a point x lying in all convex polygons containinggreater than 2n/3 points of P.
Pick the pair of polygons in C whose lowest common intersectionpoint is the highest of all pairs.
> 2n3
> 2n3
Existence of the required point
Claim: There exists a point x lying in all convex polygons containinggreater than 2n/3 points of P.
Pick the pair of polygons in C whose lowest common intersectionpoint is the highest of all pairs.
> 2n3
> 2n3> n
3
Existence of the required point
Claim: There exists a point x lying in all convex polygons containinggreater than 2n/3 points of P.
Pick the pair of polygons in C whose lowest common intersectionpoint is the highest of all pairs.
> 2n3
> 2n3> n
3
Existence of the required point
Claim: There exists a point x lying in all convex polygons containinggreater than 2n/3 points of P.
Pick the pair of polygons in C whose lowest common intersectionpoint is the highest of all pairs.
> 2n3
> 2n3> n
3
Existence of the required point
Claim: There exists a point x lying in all convex polygons containinggreater than 2n/3 points of P.
Pick the pair of polygons in C whose lowest common intersectionpoint is the highest of all pairs.
> 2n3
> 2n3> n
3
Existence of the required point
Claim: There exists a point x lying in all convex polygons containinggreater than 2n/3 points of P.
Pick the pair of polygons in C whose lowest common intersectionpoint is the highest of all pairs.
> 2n3
> 2n3> n
3
Existence of the required point
Claim: There exists a point x lying in all convex polygons containinggreater than 2n/3 points of P.
Pick the pair of polygons in C whose lowest common intersectionpoint is the highest of all pairs.
> 2n3
> 2n3> n
3
Existence of the required point
Claim: There exists a point x lying in all convex polygons containinggreater than 2n/3 points of P.
Pick the pair of polygons in C whose lowest common intersectionpoint is the highest of all pairs.
> 2n3
> 2n3> n
3
> 2n3
A generalization
Given a set P of n points, we can pick a point x such that any convexpolygon that contains greater than 2n/3 points of P must contain x .
This implies the centerpoint theorem.
What if we want to hit polygons containing fewer points?
I Then we have to use more points!
Generalization of centerpoint theorem
Given a set P of n points in R2, one can find two points such that anyconvex polygon containing greater than 4n/7 points contains at least oneof them.
A generalization
Given a set P of n points, we can pick a point x such that any convexpolygon that contains greater than 2n/3 points of P must contain x .
This implies the centerpoint theorem.
What if we want to hit polygons containing fewer points?
I Then we have to use more points!
Generalization of centerpoint theorem
Given a set P of n points in R2, one can find two points such that anyconvex polygon containing greater than 4n/7 points contains at least oneof them.
A generalization
Given a set P of n points, we can pick a point x such that any convexpolygon that contains greater than 2n/3 points of P must contain x .
This implies the centerpoint theorem.
What if we want to hit polygons containing fewer points?
I Then we have to use more points!
Generalization of centerpoint theorem
Given a set P of n points in R2, one can find two points such that anyconvex polygon containing greater than 4n/7 points contains at least oneof them.
A generalization
Given a set P of n points, we can pick a point x such that any convexpolygon that contains greater than 2n/3 points of P must contain x .
This implies the centerpoint theorem.
What if we want to hit polygons containing fewer points?
I Then we have to use more points!
Generalization of centerpoint theorem
Given a set P of n points in R2, one can find two points such that anyconvex polygon containing greater than 4n/7 points contains at least oneof them.
A generalization
Given a set P of n points, we can pick a point x such that any convexpolygon that contains greater than 2n/3 points of P must contain x .
This implies the centerpoint theorem.
What if we want to hit polygons containing fewer points?
I Then we have to use more points!
Generalization of centerpoint theorem
Given a set P of n points in R2, one can find two points such that anyconvex polygon containing greater than 4n/7 points contains at least oneof them.
A further generalization
Three points?
Then 4n/7 can be decreased to what?
Lower-bound: There exists a point set P such that no three points can hitall convex objects containing 5n/11 points.
Generalization
Given a set P of n points in R2, ∃ three points such that any convexpolygon containing greater than 8n/15 points contains one of them.
“An Optimal Generalization of the Centerpoint Theorem”, by N. Mustafaand S. Ray. In Computational Geometry: Theory andApplications, 2009.
Open Problem
If we want to hit with three points, what is the exact number of pointseach convex polygon must contain?
A further generalization
Three points? Then 4n/7 can be decreased to what?
Lower-bound: There exists a point set P such that no three points can hitall convex objects containing 5n/11 points.
Generalization
Given a set P of n points in R2, ∃ three points such that any convexpolygon containing greater than 8n/15 points contains one of them.
“An Optimal Generalization of the Centerpoint Theorem”, by N. Mustafaand S. Ray. In Computational Geometry: Theory andApplications, 2009.
Open Problem
If we want to hit with three points, what is the exact number of pointseach convex polygon must contain?
A further generalization
Three points? Then 4n/7 can be decreased to what?
Lower-bound: There exists a point set P such that no three points can hitall convex objects containing 5n/11 points.
Generalization
Given a set P of n points in R2, ∃ three points such that any convexpolygon containing greater than 8n/15 points contains one of them.
“An Optimal Generalization of the Centerpoint Theorem”, by N. Mustafaand S. Ray. In Computational Geometry: Theory andApplications, 2009.
Open Problem
If we want to hit with three points, what is the exact number of pointseach convex polygon must contain?
A further generalization
Three points? Then 4n/7 can be decreased to what?
Lower-bound: There exists a point set P such that no three points can hitall convex objects containing 5n/11 points.
Generalization
Given a set P of n points in R2, ∃ three points such that any convexpolygon containing greater than 8n/15 points contains one of them.
“An Optimal Generalization of the Centerpoint Theorem”, by N. Mustafaand S. Ray. In Computational Geometry: Theory andApplications, 2009.
Open Problem
If we want to hit with three points, what is the exact number of pointseach convex polygon must contain?
A further generalization
Three points? Then 4n/7 can be decreased to what?
Lower-bound: There exists a point set P such that no three points can hitall convex objects containing 5n/11 points.
Generalization
Given a set P of n points in R2, ∃ three points such that any convexpolygon containing greater than 8n/15 points contains one of them.
“An Optimal Generalization of the Centerpoint Theorem”, by N. Mustafaand S. Ray. In Computational Geometry: Theory andApplications, 2009.
Open Problem
If we want to hit with three points, what is the exact number of pointseach convex polygon must contain?
A further generalization
Three points? Then 4n/7 can be decreased to what?
Lower-bound: There exists a point set P such that no three points can hitall convex objects containing 5n/11 points.
Generalization
Given a set P of n points in R2, ∃ three points such that any convexpolygon containing greater than 8n/15 points contains one of them.
“An Optimal Generalization of the Centerpoint Theorem”, by N. Mustafaand S. Ray. In Computational Geometry: Theory andApplications, 2009.
Open Problem
If we want to hit with three points, what is the exact number of pointseach convex polygon must contain?
A different way of looking at the same question
So far, we have fixed the number of points we want to use:
I One point hits all convex polygons containing > 2n/3 points.
I Two points hit all convex polygons containing > 4n/7 points.
I Three points hit all convex polygons containing > 8n/15 points.
I And so on...
What if we first fix how much we want to hit:
I Polygons containing > 23 · n points: One point.
I Polygons containing > 47 · n points: Two points.
I And so on . . .
Question: How many points to hit all convex polygons containing > ε · npoints, where 0 < ε ≤ 1 is a given real parameter.
A different way of looking at the same question
So far, we have fixed the number of points we want to use:
I One point hits all convex polygons containing > 2n/3 points.
I Two points hit all convex polygons containing > 4n/7 points.
I Three points hit all convex polygons containing > 8n/15 points.
I And so on...
What if we first fix how much we want to hit:
I Polygons containing > 23 · n points: One point.
I Polygons containing > 47 · n points: Two points.
I And so on . . .
Question: How many points to hit all convex polygons containing > ε · npoints, where 0 < ε ≤ 1 is a given real parameter.
A different way of looking at the same question
So far, we have fixed the number of points we want to use:
I One point hits all convex polygons containing > 2n/3 points.
I Two points hit all convex polygons containing > 4n/7 points.
I Three points hit all convex polygons containing > 8n/15 points.
I And so on...
What if we first fix how much we want to hit:
I Polygons containing > 23 · n points: One point.
I Polygons containing > 47 · n points: Two points.
I And so on . . .
Question: How many points to hit all convex polygons containing > ε · npoints, where 0 < ε ≤ 1 is a given real parameter.
A different way of looking at the same question
So far, we have fixed the number of points we want to use:
I One point hits all convex polygons containing > 2n/3 points.
I Two points hit all convex polygons containing > 4n/7 points.
I Three points hit all convex polygons containing > 8n/15 points.
I And so on...
What if we first fix how much we want to hit:
I Polygons containing > 23 · n points: One point.
I Polygons containing > 47 · n points: Two points.
I And so on . . .
Question: How many points to hit all convex polygons containing > ε · npoints, where 0 < ε ≤ 1 is a given real parameter.
A different way of looking at the same question
So far, we have fixed the number of points we want to use:
I One point hits all convex polygons containing > 2n/3 points.
I Two points hit all convex polygons containing > 4n/7 points.
I Three points hit all convex polygons containing > 8n/15 points.
I And so on...
What if we first fix how much we want to hit:
I Polygons containing > 23 · n points: One point.
I Polygons containing > 47 · n points: Two points.
I And so on . . .
Question: How many points to hit all convex polygons containing > ε · npoints, where 0 < ε ≤ 1 is a given real parameter.
A different way of looking at the same question
So far, we have fixed the number of points we want to use:
I One point hits all convex polygons containing > 2n/3 points.
I Two points hit all convex polygons containing > 4n/7 points.
I Three points hit all convex polygons containing > 8n/15 points.
I And so on...
What if we first fix how much we want to hit:
I Polygons containing > 23 · n points: One point.
I Polygons containing > 47 · n points: Two points.
I And so on . . .
Question: How many points to hit all convex polygons containing > ε · npoints, where 0 < ε ≤ 1 is a given real parameter.
A different way of looking at the same question
So far, we have fixed the number of points we want to use:
I One point hits all convex polygons containing > 2n/3 points.
I Two points hit all convex polygons containing > 4n/7 points.
I Three points hit all convex polygons containing > 8n/15 points.
I And so on...
What if we first fix how much we want to hit:
I Polygons containing > 23 · n points: One point.
I Polygons containing > 47 · n points: Two points.
I And so on . . .
Question: How many points to hit all convex polygons containing > ε · npoints, where 0 < ε ≤ 1 is a given real parameter.
A different way of looking at the same question
So far, we have fixed the number of points we want to use:
I One point hits all convex polygons containing > 2n/3 points.
I Two points hit all convex polygons containing > 4n/7 points.
I Three points hit all convex polygons containing > 8n/15 points.
I And so on...
What if we first fix how much we want to hit:
I Polygons containing > 23 · n points: One point.
I Polygons containing > 47 · n points: Two points.
I And so on . . .
Question: How many points to hit all convex polygons containing > ε · npoints, where 0 < ε ≤ 1 is a given real parameter.
A different way of looking at the same question
So far, we have fixed the number of points we want to use:
I One point hits all convex polygons containing > 2n/3 points.
I Two points hit all convex polygons containing > 4n/7 points.
I Three points hit all convex polygons containing > 8n/15 points.
I And so on...
What if we first fix how much we want to hit:
I Polygons containing > 23 · n points: One point.
I Polygons containing > 47 · n points: Two points.
I And so on . . .
Question: How many points to hit all convex polygons containing > ε · npoints, where 0 < ε ≤ 1 is a given real parameter.
A different way of looking at the same question
So far, we have fixed the number of points we want to use:
I One point hits all convex polygons containing > 2n/3 points.
I Two points hit all convex polygons containing > 4n/7 points.
I Three points hit all convex polygons containing > 8n/15 points.
I And so on...
What if we first fix how much we want to hit:
I Polygons containing > 23 · n points: One point.
I Polygons containing > 47 · n points: Two points.
I And so on . . .
Question: How many points to hit all convex polygons containing > ε · npoints, where 0 < ε ≤ 1 is a given real parameter.
Weak ε-netsGoal: Given a set P of n points in R2, and a real number ε > 0, would liketo find a small set Q ⊆ R2 such that any convex polygon containinggreater than ε · n points of P is hit by Q.
Such a Q is called a weak ε-net.
n = 16ε = 1/4
Weak ε-netsGoal: Given a set P of n points in R2, and a real number ε > 0, would liketo find a small set Q ⊆ R2 such that any convex polygon containinggreater than ε · n points of P is hit by Q.
Such a Q is called a weak ε-net.
n = 16ε = 1/4
Weak ε-netsGoal: Given a set P of n points in R2, and a real number ε > 0, would liketo find a small set Q ⊆ R2 such that any convex polygon containinggreater than ε · n points of P is hit by Q.
Such a Q is called a weak ε-net.
n = 16ε = 1/4
A suboptimal boundHere is one simple algorithm, starting with Q = ∅:
1 Let C1 be any convex polygon not hit by Q.2 Add the centerpoint of C1 ∩ P to Q.3 If Q is a weak ε-net, we’re done.4 Otherwise, repeat Step 1.
A suboptimal boundHere is one simple algorithm, starting with Q = ∅:
n = 16ε = 1/4
1 Let C1 be any convex polygon not hit by Q.2 Add the centerpoint of C1 ∩ P to Q.3 If Q is a weak ε-net, we’re done.4 Otherwise, repeat Step 1.
A suboptimal boundHere is one simple algorithm, starting with Q = ∅:
n = 16ε = 1/4
1 Let C1 be any convex polygon not hit by Q.2 Add the centerpoint of C1 ∩ P to Q.3 If Q is a weak ε-net, we’re done.4 Otherwise, repeat Step 1.
A suboptimal boundHere is one simple algorithm, starting with Q = ∅:
n = 16ε = 1/4
1 Let C1 be any convex polygon not hit by Q.2 Add the centerpoint of C1 ∩ P to Q.3 If Q is a weak ε-net, we’re done.4 Otherwise, repeat Step 1.
A suboptimal boundHere is one simple algorithm, starting with Q = ∅:
n = 16ε = 1/4
1 Let C1 be any convex polygon not hit by Q.2 Add the centerpoint of C1 ∩ P to Q.3 If Q is a weak ε-net, we’re done.4 Otherwise, repeat Step 1.
A suboptimal boundHere is one simple algorithm, starting with Q = ∅:
n = 16ε = 1/4
1 Let C1 be any convex polygon not hit by Q.2 Add the centerpoint of C1 ∩ P to Q.3 If Q is a weak ε-net, we’re done.4 Otherwise, repeat Step 1.
A suboptimal boundHere is one simple algorithm, starting with Q = ∅:
n = 16ε = 1/4
1 Let C1 be any convex polygon not hit by Q.2 Add the centerpoint of C1 ∩ P to Q.3 If Q is a weak ε-net, we’re done.4 Otherwise, repeat Step 1.
A suboptimal boundHere is one simple algorithm, starting with Q = ∅:
n = 16ε = 1/4
1 Let C1 be any convex polygon not hit by Q.2 Add the centerpoint of C1 ∩ P to Q.3 If Q is a weak ε-net, we’re done.4 Otherwise, repeat Step 1.
A suboptimal boundHere is one simple algorithm, starting with Q = ∅:
n = 16ε = 1/4
1 Let C1 be any convex polygon not hit by Q.2 Add the centerpoint of C1 ∩ P to Q.3 If Q is a weak ε-net, we’re done.4 Otherwise, repeat Step 1.
A suboptimal boundHere is one simple algorithm, starting with Q = ∅:
n = 16ε = 1/4
1 Let C1 be any convex polygon not hit by Q.2 Add the centerpoint of C1 ∩ P to Q.3 If Q is a weak ε-net, we’re done.4 Otherwise, repeat Step 1.
A suboptimal boundHere is one simple algorithm, starting with Q = ∅:
n = 16ε = 1/4
1 Let C1 be any convex polygon not hit by Q.2 Add the centerpoint of C1 ∩ P to Q.3 If Q is a weak ε-net, we’re done.4 Otherwise, repeat Step 1.
A suboptimal boundHere is one simple algorithm, starting with Q = ∅:
n = 16ε = 1/4
1 Let C1 be any convex polygon not hit by Q.2 Add the centerpoint of C1 ∩ P to Q.3 If Q is a weak ε-net, we’re done.4 Otherwise, repeat Step 1.
A suboptimal boundHere is one simple algorithm, starting with Q = ∅:
n = 16ε = 1/4
1 Let C1 be any convex polygon not hit by Q.2 Add the centerpoint of C1 ∩ P to Q.3 If Q is a weak ε-net, we’re done.4 Otherwise, repeat Step 1.
A suboptimal boundHere is one simple algorithm, starting with Q = ∅:
n = 16ε = 1/4
1 Let C1 be any convex polygon not hit by Q.2 Add the centerpoint of C1 ∩ P to Q.3 If Q is a weak ε-net, we’re done.4 Otherwise, repeat Step 1.
A suboptimal boundHere is one simple algorithm, starting with Q = ∅:
n = 16ε = 1/4
1 Let C1 be any convex polygon not hit by Q.2 Add the centerpoint of C1 ∩ P to Q.3 If Q is a weak ε-net, we’re done.4 Otherwise, repeat Step 1.
A diversionWhy does the algorithm work well?
To see that, we need thefollowing wonderful property of centerpoints:
First selection lemma in R2
Given a set P of n points in R2, there exists a centerpoint of P that hits≥ (n/3)3 triangles of P, out of a total of
(n3
)possible triangles.
A diversionWhy does the algorithm work well? To see that, we need thefollowing wonderful property of centerpoints:
First selection lemma in R2
Given a set P of n points in R2, there exists a centerpoint of P that hits≥ (n/3)3 triangles of P, out of a total of
(n3
)possible triangles.
A diversionWhy does the algorithm work well? To see that, we need thefollowing wonderful property of centerpoints:
First selection lemma in R2
Given a set P of n points in R2, there exists a centerpoint of P that hits≥ (n/3)3 triangles of P, out of a total of
(n3
)possible triangles.
A diversionWhy does the algorithm work well? To see that, we need thefollowing wonderful property of centerpoints:
First selection lemma in R2
Given a set P of n points in R2, there exists a centerpoint of P that hits≥ (n/3)3 triangles of P, out of a total of
(n3
)possible triangles.
A diversionWhy does the algorithm work well? To see that, we need thefollowing wonderful property of centerpoints:
1
First selection lemma in R2
Given a set P of n points in R2, there exists a centerpoint of P that hits≥ (n/3)3 triangles of P, out of a total of
(n3
)possible triangles.
A diversionWhy does the algorithm work well? To see that, we need thefollowing wonderful property of centerpoints:
2
First selection lemma in R2
Given a set P of n points in R2, there exists a centerpoint of P that hits≥ (n/3)3 triangles of P, out of a total of
(n3
)possible triangles.
A diversionWhy does the algorithm work well? To see that, we need thefollowing wonderful property of centerpoints:
3
First selection lemma in R2
Given a set P of n points in R2, there exists a centerpoint of P that hits≥ (n/3)3 triangles of P, out of a total of
(n3
)possible triangles.
A diversionWhy does the algorithm work well? To see that, we need thefollowing wonderful property of centerpoints:
4
First selection lemma in R2
Given a set P of n points in R2, there exists a centerpoint of P that hits≥ (n/3)3 triangles of P, out of a total of
(n3
)possible triangles.
A diversionWhy does the algorithm work well? To see that, we need thefollowing wonderful property of centerpoints:
4
First selection lemma in R2
Given a set P of n points in R2, there exists a centerpoint of P that hits≥ (n/3)3 triangles of P, out of a total of
(n3
)possible triangles.
A diversion
First selection lemma in R2
Given a set P of n points in R2, there exists a centerpoint of P that hits(n/3)3 triangles of P.
Question: What about a similar statement in R3?
First selection lemma in R3?
Given a set P of n points in R3, there exists a point that hits(n/4)4 = 0.0039 · n4 tetrahedra of P.
It is a nice conjecture, but no one has been able to prove it!
Current best result: a point hitting 0.0022 · n4 tetrahedra.
“An Improved Bound on First Selection Lemma in R3”, by A. Basit, N.Mustafa, S. Ray and S. Raza. In 26th Symposium onComputational Geometry, 2010.
A diversion
First selection lemma in R2
Given a set P of n points in R2, there exists a centerpoint of P that hits(n/3)3 triangles of P.
Question: What about a similar statement in R3?
First selection lemma in R3?
Given a set P of n points in R3, there exists a point that hits(n/4)4 = 0.0039 · n4 tetrahedra of P.
It is a nice conjecture, but no one has been able to prove it!
Current best result: a point hitting 0.0022 · n4 tetrahedra.
“An Improved Bound on First Selection Lemma in R3”, by A. Basit, N.Mustafa, S. Ray and S. Raza. In 26th Symposium onComputational Geometry, 2010.
A diversion
First selection lemma in R2
Given a set P of n points in R2, there exists a centerpoint of P that hits(n/3)3 triangles of P.
Question: What about a similar statement in R3?
First selection lemma in R3?
Given a set P of n points in R3, there exists a point that hits(n/4)4 = 0.0039 · n4 tetrahedra of P.
It is a nice conjecture, but no one has been able to prove it!
Current best result: a point hitting 0.0022 · n4 tetrahedra.
“An Improved Bound on First Selection Lemma in R3”, by A. Basit, N.Mustafa, S. Ray and S. Raza. In 26th Symposium onComputational Geometry, 2010.
A diversion
First selection lemma in R2
Given a set P of n points in R2, there exists a centerpoint of P that hits(n/3)3 triangles of P.
Question: What about a similar statement in R3?
First selection lemma in R3?
Given a set P of n points in R3, there exists a point that hits(n/4)4 = 0.0039 · n4 tetrahedra of P.
It is a nice conjecture, but no one has been able to prove it!
Current best result: a point hitting 0.0022 · n4 tetrahedra.
“An Improved Bound on First Selection Lemma in R3”, by A. Basit, N.Mustafa, S. Ray and S. Raza. In 26th Symposium onComputational Geometry, 2010.
A diversion
First selection lemma in R2
Given a set P of n points in R2, there exists a centerpoint of P that hits(n/3)3 triangles of P.
Question: What about a similar statement in R3?
First selection lemma in R3?
Given a set P of n points in R3, there exists a point that hits(n/4)4 = 0.0039 · n4 tetrahedra of P.
It is a nice conjecture, but no one has been able to prove it!
Current best result: a point hitting 0.0022 · n4 tetrahedra.
“An Improved Bound on First Selection Lemma in R3”, by A. Basit, N.Mustafa, S. Ray and S. Raza. In 26th Symposium onComputational Geometry, 2010.
Coming back to a suboptimal bound
1 Let C1 be any polygon not hit by Q.
2 Add the centerpoint of C1 ∩ P to Q.
3 If Q is a weak ε-net, we’re done.
4 Otherwise, repeat Step 1.
At each step, the centerpoint hits (εn/3)3 new triangles.
Since there are(n3
)total triangles, total steps:(n
3
)(εn/3)3
= Θ(1
ε3)
At each step, one point added to Q, so Q has the same size ofO(1/ε3).
Coming back to a suboptimal bound
1 Let C1 be any polygon not hit by Q.
2 Add the centerpoint of C1 ∩ P to Q.
3 If Q is a weak ε-net, we’re done.
4 Otherwise, repeat Step 1.
At each step, the centerpoint hits (εn/3)3 new triangles.
Since there are(n3
)total triangles, total steps:(n
3
)(εn/3)3
= Θ(1
ε3)
At each step, one point added to Q, so Q has the same size ofO(1/ε3).
Coming back to a suboptimal bound
1 Let C1 be any polygon not hit by Q.
2 Add the centerpoint of C1 ∩ P to Q.
3 If Q is a weak ε-net, we’re done.
4 Otherwise, repeat Step 1.
At each step, the centerpoint hits (εn/3)3 new triangles.
Since there are(n3
)total triangles, total steps:(n
3
)(εn/3)3
= Θ(1
ε3)
At each step, one point added to Q, so Q has the same size ofO(1/ε3).
Coming back to a suboptimal bound
1 Let C1 be any polygon not hit by Q.
2 Add the centerpoint of C1 ∩ P to Q.
3 If Q is a weak ε-net, we’re done.
4 Otherwise, repeat Step 1.
At each step, the centerpoint hits (εn/3)3 new triangles.
Since there are(n3
)total triangles, total steps:(n
3
)(εn/3)3
= Θ(1
ε3)
At each step, one point added to Q, so Q has the same size ofO(1/ε3).
Weak ε-netsWhat we have proved so far:
A suboptimal bound
Given a set P of n points in R2, there exists a set Q of size O(1/ε3) suchthat any convex polygon containing at least ε · n points of P must containat least one point of Q.
With some other clever ideas, MW[05] improved this bound to O(1/ε2) inR2, and O(1/εd) for Rd .
A Big Conjecture
Given a set P of n points in R2, there exists a set Q of size o(1/ε2) suchthat any convex polygon containing at least ε · n points of P must containat least one point of Q.
Needs someone with a clever idea to solve this!!
Weak ε-netsWhat we have proved so far:
A suboptimal bound
Given a set P of n points in R2, there exists a set Q of size O(1/ε3) suchthat any convex polygon containing at least ε · n points of P must containat least one point of Q.
With some other clever ideas, MW[05] improved this bound to O(1/ε2) inR2, and O(1/εd) for Rd .
A Big Conjecture
Given a set P of n points in R2, there exists a set Q of size o(1/ε2) suchthat any convex polygon containing at least ε · n points of P must containat least one point of Q.
Needs someone with a clever idea to solve this!!
Weak ε-netsWhat we have proved so far:
A suboptimal bound
Given a set P of n points in R2, there exists a set Q of size O(1/ε3) suchthat any convex polygon containing at least ε · n points of P must containat least one point of Q.
With some other clever ideas, MW[05] improved this bound to O(1/ε2) inR2, and O(1/εd) for Rd .
A Big Conjecture
Given a set P of n points in R2, there exists a set Q of size o(1/ε2) suchthat any convex polygon containing at least ε · n points of P must containat least one point of Q.
Needs someone with a clever idea to solve this!!
Weak ε-netsWhat we have proved so far:
A suboptimal bound
Given a set P of n points in R2, there exists a set Q of size O(1/ε3) suchthat any convex polygon containing at least ε · n points of P must containat least one point of Q.
With some other clever ideas, MW[05] improved this bound to O(1/ε2) inR2, and O(1/εd) for Rd .
A Big Conjecture
Given a set P of n points in R2, there exists a set Q of size o(1/ε2) suchthat any convex polygon containing at least ε · n points of P must containat least one point of Q.
Needs someone with a clever idea to solve this!!
Weak ε-nets
No one has been able to solve the conjecture so far.
The result that comes closest to it is as follows.
Existence of Small Basis
Given a set P of n points in Rd , and a parameter ε > 0, there exists a setof O(1/ε log 1/ε) points of P, called its basis, from which one canconstruct a weak ε-net Q of P (but |Q| still increases exponentially withd).
“Weak ε-nets Have a Basis of Size O(1/ε log 1/ε) in any dimension”, byN. Mustafa and S. Ray. In Computational Geometry: Theory andApplications, 2008.
Weak ε-nets
No one has been able to solve the conjecture so far.
The result that comes closest to it is as follows.
Existence of Small Basis
Given a set P of n points in Rd , and a parameter ε > 0, there exists a setof O(1/ε log 1/ε) points of P, called its basis, from which one canconstruct a weak ε-net Q of P (but |Q| still increases exponentially withd).
“Weak ε-nets Have a Basis of Size O(1/ε log 1/ε) in any dimension”, byN. Mustafa and S. Ray. In Computational Geometry: Theory andApplications, 2008.
Weak ε-nets
No one has been able to solve the conjecture so far.
The result that comes closest to it is as follows.
Existence of Small Basis
Given a set P of n points in Rd , and a parameter ε > 0, there exists a setof O(1/ε log 1/ε) points of P, called its basis, from which one canconstruct a weak ε-net Q of P (but |Q| still increases exponentially withd).
“Weak ε-nets Have a Basis of Size O(1/ε log 1/ε) in any dimension”, byN. Mustafa and S. Ray. In Computational Geometry: Theory andApplications, 2008.
Weak ε-nets
No one has been able to solve the conjecture so far.
The result that comes closest to it is as follows.
Existence of Small Basis
Given a set P of n points in Rd , and a parameter ε > 0, there exists a setof O(1/ε log 1/ε) points of P, called its basis, from which one canconstruct a weak ε-net Q of P (but |Q| still increases exponentially withd).
“Weak ε-nets Have a Basis of Size O(1/ε log 1/ε) in any dimension”, byN. Mustafa and S. Ray. In Computational Geometry: Theory andApplications, 2008.
Other work
There are several interesting questions related to centerpoints.
Algorithms for centerpoints:
“Location Depth Using the Graphics Hardware”, by S. Krishnan, N.Mustafa and S. Venkat. In 12th Symposium on DiscreteAlgorithms, 2002.
“Statistical Data Depth and the Graphics Hardware”, by S. Krishnan, N.Mustafa and S. Venkat. In Data Depth: Robust MultivariateAnalysis, Computational Geometry and Applications,DIMACS Series on Mathematics and Theoretical ComputerScience, 2006.
Generalization to different objects:
“Centerdisks and Tverberg’s Technique”, by A. Basit, N. Mustafa, S. Rayand S. Raza. In Computational Geometry: Theory andApplications, 2010.
Other work
There are several interesting questions related to centerpoints.
Algorithms for centerpoints:
“Location Depth Using the Graphics Hardware”, by S. Krishnan, N.Mustafa and S. Venkat. In 12th Symposium on DiscreteAlgorithms, 2002.
“Statistical Data Depth and the Graphics Hardware”, by S. Krishnan, N.Mustafa and S. Venkat. In Data Depth: Robust MultivariateAnalysis, Computational Geometry and Applications,DIMACS Series on Mathematics and Theoretical ComputerScience, 2006.
Generalization to different objects:
“Centerdisks and Tverberg’s Technique”, by A. Basit, N. Mustafa, S. Rayand S. Raza. In Computational Geometry: Theory andApplications, 2010.
Other work
There are several interesting questions related to centerpoints.
Algorithms for centerpoints:
“Location Depth Using the Graphics Hardware”, by S. Krishnan, N.Mustafa and S. Venkat. In 12th Symposium on DiscreteAlgorithms, 2002.
“Statistical Data Depth and the Graphics Hardware”, by S. Krishnan, N.Mustafa and S. Venkat. In Data Depth: Robust MultivariateAnalysis, Computational Geometry and Applications,DIMACS Series on Mathematics and Theoretical ComputerScience, 2006.
Generalization to different objects:
“Centerdisks and Tverberg’s Technique”, by A. Basit, N. Mustafa, S. Rayand S. Raza. In Computational Geometry: Theory andApplications, 2010.
Other work
There are several interesting questions related to centerpoints.
Algorithms for centerpoints:
“Location Depth Using the Graphics Hardware”, by S. Krishnan, N.Mustafa and S. Venkat. In 12th Symposium on DiscreteAlgorithms, 2002.
“Statistical Data Depth and the Graphics Hardware”, by S. Krishnan, N.Mustafa and S. Venkat. In Data Depth: Robust MultivariateAnalysis, Computational Geometry and Applications,DIMACS Series on Mathematics and Theoretical ComputerScience, 2006.
Generalization to different objects:
“Centerdisks and Tverberg’s Technique”, by A. Basit, N. Mustafa, S. Rayand S. Raza. In Computational Geometry: Theory andApplications, 2010.
Other work
There are several interesting questions related to centerpoints.
Algorithms for centerpoints:
“Location Depth Using the Graphics Hardware”, by S. Krishnan, N.Mustafa and S. Venkat. In 12th Symposium on DiscreteAlgorithms, 2002.
“Statistical Data Depth and the Graphics Hardware”, by S. Krishnan, N.Mustafa and S. Venkat. In Data Depth: Robust MultivariateAnalysis, Computational Geometry and Applications,DIMACS Series on Mathematics and Theoretical ComputerScience, 2006.
Generalization to different objects:
“Centerdisks and Tverberg’s Technique”, by A. Basit, N. Mustafa, S. Rayand S. Raza. In Computational Geometry: Theory andApplications, 2010.
Other work
There are several interesting questions related to centerpoints.
Algorithms for centerpoints:
“Location Depth Using the Graphics Hardware”, by S. Krishnan, N.Mustafa and S. Venkat. In 12th Symposium on DiscreteAlgorithms, 2002.
“Statistical Data Depth and the Graphics Hardware”, by S. Krishnan, N.Mustafa and S. Venkat. In Data Depth: Robust MultivariateAnalysis, Computational Geometry and Applications,DIMACS Series on Mathematics and Theoretical ComputerScience, 2006.
Generalization to different objects:
“Centerdisks and Tverberg’s Technique”, by A. Basit, N. Mustafa, S. Rayand S. Raza. In Computational Geometry: Theory andApplications, 2010.
Most of this (and on-going) work in collaboration with
Amirali Abdullah (RA, going for PhD at Utah University)
Abdul Basit (RA, going for PhD at Rutgers University)
Saurabh Ray (PhD Student, Max-Planck Institute)
Sarfraz Raza (PhD Student, LUMS)
Thank you for listening!
Most of this (and on-going) work in collaboration with
Amirali Abdullah (RA, going for PhD at Utah University)
Abdul Basit (RA, going for PhD at Rutgers University)
Saurabh Ray (PhD Student, Max-Planck Institute)
Sarfraz Raza (PhD Student, LUMS)
Thank you for listening!