Phone Vision 06 – RGB Color Intensities

21 01 2011

I enjoyed our last blog’s bit of history, but how can we make something more useful than a spinning top?  It turns out that the fundamental principle Maxwell described for color photography is still in use today.  Though the medium is different, for all intents and purposes we are still taking three pictures – one with a red filter, one with a green filter, and one with a blue filter.  Those images are then combined to form a color photograph. 

Color Intensity

When we take a picture, we are just measuring the intensity of light at a given point.  With a sensor alone we would only be capable of creating a map of intensities.  This is definitely useful in some situations, but it ignores oodles of information humans can decipher naturally – color.

Using a filter that only lets through a specific color, we  can measure the intensity of that color.  Remember, we need at least three distinct colors to create the gamut of colors our eyes can distinguish.  To capture three different colors we need three different filters.

In the tender, early years of color photography, three separate pictures had to be taken.  I can’t imagine it was easy to capture a moving target.  Some enterprising people came up with a way to split the light into the three different wavelengths and then aim those beams at three separate light sensitive plates.  Clever as it may be, it can also be expensive and bulky.  There has to be a better way.

Bayer Filter

Conceptually, this next idea is very simple.  From a manufacturing standpoint, I’m not so certain.   Regardless, most modern digital cameras use a single image sensor covered by a grid of color filters arranged in what’s called a Bayer pattern (named after its inventor Bryce Bayer).

BayerFilter

The astute observer might notice that there are twice as many green filters as there are red or blue.  Wikipedia says this is because humans have a heck of a lot of rod cells (light sensing) and they are most sensitive to green light (~498 nm if you care).

In its raw form, the output of the camera above could be thought of as 3 separate images with with missing information:

red_mosaic_question

green_mosaic_questionblue_mosaic_question

Converting this raw data into a human friendly image is a process called demosaicing.  There are lots and lots and lots of ways to accomplish this.  They range from the very simple – like treating a 4×4 group as one pixel by averaging the green values – to a little more complex – like filling in the gaps by averaging neighboring pixels – to way beyond the scope of this blog.

Artificial Bayer Filter

The images we deal with are already demosaiced.  For this lesson we want to mess around with a raw image, but we don’t have one.  With that in mind, we are just going to have to “de-demosaic” an image.  For the color [255,128,64], the “de-demosaic” transform looks something like this:

pixel-255-128-64_to_bayer

 

Notice that our width will be twice the original width and our height will be twice the original height.  This will quadruple the size of the image.  Another important piece is the original coordinate (x,y) becomes 4 separate coordinates in the new image:

image

Let’s see what we can do.

//This assumes you already have the image in

//WriteableBitmap form. Refer to earlier lessons

//if you’re not sure how to do this.

WriteableBitmap ToBayerRaw(WriteableBitmap bmp)

{

    //we need to double the size of the target bitmap

    WriteableBitmap bayerRaw =

        new WriteableBitmap(bmp.PixelWidth * 2, bmp.PixelHeight * 2);

           

    //we loop through every column, row by row.

    //the two loops are not necessary, but

    //they demonstrate the concept easier

    //than a single loop

    for (int y = 0; y < bmp.PixelHeight; y++)

    {

        for(int x = 0; x < bmp.PixelWidth; x++)

        {

            //first recover the RGB pixel data

            int pixel = bmp.Pixels[x + y * bmp.PixelWidth];

                   

            // remember that (x,y) translates to 4 coordinates

            // g1 = (2x, 2y)

            // b = (2x+1, 2y)

            // r = (2x, 2y+1)

            // g2 = (2x+1, 2y+1)

            // also note that we are using the new pixel width

            // as our offset

            // notice the masks in use as well

 

            //upper left green

            bayerRaw.Pixels

                [2*x + 2 * y * (bayerRaw.PixelWidth)]

                = (int)(pixel & 0xFF00FF00);

                   

            //upper right blue

            bayerRaw.Pixels

                [2 * x + 1 + 2 * y * (bayerRaw.PixelWidth)]

                = (int)(pixel & 0xFF0000FF);

                   

            //lower left red

            bayerRaw.Pixels

                [2*x + (2 * y + 1) * (bayerRaw.PixelWidth)]

                = (int)(pixel & 0xFFFF0000);

 

            //lower right green

            bayerRaw.Pixels

                [2*x + 1 + (2 * y + 1) * (bayerRaw.PixelWidth)]

                = (int)(pixel & 0xFF00FF00);

        }

    }

    return bayerRaw;

}

 

image

image

Since half of the pixels are green and only a quarter are red and a quarter are blue, we would expect a green tint.  Matrixesque, am I right?

Interpolation

So now we have a raw file to work with, let’s see if we can fill in some of the “gaps” we’ve created.  There are quite a few techniques to accomplish this, but we are going to ignore them and try to figure them out on our own.   Let’s start with the red pixels:

red_mosaic_question

The first thing I notice is that some of ? pixels have immediate neighbors with values so let’s tackle those first.  My gut says that averaging them would yield a decent approximation:

red_average_01

This gives us value for 8 of the missing pixels leaving 4 behind:

image

If we ran through the image one more time, we could fill in the remaining gaps.  On the second run the new top, bottom, left, and right values would determine the new value of the missing pixels:

image

Wait a minute… Top, bottom, left and right were also the average of the original pixels.  Can we use that?  Let’s label a set of original pixels w, x, y, and z.

image

That means that top, bottom, left, and right are defined as:

image

image

image

image

This simplifies rather nicely (I left out a couple steps):

image

What I like to call the average of the original four pixels.  This is intuitive for a lot of us, but it’s nice to have some math to back it up.

Putting it all Together

WriteableBitmap Demosaic(WriteableBitmap bmp)

{

    WriteableBitmap interpolatedBmp =

        new WriteableBitmap(bmp.PixelWidth, bmp.PixelHeight);

    // we are going to cheat and ignore the boundaries

    // dealing with the boundaries is not difficult,

    // but the code is long enough as it is

    for (int y = 1; y < bmp.PixelHeight – 1; y++)

    {

        for (int x = 1; x < bmp.PixelWidth – 1; x++)

        {

            //first we are going to recover the pixel neighborhood.

            //the mask at the end is simply to get rid of the alpha

            //channel

            //middlecenter is the pixel we are working with (x,y)

            int topleft =

                bmp.Pixels[x – 1 + (y – 1) * bmp.PixelWidth] & 0xFFFFFF;

            int topcenter =

                bmp.Pixels[x + (y – 1) * bmp.PixelWidth] & 0xFFFFFF;

            int topright =

                bmp.Pixels[x + 1 + (y – 1) * bmp.PixelWidth] & 0xFFFFFF;

 

            int middleleft =

                bmp.Pixels[x – 1 + y * bmp.PixelWidth] & 0xFFFFFF;

            int middlecenter =

                bmp.Pixels[x + y * bmp.PixelWidth] & 0xFFFFFF;

            int middleright =

                bmp.Pixels[x + 1 + y * bmp.PixelWidth] & 0xFFFFFF;

 

            int bottomleft =

                bmp.Pixels[x – 1 + (y + 1) * bmp.PixelWidth] & 0xFFFFFF;

            int bottomcenter =

                bmp.Pixels[x + (y + 1) * bmp.PixelWidth] & 0xFFFFFF;

            int bottomright =

                bmp.Pixels[x + 1 + (y + 1) * bmp.PixelWidth] & 0xFFFFFF;

 

            int blue = 0;

            int red = 0;

            int green = 0;

 

            // if we are on an even row and an even column

            // like (2, 2) then we are on a green pixel

            // we need to average top center and bottom

            // center for red we need to average left

            // middle and right middle for blue

            if (y % 2 == 0 && x % 2 == 0)

            {

                red = (((topcenter >> 16) +

                    (bottomcenter >> 16)) / 2) << 16;

                blue = (middleleft + middleright) / 2;

                green = middlecenter;

            }

 

            // if we are on an even row and an odd column

            // like (2, 1) then we are on a blue pixel

            // red is an average of top left, top right,

            // bottom left, and bottom right

            // green is top center, bottom center,

            // left middle, and right middle

            else if (y % 2 == 0 && x % 2 == 1)

            {

                red = (((topleft >> 16) +

                    (topright >> 16) +

                    (bottomleft >> 16) +

                    (bottomright >> 16)) / 4) << 16;

                blue = middlecenter;

                green = (((topcenter >> 8 ) +

                    (bottomcenter >> 8 ) +

                    (middleleft >> 8 ) +

                    (middleright >> 8 )) / 4) << 8;

            }

 

            // if we are on an odd row and an even column

            // like (1, 2) then we are on a red pixel

            // blue is an average of top left, top right,

            // bottom left, and bottom right

            // green is top center, bottom center,

            // left middle, and right middle

            else if (y % 2 == 1 && x % 2 == 0)

            {

                red = middlecenter;

                blue = (topleft +

                    topright +

                    bottomleft +

                    bottomright) / 4;

                green = (((topcenter >> 8 ) +

                    (bottomcenter >> 8 ) +

                    (middleleft >> 8 ) +

                    (middleright >> 8 )) / 4) << 8;

            }

 

            // otherwise we are on an odd row and odd column

            // like (1,1). this is a green pixel

            // red left middle + right middle

            // blue is top center + bottom center

            else

            {

                red = (((middleleft >> 16) +

                    (middleright >> 16)) / 2) << 16;

                blue = (topcenter + bottomcenter) / 2;

                green = middlecenter;

            }

 

            interpolatedBmp.Pixels[x + y * interpolatedBmp.PixelWidth]

                = (255 << 24) | red | green | blue;

        }

    }

    return interpolatedBmp;

}

 

With a little luck, our image should look pretty close to the original.

image

image

image

On the left you’ll find the original followed by the Bayer filter in the middle and, finally the demosaiced is on the right.  I think we did alright, eh?  This image is large, much larger than the viewing space.  If, however, you were to perform the same procedure on an image that is smaller than the viewing space, you would most likely see artifacts from the resizing.

Summary

Hopefully you learned a little about how cameras actually work and maybe a little math.  It wasn’t too hard, but trying to keep all those colors straight can make the code messy.

Download Code

http://cid-88e82fb27d609ced.office.live.com/embedicon.aspx/Blog%20Files/PhoneVision/PhoneVision%2006%20-%20RGBColorIntensities.zip

Up next: Grayscale

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2 responses

21 01 2011
Phone Vision 05 – A Brief History of Color «

[…] Up Next: RGB Color Intensities […]

24 01 2011
Phone Vision – 07 Grayscale «

[…] processing concepts are simpler and faster without color information, so now that you have a solid grasp of color, let’s remove it and take a look at the world in […]

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