The methode de batons, or rod method was developed in- dependently by Knox, Candela, and - no doubt - others. It can be used against unsteckered Enigmas when the rotor wiring is known. During the Spanish Civil War all sides used the unsteckered commercial Enigmas. Unsteckered Enigmas were also used during WWII by the Swiss and the Abwehr (German espionage service). For this example we guess (or know) that the fast rotor is 'ROTOR I'. This rotor provides the alphabets listed below. Write N_k for the permutation at step k. Then we have the relations N_k.C = C.N_(k-1) where C is the cyclic permutation (a b c ... y z). Recall that we write our functions on the right. 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 step 0 E K M F L G D Q V Z N T O W Y H X U S P A I B R C J 1 J L E K F C P U Y M S N V X G W T R O Z H A Q B I D 2 K D J E B O T X L R M U W F V S Q N Y G Z P A H C I 3 C I D A N S W K Q L T V E U R P M X F Y O Z G B H J 4 H C Z M R V J P K S U D T Q O L W E X N Y F A G I B R 5 B Y L Q U I O J R T C S P N K V D W M X E Z F H A G O 6 X K P T H N I Q S B R O M J U C V L W D Y E G Z F A T 7 J O S G M H P R A Q N L I T B U K V C X D F Y E Z W O 8 N R F L G O Q Z P M K H S A T J U B W C E X D Y V I R 9 Q E K F N P Y O L J G R Z S I T A V B D W C X U H M 10 D J E M O X N K I F Q Y R H S Z U A C V B W T G L P I 11 I D L N W M J H E P X Q G R Y T Z B U A V S F K O C 12 C K M V L I G D O W P F Q X S Y A T Z U R E J N B H 13 J L U K H F C N V O E P W R X Z S Y T Q D I M A G B 14 K T J G E B M U N D O V Q W Y R X S P C H L Z F A I 15 S I F D A L T M C N U P V X Q W R O B G K Y E Z H J 16 H E C Z K S L B M T O U W P V Q N A F J X D Y G I R 17 D B Y J R K A L S N T V O U P M Z E I W C X F H Q G 18 A X I Q J Z K R M S U N T O L Y D H V B W E G P F C 19 W H P I Y J Q L R T M S N K X C G U A V D F O E B Z 20 G O H X I P K Q S L R M J W B F T Z U C E N D A Y V 21 N G W H O J P R K Q L I V A E S Y T B D M C Z X U F 22 F V G N I O Q J P K H U Z D R X S A C L B Y W T E M 23 U F M H N P I O J G T Y C Q W R Z B K A X V S D L E 24 E L G M O H N I F S X B P V Q Y A J Z W U R C K D T 25 K F L N G M H E R W A O U P X Z I Y V T Q B J C S D We have a crib and a possible entry: r e c o n n a i s s a n c e U P Y T E Z O J Z E G B O T Lay out the alphabets for the plain text and the cipher text. The idea is that for plain text p_i and corresponding cipher text c_i we have (c_i)N_(i+s) = (p_i)N_(i+s)U for some fixed step s and some fixed involution U. step r e c o n n a i s s a n c e U P Y T E Z O J Z E G B O T 0 u L M Y W W E V S S E W M L a H C P L J Y Z J L D K Y P 1 R f E G X X J Y O O J X E F H w I Z F D G M D F P L G Z 2 N B j V F F K L Y Y K F J B Z S c G B I V R I B T D V G 3 X N D r U U C Q F F C U D N O P H y N J R L J N W I R Y 4 E R Z O q Q H K X X H Q Z R Y L I N r B O S B R J C O N If we try step s=0 we 5 W U L K N n B R M M B N L U E V A X U g K T G U O Y K X get the following facts about 6 L H P U J J x S W W X J P H Y C F D H A u B A H I K U D the involution U 7 V M S B T T J a C C J T S M D U Z X M W B q W M P O B X 8 B G F T A A N P w W N A F G E J V C G I T M i G Q R T C u f j r q n x a w b d r m h 9 V N K I S S Q L B b Q S K N W T H D N M I J M n Y E I D a w c y r g u q i n n d s q 10 A O E S H H D I C C d H E O B Z L V O P S F P O n J S V 11 B W L Y R R I E U U I r L W V T O A W C Y P C W J d Y A This doesn't work since, for 12 T L M S X X C O Z Z C X m L R Y B U L H S W H L G K s U example, it has U swapping 13 Y H U X R R J V T T J R U h D Z G Q H B X O B H C L X q 'w' with both 'i' and 'f'. 14 S E J Y W W K N P P K W J E H R A C E I Y D I E M T Y C 15 O A F Q X X S C B B S X F A K W H G A J Q N J A T I Q G 16 A K C V P P H M F F H P C K X Q I J K R V T R K L E V J 17 E R Y P U U D S I I D U Y R C M Q W R G P N G R A B P W 18 H J I L O O A M V V A O I J W Y F B J C L S C J K X L B 19 U Y P X K K W R A A W K P Y D C B V Y Z X T Z Y Q H X V 20 Z I H B W W G S U U G W H I E F Y C I V B L V I K O B C etc. 21 T O W E A A N K B B N A W O M S U D O F E Q F O P G E D 22 A I G R D D F P C C F D G I B X E L I M R K M I Q V R L 23 B N M W Q Q U J K K U Q M N X R L A N E W G E N I F W A 24 j O G Q V V E F Z Z E V G O u Y D W O T Q S T O N L Q W 25 Y g L X P P K R V V K P L G Q z S T G D X W D G H F X T 0 U L m Y W W E V S S E W M L A H c P L J Y Z J L D K Y P But if we try step s=24 1 R F E g X X J Y O O J X E F H W I z F D G M D F P L G Z we get the following facts 2 N B J V f F K L Y Y K F J B Z S C G b I V R I B T D V G about the involution U 3 X N D R U u C Q F F C U D N O P H Y N j R L J N W I R Y 4 E R Z O Q Q h K X X H Q Z R Y L I N R B o S B R J C O N j g m g f u h r w c n s e w 5 W U L K N N B r M M B N L U E V A X U G K t G U O Y K X u z c z b j o t a m q e s a 6 L H P U J J X S w W X J P H Y C F D H A U B a H I K U D 7 V M S B T T J A C c J T S M D U Z X M W B Q W m P O B X This works just fine, and 8 B G F T A A N P W W n A F G E J V C G I T M I G q R T C we have the starting position 9 V N K I S S Q L B B Q s K N W T H D N M I J M N Y e I D of the fast rotor. We also 10 A O E S H H D I C C D H e O B Z L V O P S F P O N J s V know that U contains the 11 B W L Y R R I E U U I R L w V T O A W C Y P C W J D Y a transpositions 12 T L M S X X C O Z Z C X M L R Y B U L H S W H L G K S U 13 Y H U X R R J V T T J R U H D Z G Q H B X O B H C L X Q (ju)(gz)(mc)(fb)(ho)(rt)(aw) 14 S E J Y W W K N P P K W J E H R A C E I Y D I E M T Y C (cm)(nq)(es) 15 O A F Q X X S C B B S X F A K W H G A J Q N J A T I Q G 16 A K C V P P H M F F H P C K X Q I J K R V T R K L E V J The middle and slow rotors 17 E R Y P U U D S I I D U Y R C M Q W R G P N G R A B P W along with their ring set- 18 H J I L O O A M V V A O I J W Y F B J C L S C J K X L B tings can now be determined 19 U Y P X K K W R A A W K P Y D C B V Y Z X T Z Y Q H X V by looking up the above 20 Z I H B W W G S U U G W H I E F Y C I V B L V I K O B C involution in a catalogue 21 T O W E A A N K B B N A W O M S U D O F E Q F O P G E D containing (2)(26)(26)=1352 22 A I G R D D F P C C F D G I B X E L I M R K M I Q V R L entries. 23 B N M W Q Q U J K K U Q M N X R L A N E W G E N I F W A 24 J O G Q V V E F Z Z E V G O U Y D W O T Q S T O N L Q W 25 Y G L X P P K R V V K P L G Q Z S T G D X W D G H F X T If the middle rotor turns during a long crib we can determine the position of the middle rotor without recourse to a catalog. Here's an example from Deavours by way of Bauer. g e n e r a l f e l d m a r s c h a l l k e s s e l r i n g L S H X B T F W U I O V B C A R X S N C V Z Y X N J B F W B Suppose we have a hit in the first 10 positions with initial step s=17 (so we have alphabets 17 through 26). g e n e r a l f e l L S H X B T F W U I and that this provides the following 2-cycles as parts of U1 (E M)(R X)(H O)(F V)(D L)(T Y)(G U) The remaining text is also linked up (with alphabets 1,2,...) and provide the 2-cycles as parts of U2. (Note: U2 not U1, the middle rotor has turned.) (B N)(D R)(A W)(Q X)(J L)(S M)(P Y)(F H)(C I)(E O)(T Z) The key is to notice that, for some shift s of the middle rotor M and some fixed involution U, U1.M_s = M_s.U U2.M_(s+1) = M_(s+1).U Or looking at it another way, conjugating U1 by M_s and U2 by M_(s+1) produce the same involution. Suppose the shifted alphabets of the middle rotor M are given by the table below. step 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 0 L W F T B A X J D S C K P R Z Q Y O E H U G M I V N 1 O M X G U C B Y K E T D L Q S A R Z P F I V H N J W 2 X P N Y H V D C Z L F U E M R T B S A Q G J W I O K 3 L Y Q O Z I W E D A M G V F N S U C T B R H K X J B : : : : : : : : : : : : : : : : : : : : : : : : : : : -------------------------------------------------------------- We'll try s=0. Conjugating the 2-cycles of U1 by M_0 produces (B P)(O I)(J Z)(A G)(T K)(H V)(X U) Conjugating the 2-cycles of U2 by M_1 produces (W R)(G Z)(O H)(R N)(E D)(P L)(A J)(C Y)(X K)(U S)(F W) These don't match. In particular (O I) and (O H) conflict, as do (A G) and (A J), (J Z) and (G Z), etc. -------------------------------------------------------------- Now try s=1. Conjugating the 2-cycles of U1 by M_1 produces (U L)(Z N)(Y S)(C V)(G D)(F J)(B I) Conjugating the 2-cycles of U2 by M_2 produces (P M)(Y S)(X W)(B I)(L U)(A E)(T O)(V C)(N Z)(H R)(Q K) Five matches, no conflicts. I'm convinced. We have the middle rotor lined up. --------------------------------------------------------------