| Break an Enigma Key with Bombe |
IntroductionThe machine room (MR) verify each good stop received from Hut 1 and then tries to find the full key (all steckers and Ringstellung). The details of the operations1. Test again the stopFor each good stop received from Hut 1, an MR cryptologist tests the stop again. This allows him to check if the operator of Hut 1 did not make a mistake but also to find the steckers deduced during the stop check. If this stop corresponds to the right key, we therefore have the Walzenlage and some of the steckers.2. Deciphering of letters outside the menuThen, the cryptologist tries to decipher the letters of the cryptogram which are not part of the menu. If he succeeds, not only he will be able to know if the right key has been found but also he will find the missing steckers. He will also be able to find the turnover of the right rotor and possibly (if the message is long) the turnover of the middle rotor.3. Searching of the complete RingstellungBy knowing the turnover of the right rotor as well as the name of the rotor, he can deduce the third letter of the ringstellung. By trying all the possible ringstellungs (26x26) corresponding to all the possible values of the first two letters, he can, each time, decipher the indicator and find the message key. If this allows the message to be decoded, the ringstellung has been found.ExampleExample of a Crib and a MenuFor the Crib and the following menu:
fx?/fy? vwxyz abcde fghij klmno pqrst u (1st Stop)
on?/oo? jklmn opqrs tuvwx yzabc defgh i (2nd Stop)
zz?/za? abcde fghij klmno pqrst uvwxy z (The starting position)
Crypto WCGLV EUTCF ZTQSA WJCSR FQOOZ Z
Crib KEINE BESON DEREN VORKO MMISS E
**** ***** (turnover) ***** * (letters of the menu)
Menu:
B V U zd zj au av
Input 1\ze|10/11 L----N----F----M----Q
letter: E zf\ | /zg 12 13 14 15
zb \|/ az ay zh
C-------E-------Z-------S-------T
1 loop \ 2 6 5 /ax 7
1 aux.chain \ zi 4/ aw zc
16 letters \-----------------O-------I-------G
3 8 9
1) Test the good Stop: 134, FXA, Stecker EEMenu letters Key->Lamp Grund Stecker EE self-stecker 1 B E->K (zf)(fxa) BK 2 E E->W (zb)(fxw) CW 3 C W->O (zi)(fxd) OO self-stecker 4 O O->J (ax)(fys) SJ 5 S J->Y (ay)(fyt) ZY 6 Z Y->E (az)(fyu) EE closure confirmation 7 T J->Q (zh)(fxe) TQ 8 I O->P (aw)(fyr) IP 9 G P->A (zc)(fxx) GA 10 V E->V (ze)(fxz) VV self-stecker 11 U E->U (zg)(fxb) UU self-steckerWe know the following steckers: BK, CW, GA, IP, TQ, ZY, SJ et EE, OO, VV, UU At the position "at" (fyo), the letter O (self-steckered) is ciphered to D, then this letter is steckered by R (DR). At the position "as" (fyn), the letter S is steckered to J. The encryption of J gives L, then we have the stecker K. This is impossible because we have already the stecker BK. Perhaps this is the position of the turnover. Then the position is as "zs" (fxn). The letter S is steckered to J. The encryption of J gives A, then we have the stecker AK, but we have already the stecker GA, then it is impossible. This stop is bad. 2) Test the good Stop: 413, ONO, Stecker EHMenu letters Key->Lamp Grund Stecker EH 1 B H->B (zf)(ono) BB Self-stecker 2 E H->A (zb)(onk) AC 3 C A->J (zi)(onr) OJ 4 O J->Z (ax)(oog) SZ 5 S Z->S (ay)(ooh) SZ Stecker confirmation 6 Z S->H (az)(ooi) EH closure confirmation 7 T Z->Y (zh)(onq) TY 8 I J->V (aw)(oof) IV 9 G V->G (zc)(onl) GG Self-stecker 10 V H->I (ze)(onn) IV Stecker confirmation 11 U H->X (zg)(onp) UXWe have obtained the following steckers: BB, GG, EH, AC, OJ, SZ, TY, IV, UX. At the position "at"(ooc), the letter O is steckered to J. The encryption of J gives M, then we have the stecker MR. At the position "as"(oob), the letter S is steckered to Z. The encryption of Z gives K, then K is self-steckered. At the position "ar"(ooa), the letter C is steckered to A. The encryption of A gives M, then we have again the stecker MR. This is a confirmation. Let’s go: At the position "aq"(ooz), the letter O is steckered to J. The encryption of J gives O, then we have the stecker OJ. It is again a confirmation. Let’s go: At the position "ap"(ooy), the letter V is steckered to I. The encryption of I gives N, then we have the stecker NW. It is possible. At the position "ao"(oox), the letter A is steckered to C. The encryption of C gives W, we have again the stecker NW: another confirmation. At the position "an"(oow), the letter E is steckered to H. The encryption of H gives Z, then we have again the stecker SZ. At the position "am"(oov), the letter R is steckered to M. The encryption of M gives Q, then Q is self-steckered. At the position "al"(oou), the letter E is steckered to H. The encryption of H gives Y, then we have again the stecker TY. At the position "ak" (oot), the letter Z is steckered to S. The encryption of S gives I, then we have the stecker ID. This is impossible. Perhaps we have a turnover? If the position is ONK, the encryption of S gives D, then D is self-steckered. We can now explore the letters that belong to the outer string. At the position "zd"(onm), the letter N is steckered to W. The encryption of W gives F, then we have the stecker FL, the last stecker. Now, we have the key almost entirely:
Then we try to decipher the cryptogram with the following partial key: (Walz: IV,I,III; Stecjers: EH, AC, OJ, SZ, TY, IV, UX, NW, MR, FL; Ring: AAC and grund: ONK): Cryptogram and plain: WCGLV EUTCF ZTQSA WJCSR FQOOZ Z KEINE BESON DEREN VORKO MMISS E Marvelous! It is correct. We need to know the beginning of the Ringstellung. Then, we try all combination of ringstellung (26x26). For each combination, we try to decipher the message key (grund: ROM, indicator: LNS) and then we try to decipher the beginning of the message with the message key found.
If the ring is AAC, then the message key is PED, then the beginning of message is GVZ... Yeh! We have found the whole key:
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