Rail fence cipher

Python 133 bytes

def cipher(t,r):
 m=r*2-2;o='';j=o.join
 for i in range(r):s=t[i::m];o+=i%~-r and j(map(j,zip(s,list(t[m-i::m])+[''])))or s
 return o

Sample usage:

>>> print cipher('FOOBARBAZQUX', 3)
FAZOBRAQXOBU

>>> print cipher('ABCDEFGHIJKLMNOPQRSTUVWXYZ', 4)
AGMSYBFHLNRTXZCEIKOQUWDJPV

>>> print cipher('ABCDEFGHIJKLMNOPQRSTUVWXYZ', 5)
AIQYBHJPRXZCGKOSWDFLNTVEMU

>>> print cipher('ABCDEFGHIJKLMNOPQRSTUVWXYZ', 6)
AKUBJLTVCIMSWDHNRXEGOQYFPZ

Note: the results from even rail counts are different than for the code you provided, but they seem to be correct. For example, 6 rails:

A         K         U
 B       J L       T V
  C     I   M     S   W
   D   H     N   R     X
    E G       O Q       Y
     F         P         Z

corresponds to AKUBJLTVCIMSWDHNRXEGOQYFPZ, and not AKUTBLVJICMSWXRDNHQYEOGZFP as your code produces.

The basic idea is that each rail can be found directly by taking string slices [i::m], where i is the rail number (0-indexed), and m is (num_rails - 1)*2. The inner rails additionally need to be interwoven with [m-i::m], achieved by zipping and joining the two sets of characters. Because the second of these can potentially be one character shorter, it is padded with a character assumed not to appear anywhere (_), and then that character is stripped off if necessary it is converted to a list, and padded with an empty string.

---

A slightly more human readable form:

def cipher(text, rails):
  m = (rails - 1) * 2
  out = ''
  for i in range(rails):
    if i % (rails - 1) == 0:
      # outer rail
      out += text[i::m]
    else:
      # inner rail
      char_pairs = zip(text[i::m], list(text[m-i::m]) + [''])
      out += ''.join(map(''.join, char_pairs))
  return out

APL 52 41

i←⍞⋄n←⍎⍞⋄(,((⍴i)⍴(⌽⍳n),1↓¯1↓⍳n)⊖(n,⍴i)⍴(n×⍴i)↑i)~' '

If the input text string i and the key number n are preinitialised the solution can be shortened by 9 characters. Running the solution against the examples given by primo gives identical answers:

FOOBARBAZQUX
3
FAZOBRAQXOBU
      
ABCDEFGHIJKLMNOPQRSTUVWXYZ
4
AGMSYBFHLNRTXZCEIKOQUWDJPV
      
ABCDEFGHIJKLMNOPQRSTUVWXYZ
5
AIQYBHJPRXZCGKOSWDFLNTVEMU
  
ABCDEFGHIJKLMNOPQRSTUVWXYZ
6
AKUBJLTVCIMSWDHNRXEGOQYFPZ

On further reflection there appears to be a shorter index based solution:

i[⍋+\1,(y-1)⍴((n←⍎⍞)-1)/1 ¯1×1 ¯1+y←⍴i←⍞]