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66 CHAPTER 4. THE EUCLIDEAN ALGORITHM
As we saw in Chapter 3 this is not a Caesar cipher; none of the standard hill and
valley patterns can be found in the frequency count. Are there any new patterns
that might replace our old ones To find out, study carefully the following two
plaintext/ciphertext alphabet pairs:
Key k = 5 :
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
Key k = 3m + 4:
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 H K N Q T W Z C F I L O R U X A D G J M P S V Y B
In the first, we have a → E, b→ J, c→ O, and while abc are consecutive, EJO
are not; they are 5 letters apart. In fact, any two consecutive plaintext letters
are sent to ciphertext letters five letters apart. The multiplicative key of 5
apparently spreads out the letters by a factor of 5. Why is this Two consecutive
plaintext letters will have plainnumbers that differ by one, say α and α + 1.
When enciphered the (unreduced) ciphernumbers are 5α and 5α + 5, which
differ by 5.
Does the same work for the linear cipher k = 3m + 4 (Check it and
see!) 6 At first glance it seems that Decimation and Linear ciphers do a good
job of “unordering” the plain alphabet, mixing it up well. But these examples
should convince us that this is not so. In fact, the rule is that the distance
consecutive plain alphabet letters are spread apart equals the multiplicative key
of the cipher.
We can use this idea to decrypt our ciphertext. In our unsolved cipher, B
and U are the two most common letters in the ciphertext: perhaps they are e
and t, respectively Since a linear cipher has the formula mk + c, so guessing
that B = e tells us 5k + c ≡ 2 and guessing U = t gives us 2km + c ≡ 21, both
equations modulo 26. So c ≡ 2 − 5k and c ≡ 21 − 20k. Setting these equal gives
20k − 21 ≡ 5k − 2 or 15k ≡ 19. From Figure 3.2 the inverse of 15 is 7. So if
we multiply both sides of 15k ≡ 19 by 7, since 15 × 7 ≡ 1 and 19 × 7 ≡ 3, we
are left with k ≡ 3. Finally, from 5k + c ≡ 2 and k = 3, we can substitute and
solve to see that c = 13. Thus the original cipher was a linear cipher with rule
3k + 13. We can then simply decipher the message. 7
Now this only worked because we were lucky enough to guess both e and
t. However, a cipher that is easily broken once your enemy correctly guesses
only two letters is probably not a very strong one. Just as Caesar ciphers
proved to be weak because they do not separate adjacent letters in the plaintext
alphabet, Decimation and Linear ciphers are weak because the plaintext letters
are enciphered using an easily recognized pattern.
6 Yes. The letters are spread out by a factor of 3 and then shifted 4 more letters down the
alphabet.
7 We can only say that the decryptment of any cipher, even the simplest, will
at times include a number of wonderings. Helen Fouché Gaines.