On Jun 4, 10:56 pm, Feng [EMAIL PROTECTED] wrote:
Hi Kunzmilan, thanks for your idea of using distance matrices. But one
of my friends came up with a seemly counter-example:
Take 3 collinear points in 2D: (0,0), (1,0), (2,0).
The distance matrix is:
0 1 4
1 0 1
4 1 0,
whose eigenvalues
On Jun 5, 1:45 am, Feng [EMAIL PROTECTED] wrote:
Hi BiGYaN, triangle is a 2D object which can be formed by any 3 non-
collinear points. It can exist in 4D, just like a point existing in
3D.
On May 31, 2:11 am, Victor Carvalho [EMAIL PROTECTED] wrote:
Feng, how can form a triangle in
Hi BiGYaN, triangle is a 2D object which can be formed by any 3 non-
collinear points. It can exist in 4D, just like a point existing in
3D.
On May 31, 2:11 am, Victor Carvalho [EMAIL PROTECTED] wrote:
Feng, how can form a triangle in four dimensions???
2007/5/29, BiGYaN [EMAIL PROTECTED]:
use the cross product to examine whether they're collinear.
points A,B,C:
AB*BC =? 0
On 5/27/07, Feng [EMAIL PROTECTED] wrote:
Hi all!
Given 3 points in 3D, what is the fast and numerically stable way to
test if they form a triangle?
I am thinking computing the determinant of the square
Feng, how can form a triangle in four dimensions???
2007/5/29, BiGYaN [EMAIL PROTECTED]:
Just test whether they are collinear or not i.e. get the slopes,
m1 from 1st and 2nd point
m2 from 2nd and 3rd point
if m1==m2 then they do not form a triangle
else they do
Computing the area
Just test whether they are collinear or not i.e. get the slopes,
m1 from 1st and 2nd point
m2 from 2nd and 3rd point
if m1==m2 then they do not form a triangle
else they do
Computing the area of the triangle and testing for 0 might also
work but I feel that the computation will be
In 3D, we can test |(p2-p1)*(p3-p1)|==0, where p1,p2 and p3 are
vectors.
On 5月27日, 上午7时22分, Feng [EMAIL PROTECTED] wrote:
Hi all!
Given 3 points in 3D, what is the fast and numerically stable way to
test if they form a triangle?
I am thinking computing the determinant of the square matrix
in 3D,we can test |(p2-p1)*(p3-p1)|==0,where p1,p2,p3 are 3D-vectors
that represent the three points.
in n-dimension,i think we can let A=(a1,a2,...,an)=p2-p1, and
B=(b1,b2,...,bn)=p3-p1, and test every elements of the matrix (ATB-
BTA). That is ai*bj-aj*bi.
On 5月27日, 上午7时22分, Feng [EMAIL
