Background
The Baconian cipher is one of the earliest examples of what is now called a ‘fractionated’ cipher. Essentially, individual letters are enciphered as 5 character groups. Each letter is represented by a unique 5 character group of a/b characters (Note that this is essentially a binary system with each letter represented by a binary number). The ACA uses the following Baconian table:
| A | aaaaa | E | aabaa | I/J | abaaa | N | abbaa | R | baaaa | W | babaa | |||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| B | aaaab | F | aabab | K | abaab | O | abbab | S | baaab | X | babab | |||||
| C | aaaba | G | aabba | L | ababa | P | abbba | T | baaba | Y | babba | |||||
| D | aaabb | H | aabbb | M | ababb | Q | abbbb | U/V | baabb | Z | babbb |
The Encipherment System:
PT: H E L P CT: aabbb aabaa ababa abbba
The a/b units can be further concealed by assigning an a/b value to each letter of the alphabet. For example, consider the following cipher ‘alphabet’ and the above message:
P: ABCDEFGHIJKLMNOPQRSTUVWXYZ V: aabbbbabbbbbaabbaaaaababab
PT: H E L P a/b: aabbb aabaa ababa abbba CT: QUICK BROWN TIGER SHEDS
In this case, a double encipherment has been used to hide the simple message ‘help’. The cipher text will read ‘QUICK BROWN TIGER SHEDS’. By replacing the cipher text letters with their a/b equivalents, the underlaying letters can be determined and the plain text recovered.
Note that a Baconian can be used in many other ways. For instance, only the first letters of each word may represent an a/b. Symbols or pictures can also be used. For this reason, Baconian ciphers can be popular for use with ornamental ciphers. However, this cipher is a bit impractical for everyday use since the cipher text is at least 5 times larger than the plain text.
Example:
We will be using JA05 E-05 as an example. This is a five alphabet Baconian, but we will only work with the first encoded message. The cipher text is:
QNGBU FAARA DSHBB NARKT NBXUM BYFTI CKBMK NELFH NWWID HCIVI BKPBK BFLGA BWKRH VBBUA NEZSK CATCO IHRLK NMCHE CBVBK BHVJY NAQBB KRJAB NRWNH PKWOG BBDZR BHXIG OFHAM
We are given the crib ‘Said’.
Start with the crib. The Baconian system would encipher this as follows:
PT: S A I D a/b: baaab aabaa abaaa aaabb
This a/b pattern must be hidden in the cipher text. The job of the cryptanalyst is to use this pattern to begin to uncover the a/b values assigned to the cipher text letters. This is easier than it first appears. The primary tool used here is to compare the a/b pattern of the crib with the cipher text and look for contradictions. For example, if the crib were placed at position 2, the following would result:
PT S A I D CT FAARA DSHBB NARKT NBXUM a/b baaab aaaaa abaaa aaabb
A quick look at the repeated letters in the cipher text shows that the A’s in FAARA have been assigned both the values of ‘a’ and ‘b’. This is a contradiction since A can have only one value. Therefore, the crib can not be placed at position 2.
In fact, attempting to fit the crib in all possible positions in the cipher shows that it fits only position 1 provides no contradictions. Therefore, the crib must be placed at the start of the cipher. This provides:
PT S A I D CT QNGBU FAARA DSHBB NARKT a/b baaab aaaaa abaaa aaabb
P: ABCDEFGHIJKLMNOPQRSTUVWXYZ V: aa*a*aaa**b**a**babbb*****
s a i d * * * * * * * QNGBU FAARA DSHBB NARKT NBXUM BYFTI CKBMK NELFH NWWID HCIVI BKPBK baaab aaaaa abaaa aaabb aa*b* a*ab* *ba*b a**aa a***a a**** ab*ab * * * * * * * * * e * BFLGA BWKRH VBBUA NEZSK CATCO IHRLK NMCHE CBVBK BHVJY NAQBB KRJAB aa*aa a*baa *aaba a**bb *ab** *aa*b a**a* *a*ab aa*** aabaa ba*aa * * * * * NRWNH PKWOG BBDZR BHXIG OFHAM aa*aa *b**a aaa*a aa**a *aaa*
Now note that CKBMK = *ba*b and that PKWOG = *b**a. Since no letter is encoded in the form bb***, we now can identify C = a and P = a. This provides:
P: ABCDEFGHIJKLMNOPQRSTUVWXYZ V: aaaa*aaa**b**a*ababbb*****
s a i d * * * * * * k QNGBU FAARA DSHBB NARKT NBXUM BYFTI CKBMK NELFH NWWID HCIVI BKPBK baaab aaaaa abaaa aaabb aa*b* a*ab* aba*b a**aa a***a aa*** abaab * * * * * * * * * e * BFLGA BWKRH VBBUA NEZSK CATCO IHRLK NMCHE CBVBK BHVJY NAQBB KRJAB aa*aa a*baa *aaba a**bb aaba* *aa*b a*aa* aa*ab aa*** aabaa ba*aa * * * * * NRWNH PKWOG BBDZR BHXIG OFHAM aa*aa ab**a aaa*a aa**a *aaa*
We now have to make some guesses to advance the analysis. Note the partial letters VBBUA = *aaba, and CBVBK = aa*ab. If V= a, these letters are ‘c’ and ‘b’ respectively. If V = b, these letters are ‘t’ and ‘f’ respectively. Since the second set of letters is more common than the first, we guess V = b. This provides:
P: ABCDEFGHIJKLMNOPQRSTUVWXYZ V: aaaa*aaa**b**a*ababbbb****
s a i d * * * * * * k QNGBU FAARA DSHBB NARKT NBXUM BYFTI CKBMK NELFH NWWID HCIVI BKPBK baaab aaaaa abaaa aaabb aa*b* a*ab* aba*b a**aa a***a aa*b* abaab * * t * * * * f * e * BFLGA BWKRH VBBUA NEZSK CATCO IHRLK NMCHE CBVBK BHVJY NAQBB KRJAB aa*aa a*baa baaba a**bb aab** *aa*b a*aa* aabab aab** aabaa ba*aa * * * * * NRWNH PKWOG BBDZR BHXIG OFHAM aa*aa ab**a aaa*a aa**a *aaa*
Now the fragment following the ‘t’ is very interesting. Since the most common word in English is ‘the’, it might be worth while trying to fit this word at this point. The cipher text now provides:
t * * VBBUA NEZSK CATCO baaba a**bb aaba*
To fit the word ‘the’ would require:
t h e VBBUA NEZSK CATCO baaba aabbb aabaa
Since this would produce no contradictions, we set E = a, Z = b, and O = a. This provides:
P: ABCDEFGHIJKLMNOPQRSTUVWXYZ V: aaaaaaaa**b**aaababbbb***b
s a i d * * * * * * k QNGBU FAARA DSHBB NARKT NBXUM BYFTI CKBMK NELFH NWWID HCIVI BKPBK baaab aaaaa abaaa aaabb aa*b* a*ab* aba*b aa*aa a***a aa*b* abaab * * t h e * * f * e * BFLGA BWKRH VBBUA NEZSK CATCO IHRLK NMCHE CBVBK BHVJY NAQBB KRJAB aa*aa a*baa baaba aabbb aabaa *aa*b a*aaa aabab aab** aabaa ba*aa * * c * * NRWNH PKWOG BBDZR BHXIG OFHAM aa*aa ab*aa aaaba aa**a aaaa*
Note the partial letter KRJAB = ba*aa. If J= a, this letters is ‘r’. If J = b, this letters is ‘w’. Since ‘r’ is a more common letter than ‘w’, we guess J = a. This provides:
P: ABCDEFGHIJKLMNOPQRSTUVWXYZ V: aaaaaaaa*ab**aaababbbb***b
s a i d * * * * * * k QNGBU FAARA DSHBB NARKT NBXUM BYFTI CKBMK NELFH NWWID HCIVI BKPBK baaab aaaaa abaaa aaabb aa*b* a*ab* aba*b aa*aa a***a aa*b* abaab * * t h e * * f * e r BFLGA BWKRH VBBUA NEZSK CATCO IHRLK NMCHE CBVBK BHVJY NAQBB KRJAB aa*aa a*baa baaba aabbb aabaa *aa*b a*aaa aabab aaba* aabaa baaaa * * c * * NRWNH PKWOG BBDZR BHXIG OFHAM aa*aa ab*aa aaaba aa**a aaaa*
Note the partial letter BHVJY = aaba*. If Y= a, this letters is ‘e’. If j = b, this letters is ‘f’. Since ‘ffer’ is a more common word fragment than ‘feer’, we guess Y = b. This provides:
P: ABCDEFGHIJKLMNOPQRSTUVWXYZ V: aaaaaaaa*ab**aaababbbb**bb
s a i d * * * * * * k QNGBU FAARA DSHBB NARKT NBXUM BYFTI CKBMK NELFH NWWID HCIVI BKPBK baaab aaaaa abaaa aaabb aa*b* abab* aba*b aa*aa a***a aa*b* abaab * * t h e * * f f e r BFLGA BWKRH VBBUA NEZSK CATCO IHRLK NMCHE CBVBK BHVJY NAQBB KRJAB aa*aa a*baa baaba aabbb aabaa *aa*b a*aaa aabab aabab aabaa baaaa * * c * * NRWNH PKWOG BBDZR BHXIG OFHAM aa*aa ab*aa aaaba aa**a aaaa*
Note the partial letters NRWNH = aa*aa, and PKWOG = ab*aa. If W = a, these letters are ‘a’ and ‘i/j’ respectively. If W = b, these letters are ‘e’ and ‘n’ respectively. Since the word fragment ‘fferenc’ is more common than the word fragment ‘fferaic’, we guess W = b. This provides:
P: ABCDEFGHIJKLMNOPQRSTUVWXYZ V: aaaaaaaa*ab**aaababbbbb*bb
s a i d * * * * * * k QNGBU FAARA DSHBB NARKT NBXUM BYFTI CKBMK NELFH NWWID HCIVI BKPBK baaab aaaaa abaaa aaabb aa*b* abab* aba*b aa*aa abb*a aa*b* abaab * n t h e * * f f e r BFLGA BWKRH VBBUA NEZSK CATCO IHRLK NMCHE CBVBK BHVJY NAQBB KRJAB aa*aa abbaa baaba aabbb aabaa *aa*b a*aaa aabab aabab aabaa baaaa e n c * * NRWNH PKWOG BBDZR BHXIG OFHAM aabaa abbaa aaaba aa**a aaaa*
Note the partial word ‘fference’ suggests ‘difference’. To match this, we set I=a, L=b, M=b, and X=b. This provides:
P: ABCDEFGHIJKLMNOPQRSTUVWXYZ V: aaaaaaaaaabbbaaababbbbbbbb
s a i d h l m e n c k QNGBU FAARA DSHBB NARKT NBXUM BYFTI CKBMK NELFH NWWID HCIVI BKPBK baaab aaaaa abaaa aaabb aabb* abab* ababb aabaa abbaa aaaba abaab e n t h e d i f f e r BFLGA BWKRH VBBUA NEZSK CATCO IHRLK NMCHE CBVBK BHVJY NAQBB KRJAB aabaa abbaa baaba aabbb aabaa aaabb a*aaa aabab aabab aabaa baaaa e n c e b NRWNH PKWOG BBDZR BHXIG OFHAM aabaa abbaa aaaba aabaa aaaab
The plain text therefore reads:
Said H.L. Mencken, the difference b
The remaining four parts of this message are decrypted in the same manner. For each part a separate crib is given. The five parts placed together should create one continuous message. The fragment ‘b’ at the end of message one suggests the word ‘between’ may be split by messages 1 and 2.
Good luck with the rest of the sections!