A question about polygonsProblem backgroundAlternative geometry?A simpler problemTriangle algorithmThe algorithm ideaCartesian AnalogyDeterminant function: 2x2Constructing a regular hexagonSubdividing the hexagonSame approach for n-gonsExtension to a tetrahedronCramer’s Rule in 3DIs point in tetrahedron?Details of Cramer RuleConvex polyhedronA question about polygonsDevising a computational method for determining if a point lies in a plane convex polygonProblem 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?•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 tabletopA simpler problem•How can we tell if a point lies in a triangle?abcpqPoint p lies inside triangle Δabc Point q lies outside triangle ΔabcTriangle 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 ideaabcpap = c1*ab + c2*acc1 = det( ap, ac )/det( ab, ac )c2 = det( ab, ap )/det( ab, ac )Cartesian Analogyp = (c1,c2)x-axisy-axisx + y = 1p lies outside the triangle if c1 < 0 or c2 < 0 or c1+c2 > 1Determinant function: 2x2typedef float scalar_t;typedef struct { scalar_t x, y; } vector_t;scalar_t det( vector_t a, vector_t b ){return a.x * b.y – b.x * a.y;}Constructing a regular hexagon( cos(0*theta), sin(0*theta) )( cos(1*theta), sin(1*theta) )( cos(2*theta), sin(2*theta) )( cos(3*theta), sin(3*theta) )( cos(4*theta), sin(4*theta) )( cos(5*theta), sin(5*theta) )theta = 2*PI / 6Subdividing the hexagonA point p lies out side the hexagon -- unless it lies inside one of these four sub-trianglesSame 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 septagonExtension 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 triangleoabcCramer’s Rule in 3Dtypedef 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_t sum = 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;return sum;}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 RuleDetails of Cramer Rulec1 = 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 – unlessc1 < 0 or c2 < 0 or c3 < 0 or c1 + c2 + c3 > 1Convex 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
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