check points inside triangle or on edge with example

Good evening everyone

function or coding for finding point is inside triangle or sub triangle or its on edges

thanks for involving your knowledge to be share to answer question

[John D'Errico]

help inpolygon

If this is homework (as I expect from your statement) then why not make an effort? We are not here to provide detailed code for your every homework problem.

[Redwan Dan]

i check that one too, is not assignment and i check isinterior function too but doesn't do what i want ?

a = [17 14 2 15 16 12 17 8 12 10; 13 1 12 12 9 6 14 8 9 10 ]; 

   p0=[a(1,p),a(2,q)]  
     scatter(xmax(1)+5,ylow-10,'*')
  px = (xmax(1)+5)
  py =(ylow-10)
  p1= [px,py]

   scatter(xlow-10,ymax,'*')
   ppx = (xlow-10)
   ppy = ymax
   p2=[ppx,ppy]

   px0 = a(1,p)
   py0=a(2,q)

x=  plot([ px ppx ],[py ppy])

   y= plot([px0 px],[py0 py])

   z= plot([px0 ppx],[py0 ppy])

   T1=[x y z]

   T= [T1]

xv=(px0+px+ppx)/3

yv=(py0+py+ppy)/3

xq=a(1,1)

yq=a(2,1)

[in,on] = inpolygon(xq,yq,xv,yv)

numel(xq(in))

numel(xq(on))

[John D'Errico]

You don't actually say what it is that you expect to happen. Anyway, your code does not run at all, since it is missing the values of p, q, xmax, and ylow to start with. Possibly others, but I've not bothered to look past the first few lines. We cannot decipher uncommented code that will not even execute.

So why did inpolygon NOT suffice? Your question is apparently about a triangle, given the title, yet it is not at all obvious from the code what you are doing. If you want an answer, you need to provide some explanation.

[Roger Stafford]

Even if 'inpolygon' is not used for some reason, checking that a point lies inside a triangle is very easily done. It requires three inequalities to be satisfied. Each one checks that a corresponding vertex lies on the same side of the line of the other two vertices as the given point. This is a comparison between two linear expressions. The point lies inside the triangle if and only if all three are true.

[Redwan Dan]

Dear Roger

you mean by this check sides with dot product method but i think this method is slow to process what do you think about barycentric coordinate system method isn't better than previous one ?

[Roger Stafford]

I've given the necessary expression as an "Answer" here. It doesn't seem like a very slow method to me.

[Walter Roberson]

Redwan Dan comments to John D'Errico:

am not here to prove you any thing and the way your answer and close is not nice treat with me if you don't know just don't comment or close

Answer 1

[Roger Stafford]

Suppose P1 = [x1,y1], P2 = [x2,y2], and P3 = [x3,y3] are row vectors giving the coordinates of the three vertices of a triangle, and P = [x,y] is a row vector for the coordinates of a point P. To determine whether P lies inside the triangle P1P2P3 do this:

   P12 = P1-P2; P23 = P2-P3; P31 = P3-P1;
   t = sign(det([P31;P23]))*sign(det([P3-P;P23])) >= 0 & ...
       sign(det([P12;P31]))*sign(det([P1-P;P31])) >= 0 & ...
       sign(det([P23;P12]))*sign(det([P2-P;P12])) >= 0 ;

Point P lies within the triangle if and only if t is true.

[Roger Stafford]

The following is simplified from the above:

   s = det([P1-P2;P3-P1]);
   t = s*det([P3-P;P2-P3])>=0 & s*det([P1-P;P3-P1])>=0 & s*det([P2-P;P1-P2])>=0;

Again point P lies within the triangle P1P2P3 if and only if t is true.

[Roger Stafford]

You mentioned barycentric coordinates. It may interest you to know that the previous computations are closely related to barycentric coordinates:

   s =  det([P1-P2;P3-P1]);
   w1 = det([P3-P;P2-P3])/s;
   w2 = det([P1-P;P3-P1])/s
   w3 = det([P2-P;P1-P2])/s;

The numbers w1, w2, and w3 are the barycentric coordinate of P and the equation

   P = w1*P1 + w2*P2 + w3*P3

holds true (within round-off accuracy of course.)

[Redwan Dan]

thanks for both of you to involve and share your knowledge ..

[T A]

UPDATED FOR CLARITY

With regard to assessing whether a point is on an edge/edges using the conditional statements given in Roger's answer,

• P is on a triangle edge when one of the three conditional statements is zero
• P is on a triangle vertex when two of the three conditional statements are zero

With regard to computational efficiency, the process can be costly if you're searching across many triangles. In such a case, the easiest thing to do is to only perform the calculations when P is within the bounding box of the current triangle:

Ptri=[P1;P2;P3];
if P(1)<=max(Ptri(:,1))&&P(1)>=min(Ptri(:,1))...
 &&P(2)<=max(Ptri(:,2))&&P(2)>=min(Ptri(:,2))
   %do the suggested calculations...
end

Evaluating this conditional statement is very, very cheap.

[Muhamad Amirulfaris Abdullah]

Hi there. what was the conditional statement that you were reffering to?

Answer 2

[Gary Bikini]

You can use the built-in function

[in,on]=inpolygon(xq,yq,xv,yv)