Perhaps the simplest way to represent an image is with a grid of pixels (i.e., dots), each of which can be of a different color. For black-and-white images, we thus need 1 bit per pixel, as 0 could represent black and 1 could represent white, as in the below.
In this sense, then, is an image just a bitmap (i.e., a map of bits). For more colorful images, you simply need more bits per pixel. A file format (like BMP, JPEG, or PNG) that supports “24-bit color” uses 24 bits per pixel. (BMP actually supports 1-, 4-, 8-, 16-, 24-, and 32-bit color.)
A 24-bit BMP uses 8 bits to signify the amount of red in a pixel’s color, 8 bits to signify the amount of green in a pixel’s color, and 8 bits to signify the amount of blue in a pixel’s color. If you’ve ever heard of RGB color, well, there you have it: red, green, blue.
If the R, G, and B values of some pixel in a BMP are, say, 0xff, 0x00, and 0x00 in hexadecimal, that pixel is purely red, as 0xff (otherwise known as 255 in decimal) implies “a lot of red,” while 0x00 and 0x00 imply “no green” and “no blue,” respectively.
Recall that a file is just a sequence of bits, arranged in some fashion. A 24-bit BMP file, then, is essentially just a sequence of bits, (almost) every 24 of which happen to represent some pixel’s color. But a BMP file also contains some “metadata,” information like an image’s height and width. That metadata is stored at the beginning of the file in the form of two data structures generally referred to as “headers,” not to be confused with C’s header files. (Incidentally, these headers have evolved over time. This problem uses the latest version of Microsoft’s BMP format, 4.0, which debuted with Windows 95.)
The first of these headers, called BITMAPFILEHEADER, is 14 bytes long. (Recall that 1 byte equals 8 bits.) The second of these headers, called BITMAPINFOHEADER, is 40 bytes long. Immediately following these headers is the actual bitmap: an array of bytes, triples of which represent a pixel’s color. However, BMP stores these triples backwards (i.e., as BGR), with 8 bits for blue, followed by 8 bits for green, followed by 8 bits for red. (Some BMPs also store the entire bitmap backwards, with an image’s top row at the end of the BMP file. But we’ve stored this problem set’s BMPs as described herein, with each bitmap’s top row first and bottom row last.) In other words, were we to convert the 1-bit smiley above to a 24-bit smiley, substituting red for black, a 24-bit BMP would store this bitmap as follows, where 0000ff signifies red and ffffff signifies white; we’ve highlighted in red all instances of 0000ff.
Because we’ve presented these bits from left to right, top to bottom, in 8 columns, you can actually see the red smiley if you take a step back.
To be clear, recall that a hexadecimal digit represents 4 bits. Accordingly, ffffff in hexadecimal actually signifies 111111111111111111111111 in binary.
Notice that you could represent a bitmap as a 2-dimensional array of pixels: where the image is an array of rows, each row is an array of pixels. Indeed, that’s how we’ve chosen to represent bitmap images in this problem.
What does it even mean to filter an image? You can think of filtering an image as taking the pixels of some original image, and modifying each pixel in such a way that a particular effect is apparent in the resulting image.
One common filter is the “grayscale” filter, where we take an image and want to convert it to black-and-white. How does that work?
Recall that if the red, green, and blue values are all set to 0x00 (hexadecimal for 0), then the pixel is black. And if all values are set to 0xff (hexadecimal for 255), then the pixel is white. So long as the red, green, and blue values are all equal, the result will be varying shades of gray along the black-white spectrum, with higher values meaning lighter shades (closer to white) and lower values meaning darker shades (closer to black).
So to convert a pixel to grayscale, we just need to make sure the red, green, and blue values are all the same value. But how do we know what value to make them? Well, it’s probably reasonable to expect that if the original red, green, and blue values were all pretty high, then the new value should also be pretty high. And if the original values were all low, then the new value should also be low.
In fact, to ensure each pixel of the new image still has the same general brightness or darkness as the old image, we can take the average of the red, green, and blue values to determine what shade of grey to make the new pixel.
If you apply that to each pixel in the image, the result will be an image converted to grayscale.
Most image editing programs support a “sepia” filter, which gives images an old-timey feel by making the whole image look a bit reddish-brown.
An image can be converted to sepia by taking each pixel, and computing new red, green, and blue values based on the original values of the three.
There are a number of algorithms for converting an image to sepia, but for this problem, we’ll ask you to use the following algorithm. For each pixel, the sepia color values should be calculated based on the original color values per the below.
sepiaRed = .393 * originalRed + .769 * originalGreen + .189 * originalBlue
sepiaGreen = .349 * originalRed + .686 * originalGreen + .168 * originalBlue
sepiaBlue = .272 * originalRed + .534 * originalGreen + .131 * originalBlue
Of course, the result of each of these formulas may not be an integer, but each value could be rounded to the nearest integer. It’s also possible that the result of the formula is a number greater than 255, the maximum value for an 8-bit color value. In that case, the red, green, and blue values should be capped at 255. As a result, we can guarantee that the resulting red, green, and blue values will be whole numbers between 0 and 255, inclusive.
Some filters might also move pixels around. Reflecting an image, for example, is a filter where the resulting image is what you would get by placing the original image in front of a mirror. So any pixels on the left side of the image should end up on the right, and vice versa.
Note that all of the original pixels of the original image will still be present in the reflected image, it’s just that those pixels may have rearranged to be in a different place in the image.
There are a number of ways to create the effect of blurring or softening an image. For this problem, we’ll use the “box blur,” which works by taking each pixel and, for each color value, giving it a new value by averaging the color values of neighboring pixels.
Consider the following grid of pixels, where we’ve numbered each pixel.
The new value of each pixel would be the average of the values of all of the pixels that are within 1 row and column of the original pixel (forming a 3x3 box). For example, each of the color values for pixel 6 would be obtained by averaging the original color values of pixels 1, 2, 3, 5, 6, 7, 9, 10, and 11 (note that pixel 6 itself is included in the average). Likewise, the color values for pixel 11 would be be obtained by averaging the color values of pixels 6, 7, 8, 10, 11, 12, 14, 15 and 16.
For a pixel along the edge or corner, like pixel 15, we would still look for all pixels within 1 row and column: in this case, pixels 10, 11, 12, 14, 15, and 16.
$ ./filter -r image.bmp reflected.bmp