Chapter 4: Analysis of a three wheels Hagelin Cipher Machine ---- Explanations Problem 1 ========= 1) First step : find the number of lugs K = ( C + P + 1) % 26 K[0] = ("Y" + "B" + 1i)%26 = 24 + 1 + 1 = 26 % 26 = 0 i 0 1 2 3 4 5 6 7 8 9 10 11 12 Cipher : Y J D O D V Q L K D O E E Plain  : B Y Z T H I S Z T I M E Z Key  : 0 8 3 8 11 4 9 11 4 12 1 9 4 Different key values : Theorical values : 0, x, y, z, x+y, x+z, y+z, x+y+z Experimental values : 0, 1, 3, 4, 8, 9, 11, 12 If x+y+z=12 and x=1 and y=3 then z=12-3-1=8 The only solution that explains these values is: x=1, y=3, z=8 There aren't any overlap. This hypothesis explains all key values: 11=z+y, 9=y+z, 4=x+y 2) Second step : find the number of lugs in front of each wheel P = (K -1 -C)%26 First hypothesis: - The number of lugs x is associated with Wheel 21 - The number of lugs y is associated with Wheel 19 - The number of lugs z is associated with Wheel 17 i 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 Cipher Y J D O D V Q L K D O E E Z X O U D Z I X O W R K 14221710 x(21) - - - - - 1 1 – 1 1 1 1 1 ]- - - - y(19) - - 3 – 3 3 – 3 3 3 - - 3 ]- - 3 – 3 3 z(17) - 8 – 8 8 – 8 8 – 8 – 8 - ]- 8 – 8 8 – 8 8 Key 0 8 3 811 4 911 412 1 9 4 11 01111 Plain B Y Z T H I S Z T I M E Z W D T A This hypothesis is not probable Second hypothesis: - The number of lugs x is associated with Wheel 17 - The number of lugs y is associated with Wheel 19 - The number of lugs z is associated with Wheel 21 i 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 Cipher Y J D O D V Q L K D O E E Z X O U D Z I X O W R K 14221710 x(17) - - - - - 1 1 – 1 1 1 1 1 ]- - - - - 1 1 - y(19) - - 3 – 3 3 – 3 3 3 - - 3 ]- - 3 – 3 3 z(21) - 8 – 8 8 – 8 8 – 8 – 8 - ]- 8 - 8 Key 0 8 3 811 4 911 412 1 9 4 3 9 411 Plain B Y Z T H I S Z T I M E Z O M M A The string "OMMA" smells good! We don't try the other combinations. 3) Third step: find all Pins. We search the following Pins: I 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 Cipher V A H U I I W H W A J V A T O H A D P A U H D U C 21 0 720 8 822 722 0 x(17) 1 1 1 1 1 y(19) - 3 3 3 - - 3 z(21) 8 – 8 8 – 8 – 8 - Key 9 41212 1 Plain N D E R S The unique word possible is COMMANDERS. Then we can place the string ZCOMMANDERSZ in the plain text. then Plain[20]=C and Wheel_21[20]= 0 Then we search new Pins in another place: I 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 Cipher V A H U I I W H W A J V A T O H A D P A U H D U C 21 0 720 8 822 722 0 0 315 020 7 320 2 x(17) 1 1 1 1 1 - ]- - - - - 1 1 – 1 1 1 1 1 y(19) - 3 3 3 - - 3 ]- - 3 – 3 3 – 3 3 3 - - z(21) 8 – 8 8 – 8 – 8 - -]- 8 – 8 8 – 8 8 Key 9 41212 1 8 0 412 11212 Plain N D E R S Z Z A W A R E Very few words begin by the string "AWARE", then the next letter is Z (word separator), or N or R or S, then the Wheel_17[46%17=13]=0 Continue to search other Pins: I 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 Cipher D O S D D R Z O R Y W Y K B C Q P H H C T M I E J 10 1 2 7 2 x(17) ]- - - - - 1 1 – 1 1 1 1 1 0 ]- - - - - 1 1 y(19) ]- - 3 – 3 3 – 3 3 3 - - 3 z(21) -]- 8 – 8 8 – 8 8 – 8 – 8 Key 4 111 011 Plain T Z I S I There are only 31 words begining by "I", with the string "SI" three letters away. The most probable word is "INVASION" in military context. Try this hyptothesis: I 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 Cipher D O S D D R Z O R Y W Y K B C Q P H H C T M I E J 10 1 21615 7 7 21912 8 4 9 x(17) ]- - - - - 1 1 – 1 1 1 1 1 - 1 - -]- - - - - 1 1 y(19) ]- - 3 – 3 3 – 3 3 3 - - 3 - - - 3 - z(21) -]- 8 – 8 8 – 8 8 – 8 – 8 Key 4 111 412 8 011 8 0 8 4 9 Plain T Z I N V A S I O N Z Z Z We put the new Pins found in place: i 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 Cipher Y J D O D V Q L K D O E E Z X O U D Z I X O W R K 82314221710 x(17) - - - - - 1 1 – 1 1 1 1 1 – 1 - -]- - - - - 1 1 - y(19) - - 3 – 3 3 – 3 3 3 - - 3 - - - 3 - ]- - 3 – 3 3 z(21) - 8 – 8 8 – 8 8 – 8 – 8 - 8 -]- 8 - 8 Key 0 8 3 811 4 911 412 1 9 4 8 0 3 9 411 Plain B Y Z T H I S Z T I M E Z Z C O M M A I 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 Cipher V A H U I I W H W A J V A T O H A D P A U H D U C 21 0 720 8 822 722 0 921 01914 7 0 315 020 7 320 2 x(17) 1 1 1 1 1 – 1 - -]- - - - - 1 1 – 1 1 1 1 1 – 1 - y(19) - 3 3 3 - - 3 - - - 3 - ]- - 3 – 3 3 – 3 3 3 - - z(21) 8 – 8 8 – 8 – 8 - -]- 8 – 8 8 – 8 8 Key 9 41212 1 8 4 8 0 0 412 11212 Plain N D E R S Z H A D Z A W A R E After the word "HAD" the letter "Z" is the most probable. The most pins have been found. We can try the values 0 or 8 for the last pins of wheel 21. I 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 Cipher V A H U I I W H W A J V A T O H A D P A U H D U C 21 0 720 8 822 722 0 921 01914 7 0 315 020 7 320 2 x(17) 1 1 1 1 1 – 1 - -]- - - - - 1 1 – 1 1 1 1 1 – 1 - y(19) - 3 3 3 - - 3 - - - 3 - ]- - 3 – 3 3 – 3 3 3 - - z(21) 8 – 8 8 – 8 – 8 - ? ? ? ? ? ? ?]- 8 – 8 8 – 8 8 Key(0) 9 41212 1 8 4 8 0 0 3 0 0 1 4 0 412 11212 Key(8) 811 8 8 912 Plain N D E R S Z H A D Z B E G M W Z A W A R E H U M 0 U E A crossword fan can guess that the missing word is "BECOME". A computer scientist can scan an English dictionnary with the following regular expression: "^[BU][EM].[GO][MU][WC]$" 4) Fourth step (final step) : recover the plain text All pins have been recovered! It is easy to find the plain text. Problem 2 ========= No explanation (the method used is the same as the previous one). Problem 3 ========= 1) First step: Find the Lugs setting (The "cage") i : 0 1 2 3 4 5 6 7 8 9 Cipher : G V V I K U M V M N Plain  : T H E Z E N E M Y Z Key  : 0 3 0 8 15 8 17 8 11 13 Different key values : { 0, 3, 8, 11, 13, 15, 17 } Only two cages can produce these values : (1) x=3, y=5, z=11, overlaps_xz=1, overlaps_yz=1, Differents values : { 0, 3, 5, 8, 11, 13, 15, 17 } (2) x=3, y=8, z=11, overlaps_xz=1, overlaps_yz=4, Differents values : { 0, 3, 8, 11, 13, 15, 17 } Remark: the two sets differ only by the value 5 which appears in the first cage. 2) Second step: find the following text (after "THEzENEMYz") i : 10 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 Cipher : W S Z J I I I B L A A E H O A I H W W O K= 0 D H A Q R R R Y O Z Z V S L Z R S D D L K= 3 G K D T U U U B R C C Y V O C U V G G O K= 5 I M F V W W W D T E E A X Q E W X I I Q K= 8 L P I Y Z Z Z G W H H D A T H Z A L L T K=11 O S L B C C C J Z K K G D W K C D O O W K=13 Q U N D E E E L B M M I F Y M E F Q Q Y K=15 S W P F G G G N D O O K H A O G H S S A K=17 U Y R H I I I P F Q Q M J C Q I J U U C Plain : S U F F E R E D Z H E A V Y Z Key : 1513 51513 013 511 8 5 5 313 0 We guess only one word  composed of 4 letters: ISNT There are several verbs of 5 letters but none which ends by the letter "S" There is only one word of 6 letters : GUNFIRE. There are several words of 7 letters : suffired, quivered, spidered. The word "SUFFERED" is the most probable. Then there are several possible words : COAST, MEDAL or HEAVY. The last one ("HEAVY") smells good Remark: qualifications at this stage are those of a good crossword fan. 3) Third step: we can rebuild Pins setting i  : 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 Cipher : G V V I K U M V M N W S Z J I I I B L A A E H O A Plain  : T H E Z E N E M Y Z S U F F E R E D Z H E A V Y Z Key  : 0 3 0 815 817 811131513 51513 013 511 8 5 5 313 0 W?  : - 3 - 3 – 3 3 3 – 3 – 3 - - 3 – 3 - - 3 - - 3 3 -] W ?  : - - - 5 5 5 5 5 - - 5 – 5 5 - - - 5 – 5 5 5 -]- - W ?  : - - - -11 -11 -11111111 -1111 -11 -11]- - - -11 - Overl  : 1 2 1 1 1 1 1 1 1 The Wheel#19 has 11 lugs, The Wheel#23 has 5 lugs and Wheel#25 has 3 lugs. 4) Fourth step: Recover plain text :it is easy to solve the cryptogram if we know the key. Problem 4 ========= No explanation (the method used is the same as the previous). Problem 5 ========= 1) We search plain text from the three messages (but the latter will not be full decrypted) Remark: we use known methods. 2) We get the key (only the first seventies values) Cipher text of first message: LZVXI QAXDF QVXCB GVEHT WOGVV TTDTZ VILXH GCJIY HVNPH NWZDO JBLVH LJPFT USAOY YTVHC Plain text of first message: WHENZ THEYZ RECEI VEDZI NTELL IGENC EZTHA TZAZM AJORZ OFFEN SIVEZ WASZI MPEND INGZZ Key : 8 7 0 11 8 – 10 8 2 2 5 - 8 0 0 7 10 – 2 0 8 7 2 10 8 11 7 7 _ 2 0 8 7 2 - 0 8 5 5 8 - 0 2 10 8 11 8 5 2 7 7 _ 2 2 5 8 2 – 2 10 7 0 7 – 8 10 8 5 2 7 8 5 2 2 _ 7 7 2 7 2 3) Find the Lugs settings (The "cage") Different key values: 0, 2, 5, 7, 8, 10, 11 Hypothesis: x = 2, y = 5, z = 7 , Then: x + z – o_xz = 8 => o_xz = 1 y + z – o_yz = 10 => o_yz = 2 x + y + z - (o_xy + o_xz + o_yz) = 11 => o_xy = 0 But we have: 7 = z or = x + y 4) Find the Pins settings and number of lugs of each wheel 1 2 3 i 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 k 8 7 011 810 8 2 2 5 8 0 0 710 2 0 8 7 210 811 7 7 2 0 8 7 2 0 ka 2 ? - 2 2 – 2 2 2 – 2 - - ? - 2 – 2 ? 2 – 2 2 ? ? 2 – 2 ? 2 - kb - ? - 5 – 5 - - - 5 - - - ? 5 - - - ? - 5 – 5 ? ? - - - ? - - kc 7 ? – 7 7 7 7 - - - 7 - - ? 7 - - 7 ? – 7 7 7 ? ? - - 7 ? - - ov -1 -3-1-2-1 -1 -2 -2-1-3 -1 If kc is tied to wheel#17, then kc[30%17=13]=0 and ka[13] = 2 and kb[13] = 5 We can deduce the indeterminate Pins (and lugs): Then Wheel#16 have 7 lugs front of it. i 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 k 8 7 011 810 8 2 2 5 8 0 0 710 2 0 8 7 210 811 7 7 2 0 8 7 2 0 k17 7 - – 7 7 7 7 - - - 7 - - - 7 - -]7 - – 7 7 7 7 - - - 7 - - - k19 - 5 - 5 – 5 - - - 5 - - - 5 5 - - - 5]- 5 – 5 – 5 - - - 5 - - k21 2 2 - 2 2 – 2 2 2 – 2 - - 2 - 2 – 2 2 2 –]2 2 – 2 2 – 2 2 2 - 5) We calculate the plain text which corresponds of the last crypto P = K – 1 - C i 70 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 c V R H W Y R X M I Q W H F L P Y F O Z H K O T L T 2117 72224172312 81622 7 5111524 51425 71014191119 k17 – 7 7 7 7 - - - 7 - - - 7 - -]7 - - 7 7 7 7 - - - k19 5 5 - - - 5]- 5 – 5 – 5 - - - 5 - - - 5 5 - - - 5 k21 2 2 - 2 - - 2 – 2 - 2 2 2 -]2 2 – 2 2 – 2 2 2 - 2 ov -3 -1 -1 -1 -3 -1-2-3-1 k 711 7 8 7 5 2 5 8 5 2 7 8 0 211 0 2 81011 8 2 0 7 P 11192511 813 4182514 525 21412122013 8 2 019 81413 L T Z L I N E S Z O F Z C O M M U N I C A T I O N Problem 6 ========= The two messages are identical until the 30th letter. We can suspect a "Stagger". Then, the following text of both messages are also identical except a slight difference. Perhaps the 31th of the first plain text is equal to letter "Z" and the cryptographer has forgotten to space words in the second message. P1[i]: The i th letter of first Plain Text. P2[i]: The i th letter of second Plain Text. C1[i]: The i th letter of first Cipher Text. C2[i]: The i th letter of second Cipher Text. P2[i] = P1[i] + (C1[i]-C2[i]), P1[31] = "Z" => P2[31] = Z + ("A"-"L") P1[31] = 25 + (0 – 11) = 14 = "O"  => P1[32]= "O" P2[2] = "O" + ("U"-"D") = 14 + (20-3) = 31 = 5 = "F" => P1[33]="F" etc... i 01234 56789 C1[i] VAUWB ZWNVV P1[i] ZOF. ... C2[i] VLDCS UQCOV P2[i] OF.. .. It smells good! It is easy to decipher the rest of the message. Problem 7 ========= We don’t use new methods. Problem 8 ========= It is assumed that the end of the cipher text ends with two letters "Z". k = c + p + 1 if n is the number of the last letter, then: k[n-2]= G+Z+1 = 8+25+1 = 8 k[n-1] = R+Z+1 = 17 We scan the text with the probable word ("THEzRESTzOF") 0: OABRQVFRHRAGD [14, 20, 9, 22, 16, 13, 10, 10, 1, 17, 15, 12, 3] 1: ABRQVFRHRAGDB [0, 21, 25, 21, 21, 23, 22, 0, 11, 0, 21, 9, 1] 2: BRQVFRHRAGDBT [1, 11, 24, 0, 5, 9, 12, 10, 20, 6, 18, 7, 19] 3: RQVFRHRAGDBTK [17, 10, 3, 10, 17, 25, 22, 19, 0, 3, 16, 25, 10] ... Then we retain only the solutions that contain key 8 and 17. The presence of key equal to 0 is desirable (because very probable). 10: AGDBTKBYMIEWA [0, 0, 11, 6, 19, 2, 6, 17, 6, 8, 19, 2, 0] 15: KBYMIEWAKNGFU [10, 21, 6, 17, 8, 22, 1, 19, 4, 13, 21, 11, 20] 41: QOPNHIMXOBNIN [16, 8, 23, 18, 7, 0, 17, 16, 8, 1, 2, 14, 13] 48: XOBNINBLXCOFV [23, 8, 9, 18, 8, 5, 6, 4, 17, 2, 3, 11, 21] The first solution (the 10th) seems the more likely. Different key values: - Experimental values: 0, 2, 6, 8, 11, 17, 19 - Theorical values: 0, x, y, z, x+y, x+z, y+z, x+y+z If we make the following assumption: x=2, y=6, z=11, x+y=8, x+z=13, y+z=17, x+y+z=19 Then we have the following key values: 0,2,6,8,11,13,17,19 Apart from the value of 13, the result is the same. There is a good chance that we have found the right cage. Then we use known methods. Problem 9 ========= We don’t use new methods. Problem 10 ========== 1) First step: find the cage (the lugs in front of each wheel) c2 = k2 – 1 -p c1 = k1 – 1 -p Delta = c2 – c1 = k2 – k1 -1 + 1 -p + p = k2 – k1 c2 – c1 = k2 – k1 => Delta_cipher = Delta_key Delta_key[i] From i=0 To i=n-1 (n == message length): 18 23 11 16 0 18 0 5 24 18 7 7 25 0 20 7 7 19 21 6 20 1 22 3 3 0 20 8 3 23 0 18 7 8 19 19 7 18 25 23 1 7 0 21 10 5 1 25 19 4 3 19 20 7 19 0 18 25 8 1 22 7 6 0 4 23 3 7 19 18 Delta_key = {0,1,2,3,4,5,6,7,8,10,11,16,18,19,20,21,22,23,24,25} We note the absence of the following key: 9,12,13,14,15,17 Where there were only two wheels, the number of Delta_key value was limited. With three wheels, it's huge. Delta_key = { 0,x,y,z,-x,-z,-z,x+y,x+z,y+z,x+y+z,-(x+y),...} But (x) value and (-x) can be guessed: They are respectively the minimum and maximum values. The experimental minimal value is (1) and maximal value is (25), then x = 1. It is a near certainty (except if (x+y+z)%26=1) For (y), it's less likely, but if (z) is greater than (y) and if (y) is very small, (y) can be guessed. The smallest value (except x) and the highest value (except -x) are respectively 3 and 23. Then y = 3 with an average certainty. x+y = 1+3=4, -(x+y)= 22, x-y=24, -x+y= 2, So far we have found the following values: Delta_key = {0,1,2,3,4,22,23,24,25, ???? } It remains to explain the following values: Delta_key = { ..., 5, 6, 7, 8, 10, 11, 16, 18, 19, 20, 21 } Find (z) is not simple. Suppose that z <= 16 If z=5, x+y+z=9 not very probable (there is not any 9) If z=6, -y-z=17 not very probable (there is not any 17) If z=8, -(x+y)-z=14 not very probable (there is not any 14) If z=10, x+y+z=14 it is the same If z=11, -y-z=12 not very probable (there is not any 12) There is z = 7. Let's try this value to explain the various remaining keys: z=7, -z=19, x+z=8, y+z=10, x+y+z=11, -(x+z)=18, -(y+z)=16, x-z=20, y-z=22, (x+y)-z=23, z-x=6, z-y=4, (x+z)-y=5, y-(x+z)=21, ... There are a few other values, but we have explained all possible values. We are not about to have found the right solution, but what we found is likely. x=1, y=3, z=7 3) Second step: Find the Pins settings Unlike the case with 2 wheels, you can not use anagraming to find the plain text, there are too many combinations. But attack by probable word is possible. Furthermore in this case, we can get two key sequences. Being given the word probable EVASION: 0 : RZVLKFWRQL [17, 8, 9, 7, 11, 24, 5, 6, 4, 11] 1 : ZVLKFWRQLG [25, 4, 25, 6, 6, 15, 0, 5, 25, 6] 2 : VLKFWRQLGS [21, 20, 24, 1, 23, 10, 25, 0, 20, 18] 3 : LKFWRQLGSG [11, 19, 19, 18, 18, 9, 20, 21, 6, 6] 4 : KFWRQLGSGE [10, 14, 10, 13, 17, 4, 15, 7, 20, 4] 5 : FWRQLGSGEB [5, 5, 5, 12, 12, 25, 1, 21, 18, 1] 6 : WRQLGSGEBM [22, 0, 4, 7, 7, 11, 15, 19, 15, 12] 7 : RQLGSGEBMH [17, 25, 25, 2, 19, 25, 13, 16, 0, 7] 8 : QLGSGEBMHK [16, 20, 20, 14, 7, 23, 10, 1, 21, 10] 9 : LGSGEBMHKP [11, 15, 6, 2, 5, 20, 21, 22, 24, 15] 10 : GSGEBMHKPS [6, 1, 20, 0, 2, 5, 16, 25, 3, 18] 11 : SGEBMHKPSS [18, 15, 18, 23, 13, 0, 19, 4, 6, 18] 12 : GEBMHKPSSW [6, 13, 15, 8, 8, 3, 24, 7, 6, 22] 13 : EBMHKPSSWH [4, 10, 0, 3, 11, 8, 1, 7, 10, 7] ... 0 : JWGBKXWWOD [9, 5, 20, 23, 11, 16, 5, 11, 2, 3] 1 : WGBKXWWODN [22, 15, 15, 6, 24, 15, 5, 3, 17, 13] 2 : GBKXWWODNZ [6, 10, 24, 19, 23, 15, 23, 18, 1, 25] 3 : BKXWWODNZF [1, 19, 11, 18, 23, 7, 12, 2, 13, 5] 4 : KXWWODNZFE [10, 6, 10, 18, 15, 22, 22, 14, 19, 4] 5 : XWWODNZFEV [23, 5, 10, 10, 4, 6, 8, 20, 18, 21] 6 : WWODNZFEVT [22, 5, 2, 25, 14, 18, 14, 19, 9, 19] 7 : WODNZFEVTO [22, 23, 17, 9, 0, 24, 13, 10, 7, 14] 8 : ODNZFEVTOD [14, 12, 1, 21, 6, 23, 4, 8, 2, 3] 9 : DNZFEVTODK [3, 22, 13, 1, 5, 14, 2, 3, 17, 10] 10 : NZFEVTODKY [13, 8, 19, 0, 22, 12, 23, 18, 24, 24] 11 : ZFEVTODKYM [25, 14, 18, 17, 20, 7, 12, 25, 12, 12] 12 : FEVTODKYMX [5, 13, 9, 15, 15, 22, 19, 13, 0, 23] 13 : EVTODKYMXD [4, 4, 7, 10, 4, 3, 7, 1, 11, 3] ... If Keys = {0,x,y,z,x+y,x+z,y+z,x+y+z}= {0,1,3,7,4,8,10,11} We are looking for clues that give likely keys for both Cryptograms: Only position 13 gives good keys in both messages. Now we can find the Pins settings: ---- First Message: i 13 14 15 16 17 18 19 20 21 22 Key 4 10 0 3 11 8 1 7 10 7 K? 1 - - - 1 1 1 - - - K? 3 3 - 3 3 - - - 3 - K? - 7 - - 7 7 - 7 7 7 ---- Second Message: i 13 14 15 16 17 18 19 20 21 22 Key 4 4 7 10 4 3 7 1 11 3 K? 1 1 - - 1 - - 1 1 - K? 3 3 - 3 3 3 - - 3 3 K? - - 7 7 - - 7 - 7 - How do keys complement? Hypothesis: 33-333--33 11--1--11- --77--7-7- 33-33---3- ??? 1---111--- -7-77-777-- It is impossible that Wheel 17 have 1 or 3 lugs front of it. Then there are 7 lugs front of it. For now, we cannot conclude for Wheel 19 and 21, but there are only two possibilities. p = k -1 -c (Second message) i 23 24 25 26 27 28 29 30 K? - - 1 1 1 - - - K? 3 - 3 3 - - - 3 - K? - 7 7 - 7 7 7 - - K 3 7 11 4 8 7 7 3 -1 -1 -1 -1 -1 -1 -1 -1 C G G S E E S S Q -c 20 20 8 22 22 8 8 10 P 19 26 15 25 3 14 14 12 P W A S Z D O O M A priori it is one of these words: DOOM, DOOMFUL, DOOMED, DOOMSTER ... The most probable word is DOOMED k = c + p + 1 i 30 31 32 P E D Z 4 3 25 C V E L 21 4 11 K 0 8 11 K? - 1 1 K? - - 3 K17 - 7 7 Then: 012345678901234567890 0123456789012345678 33-333--333-33---3--3 11--1--11--111---11 Then The wheel#19 can't be associated with 3 Then Key: 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 1 1 - - 1 - - 1 1 - 13 14 15 16 17 18 19 20 21 22 Key 4 10 0 3 11 8 1 7 10 7 K? 1 - - - 1 1 1 - - - K? 3 3 - 3 3 - - - 3 - K? - 7 - - 7 7 - 7 7 7 Now, the problem is easy to solve.