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Home Page Hebern's machines The 5 rotors machine home page |
IntroductionJust knowing the wiring of the rotor in position V (right rotor, fast rotor), alone allows you to decipher several lines of text of 26 letters. Which in turn allows you to find the wiring of the rotor in position I (left rotor, medium rotor). This page describes the method to achieve this. Then, this knowledge accelerates decryption and also makes it possible to reconstruct the stators S equivalent to rotors II, III and IV for a message or part of a message. This knowledge in turn allows almost all cryptograms to be deciphered relatively easily. Example of reconstruction of rotor I wiringWe start from the following assumptions: for a cryptogram of several hundred letters, we know the cryptogram, the plain text corresponding (or at least between five to ten lines), the input and output permutations (Keyboard and Lampboard), the wiring of the Rotor V as well as the external key (position of the rotors). If we know the external key, we can cut the cryptogram into slices of 26 letters, each using a stationary R1 rotor. In addition, for the entire part of the cryptogram that we choose, we can have the rotors R2, R3 and R4 stationary (we know when R3 is moving). Note: to facilitate my reasoning, I did not take a realistic example, but I invented one to be able to be sure of my deductions. The chapter "Find the rotor in position I" (link) whose subject is the search for the rotor which is in position I, the wiring of the rotors being known, follows a realistic example (taken from the book by Deavors & Kruh). These two problems, find the rotor in position I and finding the rotor wiring in position I indeed follow the same approach. On the other hand, searching for the rotor in position I requires much less data. We create the data (plain text, cryptogram, etc.)- Here is the plain text: Note: It corresponds to the reverse permutation of the Keyboard (LFS) repeated several times. C:\H5_TOOLS> more keyb_inv.pln BSXRZTKDNGCHMVOLYQEUPWJAIF BSXRZTKDNGCHMVOLYQEUPWJAIF BSXRZTKDNGCHMVOLYQEUPWJAIF BSXRZTKDNGCHMVOLYQEUPWJAIF- Here is the cryptogram: Note: I use the debug mode of my simulator. So I know the details for each plain text letter (p): its value after the input permutation (pK), its encryption by Rotor R1, then the action of the other rotors R2, R3, R4, then the action of the rotor R5, and finally the action of the output permutation, the Lampboard (L). We finally obtain the encrypted letter (c). The program then displays the external key (the left ratchet, the position of the 5 rotors, the right ratchet). We can see that the rotors R2, R3, R4 are not moving forward (position AAA). C:\H5_TOOLS> python hebern5_tui.py -E ZZAAAZN -D < MSGS\keyb_inv.pln 001: B ->AGUSUK-> Q AAAAAAO 002: S ->BAIUSQ-> S AAAAABP 003: X ->CDCXDY-> G AAAAACQ 004: R ->DBZDAD-> E AAAAADR 005: Z ->EOLBLD-> E AAAAAES 006: T ->FCNHCR-> H AAAAAFT 007: K ->GTMMNL-> V AAAAAGU 008: D ->HKEWEB-> Y AAAAAHV 009: N ->INWQTI-> F AAAAAIW 010: G ->JUGTPV-> R AAAAAJX 011: C ->KZRNYS-> C AAAAAKY 012: H ->LXYZZD-> E AAAAALZ 013: M ->MIPOIY-> G AAAAAMA 014: V ->NWAPOD-> E AAAAANB 015: O ->OHDFGW-> Z AAAAAOC 016: L ->PFKYHD-> E AAAAAPD 017: Y ->QQHAFE-> U AAAAAQE 018: Q ->RYBJQF-> M AAAAARF 019: E ->SJJGWX-> A AAAAASG 020: U ->TVQVKN-> W AAAAATH 021: P ->UPFLXH-> D AAAAAUI 022: W ->VMORJC-> O AAAAAVJ 023: J ->WETKBP-> N AAAAAWK 024: A ->XLVERS-> C AAAAAXL 025: I ->YSSCVT-> I AAAAAYM 026: F ->ZRXIMJ-> J AAAAAZN 027: B ->AZRNYE-> U BBAAAAO 028: S ->BCNHCF-> M BBAAABP 029: X ->CAIUSI-> F BBAAACQ 030: R ->DNWQTP-> N BBAAADR 031: Z ->EBZDAT-> I BBAAAES 032: T ->FSSCVA-> T BBAAAFT 033: K ->GJJGWN-> W BBAAAGU 034: D ->HMORJR-> H BBAAAHV 035: N ->ITMMNM-> K BBAAAIW 036: G ->JYBJQD-> E BBAAAJX 037: C ->KWAPOU-> L BBAAAKY 038: H ->LHDFGQ-> S BBAAALZ 039: M ->MVQVKG-> X BBAAAMA 040: V ->NGUSUJ-> J BBAAANB 041: O ->OETKBT-> I BBAAAOC 042: L ->PPFLXG-> X BBAAAPD 043: Y ->QXYZZR-> H BBAAAQE 044: Q ->RIPOIV-> R BBAAARF 045: E ->SUGTPE-> U BBAAASG 046: U ->TOLBLE-> U BBAAATH 047: P ->ULVERO-> B BBAAAUI 048: W ->VDCXDJ-> J BBAAAVJ 049: J ->WKEWEJ-> J BBAAAWK 050: A ->XRXIMM-> K BBAAAXL 051: I ->YQHAFI-> F BBAAAYM 052: F ->ZFKYHO-> B BBAAAZN 053: B ->ABZDAF-> M CCAAAAO 054: S ->BZRNYL-> V CCAAABP 055: X ->CMORJG-> X CCAAACQ 056: R ->DAIUSR-> H CCAAADR 057: Z ->ERXIMU-> L CCAAAES 058: T ->FIPOIU-> L CCAAAFT 059: K ->GLVERF-> M CCAAAGU 060: D ->HSSCVM-> K CCAAAHV 061: N ->IXYZZO-> B CCAAAIW 062: G ->JVQVKI-> F CCAAAJX 063: C ->KGUSUN-> W CCAAAKY 064: H ->LUGTPU-> L CCAAALZ 065: M ->MFKYHF-> M CCAAAMA 066: V ->NDCXDL-> V CCAAANB 067: O ->OOLBLY-> G CCAAAOC 068: L ->PWAPOR-> H CCAAAPD 069: Y ->QHDFGC-> O CCAAAQE 070: Q ->RTMMNG-> X CCAAARF 071: E ->SNWQTQ-> S CCAAASG 072: U ->TKEWES-> C CCAAATH 073: P ->UCNHCY-> G CCAAAUI 074: W ->VJJGWG-> X CCAAAVJ 075: J ->WQHAFU-> L CCAAAWK 076: A ->XPFLXN-> W CCAAAXL 077: I ->YETKBO-> B CCAAAYM 078: F ->ZYBJQI-> F CCAAAZN 079: B ->AYBJQY-> G DDAAAAO 080: S ->BLVERC-> O DDAAABP 081: X ->CZRNYD-> E DDAAACQ 082: R ->DQHAFZ-> P DDAAADR 083: Z ->EHDFGK-> Q DDAAAES 084: T ->FKEWEE-> U DDAAAFT 085: K ->GRXIMX-> A DDAAAGU 086: D ->HWAPON-> W DDAAAHV 087: N ->IUGTPD-> E DDAAAIW 088: G ->JFKYHP-> N DDAAAJX 089: C ->KTMMNB-> Y DDAAAKY 090: H ->LETKBK-> Q DDAAALZ 091: M ->MCNHCD-> E DDAAAMA 092: V ->NNWQTA-> T DDAAANB 093: O ->OVQVKQ-> S DDAAAOC 094: L ->PGUSUU-> L DDAAAPD 095: Y ->QSSCVS-> C DDAAAQE 096: Q ->RMORJO-> B DDAAARF 097: E ->SJJGWX-> A DDAAASG 098: U ->TBZDAY-> G DDAAATH 099: P ->UIPOIZ-> P DDAAAUI 100: W ->VPFLXI-> F DDAAAVJ 101: J ->WOLBLA-> T DDAAAWK 102: A ->XDCXDI-> F DDAAAXL 103: I ->YXYZZN-> W DDAAAYM 104: F ->ZAIUSC-> O DDAAAZN Show data as lines with rotor I stationary- We represent the previous data in the form of lines made of 26 letters: Each line of 26 letters in the cryptogram (line “c”) corresponds to a slice of cipher text for which the rotor R1 (here named R) does not advance. On the other hand, when we move to the next line c, rotor R1 has advanced one step (i=0 to 3). For the entire cryptogram, the permutation S (R2.R3.K4) is the same because these rotors do not move forward.
Note: In my reasoning, I do not use the pKR lines (the result of
the encryption by the rotor R1) nor the value m which corresponds
to
the index of the offset array. I only used it to check my
reasoning. Note that for example the different lines pKR(i)
correspond to the values of the Rotor I wiring, as they appear
in the encryption table (GADBO...),
(ZCANBS...).... The pK lines
correspond to the alphabet (ABC...). m,(i=0) 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 6 7 8 920 1 2 3 4 5 p (plain) B S X R Z T K D N G C H M V O L Y Q E U P W J A I F c (cipher) Q S G E E H V Y F R C E G E Z E U M A W D O N C I J pK (Keyb) A B C D E F G H I J K L M N O P Q R S T U V W X Y Z pKRS (c’) U S D A L C N E T P Y Z I O G H F Q W K X J B R V M pKR (R1(0)) G A D B O C T K N U Z X I W H F Q Y J V P M E L S R m,(i=1) 1 2 3 4 5 6 7 8 910 1 2 3 4 5 6 7 8 920 1 2 3 4 5 6 p (plain) B S X R Z T K D N G C H M V O L Y Q E U P W J A I F c (cipher) U M F N I T W H K E L S X J I X H R U U B J J K F B pK (Keyb) A B C D E F G H I J K L M N O P Q R S T U V W X Y Z pKRS (c’) Y C S T A V W J N Q O G K U B X Z I P L R D E M F H pKR (R1(1)) Z C A N B S J M T Y W H V G E P X I U O L D K R Q F m,(i=2) 2 3 4 5 6 7 8 910 1 2 3 4 5 6 7 8 920 1 2 3 4 5 6 7 p (plain) B S X R Z T K D N G C H M V O L Y Q E U P W J A I F c (cipher) M V X H L L M K B F W L M V G H O X S C G X L W B F pK (Keyb) A B C D E F G H I J K L M N O P Q R S T U V W X Y Z pKRS (c’) A Y J S M I R V Z K U P H D L O G N T E C W F X B Q pKR (R1(2)) B Z M A R I L S X V G U F D O W H T N K C J Q P E Y m,(i=3) 3 4 5 6 7 8 910 1 2 3 4 5 6 7 8 920 1 2 3 4 5 6 7 8 p (plain) B S X R Z T K D N G C H M V O L Y Q E U P W J A I F c (cipher) G O E P Q U A W E N Y Q E T S L C B A G P F T F W O pK (Keyb) A B C D E F G H I J K L M N O P Q R S T U V W X Y Z pKRS (c’) Q R Y F G E M O P H N B C T K U V J W A I X L D Z S pKR (R1(3)) Y L Z Q H K R W U F T E C N V G S M J B I P O D X A Let p: plain letter, c: cipher letter, K: keyboard permutation, R1, R2, R3, R4, R5: permutations of rotors, L: lampboard permutation. Here is the encryption formula (cf. description of the 5 rotors machine): p.K.R1(i).R2(j).R3(k).R4(l).R5(m).L = c If we know the external key, we can cut the cryptogram into slices of 26 letters which use a fixed rotor R1 for each slice. In addition, for the entire part of the cryptogram that we choose, we can have the rotors R2, R3 and R4 stationary. We will be able to simplify the previous formula: Let c' be the encrypted letter having removed the influence of lampboard and rotor R5 (R5(m).L). Let R be the permutation R1. Let S be the permutation R2(j)R3(k)R4(l), the previous formula simplifies: pKR(i)S = c’Let c’’ be the cipher letter where the influence of S is removed, we obtain: pKR(i) = c’’Although p, K and c' are known, we do not know c’'. On the other hand, if we take several ciphered letters which give the same value of c’, we can arbitrarily fix c’’ which corresponds to c’S_inverse. For example, we can set c’’=c’. On the other hand, if we apply the following formula (cf. the concept of rotor) which corresponds to the value of the encrypted letter y(i) in relation to the plain letter x(i) and the offset D[m]: y(i) = x(i) + D[ x(i) + i ]We obtain the final formula (in the case where c' has the same value): c’ = pK + D[ pk + i ]1) In the case where c’ == N, we have: (i=0) N = G + D[ G + 0 ], 13 = 6 + D[6] => D[6] = 7 (i=1) N = I + D[ I + 1 ], 13 = 8 + D[8+1] => D[9] = 9 (i=2) N = R + D[ R + 2 ], 13 = 17 + D[17+2] => D[19] = 22 (i=3) N = K + D[ K + 3 ], 13 = 10 + D[10+3] => D[13] = 32) In the case where c’ == O, we have: (i=0) O = N + D[ N + 0 ], 14 = 13 + D[13] => D[13] = 1 (i=1) O = K + D[ K + 1 ], 14 = 10 + D[10+1] => D[11] = 4 (i=2) O = P + D[ P + 2 ], 14 = 15 + D[15+2] => D[17] = 25 (i=3) O = H + D[ H + 3 ], 14 = 7 + D[7+3] => D[10] = 7 We still do not know S, but we know two values of its wiring (relatively), if c' == N, then for the value O, c’[N] = c’[O] + 2 We can aggregate the two series of movements by taking the offsets where c’= N as reference: D[6]=7, D[9]=9, D[10] = 7+2 = 9, D[11]= 4+2 = 6, D[13]=3=1+2, D[15] = 16+2= 18, D[17] ) 25+2 = 1 If we continue with other series where the value of c' is identical, we can reconstruct all the shifts... but in relative terms! We can also reconstruct S but it is simpler to do so after we have finished reconstituting the rotor R1. We obtain the following displacements for permutation R (that of Rotor I). For information, we indicate the shifts deduced from the GADBO rotor wiring... (grs = genuine Rotor shifts). We can see that there is a gap of 6 between the two series. m 0 1 2 3 4 5 6 7 8 9 10 11 12 R 0 19 21 18 4 17 7 23 25 5 9 6 16 grs 6 25 1 24 10 23 13 3 5 11 15 12 22 m 13 14 15 16 17 18 19 20 21 22 23 24 25 R 3 13 10 20 1 11 22 15 11 2 8 14 12 grs 9 19 16 0 7 17 2 21 17 8 14 20 18 From the formula PI[x] = x + D[x] (cf. The concept of rotor), we can reconstruct the wiring for both series: R A U X V I W N E H O T R C Q B Z K S D P J G Y F M L grs G A D B O C T K N U Z X I W H F Q Y J V P M E L S R Note: The grs line gives us the rotor wiring and the R line corresponds to the G line of the encryption table for column U. We also notice that the pKR lines correspond to the first lines of the following encryption table: A B C D E F G H I J K L M N O P Q R S T U V W X Y Z A G A D B O C T K N U Z X I W H F Q Y J V P M E L S R B Z C A N B S J M T Y W H V G E P X I U O L D K R Q F C B Z M A R I L S X V G U F D O W H T N K C J Q P E Y D Y L Z Q H K R W U F T E C N V G S M J B I P O D X A E K Y P G J Q V T E S D B M U F R L I A H O N C W Z X F X O F I P U S D R C A L T E Q K H Z G N M B V Y W J G N E H O T R C Q B Z K S D P J G Y F M L A U X V I W H D G N S Q B P A Y J R C O I F X E L K Z T W U H V M I F M R P A O Z X I Q B N H E W D K J Y S V T G U L C J L Q O Z N Y W H P A M G D V C J I X R U S F T K B E K P N Y M X V G O Z L F C U B I H W Q T R E S J A D K L M X L W U F N Y K E B T A H G V P S Q D R I Z C J O M W K V T E M X J D A S Z G F U O R P C Q H Y B I N L N J U S D L W I C Z R Y F E T N Q O B P G X A H M K V O T R C K V H B Y Q X E D S M P N A O F W Z G L J U I P Q B J U G A X P W D C R L O M Z N E V Y F K I T H S Q A I T F Z W O V C B Q K N L Y M D U X E J H S G R P R H S E Y V N U B A P J M K X L C T W D I G R F Q O Z S R D X U M T A Z O I L J W K B S V C H F Q E P N Y G T C W T L S Z Y N H K I V J A R U B G E P D O M X F Q U V S K R Y X M G J H U I Z Q T A F D O C N L W E P B V R J Q X W L F I G T H Y P S Z E C N B M K V D O A U W I P W V K E H F S G X O R Y D B M A L J U C N Z T Q X O V U J D G E R F W N Q X C A L Z K I T B M Y S P H Y U T I C F D Q E V M P W B Z K Y J H S A L X R O G N Z S H B E C P D U L O V A Y J X I G R Z K W Q N F M T ConclusionWe can see that it is possible, after having reconstituted the rotor which is in position V, to reconstitute the Rotor which is in position I. On the other hand, many lines of 26 letters are required for which the plain text and the cipher text are known, between five and ten lines. Note: in the example, four lines may be enough to reconstruct the wiring of the rotor in position I. In fact, each line contains all the letters of the alphabet. But in ordinary text, this is not the case. Indeed, for ordinary plain text certain letters (E,T,A,O,N) are very frequent and others are completely absent. Web Links
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