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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 am trying to wrap my head around Perlin noise.
This article has helped and I have been trying to recreate the cloud type images that it provides.
My noise code is as follows:
#include "terrain_generator.hpp"
using namespace std;
#define PI 3.1415927;
float noise(int x, int y)
{
int n = x + y * 57;
n = (n<<13) ^ n;
return (1.0 - ( (n * ((n * n * 15731) + 789221) + 1376312589) & 0x7fffffff) / 1073741824.0);
}
float cosine_interpolate(float a, float b, float x)
{
float ft = x * PI;
float f = (1 - cos(ft)) * 0.5;
float result = a*(1-f) + b*f;
return result;
}
float smooth_noise_2D(float x, float y)
{
float corners = ( noise(x-1, y-1)+noise(x+1, y-1)+noise(x-1, y+1)+noise(x+1, y+1) ) / 16;
float sides = ( noise(x-1, y) +noise(x+1, y) +noise(x, y-1) +noise(x, y+1) ) / 8;
float center = noise(x, y) / 4;
return corners + sides + center;
}
float interpolated_noise(float x, float y)
{
int x_whole = (int) x;
float x_frac = x - x_whole;
int y_whole = (int) y;
float y_frac = y - y_whole;
float v1 = smooth_noise_2D(x_whole, y_whole);
float v2 = smooth_noise_2D(x_whole, y_whole+1);
float v3 = smooth_noise_2D(x_whole+1, y_whole);
float v4 = smooth_noise_2D(x_whole+1, y_whole+1);
float i1 = cosine_interpolate(v1,v3,x_frac);
float i2 = cosine_interpolate(v2,v4,x_frac);
return cosine_interpolate(i1, i2, y_frac);
}
float perlin_noise_2D(float x, float y)
{
int octaves=5;
float persistence=0.5;
float total = 0;
for(int i=0; i<octaves-1; i++)
{
float frequency = pow(2,i);
float amplitude = pow(persistence,i);
total = total + interpolated_noise(x * frequency, y * frequency) * amplitude;
}
return total;
}
To actually implement the algorithm, I am trying to make the clouds he depicted in the article.
I am using openGL and I am making my own texture and pasting it onto a quad that covers the screen. That is irrelevant though. In the code below, just know that the set pixel function works correctly and that its parameters are (x, y, red, green, blue).
This is essentially my draw loop:
for(int y=0; y<texture_height; y++)
{
for(int x=0; x<texture_width; x++)
{
seed2+=1;
float Val=perlin_noise_2D(x,y);
Val = Val/2.0;
Val = (Val + 1.0) / 2.0;
setPixel(x,y,Val,Val,Val);
}
}
What I get is the following:
How can I manipulate my algorithm to achieve the effect I am looking for? changing the persistence or number of octaves doesn't seem to do much at all.
As your result looks almost like white noise, your samples are probably too far apart within the perlin noise. Try using something smaller than the pixel coordinates to evaluate the noise at.
Something similar to this:
perlin_noise_2D((float)x/texture_width,(float)y/texture_height);
I have points in an image. I need to detect the most collinear points. The fastest method is to use Hough transform, but I have to modify the opencv method. Actually I need that the semi collinear points to be returned with detected line, for this reason I modified the polar line struct. A tolerance threshold is also needed to detect nearly detected points as shown in the image. Can someone help in how to tune this threshold?
I need at least four semi collinear points to detect the line to which they belong.
The points of first image were detected by 6 overlapped lines
the point of middle images were detected by nothing
the third's points
were detected by three lines
Which is the best way to get rid from the overlapped liens?? Or how to tune the tolerance threshold to detect the semi collinear points by only one line?
the is my own function call:
vector<CvLinePolar2> lines;
CvMat c_image = source1; // loaded image
HoughLinesStandard(&c_image,1,CV_PI/180,4,&lines,INT_MAX);
typedef struct CvLinePolar2
{
float rho;
float angle;
vector<CvPoint> points;
};
void HoughLinesStandard( const CvMat* img, float rho, float theta,
int threshold, vector<CvLinePolar2> *lines, int linesMax= INT_MAX )
{
cv::AutoBuffer<int> _accum, _sort_buf;
cv::AutoBuffer<float> _tabSin, _tabCos;
const uchar* image;
int step, width, height;
int numangle, numrho;
int total = 0;
int i, j;
float irho = 1 / rho;
double scale;
vector<vector<CvPoint>> lpoints;
CV_Assert( CV_IS_MAT(img) && CV_MAT_TYPE(img->type) == CV_8UC1 );
image = img->data.ptr;
step = img->step;
width = img->cols;
height = img->rows;
numangle = cvRound(CV_PI / theta);
numrho = cvRound(((width + height) * 2 + 1) / rho);
_accum.allocate((numangle+2) * (numrho+2));
_sort_buf.allocate(numangle * numrho);
_tabSin.allocate(numangle);
_tabCos.allocate(numangle);
int *accum = _accum, *sort_buf = _sort_buf;
float *tabSin = _tabSin, *tabCos = _tabCos;
memset( accum, 0, sizeof(accum[0]) * (numangle+2) * (numrho+2) );
//memset( lpoints, 0, sizeof(lpoints) );
lpoints.resize(sizeof(accum[0]) * (numangle+2) * (numrho+2));
float ang = 0;
for(int n = 0; n < numangle; ang += theta, n++ )
{
tabSin[n] = (float)(sin(ang) * irho);
tabCos[n] = (float)(cos(ang) * irho);
}
// stage 1. fill accumulator
for( i = 0; i < height; i++ )
for( j = 0; j < width; j++ )
{
if( image[i * step + j] != 0 )
{
CvPoint pt;
pt.x = j; pt.y = i;
for(int n = 0; n < numangle; n++ )
{
int r = cvRound( j * tabCos[n] + i * tabSin[n] );
r += (numrho - 1) / 2;
int ind = (n+1) * (numrho+2) + r+1;
int s = accum[ind];
accum[ind]++;
lpoints[ind].push_back(pt);
}
}
}
// stage 2. find local maximums
for(int r = 0; r < numrho; r++ )
for(int n = 0; n < numangle; n++ )
{
int base = (n+1) * (numrho+2) + r+1;
if( accum[base] > threshold &&
accum[base] > accum[base - 1] && accum[base] >= accum[base + 1] &&
accum[base] > accum[base - numrho - 2] && accum[base] >= accum[base + numrho + 2] )
sort_buf[total++] = base;
}
// stage 3. sort the detected lines by accumulator value
icvHoughSortDescent32s( sort_buf, total, accum );
// stage 4. store the first min(total,linesMax) lines to the output buffer
linesMax = MIN(linesMax, total);
scale = 1./(numrho+2);
for( i = 0; i < linesMax; i++ )
{
CvLinePolar2 line;
int idx = sort_buf[i];
int n = cvFloor(idx*scale) - 1;
int r = idx - (n+1)*(numrho+2) - 1;
line.rho = (r - (numrho - 1)*0.5f) * rho;
line.angle = n * theta;
line.points = lpoints[idx];
lines->push_back(line);
}
}
One approach is non-maximal suppression to thin out the candidate set for potential lines. Once you've identified the thinned potential lines you could then compute an average of the remaining lines that would satisfy some angular or spatial difference threshold.
Try HoughLinesP..opencv reference
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.
I'd like to implement a Bézier curve. I've done this in C# before, but I'm totally unfamiliar with the C++ libraries. How should I go about creating a quadratic curve?
void printQuadCurve(float delta, Vector2f p0, Vector2f p1, Vector2f p2);
Clearly we'd need to use linear interpolation, but does this exist in the standard math library? If not, where can I find it?
I'm using Linux.
Recently I ran across the same question and wanted to implemented it on my own.
This image from Wikipedia helped me:
The following code is written in C++ and shows how to compute a quadratic bezier.
int getPt( int n1 , int n2 , float perc )
{
int diff = n2 - n1;
return n1 + ( diff * perc );
}
for( float i = 0 ; i < 1 ; i += 0.01 )
{
// The Green Line
xa = getPt( x1 , x2 , i );
ya = getPt( y1 , y2 , i );
xb = getPt( x2 , x3 , i );
yb = getPt( y2 , y3 , i );
// The Black Dot
x = getPt( xa , xb , i );
y = getPt( ya , yb , i );
drawPixel( x , y , COLOR_RED );
}
With (x1|y1), (x2|y2) and (x3|y3) being P0, P1 and P2 in the image. Just for showing the basic idea...
For the ones who ask for the cubic bezier, it just works analogue (also from Wikipedia):
This answer provides Code for it.
Here is a general implementation for a curve with any number of points.
vec2 getBezierPoint( vec2* points, int numPoints, float t ) {
vec2* tmp = new vec2[numPoints];
memcpy(tmp, points, numPoints * sizeof(vec2));
int i = numPoints - 1;
while (i > 0) {
for (int k = 0; k < i; k++)
tmp[k] = tmp[k] + t * ( tmp[k+1] - tmp[k] );
i--;
}
vec2 answer = tmp[0];
delete[] tmp;
return answer;
}
Note that it uses heap memory for a temporary array which is not all that efficient. If you only need to deal with a fixed number of points you could hard-code the numPoints value and use stack memory instead.
Of course, the above assumes you have a vec2 structure and operators for it like this:
struct vec2 {
float x, y;
vec2(float x, float y) : x(x), y(y) {}
};
vec2 operator + (vec2 a, vec2 b) {
return vec2(a.x + b.x, a.y + b.y);
}
vec2 operator - (vec2 a, vec2 b) {
return vec2(a.x - b.x, a.y - b.y);
}
vec2 operator * (float s, vec2 a) {
return vec2(s * a.x, s * a.y);
}
You have a choice between de Casteljau's method, which is to recursively split the control path until you arrive at the point using a linear interpolation, as explained above, or Bezier's method which is to blend the control points.
Bezier's method is
p = (1-t)^3 *P0 + 3*t*(1-t)^2*P1 + 3*t^2*(1-t)*P2 + t^3*P3
for cubics and
p = (1-t)^2 *P0 + 2*(1-t)*t*P1 + t*t*P2
for quadratics.
t is usually on 0-1 but that's not an essential - in fact the curves extend to infinity. P0, P1, etc are the control points. The curve goes through the two end points but not usually through the other points.
Did you use a C# library earlier?
In C++, no standard library function for Bezier curves is available (yet). You can of course roll your own (CodeProject sample) or look for a math library.
This blogpost explains the idea nicely but in Actionscript. Translation should not be much of a problem.
To get an individual point (x, y) along a cubic curve at a given percent of travel (t), with given control points (x1, y1), (x2, y2), (x3, y3), and (x4, y4) I expanded De Casteljau’s algorithm and rearranged the equation to minimize exponents:
//Generalized code, not C++
variables passed to function: t, x1, y1, x2, y2, x3, y3, x4, y4
variables declared in function: t2, t3, x, y
t2 = t * t
t3 = t * t * t
x = t3*x4 + (3*t2 - 3*t3)*x3 + (3*t3 - 6*t2 + 3*t)*x2 + (3*t2 - t3 - 3*t + 1)*x1
y = t3*y4 + (3*t2 - 3*t3)*y3 + (3*t3 - 6*t2 + 3*t)*y2 + (3*t2 - t3 - 3*t + 1)*y1
(t) is a decimal value between 0 and 1 (0 <= t <= 1) that represents percent of travel along the curve.
The formula is the same for x and y, and you can write a function that takes a generic set of 4 control points or group the coefficients together:
t2 = t * t
t3 = t * t * t
A = (3*t2 - 3*t3)
B = (3*t3 - 6*t2 + 3*t)
C = (3*t2 - t3 - 3*t + 1)
x = t3*x4 + A*x3 + B*x2 + C*x1
y = t3*y4 + A*y3 + B*y2 + C*y1
For quadratic functions, a similar approach yields:
t2 = t * t
A = (2*t - 2*t2)
B = (t2 - 2*t + 1)
x = t2*x3 + A*x2 + B*x1
y = t2*y3 + A*y2 + B*y1
If you just want to display a Bezier curve, you can use something like PolyBezier for Windows.
If you want to implement the routine yourself, you can find linear interpolation code all over the Intarnetz.
I believe the Boost libraries have support for this. Linear interpolation, not Beziers specifically. Don't quote me on this, however.
I made an implementation based on this example https://stackoverflow.com/a/11435243/15484522 but for any amount of path points
void bezier(int [] arr, int size, int amount) {
int a[] = new int[size * 2];
for (int i = 0; i < amount; i++) {
for (int j = 0; j < size * 2; j++) a[j] = arr[j];
for (int j = (size - 1) * 2 - 1; j > 0; j -= 2)
for (int k = 0; k <= j; k++)
a[k] = a[k] + ((a[k+2] - a[k]) * i) / amount;
circle(a[0], a[1], 3); // draw a circle, in Processing
}
}
Where arr is array of points {x1, y1, x2, y2, x3, y3... xn, yn}, size is amount of points (twice smaller than array size), and amount is number of output points.
For optimal calculations you can use 2^n amount and bit shift:
void bezier(int [] arr, int size, int dense) {
int a[] = new int[size * 2];
for (int i = 0; i < (1 << dense); i++) {
for (int j = 0; j < size * 2; j++) a[j] = arr[j];
for (int j = (size - 1) * 2 - 1; j > 0; j -= 2)
for (int k = 0; k <= j; k++)
a[k] = a[k] + (((a[k+2] - a[k]) * i) >> dense);
circle(a[0], a[1], 3); // draw a circle, in Processing
}
}
This implementation on github shows how to calculate a simple cubic bezier, with normal and tangent values for values of 't' from 0->1.
It is a direct transposition of the formulas at wikipedia.