ShuklaSannidhya - 4 days ago 5
Python Question

# Rail Fence Cipher- Looking for a better solution

I have coded Rail Fence Cipher in Python. I was wondering if there could be a better solution.

For those who don't know what rail fence cipher is, it is basically a method of writing plain text in a way it creates linear pattern in a spiral way.
Example - when "FOOBARBAZ" rail-fenced using key of 3.

``````F . . . A . . . Z . . .
. O . B . R . A . Q . X
. . O . . . B . . . U .
``````

Reading the above spiral line-by-line, the cipher text becomes "FAZOBRAQXOBU". Read more at - Rail fence - Wikipedia.

``````def cipher(s, key, graph=False) :
down=True
raw_out=[]
out=''
i=0
for x in range(key) :
raw_out.append({})
for pos in range(len(s)) :
raw_out[i][pos]=s[pos]
if i==key-1 :
down=False
if i==0 :
down=True
if down :
i=i+1
else :
i=i-1
for p in raw_out :
for q in p :
out+=p[q]
if graph :
return raw_out
return out

def decipher(s, key) :
map_list=cipher(s, key, True) #CREATING JUST FOR MAPPING - WHICHth CHARACTER OF THE STRING - IS WHICHth CHARACTER OF THE CIPHER
new={}
out=''
s_counter=0
for x in map_list :
for y in x :
new[y]=s[s_counter]
s_counter+=1
for p in new :
out+=new[p]
return map_list
``````

I was wondering if there was any better way of doing this, since my procedure is very costly, it a uses couple of dictionaries.

Code in any language is welcomed.

Just for the heck of it...

``````def fence(lst, numrails):
fence = [[None] * len(lst) for n in range(numrails)]
rails = range(numrails - 1) + range(numrails - 1, 0, -1)
for n, x in enumerate(lst):
fence[rails[n % len(rails)]][n] = x

if 0: # debug
for rail in fence:
print ''.join('.' if c is None else str(c) for c in rail)

return [c for rail in fence for c in rail if c is not None]

def encode(text, n):
return ''.join(fence(text, n))

def decode(text, n):
rng = range(len(text))
pos = fence(rng, n)
return ''.join(text[pos.index(n)] for n in rng)

z = encode('ATTACK.AT.DAWN', 3)
print z # ACTWTAKA.ANT.D
y = decode(z, 3)
print y # ATTACK.AT.DAWN
``````