To get the center, I have tried, for each vertex, to add to the total, divide by the number of vertices.
I've also tried to find the topmost, bottommost -> get midpoint... find leftmost, rightmost, find the midpoint.
Both of these did not return the perfect center because I'm relying on the center to scale a polygon.
I want to scale my polygons, so I may put a border around them.
What is the best way to find the centroid of a polygon given that the polygon may be concave, convex and have many many sides of various lengths?
The formula is given here for vertices sorted by their occurance along the polygon's perimeter.
For those having difficulty understanding the sigma notation in those formulas, here is some C++ code showing how to do the computation:
#include <iostream>
struct Point2D
{
double x;
double y;
};
Point2D compute2DPolygonCentroid(const Point2D* vertices, int vertexCount)
{
Point2D centroid = {0, 0};
double signedArea = 0.0;
double x0 = 0.0; // Current vertex X
double y0 = 0.0; // Current vertex Y
double x1 = 0.0; // Next vertex X
double y1 = 0.0; // Next vertex Y
double a = 0.0; // Partial signed area
// For all vertices except last
int i=0;
for (i=0; i<vertexCount-1; ++i)
{
x0 = vertices[i].x;
y0 = vertices[i].y;
x1 = vertices[i+1].x;
y1 = vertices[i+1].y;
a = x0*y1 - x1*y0;
signedArea += a;
centroid.x += (x0 + x1)*a;
centroid.y += (y0 + y1)*a;
}
// Do last vertex separately to avoid performing an expensive
// modulus operation in each iteration.
x0 = vertices[i].x;
y0 = vertices[i].y;
x1 = vertices[0].x;
y1 = vertices[0].y;
a = x0*y1 - x1*y0;
signedArea += a;
centroid.x += (x0 + x1)*a;
centroid.y += (y0 + y1)*a;
signedArea *= 0.5;
centroid.x /= (6.0*signedArea);
centroid.y /= (6.0*signedArea);
return centroid;
}
int main()
{
Point2D polygon[] = {{0.0,0.0}, {0.0,10.0}, {10.0,10.0}, {10.0,0.0}};
size_t vertexCount = sizeof(polygon) / sizeof(polygon[0]);
Point2D centroid = compute2DPolygonCentroid(polygon, vertexCount);
std::cout << "Centroid is (" << centroid.x << ", " << centroid.y << ")\n";
}
I've only tested this for a square polygon in the upper-right x/y quadrant.
If you don't mind performing two (potentially expensive) extra modulus operations in each iteration, then you can simplify the previous compute2DPolygonCentroid function to the following:
Point2D compute2DPolygonCentroid(const Point2D* vertices, int vertexCount)
{
Point2D centroid = {0, 0};
double signedArea = 0.0;
double x0 = 0.0; // Current vertex X
double y0 = 0.0; // Current vertex Y
double x1 = 0.0; // Next vertex X
double y1 = 0.0; // Next vertex Y
double a = 0.0; // Partial signed area
// For all vertices
int i=0;
for (i=0; i<vertexCount; ++i)
{
x0 = vertices[i].x;
y0 = vertices[i].y;
x1 = vertices[(i+1) % vertexCount].x;
y1 = vertices[(i+1) % vertexCount].y;
a = x0*y1 - x1*y0;
signedArea += a;
centroid.x += (x0 + x1)*a;
centroid.y += (y0 + y1)*a;
}
signedArea *= 0.5;
centroid.x /= (6.0*signedArea);
centroid.y /= (6.0*signedArea);
return centroid;
}
The centroid can be calculated as the weighted sum of the centroids of the triangles it can be partitioned to.
Here is the C source code for such an algorithm:
/*
Written by Joseph O'Rourke
orourke#cs.smith.edu
October 27, 1995
Computes the centroid (center of gravity) of an arbitrary
simple polygon via a weighted sum of signed triangle areas,
weighted by the centroid of each triangle.
Reads x,y coordinates from stdin.
NB: Assumes points are entered in ccw order!
E.g., input for square:
0 0
10 0
10 10
0 10
This solves Exercise 12, p.47, of my text,
Computational Geometry in C. See the book for an explanation
of why this works. Follow links from
http://cs.smith.edu/~orourke/
*/
#include <stdio.h>
#define DIM 2 /* Dimension of points */
typedef int tPointi[DIM]; /* type integer point */
typedef double tPointd[DIM]; /* type double point */
#define PMAX 1000 /* Max # of pts in polygon */
typedef tPointi tPolygoni[PMAX];/* type integer polygon */
int Area2( tPointi a, tPointi b, tPointi c );
void FindCG( int n, tPolygoni P, tPointd CG );
int ReadPoints( tPolygoni P );
void Centroid3( tPointi p1, tPointi p2, tPointi p3, tPointi c );
void PrintPoint( tPointd p );
int main()
{
int n;
tPolygoni P;
tPointd CG;
n = ReadPoints( P );
FindCG( n, P ,CG);
printf("The cg is ");
PrintPoint( CG );
}
/*
Returns twice the signed area of the triangle determined by a,b,c,
positive if a,b,c are oriented ccw, and negative if cw.
*/
int Area2( tPointi a, tPointi b, tPointi c )
{
return
(b[0] - a[0]) * (c[1] - a[1]) -
(c[0] - a[0]) * (b[1] - a[1]);
}
/*
Returns the cg in CG. Computes the weighted sum of
each triangle's area times its centroid. Twice area
and three times centroid is used to avoid division
until the last moment.
*/
void FindCG( int n, tPolygoni P, tPointd CG )
{
int i;
double A2, Areasum2 = 0; /* Partial area sum */
tPointi Cent3;
CG[0] = 0;
CG[1] = 0;
for (i = 1; i < n-1; i++) {
Centroid3( P[0], P[i], P[i+1], Cent3 );
A2 = Area2( P[0], P[i], P[i+1]);
CG[0] += A2 * Cent3[0];
CG[1] += A2 * Cent3[1];
Areasum2 += A2;
}
CG[0] /= 3 * Areasum2;
CG[1] /= 3 * Areasum2;
return;
}
/*
Returns three times the centroid. The factor of 3 is
left in to permit division to be avoided until later.
*/
void Centroid3( tPointi p1, tPointi p2, tPointi p3, tPointi c )
{
c[0] = p1[0] + p2[0] + p3[0];
c[1] = p1[1] + p2[1] + p3[1];
return;
}
void PrintPoint( tPointd p )
{
int i;
putchar('(');
for ( i=0; i<DIM; i++) {
printf("%f",p[i]);
if (i != DIM - 1) putchar(',');
}
putchar(')');
putchar('\n');
}
/*
Reads in the coordinates of the vertices of a polygon from stdin,
puts them into P, and returns n, the number of vertices.
The input is assumed to be pairs of whitespace-separated coordinates,
one pair per line. The number of points is not part of the input.
*/
int ReadPoints( tPolygoni P )
{
int n = 0;
printf("Polygon:\n");
printf(" i x y\n");
while ( (n < PMAX) && (scanf("%d %d",&P[n][0],&P[n][1]) != EOF) ) {
printf("%3d%4d%4d\n", n, P[n][0], P[n][1]);
++n;
}
if (n < PMAX)
printf("n = %3d vertices read\n",n);
else
printf("Error in ReadPoints:\too many points; max is %d\n", PMAX);
putchar('\n');
return n;
}
There's a polygon centroid article on the CGAFaq (comp.graphics.algorithms FAQ) wiki that explains it.
boost::geometry::centroid(your_polygon, p);
Here is Emile Cormier's algorithm without duplicated code or expensive modulus operations, best of both worlds:
#include <iostream>
using namespace std;
struct Point2D
{
double x;
double y;
};
Point2D compute2DPolygonCentroid(const Point2D* vertices, int vertexCount)
{
Point2D centroid = {0, 0};
double signedArea = 0.0;
double x0 = 0.0; // Current vertex X
double y0 = 0.0; // Current vertex Y
double x1 = 0.0; // Next vertex X
double y1 = 0.0; // Next vertex Y
double a = 0.0; // Partial signed area
int lastdex = vertexCount-1;
const Point2D* prev = &(vertices[lastdex]);
const Point2D* next;
// For all vertices in a loop
for (int i=0; i<vertexCount; ++i)
{
next = &(vertices[i]);
x0 = prev->x;
y0 = prev->y;
x1 = next->x;
y1 = next->y;
a = x0*y1 - x1*y0;
signedArea += a;
centroid.x += (x0 + x1)*a;
centroid.y += (y0 + y1)*a;
prev = next;
}
signedArea *= 0.5;
centroid.x /= (6.0*signedArea);
centroid.y /= (6.0*signedArea);
return centroid;
}
int main()
{
Point2D polygon[] = {{0.0,0.0}, {0.0,10.0}, {10.0,10.0}, {10.0,0.0}};
size_t vertexCount = sizeof(polygon) / sizeof(polygon[0]);
Point2D centroid = compute2DPolygonCentroid(polygon, vertexCount);
std::cout << "Centroid is (" << centroid.x << ", " << centroid.y << ")\n";
}
Break it into triangles, find the area and centroid of each, then calculate the average of all the partial centroids using the partial areas as weights. With concavity some of the areas could be negative.
Related
I am attempting to generate noise similar to perlin or value noise.
I am using stb_image_write library from here to write to image file (i write to disk as hdr and converted to png with GIMP to post on here).
This code requires C++ 20 because I am using std::lerp();
Because I am generating a linear gradient as testing I am expecting a linear gradient as output.
I know there are more steps to generating the desired noise but, this is where I'm having issues.
#include <cmath>
template <size_t size>
void noise(float seed[size][size], float output[size][size], size_t octave);
int main()
{
// generate test gradient
float seedNoise[64][64] = {};
for ( size_t x = 0; x < 64; x++ )
{
for ( size_t y = 0; y < 64; y++ )
{
seedNoise[x][y] = ((((float)x) / 64.0f + ((float)y / 64.0f)) / 2.0f);
}
}
float _map[64][64] = { 0 };
noise<64>(seedNoise, _map, 4);
}
template <size_t size>
void noise(float seed[size][size], float output[size][size], size_t octave)
{
size_t step = size / octave; // went back to this
// size_t step = (size - 1) / octave; // took this back out
for ( size_t x = 0; x <= size - step; x += step )
{
for ( size_t y = 0; y <= size - step; y += step )
{ // each octave
// extract values at corners octave from seed
float a = seed[x][y];
float b = seed[x + (step - 1)][y]; // changed this
float c = seed[x][y + (step - 1)]; // this
float d = seed[x + (step - 1)][y + (step - 1)]; // and this
for ( size_t u = 0; u < step; u++ )
{
float uStep = ((float)u) / ((float)step); // calculate x step
for ( size_t v = 0; v < step; v++ )
{ // each element in each octave
float vStep = ((float)v) / ((float)step); // calculate y step
float x1 = std::lerp(a, b, uStep); // interpolate top edge
float x2 = std::lerp(c, d, uStep); // interpolate bottom edge
float y1 = std::lerp(a, c, vStep); // interpolate left edge
float y2 = std::lerp(b, d, vStep); // interpolate right edge
float x3 = std::lerp(x1, x2, vStep); // interpolate between top and bottom edges
float y3 = std::lerp(y1, y2, uStep); // interpolate between left and right edges
float odat = (x3 + y3) / 2; // average top/bottom and left/right interpolations
output[x + u][y + v] = odat;
}
}
}
}
}
Source gradient I think this should be similar to what the output should be.
Output As you can see here the right and bottom of the output is all messed up.
new output
I think you're accessing the imago outside it's boundaries.
X and y can go up to 60 in the loops:
for ( size_t x = 0; x <= size - step; x += step )
And the you are accessing position y+step and x+step, which gives 64.
I have two points on XZ plane, larger/taller point is L=(XL, ZL) and smaller/shorter point is S=(XS, ZS)
By connecting L and S points to Z=0 line, I have two lines
I intend to connect my two lines by zigzag diagonal cross lines
I need to find points L0, L1, Lk, ... LN-1 and also S0, S1, Sk, ... SN-1, SN.
I already know two points:
S0 = S = (XS, ZS)
SN = (XS, 0)
So far, I have implemented this algorithm:
float lX, lZ = ... // "L" (larger/taller) point coordinate on (X, Z) plane
float sX, sZ = ... // "S" (smaller/shorter) point coordinate on (X, Z) plane
size_t N = 5; // N sections below S
float sZsectionLength = sZ / N; // length of each section below S
std::vector<float> sZ_dots(N+1, 0.0); // N+1 points/dots below S
for (size_t k = 0; k < N+1; ++k) {
sZ_dots[k] = sZ - k * sZsectionLength;
}
std::vector<float> lZ_dots(N, 0.0); // N points/dots below L
for (size_t k = 0; k < N; ++k) {
// // Each point below L is average of two points below S
lZ_dots[k] = ( sZ_dots[k] + sZ_dots[k+1] ) / 2.0f;
}
for (size_t k = 0; k < N; ++k) {
Line *zig = new Line();
zig->setStartDot(sX, sZ_dots[k]);
zig->setCloseDot(lX, lZ_dots[k]);
linesContainer.append(zig);
Line *zag = new Line();
zag->setStartDot(lX, lZ_dots[k]);
zag->setCloseDot(sX, sZ_dots[k+1]);
linesContainer.append(zag);
}
The above algorithm generates the zig zags just fine. However, I wonder if there is any faster algorithm to generate the zig zag cross lines. Anything which I'm missing?
I would implement it like this:
struct Line
{
Line(float x1, float z1, float x2, float z2)
:
m_x1(x1),
m_z1(z1),
m_x2(x2),
m_z2(z2)
{}
float m_x1;
float m_z1;
float m_x2;
float m_z2;
};
using LineContainer = std::vector<Line>;
LineContainer getZigZag(float lx, float sx, float sz, size_t sectionCount)
{
assert(lx < sx && sz > 0.0f);
LineContainer lines;
auto sectionHeight = sz / sectionCount;
for (auto i = 0; i < sectionCount; ++i)
{
auto sz1 = sz - sectionHeight * i;
auto sz2 = sz - sectionHeight * (i + 1);
auto lz = sz1 - (sz1 - sz2) / 2.0f;
// A section.
//
// From S to L
lines.emplace_back(sx, sz1, lx, lz);
// From L to S
lines.emplace_back(lx, lz, sx, sz2);
}
return lines;
}
and use the function like this:
int main()
{
auto zigzag = getZigZag(1.0f, 2.0f, 4.0f, 2);
[..]
As you probably noticed, I replaced three loops with a single one that creates two lines (a single section) on each iteration.
I'm trying to implement the Delaunay triangulation in C++. Currently it's working, but I'm not getting the correct amount of triangles.
I try it with 4 points in a square pattern : (0,0), (1,0), (0,1), (1,1).
Here's the algorithm I use :
std::vector<Triangle> Delaunay::triangulate(std::vector<Vec2f> &vertices) {
// Determinate the super triangle
float minX = vertices[0].getX();
float minY = vertices[0].getY();
float maxX = minX;
float maxY = minY;
for(std::size_t i = 0; i < vertices.size(); ++i) {
if (vertices[i].getX() < minX) minX = vertices[i].getX();
if (vertices[i].getY() < minY) minY = vertices[i].getY();
if (vertices[i].getX() > maxX) maxX = vertices[i].getX();
if (vertices[i].getY() > maxY) maxY = vertices[i].getY();
}
float dx = maxX - minX;
float dy = maxY - minY;
float deltaMax = std::max(dx, dy);
float midx = (minX + maxX) / 2.f;
float midy = (minY + maxY) / 2.f;
Vec2f p1(midx - 20 * deltaMax, midy - deltaMax);
Vec2f p2(midx, midy + 20 * deltaMax);
Vec2f p3(midx + 20 * deltaMax, midy - deltaMax);
// Add the super triangle vertices to the end of the vertex list
vertices.push_back(p1);
vertices.push_back(p2);
vertices.push_back(p3);
// Add the super triangle to the triangle list
std::vector<Triangle> triangleList = {Triangle(p1, p2, p3)};
// For each point in the vertex list
for(auto point = begin(vertices); point != end(vertices); point++)
{
// Initialize the edges buffer
std::vector<Edge> edgesBuff;
// For each triangles currently in the triangle list
for(auto triangle = begin(triangleList); triangle != end(triangleList);)
{
if(triangle->inCircumCircle(*point))
{
Edge tmp[3] = {triangle->getE1(), triangle->getE2(), triangle->getE3()};
edgesBuff.insert(end(edgesBuff), tmp, tmp + 3);
triangle = triangleList.erase(triangle);
}
else
{
triangle++;
}
}
// Delete all doubly specified edges from the edge buffer
// Black magic by https://github.com/MechaRage
auto ite = begin(edgesBuff), last = end(edgesBuff);
while(ite != last) {
// Search for at least one duplicate of the current element
auto twin = std::find(ite + 1, last, *ite);
if(twin != last)
// If one is found, push them all to the end.
last = std::partition(ite, last, [&ite](auto const &o){ return !(o == *ite); });
else
++ite;
}
// Remove all the duplicates, which have been shoved past "last".
edgesBuff.erase(last, end(edgesBuff));
// Add the triangle to the list
for(auto edge = begin(edgesBuff); edge != end(edgesBuff); edge++)
triangleList.push_back(Triangle(edge->getP1(), edge->getP2(), *point));
}
// Remove any triangles from the triangle list that use the supertriangle vertices
triangleList.erase(std::remove_if(begin(triangleList), end(triangleList), [p1, p2, p3](auto t){
return t.containsVertex(p1) || t.containsVertex(p2) || t.containsVertex(p3);
}), end(triangleList));
return triangleList;
}
And here's what I obtain :
Triangle:
Point x: 1 y: 0
Point x: 0 y: 0
Point x: 1 y: 1
Triangle:
Point x: 1 y: 0
Point x: 1 y: 1
Point x: 0 y: 1
Triangle:
Point x: 0 y: 0
Point x: 1 y: 1
Point x: 0 y: 1
While this would be the correct output :
Triangle:
Point x: 1 y: 0
Point x: 0 y: 0
Point x: 0 y: 1
Triangle:
Point x: 1 y: 0
Point x: 1 y: 1
Point x: 0 y: 1
I have no idea why there is a triangle with the (0, 0) and the (1, 1).
I need an outside eye to review the code and find out what's going wrong.
All the sources are on my Github repo. Feel free to fork it and to PR your code.
Thanks!
what about this implementation of Paul Bourke's Delaunay triangulation algorithm. Take a look at Triangulate() I have used this source many times without any complains
#include <iostream>
#include <stdlib.h> // for C qsort
#include <cmath>
#include <time.h> // for random
const int MaxVertices = 500;
const int MaxTriangles = 1000;
//const int n_MaxPoints = 10; // for the test programm
const double EPSILON = 0.000001;
struct ITRIANGLE{
int p1, p2, p3;
};
struct IEDGE{
int p1, p2;
};
struct XYZ{
double x, y, z;
};
int XYZCompare(const void *v1, const void *v2);
int Triangulate(int nv, XYZ pxyz[], ITRIANGLE v[], int &ntri);
int CircumCircle(double, double, double, double, double, double, double, double, double&, double&, double&);
using namespace std;
////////////////////////////////////////////////////////////////////////
// CircumCircle() :
// Return true if a point (xp,yp) is inside the circumcircle made up
// of the points (x1,y1), (x2,y2), (x3,y3)
// The circumcircle centre is returned in (xc,yc) and the radius r
// Note : A point on the edge is inside the circumcircle
////////////////////////////////////////////////////////////////////////
int CircumCircle(double xp, double yp, double x1, double y1, double x2,
double y2, double x3, double y3, double &xc, double &yc, double &r){
double m1, m2, mx1, mx2, my1, my2;
double dx, dy, rsqr, drsqr;
/* Check for coincident points */
if(abs(y1 - y2) < EPSILON && abs(y2 - y3) < EPSILON)
return(false);
if(abs(y2-y1) < EPSILON){
m2 = - (x3 - x2) / (y3 - y2);
mx2 = (x2 + x3) / 2.0;
my2 = (y2 + y3) / 2.0;
xc = (x2 + x1) / 2.0;
yc = m2 * (xc - mx2) + my2;
}else if(abs(y3 - y2) < EPSILON){
m1 = - (x2 - x1) / (y2 - y1);
mx1 = (x1 + x2) / 2.0;
my1 = (y1 + y2) / 2.0;
xc = (x3 + x2) / 2.0;
yc = m1 * (xc - mx1) + my1;
}else{
m1 = - (x2 - x1) / (y2 - y1);
m2 = - (x3 - x2) / (y3 - y2);
mx1 = (x1 + x2) / 2.0;
mx2 = (x2 + x3) / 2.0;
my1 = (y1 + y2) / 2.0;
my2 = (y2 + y3) / 2.0;
xc = (m1 * mx1 - m2 * mx2 + my2 - my1) / (m1 - m2);
yc = m1 * (xc - mx1) + my1;
}
dx = x2 - xc;
dy = y2 - yc;
rsqr = dx * dx + dy * dy;
r = sqrt(rsqr);
dx = xp - xc;
dy = yp - yc;
drsqr = dx * dx + dy * dy;
return((drsqr <= rsqr) ? true : false);
}
///////////////////////////////////////////////////////////////////////////////
// Triangulate() :
// Triangulation subroutine
// Takes as input NV vertices in array pxyz
// Returned is a list of ntri triangular faces in the array v
// These triangles are arranged in a consistent clockwise order.
// The triangle array 'v' should be malloced to 3 * nv
// The vertex array pxyz must be big enough to hold 3 more points
// The vertex array must be sorted in increasing x values say
//
// qsort(p,nv,sizeof(XYZ),XYZCompare);
///////////////////////////////////////////////////////////////////////////////
int Triangulate(int nv, XYZ pxyz[], ITRIANGLE v[], int &ntri){
int *complete = NULL;
IEDGE *edges = NULL;
IEDGE *p_EdgeTemp;
int nedge = 0;
int trimax, emax = 200;
int status = 0;
int inside;
int i, j, k;
double xp, yp, x1, y1, x2, y2, x3, y3, xc, yc, r;
double xmin, xmax, ymin, ymax, xmid, ymid;
double dx, dy, dmax;
/* Allocate memory for the completeness list, flag for each triangle */
trimax = 4 * nv;
complete = new int[trimax];
/* Allocate memory for the edge list */
edges = new IEDGE[emax];
/*
Find the maximum and minimum vertex bounds.
This is to allow calculation of the bounding triangle
*/
xmin = pxyz[0].x;
ymin = pxyz[0].y;
xmax = xmin;
ymax = ymin;
for(i = 1; i < nv; i++){
if (pxyz[i].x < xmin) xmin = pxyz[i].x;
if (pxyz[i].x > xmax) xmax = pxyz[i].x;
if (pxyz[i].y < ymin) ymin = pxyz[i].y;
if (pxyz[i].y > ymax) ymax = pxyz[i].y;
}
dx = xmax - xmin;
dy = ymax - ymin;
dmax = (dx > dy) ? dx : dy;
xmid = (xmax + xmin) / 2.0;
ymid = (ymax + ymin) / 2.0;
/*
Set up the supertriangle
his is a triangle which encompasses all the sample points.
The supertriangle coordinates are added to the end of the
vertex list. The supertriangle is the first triangle in
the triangle list.
*/
pxyz[nv+0].x = xmid - 20 * dmax;
pxyz[nv+0].y = ymid - dmax;
pxyz[nv+1].x = xmid;
pxyz[nv+1].y = ymid + 20 * dmax;
pxyz[nv+2].x = xmid + 20 * dmax;
pxyz[nv+2].y = ymid - dmax;
v[0].p1 = nv;
v[0].p2 = nv+1;
v[0].p3 = nv+2;
complete[0] = false;
ntri = 1;
/*
Include each point one at a time into the existing mesh
*/
for(i = 0; i < nv; i++){
xp = pxyz[i].x;
yp = pxyz[i].y;
nedge = 0;
/*
Set up the edge buffer.
If the point (xp,yp) lies inside the circumcircle then the
three edges of that triangle are added to the edge buffer
and that triangle is removed.
*/
for(j = 0; j < ntri; j++){
if(complete[j])
continue;
x1 = pxyz[v[j].p1].x;
y1 = pxyz[v[j].p1].y;
x2 = pxyz[v[j].p2].x;
y2 = pxyz[v[j].p2].y;
x3 = pxyz[v[j].p3].x;
y3 = pxyz[v[j].p3].y;
inside = CircumCircle(xp, yp, x1, y1, x2, y2, x3, y3, xc, yc, r);
if (xc + r < xp)
// Suggested
// if (xc + r + EPSILON < xp)
complete[j] = true;
if(inside){
/* Check that we haven't exceeded the edge list size */
if(nedge + 3 >= emax){
emax += 100;
p_EdgeTemp = new IEDGE[emax];
for (int i = 0; i < nedge; i++) { // Fix by John Bowman
p_EdgeTemp[i] = edges[i];
}
delete []edges;
edges = p_EdgeTemp;
}
edges[nedge+0].p1 = v[j].p1;
edges[nedge+0].p2 = v[j].p2;
edges[nedge+1].p1 = v[j].p2;
edges[nedge+1].p2 = v[j].p3;
edges[nedge+2].p1 = v[j].p3;
edges[nedge+2].p2 = v[j].p1;
nedge += 3;
v[j] = v[ntri-1];
complete[j] = complete[ntri-1];
ntri--;
j--;
}
}
/*
Tag multiple edges
Note: if all triangles are specified anticlockwise then all
interior edges are opposite pointing in direction.
*/
for(j = 0; j < nedge - 1; j++){
for(k = j + 1; k < nedge; k++){
if((edges[j].p1 == edges[k].p2) && (edges[j].p2 == edges[k].p1)){
edges[j].p1 = -1;
edges[j].p2 = -1;
edges[k].p1 = -1;
edges[k].p2 = -1;
}
/* Shouldn't need the following, see note above */
if((edges[j].p1 == edges[k].p1) && (edges[j].p2 == edges[k].p2)){
edges[j].p1 = -1;
edges[j].p2 = -1;
edges[k].p1 = -1;
edges[k].p2 = -1;
}
}
}
/*
Form new triangles for the current point
Skipping over any tagged edges.
All edges are arranged in clockwise order.
*/
for(j = 0; j < nedge; j++) {
if(edges[j].p1 < 0 || edges[j].p2 < 0)
continue;
v[ntri].p1 = edges[j].p1;
v[ntri].p2 = edges[j].p2;
v[ntri].p3 = i;
complete[ntri] = false;
ntri++;
}
}
/*
Remove triangles with supertriangle vertices
These are triangles which have a vertex number greater than nv
*/
for(i = 0; i < ntri; i++) {
if(v[i].p1 >= nv || v[i].p2 >= nv || v[i].p3 >= nv) {
v[i] = v[ntri-1];
ntri--;
i--;
}
}
delete[] edges;
delete[] complete;
return 0;
}
int XYZCompare(const void *v1, const void *v2){
XYZ *p1, *p2;
p1 = (XYZ*)v1;
p2 = (XYZ*)v2;
if(p1->x < p2->x)
return(-1);
else if(p1->x > p2->x)
return(1);
else
return(0);
}
I didn't go with a debugger, but from the resulting triangles it seems that this is an accuracy/ambiguity problem.
When you are triangulating a square there are two ways to split it into triangles and both are OK from Delaunay criteria (circumscribed circle center is on border of triangle).
So if you evaluate every triangle independently you may sometimes get even 4 triangles (depending on implementation).
Normally in such cases I recommend to build algorithm as a series of questions which cannot produce contradicting answers. In this case the question is "to which point goes triangle based on edge (1,0)-(1,1)". But often this requires significant changes to the algorithm.
A quick fix usually involves adding some tolerances for comparisons and extra checks (like non-intersecting triangles). But usually it just makes problems rarer.
Most likely you didn't delete all the double edges, especially not the edges from same triangles but with vertices only in another order. The correct function is in the answer from #cMinor.
I have two points and I would like to compute n evenly distributed points on top of the line created by the given line. How could I perform this in c++?
Linear interpolation (affectionately called lerp by the Graphics community) is what you want. Given the end points it can generate the points lying in between with a parameter t.
Let the end points be A (Ax, Ay) and B (Bx, By). The vector spanning from A to B would be given by
V = B − A = <Vx, Vy>
L(t) = A + tV
This essentially means that starting from the point A, we scale the vector V with the scalar t; the point A is displaced by this scaled vector and thus the point we get depends on the value of t, the parameter. When t = 0, we get back A, if t = 1 we get B, if it's 0.5 we get the point midway between A and B.
line A----|----|----|----B
t 0 ¼ ½ ¾ 1
It works for any line (slope doesn't matter). Now coming to your problem of N stops. If you need N to be 10, then you'd have t vary by 1/N, so t = i/10, where i would be the loop iterator.
i = 0, t = 0
i = 1, t = 0.1
i = 2, t = 0.2
⋮
i = 9, t = 0.9
i = 10, t = 1.0
Here's one way to implement it:
#include <iostream>
struct Point {
float x, y;
};
Point operator+ (Point const &pt1, Point const &pt2) {
return { pt1.x + pt2.x, pt1.y + pt2.y };
}
Point operator- (Point const &pt1, Point const &pt2) {
return { pt1.x - pt2.x, pt1.y - pt2.y };
}
Point scale(Point const &pt, float t) {
return { pt.x * t, pt.y * t };
}
std::ostream& operator<<(std::ostream &os, Point const &pt) {
return os << '(' << pt.x << ", " << pt.y << ')';
}
void lerp(Point const &pt1, Point const &pt2, float stops) {
Point const v = pt2 - pt1;
float t = 0.0f;
for (float i = 0.0f; i <= stops; ++i) {
t = i / stops;
Point const p = pt1 + scale(v, t);
std::cout << p << '\n';
}
}
int main() {
lerp({0.0, 0.0}, {5.0f, 5.0f}, 5.0f);
}
Output
(0, 0)
(1, 1)
(2, 2)
(3, 3)
(4, 4)
(5, 5)
Aside
Notice that on every iteration t gets incremented by Δt = 1 / N. Thus another way to update t in a loop would be
t₀ = 0
t₁ = t₀ + Δt
t₂ = t₁ + Δt
⋮
t₉ = t₈ + Δt
t₁₀ = t₉ + Δt
However, this isn't very parallelizable since every iteration of the loop would depend on the previous iteration.
You can use the following give_uniform_points_between(M, N, num_points) which gives a number of #num_points points between M and N. I assume here that the line is not vertical (see below if the line can be vertical).
std::vector<Point> give_uniform_points_between(const Point& M, const Point& N, const unsigned num_points) {
std::vector<Point> result;
// get equation y = ax + b
float a = (N.y - M.y) / (N.x - M.x);
float b = N.y - a * N.x;
float step = std::fabs(M.x - N.x) / num_points;
for (float x = std::min(M.x, N.x); x < std::max(M.x, N.x); x += step) {
float y = a*x+b;
result.push_back(Point{x,y});
}
return result;
}
Demo : Live on Coliru
and result is :
(-3,9);(-2.3,7.6);(-1.6,6.2);(-0.9,4.8);(-0.2,3.4);(0.5,2);(1.2,0.6);(1.9,-0.8);(2.6,-2.2);(3.3,-3.6);
Explanation
From two points (x1,y1) and (x2,y2) you can guess the line equation which pass through these points.
This equation takes the form a*x + b*y + c = 0 or simply y = a*x + b if you cannot have vertical line
where a = (y2 - y1) / (x2 - x1) and you deduce b as shown in the code.
Then you can just vary x or y along your line starting for the point with a minimum value coordinate.
All these (x,y) points you find are on your line and should be uniformely distributed (thanks to the fixed step).
View the line as (x1,y1) + λ(x2-x1,y2-y1), i.e. the first point, plus a multiple of the vector between them.
When λ=0 you have the first point and when λ=1 you have the second.
So you just want to take n equally distributed values of λ between 0 and 1.
How you do this depends on what you mean by between: are the end points included or not?
For example you could take λ=0/(n-1), λ=1/(n-1), λ=2/(n-1), ... λ=(n-1)/(n-1).
That would give n equally distributed points including the endpoints.
Or you could take λ=1/(n+1), λ=2/(n+1), ... λ=n/(n+1).
That would give n equally distributed points not including the endpoints.
Not so mcuh math though...
vector<Rect> Utils::createReactsOnLine(Point pt1, Point pt2, int numRects, int height, int width){
float x1 = pt1.x;
float y1 = pt1.y;
float x2 = pt2.x;
float y2 = pt2.y;
float x_range = std::abs(x2 - x1);
float y_range = std::abs(y2 - y1);
// Find center points of rects on the line
float x_step_size = x_range / (float)(numRects-1);
float y_step_size = y_range / (float)(numRects-1);
float x_min = std::min(x1,x2);
float y_min = std::min(x1,x2);
float x_max = std::max(x1,x2);
float y_max = std::max(x1,x2);
cout << numRects << endl;
float next_x = x1;
float next_y = y1;
cout << "Next x, y: "<< next_x << "," << next_y << endl;
for(int i = 0; i < numRects-1; i++){
if (x1 < x2)
next_x = next_x + x_step_size;
else
next_x = next_x - x_step_size;
if (y1 < y2)
next_y = next_y + y_step_size;
else
next_y = next_y - y_step_size;
cout << "Next x, y: "<< next_x << "," << next_y << endl;
}
return vector<Rect>();
}
Given the lengths of the sides of 2 triangles. Determine if the second triangle can fit inside the first triangle?
For more detailed info read the full problem statement below:
http://acm.timus.ru/problem.aspx?space=1&num=1566&locale=en
My implementation below tries all the (3!)^2 possible combinations of aligning the bases of the triangles. It then tries to shift the second triangle inside the first triangle while checking that the base of the second triangle doesn't exceed the base of the first triangle.
But I keep getting Wrong Answer(WA) #16.
The case I gave is the second image. It is obvious that if you rotate PQR to align the sides of length 2.77 and 3.0 the third vertex will not be inside triangle ABC. The side of length 4.2 can only be aligned along the side of len 5. Thus this case is satisfied only in the configuration show in the second image.
Can you help me find the bug, suggest some test cases where my algorithm breaks down. Alternative algorithms are also welcome.
#include <cmath>
#include <iostream>
using namespace std;
const double PI = atan(1.0)* 4;
// Traingle ABC (Envelope)
double xa, ya, xb, yb, xc, yc;
// Traingle PQR (Postcard)
double xp, yp, xq, yq, xr, yr;
// Angle between sides AB and AC
double theta;
double signWrtLine(double x1, double y1, double x2, double y2, double x, double y)
{
double A = y2 - y1;
double B = x1 - x2;
double C = -(A * x1 + B * y1);
return (A * x + B * y + C);
}
bool fit()
{
if ((xr > xc) || (yq > yb)) return false;
if (signWrtLine(xa, ya, xb, yb, xq, yq) < 0) {
double d = (yq / tan(theta)) - xq;
return (xr + d <= xc);
}
return (signWrtLine(xa, ya, xb, yb, xq, yq) >= 0 &&
signWrtLine(xb, yb, xc, yc, xq, yq) >= 0 &&
signWrtLine(xc, yc, xa, ya, xq, yq) >= 0);
}
bool fit(double a[], double b[])
{
// generate the 3! permutations of the envelope
// loops i,k
for (int i = 0; i < 3; i++) {
double angle;
double u = a[i], v = a[(i + 1) % 3], w = a[(i + 2) % 3];
for (int k = 0; k < 2; k++) {
switch (k) {
case 0:
xa = 0, ya = 0;
angle = theta = acos((u * u + v * v - w * w) / (2 * u * v));
xb = v * cos(angle), yb = v * sin(angle);
xc = u, yc = 0;
break;
case 1:
// reflect envelope
swap(v, w);
angle = theta = acos((u * u + v * v - w * w) / (2 * u * v));
xb = v * cos(angle), yb = v * sin(angle);
break;
}
// generate the 3! permutations of the postcard
// loops j,k
for (int j = 0; j < 3; j++) {
double angle;
double u = b[j], v = b[(j + 1) % 3], w = b[(j + 2) % 3];
for (int k = 0; k < 2; k++) {
switch (k) {
case 0:
xp = 0, yp = 0;
angle = acos((u * u + v * v - w * w) / (2 * u * v));
xq = v * cos(angle), yq = v * sin(angle);
xr = u, yr = 0;
break;
case 1:
// reflect postcard
swap(v, w);
angle = acos((u * u + v * v - w * w) / (2 * u * v));
xq = v * cos(angle), yq = v * sin(angle);
break;
}
if (fit()) return true;
}
}
}
}
return false;
}
int main()
{
double a[3], b[3];
for (int i = 0; i < 3; i++) cin >> a[i];
for (int i = 0; i < 3; i++) cin >> b[i];
if(fit(a, b)) cout << "YES" << endl;
else cout << "NO" << endl;
return 0;
}
Barycentric coordinates! In detail:
Let the "envelope" triangle have vertices A, B, C; without loss of generality you can place vertex A at the origin and align the side AB with the +x axis. Use the edge lengths of the envelope triangle to find the angle at vertex A, i.e., the angle between the sides AB and AC. Using this angle, you define a new coordinate system (u,v); in this coordinate system the vertex coordinates are A=(0,0), B=(1,0) and C=(0,1).
Now, take the other triangle with vertices A',B',C', and find first the XY coordinates of the 3 vertices for each case of: (A'B', B'C', A'C') aligned with +x coordinate axis. For each such alignment, transform the other two vertices to the UV-coordinate system determined by the envelope triangle. If it happens that both of the other vertices have (u,v) coordinates with 0 <= u,v <= 1 with u+v<=1, the triangle fits within the envelope triangle.
Angle between two sides can be obtained through the sine theorem for planar triangles; though you have to be a bit careful if the angle at a vertex is obtuse (> PI/2) since the sine function is symmetric around PI/2 on the interval [0,PI]. To check whether the angle is obtuse, you also need to use the cosine theorem, though you don't need to calculate the cosine itself: if |AB|^2 + |AC|^2 > |BC|^2, the angle at A is obtuse.
I think that about sums it up.
//23/08/11 13:56
//determine if a triangle will fit inside a second triangle
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
const double pi= 3.1414926;
const double tri=180;//deg in triangle
double find_B_ang(double a,double b,double c);
double find_C_ang(double a,double b,double c);
double movetri_r , tc_ghosthor_B;
int main()
{double a=0.0,b=0.0,c=0.0,x=0.0,y=0.0,z=0.0;
double A=0.0,B=0.0,C=0.0,A1=0.0,B1=0.0,C1=0.0;// L&R base angles
double te_vert_B=0.0,te_hor_B=0.0,te_hor_C=0.0;
double tc_vert_B=0.0,tc_hor_B=0.0,tc_hor_C=0.0;
//points B and B1 are considered co-incedent
cout<<"\n\ndetermine if a triangular card will fit inside\n"
<<"a triangular envelope\n";
//envelope dimensions
cout<<"\nenter lengths of the sides of envelope (space between)\n";
cout<<"ensure longest of them is less than sum of other two\n";
do
{
cin>>a>>b>>c;//the e sides
if(c>a)swap(a,c);//sort sides in decending order
if(b>a)swap(a,b);
if(c>b)swap(b,c);
if(a >(b+c))
cout<<"WRONG...b+c must be greater than a";
}while(a >(b+c));
cout<<"\nthe sides of the envelope are "<<a<<','<<b<<','<<c<<endl;
B=find_B_ang(a,b,c);
C=find_C_ang(a,b,c);
te_vert_B=c*sin(B*pi/tri);//apex to base vertical line
te_hor_B=te_vert_B/tan(B*pi/tri);//B to vertical line
te_hor_C=a-te_hor_B;//C to vertical line
cout<<"-------------------------------------------\n";
//card dimensions
do
{
cout<<"\nenter lengths of sides of card (space between) \n";
cout<<"ensure longest of them is less than sum of other two\n";
do
{
cin>>x>>y>>z;//the c sides
if(z>x)swap(z,x);//sort sides in decending order
if(y>x)swap(y,x);
if(z>y)swap(y,z);
if(x>(y+z))
cout<<"WRONG...y+z must be greater than x\n";
}while(x>(y+z));
cout<<"\nthe sides of card are "<<x<<','<<y<<','<<z<<endl;//x is base
B1=find_B_ang(x,y,z);
C1=find_C_ang(x,y,z);
tc_vert_B=z*sin(B1*pi/tri);//apex to base vertical line
tc_hor_B=tc_vert_B/tan(B1*pi/tri);//B to vertical line
tc_hor_C=x-tc_hor_B;//C to vertical line
tc_ghosthor_B=tc_vert_B/tan(B*pi/tri);
movetri_r= abs(tc_ghosthor_B-tc_hor_B);
cout<<"------------------------------------------------\n";
//determine and advise if card fits within envelope
if(B1<B && tc_vert_B <(tc_hor_C + a-x)*tan(C*pi/tri))cout<<"\ntrue";
else if(B1<B && tc_hor_B< te_hor_B && tc_vert_B<te_vert_B)cout<<"true";
else if(B1>B && movetri_r<a-x && tc_vert_B<te_vert_B)cout<<"true";
else cout<<"\nfalse";
} while(x>0);
cout<<"\npress any key...";
cin.ignore();
cin.get();
return 0;
}
double find_B_ang(double a,double b,double c)
{
double X=0.0;
X=((a*a)+(c*c)-(b*b));
X/=2*a*c;
X=acos(X);
X*=tri/pi;
return X; //degrees
}
double find_C_ang(double a,double b,double c)
{
double X=0.0;
X=((a*a)+(b*b)-(c*c));
X/=2*a*b;
X=acos(X);
X*=tri/pi;
return X;//degrees
}
Use epsilon (1e-10) when comparing doubles!
Might try from here - Link. It seems that the problem is unsolved so far, so best bet to go with some heuristics to get simple cases (like checks for inscribed / circumscribed circles, aligning borders, etc) and hope for the best.