When checking this method, I was expecting for red, green and blue to be in the 0-255 range. Instead, it's in 0-1.
Am I the only one who thinks this is weird?
Is there any reason not o use the more common 0-255 values for RGB, or even hex numbers (as in html)?
In my opinion this is not weird. Both 0-255 and 0.0-1.0 levels are widely used in different platforms. You can always convert that by using something like this:
#define FLOAT_COLOR_VALUE(n) (n)/255.0
The reason sometimes RGB values are represented as float values rather than 0 to 255 is because 0 to 255 assumes you are using 8 bits to represent each colour component and hence have to use 24 bits for each colour in your frame buffers. This may not be the case if you are using displays that only support 256 colours in total or more than 16 million etc.
In theory then can be an infinite number of shades of red, green or blue. The number of bits you use to represent them depends on how accurate you need to represent colour and how much memory you have on graphics cards to represent images etc.
For many cases 0 to 255 is fine. But there is another world out there where it isn't fine, and for those devices / accurate rendering requirements, floating point numbers provide a much needed alternative.
Related
I have the task to simulate a camera with a full well capacity of 10.000 Photons per sensor element
in numpy. My first Idea was to do it like that:
camera = np.random.normal(0.0,1/10000,np.shape(img))
Imgwithnoise= img+camera
but it hardly shows an effect.
Has someone an idea how to do it?
From what I interpret from your question, if each physical pixel of the sensor has a 10,000 photon limit, this points to the brightest a digital pixel can be on your image. Similarly, 0 incident photons make the darkest pixels of the image.
You have to create a map from the physical sensor to the digital image. For the sake of simplicity, let's say we work with a grayscale image.
Your first task is to fix the colour bit-depth of the image. That is to say, is your image an 8-bit colour image? (Which usually is the case) If so, the brightest pixel has a brightness value = 255 (= 28 - 1, for 8 bits.) The darkest pixel is always chosen to have a value 0.
So you'd have to map from the range 0 --> 10,000 (sensor) to 0 --> 255 (image). The most natural idea would be to do a linear map (i.e. every pixel of the image is obtained by the same multiplicative factor from every pixel of the sensor), but to correctly interpret (according to the human eye) the brightness produced by n incident photons, often different transfer functions are used.
A transfer function in a simplified version is just a mathematical function doing this map - logarithmic TFs are quite common.
Also, since it seems like you're generating noise, it is unwise and conceptually wrong to add camera itself to the image img. What you should do, is fix a noise threshold first - this can correspond to the maximum number of photons that can affect a pixel reading as the maximum noise value. Then you generate random numbers (according to some distribution, if so required) in the range 0 --> noise_threshold. Finally, you use the map created earlier to add this noise to the image array.
Hope this helps and is in tune with what you wish to do. Cheers!
I'm unclear as to what value to set for NSOpenGLPFAColorSize when creating an NSOpenGLPixelFormat. From the documentation it states:
Value is a nonnegative buffer size specification. A color buffer that most closely matches the specified size is preferred. If unspecified, OpenGL chooses a color size that matches the screen.
But does this mean the number of bits per pixel? Or bits per component? For example, if it were set 24 and interpreted as bits per pixel then that would mean that each RGBA color would have 6-bits per component for a total of 24-bits for the entire RGBA pixel.
However, if it is to be interpreted as bits per component then that would mean 24-bits for each of the red, green, blue and alpha components to make a 96-bit RGBA pixel.
I'm inclined to believe that it means bits per component as the values I've seen set in sample code ranges from 8, 16, 24, 32 and everything but 24 makes sense when interpreted as bits per component. It would be nice though to have some definitive answer.
Note: Edited to reflect that pixels in OpenGL are RGBA not RGB.
After scouring the documentation further I came across the NSOpenGLPFAColorFloat attribute, which according to the documentation:
A Boolean attribute. If present, this attribute indicates that only renderers that are capable using buffers storing floating point pixels are considered. This should be accompanied by a NSOpenGLPFAColorSize of 64 (for half float pixel components) or 128 (for full float pixel components). Note, not all hardware supports floating point color buffers thus the returned pixel format could be NULL.
With that additional information it must mean bits per pixel.
I did some experimenting as well, setting NSOpenGLPFAColorSize to each of 8, 16, 24 & 32 and then checking what I got back. In each case I was returned a pixel format with NSOpenGLPFAColorSize set to 32 - meaning 32-bits per RGBA pixel. Just passing NSOpenGLPFAColorFloat with nothing set for the Color Size is enough to get back a pixel format with 64-bits per pixel.
I want to know how much data can be embedded into an image of different sizes.
For example in 30kb image file how much data can be stored without distortion of the image.
it depends on the image type , algoridum , if i take a example as a 24bitmap image to store ASCII character
To store a one ASCII Character = Number of Pixels / 8 (one ASCII = 8bits )
It depends on two points:
How much bits per pixel in your image.
How much bits you will embed in one pixel .
O.K lets suppose that your color model is RGB and each pixel = 8*3 bits (one byte for each color), and you want embed 3 bits in one pixel.
data that can be embedded into an image = (number of pixels * 3) bits
If you would use the LSB to hide your information this would give 30000Bits of available space to use. 3750 bytes.
As the LSB represents 1 or 0 into a byte that gets values from 0-256 this gives you in the worst case scenario that you are going to modify all the LSBs distortion of 1/256 that equals 0,4%.
In the statistical average scenario you would get 0,2% distortion.
So depends on which bit of the byte you are going to change.
This question comes from a homework assignment I was given. You can base your storage system off of one of the three following formats:
DD MM SS.S
DD MM.MMM
DD.DDDDD
You want to maximize the amount of data you can store by using as few bytes as possible.
My solution is based off the first format. I used 3 bytes for latitude: 8 bits for the DD (-90 to 90), 6 bits for the MM (0-59), and 10 bits for the SS.S (0-59.9). I then used 25 bits for the longitude: 9 bits for the DDD (-180 to 180), 6 bits for the MM, and 10 for the SS.S. This solution doesn't fit nicely on a byte border, but I figured the next reading can be stored immediately following the previous one, and 8 readings would use only 49 bytes.
I'm curious what methods others can come up. Is there a more efficient method to storing this data? As a note, I considered an offset based storage, but the problem gave no indication of how much the values may change between readings, so I'm assuming any change is possible.
Your suggested method is not optimal. You are using 10 bits (1024 possible values) to store a value in the range (0..599). This is a waste of space.
If you'll use 3 bytes for latitude, you should map the range [0, 2^24-1] to the range [-90, 90]. Hence each of the 2^24 values represents 180/2^24 degrees, which is 0.086 seconds.
If you want only 0.1 second accuracy, you'll need 23 bits for latitudes and 24 bits for longitudes (you'll get 0.077 seconds accuracy). That's 47 bit total instead of your 49 bits, with better accuracy.
Can we do even better?
The exact number of bits needed for 0.1 second accuracy is log2(180*60*60*10 * 360*60*60*10) < 46.256. Which means that you can use 46256 bits (5782 bytes) to store 1000 (lat,lon) pairs, but the mathematics involved will require dealing with very large integers.
Can we do even better?
It depends. If your data set has concentrations, you can store only some points and relative distances from these points, using less bits. Clustering algorithms should be used.
Sticking to existing technology:
If you used half precision floating point numbers to store only the DD.DDDDD data, you can be a lot more space-efficent, but you'd have to accept an exponent bias of 15, which means: The coordinates stored might not be exact, but at an offset from the original value.
This is due to the way floating point numbers are stored, essentially: A normalized significant is multiplied by an exponent to result in a number, instead of just storing a single value (as in integer numbers, the way you calculated the numbers for your solution).
The next highest commonly used floating point number mechanism uses 32 bits (the type "float" in many programming languages) - still efficient, but larger than your custom format.
If, however, you would design your own custom floating point type as well, and you gradually added more bits, your results would become more exact and it would STILL be more efficient than the solution you first found. Just play around with the number of bits used for significant and exponent, and find out how close your fp approximations come to the desired result in degrees!
Well, if this is for a large number of readings, then you may try a differential approach. Start with an absolute location, and then start saving incremental changes, which should ideally require less bits, depending on the nature of the changes. This is effectively compressing the stream. But somehow I don't think that's what this homework is about.
I need some quick advice.
I would like to simulate a cellular automata (from A Simple, Efficient Method
for Realistic Animation of Clouds) on the GPU. However, I am limited to OpenGL ES 2.0 shaders (in WebGL) which does not support any bitwise operations.
Since every cell in this cellular automata represents a boolean value, storing 1 bit per cell would have been the ideal. So what is the most efficient way of representing this data in OpenGL's texture formats? Are there any tricks or should I just stick with a straight-forward RGBA texture?
EDIT: Here's my thoughts so far...
At the moment I'm thinking of going with either plain GL_RGBA8, GL_RGBA4 or GL_RGB5_A1:
Possibly I could pick GL_RGBA8, and try to extract the original bits using floating point ops. E.g. x*255.0 gives an approximate integer value. However, extracting the individual bits is a bit of a pain (i.e. dividing by 2 and rounding a couple times). Also I'm wary of precision problems.
If I pick GL_RGBA4, I could store 1.0 or 0.0 per component, but then I could probably also try the same trick as before with GL_RGBA8. In this case, it's only x*15.0. Not sure if it would be faster or not seeing as there should be fewer ops to extract the bits but less information per texture read.
Using GL_RGB5_A1 I could try and see if I can pack my cells together with some additional information like a color per voxel where the alpha channel stores the 1 bit cell state.
Create a second texture and use it as a lookup table. In each 256x256 block of the texture you can represent one boolean operation where the inputs are represented by the row/column and the output is the texture value. Actually in each RGBA texture you can represent four boolean operations per 256x256 region. Beware texture compression and MIP maps, though!