Francis Bacon’s Cipher

Francis Bacon’s Cipher or Baconian Cipher is strictly speaking a steganographic code and not a true cipher. To encode your message with Baconian cipher you replace each letter of your original text with a group of five ‘A’s and ‘B’s. There are two versions of the Baconian alphabet, in the first version the letters ‘I’ and ‘J’ have the same code and the same applies to ‘U’ and ‘V’. In the second version each letter has its own code.

Version 1:

A   AAAAA   G     AABBA   N    ABBAA   T     BAABA
B   AAAAB   H     AABBB   O    ABBAB   U/V   BAABB
C   AAABA   I/J   ABAAA   P    ABBBA   W     BABAA
D   AAABB   K     ABAAB   Q    ABBBB   X     BABAB
E   AABAA   L     ABABA   R    BAAAA   Y     BABBA
F   AABAB   M     ABABB   S    BAAAB   Z     BABBB

Version 2:

A   AAAAA   H     AABBB   O    ABBBA   V     BABAB
B   AAAAB   I     ABAAA   P    ABBBB   W     BABBA
C   AAABA   J     ABAAB   Q    BAAAA   X     BABBB
D   AAABB   K     ABABA   R    BAAAB   Y     BBAAA
E   AABAA   L     ABABB   S    BAABA   Z     BBAAB
F   AABAB   M     ABBAA   T    BAABB
G   AABBA   N     ABBAB   U    BABAA

If you want to decode a message encoded with Baconian cipher, you split the text into groups of five letters and you match the five-letter codes with the original letters based on the Baconian alphabet. Be careful, if you choose the wrong version your message will not make sense. You see ‘A’s and ‘B’s in different order and it might be confusing but it is not, it is basically very simple. In order to read further you should know a little about the binary system.

First of all, you must be wondering (or not), why 5 letters? Why not 4 or 6? The letters of the alphabet are 26 and in order to represent 26 different things (in our case letters) by using two symbols (‘A’ and ‘B’) you need at least a sequence of 5 symbols. In general, if you want to use two symbols, with a sequence of n symbols you can distinguish 2ⁿ things. The choice of 5 letters is the right decision, it is the minimum number to represent the letters with ‘A’s and ‘B’s.

Lets take a look to the version 2 of the Baconian alphabet where all letters have their own unique code. We will replace ‘A’s with ‘0’s and ‘B’s with ‘1’s. With this replacement we will get for each letter, a number in the binary system. For example A is AAAAA and is turned into 00000 which is 0 in decimal system (the one we use everywhere all day). If we do the same with the other letter we get B=00001=1, C=2, D=3 and so on. If you noticed it is their place in the alphabet minus one. With that in mind the cipher does not look so complex, does it?