Golfscript, 59 51 50 chars
Man each character is extremely difficult to lose:
0[2.{).,2>{\.@%!},{.2$-.4$>{].p~\[}{;\;}if..}or}do
Output:
[2 3 1]
[3 5 2]
[7 11 4]
[23 29 6]
[89 97 8]
[113 127 14]
...
Explanation:
The stack is set up so each iteration starts with the stack like this, the top being to the right. The [ indicates the current array marker, meaning when the interpreter encounters a ], everything on the stack from the mark to the top is put into an array.
g [ last | cur
g is the maximum gap so far. From the top down:
command | explanation
-----------------+----------------------------------------
0[2. | initialize vars g=0, last=2, cur=2
{...}do | loop forever...
Inside the loop:
) | cur += 1
.,2>{\.@%!}, | put all divisors of cur into a list
{...}or | if the list is empty, cur is prime, so
| the block is executed. otherwise,
| 'do' consumes the stack, sees it is truthy,
| and loops again
How does it put all divisors into a list? Let's do it step by step
Command | explanation | stack
-----------------+----------------------------------------------+----------------
| initial stack | n
., | make list of 0..n-1 | n [0,1,...,n-1]
2> | take elements at index 2 and greater | n [2,3,...,n-1]
{...}, | take list off stack, then iterate through |
| the list. on each iteration, put the current |
| element on the stack, execute the block, and |
| pop the top of the stack. if the top is |
| true then keep the element, else drop it. |
| when done, push list of all true elements |
| So, for each element... | n x
\. | Swap & dup | x n n
@ | Bring x around | n n x
% | Modulo | n (n%x)
! | Boolean not. 0->1, else->0. Thus this is 1 |
| if x divides n. | n (x divides n)
| So only the divisors of n are kept | n [divisors of n]
What does it do if the divisors are empty?
Command | explanation | stack
-----------------+----------------------------------------------+----------------
| initial stack | g [ last | cur
. | dup | g [ l | c | c
2$ | copy 3rd down | g [ l | c | c | l
- | sub. This is the current gap, cur-last | g [ l | c | c-l
. | dup | g [ l | c | c-l | c-l
4$ | copy 4th down | g [ l | c | c-l | c-l | g
> | is cur gap > max gap so far? | g [ l | c | c-l | c-l>g
{#1}{#2}if.. | #1 if c-l > g, #2 otherwise, and do ".." in | ... | g [ c | c | c
| either situation |
Two paths: yes and no. If yes (note that if consumes the top value on the stack):
Command | explanation | stack
-----------------+----------------------------------------------+----------------
| initial stack. note that now the old `g` is | XX [ l | c | g
| garbage and `c-l` is the new `g`. |
] | close the array | XX [l, c, g]
.p | duplicate it and print it, consuming the dup | XX [l, c, g]
~ | pump array back onto the stack. Note now the | XX | l | c | j
| array marker [ is gone. |
\ | swap. | XX | l | g | c
[ | mark the array | XX | l | g | c [
. | this is the part after the if. dups the top, | XX | l | g [ c | c
| but it does this in two steps, first popping |
| c then putting two copies on top, so the |
| array marker moves |
. | dup again | XX | l | g [ c | c | c
If no:
Command | explanation | stack
-----------------+----------------------------------------------+----------------
| initial stack. In this case g is still the | g [ l | c | c-l
| max gap so far |
;\; | dump top of stack, swap, and dump again | g [ c
.. | the part after the if. dup twice | g [ c | c | c
Note in either case, our stack is now in the form ... | g [ c | c | c.
Now the do pops the top value off the stack - always c - and loops if it is positive. Since c always increasing, this is always true, so we loop forever.
Once popped, the top of the stack is g [ c | c, meaning last has been updated to c, the array mark is in the same place, and g is still where we expect it.
These are the convoluted operations of GolfScript. I hope you enjoyed following along!
