Cryptology - Unofficial St. Mary's College of California Web Site

1.3. FREQUENCY ANALYSIS 19
Examples: Use Frequency Analysis to decrypt the following Caesar ciphers.
(1) ZU SGQK G VXUVKX YKTZKTIK KBKXE RKZZKX SAYZ VRGE OZY VXUVKX XURK
We begin by making a frequency count:
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
The giant peak at K surrounded by a vast wasteland jumps out at us.
If we guess that the six letters LMNOPQ are uvwxyz, then the peak of K
is t, which would be fine, and RVXYZ=aeghi. Putting our guess at the
plaintext alphabet next to the ciphertext alphabet we may compare.
plaintext
ciphertext
\
\ \ \ \
j k l m n o p q r s t u v w x y z a b c d e f g h i
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
The peaks and valleys seem reasonable. To test, we decipher the message.
Since our prediction is that a became R, we use the key of R. Unfortunately,
the message then begins id bpzt p edgetg. This is not right. We must
try again.
Perhaps the low sextet belongs at the beginning of the ciphertext alphabet.
If we guess that K = e, then a must be G, and before G is a sextet of
nothingness. This is likely uvwxyz. Let’s again add the plaintext alphabet
to compare:
plaintext
ciphertext
\
\ \ \ \
u v w x y z a b c d e f g h i j k l m n o p q r s t
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
The lmnop and rst sets are both relatively high since in such a short
cipher appearing even three or four times counts as “many”. We ought to
be quite confident of our plaintext placement. As always, the key is the
letter a becomes, which is G. To finish, we decipher the first couple letters
using a key G. This gives to make a proper, which seems correct. We’ve
found the key! (Now finish up the decryption by deciphering the rest of
the message using the key G.)