I've submitted this as a potential bug on OpenImaj's github issues board, but wanted to reach out on SO and see if this has been encountered before (it's not an obscure class / method and I would be surprised if I am the first to come across this issue). I realize this is less of a question and more of a raised issue, but this would warrant comment from others in their experience with this method.
https://github.com/openimaj/openimaj/issues/86
While attempting to resize photo's of various uncontrolled sizes, I encountered a divide by zero exception. The relevant portion of the stacktrace is below:
Exception in thread "main" java.lang.ArithmeticException: / by zero
at org.openimaj.image.processing.resize.ResizeProcessor.resizeMaxArea (ResizeProcessor.java:443)
Upon further review, it appears to only occur when the image height is greater than the width. In such a circumstance, the whRatio float ends up being 0 as it performs integer division on line 441. That then causes newWidth on line 442 to be 0 which then causes the exception on line 443.
I don't know anything about the algorithm in general, so I don't know if this change produces correct values (I'm inclined to think not due to changes made to images via this method result in loss of aspect ratio), however if you cast the width / height division to float prior to setting the result to whRatio, it no longer produces a 0 value (see code below).
final int width = 2687;
final int height = 3356;
final int area = width * height;
final int maxArea = 3700000;
final float whRatio = (float) width / height;
System.out.println("whRatio: " + whRatio);
final double newWidth = Math.sqrt(maxArea * whRatio);
System.out.println("New width: " + newWidth);
Related
I'm trying to create a randomly generated "planet" (circle), and I want the areas of water, land and foliage to be decided by perlin noise, or something similar. Currently I have this (psudo)code:
for (int radius = 0; radius < circleRadius; radius++) {
for (float theta = 0; theta < TWO_PI; theta += 0.1) {
float x = radius * cosine(theta);
float y = radius * sine(theta);
int colour = whateverFunctionIMake(x, y);
setPixel(x, y, colour);
}
}
Not only does this not work (there are "gaps" in the circle because of precision issues), it's incredibly slow. Even if I increase the resolution by changing the increment to 0.01, it still has missing pixels and is even slower (I get 10fps on my mediocre computer using Java (I know not the best) and an increment of 0.01. This is certainly not acceptable for a game).
How might I achieve a similar result whilst being much less computationally expensive?
Thanks in advance.
Why not use:
(x-x0)^2 + (y-y0)^2 <= r^2
so simply:
int x0=?,y0=?,r=?; // your planet position and size
int x,y,xx,rr,col;
for (rr=r*r,x=-r;x<=r;x++)
for (xx=x*x,y=-r;y<=r;y++)
if (xx+(y*y)<=rr)
{
col = whateverFunctionIMake(x, y);
setPixel(x0+x, y0+y, col);
}
all on integers, no floating or slow operations, no gaps ... Do not forget to use randseed for the coloring function ...
[Edit1] some more stuff
Now if you want speed than you need direct pixel access (in most platforms Pixels, SetPixel, PutPixels etc are slooow. because they perform a lot of stuff like range checking, color conversions etc ... ) In case you got direct pixel access or render into your own array/image whatever you need to add clipping with screen (so you do not need to check if pixel is inside screen on each pixel) to avoid access violations if your circle is overlapping screen.
As mentioned in the comments you can get rid of the x*x and y*y inside loop using previous value (as both x,y are only incrementing). For more info about it see:
32bit SQRT in 16T without multiplication
the math is like this:
(x+1)^2 = (x+1)*(x+1) = x^2 + 2x + 1
so instead of xx = x*x we just do xx+=x+x+1 for not incremented yet x or xx+=x+x-1 if x is already incremented.
When put all together I got this:
void circle(int x,int y,int r,DWORD c)
{
// my Pixel access
int **Pixels=Main->pyx; // Pixels[y][x]
int xs=Main->xs; // resolution
int ys=Main->ys;
// circle
int sx,sy,sx0,sx1,sy0,sy1; // [screen]
int cx,cy,cx0, cy0 ; // [circle]
int rr=r*r,cxx,cyy,cxx0,cyy0; // [circle^2]
// BBOX + screen clip
sx0=x-r; if (sx0>=xs) return; if (sx0< 0) sx0=0;
sy0=y-r; if (sy0>=ys) return; if (sy0< 0) sy0=0;
sx1=x+r; if (sx1< 0) return; if (sx1>=xs) sx1=xs-1;
sy1=y+r; if (sy1< 0) return; if (sy1>=ys) sy1=ys-1;
cx0=sx0-x; cxx0=cx0*cx0;
cy0=sy0-y; cyy0=cy0*cy0;
// render
for (cxx=cxx0,cx=cx0,sx=sx0;sx<=sx1;sx++,cxx+=cx,cx++,cxx+=cx)
for (cyy=cyy0,cy=cy0,sy=sy0;sy<=sy1;sy++,cyy+=cy,cy++,cyy+=cy)
if (cxx+cyy<=rr)
Pixels[sy][sx]=c;
}
This renders a circle with radius 512 px in ~35ms so 23.5 Mpx/s filling on mine setup (AMD A8-5500 3.2GHz Win7 64bit single thread VCL/GDI 32bit app coded by BDS2006 C++). Just change the direct pixel access to style/api you use ...
[Edit2]
to measure speed on x86/x64 you can use RDTSC asm instruction here some ancient C++ code I used ages ago (on 32bit environment without native 64bit stuff):
double _rdtsc()
{
LARGE_INTEGER x; // unsigned 64bit integer variable from windows.h I think
DWORD l,h; // standard unsigned 32 bit variables
asm {
rdtsc
mov l,eax
mov h,edx
}
x.LowPart=l;
x.HighPart=h;
return double(x.QuadPart);
}
It returns clocks your CPU has elapsed since power up. Beware you should account for overflows as on fast machines the 32bit counter is overflowing in seconds. Also each core has separate counter so set affinity to single CPU. On variable speed clock before measurement heat upi CPU by some computation and to convert to time just divide by CPU clock frequency. To obtain it just do this:
t0=_rdtsc()
sleep(250);
t1=_rdtsc();
fcpu = (t1-t0)*4;
and measurement:
t0=_rdtsc()
mesured stuff
t1=_rdtsc();
time = (t1-t0)/fcpu
if t1<t0 you overflowed and you need to add the a constant to result or measure again. Also the measured process must take less than overflow period. To enhance precision ignore OS granularity. for more info see:
Measuring Cache Latencies
Cache size estimation on your system? setting affinity example
Negative clock cycle measurements with back-to-back rdtsc?
How do you get the pixel location and image size inside the fragment shader for the WebGL version of a FabricJS filter? If you just have to pass it in as a uniform is the information available inside the filter or does it need to be passed to the filter as an option?
Ok so, i ll try to navigate in the code and remember, i do not touch that part of code often.
In the base filter class there is a method called createProgram.
This method is going to return an object that contains a property called uniformLocations.
This property holds an object that is returned from each filter getUniformLocations method. This method again returns the object using the gl.getUniformLocation from the webgl context.
Now fabric has a convention that in the uniforms uStepW and uStepH you will have what percentage of the unit is a pixel.
In the webgl context the width and height are 1, a pixel is like uStepH by uStepW, that are tipically float like 0.001231432432 and i think the image is in pixels:
width: 1 / uStepW
height: 1 / uStepH
fabric also uses vTexCoord varying to indicate the position (i think because is pretty standard in webgl tutorials all over the internet). So the pixel position is likely:
x: vTextCoord.x / uStepW
y: vTextCoord.y / uStepH
You can look at the webgl filter pixelate in the fabricJS repo to better understand how to access those variables in your own filter.
In that filter the method getUniformLocations is returning uStepW and uStepH, but that is unnecessary since the basic filter is going to do the same for us.
So those informations are always available and a fabricjs fragment shader that uses them looks like this:
/**
* Fragment source for the Pixelate program
*/
fragmentSource: 'precision highp float;\n' +
'uniform sampler2D uTexture;\n' +
'uniform float uBlocksize;\n' +
'uniform float uStepW;\n' +
'uniform float uStepH;\n' +
'varying vec2 vTexCoord;\n' +
'void main() {\n' +
'float blockW = uBlocksize * uStepW;\n' +
'float blockH = uBlocksize * uStepW;\n' +
'int posX = int(vTexCoord.x / blockW);\n' +
'int posY = int(vTexCoord.y / blockH);\n' +
'float fposX = float(posX);\n' +
'float fposY = float(posY);\n' +
'vec2 squareCoords = vec2(fposX * blockW, fposY * blockH);\n' +
'vec4 color = texture2D(uTexture, squareCoords);\n' +
'gl_FragColor = color;\n' +
'}',
Now, said so, i do not think you need to know those things in a webgl filter, unless you are doing something particular.
This kind of question to me smells of the XY problem, but since i'm not a WEBGL expert at all, i may be wrong. If you are trying to build a filter and you think you need these informations to progress further, you could do better at explaining also what are you trying to build.
I want to have a fullscreen mode that keeps the aspect ratio by adding black bars on either side. I tried just creating a display mode, but I can't make it fullscreen unless it's a pre-approved resolution, and when I use a bigger diaplay than the native resolution the pixels become messed up, and lines appeared between all of the tiles in the game for some reason.
I think I need to use FBOs to render the scenario to a texture instead of the window, and then just use a fullscreen approved resolution and render the texture properly stretched out in the center of the screen, but I just don't understand how to render to a texture in order to do that, or how to stretch an image. Could someone please help me?
EDIT
I got fullscreen working, but it makes everything all broken looking There are random lines on the edges of anything that's written to the window. There are no glitchy lines when it's in native resolution though. Here's my code:
Display.setTitle("Mega Man");
try{
Display.setDisplayMode(Display.getDesktopDisplayMode());
Display.create();
}catch(LWJGLException e){
e.printStackTrace();
}
glMatrixMode(GL_PROJECTION);
glLoadIdentity();
glOrtho(0,WIDTH,HEIGHT,0,1,-1);
glMatrixMode(GL_MODELVIEW);
glEnable(GL_TEXTURE_2D);
glEnable(GL_BLEND);
glBlendFunc(GL_SRC_ALPHA, GL_ONE_MINUS_SRC_ALPHA);
glHint(GL_PERSPECTIVE_CORRECTION_HINT, GL_NICEST);
glHint(GL_LINE_SMOOTH_HINT, GL_NICEST);
try{Display.setFullscreen(true);}catch(Exception e){}
int sh=Display.getHeight();
int sw=WIDTH*sh/HEIGHT;
GL11.glViewport(Display.getWidth()/2-sw/2, 0, sw, sh);
Screenshot of the glitchy fullscreen here: http://sta.sh/021fohgnmxwa
EDIT
Here is the texture rendering code that I use to draw everything:
public static void DrawQuadTex(Texture tex, int x, int y, float width, float height, float texWidth, float texHeight, float subx, float suby, float subd, String mirror){
if (tex==null){return;}
if (mirror==null){mirror = "";}
//subx, suby, and subd are to grab sprites from a sprite sheet. subd is the measure of both the width and length of the sprite, as only images with dimensions that are the same and are powers of 2 are properly displayed.
int xinner = 0;
int xouter = (int) width;
int yinner = 0;
int youter = (int) height;
if (mirror.indexOf("h")>-1){
xinner = xouter;
xouter = 0;
}
if (mirror.indexOf("v")>-1){
yinner = youter;
youter = 0;
}
tex.bind();
glTranslatef(x,y,0);
glBegin(GL_QUADS);
glTexCoord2f(subx/texWidth,suby/texHeight);
glVertex2f(xinner,yinner);
glTexCoord2f((subx+subd)/texWidth,suby/texHeight);
glVertex2f(xouter,yinner);
glTexCoord2f((subx+subd)/texWidth,(suby+subd)/texHeight);
glVertex2f(xouter,youter);
glTexCoord2f(subx/texWidth,(suby+subd)/texHeight);
glVertex2f(xinner,youter);
glEnd();
glLoadIdentity();
}
Just to keep it clean I give you a real answer and not just a comment.
The aspect ratio problem can be solved with help of glViewport. Using this method you can decide which area of the surface that will be rendered to. The default viewport will cover the whole surface.
Since the second problem with the corrupt rendering (also described here https://stackoverflow.com/questions/28846531/sprite-game-in-full-screen-aliasing-issue) appeared after changing viewport I will give my thought about it in this answer as well.
Without knowing exactly how the rendering code for the tile background looks. I would guess that the problem is due to any differences in the resolution between the glViewport and glOrtho calls.
Example: If the glOrtho resolution is half the viewport resolution then each openGL unit is actually 2 pixels. If you then renders a tile between x=0 and x=9 and then the next one between x=10 and x=19 you will get an empty space between them.
To solve this you can change the resolution so that they are the same. Or you can render the tile to overlap, first one x=0 to x=10 second one x=10 to x=20 and so on.
Without seeing the tile rendering code I can't verify it this is the problem though.
The file is about 24mb, and it's held in a DataBase so I convert it to a bit array and then, after multiple suggestions, I use bitconverter.tosingle(,) and this is giving me bad results, here's my code:
byte[] imgData = prod.ImageData;
float myFloat = BitConverter.ToSingle(imgData, 0);
float mb = (myFloat / 1024f) / 1024f;
When I debug, I get these results:
byte[24786273]
myFloat = 12564.0361
mb = 0.0119819986
what is weird is that he size of the array is exactly as the file should be. How do I correctly convert this to float and then so it shows as mb?
EDIT: I tried setting up myFloat as imgData.Length, then the size is correct, however is this a correct way to do it, and can it cause a problem in the future with bigger values?
You are taking the first four bytes out of the image and converting it to an IEEE floating point. I'm not an expert on image files so I'm not sure if the first four bytes are always the length, even if this is the case it would still not be correct (see the specification). However the length of the file is already known through the length of the array, so an easier way to get the size is:
byte[] imgData = prod.ImageData;
float mb = (imgData.Length / 1024f) / 1024f;
To address your concerns: this will still work for large files, consider a 24TB example.
var bytes = 24L * 1024 * 1024 * 1024 * 1024;
var inMb = (bytes / 1024.0F / 1024.0F);
I made some simple shading in GLSL of a checkers board:
f(P) = [ floor(Px)+floor(Py)+floor(Pz) ] mod 2
It seems to work well except the fact that i see the interior of the objects but i want to see only the front face.
Any ideas how to fix this? Thanks!
Teapot (glutSolidTeapot()):
Cube (glutSolidCube):
The vertex shader file is:
varying float x,y,z;
void main(){
gl_Position = gl_ProjectionMatrix * gl_ModelViewMatrix * gl_Vertex;
x = gl_Position.x;
y = gl_Position.y;
z = gl_Position.z;
}
And the fragment shader file is:
varying float x,y,z;
void main(){
float _x=x;
float _y=y;
float _z=z;
_x=floor(_x);
_y=floor(_y);
_z=floor(_z);
float sum = (_x+_y+_z);
sum = mod(sum,2.0);
gl_FragColor = vec4(sum,sum,sum,1.0);
}
The shaders are not the problem - the face culling is.
You should either disable the face culling (which is not recommended, since it's bad for performance reasons):
glDisable(GL_CULL_FACE);
or use glCullFace and glFrontFace to set the culling mode, i.e.:
glEnable(GL_CULL_FACE); // enables face culling
glCullFace(GL_BACK); // tells OpenGL to cull back faces (the sane default setting)
glFrontFace(GL_CW); // tells OpenGL which faces are considered 'front' (use GL_CW or GL_CCW)
The argument to glFrontFace depends on application conventions, i.e. the matrix handedness.