How we Found the Solution

Having decided upon the outlined attack, we tried it by creating an Enigma simulator, encrypting English texts, and letting the program decrypt them. The success rate was fairly high; the wheel and ring settings were always correct, but the plugboard settings were sometimes slightly off.

On October 7, Gunnar and Staffan felt ready to go and let a group of workstations try all the possible settings, storing the 100 most likely decryptions for manual inspection. In the morning, we were disappointed: None of the candidate decryptions looked even remotely like text in any language. What had gone wrong? An intense testing and debugging pass commenced, but no bugs were found.

The explanation came a couple of days later when Stage 7 was cracked: The plaintext for that stage revealed that Singh had been mischievous when constructing this stage: The first two wheels should actually be exchanged! Confident that Stage 8 would succumb immediately, the three team members who found the solution to Stage 7 pulled an all-nighter running the Enigma cracker with the correct wheel settings. Again, nothing sensible was produced by the program.

At this point we knew that we had the correct wheel setting, but our program still failed to find the solution. So what could be wrong? We looked for bugs in the program, but could not find any. Turning to the algorithmic approach, we decided on changing scoring method and removed the assumption that the plaintext was in German, instead trying the less efficient but more general method of assuming that the correct plaintext had the most skewed frequency distribution. Again, the workstations were set to their work during the night.

When Gunnar came to work the next morning, he found the following output from the program.

Highest scores:
  10356.00  OUA  FA  E->I  SBTDEFGHINLKUJWPQRACMVOXYZ
  10354.00  OUA  FP  E->I  SBTDEFGHINLKUJWPQRACMVOXYZ
  10232.00  OUA  FA  E->X  SQCDEFGHRNLKUJTPBIAOMVWXYZ
  10220.00  OUA  FP  E->X  SQCDEFGHRNLKUJTPBIAOMVWXYZ
  10206.00  OUB  FB  E->I  SBTDEFGHINLKUJWPQRACMVOXYZ
  10168.00  OUA  EA  E->I  SBTDEFGHINLKUJWPQRACMVOXYZ
  10168.00  OUA  EP  E->I  SBTDEFGHINLKUJWPQRACMVOXYZ
  10150.00  OUB  FC  E->I  SBTDEFGHINLKUJWPQRACMVOXYZ
  10112.00  OUA  EA  E->X  SQCDEFGHRNLKUJTPBIAOMVWXYZ
  10104.00  OUA  EP  E->X  SQCDEFGHRNLKUJTPBIAOMVWXYZ

The second column gives the initial position of the wheels, the third column gives the ring settings, the fourth column the mapping of the letter E, and the sixth column the rest of the plugboard settings. The best solution corresponds to a plugboard with 7 swaps: E $\leftrightarrow$ I, A $\leftrightarrow$ S, C $\leftrightarrow$ T, J $\leftrightarrow$ N, K $\leftrightarrow$ L, M $\leftrightarrow$ U, O $\leftrightarrow$ W. When Gunnar used this plugboard to decrypt the ciphertext, he got the following plaintext.

DASXATESUNGKOCRWXISWJBLHWCXXSWEXEXNEUIXELWHAELAKEINEHJIWWEIL
KRVXCIEXVIWXXESXEAWKCDIENOXISWXXITHXHABEXVASXLINKSSWEHEMDEXB
YWEXDEXESTHLUESSEPSXENJDETKWXXESXISWIEINSXNIESXGERCXEHNSXZER
MXZERCXEVESXEIOSXEINSXEITHEPRCGRAMMIERWEXDESXUNDXENWDETKWEXD
ASSXDASXOCRWXYEBUGGERXCEDNNESYMIWXOEMXUNWEBSWEHENDENXSTHJUEU
SELXENWKCDIERBXONRDFALJIRESULWAWXDIEXUNWENSWEHENDENXSTHRISWZ
EITHENXHAW

As we can see this look almost like German. Some of the plugboard connections are however wrong. If we look carefully we see the ``words'' STHLUESSE, PRCGRAMMIER, and RESULWAW in the plaintext. It looks as if C should not be swapped at all, and that instead O and T should be swapped. If we then leave W unmodified we get 6 swaps altogether: E $\leftrightarrow$ I, A $\leftrightarrow$ S, J $\leftrightarrow$ N, K $\leftrightarrow$ L, M $\leftrightarrow$ U, O $\leftrightarrow$ T.

Using these settings the complete plaintext emerged along with the codeword PLUTO:

DASXLOESUNGSWORTXISTXPLUTOXXSTUFEXNEUNXENTHAELTXEINEXMITTEIL
UNGXDIEXMITXDESXENTKODIERTXISTXXICHXHABEXDASXLINKSSTEHENDEXB
YTEXDESXSCHLUESSELSXENTDECKTXXESXISTXEINSXEINSXZEROXEINSXZER
OXZEROXEINSXEINSXEINSXXICHXPROGRAMMIERTEXDESXUNDXENTDECKTEXD
ASSXDASXWORTXDEBUGGERXWENNXESXMITXDEMXUNTENSTEHENDENXSCHLUES
SELXENTKODIERTXWIRDXALSXRESULTATXDIEXUNTENSTEHENDENXSCHRIFTZ
EICHENXHAT

The plaintext was in German, but the program still could not recognize the right plaintext when using frequency tables for German. What was the explanation for this? Inspecting the plaintext, it is clear that X was used as word delimiter and XX as sequence delimiter. As X is rare in most German texts, the approach based on frequency tables failed to recognize the correct plaintext.