Using Emacs for Fun & Profit. Part 0 : align-regexp

Summary
Column alignment of text in emacs is discussed briefly.

If you're a neat freak about your code then you probably prefer related items to line up so that they can be read easily.  For example the C declarations

RGBA zx_plane_color = {RED,    0.1};     
RGBA zy_plane_color = {GREEN,  0.1};      
RGBA xy_plane_color = {BLUE,   0.1};      
RGBA plane_grid_color = {GREY,   0.5};
RGBA background_color = {WHITE,  1};

are a lot easier to read as when aligned up on the "=" sign

RGBA zx_plane_color   = {RED,    0.1};     
RGBA zy_plane_color   = {GREEN,  0.1};      
RGBA xy_plane_color   = {BLUE,   0.1};      
RGBA plane_grid_color = {GREY,   0.5};
RGBA background_color = {WHITE,  1};

The Emacs command align-regexp can handle situations like this and more. The alignment above is accomplished with the sequence of commands below (assuming that the region covers the declarations to align)

Keystrokes	Action
M-x align-regexp	Emacs responds with "Align regexp:" in the minibuffer.
=	Declarations are aligned

To align the after the commas, repeat the same commands but type "," when prompted for a regexp.

Unfortunately, the default is to align on the first occurrence of the regexp.  In many cases, you want to format a table of data such as

12, 34, 23123,     3939
2343, 344,
343, 33, 232,          2323,  zzzz

The sequence of commands then is

Keystrokes	Action
C-u M-x align-regexp	Emacs responds with "Complex align using regexp: \(\s-*\)" in the minibuffer.
Erase the "\(\s-*\)" and replace with "\(\s-+\)"	Emacs responds with "Parenthesis group to modify (justify if negative) : 1"
Press ENTER	Emacs responds with "Amount of spacing (or column if negative): 1"
Press ENTER	Emacs responds with "Repeat throughout line? (y or n) "
Type "Y" then press ENTER	Text is aligned.

The result is

12,   34,  23123, 3939
2343, 344, 
343,  33,  232,   2323, zzzz

Notice that the number of items in each row can differ.
align-regexp can handle much more complex situations than the ones covered here but I believe that these describe the two most common cases in general.

The Coffee Can Problem

Summary
A simple problem involving invariants and termination condition of an iterative process is solved.  A simple property of this iterative process is shown.

This problem appears David Gries's The Science of Programming:
You are given a coffee can that contains beans painted either black or white. Next to the can is a large pile of black beans. You perform the following algorithm:
Let \(N\) be the number of beans in the can at the current instant of time.

1. If \(N=1\), then stop.
2. Randomly draw two beans from the can. Call them \(B_0\) and \(B_1\).
3. If color(\(B_0\)) \(\neq\) color(\(B_1\)), then return white bean to the can and throw away the black bean.
4. If color(\(B_0\)) \(=\) color(\(B_1\)), then throw away both beans and place a black bean from the pile into the can.
5. Go to step 1.

Does the process terminate? If the process does terminate, then what can be said about the remaining bean in the can in terms of the number of black and white beans which were in the can to start with?
Proving that this terminates is easy. Let \(N_b\) be the number of BLACK beans in the can at the current instant of time and let \(N_w\) be the number of WHITE beans in the can at the current instant of time. Also, let \(N'\) be the number of beans in the can after either steps 3 or 4. The following is then true:

1. color(\(B_0\)) \(\neq\) color(\(B_1\)) \(\Rightarrow\)
        \(N'= (N_b - 1) + N_w = N - 1\)
   
2. color(\(B_0\)) \(=\) color(\(B_1\)) \(=\) WHITE \(\Rightarrow\)
        \(N' = (N_b + 1) + (N_w - 2) = N - 1\)
   
3. color(\(B_0\)) \(=\) color(\(B_1\)) \(=\) BLACK \(\Rightarrow\)
        \(N' = (N_b - 1) + N_w = N - 1\)
   

The total number of beans in the can is reduced by one for every iteration of the algorithm and therefore the process terminates when the condition in step 1 is satisfied. (Most people can just work this out in their heads)
"Saying something" about the last remaining bean in terms of the initial number of black and white beans is more open-ended and slightly harder. Looking at the three statements above shows that the number of white beans in the can is always reduced by two. So if \(N_w\) is the number of white beans in the can to begin with and \(B\) is the last remaining bean in the can then:
odd(\(N_w\)) \(\Rightarrow\)
    color(\(B\)) \(=\) WHITE

What about the reverse? Does:
even(\(N_w\)) \(\Rightarrow\)
    color(\(B\)) \(=\) BLACK

Since the number of beans is always reduced by one at each step and since the number of white beans is always reduced by two in step 2 above, then an even number of white beans to begin with will eventually result in a can filled with black beans and therefore step 3 of the algorithm will never be performed after that point. The algorithm will terminate with one black bean.
The statements can then be made:
odd(\(N_w\)) \(\Leftrightarrow\)
    color(\(B\)) \(=\) WHITE

even(\(N_w\)) \(\Leftrightarrow\)
    color(\(B\)) \(=\) BLACK