cryptanalysis, affine cipher

An affine cipher applies the affine tranform x -> ax + b mod 26 to the letters of the plaintext. a must be prime to 26 for this map to be one-to-one and hence the cipher to be decodable (which it needs to be, for otherwise Bob is going to have a serious problem!).

To crytanalyse an affine cipher, one tries to identify the preimages of two distinct letters in the ciphertext, using frequency analysis. Not every such assignment turns out to work, for instance due to the condition to a. Sometimes, one such assignment leads to two possible keys, which are then tried out both. This is because arithmetic mod 26 isn't over a field, and hence congruences may have several solutions.

You may try cryptanalyzing the following texts, which were obtained from an english text by an affine cipher. The messages need not be long for cryptanalysis to succeed. This tells us something about the security of affine ciphers: it is low.

task 1

Einstein about mathematics:

tpctitpwqvstrpncztwqvztwxhpivcviwnivtsxwfwqvftivgnwhviwtxgtgotpctitpwqvftivhviwtxgwqvfongnwivcviwnivtsxwf

task 2

John von Neumann about mathematics:

pwfjuyzfjuprdbnlinwulwizmdujwiuypwhdbnlglduhzuldziunuyzf

task 3

Dijkstra about an old fashioned programming language :

qgxtnxbarbobsrkjeesxnqgxvjyujqnqxlrgjydngbtsuqgxkxabkxoxkxdlkuxulnlrkjvjylsbaaxynx

task 4

Adlai Stevenson on the US:

zgtpjwzbttglkxslbtgkjowjnzsjgeeitenxgjxaeijwznhnlxvethj

task 5

task 6

probabilities of letters in English text:
a .082
b .015
c .028
d .043
e .127
f .022
g .020
h .061
i .070
j .002
k .008
l .040
m .024
n .067
o .075
p .019
q .011
r .060
s .063
t .091
u .028
v .010
w .023
x .001
y .020
z .001

sorted:
e .127 vowel
t .091
a .082 vowel
o .075 vowel
i .070 vowel
n .067
s .063
h .061
r .060
d .043
l .040
c .028
u .028 vowel
m .024
w .023
f .022
g .020
y .020
p .019
b .015
q .011
v .010
k .008
j .002
x .001
z .001

created by A. Betten, August 23, 2003, last change August 23, 2003