I'm following the book Ray Tracing in on Weekend in which the author produces a small Ray Tracer using plain C++ and the result is a PPM image.
The author's code
Which produces this PPM image.
So the author suggests as an exercise to make it so the program produces a JPG image via the stb_image library. So far I tried by changing the original code like this :
#include <fstream>
#define STB_IMAGE_WRITE_IMPLEMENTATION
#include "stb_image_write.h"
struct RGB{
unsigned char R;
unsigned char G;
unsigned char B;
};
int main(){
int nx = 200;
int ny = 100;
struct RGB data[nx][ny];
for(int j = ny - 1 ; j >= 0 ; j-- ){
for(int i = 0; i < nx ; i++){
float r = float(i) / float(nx);
float g = float(j) / float(ny);
float b = 0.2;
int ir = int(255.99 * r);
int ig = int(255.99 * g);
int ib = int(255.99 * b);
data[i][j].R = ir;
data[i][j].G = ig;
data[i][j].B = ib;
}
}
stbi_write_jpg("image.jpg", nx, ny, 3, data, 100);
}
And this is the result:
As you can see my result is slightly different and I don't know why.
The main problems are:
That the black color shows at the top left of the screen, and in general the colors don't show in the correct order left to right, top to bottom.
The image is "split" in half and the result is actually the original image of the author but produced in a pair ????
Probably I'm misunderstanding something about the way STB_IMAGE_WRITE is supposed to be used so if anyone experienced with this library can tell me what's going on I would be grateful.
EDIT 1 I implemented the changes suggested by #1201ProgramAlarm in comments plus I changed struct RGB data[nx][ny] to struct RGB data[ny][nx] ,so the result now is this.
The library should be working as intended. The problem is how you give the data and what you should do is invert the y axis. So, when you are at index 4 from the start, you should give the color of the index 4 from the end.
Taking the result from your edit, just change the line:
float g = float(j) / float(ny);
to
float g = float(ny - 1 - j) / float(ny);
You've got your indexes wrong for data. The inner loop variable should be the second subscript (data[j][i]).
Related
Unexpectedly for me I faced strange issue:
Here is example of LOGICAL implementation of trivial linear Lagrange interpolation :
unsigned char mix(unsigned char x0, unsigned char x1, float position){
// LOGICALLY must be something like (real implementation should be
// with casts)...
return x0 + (x1 - x0) * position;
}
Arguments x0, x1 are always in range 0 - 255.
Argument position is always in range 0.0f - 1.0f.
Really I tried huge amount of implementations (with different casts and etc.) but it doesn't work in my case! It returns incorrect results (looks like variable overflow or something similar. After looking for solution in internet for a whole week i decided to ask. May be someone has faced similar issues.
I'm using MSVC 2017 compiler (most of parameters are default except language level).
OS - Windows 10x64, Little Endian.
What do i do wrong and what is possible source of the issue?
UPDATED:
It looks like this issue is more deep than I expected (thanks for your responses).
Here is the link to tiny github project which demonstrates my issue:
https://github.com/elRadiance/altitudeMapVisualiser
Output bmp-file should contain smooth altitude map. Instead of it, it contains garbage. If I use just x0 or x1 as result of interpolation function (without interpolation) it works. Without it - doesn't (produces garbage).
Desired result (as here, but in interpolated colors, smooth)
Actual result (updated, best result achieved)
Main class to run it:
#include "Visualiser.h"
int main() {
unsigned int width = 512;
unsigned int height = 512;
float* altitudes = new float[width * height];
float c;
for (int w = 0; w < width; w++) {
c = (2.0f * w / width) - 1.0f;
for (int h = 0; h < height; h++) {
altitudes[w*height + h] = c;
}
}
Visualiser::visualiseAltitudeMap("gggggggggg.bmp", width, height, altitudes);
delete(altitudes);
}
Thank you in advance!
SOLVED: Thankfully #wololo. Mistake in my project was not in calculations.
I should open file with option "binary":
file.open("test.bin", std::ios_base::out | std::ios_base::trunc | std::ios_base::binary);
Without it in some point in data can be faced byte with value 10
In Windows environment it can be processed like LineFeed and changed to 13.
I have a 13 x 13 array of pixels, and I am using a function to draw a circle onto them. (The screen is 13 * 13, which may seem strange, but its an array of LED's so that explains it.)
unsigned char matrix[13][13];
const unsigned char ON = 0x01;
const unsigned char OFF = 0x00;
Here is the first implementation I thought up. (It's inefficient, which is a particular problem as this is an embedded systems project, 80 MHz processor.)
// Draw a circle
// mode is 'ON' or 'OFF'
inline void drawCircle(float rad, unsigned char mode)
{
for(int ix = 0; ix < 13; ++ ix)
{
for(int jx = 0; jx < 13; ++ jx)
{
float r; // Radial
float s; // Angular ("theta")
matrix_to_polar(ix, jx, &r, &s); // Converts polar coordinates
// specified by r and s, where
// s is the angle, to index coordinates
// specified by ix and jx.
// This function just converts to
// cartesian and then translates by 6.0.
if(r < rad)
{
matrix[ix][jx] = mode; // Turn pixel in matrix 'ON' or 'OFF'
}
}
}
}
I hope that's clear. It's pretty simple, but then I programmed it so I know how it's supposed to work. If you'd like more info / explanation then I can add some more code / comments.
It can be considered that drawing several circles, eg 4 to 6, is very slow... Hence I'm asking for advice on a more efficient algorithm to draw the circles.
EDIT: Managed to double the performance by making the following modification:
The function calling the drawing used to look like this:
for(;;)
{
clearAll(); // Clear matrix
for(int ix = 0; ix < 6; ++ ix)
{
rad[ix] += rad_incr_step;
drawRing(rad[ix], rad[ix] - rad_width);
}
if(rad[5] >= 7.0)
{
for(int ix = 0; ix < 6; ++ ix)
{
rad[ix] = rad_space_step * (float)(-ix);
}
}
writeAll(); // Write
}
I added the following check:
if(rad[ix] - rad_width < 7.0)
drawRing(rad[ix], rad[ix] - rad_width);
This increased the performance by a factor of about 2, but ideally I'd like to make the circle drawing more efficient to increase it further. This checks to see if the ring is completely outside of the screen.
EDIT 2: Similarly adding the reverse check increased performance further.
if(rad[ix] >= 0.0)
drawRing(rad[ix], rad[ix] - rad_width);
Performance is now pretty good, but again I have made no modifications to the actual drawing code of the circles and this is what I was intending to focus on with this question.
Edit 3: Matrix to polar:
inline void matrix_to_polar(int i, int j, float* r, float* s)
{
float x, y;
matrix_to_cartesian(i, j, &x, &y);
calcPolar(x, y, r, s);
}
inline void matrix_to_cartesian(int i, int j, float* x, float* y)
{
*x = getX(i);
*y = getY(j);
}
inline void calcPolar(float x, float y, float* r, float* s)
{
*r = sqrt(x * x + y * y);
*s = atan2(y, x);
}
inline float getX(int xc)
{
return (float(xc) - 6.0);
}
inline float getY(int yc)
{
return (float(yc) - 6.0);
}
In response to Clifford that's actually a lot of function calls if they are not inlined.
Edit 4: drawRing just draws 2 circles, firstly an outer circle with mode ON and then an inner circle with mode OFF. I am fairly confident that there is a more efficient method of drawing such a shape too, but that distracts from the question.
You're doing a lot of calculations that aren't really needed. For example, you're calculating the angle of the polar coordinates, but never use it. The square root can also easily be avoided by comparing the square of the values.
Without doing anything fancy, something like this should be a good start:
int intRad = (int)rad;
int intRadSqr = (int)(rad * rad);
for (int ix = 0; ix <= intRad; ++ix)
{
for (int jx = 0; jx <= intRad; ++jx)
{
if (ix * ix + jx * jx <= radSqr)
{
matrix[6 - ix][6 - jx] = mode;
matrix[6 - ix][6 + jx] = mode;
matrix[6 + ix][6 - jx] = mode;
matrix[6 + ix][6 + jx] = mode;
}
}
}
This does all the math in integer format, and takes advantage of the circle symmetry.
Variation of the above, based on feedback in the comments:
int intRad = (int)rad;
int intRadSqr = (int)(rad * rad);
for (int ix = 0; ix <= intRad; ++ix)
{
for (int jx = 0; ix * ix + jx * jx <= radSqr; ++jx)
{
matrix[6 - ix][6 - jx] = mode;
matrix[6 - ix][6 + jx] = mode;
matrix[6 + ix][6 - jx] = mode;
matrix[6 + ix][6 + jx] = mode;
}
}
Don't underestimate the cost of even basic arithmetic using floating point on a processor with no FPU. It seems unlikely that floating point is necessary, but the details of its use are hidden in your matrix_to_polar() implementation.
Your current implementation considers every pixel as a candidate - that is also unnecessary.
Using the equation y = cy ± √[rad2 - (x-cx)2] where cx, cy is the centre (7, 7 in this case), and a suitable integer square root implementation, the circle can be drawn thus:
void drawCircle( int rad, unsigned char mode )
{
int r2 = rad * rad ;
for( int x = 7 - rad; x <= 7 + rad; x++ )
{
int dx = x - 7 ;
int dy = isqrt( r2 - dx * dx ) ;
matrix[x][7 - dy] = mode ;
matrix[x][7 + dy] = mode ;
}
}
In my test I used the isqrt() below based on code from here, but given that the maximum r2 necessary is 169 (132, you could implement a 16 or even 8 bit optimised version if necessary. If your processor is 32 bit, this is probably fine.
uint32_t isqrt(uint32_t n)
{
uint32_t root = 0, bit, trial;
bit = (n >= 0x10000) ? 1<<30 : 1<<14;
do
{
trial = root+bit;
if (n >= trial)
{
n -= trial;
root = trial+bit;
}
root >>= 1;
bit >>= 2;
} while (bit);
return root;
}
All that said, on such a low resolution device, you will probably get better quality circles and faster performance by hand generating bitmap lookup tables for each radius required. If memory is an issue, then a single circle needs only 7 bytes to describe a 7 x 7 quadrant that you can reflect to all three quadrants, or for greater performance you could use 7 x 16 bit words to describe a semi-circle (since reversing bit order is more expensive than reversing array access - unless you are using an ARM Cortex-M with bit-banding). Using semi-circle look-ups, 13 circles would need 13 x 7 x 2 bytes (182 bytes), quadrant look-ups would be 7 x 8 x 13 (91 bytes) - you may find that is fewer bytes that the code space required to calculate the circles.
For a slow embedded device with only a 13x13 element display, you should really just make a look-up table. For example:
struct ComputedCircle
{
float rMax;
char col[13][2];
};
Where the draw routine uses rMax to determine which LUT element to use. For example, if you have 2 elements with one rMax = 1.4f, the other = 1.7f, then any radius between 1.4f and 1.7f will use that entry.
The column elements would specify zero, one, or two line segments per row, which can be encoded in the lower and upper 4 bits of each char. -1 can be used as a sentinel value for nothing-at-this-row. It is up to you how many look-up table entries to use, but with a 13x13 grid you should be able to encode every possible outcome of pixels with well under 100 entries, and a reasonable approximation using only 10 or so. You can also trade off compression for draw speed as well, e.g. putting the col[13][2] matrix in a flat list and encoding the number of rows defined.
I would accept MooseBoy's answer if only he explained the method he proposes better. Here's my take on the lookup table approach.
Solve it with a lookup table
The 13x13 display is quite small, and if you only need circles which are fully visible within this pixel count, you will get around with a quite small table. Even if you need larger circles, it should be still better than any algorithmic way if you need it to be fast (and have the ROM to store it).
How to do it
You basically need to define how each possible circle looks like on the 13x13 display. It is not sufficient to just produce snapshots for the 13x13 display, as it is likely you would like to plot the circles at arbitrary positions. My take for a table entry would look like this:
struct circle_entry_s{
unsigned int diameter;
unsigned int offset;
};
The entry would map a given diameter in pixels to offsets in a large byte table containing the shape of the circles. For example for diameter 9, the byte sequence would look like this:
0x1CU, 0x00U, /* 000111000 */
0x63U, 0x00U, /* 011000110 */
0x41U, 0x00U, /* 010000010 */
0x80U, 0x80U, /* 100000001 */
0x80U, 0x80U, /* 100000001 */
0x80U, 0x80U, /* 100000001 */
0x41U, 0x00U, /* 010000010 */
0x63U, 0x00U, /* 011000110 */
0x1CU, 0x00U, /* 000111000 */
The diameter specifies how many bytes of the table belong to the circle: one row of pixels are generated from (diameter + 7) >> 3 bytes, and the number of rows correspond to the diameter. The output code of these can be made quite fast, while the lookup table is sufficiently compact to get even larger than the 13x13 display circles defined in it if needed.
Note that defining circles this way for odd and even diameters may or may not appeal you when output by a centre location. The odd diameter circles will appear to have a centre in the "middle" of a pixel, while the even diameter circles will appear to have their centre on the "corner" of a pixel.
You may also find it nice later to refine the overall method so having multiple circles of different apparent sizes, but having the same pixel radius. Depends on what is your goal: if you want some kind of smooth animation, you may get there eventually.
Algorithmic solutions I think mostly will perform poorly here, since with this limited display surface really every pixel's state counts for the appearance.
I am trying to implement some function like below
For this I am trying to use Cubic interpolation and Catmull interpolation ( check both separately to compare the best result) , what i am not understanding is what impact these interpolation show on image and how we can get these points values where we clicked to set that curve ? and do we need to define the function these black points on the image separately ?
I am getting help from these resources
Source 1
Source 2
Approx the same focus
Edit
int main (int argc, const char** argv)
{
Mat input = imread ("E:\\img2.jpg");
for(int i=0 ; i<input.rows ; i++)
{
for (int p=0;p<input.cols;p++)
{
//for(int t=0; t<input.channels(); t++)
//{
input.at<cv::Vec3b>(i,p)[0] = 255*correction(input.at<cv::Vec3b>(i,p)[0]/255.0,ctrl,N); //B
input.at<cv::Vec3b>(i,p)[1] = 255*correction(input.at<cv::Vec3b>(i,p)[1]/255.0,ctrl,N); //G
input.at<cv::Vec3b>(i,p)[2] = 255*correction(input.at<cv::Vec3b>(i,p)[2]/255.0,ctrl,N); //R
//}
}
}
imshow("image" , input);
waitKey();
}
So if your control points are always on the same x coordinate
and linearly dispersed along whole range then you can do it like this:
//---------------------------------------------------------------------------
const int N=5; // number of control points (must be >= 4)
float ctrl[N]= // control points y values initiated with linear function y=x
{ // x value is index*1.0/(N-1)
0.00,
0.25,
0.50,
0.75,
1.00,
};
//---------------------------------------------------------------------------
float correction(float col,float *ctrl,int n)
{
float di=1.0/float(n-1);
int i0,i1,i2,i3;
float t,tt,ttt;
float a0,a1,a2,a3,d1,d2;
// find start control point
col*=float(n-1);
i1=col; col-=i1;
i0=i1-1; if (i0< 0) i0=0;
i2=i1+1; if (i2>=n) i2=n-1;
i3=i1+2; if (i3>=n) i3=n-1;
// compute interpolation coefficients
d1=0.5*(ctrl[i2]-ctrl[i0]);
d2=0.5*(ctrl[i3]-ctrl[i1]);
a0=ctrl[i1];
a1=d1;
a2=(3.0*(ctrl[i2]-ctrl[i1]))-(2.0*d1)-d2;
a3=d1+d2+(2.0*(-ctrl[i2]+ctrl[i1]));
// now interpolate new colro intensity
t=col; tt=t*t; ttt=tt*t;
t=a0+(a1*t)+(a2*tt)+(a3*ttt);
return t;
}
//---------------------------------------------------------------------------
It uses 4-point 1D interpolation cubic (from that link in my comment above) to get new color just do this:
new_col = correction(old_col,ctrl,N);
this is how it looks:
the green arrows shows derivation error (always only on start and end point of whole curve). It can be corrected by adding 2 more control points one before and one after all others ...
[Notes]
color range is < 0.0 , 1.0 > so if you need other then just multiply the result and divide the input ...
[edit1] the start/end derivations fixed a little
float correction(float col,float *ctrl,int n)
{
float di=1.0/float(n-1);
int i0,i1,i2,i3;
float t,tt,ttt;
float a0,a1,a2,a3,d1,d2;
// find start control point
col*=float(n-1);
i1=col; col-=i1;
i0=i1-1;
i2=i1+1; if (i2>=n) i2=n-1;
i3=i1+2;
// compute interpolation coefficients
if (i0>=0) d1=0.5*(ctrl[i2]-ctrl[i0]); else d1=ctrl[i2]-ctrl[i1];
if (i3< n) d2=0.5*(ctrl[i3]-ctrl[i1]); else d2=ctrl[i2]-ctrl[i1];
a0=ctrl[i1];
a1=d1;
a2=(3.0*(ctrl[i2]-ctrl[i1]))-(2.0*d1)-d2;
a3=d1+d2+(2.0*(-ctrl[i2]+ctrl[i1]));
// now interpolate new colro intensity
t=col; tt=t*t; ttt=tt*t;
t=a0+(a1*t)+(a2*tt)+(a3*ttt);
return t;
}
[edit2] just some clarification on the coefficients
they are all derived from this conditions:
y(t) = a0 + a1*t + a2*t*t + a3*t*t*t // direct value
y'(t) = a1 + 2*a2*t + 3*a3*t*t // first derivation
now you have points y0,y1,y2,y3 so I chose that y(0)=y1 and y(1)=y2 which gives c0 continuity (value is the same in the joint points between curves)
now I need c1 continuity so i add y'(0) must be the same as y'(1) from previous curve.
for y'(0) I choose avg direction between points y0,y1,y2
for y'(1) I choose avg direction between points y1,y2,y3
These are the same for the next/previous segments so it is enough. Now put it all together:
y(0) = y0 = a0 + a1*0 + a2*0*0 + a3*0*0*0
y(1) = y1 = a0 + a1*1 + a2*1*1 + a3*1*1*1
y'(0) = 0.5*(y2-y0) = a1 + 2*a2*0 + 3*a3*0*0
y'(1) = 0.5*(y3-y1) = a1 + 2*a2*1 + 3*a3*1*1
And solve this system of equtions (a0,a1,a2,a3 = ?). You will get what I have in source code above. If you need different properties of the curve then just make different equations ...
[edit3] usage
pic1=pic0; // copy source image to destination pic is mine image class ...
for (y=0;y<pic1.ys;y++) // go through all pixels
for (x=0;x<pic1.xs;x++)
{
float i;
// read, convert, write pixel
i=pic1.p[y][x].db[0]; i=255.0*correction(i/255.0,red control points,5); pic1.p[y][x].db[0]=i;
i=pic1.p[y][x].db[1]; i=255.0*correction(i/255.0,green control points,5); pic1.p[y][x].db[1]=i;
i=pic1.p[y][x].db[2]; i=255.0*correction(i/255.0,blue control points,5); pic1.p[y][x].db[2]=i;
}
On top there are control points per R,G,B. On bottom left is original image and on bottom right is corrected image.
I am working with depth images retrieved from kinect which are 16 bits. I found some difficulties on making my own filters due to the index or the size of the images.
I am working with Textures because allows to work with any bit size of images.
So, I am trying to compute an easy gradient to understand what is wrong or why it doesn't work as I expected.
You can see that there is something wrong when I use y dir.
For x:
For y:
That's my code:
typedef concurrency::graphics::texture<unsigned int, 2> TextureData;
typedef concurrency::graphics::texture_view<unsigned int, 2> Texture
cv::Mat image = cv::imread("Depth247.tiff", CV_LOAD_IMAGE_ANYDEPTH);
//just a copy from another image
cv::Mat image2(image.clone() );
concurrency::extent<2> imageSize(640, 480);
int bits = 16;
const unsigned int nBytes = imageSize.size() * 2; // 614400
{
uchar* data = image.data;
// Result data
TextureData texDataD(imageSize, bits);
Texture texR(texDataD);
parallel_for_each(
imageSize,
[=](concurrency::index<2> idx) restrict(amp)
{
int x = idx[0];
int y = idx[1];
// 65535 is the maxium value that can take a pixel with 16 bits (2^16 - 1)
int valX = (x / (float)imageSize[0]) * 65535;
int valY = (y / (float)imageSize[1]) * 65535;
texR.set(idx, valX);
});
//concurrency::graphics::copy(texR, image2.data, imageSize.size() *(bits / 8u));
concurrency::graphics::copy_async(texR, image2.data, imageSize.size() *(bits) );
cv::imshow("result", image2);
cv::waitKey(50);
}
Any help will be very appreciated.
Your indexes are swapped in two places.
int x = idx[0];
int y = idx[1];
Remember that C++AMP uses row-major indices for arrays. Thus idx[0] refers to row, y axis. This is why the picture you have for "For x" looks like what I would expect for texR.set(idx, valY).
Similarly the extent of image is also using swapped values.
int valX = (x / (float)imageSize[0]) * 65535;
int valY = (y / (float)imageSize[1]) * 65535;
Here imageSize[0] refers to the number of columns (the y value) not the number of rows.
I'm not familiar with OpenCV but I'm assuming that it also uses a row major format for cv::Mat. It might invert the y axis with 0, 0 top-left not bottom-left. The Kinect data may do similar things but again, it's row major.
There may be other places in your code that have the same issue but I think if you double check how you are using index and extent you should be able to fix this.
In my current program I would like to be able to draw "DNA Shapes". I have written a "DrawPixel(x,y,r,g,b)" function which can draw a pixel on the screen. Moving on from this point I implemented the Bresenham line algorithm to draw a line as: "DrawLine(x1,y1,x2,y2,r,g,b)".
Now I realized that using an image for the DNA shapes would be a very bad choice (in multiple aspects). So I tried making a function to draw a DNA shape (As I couldn't find an algorithm). This is currently based on a circle drawing algorithm (Midpoint Circle Algorithm):
void D3DGHandler::DrawDNAShape(int x1, int y1, int length, int curves, int dir
int off, int r, int g, int b){
int x2Pos = sin(dir)*length+x1;
int y2Pos = cos(dir)*length+y1;
for (int i = 0; i < curves; i++) {
int xIncrease = (x2Pos / curves) * i;
int yIncrease = (y2Pos / curves) * i;
int rSquared = off * off;
int xPivot = (int)(off * 0.707107 + 0.5f);
for (int x = 0; x <= xPivot; x++) {
int y = (int)(sqrt((float)(rSquared - x*x)) + 0.5f);
DrawPixel(x1+x+xIncrease,y1+y+yIncrease,r,g,b);
DrawPixel(x1-x+xIncrease,y1+y+yIncrease,r,g,b);
DrawPixel(x1+x+xIncrease,y1-y+yIncrease,r,g,b);
DrawPixel(x1-x+xIncrease,y1-y+yIncrease,r,g,b);
DrawPixel(x1+y+xIncrease,y1+x+yIncrease,r,g,b);
DrawPixel(x1-y+xIncrease,y1+x+yIncrease,r,g,b);
DrawPixel(x1+y+xIncrease,y1-x+yIncrease,r,g,b);
DrawPixel(x1-y+xIncrease,y1-x+yIncrease,r,g,b);
}
}
}
This implementation is currently getting me some completely new functionality I was not looking for.
Along the lines of:
I would be very happy to hear any information you can give me!
Update
Expected result:
Expected Result
But then way more line-like.
You can draw two sinusoidal graphs with some offset will give you the required shape.
Eg. In R
x=(1:100)/10.0
plot(sin(x),x)
points(sin(x+2.5),x)