Classical Cryptography Course,
Volumes I and II from Aegean Park Press

By Randy Nichols (LANAKI)
President of the American Cryptogram Association from 1994-1996.
Executive Vice President from 1992-1994

Table of Contents
  • Lesson 1
  • Lesson 2
  • Lesson 3
  • Lesson 4
  • Lesson 5
  • Lesson 6
  • Lesson 7
  • Lesson 8
  • Lesson 9
  • Lesson 10
  • Lesson 11
  • Lesson 12


    April 6, 1996

    Revision 0
    COPYRIGHT 1996
    LECTURE 10




    In Lecture 10, we return to our course schedule with a study offascinating cipher systems based on multiple alphabets-Polyalphabetic Substitution systems. What is amazing aboutthese systems is how long they remained secure. The Viggysystems (my name for Vigenere) was considered unbreakable forover 200 years. Along comes Major Kasiski, and poof, we haverecreational cryptography.

    I think the best way to introduce the subject is via anoverview based on the Op-20-GYT course notes (Office of ChiefOf Naval Operations, Washington) [OP20]. From there, I willbring in MASTERTON's dissolution of QUAGMIRES I-IV. [MAST]

    In Lecture 11, we will revisit polyalphabetic cipher systemsand the polygraphic cases using Friedman's detailed analysis.We will cover the PORTA system and other family members. Iwill cover decimation processes in detail. [FRE4], [FRE5],FRE6], [FRE7], [FRE8]

    In Lecture 12, we will describe the aperiodic polyalphabeticcase and give a diagram of topics considered in Lectures 10 -12. [FR3]

    I have updated our Resources Section with many references onthese systems - focusing on the cryptanalytic attack and thoseof historical interest. Kahn has some interesting stories aboutthe Viggy family. [KAHN]


    A cipher system which employs two or more cipher alphabets andincludes a method for designating which cipher alphabet is tobe used for the encipherment of each plain-text letter, iscalled a polyalphabetic substitution system. Cipher systemsemploying variant values may appear to use more than onealphabet, but they have characteristics of mono-alphabeticsubstitution and are properly classified as such.

    Polyalphabetic substitution systems consists of two generaltypes; periodic and non-periodic.

    (a) In the periodic type the text of a message is dividedinto definite, regular groups or cycles of letters which areenciphered with identical portions of the key. Periodicsystems are further subdivided as follows:

    The cipher alphabets employed in multiple alphabet substitutionsystems may be constructed by any number of methods. As anexample, the QUAGMIRE IV uses both vertical and horizontalkeywords.


    Plain      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 ZCipher  1  R T U V W X Y Z P E N C I L S A B D F G H J K M O Q  "     2  E N C I L S A B D F G H J K M O Q R T U V W X Y Z P  "     3  D F G H J K M O Q R T U V W X Y Z P E N C I L S A B
    Here the plain component is a normal sequence, and the ciphercomponent are identical keyword sequences. The same keywordsequences may be used in both the plain cipher components, ordifferent sequences may be used. The key which determines thesetting of the cipher alphabets against the plain component(RED) may be any prearranged word or phrase. Also, each cipheralphabet may be assigned a number and the alphabets used inaccordance with a prearranged numerical key.

    The process of enciphering a message with the multiple alphabetsystem above would appear as follows:

    Cipher Alphabet No.          1-2-3-1-2-3-1-2-3-1-2-3-1-2-3-1-2-3-1-2-3-1-2-3Plain  -  M Y C O U R S E Z E R O T H R E E Z E R O A T TCipher -  I Z G S V P F L B W R X G B P W L B W R X R U N          1-2-3-1-2-3-1-2-3-1-2-3-1-2-3-1-2-3Plain  -  H I R T E E N T H I R T Y T H R E ECipher -  Z D P G L J L U O P R N O U O D L J
    In order to reduce the chances of encipherment by the wrongalphabet, the plain text is often written so that the lettersdesignated by the key for encipherment by each alphabet areplaced in the same vertical column.

    Note the repetitions in the plain text which begin at the samepoint in the key produce repetitions in the cipher text, whileothers [may not] do not. Friedman discusses accidentalrepetitions in [FR7].


    Major Friedrich W. Kasiski (1805-1881) was a career officer inEast Prussia's 33 Infantry Regiment. He is credited with arevolutionary insight regarding polyalphabetic repeating keysystems - that the conjuction of a repeated portion of the keywith the repetition in the plaintext produces a repetition inthe ciphertext. Like causes produce like effects. Theinterval between plaintext or ciphertext repetitions is notedthroughout the cryptogram, factored and the commonality of thefactor is a good indication of the key and number of alphabetsused to encipher the original methods. The fall of theVigenere family is attributed to Kasiski's examination. [KASI][KAS1], [KAHN]

    If there are several long repetitions in the cipher text of anunknown system, the intervals between the initial letters ofthese repetition have a common factor, this factor representsthe number of alphabets used to encipher the message and theexact number of repetitions of the key.

    A simple example:   Given the cryptogram:      IZGSV   PFLBW   RXGBP   WLBWR   XRUNZ      DPGLJ   LUOPR   NOUOD   LJFactoring:      Repetition   Interval  Factors   Common Factor(s)      LBWRX          9         3,3            3      LJ            12         2,2,3          3      UO             6         2,3            3
    The "period" or common factor is three and this is the numberof alphabets employed.

    Digraph and trigraph repetitions may be the result of chanceinstead of plain text repetitions. [FR7] discusses in detail.

    When factoring results in more than one common factor we shalluse the highest common factor and check with frequencies of theexpected alphabets to see how close to normal they are. Onlyshort messages fail to lead to the correct determination of thenumber of cipher alphabets employed in the system. Whenfactoring fails on a longer message, an aperiodic cipher mayhave been employed.


    Phamplet Number 7, Office of Operations Cryptanalysis, Officeof the Chief of Naval Operations, Washington, 1930 [OP20]prepared this problem for discussion.

    From:           A B  (Black Force Commander)To:             CD, EF, GH, IJ (Black Ships)Time Groups:    0013-2300 April 1930Remarks:        Cruiser transmitter.Cryptogram written out in worksheet format:Alpha. -  1 2 3 4 5 6 7 8 9 10  Alpha. - 1 2 3 4 5 6 7 8 9 101         K P T X S L I C T M    16      M V H A W A D G G Z2         I A M C B B N M S Z    17      Y F A R Q V K M M Q3         M J K A Q J B F Z A    18      K F M P S L G X A H4         J G M B S L N P H H    19      E F W K G C B F T H5         E E J Z W N C L O W    20      S V C B B U A H S S6         Z F S A A S Z D E P    21      K P K D E C G O H Z7         Z X C D J D D H A J    22      L V O D S C O C H A8         O D B K A H P L G H    23      G V W B Z C A M O Z9         A J M K T V A M K H    24      M J K A Q J B F J H10        M B C A A C N W S Z    25      X B H A A V A K O S11        Z D W I J K G M C X    26      K P K G U L T J O Q12        M V X X U N B W Z T    27      D F Q Q J K K M H Z13        I Y N C P O G H H W    28      H V H A E P Z W Q R14        L G T B W P L V T T    29      O P L A U L B M O Z15        O B O X J L R M H Z    30      M J K A Q J B F
    Collateral Information:

    The Black and Blue Fleets are engaged in war maneuvers in theCaribbean Sea. The Fleets are not in contact. The location ofthe enemy (the Black Fleet) is unknown. The message inquestion was intercepted by the Blue Flagship at 0015 on 14April 1930. The operator had reason to believe that a cruisersent the message.

    The composition of the Black Fleet is as follows:

    Battleships                   CruisersWest Virginia (flag)          Trenton (flag)Maryland                      MarbleheadTennessee                     RichmondNew Mexico                    MemphisMississippiCaliforniaDestroyers                    Air ForceLitchfield (flag)             Saratoga (flag)Preble                        LangleyPruitt                        GannetNoaDecatur                       Submarine ForceSicardHulbert                       Argonne (flag and tender)                              V-1, V-2, V-3William B. PrestonFactoring:Repetition       Interval          FactorsZMJKAQJBF          210             2,3,5,7,10ZMJKAQJBF          270             2,3,3,5,10ZMJKAQJBF           60             2,2,3,5,10MHZMVHA            120             2,2,2,3,5,10ZMV                 40             2,2,2,5,10ZMV                160             2,2,2,2,2,5,10KPK                 50             2,5,5,10
    The highest common factor is 10; the period and number ofalphabets used is 10, so the sequence repeats itself aftereach 10 letters.

    "Lining-up" is one of the basic operations of solution.We group the message in lines of ten letters. The letters ineach column are enciphered by the same alphabet. Checking thefrequency tables, each alphabet resembles a single alphabet.

                            Frequency Tables#1     #2    #3   #4    #5    #6    #7    #8   #9    #10A 1    A 1   A 1  A 9   A 4   A 1   A 4   A    A 2   A 2B      B 3   B 1  B 4   B 2   B 1   B 6   B    B     BC      C     C 3  C 2   C     C 5   C 1   C 2  C 1   CD 1    D 2   D    D 3   D     D 1   D 2   D 1  D     DE 2    E 1   E    E     E 2   E     E     E    E 1   EF      F 5   F    F     F     F     F     F 4  F     FG 1    G 2   G    G 1   G 1   G     G 4   G 1  G 2   GH      H     H 3  H     H     H 1   H     H 3  H 6   H 6I 2    I     I    I 1   I     I     I 1   I    I     IJ 1    J 4   J 1  J     J 4   J 3   J     J 1  J 1   J 1K 4    K     K 5  K 1   K     K 2   K 2   K 1  K 1   KL 2    L     L 1  L 1   L     L 6   L 1   L 2  L     LM 7    M     M 4  M     M     M     M     M 8  M 1   M 1N      N     N 1  N     N     N 2   N 3   N    N     NO 3    O     O 2  O     O     O 1   O 1   O 1  O 5   OP      P 4   P    P 1   P 1   P 2   P 1   P 1  P     P 1Q      Q     Q 1  Q 1   Q 4   Q     Q     Q    Q 1   Q 2R      R     R    R 1   R     R     R 1   R    R     R 1S 1    S     S 1  S     S 4   S 1   S     S    S 3   S 2T      T     T 2  T     T 1   T     T 1   T    T 3   T 2U      U     U    U     U 3   U 1   U     U    U     UV      V 6   V    V     V     V 3   V     V 1  V     VW      W     W 3  W     W 3   W     W     W 3  W     W 2X 1    X 1   X 1  X 3   X     X     X     X 1  X     XY 1    Y 1   Y    Y     Y     Y     Y     Y    Y     YZ 3    Z     Z    Z 1   Z 1   Z     Z 2   Z    Z 2   Z 9  30     30    30   30    30    30    30    30   29    29


    When ample collateral information is available, the known-wordattack is the easiest and potentially the quickest method ofsolution. From the given data, the message is presumably fromthe Commander of a cruiser division to his four cruisers,giving orders for scouting operations of the cruiser division.

    The words most likely to appear are:

    Scouting    Scouting line       Trenton      LatitudeCourse      Scouting course     Marblehead   LongitudeSpeed       Scouting speed      Richmond     HundredDistance    Scouting distance   Memphis      NumbersPosition    Commence scouting   Enemy        Times/Dates
    Our concern is not with guessing words but standardizing thesolution.

    The Known-Word" method applied in two ways:

    The best method to use depends on the circumstances. In thisproblem both methods apply.


    The long repetitions are words or phrases, important to thesubject of the message, and may be known-words. They areexcellent points of attack. The beginning of the message orthe end of the message are usually good points of attack.

    The second longest repetition is the right length for Trenton,Memphis, or Hundred; furthermore it links in the letters of thelongest repetition.

    Original Assumptions -MHZ MVHA     lines 15-27  TRENTON is best assumption.TRE NTONMEM PHISHUN DREDCheckMOZ MJKAQJBF  lines 24, 30     MOZ MJKAQJBF  could beT E N  N      Excellent        TEE NHUNDRED  excellentM M P  S      Poor             THE E--N ---  poorH N D  D      PoorCheckMCZ MVX       lines 1-12TWE NTY       excellentM M PH        poorH V DP        poorCheck the values of TEEN HUNDRED and TRENTONLine 2-3       12345678910     12345678910               IAMCBBNMSZ      MJKAQJBFZA                      T E      NHUNDRED suggests            ATTE      NHUNDREDLine 23-24     GVWBZCAMOZ      MJKAQJBFDI                T     TEE      NHUNDRED suggests         THIR                  FOUR                   FIF                   SIX               ATSEVEN                  EIGHLines 29-30    OPLAULBMOZ      MJKAQJBF--                  N  ETEE      NHUNDREDsuggests          NINETEE      NHUNDRED
    It is possible that all the above assumptions are incorrect butthey are too good to ignore. We enter the above values intothe cryptogram to see if skeletons of words appear.

    Possibilities are indicated below:Lines 19-20     12345678910     12345678910                EFWKGCBFTH      SVCBBUAHSS                      ED         T      T                   SPEEDFI      FTEENKNOTS                        SI      XLine 19 ED could be Speed.. building on that we have otherpossibilities.Lines 21-22     KPKDECGOHZ      LVODSCOCHA                  U     RE       T      R                COURSETHRE      ETHREEZEROLines 11-12     ZEWIJKGMCZ      MVXXUNBWZT                       T E      NT    E                       TWE      NTYMILES                      T             THREE                                   FIVE
    TRENTON is the most obvious break. Check letter-combinationsof frequencies to see which of the three chosen words fittedbest.

    HZ =1      ZMV=1    ZM =4   HA=1RE         ENT      EN      ON      Trenton is only assumptionEM         MPH      MP      ISUN         NDR      ND      EDFrequency    869  7639Cipher       MHZ  MVHAFrequency    XXX  XXXX      X = high frequencyPlain        TRE  NTON                            - = intermediate frequencyFrequency    -X-  --XXPlain        MEM  PHIS      O + low frequencyFrequency    --X  -XX-Plain        HUN  DRED


    One method of fixing the location of an obvious word is byfrequencies, provided the obvious word has one or more lettersof very low frequency. The word should be 10 or more lettersto be practical.

    First, frequencies are written over each letter of thecryptogram. The Known-word is put on a card and slid over thecryptogram until it fits with the very low frequency lettersand neighbors. This method is rather tedious and painful, butgood in a pinch.


    Location of words by symmetry is commonly employed when dealingwith single key ciphers. With double key ciphers itsapplication depends much on chance. If the alphabets arerepeated in the key or the key is short, we employ a limitedform of symmetry.

    With a non repeating key or very long key, this method fails.With a fairly short key we employ this method provided:

    For our sample problem, one of our choices might be:
    Therefore, any place in the cryptogram where two successivelines have common letters in the same column is a possiblelocation of our word. Failure to find this location,eliminates the possibility of this word.

    Table one partially shows the ciphertext where repeated lettersare ten spaces apart. Of the twelve possibilities for the word"SCOUTINGDISTANCE" some are eliminated by frequencies of theletters C,G,C, others by letter combinations and the balance bytest. All fail.

    Our Navy students would try the scouting line of cruisers as:

      4             3                   1          2MEMPHIS       RICHMOND           TRENTON    MARBLEHEAD  2             1          OR       3          4MARBLEHEAD    TRENTON            RICHMOND   MEMPHIS               (flag)These names might appear as follows:       MEMPHISRIC                 MARBLEHEAD       HMONDTRENT      OR         TRENTONRIC       ONMARBLEHE                 HMONDMEMPH       AD                         IS
    These can be checked against Table I and cross checked byfrequency or digram analysis.

    We have a little luck at Line 14 - 15 - 16

     Line 14             LGTBWPLVTT                     --MEMPHISR Line 15             OBOXJLRMHZ                     ICHMONDTRE Line 16             MVHAWADGGZ                     NTONMARBLE check Line 29             OPLAULDMOZ         Line 11        MOZ                     I  N N T E                        I E                        NINETEE                        TWE Line 30             MJKAQJBF           Line 12        MVX                     NHUNDRED                          NT                                                       NTY


    Table I gives a list of obvious locations. We suspectthe word COURSE followed by a ZERO and ONE TWO or THREE.

    Some possibilities are:

    COURSEZERO             COURSETHREFOUR                   EZEROCOURSEONET             COURSETHREWO                     EONECOURSEZERO             (promising but no check)FOURCOURSETHREETHREE                 (checks with #9 in Table I)AssumptionLine 21         KPKDECGOHZ        Line 26   S KPKGULT                                              COU                                            S COUTINGLine 22         LVODSCOCHA                ETHREEZEROBoth assumptions are entered into the cryptogram.                        TABLE ILines                                         Reference6-7              ZFSAASZDEPZXCDJD                18-9              KAHPLGHAJMKTVAMK                28-9              HAJMKTVAMKHMBCAA                310-11            ZZDWIJKGMCZMVXXU                415-16            ZMVHAWADGGZYFARQ                517-18            FARQVKMMQKFMPSLG                618-19            FPMSLGXAHEFWKGCB                718-19            HEFWKGCBFTHSVCBB                821-22            DECGOHZLVODSCOCH                921-22            CGOHZLVODSCOCHAG                1021-22            HZLVODSCOCHAGVWB                1122-23            VCDSCOCHAGVWBZCA                1222-23            COCHAGVWBZCAMOZM                1324-25            AQJBFJHXBHAAVAKO                1425-26            OSKPKGULTJOQDFQQ                1528-29            AEPZWQROPLAULBMO                1629-30            AVLBMOZMJKAQJBF                 17                            TABLE II12345678910     12345678910     12345678910    12345678910COURSEZERO      COURSETHRE      COURSEONE      COURSETWOZERO            EZERO           ERO      Z     ERO      ZONE              ONE            NE       O     NE       OTWO              TWO            WO       T     WO       TTHREE            THREE          HREE     T     HREE     TFOUR             FOUR           OUR      F     OUR      FFIVE             FIVE           IVE      F     IVE      FSIX              SIX            IX       S     IX       SSEVEN            SEVEN          EVEN     S     EVEN     SEIGHT            EIGHT          IGHT     E     IGHT     ENINE             NINE           INE      N     INE      NCOURSEZERO       COURSETHRE     COURSEONET     COURSETWOTFOUR             EZER           WO             WO                 EONE                 ETHREEDISCOVERY OF THE SYSTEMWe study the values assumed previously:Value      Alphabets       Value        AlphabetSC=E        3,6,8           H=O, O=H     3,6,8O=H        3,8             N=L,L=N      3,6,8H=O        3,8             K=U, U=K     3,6,8B=E        4,7             N=A,A=N      4,7A=N        4,7             S=E,E=S      5The common values indicate that alphabets 3,6, and 8 areidentical and similarly so are 4 and 7.   Five reciprocalvalues are noted without inconsistencies.  Seven differentalphabets are used.  The alphabets are probably reciprocal.If the seven alphabets are Secondary (derived from the samecipher component set against the same plaintext but indifferent alignments) a short cut solution is possible. We cannext combine the alphabets into one system.We have enough clear text to solve the cryptogram - I leave thebalance to the student.Alpha. -  1 2 3 4 5 6 7 8 9 10  Alpha. - 1 2 3 4 5 6 7 8 9 101         K P T X S L I C T M    16      M V H A W A D G G Z          C O   M E N   E                N T O N     R     E2         I A M C B B N M S Z    17      Y F A R Q V K M M Q              T   N   A T T E                    D   S T3         M J K A Q J B F Z A    18      K F M P S L G X A H          N H U N D R E D   O            C   T   E N T Y   I4         J G M B S L N P H H    19      E F W K G C B F T H              T E E N A   R I                S S P E E D   I5         E E J Z W N C L O W    20      S V C B B U A H S S              R     L   N E                T E E N K N O T S6         Z F S A A S Z D E P    21      K P K D E C G O H Z                N                        C O U R S E T H R E7         Z X C D J D D H A J    22      L V O D S C O C H A              E R     R O                E T H R E E Z E R O8         O D B K A H P L G H    23      G V W B Z C A M O Z                S   O   N   I            A T S E V E N T E E9         A J M K T V A M K H    24      M J K A Q J B F J H            H T S     N T   I            N H U N D R E D   I10        M B C A A C N W S Z    25      X B H A A V A K O S          N   E N   E A S T E                O N     N U E S11        Z D W I J K G M C X    26      K P K G U L T J O Q              S     U T T W E            C O U T I N   R E12        M V X X U N B W Z T    27      D F Q Q J K K M H Z          N T Y M I L E S                          U S T R E13        I Y N C P O G H H W    28      H V H A E P Z W Q R              I     H T O R              N T O N S     S14        L G T B W P L V T T    29      O P L A U L B M O Z          E     E                          O N N I N E T E E15        O B O X J L R M H Z    30      M J K A Q J B F              H M   N   T R E            N H U N D R E D                           TABLE III                       DECIPHERING TABLEPLAIN-  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 Z1       G   K   L                 M2                     J             P         V3               C     O             H     J W   K       X4               B               X A       D K5             Q S       U         B   G     E     Z6               C           U N   L7         N     B                 A           G           O8             F C     O             H       W M9               O                         H   S        C10              Z       H           A       S                            TABLE IV                       ENCIPHERING TABLEPLAIN-  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 Z1       G   K   L                 M2                     J             P         V3-6-8         F C     O     U N   L H     J W M K       X4-7     N       B               X A       D K G           O5             Q S       U         B   G     E     Z9               O                         H   S        C10              Z       H           A       S
    Op-20-G gives us the quick and dirty of the problem. We needto understand what equivalent cipher alphabets are and how themultiple alphabet system lends itself to reconstruction.


    Any sequence containing 26 letters may be rearranged so thatall the letters which are originally separated by equalintervals will also be spaced at equal intervals in the newrelated sequences. Including the original sequence, a total ofof six related sequences may be constructed. [Friedman expandson this principle in FR7.]

    Example:     1   3   5   7   9  111  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 Z2  A D G J M P S V Y B E H K N Q T W Z C F I L O R U X3  A F K P U Z E J O T Y D I N S X C H M R W B G L Q V4  A H O V C J Q X E L S Z G N U B I P W D K R Y F M T5  A J S B K T C L U D M V E N W F O X G P Y H Q Z I R6  A L W H S D O Z K V G R C N Y J U F Q B M X I T E P
    In this example, a normal alphabet sequence has been re-spacedto form five related sequences. In constructing them, theoriginal sequence is regarded as a circle and the letters arecounted off in equal intervals, then written in adjacentpositions to form a related sequence.

    Only the odd intervals from 3 - 11 can be used in re-spacing a26 letter sequence to form different related sequences.{primes} Even intervals will produce only 13 letter sequences,and the interval 13 can not be used. Odd intervals from 15-25will produce identical sequences with those from 1-11 but inreversed direction. (like the Porta)

    Cipher alphabets may be re-spaced to form equivalent cipheralphabets by the same process as that applied to constructrelated sequences.

    Example:                    Original Cipher AlphabetPlain  - D I P L O M A C Y B E F G H J K N Q R S T U V W X ZCipher - V W X Z T H U R S D A Y B C E F G I J K L M N O P Q                   Equivalent Cipher AlphabetPlain  - D L A B G K R U X I O C E H N S V Z P M Y F J Q T WCipher - V Z U D B F J M P W T R A C G K N Q X H S Y E I L O
    An equivalent cipher alphabet can not be distinguished from theoriginal cipher alphabet unless a systematic construction orsome outside information is available to identify the originalone. The secondary alphabets generated by shifting the pointsof coincidence of the plain and cipher components are the samealphabets regardless of which equivalent cipher alphabet hasbeen shifted.

    Example:                    Original Cipher AlphabetPlain  - D I P L O M A C Y B E F G H J K N Q R S T U V W X ZCipher - X Z T H U R S D A Y B C E F G I J K L M N O P Q V W                   Equivalent Cipher AlphabetPlain  - D L A B G K R U X I O C E H N S V Z P M Y F J Q T WCipher - X H S Y E I L O V Z U D B F J M P W T R A C G K N Q
    The secondary alphabet of this example has been derived byshifting the cipher component of the original alphabet of theprevious paragraph, and the equivalent secondary cipheralphabet by shifting the cipher component of the equivalentalphabet of the previous paragraph.

    The number of spaces each cipher component has been shifted isnot the same in each case, yet the plain and cipher valuescorrespond exactly. This illustrates the most importantprinciple of symmetry in the secondary alphabets.


    When the same sequence has been used for each of the ciphercomponents of a multiple alphabet system, there are definiterelationships between the individual cipher values which may beused in recovering other cipher values after a few have beenidentified through analysis.

    The principles are explained by another example in which theplain and cipher components are different mixed sequences:
    Plain  0 - D I P L O M A C Y B E F G H J K N Q R S T U V W X ZCipher 1 - O P Q V W X Z T H U R S D A Y B C D F G I J K L M N       2 - N O P Q V W X Z T H U R S D A Y B C E F G I J K L M       3 - E F G I J K L M N O P Q V W X Z T H U R S D A Y B C 
    The interval between letters of two cipher components, letterswhich occur in the same vertical column, is equal to the amountof displacement of one component from the other.

    O (1) To N(2) is an interval of one, the amount of shiftbetween the cipher components (1) and (2).

    E (3) to O (1) is the same interval as O (3) to U (1), and isthe same interval as U (3) to F (1), etc.

    Thus a chain of letters, EOUF with current relative spacingscould be made from the vertical relationship alone, when theorder of plain component sequence is unknown. A set ofequivalent alphabets might be the result of construction bythis means, but the original in this case would be recognizedwhen the proper spacing is found.

    If the vertical relationship is used between components whichare displaced an even number of letters, such as ciphers (2)and (3), a chain of 13 letters will result, and if thecomponents were originally displaced 13 letters, they wouldshow only reciprocal relationships.


    Suppose the Enciphering table obtained during the solution of acryptogram appeared as follows:

    Plain  0 - 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 ZCipher 1 - Z U T   R   D A P     V   C W       G I         H       2 - X H Z N U     D O       W B V     E F G         T       3 - L     E P     W F     I K T J     U R S
    Since the interval between R and P in the cipher sequence isthe same as that between P and F, we may arbitrarily assumethis interval to be one and build up a cipher sequenceaccordingly.

    The vertical columns remain unchanged. We write:

    0    E I       R in the third cipher           S E I1    R P F     component appears under         G R P F U O2    U O       S plain, so we continue   G R P F U O3  R P F                                     G R P F U O
    The progress of adding values to the plain and cipher sequencesprogresses through the various stages:
    0               T     S E I R B     Y1               I S   G R P F U O E H      T2        I S    G R P F U O E H     T3             I S   G R P F U O E H      T0          O     L T     S E I R B     Y   N C1          W J   V I S   G R P F U O E H   C T   B Z2    W J   V I S   G R P F U O E H   C T   B Z3        W J   V I S   G R P F U O E H   C T   B Z0        M   H O   G L T     S E I R B     Y   N C     A1    L   X K A W J D V I S   G R P F U O E H   C T   B Z2    K A W J D V I S   G R P F U O E H   C T   B Z L   X3      X K A W J D V I S   G R P F U O E H   C T   B Z L
    The intervals between E, F, G and between V, W, X in the ciphersequence obtained above, indicate the equivalent alphabets havebeen recovered which should be re-spaced by counting off everythird letter in the reverse direction.

    0      I   L O M A C Y B E   G H     N   R S T1    O P   V W X Z T H U R S D A   B C E F G I J K L2      O P   V W X Z T H U R S D A   B C E F G I J K L3    E F G I J K L     O P   V W X Z T H U R S D A  B C


    A few more values are necessary in Table IV in order tocompletely reconstruct the system used.

    Line 1                          Line 18Alpha    1 2 3 4 5 6 7 8 9 10   Alpha    1 2 3 4 5 6 7 8 9 10Cipher   K P T X S L I C        Cipher   K F M P S L G X A HPlain    C O   M E N   E        Plain    C   T   E N T Y   INew          M       C          New            WLine 3 to 5Alpha    1 2 3 4 5 6 7 8 9 10   1 2 3 4 5 6 7 8 9 10    1Cipher   M J K A Q J B F Z A    J G M B S L N P H H     EPlain    N H U N D R E D   O        T E E N A   R INew                      F      U R           P         L
    Adding these new values to Table IV gives the following tablefor use in reconstruction of the system:
                                TABLE IV                            Revised                       ENCIPHERING TABLEPLAIN-  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 Z1       G   K   L             E   M             J2                     J             P         V3-6-8         F C     O     U N T L H P   J W M K       X4-7     N   I   B               X A       D K G     P     O5             Q S       U         B   G     E     Z9               O Z                       H   S        C10              Z       H           A       S
    The reciprocal relationship will be ignored.

    On account of L and B being found in two vertical columns, agood starting point is to assume that L and B are adjacent inthe cipher component. Then we would have the following in thecipher component: GN, KI, MA, FQ, CS, PQ, AND WE.

    Using the PGN sequence in the first three cipher components,partial reconstruction can be made:

    PLAIN-          W T A           O R         P   L1                 P G N                       W E2                 V             P G N3-6-8             M A           H J         P G N4-7             P G N             D5                                         P G N9               C S               H J10                            M A
    Since HJ appears with the same interval as LB, then OC and SMare also adjacent in the cipher sequence being constructed.
    PLAIN-  H E W T A      S   O R       Z       N   P   L   U1         L B P G N                    O C S M A   W E H J2     H J     V        L B P G N3-6-8   O C S M A      W E H J    V          L B P G N   K4-7     L B P G N      K I   D       O C S M A5     O C S M A      W E H J     V         L B P G N9         O G S M A      W E H  J   V10                 O C S M A
    We combine the three partials:
    PLAIN-  H E W T A   S   O R    Z       N   P   L   U1         L B P G N              O C S M A   W E H J2     H J     V     L B P G N3-6-8   O C S M A   W E H J      V     L B P G N   K I  D4-7     L B P G N   K I   D    O C S M A5     O C S M A   W E H J      V     L B P G N9         O G S M A   W E H  J     V10        Z     O C S M A
    I think you can see that most of the cipher sequence could beobtained without considering the fact that the plain componentis the same sequence reversed. The important point is that thecomplete system may be reconstructed from relatively few valuesobtained through analysis of the cryptogram.

    The sequence used in this problem is randomly mixed, thereforethe original one can not be distinguished from a related onewhich may be reconstructed. The ten cipher components are setwith the key GUANTANAMO under the A plain.


    The same method used in determining which cipher valuesprobably represent vowels or consonants may be applied to polyalphabetic substitution ciphers as described in Lectures 1 and2. However, the values in each alphabet must be consideredwith their respective prefixes and suffixes in adjacentalphabets, in studying the frequencies of their combinations.

    After the original sequences of a poly-alphabetic substitutionsystem are recovered, subsequent messages using these sequencesmay be solved by a modified method. The "generatrix frequency"method was developed by W. F. Friedman and is described in FR7.


    MASTERTON (Frank W. Lewis) was a personal 'pick' of William F.Friedman. His experience and book [MAST] is as insightful asit is brilliant. He takes us through the QUAGMIRE family. TheAmerican Cryptogram Association calls the class of periodicpolyalphabetic substitution QUAGMIRES I, II, III, IV after theterminology used for keying Aristocrats. QUAGMIRES have amixed alphabet in at least one of the components. QUAGMIRE Iuses a keyword-mixed plain component with a determined numberof normal cipher alphabets at different settings; QUAGMIRE IIuses a normal plain and various settings of the same mixedcipher component; QUAGMIRE III employs the same mixed alphabetfor plain and cipher (juxtaposition repeated on a cycle); andQUAGMIRE IV which has one mixed alphabet for plain and a seriesof slides of another mixed alphabet for the cipher components.[MAST] The use of normal alphabets on a cycle, either director reverse, is a weakness because the components are known andare more vulnerable to solution.


    We will take the QUAGMIRES in turn, making sure we understandthe method of encipherment and tricks of unraveling the text.

    Lets build an alphabet on the Keyword ENCIPHERMENT:

    Let us take a NORMAL alphabet, with C under the first letter ofplain sequence. This is cipher setting No 1. Slide the normalalphabet to I, under E, P, H, E, R to get:
    Plain  0 E N C I P H R M T A B D F G J K L O Q S U V W X Y ZCipher 1 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 B       2 I J K L M N O P Q R S T U V W X Y Z A B C D E F G H       3 P Q R S T U V W X Y Z A B C D E F G H I J K L M N O       4 H I J K L M N O P Q R S T U V W X Y Z A B C D E F G       5 E F G H I J K L M N O P Q R S T U V W X Y Z A B C D       6 R S T U V W X Y Z A B C D E F G H I J K L M N O P Q
    I have numbered the alphabets for ease of use. The initialcolumn keyword is standard practice.

    To encipher the word regarding: The first R is found in theplain sequence, and the letter under it in alphabet 1 is I, weuse the cipher alphabets sequentially and return to alphabet 1after using the sixth alphabet.



    The Cryptogram usually provides a tip: "ILEANDTHENREPLIED. "This will appear in the text someplace.

    The repeat method of factoring doesn't work to well on thisexample. So assume 6, 7 or 8. Write the crib based on thosecycles.

            awh                awh               awh     ILEAND            ILEANDT          ILEANDTH     THENRE            HENREPL          ENREPLIE     PLIED             IED              D
    We have added a possible text of awh to the crib. The middlecrib has the I over an I 13 letters apart and the E's intervalof 6. The stretch of cipher we want will have a repeat as:
    The stretch "glffbYoeydmihYyjjcpYYdvie" fits the bill. Werewrite the cryptogram into a cycle of seven letters either incolumns or rows. We fill in the tip and number the alphabets:
    1234567  1234567  1234567  1234567  1234567  1234567  1234567WBFWXLW  VPYWICQ  JHJYDLL  NABFJCQ  FBBHMPA  XGKIUCR  HVKYNEJ1234567  1234567  1234567  1234567  1234567  1234567  1234567OVMDEJS  PQPTGLF  FBYOEYD  MIHYYJJ  CPYYDVI  ETOFXXL  WPSCYTB               a  whILEAN  DTHENRE  PLIED1234567  1234567  1KJORCYZ  DBYDHYH  R.
    We prepare a deciphering tableux, putting the plain valuesabove the normal cipher strip and using the plain E to start.

    Plain  0         E       -----------------------------------------------------Cipher 1       2       3       4 U V W X Y Z A B C D E F G H I J K L M N O P Q R S T 5       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       6       7 F G H I J K L M N O P Q R S T U V W X Y Z A B C D E
    Since the fourth alphabet also has a plain L, we enter it onthe top line, and similarly place a plain N from the fifthalphabet. The N is confirmed by its appearance in the 7thalphabet, so we know we are on the right track.

    Since we have the plain L, the second alphabet comes in too andhence the plain H and T. This gives us the third alphabet andthe plain I. There is more help. Looking down the variouscolumns we find the Keyword COUNTRY which must have been placedunder the first letter of the plain sequence. Snowballs.

    Plain  0 A B C D E   H         R   T           P L   W I N G       -----------------------------------------------------Cipher 1 J K L M N O P Q R S T U V W X Y Z A B C D E F G H I       2 V W X Y Z A B C D E F G H I J K L M N O P Q R S T U       3 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       4 U V W X Y Z A B C D E F G H I J K L M N O P Q R S T       5 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       6 Y Z A B C D E F G H I J K L M N O P Q R S T U V W X       7 F G H I J K L M N O P Q R S T U V W X Y Z A B C D E
    The clues add up. The Keywords are PLOWING and COUNTRY.

    The RST sequence is obvious. The message reads: The cityslicker asked the farmer what's your mules name? The farmerthought awhile and replied I don't rightly know but I call himJACK.


    This polyalphabetic substitution uses a Normal plain and akeyword mixed cipher alphabet. Lets tackle a problem with thetip of 20 letters TAPHORICORTABOONATUR and also the tip"usage." Sometimes we have hunches. Assume the period is 10,and write out the tip on this basis. Nice pattern with adigraphic hit TT, OO, RR

                         TAPHORICOR                     TABOONATURe    (I have added the e                                    possibility.)and the cipher is:12345678910 12345678910 12345678910 12345678910 12345678910GJGQHJLELW  SZGGETGMQS  YVAHUOLFYN  NIRJHVKJDS  XMZVUEPETG12345678910 12345678910 1HIAHWZOTFN  HIHVWQUQDN  UENAEQMFQA  YXIOVUIVYG  NYLUJMOCVLTAPHORICOR  TABOONATUR  eRXSOTVSSMT  CIIFHVEFYA  VJLEUVDQFX  OZJHNNUHQY  EOGQDYGHEGRXVVVOBVYY  SRNow we develop the deciphering tableaux.Plain  0 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       -----------------------------------------------------Cipher 1         U                             H       2 I       3   H                           A       4               H              N       5                              W       6                            Q      Z       7 U               O       8     T                                 Q       9                              F          D      10                                   N


    We know that the plain sequence is normal. It is in the rightorder and we can base our interval analysis on the plain. Weintroduce Mr. Friedman's principle of symmetry to discover therelationships in the cipher alphabets.

    We know that the cipher text reads from left to right just aswe see it. The skeleton sequence is:

       H------V------A, Q---Z----T, U-------O, and  F-----D,
    We can fill in a few letters. The Q---Z is either QVW-Z or Q-VWZ. In No 1 Q cipher is either Y or Z and Z cipher is either Cor D. [MASTERTON jumps in with a NIO combination and VW but Ididn't see this until after the solution.] Alpha 4 puts V +6from H, transposing that to alpha 1, puts a V under the Aplain, and suggests Q V W X Z sequence with Y in the Keyword.X is pretty unpopular in keywords so we will go with thisassumption.


    Plain  0 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       -----------------------------------------------------Cipher 1 V W X Z U ? ? A T       O              H           Q       2 I       3   H           Q V W X Z U ? ? A T       O       4               H           Q V W X Z U     A T       5             H           Q V W X Z U     A T        O       6 O              H           Q V W X Z U     A T       7 U       T        O              H            Q V W X Z       8   A T       O              H           Q V W X Z U       9                              F          D      10                                   N
    So we build up alpha's 1, 3, 5, 6, 8. We can place the H'sback in them from the Q by -6. in alpha 8 and 5. We see thatU +8 = O in alpha 7. The sequence ---A starts the keyword fromalpha three. Look at the T behind the Q by -17 offset inalpha 8. Remember my assumed 'e' = U in alpha 1. We place thishunch and let it play through.

    We have U - - AT ........Y. I see the prefix UN and digram SA.The word "unsatisfactory" comes to mind but I haven't gotenough hard evidence yet. We have a U +8 to O in the 7thalpha. Fill in the alphas.


    Plain  0 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       -----------------------------------------------------Cipher 1 V W X Z U n s A T i f c O r Y b d e g H j k l m p Q       2 I       3   H           Q V W X Z U ? ? A T       O       4               H           Q V W X Z U     A T       5             H           Q V W X Z U     A T        O       6 O              H           Q V W X Z U     A T       7 U       T        O              H            Q V W X Z       8   A T       O              H           Q V W X Z U       9                              F          D      10                                   N
    I know that Y is in the keyword and could be the last letter ofit. Look at the F-----D sequence. F is in the keyword and theO-------H is the only area than can fit the F and the Y.

    Plug in my UNSATifcOrY guess. The lower letters requirechecking. Alphabet 1 fits the key as UNSATISFACTORY adjustedfor duplicate letters.

    The message reads in part: Slang is language or phrases of avigorous colorful metaphoric or taboo nature invented to ...


    The QUAGMIRE III is a very important class of ciphers becausethey introduce the one of the most important tools invented byMr. Friedman, as explained in his Riverbank papers, called"Direct and Indirect Symmetry."

    The title of this problem is "Inertia in the British LaborMarket" and has the tip "ANDTHREECALLINGFORAMANTOSTANDON."



    Note the repeat of the first three letters IBW at interval 81.If the message starts with THE and the period turns out to be 9we have found a wedge. Next place the tip in columnar line fora cycle of nine.

        A N D T H R E E C              A I K P S R T C O    A L L I N G F O R              J J W P R R V O L    A M A N T O S T A              A A A R U R J N U    N D O N t w o f e e t  ?       I X M X P Q B V U            t h e -------  ?       I B W O G P C D P                                   (also first three IBW)
    The three A's in the first column followed by the two N'sprove the period of 9. This is not accidental. My guessesof additional plain text are partially right - 'the' as youwill see later. Note the triple R's, two U's and Two I's inthe ciphertext lined up by columns in a period of 9.

    Break the ciphertext into groups of nine.

    Place the extended tip. In a QUAGMIRE III, or in any casewhere the cipher component is the same as the plain component,if one cipher -plain matches E for E, all pairs must match,for the sequence is set A to A, B to B, etc. When thishappens, we get a column of our write-out as "free plain text,"which is of considerable help.

    I can not overemphasize the next step. Because of the K3nature of the keying, the Plain component and the Cipher 1alphabet represents pairs that are the same distance removed -H to J, N to A, T to I, in this case. Similarly G to A, H toB, O to X, and R to J are equally separated - though not at thesame interval as the first pairs obtained from line 1.(Obviously, if H to J is "x" distance, H to B cannot be thesame distance.) Check this observation of Symmetry on thedecipher tableaux.


    Plain 0 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      -----------------------------------------------------Cipher1               J           A           I      2             A B             X     J      3         W A                         M      4         P                   R         X      5 P   R                             U      6 R                         Q      7       B               V J T      8 N     C               O     V      9                 L         U           O
    Let us write down all the pairs we get by going from plain tocipher in each of the alphabets in turn. We can also writedown the from the sidewise relationships. For instance, A to Con the plain sequence is the same distance P to R on Row 5. Inaddition, Row 7 to Row 8 tells us that BC is the same distanceapart as VO.

    This is a most powerful tool in solution of a sequence againstitself. You can imagine a little "square" and go up, or down,or across, to find relationships within and between both plainand cipher components.

    Plain sequence to Row 1  HJ NA TI                      2  GA HB OX RJ                      3  EW FA SM                      4  EP OR TX                      5  AP CRU      (CR-RU)                      6  AR NQ                      7  DB LV MJ NT                      8  AN DC LOV    (LO-OV)                      9  IL NU TO>From Plain A to C        AC PR>From Row 7 to 8          BC VO
    There are a lot of relationships. I have not listed thesidewise ones like Plain to Row 1 - H to N and J to A.

    MASTERTON points out that Row 1 is the reverse of Row 8.[MAST] I didn't see this "little" jump.

    But I did make sense of the three letter chains; if L-O is thesame as O-V we have a three letter segment. Do you see thatthe pairs in the listing above are separated by one letter in asequence obtained from the next set, as evidenced by LV in 7and LOV in 8? We can add the two together:

    Look at the fragments, and realize that we have found some goodinformation about the sequence. First of all the sequences arereversed alphabets. The sequence has BCD, VOL, JKM since wehave used L and T-NA in it? [We can also look at a processcalled decimination to bring the sequence to bear. We will dothat in the Friedman section.] Remember the very importantpart of the tool of symmetry - that because the plain and allthe cipher alphabets are the same, we can associated pairs inthe straight, sideways, down etc as we find them, using theplain or all nine cipher alphabets. In a QUAGMIRE IV, wecannot use the plain sequence in this way because of adifferent key.

    We continue our recovery with A to N plain as the samedistance as R to Q in alpha 6. We add QR to our line.

                VOL   TINA      BCD     HJKM   QR
    Notice the H to B and G to A in the plain to alphabet 2relationship. This tells us to put G ahead of H, then A goesbehind B as we expect. Since O is in VOL and N is in TINA

    the only missing element is P which we place as follows:

              ku  VOL/?/TINABCD (f)GHJMPQR swxyz
    missing elements at this stage are e, k, u, w , x , y , z whichlikely the E and U are in the Keyword.


    Plain 0   V O L         T I N A B C D F G H J M P Q R S      -----------------------------------------------------Cipher1 V O L         T I N A B C D F G H J M P Q R S w      2     X                     T I N A B C D F G H J M P Q      3                             T I N A B C D F G H J M P      4   Q R S W? X      5      6      7      8     V O L            T I N A B C F G H J M P Q R S      9
    The line ups are not correct. We can find where alphabets 1,2 and 3 start by putting the low frequency X in the rightspot. I leave this part of the work to you all. [ Hint:compress the V O L -----T I N A space and what keyword will fitinto - V O L u? T I (O)N. and place the E in the beginning.]

    The answer is with Keywords EVOLUTION and BLUEPRINT:


    Plain 0 E V O L U T I N A B C D F G H J K M P Q R S W X Y Z      -----------------------------------------------------Cipher1 V O L U T I N A B C D F G H J K M P Q R S W X Y Z E      2 S W X Y Z E V O L U T I N A B C D F G H J K M P Q R      3 W X Y Z E V O L U T I N A B C D F G H J K M P Q R S      4 P Q R S W X Y Z E V O L U T I N A B C D F G H J K M      5 C D F G H J K M P Q R S W X Y Z E V O L U T I N A B      6 F G H J K M P Q R S W X Y Z E V O L U T I N A B C D      7 Y Z E V O L U T I N A B C D F G H J K M P Q R S W X      8 Z E V O L U T I N A B C D F G H J K M P Q R S W X Y      9 X Y Z E V O L U T I N A B C D F G H J K M P Q R S W
    The message reads: The British created a civil service job ineighteen hundred and three calling for a man to stand on thecliffs of Dover with a spyglass...


    The QUAGMIRE IV is probably the most difficult of the QUAGMIRESbecause we need to recover two keyworded alphabets and directsymmetry will not work with the plain.

    We are given:

    MWQYD  KMCAO  KHSEE  YULIH  WYTEW  YRLHG  LMEJC  ZHAKE  NYWUPthegr  reatQSQSO  ESYEP  BIZEW  QYPKZ  FHAAM  GWPTR  XNYWR  LKSQE  XHGRAQCWAV  JNCPM  HDHZT  BCBHR  AMXUE  OLTWR  RIKNQ  AKKDZ  VJOYW                                                          bet?WHQJR  FGYVP  GILWV  WGPTF  MLYKX  TAKOZ  ATFGL  AUT.weenl  atese  ptemb  erand  decem  berof  thaty  ear


    The Title is "Lost Horsepower", the tips are starts with THEGREAT and has WEENLATESEPTEMBERANDDECEMBEROFTHATYEAR in thetext. The letters bet?WEEN might be inferred.

    Finding the cycle is our first challenge.

    The WQY is +58, a discouraging number for factors. The cribsare pretty generous, so looking at them we might findsomething. Obviously, a plain hit at the correct interval ofthe cycle would result in a cipher coincidence at the sameinterval. Two occurrences of a plain letter at some intervalother than the period or multiple of the cycle, the cipherscannot be the same. MASTERTON describes a graphical techniquefor knocking out intervals. [MAST]

      OYWWHQJRFGYVPGILWVWGPTFMLYKXTAKOZATFGLAUT  betweenlateseptemberanddecemberofthatyear   * --9--  *
    Thus the Y over E and H and Q over E "knock out" the intervals3, 4 which are too short anyway, and also 11 because of the Yover P. Note the +9 hit for Y over E. So we write out thecipher in a period of nine:
    123456789  123456789  123456789  123456789  123456789MWQYDKMCA  OKHSEEYUL  IHWYTEWYR  LHGLMEJCZ  HAKENYWUPthegreatE              E GH EE                    E AQSQSOESYE  PBIZEWQYP  KZFHAAMGW  PTRXNYWRL  KSQEXHGRA  E  ?HE   E T    EA  R      RT  ER    E    R E     EQCWAVJNCP  MHDHZTBCB  HRAMXUEOL  TWRRIKNQA  KKDZVJOYW   T    A  TE                    NH  E   E  R     betWHQJRFGYV  PGILWVWGP  TFMLYKXTA  KOZATFGLA  UT.weenlates  eptembera  nddecembe  rofthatye  ar
    Even with all the help and correct hits, the message is not agive a way.


    Plain 0 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      -----------------------------------------------------Cipher1 U       P                 T       K   M     W      2       F H     W             O G   T      3       M Q Z                           I      4         L   Y             J           A      5     Y         T       R W         D      6 F V     K      7 M O     W               X             G      8   T     Y                         G   C      9 P       A   C                       V W
    Since the alphabets are different we can not chain from theplain to cipher. However, WITHIN the cipher, the same rulesapply as before - except their isn't nearly as muchinformation. In Cipher 1 row we see that U to P is the samedistance as F to K , M to W and P to A. Ok. Remember that weare dealing with unknown decimations, so the relationshipsbetween UPA, PK and PT is unknown.

    By decimation I mean the process of selection of elements froma sequence according to some fixed interval. For example, thesequence A E I M is derived, by decimation , from a normalalphabet by selecting every fourth letter. It is the key toSymmetry solutions because the latent relationships in a cipheralphabet can be made patent by decimation. Lecture 11 willgive two methods of decimation in detail.

    Table of Relationships in foregoing example:

    UPA FK  MW         Plain A to E and Rows 1 to 9PT  LJ              "    E to NPK  HT  YG          "    E to R and Rows 1 to 6  adding UFPM  QI  LAWG  YC    "    E to T and Rows 9 to 7 and 4 to 9UMG  PW             "    A to T and Rows 1 to 7TM  JA              "    N TO TFH  MQ              "    D to EWTD                 "    H to R and Rows 2 to 5FV  MO              "    A to BVK  OW  TY          "    B to EOG  TC              "    B to TPH  KT             Rows  1 to 2PQ  MI             Rows  1 to 3PL  TJ  MA         Rows  1 to 4PY  KG  MC         Rows  1 to 8FM  HQ  KW  VO     Rows  2 to 0HY  TG             Rows  2 to 9QL  IA             Rows  3 to 4QW  IG             Rows  3 to 7QY  IC             Rows  3 to 8QA  IW             Rows  3 to 9LW  AG             Rows  4 to 7LY  AC             Rows  4 to 8   and Plain A to G adding                                  Cipher C under Plain G on RowFP  KA             Rows  6 to 9   9OT  WY  GC         Rows  7 to 8YA  CW             Rows  8 to 9
    Row 2 to 3 and 6 to 7 are combined. S and T in plain are mostlikely adjacent from VW in Cipher 9. Partials FH and MQ lookgood without an intervening letter.

    LAWG is our best bet for the wedge. It ties together E and Tin the same decimation. So:

                   Plain         E T               Cipher        P M                             H                             Q I                             L A W G                                K                            L A W G                                Y C                              L A W G
    If FH and MQ are the right order, P is in the keyword, sincethe reverse bits of above (MP, IQ, GWAL) would not beconsistent with MPQ. Unfortunately, we have run out of gas andmust guess more plain. The plain E-gh-EE most likely isEighteen and since they are talking about years, why notSeventy, since so many E's are fitting? The plain T of seventyis confirmed. The plain V may not produce much but the cipherG might be a bonanza. These new values add KE and JR to thechain.

    123456789  123456789  123456789  123456789  123456789MWQYDKMCA  OKHSEEYUL  IHWYTEWYR  LHGLMEJCZ  HAKENYWUPthegreatE       T      EIGHTEEN  SEVENTY          E AQSQSOESYE  PBIZEWQYP  KZFHAAMGW  PTRXNYWRL  KSQEXHGRA  E  THE   E T    EA  R      RT  ER    E    R E     EQCWAVJNCP  MHDHZTBCB  HRAMXUEOL  TWRRIKNQA  KKDZVJOYW   T    A  TE                    NH  E   E  R     betWHQJRFGYV  PGILWVWGP  TFMLYKXTA  KOZATFGLA  UT.weenlates  eptembera  nddecembe  rofthatye  ar


    Plain 0 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      -----------------------------------------------------Cipher1 U       P                 T       K L M     W      2       F H     W             O G   T      3       M Q Z     W                     I   G      4         L   Y             J           A      5     Y         T       R W M       D      6 F V     K                         J   E      7 M O     W               X             G      8   T     Y                         G   C      9 P       A   C                       V W
    We look at VW and LM and KLM under the plain RST. We mustconclude that G-C is correct. Rows 7 and 8 have a G and Cunder plain T, and WY under E and OT under B. This suggeststhat WXY and O-T are part of the final chain. So push thefollowing chains:

          KLM, G-C, VWXY, EA, O-TThe cipher sequence appears to go:      JKLMQVWXYZ0              A N D E   I C B F G H---------------------------------------------1              U T   P R A2                  F H J K L M Q V W X Y Z3        F H J K L M Q V W X Y Z4            F H J K L M Q V W X Y Z5      F H J K L M Q V W X Y Z6              F H J K L M Q V W X Y Z7    F H J K L M Q V W X Y Z8F H J K L M Q V W X Y Z9              P R   A
    The cipher keyword has this form O U T - P R - A I N Gwith S, E, D candidates. The keyword is SPREADING.The plain keyword can be derived as PANDEMIC and the ciphersetting key is HORSETAIL. The groundwork is left to thestudent. Notice how resistant the QUAGMIRE IV was even withloads of help.




    [updated 6 April 1996]

    REFERENCES / RESOURCES [updated 6 April 1996]

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And, it has one of the better       illustrations of the Soviet one-time pad with example,       with three errors in cipher text, that I have corrected       for the author.][MARS] Marshall, Alan, "Intelligence and Espionage in the Reign       of Charles II," 1660-1665, Cambridge University, New       York, N.Y., 1994.[MART] Martin, James,  "Security, Accuracy and Privacy in       Computer Systems," Prentice Hall, Englewood Cliffs,       N.J., 1973.[MAST] Lewis, Frank W., "Solving Cipher Problems -       Cryptanalysis, Probabilities and Diagnostics," Aegean       Park Press, Laguna Hills, CA, 1992.[MAU]  Mau, Ernest E., "Word Puzzles With Your Microcomputer,"       Hayden Books, 1990.[MAVE] Mavenel, Denis L.,  Lettres, Instructions Diplomatiques       et Papiers d' Etat du Cardinal Richelieu, Historie       Politique, Paris 1853-1877 Collection.[MAYA] Coe, M. 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