Many resources on the web deal with the

####

Code sample :

function pointInTriangle(x1, y1, x2, y2, x3, y3, x, y:Number):Boolean

{

var denominator:Number = ((y2 - y3)*(x1 - x3) + (x3 - x2)*(y1 - y3));

var a:Number = ((y2 - y3)*(x - x3) + (x3 - x2)*(y - y3)) / denominator;

var b:Number = ((y3 - y1)*(x - x3) + (x1 - x3)*(y - y3)) / denominator;

var c:Number = 1 - a - b;

return 0 <= a && a <= 1 && 0 <= b && b <= 1 && 0 <= c && c <= 1;

}

####

Code sample :

function pointInTriangle(x1, y1, x2, y2, x3, y3, x, y:Number):Boolean

{

var denominator:Number = (x1*(y2 - y3) + y1*(x3 - x2) + x2*y3 - y2*x3);

var t1:Number = (x*(y3 - y1) + y*(x1 - x3) - x1*y3 + y1*x3) / denominator;

var t2:Number = (x*(y2 - y1) + y*(x1 - x2) - x1*y2 + y1*x2) / -denominator;

var s:Number = t1 + t2;

return 0 <= t1 && t1 <= 1 && 0 <= t2 && t2 <= 1 && s <= 1;

}

####

Code sample :

function side(x1, y1, x2, y2, x, y:Number):Number

{

return (y2 - y1)*(x - x1) + (-x2 + x1)*(y - y1);

}

function pointInTriangle(x1, y1, x2, y2, x3, y3, x, y:Number):Boolean

{

var checkSide1:Boolean = side(x1, y1, x2, y2, x, y) >= 0;

var checkSide2:Boolean = side(x2, y2, x3, y3, x, y) >= 0;

var checkSide3:Boolean = side(x3, y3, x1, y1, x, y) >= 0;

return checkSide1 && checkSide2 && checkSide3;

}

####

####

The steps of the algorithm are:

Because our new algorithm will involve more computations, it is convenient to add a pre-test to reject very quickly most of the false cases.

The bounding box is simply the min/max of the x/y values among the 3 triangle's vertices, slightly inflated by the

Code sample :

const EPSILON:Number = 0.001;

function pointInTriangleBoundingBox(x1, y1, x2, y2, x3, y3, x, y:Number):Boolean

{

var xMin:Number = Math.min(x1, Math.min(x2, x3)) - EPSILON;

var xMax:Number = Math.max(x1, Math.max(x2, x3)) + EPSILON;

var yMin:Number = Math.min(y1, Math.min(y2, y3)) - EPSILON;

var yMax:Number = Math.max(y1, Math.max(y2, y3)) + EPSILON;

if ( x < xMin || xMax < x || y < yMin || yMax < y )

return false;

else

return true;

}

If the test is positive, we can trust it and stop the algorithm immediatly. But if the test is negative, maybe we face the situation with the point lying on one triangle's edge : then we need more investigations, involving methods using the

For every edge of the triangle, we compute the shortest distance between the edge and the point to evaluate. If the distance is shorter that

In detail, given 3 points

Finally, because our algorithm use distances for comparisons only, we will restrict our computations to square distances only (faster because we can omit the square root).

Code sample:

function distanceSquarePointToSegment(x1, y1, x2, y2, x, y:Number):Number

{

var p1_p2_squareLength:Number = (x2 - x1)*(x2 - x1) + (y2 - y1)*(y2 - y1);

var dotProduct:Number = ((x - x1)*(x2 - x1) + (y - y1)*(y2 - y1)) / p1_p2_squareLength;

if ( dotProduct < 0 )

{

return (x - x1)*(x - x1) + (y - y1)*(y - y1);

}

else if ( dotProduct <= 1 )

{

var p_p1_squareLength:Number = (x1 - x)*(x1 - x) + (y1 - y)*(y1 - y);

return p_p1_squareLength - dotProduct * dotProduct * p1_p2_squareLength;

}

else

{

return (x - x2)*(x - x2) + (y - y2)*(y - y2);

}

}

####

*2D point in triangle test*. This post will summarize the most famous methods to solve it and will show why in some cases they are not accurate and can lead to errors. As a conclusion, we will expose a new algorithm.
At first, we need to remember the problem. We are in 2D. We have a triangle

*T*defined by 3 points*p1(x1, y1), p2(x2, y2), p3(x3, y3)*and a single point*p(x, y)*. Does this single point*p*lies inside the triangle*T*?####
**1st method : barycentric coordinate system**

Barycentric coordinate allows to express new

*p*coordinates as a linear combination of*p1, p2, p3*. More precisely, it defines 3 scalars*a, b, c*such that :
x = a * x1 + b * x2 + c * x3

y = a * y1 + b * y2 + c * y3

a + b + c = 1

The way to compute

*a, b, c*is not difficult :
a = ((y2 - y3)*(x - x3) + (x3 - x2)*(y - y3)) / ((y2 - y3)*(x1 - x3) + (x3 - x2)*(y1 - y3))

b = ((y3 - y1)*(x - x3) + (x1 - x3)*(y - y3)) / ((y2 - y3)*(x1 - x3) + (x3 - x2)*(y1 - y3))

c = 1 - a - b

Then we just need to apply the interesting following property :

*p*lies in

*T*if and only if

*0 <= a <= 1*and

*0 <= b <= 1*and

*0 <= c <= 1*

Code sample :

function pointInTriangle(x1, y1, x2, y2, x3, y3, x, y:Number):Boolean

{

var denominator:Number = ((y2 - y3)*(x1 - x3) + (x3 - x2)*(y1 - y3));

var a:Number = ((y2 - y3)*(x - x3) + (x3 - x2)*(y - y3)) / denominator;

var b:Number = ((y3 - y1)*(x - x3) + (x1 - x3)*(y - y3)) / denominator;

var c:Number = 1 - a - b;

return 0 <= a && a <= 1 && 0 <= b && b <= 1 && 0 <= c && c <= 1;

}

####
**2nd method : parametric equations system**

Here the idea is to consider the parametric expressions of the 2 edges [

*p1, p2]*and*[p1, p3]*in*T :*
x(t1) = t1*(x2 - x1)

y(t1) = t1*(y2 - y1)

x(t2) = t2*(x3 - x1)

y(t2) = t2*(y3 - y1)

Then express

*p(x, y)*as a linear combination of them :
x = x1 + x(t1) + x(t2)

y = y1 + y(t1) + y(t2)

Solving the system, it gives to us :

t1 = (x*(y3 - y1) + y*(x1 - x3) - x1*y3 + y1*x3) / (x1*(y2 - y3) + y1*(x3 - x2) + x2*y3 - y2*x3)

t2 = (x*(y2 - y1) + y*(x1 - x2) - x1*y2 + y1*x2) / -(x1*(y2 - y3) + y1*(x3 - x2) + x2*y3 - y2*x3)

Finally, we just need to apply the interesting following property :

*p*lies in

*T*if and only if

*0 <= t1 <= 1*and

*0 <= t2 <= 1*and

*t1 + t2 <= 1*

Code sample :

function pointInTriangle(x1, y1, x2, y2, x3, y3, x, y:Number):Boolean

{

var denominator:Number = (x1*(y2 - y3) + y1*(x3 - x2) + x2*y3 - y2*x3);

var t1:Number = (x*(y3 - y1) + y*(x1 - x3) - x1*y3 + y1*x3) / denominator;

var t2:Number = (x*(y2 - y1) + y*(x1 - x2) - x1*y2 + y1*x2) / -denominator;

var s:Number = t1 + t2;

return 0 <= t1 && t1 <= 1 && 0 <= t2 && t2 <= 1 && s <= 1;

}

####
**3rd method : check sides with dot product**

Maybe the most famous method, based on dot product. We assume that

*p1, p2, p3*are ordered in counterclockwise. Then we can check if*p*lies at left of the 3 oriented edges*[p1, p2]*,*[p2, p3]*and*[p3, p1].*
For that, first we need to consider the 3 vectors

*v1, v2*and*v3*that are respectively left-orthogonal to*[p1, p2]*,*[p2, p3]*and*[p3, p1]**:*
v1 = <y2 - y1, -x2 + x1>

v2 = <y3 - y2, -x3 + x2>

v3 = <y1 - y3, -x1 + x3>

Then we get the 3 following vectors :

v1' = <x - x1, y - y1>

v2' = <x - x2, y - y2>

v3' = <x - x3, y - y3>

At last, we compute the 3 dot products :

dot1 = v1 . v1' = (y2 - y1)*(x - x1) + (-x2 + x1)*(y - y1)

dot2 = v1 . v2' = (y3 - y2)*(x - x2) + (-x3 + x2)*(y - y2)

dot3 = v3 . v3' = (y1 - y3)*(x - x3) + (-x1 + x3)*(y - y3)

Finally, we can apply the interesting property :

*p*lies in

*T*if and only if

*0 <= dot1*and

*0 <= dot2*and

*0 <= dot3*

Code sample :

function side(x1, y1, x2, y2, x, y:Number):Number

{

return (y2 - y1)*(x - x1) + (-x2 + x1)*(y - y1);

}

function pointInTriangle(x1, y1, x2, y2, x3, y3, x, y:Number):Boolean

{

var checkSide1:Boolean = side(x1, y1, x2, y2, x, y) >= 0;

var checkSide2:Boolean = side(x2, y2, x3, y3, x, y) >= 0;

var checkSide3:Boolean = side(x3, y3, x1, y1, x, y) >= 0;

return checkSide1 && checkSide2 && checkSide3;

}

These 3 methods are quite good to solve the point in triangle test. Purely mathematically speaking, they must validate any point inside the triangle and even those lying exactly on the boundary (on any edge).

####
**Accuracy problems**

Despite the strong mathematical background of our methods, in some cases they can lead to a lack of accuracy because the floating-point number system have limited size and most of the time it deals with approximations. The problem occurs sometimes when the point p should exactly on one triangle's edge ; the approximations lead to fail the test.

**Example 1*:**

We consider the triangle

*T*defined by 3 points*p1(x1, y1), p2(x2, y2), p3(x3, y3)*with values :
x1 = 1/10

y1 = 1/9

x2 = 100/8

y2 = 100/3

x3 = 100/4

y3 = 100/9

and a single point

*p(x, y)*lying exactly on the segment [p1, p2] :
If we apply the barycentric method, we get the 3 following values for

*a, b, c*:
a : 0.5714285714285715

b : 0.42857142857142855

c : -5.551115123125783e-17

Because

*c < 0*, the test fails to validate the point inside the triangle. In many applications, this situation is not really a problem because a lack of accuracy is not a tragedy.
But in some situations, it can be really annoying.

**Example 2*:**

We consider 2 triangles

*T*and*T'*defined respectively by the points*p1(x1, y1), p2(x2, y2), p3(x3, y3) and**p1'(x1', y1'), p2'(x2', y2'), p3'(x3', y3').*Values are :
x1 = 1/10

y1 = 1/9

x2 = 100/8

y2 = 100/3

x3 = 100/4

y3 = 100/9

x1' = x1

y1' = y1

x2' = x2

y2' = y2

x3' = -100/8

y3' = 100/6

and a single point

*p(x, y)*lying exactly on the segment [p1, p2] :
The situation is quite simple : 2 non-overlapping triangles sharing one edge and one single point lying on this edge. We know from the previous example that the barycentric method fails to validate

*p*inside the triangle*T*:
a : 0.5714285714285715

b : 0.42857142857142855

c : -5.551115123125783e-17

But more surprisely, the method fails again when applied to the triangle

*T'*:
a : 0.5714285714285715

b : 0.4285714285714285

c : -1.1102230246251565e-16

Mathematically speaking, the point p belongs to both triangles. In practice, we could tolerate that a lack of accuracy leads to validate only one triangle belonging. But here we face a complete invalidation and the point is detected as outside both triangles.

We could suppose that the barycentric-based method is intrinsically inaccurate and one other method will lead to satisfiable results. But in reality the problem remains.

**Example 3*:**

We consider the oriented edge

*E*defined by 2 points*p1(x1, y1)*and*p2(x2, y2)*with values :
x1 = 1/10

y1 = 1/9

x2 = 100/8

y2 = 100/3

and a single point

*p(x, y)*lying exactly on edge*E*:
x = x1 + (3/7)*(x2 - x1)

y = y1 + (3/7)*(y2 - y1)

If we apply the check side method with

*E*and*p*, we get the following dot product :
dot = (y2 - y1)*(x - x1) + (-x2 + x1)*(y - y1)

dot = -2.842170943040401e-14

Then if we apply the same method with

*E'*the reversed edge of*E*, we get the same result :
dot = (y1 - y2)*(x - x2) + (-x1 + x2)*(y - y2)

dot = -2.842170943040401e-14

It means that the check side test gives us 2 contradictory results. The point looks on the right sides of both the edge E and his reversed ; that is of course mathematically impossible. We face the same situation as using barycentric method : if the edge

*E*is shared by 2 non-overlapping triangles*T*and*T'*, despite the fact that*p*lies mathematically on*E*, our method will lead us to tragically conclude that*p*is outside both triangles*T*and*T'*.####
**Any accurate solution ?**

The answer is yes. We can write a complete algorithm leading to a safe point in triangle test by combining many familiar algorithms. The core of the method is to assume a real thickness value for the triangle's edges and vertices ; it contrasts with the original purely mathematical situation where the triangle's edges and vertices have an infinitesimal thickness. We will call it

**epsilon**, because in practice we will keep it very small (near 0.001).The steps of the algorithm are:

**1: use a bounding box as a fast pre-validation test**

Because our new algorithm will involve more computations, it is convenient to add a pre-test to reject very quickly most of the false cases.

The bounding box is simply the min/max of the x/y values among the 3 triangle's vertices, slightly inflated by the

*epsilon*value.

Code sample :

const EPSILON:Number = 0.001;

function pointInTriangleBoundingBox(x1, y1, x2, y2, x3, y3, x, y:Number):Boolean

{

var xMin:Number = Math.min(x1, Math.min(x2, x3)) - EPSILON;

var xMax:Number = Math.max(x1, Math.max(x2, x3)) + EPSILON;

var yMin:Number = Math.min(y1, Math.min(y2, y3)) - EPSILON;

var yMax:Number = Math.max(y1, Math.max(y2, y3)) + EPSILON;

if ( x < xMin || xMax < x || y < yMin || yMax < y )

return false;

else

return true;

}

**2: use any method studied prevously (barycentric, parametric or dot product)**

If the test is positive, we can trust it and stop the algorithm immediatly. But if the test is negative, maybe we face the situation with the point lying on one triangle's edge : then we need more investigations, involving methods using the

*epsilon*value.

**3: use the point to segment distance**

For every edge of the triangle, we compute the shortest distance between the edge and the point to evaluate. If the distance is shorter that

*epsilon*, we can validate definitely the test.

In detail, given 3 points

*p*,

*p1*and

*p2*, a very tricky use of the dot product allows us to check efficiently the relative position of the orthogonal projection

*p'*of

*p*on the infinite line passing through

*p1*and

*p2*. If the projection lies between

*p1*and

*p2*then we compute the distance

*p*and

*p'.*Otherwise, we compute the distance between

*p*and the nearest among

*p1*and

*p2.*

Finally, because our algorithm use distances for comparisons only, we will restrict our computations to square distances only (faster because we can omit the square root).

Code sample:

function distanceSquarePointToSegment(x1, y1, x2, y2, x, y:Number):Number

{

var p1_p2_squareLength:Number = (x2 - x1)*(x2 - x1) + (y2 - y1)*(y2 - y1);

var dotProduct:Number = ((x - x1)*(x2 - x1) + (y - y1)*(y2 - y1)) / p1_p2_squareLength;

if ( dotProduct < 0 )

{

return (x - x1)*(x - x1) + (y - y1)*(y - y1);

}

else if ( dotProduct <= 1 )

{

var p_p1_squareLength:Number = (x1 - x)*(x1 - x) + (y1 - y)*(y1 - y);

return p_p1_squareLength - dotProduct * dotProduct * p1_p2_squareLength;

}

else

{

return (x - x2)*(x - x2) + (y - y2)*(y - y2);

}

}

####
**Final sample code**

Here is the code illustrating the steps we described in the previous section.

const EPSILON:Number = 0.001;

const EPSILON_SQUARE:Number = EPSILON*EPSILON;

function side(x1, y1, x2, y2, x, y:Number):Number

{

return (y2 - y1)*(x - x1) + (-x2 + x1)*(y - y1);

}

function naivePointInTriangle(x1, y1, x2, y2, x3, y3, x, y:Number):Boolean

{

var checkSide1:Boolean = side(x1, y1, x2, y2, x, y) >= 0;

var checkSide2:Boolean = side(x2, y2, x3, y3, x, y) >= 0;

var checkSide3:Boolean = side(x3, y3, x1, y1, x, y) >= 0;

return checkSide1 && checkSide2 && checkSide3;

}

function pointInTriangleBoundingBox(x1, y1, x2, y2, x3, y3, x, y:Number):Boolean

{

var xMin:Number = Math.min(x1, Math.min(x2, x3)) - EPSILON;

var xMax:Number = Math.max(x1, Math.max(x2, x3)) + EPSILON;

var yMin:Number = Math.min(y1, Math.min(y2, y3)) - EPSILON;

var yMax:Number = Math.max(y1, Math.max(y2, y3)) + EPSILON;

if ( x < xMin || xMax < x || y < yMin || yMax < y )

return false;

else

return true;

}

function distanceSquarePointToSegment(x1, y1, x2, y2, x, y:Number):Number

{

var p1_p2_squareLength:Number = (x2 - x1)*(x2 - x1) + (y2 - y1)*(y2 - y1);

var dotProduct:Number = ((x - x1)*(x2 - x1) + (y - y1)*(y2 - y1)) / p1_p2_squareLength;

if ( dotProduct < 0 )

{

return (x - x1)*(x - x1) + (y - y1)*(y - y1);

}

else if ( dotProduct <= 1 )

{

var p_p1_squareLength:Number = (x1 - x)*(x1 - x) + (y1 - y)*(y1 - y);

return p_p1_squareLength - dotProduct * dotProduct * p1_p2_squareLength;

}

else

{

return (x - x2)*(x - x2) + (y - y2)*(y - y2);

}

}

function accuratePointInTriangle(x1, y1, x2, y2, x3, y3, x, y:Number):Boolean

{

if (! pointInTriangleBoundingBox(x1, y1, x2, y2, x3, y3, x, y))

return false;

if (naivePointInTriangle(x1, y1, x2, y2, x3, y3, x, y))

return true;

if (distanceSquarePointToSegment(x1, y1, x2, y2, x, y) <= EPSILON_SQUARE)

return true;

if (distanceSquarePointToSegment(x2, y2, x3, y3, x, y) <= EPSILON_SQUARE)

return true;

if (distanceSquarePointToSegment(x3, y3, x1, y1, x, y) <= EPSILON_SQUARE)

return true;

return false;

}

I hope you appreciated.

const EPSILON:Number = 0.001;

const EPSILON_SQUARE:Number = EPSILON*EPSILON;

function side(x1, y1, x2, y2, x, y:Number):Number

{

return (y2 - y1)*(x - x1) + (-x2 + x1)*(y - y1);

}

function naivePointInTriangle(x1, y1, x2, y2, x3, y3, x, y:Number):Boolean

{

var checkSide1:Boolean = side(x1, y1, x2, y2, x, y) >= 0;

var checkSide2:Boolean = side(x2, y2, x3, y3, x, y) >= 0;

var checkSide3:Boolean = side(x3, y3, x1, y1, x, y) >= 0;

return checkSide1 && checkSide2 && checkSide3;

}

function pointInTriangleBoundingBox(x1, y1, x2, y2, x3, y3, x, y:Number):Boolean

{

var xMin:Number = Math.min(x1, Math.min(x2, x3)) - EPSILON;

var xMax:Number = Math.max(x1, Math.max(x2, x3)) + EPSILON;

var yMin:Number = Math.min(y1, Math.min(y2, y3)) - EPSILON;

var yMax:Number = Math.max(y1, Math.max(y2, y3)) + EPSILON;

if ( x < xMin || xMax < x || y < yMin || yMax < y )

return false;

else

return true;

}

function distanceSquarePointToSegment(x1, y1, x2, y2, x, y:Number):Number

{

var p1_p2_squareLength:Number = (x2 - x1)*(x2 - x1) + (y2 - y1)*(y2 - y1);

var dotProduct:Number = ((x - x1)*(x2 - x1) + (y - y1)*(y2 - y1)) / p1_p2_squareLength;

if ( dotProduct < 0 )

{

return (x - x1)*(x - x1) + (y - y1)*(y - y1);

}

else if ( dotProduct <= 1 )

{

var p_p1_squareLength:Number = (x1 - x)*(x1 - x) + (y1 - y)*(y1 - y);

return p_p1_squareLength - dotProduct * dotProduct * p1_p2_squareLength;

}

else

{

return (x - x2)*(x - x2) + (y - y2)*(y - y2);

}

}

function accuratePointInTriangle(x1, y1, x2, y2, x3, y3, x, y:Number):Boolean

{

if (! pointInTriangleBoundingBox(x1, y1, x2, y2, x3, y3, x, y))

return false;

if (naivePointInTriangle(x1, y1, x2, y2, x3, y3, x, y))

return true;

if (distanceSquarePointToSegment(x1, y1, x2, y2, x, y) <= EPSILON_SQUARE)

return true;

if (distanceSquarePointToSegment(x2, y2, x3, y3, x, y) <= EPSILON_SQUARE)

return true;

if (distanceSquarePointToSegment(x3, y3, x1, y1, x, y) <= EPSILON_SQUARE)

return true;

return false;

}

I hope you appreciated.

* all the tests are done in Actionscript 3, typing values with the native Number type (IEEE-754 double-precision floating-point number).

Thank you! Your algorythm for checking if point is in triangle is very cool!

ReplyDeletethanks for your insights! though im still looking for how to determine if a point is on a triangle in three dimension space

ReplyDeleteOn the dot product interiority test, I don't think you have to assume that p1, p2, p3 are in counter-clockwise order. Just continue with the rest, and if the dot products /are all the same sign/ then you're on the same side of all, and thus inside. (You could not be on the /outside/ of them all is why this works.)

ReplyDeleteYou are totally right.

DeleteNice article!

ReplyDeleteI am trying out the second one (parametric equations system). I am wondering however: in the calculations the denominator could be zero. Could this happen in practise? What would this mean geometrically if this was the case?

Usually, 0 value for denominator happens for degenerate cases.

DeleteFor example, when you try to find the intersection of 2 lines using a parametric system, you face a 0 value when lines are parallel.

In our case our system try to find the p coordinates as a linear combination of [p1, p2] and {p1, p3]. So what are our degenerate cases ? For example, when p2 = p3 ! Meaning your triangle is collapsed as a flat line !

so first you have to test if the triangle is in fact a triangle?

DeleteIf you are not sure that your triangle is safe, better is to check it before perfoming calculation. Check if any 2 points don't overlap. But problems can still occur with points being too close from each other !

DeleteThat is why in my library (Daedalus Lib) I proceeded differently. I choosed to merge points that are too close each other. For that I checked if the distance is smaller than the epsilon value.

Hi,

ReplyDeleteCan you tell how can i interpolate z using method 1. I also have a z coordinate for each point. I think there will be a d along with a, b, c but don't know how to calculate it.

Thanks

Assuming you have P(x, y, z) and a triangle defined by P1(x1, y1, z1), P2(x2, y2, z2), P3(x3, y3, z3), you can try to solve:

Deletex = a * x1 + b * x2 + c * x3

y = a * y1 + b * y2 + c * y3

z = a * z1 + b * z2 + c * z3

a + b + c = 1

But having 3 free variables and 4 constraints means you are surely in trouble.

You must add a new free degree in your system. For example, you could consider to have a new axis pointing orthogonaly to the plane defined by the triangle ; let's call it Ph(xh, yh, zh). Then you consider the system:

x = a * x1 + b * x2 + c * x3 + t * xh

y = a * y1 + b * y2 + c * y3 + t * yh

z = a * z1 + b * z2 + c * z3 + t * zh

a + b + c = 1

and you solve it for a, b, c, t.

The t value can be:

t=0 if P is on the plane defined by the triangle

t>0 if P is above

t<0 if P is below

This is a great article! Very informative.

ReplyDeleteI have one question though. Wouldn't it also be possible to acchive accurate results with only the barycentric method if you use it taking EPSILON into account?

Something like this:

(0 - EPSILON) <= a <= (1 + EPSILON) and

(0 - EPSILON) <= b <= (1 + EPSILON) and

(0 - EPSILON) <= c <= (1 + EPSILON)

Computation-wise that would be faster than the distance check on each segment. Wouldn't it be just as accurate?

The problem here is you don't have the control of the acceptance distance. This distance will be variable and will depend on the size of your triangle. So you face the danger to have wrong results with large triangles.

DeleteAha I see that makes sense!

DeleteWhat I am trying to do is a is-point-on-quad-test. The quad might be distorted in whatever way the user drags the corners. So my idea was to divide the quad into two triangles and simply test if the point is on either of those. There might be an arbitrary amount of quads, so I am a little cautious about computation time.

Since the trinagles share a border is it save to assume, that the barycentric test including Epsilon will not fail, if the point lies directly on the border?

If your purpose is to identify the quad the user has clicked on, you can use your method, I think you will not face problems.

DeleteAwesome thank you very much for the help!

DeleteI will go ahead and implement it that way. Also thanks again for this great article! I will definitely come back to it, the next time I need a accurate triangle/point test!

thanks a lot

ReplyDeleteAren't the first and second methods the same thing?

ReplyDeleteYes from a computational point of view, they are very close. But the concepts involved in the reasonings are different, that's why I found interesting to expose both.

DeleteA graphical area is a semi-open interval - for consistency this is usually done with lower bound inclusive, upper bound exclusive. A point on the boundary of two regions should not be in both; it should be in the region where the point is inclusive and excluded from the region where the point is exclusive. Very few graphics packages handle this issue properly - most have a fencepost error on the boundary. It should be possible to draw two squares from (0,0) to (10,10) and from (10,0) to (20,10) and have them abut exactly, as the pixels should be drawn from 0..9 and 10..19 - in which case a point such as (10,5) would belong to the (10,0) to (20,10) square only.

ReplyDeleteThank you.

ReplyDeleteVery good article. Thanks a lot it was very usefull to me.

ReplyDeleteThank you very much.

ReplyDeleteHello, what about the winding number method [1] for the step 2?

ReplyDeletethanks!

[1] http://geomalgorithms.com/a03-_inclusion.html

Winding number method is a powerful method for n-sides polygons. But applied and optimized to 3 sides polygons, it is equivalent to the "check sides" method I exposed ;)

DeleteThanks for you reply!

DeleteOk, suppose that we have a general convex polygon, N vertices.

ReplyDeleteWe seek to place or reject a test point, P within the polygon.

We select as many test triangles as possible, N points 3 at a time, and apply the test on each triangle.

The test point is in the polygon if it rests within at least one of the [N,3] of the triangles.

Right?

Take care how you build your list of triangles. Considering ALL possible triangles made from the vertices list is not efficient. On the other hand, trying to optimize your triangles list, you could miss one or several.

DeleteYou should consider to build a Delaunay triangulation from your n-vertices polygon.

Or check for algorithms directly dedicated to n-sides polygons.

Well, for N<7 its not too bad, but the Delaunay procedure is certainly worth looking into. Can "CM" of the polygon, X=(sum Xi, i=1..N)/N, play a useful role in sorting vertices?

DeleteHi! Your article is very helpful. I am a beginner and I would like to know how to use these codes using IDL (Interactive Data Language). Thank you and more power.

ReplyDeleteHi.

DeleteYou must "translate" my AS3 code into IDL code.

It is not difficult because my code only use basic features (variable, function, if-else structure) and a few maths being common in all programming languages.

At this point I have never worked with IDL so I can't help you more.

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ReplyDeleteHi!

ReplyDeleteExcellent article!

Will this work with lat,lng coordinates (of earth, assuming perfect sphere)?

Unfortunately it doesn't work on a sphere. If you flatten your local geometry to fit an euclidean 2D space, your triangle will be made from curves, that makes things a bit more complicated.

Delete