Problem: Whodunit

* The Wikipedia articles are a bit dense; feel free to skim or skip!

This course’s philosophy on academic honesty is best stated as "be reasonable." The course recognizes that interactions with classmates and others can facilitate mastery of the course’s material. However, there remains a line between enlisting the help of another and submitting the work of another. This policy characterizes both sides of that line.

The essence of all work that you submit to this course must be your own. Collaboration on problems is not permitted (unless explicitly stated otherwise) except to the extent that you may ask classmates and others for help so long as that help does not reduce to another doing your work for you. Generally speaking, when asking for help, you may show your code or writing to others, but you may not view theirs, so long as you and they respect this policy’s other constraints. Collaboration on quizzes and tests is not permitted at all. Collaboration on the final project is permitted to the extent prescribed by its specification.

Below are rules of thumb that (inexhaustively) characterize acts that the course considers reasonable and not reasonable. If in doubt as to whether some act is reasonable, do not commit it until you solicit and receive approval in writing from your instructor. If a violation of this policy is suspected and confirmed, your instructor reserves the right to impose local sanctions on top of any disciplinary outcome that may include an unsatisfactory or failing grade for work submitted or for the course itself.

Your work on this problem set will be evaluated along four axes primarily.

To obtain a passing grade in this course, all students must ordinarily submit all assigned problems unless granted an exception in writing by the instructor.

First, curl up with Jason’s short on file I/O and Rob’s short on structs. Just keep in mind that Jason’s short happens to focus on ASCII (i.e., text) files as opposed to binary files (like images). More on those later!

Next, join Nate on a tour of valgrind, a command-line tool that will help you find "memory leaks": memory that you’ve allocated (i.e., asked the operating system for), as with malloc, but not freed (i.e., given back to the operating system).

Finally, remind yourself how GDB works if you’ve forgotten or not yet used! (It’s worth it!)

Welcome back!

As always, first open a terminal window and execute

to make sure your workspace is up-to-date.

Next, navigate to your ~/workspace/chapter4 directory. Then execute

in order to download a ZIP (i.e., compressed version) of this problem set’s distro. If you then execute

you should see that you now have a file called whodunit.zip in your ~/workspace/chapter4 directory. Unzip it by executing the below.

If you again execute

you should see that you now also have a whodunit directory. You’re now welcome to delete the ZIP file with the below.

Now dive into that whodunit directory by executing the below.

Now execute

and you should see that the directory contains the below.

How fun! A C file, a header file, and four images. Who knows what could be inside those! Let’s get started.

If you ever saw Windows XP’s default wallpaper, then you’ve seen a BMP. If you’ve ever looked at a webpage, you’ve probably seen a GIF. If you’ve ever looked at a digital photo, you’ve probably seen a JPEG. If you’ve ever taken a screenshot on a Mac, you’ve probably seen a PNG. Read up online on the BMP, GIF, JPEG, and PNG file formats. Then, open up questions.txt in your whodunit directory and tell us the below.

Next, curl up with the article from MIT at http://cdn.cs50.net/2015/fall/psets/4/garfinkel.pdf.

Though somewhat technical, you should find the article’s language quite accessible. Once you’ve read the article, answer each of the following questions in a sentence or more in ~/workspace/chapter4/whodunit/questions.txt.

Anyhow, welcome to Tudor Mansion. Your host, Mr. John Boddy, has met an untimely end—he’s the victim of foul play. To win this game, you must determine whodunit.

Unfortunately for you (though even more unfortunately for Mr. Boddy), the only evidence you have is a 24-bit BMP file called clue.bmp, pictured below, that Mr. Boddy whipped up on his computer in his final moments^[1]. Hidden among this file’s red "noise" is a drawing of whodunit.

You long ago threw away that piece of red plastic from childhood that would solve this mystery for you, and so you must attack it as a computer scientist instead. If you want to try to decode a picture of your own, check out red.cs50.net to download and red-ify your own image!

But, first, some background.

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. (Image adapted from http://www.brackeen.com/vga/bitmaps.html.)

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 GIF) that supports "8-bit color" uses 8 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 like Mr. Boddy’s 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. Given how red Mr. Boddy’s BMP is, it clearly has a lot of pixels with those RGB values. But it also has a few with other values.

Incidentally, HTML and CSS (languages in which webpages can be written) model colors in this same way. If curious, see http://en.wikipedia.org/wiki/Web_colors for more details.

Now let’s get more technical. 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 set only expects that you support version 4.0 (the latest) of Microsoft’s BMP format, 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. (In 1-, 4-, and 16-bit BMPs, but not 24- or 32-, there’s an additional header right after BITMAPINFOHEADER called RGBQUAD, an array that defines "intensity values" for each of the colors in a device’s palette.) 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.

Okay, stop! Don’t proceed further until you’re sure you understand why 0000ff represents a red pixel in a 24-bit BMP file.

Okay, let’s transition from theory to practice. Within CS50 IDE’s file browser, expand (i.e., open via the small triangle) chapter4 and then whodunit. Double-click smiley.bmp, and you should see a tiny smiley face that’s only 8 pixels by 8 pixels. Via the drop-down menu in that file’s newly opened tab, change 100% to 400% to zoom in a bit, and you should see a larger, albeit blurrier, version. (So much for "enhance," huh?) Actually, this particular image shouldn’t really be blurry, even when enlarged. CS50 IDE is simply trying to be helpful (CSI-style) by "dithering" the image (i.e., by smoothing out its edges). Below’s what the smiley looks like if you zoom in without dithering. At this zoom level, you can really see the image’s pixels (as big squares).

Okay, go ahead and return your attention to a terminal window, and navigate your way to ~/workspace/chapter4/whodunit. (Remember how?) Let’s look at the underlying bytes that compose smiley.bmp using xxd, a command-line "hex editor." Execute:

You should see the below; we’ve again highlighted in red all instances of 0000ff.

In the leftmost column above are addresses within the file or, equivalently, offsets from the file’s first byte, all of them given in hex. Note that 00000036 in hexadecimal is 54 in decimal. You’re thus looking at byte 54 onward of smiley.bmp. Recall that a 24-bit BMP’s first 14 + 40 = 54 bytes are filled with metadata. If you really want to see that metadata in addition to the bitmap, execute the command below.

If smiley.bmp actually contained ASCII characters, you’d see them in xxd's rightmost column instead of all of those dots.

So, smiley.bmp is 8 pixels wide by 8 pixels tall, and it’s a 24-bit BMP (each of whose pixels is represented with 24 ÷ 8 = 3 bytes). Each row (aka "scanline") thus takes up (8 pixels) × (3 bytes per pixel) = 24 bytes, which happens to be a multiple of 4. It turns out that BMPs are stored a bit differently if the number of bytes in a scanline is not, in fact, a multiple of 4. In small.bmp, for instance, is another 24-bit BMP, a green box that’s 3 pixels wide by 3 pixels wide. If you view it with Image Viewer (as by double-clicking it), you’ll see that it resembles the below, albeit much smaller. (Indeed, you might need to zoom in again to see it.)

Each scanline in small.bmp thus takes up (3 pixels) × (3 bytes per pixel) = 9 bytes, which is not a multiple of 4. And so the scanline is "padded" with as many zeroes as it takes to extend the scanline’s length to a multiple of 4. In other words, between 0 and 3 bytes of padding are needed for each scanline in a 24-bit BMP. (Understand why?) In the case of small.bmp, 3 bytes' worth of zeroes are needed, since (3 pixels) × (3 bytes per pixel) + (3 bytes of padding) = 12 bytes, which is indeed a multiple of 4.

To "see" this padding, go ahead and run the below.

Note that we’re using a different value for -c than we did for smiley.bmp so that xxd outputs only 4 columns this time (3 for the green box and 1 for the padding). You should see output like the below; we’ve highlighted in green all instances of 00ff00.

For contrast, let’s use xxd on large.bmp, which looks identical to small.bmp but, at 12 pixels by 12 pixels, is four times as large. Go ahead and execute the below; you may need to widen your window to avoid wrapping.

You should see output like the below; we’ve again highlighted in green all instances of 00ff00

Worthy of note is that this BMP lacks padding! After all, (12 pixels) × (3 bytes per pixel) = 36 bytes is indeed a multiple of 4.

Knowing all this has got to be useful!

Okay, xxd only showed you the bytes in these BMPs. How do we actually get at them programmatically? Well, in copy.c is a program whose sole purpose in life is to create a copy of a BMP, piece by piece. Of course, you could just use cp for that. But cp isn’t going to help Mr. Boddy. Let’s hope that copy.c does!

Go ahead and compile copy.c into a program called copy using make. (Remember how?) Then execute a command like the below.

If you then execute ls (with the appropriate switch), you should see that smiley.bmp and copy.bmp are indeed the same size. Let’s double-check that they’re actually the same! Execute the below.

If that command tells you nothing, the files are indeed identical. (Note that some programs, like Photoshop, include trailing zeroes at the ends of some BMPs. Our version of copy throws those away, so don’t be too worried if you try to copy a BMP that you’ve downloaded or made only to find that the copy is actually a few bytes smaller than the original.) Feel free to open both files in Ristretto Image Viewer (as by double-clicking each) to confirm as much visually. But diff does a byte-by-byte comparison, so its eye is probably sharper than yours!

So how now did that copy get made? It turns out that copy.c relies on bmp.h. Let’s take a look. Open up bmp.h, and you’ll see actual definitions of those headers we’ve mentioned, adapted from Microsoft’s own implementations thereof. In addition, that file defines BYTE, DWORD, LONG, and WORD, data types normally found in the world of Win32 (i.e., Windows) programming. Notice how they’re just aliases for primitives with which you are (hopefully) already familiar. It appears that BITMAPFILEHEADER and BITMAPINFOHEADER make use of these types. This file also defines a struct called RGBTRIPLE that, quite simply, "encapsulates" three bytes: one blue, one green, and one red (the order, recall, in which we expect to find RGB triples actually on disk).

Why are these structs useful? Well, recall that a file is just a sequence of bytes (or, ultimately, bits) on disk. But those bytes are generally ordered in such a way that the first few represent something, the next few represent something else, and so on. "File formats" exist because the world has standardized what bytes mean what. Now, we could just read a file from disk into RAM as one big array of bytes. And we could just remember that the byte at location [i] represents one thing, while the byte at location [j] represents another. But why not give some of those bytes names so that we can retrieve them from memory more easily? That’s precisely what the structs in bmp.h allow us to do. Rather than think of some file as one long sequence of bytes, we can instead think of it as a sequence of `struct`s.

Recall that smiley.bmp is 8 by 8 pixels, and so it should take up 14 + 40 + (8 × 8) × 3 = 246 bytes on disk. (Confirm as much if you’d like using ls.) Here’s what it thus looks like on disk according to Microsoft:

As this figure suggests, order does matter when it comes to structs' members. Byte 57 is rgbtBlue (and not, say, rgbtRed), because rgbtBlue is defined first in RGBTRIPLE. Our use, incidentally, of the attribute called packed ensures that clang does not try to "word-align" members (whereby the address of each member’s first byte is a multiple of 4), lest we end up with "gaps" in our `struct`s that don’t actually exist on disk.

Now go ahead and pull up the URLs to which BITMAPFILEHEADER and BITMAPINFOHEADER are attributed, per the comments in bmp.h. You’re about to start using MSDN (Microsoft Developer Network)!

Rather than hold your hand further on a stroll through copy.c, we’re instead going to ask you some questions and let you teach yourself how the code therein works. As always, man is your friend, and so, now, is MSDN. If not sure on first glance how to answer some question, do some quick research and figure it out! You might want to turn to stdio.h at https://reference.cs50.net/ as well.

Allow us to suggest that you also run copy within GDB while answering these questions as follows:

In ~/workspace/chapter4/whodunit/questions.txt, answer each of the following questions in a sentence or more.

Okay, back to Mr. Boddy.

Write a program called whodunit in a file called whodunit.c that reveals Mr. Boddy’s drawing.

Ummm, what?

Well, think back to childhood when you held that piece of red plastic over similarly hidden messages. (If you remember no such piece of plastic, best to ask a classmate about his or her childhood.) Essentially, the plastic turned everything red but somehow revealed those messages. Implement that same idea in whodunit. Like copy, your program should accept exactly two command-line arguments. And if you execute a command like the below, stored in verdict.bmp should be a BMP in which Mr. Boddy’s drawing is no longer covered with noise.

Allow us to suggest that you begin tackling this mystery by executing the command below.

Wink wink. You may be amazed by how few lines of code you actually need to write in order to help Mr. Boddy.

There’s nothing hidden in smiley.bmp, but feel free to test your program out on its pixels nonetheless, if only because that BMP is small and you can thus compare it and your own program’s output with xxd during development. (Or maybe there is a message hidden in smiley.bmp too. No, there’s not.^[2][3])

Rest assured that more than one solution is possible. So long as Mr. Boddy’s drawing is identifiable (by you), no matter its legibility, Mr. Boddy will rest in peace.

Because whodunit can be implemented in several ways, you won’t be able to check your implementation’s correctness with check50. And, lest it spoil your fun, the staff’s solution to whodunit is not available.

But here is Zamyla!

In ~/workspace/chapter4/whodunit/questions.txt, answer the question below.

This was Whodunit.