Chapter 2: Analysis of a Single-Wheel Hagelin Cipher machine

Explanations

1. C == Cipher text, K == Key, P == Plain text
      C = (K -1 -P) % 26
   Key : 0,9,0,...
   Plain text : "USz..."

   C = (0 -1 -20) % 26 = 5 = "F"
   C = (9 -1 -18) % 26 = 16 = "Q"
   C = (0 -1 -25) % 26 = 0 = "A"
   etc...

   Cipher text : "FQA..."

2. K = (C + P + 1) % 26
   Cipher text : "URC..."
   Plain text  : "FIE..."

   K = (20 + 5 + 1) % 26 = 0
   K = (17 + 8 + 1) % 26 = 0
   k = (2  + 4 + 1) % 26 = 7
   etc...

   Key: 0 0 7 7 0 0 7 7 7 0 0 7 0 7 0 0 7 0 0 (Wheel 19)

3. P = (K -1 -C) % 26

   Message Key: "L" => The active Pin : (11+10)%17=4 => "E"

     A B C D E F G H I J K L M N O P Q
     0 3 3 0 3 3 3 0 3 0 3 0 0 0 3 3 0 (Wheel 17)
     
   K:  3  3  3  0 ... (E F G H ...)
      -1 -1 -1 -1
   -C:-1 -2 -9 -6 ... (B C J G ...)
   ================================
   P:  1  0 19 19 ... (B A T T ...) 

   Plain text: BATTLES WERE FOUGHT FEROCIOUSLY 


4. C1 = (K1 -1) -P
   C2 = (K2 -1) -P
   =>   (C1 - C2) = (K1 - K2) = DELTA
   =>   DELTA = 0 - 0 = 0 or  X - X = 0 or X - 0 = X or 0 - X = -X 
   =>   DELTA = { 0, X, -X }
   Where "X" is the number of Lugs.

   Cipher1 = "YXXRW WYARZ..."
   Cipher2 = "YXQRW DNATS..."
   Delta[0] = (24 - 24) =  0
   Delta[2] = (23 - 16) =  7 
   Delta[5] = (22 -  3) = 19 = 26 - 7  
 
   Then, X = 7
 
   If K=0 Then:  TSUAT TXRYS OYZYI XEWST RTIEL EKYSZ
   If K=7 Then:  AZBHA AEYFZ VFGFP ELDZA YAPLS LRFZG
   Plain:        A BAT TERY  OF FI ELD A RTILL ERY

   If K=0 Then:  TSBAT MERRZ HFSFB XEWSA RTBLL XRRZS
   If K=7 Then:  AZIHA TLYYG OMZMI ELDZH YAISS EYYGZ
   Plain:        A BAT TERY  OF FI ELD A RTILL ERY   

5. The 1st message begins by "THE" (1st hint)

   PSVA = THEZ => Key: (15+19+1),(18+7+1),(21+4+1),(0+25+1)= 9,0,0,0
   
   We can deduce the beginning of the 2nd message: "OURz"

   We decipher each cipher message with key equal to 0 or equal to 9,
   and we choose one letter in each column to produce plain text :

       1st Message : K=0  KHEZS RDVZU RYQWA RQNRJ ZDECC RIEUQ
                     K=9  TQNIB AMEID AHZFJ AZWAS IMNLL ARNDZ
                  Plain   THE S AME D AY WA R WAS  DECL ARED 
                  Key     90000 99909 90900 09999 00009 99099 (Wheel 19)

       2nd Message : K=0  FURZL VXRLQ SAKTL EQZJQ NOWZF MVRQQ
                     K=9  ODAIU EGAUZ BJTCU NZISZ WXFIO VEAZZ
                  Plain   OUR L EGAL  BATTL E IS  NOW O VER

   With two messages, we can verify each hypothesis easily.  

6. The message begins by "FIRST", we can deduce the key:
   UXIHMG = FIRSTZ => Key : (20+5+1)=0, (23+8+1)=32%26=6,(8+17+1)=0,
                            (7+18+1)=0, (12+19+1)=32%26=6,(6+25+1)=6

   There are 6 lugs.

   We can decipher with two hypotheses: the K=0 or K=6 
   and we choose one letter in each column to produce plain text :

   If K=0  FCRSN TCOLP STXYZ ENDYD ZNBEZ YASNY LHZSE WTOLT
   If K=6  LIXYT ZIURV YZDEF KTJEJ FTHKF EGYTE RNFYK CZURZ
           FIRST  CORP S DEF ENDED  THE  EASTE RN SE CTOR 

7. We suppose the last letters are "Z":
   ..LLA = ..ZZZ => Key : (11+25+1)=11, (11+25+1)=11, (0+25+1)=0

   Then, the number of lugs are 11.

   If K=0  WEOUD RCESO WXSOW ANZXC TDOTN EMNOU DXWOL TSOOZ
   If K=11 HPZFO CNPDZ HIDZH LYKIN EOZEY PXYZF OIHZW EDZZK
   Plain:  HE FO RCED  HIS W AY IN TO EN EMY F OXHOL ES  

8. We suppose the last letters are "Z":
   ..IDD = ..ZZZ => Key : (8+25+1)=8,(3+25+1)=3,(3+25+1)=3

   Then, we have 3 lugs.

   If K=0  EBZQU OKEAZ THBWD UKZOK WTEEW BNBMY ZPLID FERWW
   If K=3  HECTX RNHDC WKEZG XNCRN ZWHHZ EQEPB CSOLG IHUZZ
   Plain:  HE TU RNED  THE G UN ON  THE  ENEMY  SOLD IER 
   Key:    33030 33030 00333 03003 30303 30300 03330 300 


9. We search the shift which is compatible with probable word "CORPS"

   If K=0  TUEMT WOMCB ECSZA RRMAO GZVNM DIRRC GMPON TAPGM
   CORPS ?          C. ..S                   C ..P           (we search 2 letters)
   We search         O RP                      OR.S
   If K=13 GHRZG JBZPO RPFMN EEZNB TMIAZ QVEEP TZCBA GNCTZ
   plain   THE T WO CO RPS A RE NO T IN  DIREC T CON TACTY


10. We sum the number of the most probable letters ("Z" and "E") for each 
    number of Lugs possible.

    If K= 1 GPSOU ITOUJ STIOI JNFAB BFSXR POAUG DDQTA GBRFE 
            OYPQB OTHTA TPMEX FSH   Z:0, E:3
    If K= 2 HQTPV JUPVK TUJPJ KOGBC CGTYS QPBVH EERUB HCSGF 
            PZQRC PUIUB UQNFY GTI   Z:1, E:2
    If K= 3 IRUQW KVQWL UVKQK LPHCD DHUZT RQCWI FFSVC IDTHG 
            QARSD QVJVC VROGZ HUJ   Z:2, E:0
    ...
    If K=12 RADZF TEZFU DETZT UYQLM MQDIC AZLFR OOBEL RMCQP 
            ZJABM ZESEL EAXPI QDS   Z:6, E:6   Youpee !!!

    K = 0   FORNT HSNTI RSHNH IMEZA AERWQ ONZTF CCPSZ FAQED 
    K =12   RADZF TEZFU DETZT UYQLM MQDIC AZLFR OOBEL RMCQP 
    Plain   FOR T HE FI RST T IME A MERIC AN TR OOPS  FACED             

    K = 0   NXOPA NSGSZ SOLDW ERG
    K =12   ZJABM ZESEL EAXPI QDS
    Plain    JAPA NESE  SOLDI ERS
    

11. DELTA = (C1 - C2 ) = { 0, X, -X } Where "X" is the number of Lugs.

    Cipher1 : GSVAO GGUUA KVICA CLGXF GGVWA
    Cipher2 : GZVHH GZBUA KVPCH VEGXF ZGCWH

    Delta[ 0] = (6 - 6) = 0
    Delta[ 1] = (18 - 25) = 19  = (26 - 7) = -7 % 26
    ...
    Delta[ 3] = (0 - 7) = 19  = (26 - 7) = -7 % 26
    Delta[ 4] = (14 - 7) = 7

    Then, X = 7

    No. 1 - K=0  THEZL TTFFZ PERXZ XOTCU TTEDZ
            K=7  AOLGS AAMMG WLYEG EVAJB AALKG
        Plain    THE S TAFF  WERE  EVACU ATED 
        Key      00007 07000 70070 77700 70000 
    No. 2 - K=0  TAESS TAYFZ PEKXS EVTCU ATXDS
            K=7  AHLZZ AHFMG WLREZ LCAJB HAEKZ
        Plain    THE S TAFF  WERE  EVACU ATED 
        Key      07070 00700 70777 00700 00707

12. We need to try each possible number of Lugs, with 3 lugs:
    (Count Z and Z letters is not sufficient because the message is
    too short).

  If K=0  CFBRC EWCID EQIKD ZMXOH EDZQE EWXZT FLN
  If K=3  FIEUF HZFLG HTLNG CPARK HGCTH HZACW IOQ 
  Plain:  FIERC EZFIG HTING ZMARK EDZTH EZACT ION