The 340 Character Zodiac Cipher uses 63 symbols to represent (up to) 26 letters in the alphabet, plus possibly "space", "period" or other characters, like numbers. In order to process this cipher with a computer, we need to represent it in ASCII text. We must chose one ASCII character to represent each 340-Cipher character. There are many ways to do this of course. This webpage will use the following scheme:

- 1.) 25 Normal English Letters (all But Q) are used by
Zodiac in the cipher. Represent these by themselves (in upper case):
A B C D E F G H I J K L M N O P R S T U V W X Y Z

- 2.) 10
**Reversed**English Letters appear in the Cipher:P L K D F Y C J Q B

Represent these as lower case versions of themselves:

p l k d f y c j q b

- 3.) 7 Symbols are easily represented by Typewriter Keyboard Characters.
They look like:
> ^ + . < - /

- 4.) 21 Strange Symbols, represented in ASCII according to
This Scheme

Here is an ASCII version.

Thus, the 63 Characters which form The Zodiac's "AlphaBet" for this cipher are:

**
ABCDEFGHIJKLMNOPRSTUVWXYZ123456789plkdfycjqbtz()>^+.<-/#_@\%&;:
**

(This choice of ASCII representation is of course completely arbitrary & not intended to reflect the content of the hidden message.)

Taking the Zalphabet in this order, and assuming each cipher text character in the 340 Cipher represents one cleartext letter (but each letter of clear text is represented by one or more characters of ciphertext), we can then decode the cipher using a 63 character long KEY, which is simply a listing of which cleartext character is represented by each of the 63 ciphertext characters above. By the way, it should be noted that in the 3 Part Cipher, two symbols (Triangle-Filled & Triangle-Dot) BOTH represent the same two 2 letters at different times. This is one exception to the rule that each symbol represents a unique letter.

Here is a simple example of one possible KEY, followed on the next line by the Zalphabet

If this were the key, then each ordinary letter would stand for itself, the number "1" would stand for "A", "2" for "B", etc, up to ":" standing for "M". In this simple example, the DECODED cleartext would be:

Since this text has no discernable meaning, we obviously guessed the wrong key. The challenge is finding the right key. It is easy to write a computer program which guesses a key, then spits out a solution based on that key. However, given that there are 63 characters in the key, and we have at least 26 characters to choose from (not counting " " or ".") there are a total of 63^26 possible keys, a number larger than the number of nano-seconds in the age of the universe!!!! Clearly a brute force approach, (where every key is examined) even by the fastest computer in the world would never work. However there are more elegant ways to crack a cipher. We can use statistics, intuition, various clues and the solved 3 part cipher. Also certain keys are better than others. The right key probably contains many more "E"'s and "A"'s than "W"s and "Z"'s because of the need to encode popular letters in multiple ways, as is done in the 3-Part Cipher.

Before you (or your computer program) begin guessing at the solution, note that the cipher MAY ALREADY HAVE BEEN SOLVED, at least in part.

Suppose we use the following Key (again shown above the Zalphabet characters it represents)

ASCII Version:
**
ALATEPHHISSEIHOIRSTUBSSAEVEWSAAAMEEAGMES_STKLLTHCNLLTDAOSSEBLNE
**

This means that the 1st letter of the Zalphabet (A) is translated to "A", the second letter of the Zalphabet (B) becomes "L" ...etc... up to the last leter of the Zalphabet (:) which becomes "E". Using this key, the 340 cipher becomes:

This Decipherment was kindly provided by Zander Kite as part of his ongoing efforts. Although independantly derived, it is similar to the Solution in Graysmith's book, in that both solutions require subsequent anagramming, and both suggest that (one or more) cipher characters can translate into a second character, beyond what is shown above. (ie. it is suggested that some of the ciphertext "K"'s which have been translated to "S" in fact should remain "K"'s in cleartext). Note that this solution contains several English words in plain text. Further, it has been shown that subsequently anagramming this message produces even more Zodiacesque words. This has led some to conclude that it is the correct solution or at least very close. If this is the correct solution it implies that Zodiac made his second cipher considerably harder to solve than the (quickly solved) 3 part cipher, since this cipher requires direct substitution, alternate substitution, anagramming, AND has dummy characters scattered through the text. (the 3 part only had dummy characters at the end.)

Another possible solution is presented here

There is however the possibility that there exists a solution, perhaps one as clean as the solution to the 3 part cipher, ie not requiring anagramming, which has simply not yet been found due to the complexity of this cipher.... Click Here for a statistical breakdown of the characters used in this cipher