Codebusters - Baconian

The Baconian cipher is a biliteral cipher: every plaintext letter is replaced by a fixed group of five symbols, and each symbol is one of only two kinds, usually written A and B. Francis Bacon described the system in the early 1600s. The two symbols do not have to look like code. They can be two letters, two glyphs, or two typographic styles such as upper and lower case or bold and plain, so a Baconian message can hide inside ordinary-looking text. Under the 2027 rules the Baconian is on the Regional list in both divisions, which means it can also appear at State and National tournaments.

What the rules cover

The 2027 rules define the Baconian in item 3.e.ii of the Division B manual and item 3.e.iii of the Division C manual, and the two definitions match (C says "4 letters", B "4 plaintext letters"). Each of the two values, A and B, may be carried by one or more letters, glyphs or symbols, or by a rendering difference such as bold, underline or italic. A word Baconian, where the A/B stream is hidden in ordinary-looking words, always comes with a crib of at least 4 plaintext letters. Other Baconians carry the stream in symbols or formatting and are not promised a crib, although the question may say which form is A.

Event format in 2027. Teams of up to 3 have about 50 minutes. Teams bring only writing utensils, and nothing else may be used except what the Event Supervisor provides. There are no notes and no calculators of any class, a change from last season, which allowed calculators. The exam may be on paper or on a computer, and the packet is printed single-sided so that partners can split the pages. The resource sheet with the Baconian table is your only reference, so practice with a printed copy of the table beside you.

Three other rules touch Baconian questions:

• No self-decryption (3.g). The rule that no letter decrypts to itself covers Aristocrats, Patristocrats and Xenocrypts only. It does not apply to the Baconian.
• Special bonus eligibility (3.i). Up to three questions that are not Aristocrats, Patristocrats or Xenocrypts are marked as special bonus questions, so a Baconian can be one. New in 2027, at least one special bonus question must be a Homophonic cipher, which leaves at most two slots for other ciphers such as the Baconian.
• Timed question (3.h). The timed first question is always an Aristocrat, never a Baconian.

How the cipher works

Each letter is written as a five-symbol pattern made from A and B. Two symbol types give 2^5 = 32 five-symbol patterns, while four symbols would give only 16, too few for 24 letters. That is why every Baconian letter is five symbols long.

The traditional table, and the one Codebusters resource sheets normally print, uses 24 letters: I and J share one pattern, and U and V share one pattern. Read the patterns in the order of their binary value, with A = 0 and B = 1:

If you write the 24 letters in order as A B C D E F G H I K L M N O P Q R S T U W X Y Z and number them from 0, each letter's index equals its pattern read as a binary number. For example, BAABA is 10010 in binary, which is 18, and the letter at index 18 is T.

The table is fixed. There is no keyword and no shuffled alphabet, so the cipher has no key to recover. Your unknowns are which symbols stand for A and which for B, where the groups begin, and the I/J and U/V choices.

When many symbols stand for A and B

A tournament Baconian may use a dozen different letters or glyphs for A and B. Sort them into two classes before reading any groups.

1. List the distinct characters. Write each different letter or glyph that appears in the ciphertext once, in a margin of the packet.
2. Fix the classes from the crib. Encode the crib with the table and line its A/B pattern up under the marked position. Every character the crib covers is now known: if it sits over an A, it is in the A class, and if it sits over a B, it is in the B class. A character that lands on both A and B means the alignment or the crib reading is wrong.
3. Extend the classes to the rest of the stream. Mark every other occurrence of the known characters with its class. Groups that are now fully marked decode at once, and partly marked groups often have only one or two readings that make sense.
4. Use the BB rule on unknown characters. No group may begin with BB. If an unknown character is the second symbol of a group whose first symbol is B, the unknown must be A. Words forming in the plaintext settle most of the remaining characters.
5. Check for consistency. Each character keeps one class through the whole message.

Encryption, step by step

1. Write the plaintext in capitals, and drop spaces, punctuation and digits. Replace J with I and V with U, because the 24-letter convention gives them the same pattern.
2. Look up each letter's five-symbol pattern in the table.
3. Join the patterns in order into one A/B stream. Its length is exactly five times the number of letters.
4. Hide the stream in a carrier. Write one carrier character for each symbol, using a form or character from the A set for A and one from the B set for B (for example, lowercase for A and uppercase for B, or several different letters for each value). The carrier's letters are otherwise arbitrary, and its spaces and punctuation are not symbols.

Worked example: encrypt CODE

The stream is AAABA ABBAB AAABB AABAA, which is 20 symbols for 4 letters. With lowercase for A and uppercase for B, the carrier becomes theRe sISoN notHE reAnd. The carrier is deliberately meaningless here. A real puzzle usually uses readable text.

Decryption, step by step

1. Establish the A/B assignment. The question may state which form or symbols are A. If it does not, the crib usually settles it (see the tactics below). Otherwise try one assignment and test it. If many different symbols are used, sort them into A and B classes first (see "When many symbols stand for A and B" above).
2. Remove everything that is not A or B. In a carrier text, spaces and punctuation go, but the letters stay so that their forms can be read.
3. Split the stream into groups of five, counting from the first symbol. The total must be a multiple of five. If it is not, recheck the reading before you go on.
4. Convert each group to a number with A = 0 and B = 1, then take the letter at that index in the 24-letter list ABCDEFGHIKLMNOPQRSTUWXYZ.
5. Choose I or J, and U or V, from the context of the words.
6. Read the result. If no words appear, recheck the grouping and the assignment.

Worked example: decrypt theResISoNnotHEreAnd

Here lowercase is A and uppercase is B. The plaintext is CODE. The 20 carrier letters give exactly 4 groups, so the count check passes.

Solving under tournament conditions

With no key to find, the time goes into the A/B assignment, the alignment, and the I/J and U/V choices.

Use the crib first. A word Baconian is guaranteed a crib of at least four plaintext letters, which is at least 20 symbols. Encode the crib with the table, then search the stream at the position the question marks. If the crib's symbols match one assignment and not the other, you have the assignment. It also fixes the alignment and, with many carrier symbols, the class of every character it covers.

Identify the forms before you count. Mark which form is A at the top of the question. Then write out the full A/B string before you group it, so that you are never grouping raw carrier letters in your head.

Use the sheet, and learn the index trick. The resource sheet includes the Baconian mappings, so check its convention first. If it is the 24-letter table, decoding by index is quick, and since no calculator is allowed in 2027, converting five binary digits in your head (16, 8, 4, 2, 1) is worth drilling.

Leave I/J and U/V open. Write the letters under their groups in one pass and decide those pairs only once the words appear.

Scoring

The Baconian uses the default scoring rule (4.d in both 2027 manuals): the answer is marked by the number of errors in the decoded plaintext. Two or fewer errors earn full credit. Each error beyond two costs 100 points, and the penalty cannot exceed the question's value.

The final score is the exam score plus the timing bonus plus the special bonus (4.b). If a Baconian is marked as a special bonus question, it adds to the special bonus only when it is solved with no penalty points: 150 points for one such solve, 400 for two and 750 for three (4.g). Under the default rule, errors up to two carry no penalty, but aim for a clean answer anyway, because a single extra error both costs 100 points and loses the bonus. The timing bonus, 2 × (600 − seconds), belongs to the Aristocrat in question 1 and never involves the Baconian.

Common pitfalls

Practice ciphertexts

References and further reading

• Wikipedia, "Bacon's cipher": https://en.wikipedia.org/wiki/Bacon%27s_cipher
• Francis Bacon, De Augmentis Scientiarum (1623), where the biliteral cipher is described. Digitized copies are available from major libraries.
• David Kahn, The Codebreakers (1967), a general history of cryptography that covers the biliteral cipher.
• American Cryptogram Association: https://www.cryptogram.org
• SciOly Wiki, Codebusters: https://scioly.org/wiki/index.php/Codebusters
• Science Olympiad, Codebusters event page and rules: https://www.soinc.org