Banburismus and the Weight of Evidence
Punched Paper Sheets, Decibans, and Sequential Bayesian Inference Against Naval Enigma
While Hut 6 at Bletchley Park read German Army and Luftwaffe Enigma traffic regularly by the summer of 1940, the German Navy (Kriegsmarine) remained a locked black box. Under Admiral Karl Dönitz, Atlantic U-boat wolfpacks communicated over the Heimische Gewässer (Dolphin) network using a hardened operational procedure. Whereas the Army and Air Force selected three Enigma rotors from a box of five ($5 \times 4 \times 3 = 60$ possible wheel orders), the German Navy issued eight distinct rotors. Those eight wheels created $8 \times 7 \times 6 = 336$ possible three-rotor orders. Worse, naval operators looked up message indicators in secret bigram tables (Kenngruppenbuch) before enciphering them. That step eliminated operator-chosen indicator habits.
With only a handful of electromechanical Bombes operating in late 1940 and early 1941, testing all 336 naval wheel orders by brute force was physically impossible. A single day's traffic would have consumed weeks of Bombe time. Before Hut 8 could mount a Bombe run, Alan Turing needed a statistical filter using paper and human clerks. That filter had to deduce which three rotors sat inside the U-boat machines that day, cutting the 336 wheel orders down to a small shortlist.
Turing's answer was Banburismus, named after the market town of Banbury in Oxfordshire. Printers in Banbury manufactured the long, twenty-six-column ruled paper sheets used in the process. Suppose two Enigma messages are encrypted with the same wheel order and happen to step onto the exact same three-letter rotor alignment during transmission. From that character onward, the two messages enter depth. Both machines then apply the exact same sequence of alphabetical substitutions until a middle rotor notches differently.
How can a cryptanalyst spot two ciphertexts in depth without knowing the plaintext or the keys? In natural German military prose, the probability that two independent plaintexts share the exact same letter at a given column is the index of coincidence $p_0 = \sum f_i^2 \approx \frac{1}{17} \approx 0.0588$. That elevation occurs because common letters like E and N collide frequently. In two random ciphertexts encrypted at unrelated rotor settings, every letter is equally likely. Repeats then occur only at the uniform chance rate $p_r = \frac{1}{26} \approx 0.0385$.
Inside Hut 8 and its outstation at Woburn Abbey, Women's Royal Naval Service operators punched the letters of each intercepted naval message as pinholes along a Banbury sheet. A clerk placed one punched sheet over another on a backlit glass table and slid it horizontally across every offset. At each alignment, the clerk counted single-letter repeats, two-letter dimers, and four-letter tetragrams where light shone through both sheets simultaneously.
To combine dozens of weak repeat counts with partial bigram clues from captured patrol vessels—such as the weather ship Krebs boarded near the Lofoten Islands in March 1941 and the submarine U-110 captured in May 1941—Turing and his statistical assistant I. J. Good invented a calculus of sequential Bayesian inference. Turing defined the weight of evidence in favor of a hypothesis $H$ (that two messages are in depth) against its negation $\neg H$ given observed repeat evidence $E$. He measured this weight in a logarithmic unit named the ban, and its tenth, the deciban:
$$W(H : E) = 10 \log_{10} \left( \frac{P(E \mid H)}{P(E \mid \neg H)} \right) = 10 \log_{10} \left( \frac{O(H \mid E)}{O(H)} \right)$$Here $O(H) = P(H) / P(\neg H)$ is the prior odds ratio and $O(H \mid E)$ is the posterior odds ratio after observing the repeat pattern $E$. Taking ten times the base-ten logarithm converted Bayes's multiplicative rule into simple addition on a clerk's scratchpad. The formula is identical in structure to the acoustic decibel and Claude Shannon's later bit of mutual information. Starting from prior odds of roughly $-32$ decibans (the one-in-sixteen-hundred chance of two random rotor starts colliding), each four-letter repeat added roughly $+20$ decibans. Once the total score crossed $+20$ decibans, the posterior odds reached one hundred to one. During the height of the U-boat wolfpack campaign in 1941 and 1942, those sliding paper sheets in Hut 8 reduced the number of three-rotor wheel orders that the Bombes had to test from 336 down to fewer than twenty.