Have a look at this link
http://mathworld.wolfram.com/SylvestersFour-PointProblem.html
--~--~-~--~~~---~--~~
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To post to this group, send email to algogeeks@googlegrou
On 3/9/07, Nat (Padmanabhan Natarajan) <[EMAIL PROTECTED]> wrote:
>
> 1. We cannot bound the triangle if we don't bound the space...thats the
> reason why I choose a unit square
I think we don't really need to bound anything. I think the question
as is phrased can only yield what I guess should b
On 3/9/07, Karthik Singaram L <[EMAIL PROTECTED]> wrote:
>
> "Show the chance of four points forming the apices of a reentrant
> quadrilateral is 1/4 if they are taken at random in an indefinite plane". It
Wow! As I solved it, I got the probability tending to 1/2 as the size
of the plane(SOP) ten
Interesting I should say...
Anyway a couple of points on the discussion we had so far...
1. We cannot bound the triangle if we don't bound the space...thats the
reason why I choose a unit square
2. It is true that there are a lot of points outside the triangle that you
cannot choose but they all
The question as it seems is known as the "Sylvesters Question"
I quote from the paper "Random points, convex bodies, lattices" by Imre
barany.
"Show the chance of four points forming the apices of a reentrant
quadrilateral is 1/4 if they are taken at random in an indefinite plane". It
was understo
On 3/8/07, Karthik Singaram L <[EMAIL PROTECTED]> wrote:
>
> The same triangle reasoning holds good here too right? If we choose the end
> points of the side that we are bisecting and the new point we are choosing
> on the perpendicular bisector as the triangle the other vertex will be
> inside th
(A) Hull is _not_ a quadrilateral if the 4th point lies inside the
triangle AND _many_ other points. For instance, extend the
perpendicular bisector (just as a sample case) of any one side beyond
the vertex of the triangle that it passes through.
The same triangle reasoning holds good here too rig
On 3/8/07, Karthik Singaram L <[EMAIL PROTECTED]> wrote:
>
> If the bound reasoning is correct then we would have a probability of 1 to
> get a quadrialetral when choosing random points in a plane.
>
The bound reasoning is valid under the condition that size of plane
tends to infinity.
As I've po
On 3/7/07, Nat (Padmanabhan Natarajan) <[EMAIL PROTECTED]> wrote:
> 4. Now join these three points to get a triangle, again the hull is not
> quadrilateral if it lies within this triangle. Now the probability is 1 -
> (area of triangle)/(area of Sq.) I think that we don't really lose anything
> by
why is it 1 - (area of triangle)/(area of Sq.)?
why do we need a square since what would happen is that the 4th point can be
anywhere in the space but the area of the triangle is bounded. The
probability of choosing a point outside the triangle would be 1 (bound - not
exact by reasoning as in (3))
I am not sure if I am thinking right about this problem but here is my
approach
1. Hull is not a quadrilateral if at least 3 points are collinear to begin
with
2. Pick two random points, draw a line through them
3. Pick the third point, the probability of the third point lying on the
line = 0 (bou
11 matches
Mail list logo
