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A question about polygons. Devising a computational method for determining if a point lies in a plane convex polygon . Problem background. Recall our ‘globe.cpp’ ray-tracing demo It depicted a scene showing two objects One of the objects was a square tabletop

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A question about polygons

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A question about polygons l.jpg

A question about polygons

Devising a computational method for determining if a point lies in a plane convex polygon


Problem background l.jpg

Problem background

  • Recall our ‘globe.cpp’ ray-tracing demo

  • It depicted a scene showing two objects

  • One of the objects was a square tabletop

  • Its edges ran parallel to x- and z-axes

  • That fact made it easy to determine if the spot where a ray hit the tabletop fell inside or outside of the square’s boundary edges

  • We were lucky!


Alternative geometry l.jpg

Alternative geometry?

  • What if we wanted to depict our globe as resting on a tabletop that wasn’t a nicely aligned square? For example, could we show a tabletop that was a hexagon?

  • The only change is that the edges of this six-sided tabletop can’t all be lined up with the coordinate system axes

  • We need a new approach to determining if a ray hits a spot that’s on our tabletop


A simpler problem l.jpg

A simpler problem

  • How can we tell if a point lies in a triangle?

c

q

p

b

a

Point p lies inside triangle Δabc

Point q lies outside triangle Δabc


Triangle algorithm l.jpg

Triangle algorithm

  • Draw vectors from a to b and from a to c

  • We can regard these vectors as the axes for a “skewed” coordinate system

  • Then every point in the triangle’s plane would have a unique pair of coordinates

  • We can compute those coordinates using Cramer’s Rule (from linear algebra)


The algorithm idea l.jpg

The algorithm idea

c

ap = c1*ab + c2*ac

p

b

a

c1 = det( ap, ac )/det( ab, ac )

c2 = det( ab, ap )/det( ab, ac )


Cartesian analogy l.jpg

Cartesian Analogy

y-axis

p = (c1,c2)

x-axis

x + y = 1

p lies outside the triangle if c1 < 0 or c2 < 0 or c1+c2 > 1


Determinant function 2x2 l.jpg

Determinant function: 2x2

typedef float scalar_t;

typedef struct { scalar_t x, y; } vector_t;

scalar_t det( vector_t a, vector_t b )

{

returna.x * b.y – b.x * a.y;

}


Constructing a regular hexagon l.jpg

Constructing a regular hexagon

theta = 2*PI / 6

( cos(2*theta), sin(2*theta) )

( cos(1*theta), sin(1*theta) )

( cos(3*theta), sin(3*theta) )

( cos(0*theta), sin(0*theta) )

( cos(5*theta), sin(5*theta) )

( cos(4*theta), sin(4*theta) )


Subdividing the hexagon l.jpg

Subdividing the hexagon

A point p lies outside the hexagon -- unless

it lies inside one of these four sub-triangles


Same approach for n gons l.jpg

Same approach for n-gons

  • Demo program ‘hexagon.cpp’ illustrates the use of our just-described algorithm

  • Every convex polygon can be subdivided into triangles, so the same ideas can be applied to any regular n-sided polygon

  • Exercise: modify the demo-program so it draws an octagon, a pentagon, a septagon


Extension to a tetrahedron l.jpg

Extension to a tetrahedron

  • A tetrahedron is a 3D analog of a triangle

  • It has 4 vertices, located in space (but not all vertices can lie in the same plane)

  • Each face of a tetrahedron is a triangle

c

b

o

a


Cramer s rule in 3d l.jpg

Cramer’s Rule in 3D

typedef float scalar_t;

typedef struct { scalar_t x, y, z; } vector_t;

scalar_t det( vector_t a, vector_t b, vector_t c )

{

scalar_tsum = 0.0;

sum += a.x * b.y * c.z – a.x * b.z * c.y;

sum += a.y * b.z * c.x – a.y * b.x * c.z;

sum += a.z * b.x * c.y – a.z * b.y * c.x;

returnsum;

}


Is point in tetrahedron l.jpg

Is point in tetrahedron?

  • Let o, a, b, c be vertices of a tetrahedron

  • Form the three vectors oa, ob, oc and regard them as the coordinate axes in a “skewed” 3D coordinate system

  • Then any point p in space has a unique triple of coordinates:op = c1*oa + c2*ob + c3*oc

  • These three coordinates can be computed using Cramer’s Rule


Details of cramer rule l.jpg

Details of Cramer Rule

c1 = det( op, ob, oc )/det( oa, ob, oc )

c2 = det( oa, op, oc )/det( oa, ob, oc )

c3 = det( oa, ob, op )/det( oa, ob, oc )

Point p lies inside the tetrahedron – unless

c1 < 0 or c2 < 0 or c3 < 0

orc1 + c2 + c3 > 1


Convex polyhedron l.jpg

Convex polyhedron

  • Just as a convex polygon can be divided into subtriangles, any convex polyhedron can be divided into several tetrahedrons

  • We can tell if a point lies in the polyhedron by testing to see it lies in one of the solid’s tetrahedral parts

  • An example: the regular dodecahedron can be raytraced by using these ideas!


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