I am not sure you can extend it. What you can do is run it twice, first with 3 points, and then, if they are collinear, run it again, substituting the 4th point for any of the other points.

However, I think you will see at least one solution (svd/rank) in the comments that extends fully to multiple points.

–Loren

My question is: how would this be generalized to N points? In particular I’m interested in detecting collinearity of 4 points, I fail to see how the matrix in Method 7 can be generalized.

Of course, the point of this article is to convey the point “there is more than one way to skin a cat,” which Loren and others have sufficiently exemplified. :)

It looks to me like Ilya’s solution and therefore yours are equivalent to the determinant. As Tim and others have pointed out, svd/rank solutions are more robust.

–Loren

using
cos(theta) = [dot(v1,v2) / (||v1||*||v2||) ], and getting one on the right side would mean theta = 0, and the points are collinear. However, since ||v1|| = sqrt(dot(v1,v1)), and sqrt(.) can lead to roundoff error, Ilya Rozenfeld has squared both sides of the equation above, giving
cos^2(theta) = dot(v1,v2)^2 / (sqrt(dot(v1,v1))^2 * sqrt(dot(v2,v2))^2 )
cos^2(theta) = dot(v1,v2)^2 / ( dot(v1,v1)*dot(v2,v2) ).
As long as they are collinear and theta = 0, cos^2(0)=1^2=1.
So, if the right side equals 1, then we have collinearity, which is why Ilya Rozenfeld has written such a solution.

The only thing I fear is numerical precision on the dot product, but this seems to be practical enough for me:
function colOrNot = collinear(A,B,C)
% … code to ensure all column or row vectors
v1 = A-B; v2 = B-C;
colOrNot = ( dot(v1,v2)^2/( dot(v1,v1) * dot(v2,v2) ) ) == 1;
endfunction

> collinear( [0 0], [-3 -3], [6 6] )
ans = 1
> collinear( [1e-8 1e-8]‘, [-6+10e-14 -6+10e-14]‘, [1e6+5e-21 1e6+5e-21]‘ )
ans = 1
—– But I am also subject to:
> eps
ans = 2.22044604925031e-16
> collinear( [0 3], [0 8], [0 -1e21+2e-15] )
ans = 1
> collinear( [0 3], [0 8], [0 -1e22+2e-15] )
ans = 0

Uh-oh! What’s happened? Looking back at collinear5, we see we are no longer doing just addition, subtraction, multiplication, and division. Now we’re computing the norm which involves computing a sqrt. As a result, we have more opportunity for numerical roundoff issues, even if the points are confined to be rational numbers (including integers).

~any(cross([p2-p1, 0], [p3-p1, 0]))

Anohter way could be using a dot product

d1 = p2 – p1;
d2 = p3 – p1;
dot(d1,d2)^2/dot(d1,d1)/dot(d2,d2) == 1

Thanks for the comments. Yes, I agree people should look at good references and rank is preferable to det.

Thinking about the problem in different ways (fitting a line vs. checking area), and especially to help get algorithmic robustness was my main take-away.

–Loren

rank ([p2-p1 ; p3-p1]) < 2

is what you need. Much more numerically reliable than your suggested

det ([p2-p1 ; p3-p1]) == 0

for testing for singular matrices, which is what you’re doing (See “help det”).