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3.3. DECIMATION CIPHERS 41
previous about face example shows that decimation ciphers apparently will
not have this weakness. The adjacent letters ef and tu were enciphered to YD
and VA, respectively, and the adjacent triple abc to EJO. In fact, if we go ahead
and encipher the entire alphabet with key k = 5 we find the following cipher
alphabet:
plaintext
ciphertext
a b c d e f g h i j k l m n o p q r s t u v w x y z
E J O T Y D I N S X C H M R W B G L Q V A F K P U Z
None of the letters are next to their former neighbors. 9 Decimation ciphers
destroy letter ordering!
We have not explained how to “properly” choose our keynumbers, but to see
that there must be some care taken consider enciphering anna or bob using a
decimation cipher with k = 2. The ciphertexts are BBBB and DDD, respectively.
How would anyone decipher BBBB It is impossible to tell which B is an a and
which is an n. So no one would be able to decipher a message whose enciphering
key was 2. Hence, only certain keys can be used to multiply. In this chapter
you will be given only proper keys, so we can safely put off solving this problem
until Chapter 4.
Thus far we have enciphered letters and only letters. What if we wish to
keep spacing Or to add punctuation in our messages The next examples will
do that.
Examples:
(1) Work modulo 28 with key k = 5 to encipher a space here. Use the usual
plainnumbers with the addition that = 27, where the symbol means
“space”.
We do this as we did before, being careful to find the remainders using 28
rather than 26.
plaintext a s p a c e h e r e
plain numbers 1 27 19 16 1 3 5 27 8 5 18 5
multiplied numbers 5 135 95 80 5 15 25 135 40 25 90 25
%28 5 23 11 24 5 15 25 23 12 25 6 25
ciphertext E W K X E O Y W L Y F Y
The answer is EQAXE OYWLY FY.
(2) Encipher no one here using the same method with key k = 9.
9 The careful reader may notice an oddity in the last plaintext–ciphertext alphabet pair.
(If you didn’t, try enciphering zzz using any key before reading further.) Since we think of
a as the first letter of the alphabet, z is the 26th. But 26 times any keynumber will be a
multiple of 26, and so when divided by 26 it will have a remainder of 0. Since 0 is just before
1, 0 must represent Z. Hence, z will always be enciphered into Z. Fortunately z is uncommon
enough that this doesn’t really threaten the security of this cipher.