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

    • CLASSICAL CRYPTOGRAPHY COURSE


    BY LANAKI

    March 30, 1996


    Revision 0
    COPYRIGHT 1996
    ALL RIGHTS RESERVED
    LECTURE 12
    POLYALPHABETIC SUBSTITUTION SYSTEMS III
    CRYPTANALYSIS OF VIGGY'S EXTENDED FAMILY
    DECIMATION IN DETAIL

    SUMMARY

    In Lectures 12 - 13, we continue our study of the "Viggy"cipher family or Polyalphabetic Substitution systems. Wewill cover decimation processes in detail and investigatespecial solutions for periodic ciphers. The importantprinciple of Superimposition will be introduced.

    The Resources Section has been updated with more than 50 ACApublished references on these and similar systems - focusingon the cryptanalytic attack and areas of historical interest.Thanks to PHOENIX for his help in compiling these sources.[INDE]

    "INCOMING"

    In Lecture 13, we will tackle the difficult aperiodicpolyalphabetic case and introduce auto/running key systems.We will diagram the topics covered in Lectures 10 - 13.

    Lecture 14 will be presented by LEDGE. He will cover furtherCryptarithm topics.

    Lectures 15-18 will discuss the various geometric,transposition and fractionation ciphers.

    PORTAX CIPHER

    We start with a difficult cousin of the PORTA described inLecture 11. The PORTAX uses pairs of letters as a unit forencipherment and decipherment as apart from single letters.

    A special slide is required for its operation, and a keywordis needed.

              A B C D E F G H I J K L M        (stationary)  . N O P Q R S T U V W X Y Z N O P Q R S T U V W X Y Z ...  . C E G I H M O Q S U W Y A C E G I K M O Q S .. (sliding  . D F H J L N P R T V X Z B D F H J L N P R T ..  key)
    (The above slide-setting is for G-H (key) directly under theA-indicator of the stationary alphabet.)

    To encipher the digraph RE, we take the R in the upper row ofletters (stationary slide) and the E from the lower pair ofletters (sliding), and use the opposite corners of therectangle formed to obtain the ciphertext, or PI. However,if the digram ER is to be enciphered, we take the E from thestationary alphabet at the top, and the R from the slidingalphabet at the bottom to obtain FP.

    Note that if the first letter of a digraph is in the range ofA-M, the equivalent ciphertext is dependent on where theslide is used for the key-letter; but, if the first letter ofthe digraph is in the range of N-Z, then it slides along withthe paired rows of lower letters, and therefore all suchdigraphs having the first letter in the N-Z are constant,without dependent of the key. There is an exception whenboth letters in the plaintext digraph are in the same column,in which case the key letter has to be known, for lettersappearing above the needed letters are used for theciphertext. [BRYA]

    To encipher with keyword, the plaintext is written in tworows under it; continuing to the end of the message. Whenthe final group is reached, if there are not enough lettersto make it complete (an even number), add a single null.

    For example, encipher the word INNOVATION using the keyOFTEN :

                    *                A B C D E F G H I J K L M        (stationary)  . N O P Q R S T U V W X Y Z N O P Q R S T U V W X Y Z ...  . C E G I K M O Q S U W Y A C E G I K M O Q S .. (sliding  . D F H J L N P R T V X Z B D F H J L N P R T ..  key)                *O F T E N   (keyword)---------I N N O VA T I O Ng we b---------S A R E FO U N D xu ik e
    Setting the O of the sliding pairs under the 'A' indicatorof the stationary alphabet, we encipher IA as GE (oppositecorners); then SO, continuing down the column we encipher thewhole column. We then slide the strip until E-F (key) isunder the A indicator and encipher that column.

    To find the period in the PORTAX is dependent on possiblefragments of the plaintext which are known (through the N-Zcombinations produced from the unchanged relationship ofletters). Lets partially decipher the following PORTAX:

    SNPOW  LBAMP  ISCWU  OOBXC  WKMAT  ZKTOW  JCBLN   CBJGBTAAJD  IWUKW  HHVZN  MNUFM  APBJW  PCBSX  JCJQX   TMVUBMDCBJ  CGUGR.   (90)Assuming a period of 6:          S N P O W L          B A M P I S          n   t u r                   natural ?          l   e d s        good          -----------          C W U O O B          X C W K M A            o y s            s o c          ok          -----------          T Z K T O W          J C B L N C          r o   s t o          n y   n d s      better          -----------          B J G B T A          A J D I W U                  y                  m          -----------          K W H H V Z          N M N U F M            t     p t            s     r y          -----------          A P B J W P          C B S X J C            n     r o            f     t e          -----------          J Q X T M V          U B M D C B            n   t o n            h u n   r          -----------          J C R  - -          U G R          -----------
    Note the NY-NDS which could be NYaNDS or NYeNDS. Look at thefinal group, we find -NTON -HUN-R (hundred?) We next test thekeyword by putting T in the final position and testing theprecursor letter; A C E F H I L N O P R S and U, At the Esetting, OM = TC, making -OYST/-SOCCU with R in the nextgroup confirming OCCUR. The E substitution also gives us theHUNDRED. The rest of the analysis is left for the studentfor credit.

    THE NIHILIST SUBSTITUTION CIPHER

    One of my favorite ciphers is the Nihilist SubstitutionCipher. Classified as a periodic, it employs numbers torepresent letters. Numbers are derived from a 5 x 5 PolybiusSquare.

    We set up a block of 25 letters and combine I/J in one cell.

                          Figure 12-1a                     1  2  3  4  5                  1  A  B  C  D  E                  2  F  G  H I/J K                  3  L  M  N  O  P                  4  Q  R  S  T  U                  5  V  W  X  Y  ZSo A = 11, L = 31, T = 44.  (Row by Column)
    The Polybius Square can be keyed. For example, usingUNITED STATES OF AMERICA and eliminating the duplicateletters, we have:
                          Figure 12-1b                     1  2  3  4  5                  1  U  N  I  T  E                  2  D  S  A  O  F                  3  M  R  C  B  G                  4  H  K  L  P  Q                  5  V  W  X  Y  Z
    We can also mix it up further with a little transposition.

    Use BLACKSMITH, transpose and remove the ciphertext bycolumns starting at 1:

                    B L A C K S M I T H                D E F G N O P Q R U                V W X Y Z      B D V L E W A F X C G Y K N Z S O M P I Q T R H UThe resulting square reads:                     Figure 12-1c                    1  2  3  4  5                 1  B  D  V  L  E                 2  W  A  X  F  C                 3  G  Y  K  N  Z                 4  S  O  M  P  I                 5  Q  T  R  H  U
    Figure 12-1c shows the effect of the transposition appliedfirst.

    Now the message COME AT ONCE enciphered with a keyword ofTENT (period = 4) is:

                   T-44  E-15  N-35  T-44               ----------------------               C-13  O-34  M-32  E-16               A-11  T-44  O-34  N-33               C-13  E-15   -     -
    We add the key and the plaintext equivalents together toproduce the ciphertext: COME: 57 49 65 59; ATON: 55 59 6777; CE: 57 30. Each column represents a monoalphabeticsubstitution in itself, and the reading or value of theseletters is dependent on the letters on either side of them.

    WEAKNESSES

    The lowest number of any key-letter which may be added to thelowest plaintext letter is 11, with a total of 22; thehighest combination is two 55's or 10 (110). The numbers6,7,8, or 9, are not involved in either the tens or the one'sadditions - but they may result in a sum. Cipher 22 mustequal 11 plus 11; and 10 can only mean the sum of two 55's.Zero in the one's column means that two 5's have been added.This is also true in the ten's column. If at any time we findthat a 6-7-8-9 is involved we can discard the period assumedas wrong. What we are looking for is a number in the 1-2-3-4-5 range that may be added to produce first the ten's sumand then the one's sum.

    FINDING THE PERIOD

    There are two ways to find the period - the short and thelong way.

    SHORT METHOD

    The short way of finding the period is to look for two ormore 30's. We treat them like a repeated digraph and factorthe interval between them looking for a common factor. We mayalso try the same procedure with the lowest number versus thehighest number, for example the distance between two 94's ortwo 26's.

    LONG METHOD

    The long way is to assume a 3 period and test the 1'st and4'th, 2'nd and 5'th, 3'rd and 6'th in the same manner as theshort method. When conflicts arise, discard the choice.We continue with an assumption of periods 4, 5, 6, etc. andincrease the differentials between ciphertext numbers. [BRYA]

    CRYPTANALYSIS OF THE NIHILIST SUBSTITUTION

    Gaines [ELCY] suggests that cracking this cipher parallelsthe Viggy. The period is found through repeated sequences, orin their absence, through repeated single letters, yieldingindividual frequency counts on the several alphabets of theperiod. If the arrangement of the ciphertext follows thenormal Polybius (aka Checkerboard) Square, the frequencycounts will follow the graph of the normal alphabet less oneletter. Even with the keyword mixed ciphertext alphabet,no matter how badly mixed, the frequency counts are parallel,the several alphabets combined follow one graph, and can be"lined up."

    Notice that the primary alphabet contains only the digits 1-2-3-4-5. The maximum difference is 4 and addition of anynumber to all of them does not change this fact. The maximumdifference between any two sums is still 4. Now the numberadded during encipherment is also a number containing nodigit other than 1-2-3-4-5; thus any number found in thecryptogram can be considered as carrying two separateadditions, one for tens and one for ones. The two 5's addedgive us the revealing 0; the carried digit 1 can be mentallyborrowed back, by decreasing the size of the digit precedingthe zero. If we find a 40 , we look at it as 3 tens with tenunits or finding 110, we may regard this as ten tens and tenunits. If we find the numbers 29 and 87 in the cryptogram,we know they were not enciphered by the same key. This isbecause a difference greater than 4 in the respective tensunits exists and no digit whatever added to any two digits ofthe original square can produce a difference greater than 4.Say we have 30 and 77, with no difference greater than 4, thepresence of the zero needs to be accounted for. The number30 has 2 tens and ten units; 7 - 2 >4, hence, we rejectthe same key hypothesis.

    Four giveaways are 22, 30, 102, and 110. The presence of anyone of these numbers gives away the key to the whole cipheralphabet.

    [BRYA] presents a useful aid for the standard PolybiusSquare in Table 12-1. At the top is the key-number, at theleft is the plaintext letter, and at ciphertext is found atthe intersection. Any two of the three variables yields theunknown letter/number.

                          Table 12-1      11  12  13  14  15  21  22  23  24  25  31  32       A   B   C   D   E   F   G   H I/J  K   L   MA 11  22  23  24  25  26  32  33  34  35  36  42  43B 12  23  24  25  26  27  33  34  35  36  37  43  44C 13  24  25  26  27  28  34  35  36  37  38  44  45D 14  25  26  27  28  29  35  36  37  38  39  45  46E 15  26  27  28  29  30  36  37  38  39  40  46  47F 21  32  33  34  35  36  42  43  44  45  46  52  53G 22  33  34  35  36  37  43  44  45  46  47  53  54H 23  34  35  36  37  38  44  45  46  47  48  54  55I 24  35  36  37  38  39  45  46  47  48  49  55  56K 25  36  37  38  39  40  46  47  48  49  50  56  57L 31  42  43  44  45  46  52  53  54  55  56  62  63M 32  43  44  45  46  47  53  54  55  56  57  63  64N 33  44  45  46  47  48  54  55  56  57  58  64  65O 34  45  46  47  48  49  55  56  57  58  59  65  66P 35  46  47  48  49  50  56  57  58  59  60  66  67Q 41  52  53  54  55  56  62  63  64  65  66  72  73R 42  53  54  55  56  57  63  64  65  66  67  73  74S 43  54  55  56  57  58  64  65  66  67  68  74  75T 44  55  56  57  58  59  65  66  67  68  69  75  76U 45  56  57  58  59  60  66  67  68  69  70  76  77V 51  62  63  64  65  66  72  73  74  75  76  82  83W 52  63  64  65  66  67  73  74  75  76  77  83  84X 53  64  65  66  67  68  74  75  76  77  78  84  85Y 54  65  66  67  68  69  75  76  77  78  79  85  86Z 55  66  67  68  69  70  76  77  78  79  80  86  87                      Table 12-1                       continued      33  34  35  41  42  43  44  45  51  52  53  54  55       N   O   P   Q   R   S   T   U   V   W   X   Y   ZA 11  44  45  46  52  53  54  55  56  62  63  64  65  66B 12  45  46  47  53  54  55  56  57  63  64  65  66  67C 13  46  47  48  54  55  56  57  58  64  65  66  67  68D 14  47  48  49  55  56  57  58  59  65  66  67  68  69E 15  48  49  50  56  57  58  59  60  66  67  68  69  70F 21  54  55  56  62  63  64  65  66  72  73  74  75  76G 22  55  56  57  63  64  65  66  67  73  74  75  76  77H 23  56  57  58  64  65  66  67  68  74  75  76  77  78I 24  57  58  59  65  66  67  68  69  75  76  77  78  79K 25  58  59  60  66  67  68  69  70  76  77  78  79  80L 31  64  65  66  72  73  74  75  76  82  83  84  85  86M 32  65  66  67  73  74  75  76  77  83  84  85  86  87N 33  66  67  68  74  75  76  77  78  84  85  86  87  88O 34  67  68  69  75  76  77  78  79  85  86  87  88  89P 35  68  69  70  76  77  78  79  80  86  87  88  89  90Q 41  74  75  76  82  83  84  85  86  92  93  94  95  96R 42  75  76  77  83  84  85  86  87  93  94  95  96  97S 43  76  77  78  84  85  86  87  88  94  95  96  97  98T 44  77  78  79  85  86  87  88  89  95  96  97  98  99U 45  78  79  80  86  87  88  89  90  96  97  98  99  00V 51  84  85  86  92  93  94  95  96  02  03  04  05  06W 52  85  86  87  93  94  95  96  97  03  04  05  06  07X 53  86  87  88  94  95  96  97  98  04  05  06  07  08Y 54  87  88  89  95  96  97  98  99  05  06  07  08  09Z 55  88  89  90  96  97  98  99  00  06  07  08  09  10
    Consider Edwin Linquist's challenge:
    24 66 35 77 37 77 55 59 55 45 55 88 28 66 4688 37 67 33 59 58 65 45 66 67 58 44 55 34 7944 59 55 45 42 87 28 76 43 78 46 86 26 67 2485 26 67 28 76 26 78 46 65 65 88 36 49 54 6728 65 42 88 36 49 44 89 57 58 54 66 47 67 26
    Try period = 2. Starting at the first number 24 constant wescan the line looking for differences greater than 4 using aconstant difference of 2. We come to 33 and 38 and stop.

    Try period = 3. The first comparison fails at 24 and 77.

    Try period = 4. We are able to go through the entirecryptogram, comparing numbers at an interval of 4, withoutfinding any difference in either tens or units greater than 4.We now must look at the numbers collectively in columns toverify the period is 4. We recopy the cryptogram into ablock.

                     Key = 4?               24  66  35  77               37  77  55  59               55  45  55  88               28  66  46  88               37  67  33  59               58  65  45  66               67  58  44  55               34  79  44  59               55  45  42  87               28  76  43  78               46  86  26  67               28  76  26  78               46  65  65  88               36  49  54  67               28  65  42  88               36  49  44  89               57  58  54  65               47  67  26  -
    Alphabet 1: The tens-half of the first column contains thedigit 2 and since this can only come from the addition of 1plus 1, the only possible key digit is 1. The units-half hasa range of 4-5-6-7-8, maximum range possible. The smallestdigit to result in 8 is 3, the largest digit to result in 4is also 3, that is the only digit which can result in all ofthe digits 4-5-6-7-8 is 3, so that the cipher key for thiscolumn is 13. It cannot be anything else.

    Alphabet 2: The tens-half of the second column ranges overthe full five digits 4-5-6-7-8 (key 3), and the units-halfranges over 5-6-7-8-9 (key 4). This suggests the key digitis 34.

    Alphabet 3: The tens-half of the third column contains the'giveaway' digit of 2 and the units-half also contains thedigit 2. The key digit to produce this situation is 11.

    Alphabet 4: The tens-half of the fourth column ranges onlyover the digits 5-6-7-8, with nothing to indicate whether themissing digit is 4 or 9. The key might be either 3 or 4.The units has the full range of digits 5-6-7-8-9, hence key =4. So we have either 34 o 44 for our key digit. The normalsquare suggests COAO or COAT as the key word. We use Table12-1 to good advantage and decipher this cryptogram.

    We decipher the whole cryptogram a column at a time:

         'C'    'O'   'A'   'T'      --     --    --    --      A      M     I     N      I      S     T     E      R      A     T     T      E      M     P     T      I      N     G     E      U      L     O     G      Y      I     N     A      F      U     N     E      R      A     L     S      E      R     M     O      M      W     E     H      A      V     E     H      E      R     E     O      N      L     Y     T      H      E     S     H      E      L     L     T      H      E     N     U      T      I     S     G      O      N     E
    Reads:
    A minister attempting eulogy in a funeral sermon: Wehave here only the shell, the nut has gone.

    For the most difficult case presenting multiple keypossibilities, we line up the alphabets graphically againsttheir frequency counts to eliminate the extra key digits.

    GROMARK

    MASTERTON describes a cipher called the GROMARK. The Gromarkis akin to the GRONSFELD in that the components never changetheir position relative to each other and every plain textvalues has 10 possible cipher representatives. The GROMARKuses a different keying method; encipherment is effected bymeans of a normal alphabet plain set against a mixed ciphertext alphabet. However, instead of cycles or predictableslides of the cipher component, one finds the plain value onthe top (normal) component and counts a specified number ofpositions to the right, then takes the letter in the cipheralphabet immediately below. The choice of how far to countalong the sequence is determined by the digital key. Oneessentially is adding 0 to 9 to the plain value, as in theGronsfeld, but it is on the mixed sequence, set underneath aplain sequence. The key is derived from a Fibonacci series.On some cycle (frequently 5 wide) the key is derived from astarting group, by adding the first position to the secondand placing the result in the sixth position. Similarly,positions 2 and 3 are added to make position number 7, 3, and4 to make 8, and so forth. All additions are non carrying -avery common cryptographic practice. [MAST]

    Example:
    Use the starter or "seed" of 48671, the key is:

      48671  24383  67119  382021 ...
    Solution follows the normal Viggy methods. The cribplacement can be interesting.

    Example:

    7 7 2 6 6 4 9 8 2 0 3 7 0 2 3 0 7 2 5 3 7 9 7J C N W Z Y C A C J N A Y N L Q P W W S T W P
    without knowing the cipher sequence, we are given the cribSUBSTITUTES and runs somewhere from the J to the final Pabove.

    Since the plain sequence is normal, a repeated cipher letter,with different key letters on it, must stand for plain valuesremoved from each other exactly by the difference of the twonumbers. Thus C A C with keys 9 8 2 above it implies thatthe first cipher C is M for example, the second C is sevenpositions to the right on the plain sequence, or T.

    Or:

    J K L M N O P Q R S T U V W X                        C                        *
    We prepare a difference table. We are looking for afavorable case where the differences in the cipher repeatsmatches the plain differences, at the correct interval.To match these differences, we measure them in one directionfor the plain and the reverse for the cipher. Table 12-1shows subtraction of the left hand letter from the right, andwe must look at the cipher in the other direction.Differences may be calculated modulo 26.

                              Table 12-1adjacent         19 21  2 19 20  9 20 21 20  5 19diff's            S  U  B  S  T  I  T  U  T  E  Sxx                2  7 17  1 15 11  1  25 11 14x-x                9 24 18 16  0  12  0  10x--x                  0  25  7 ...
    There is a difference of 7 with the C-C hit, but it doesn'tappear on the second row of the table. The keyword mustfirst be between A (between C's) and W.
    7 7 2 6 6 4 9 8 2 0 3 7 0 2 3 0 7 2 5 3 7 9 7J C N W Z Y C A C J N A Y N L Q P W W S T W P              S U B S T I T U T E S
    This is a good tip placement and confirmed by the N-N hit.The A---A in the cipher matches the S---T plain. We buildthe cipher component by writing the cipher component, and anormal alphabet, count along it from any given plain thenumber of steps given by the key, then write the ciphervalue. Find S on the top strip, count 8 to right, place anA. C is two spaces to the right of the position held by theU, and so on. Decipher other letters by counting backwardsthe number of steps given by the key. Cipher C ahead of thewcrib translates to N.

    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 ZA J             Y     P               Q W N C L
    Without a tip the system will fall to statistics. The numbersassociated with any given cipher letter represent a stretchof 10 consecutive values along a normal alphabet such as C toL or X to G, we could prepare a table with A to Z as the rowsand 9 to 0 as the columns. Frequencies can be combinedand a stretch such as PQRST area will show as the normal.The backwards normal sequence yields a bar graph of thesegment of the normal alphabetic frequencies.

    DECIMATION PROCESSES - FURTHER REMARKS

    In Lecture 11, we presented QUAGMIRES I-IV and solved them bya variety of methods. Inherent in their solution wasFriedman's principle of indirect symmetry. [FRE7] Primafacie to this symmetry principle is a process of alphabetdissociation called Decimation. This same process effectsall Viggy class ciphers and is important from a theoreticalpoint of view. Decimation is especially effective in solvingmixed alphabet systems like the Quagmire III & IV.Decimation is a process of selection and derivation of asequence of equivalent components according to some fixedinterval. For example, the sequence A E I M is derived bydecimation of extracting every fourth letter from a normalalphabet.

    Consider the two mixed alphabets in a QUAGMIRE III:

                    O1                 *       *Plain:           QUESTIONABLYCDFGHJKMPRVWXZCipher:  QUESTIONABLYCDFGHJKMPRVWXZQUESTIONABLYCDFGHJKMPRVWXZ                 *       *                Ok
    By setting the two sliding components against each other inthe two positions shown: A in the first set and B in thesecond set we can derive two, we can derive two differentsets of secondary alphabets based on the key letters.

                     O1 *       *Plain:            QUESTIONABLYCDFGHJKMPRVWXZCipher:  QUESTIONABLYCDFGHJKMPRVWXZQUESTIONABLYCDFGHJKMPRVWXZ                  *       *                  OkSecondary Alphabet (1)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: H J P R L V W X D Z Q K U G F E A S Y C B T I O M NSecondary Alphabet (2)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: J K R V Y W X Z F Q U M E H G S B T C D L I O N P A
    Sliding strips will yield the same results as a Viggy typetable based on the Keyword QUESTIONABLY (see a partial tablein Table 12-2.
                         Table 12-2               Partial Reconstruction            QUESTIONABLYCDFGHJKMPRVWXZ            UESTIONABLYCDFGHJKMPRVWXZQ            ESTIONABLYCDFGHJKMPRVWXZQU            STIONABLYCDFGHJKMPRVWXZQUE            TIONABLYCDFGHJKMPRVWXZQUES            IONABLYCDFGHJKMPRVWXZQUEST            ONABLYCDFGHJKMPRVWXZQUESTI            NABLYCDFGHJKMPRVWXZQUESTIO            ABLYCDFGHJKMPRVWXZQUESTION            BLYCDFGHJKMPRVWXZQUESTIONA            LYCDFGHJKMPRVWXZQUESTIONAB            YCDFGHJKMPRVWXZQUESTIONABL            CDFGHJKMPRVWXZQUESTIONABLY            .                        .
    Superficially secondary alphabets (1) and (2) show noresemblance of symmetry despite the fact that they were bothcreated from the same primary alphabet. We do find a LatentSymmetry Of Position (aka Indirect Symmetry of Position).This phenomenon has widespread use in the Viggy family.Consider alphabet (2):
    Secondary Alphabet (2)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: J K R V Y W X Z F Q U M E H G S B T C D L I O N P A
    We construct a chain of alternating plaintext and ciphertextequivalents, beginning at any point and continuing until thechain is completed. We start Aplain = Jcipher, Jplain =Qcipher, Qplain = Bcipher...., dropping the common letterswe have A J Q B. The complete sequence of letters is:
    When slid against itself it will produce exactly the samesecondary alphabets as do the primary components based uponthe word QUESTIONABLY. For example, compare the secondaryalphabets given by the two settings of the externallydifferent components below:
                      *        *Plain:            QUESTIONABLYCDFGHJKMPRVWXZCipher:  QUESTIONABLYCDFGHJKMPRVWXZQUESTIONABLYCDFGHJKMPRVWXZ                  *        *Secondary Alphabet (1)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: J K R V Y W X Z F Q U M E H G S B T C D L I O N P A         *  *Plain:   AJQBKULMEYPSCRTDVIFWOGXNHZCipher: AJQBKULMEYPSCRTDVIFWOGXNHZAJQBKULMEYPSCRTDVIFWOGXNHZ         *  *Secondary Alphabet (2)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: J K R V Y W X Z F Q U M E H G S B T C D L I O N P A
    Since the sequence A J Q B K ... gives exactly the sameequivalents in the secondary alphabets as does the sequenceQUEST......XZ, the former is cryptographically equivalent tothe latter sequence. For this reason the A J Q B K ..sequence is termed an equivalent primary component. If thereal or original primary component is a keyword mixedsequence, it is hidden or latent within the equivalentprimary sequence; it can also be made patent by the processof decimation of the equivalent primary component.

    Friedman in [FRE7] describes the process as follows: findthree letters in the equivalent primary component that are alikely unbroken sequence in the original primary component,and see if the interval between the first and second is thesame as that of the second and third. Try X, Y, Z in theequivalent primary component above. Note the sequence ..W OG X N H Z...; the distance or interval between W X Z is threeletters. Continuing the chain by adding letters threeintervals removed, the latent original primary component ismade patent.

    1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 WX Z Q U E S T I  O  N  A  B  L  Y  C  D  F  G  H  J  K  M24 25 26 P  R  V

    KEYWORD - MIXED SEQUENCE

    We can combine the previous steps into one operation.Starting with any pair of letters in the cipher component ofthe secondary alphabets, likely to be sequent in the keyword-mixed sequence, such as JK, the following chains of digraphsmay be produced. Thus JK plain stand over QU cipherrespectively, QU in the plain stand over BL in the cipher,respectively, etc. Connecting the pairs:

    JK>QU>BL>KM>UE>LY>MP>ES>YC>PR>ST>CD>RV>TI>DF>VW>IO>FG>WX>ON>GH>XZ>NA>HJ>ZQ>AB>JK.....We then unite by common letters:JK>KM>MP>PR>RV>VW>WX>XZ>ZQ>QU>UE>ES>ST>TI>IO>ON>NA>AB>BL>LY>YC>CD>DF>FG>GH>HJ>JK.....or:JKMPRVWXZ-QUESTIONABLY-CDFGH

    HALF CHAINS

    Only 12 /26 alphabets will yield a complete equivalentprimary component, as shown above. Even number of intervalsfor sliding the alphabets will yield half chains or 13 letterchains. Friedman [FRE7] describes several methods to combinethe half chains into fully equivalent primary components.

    FRIEDMAN'S OBSERVATIONS

    Friedman observed that in the case of a 26-element componentsliding against itself (both components proceeding in thesame direction), it is only the secondary alphabets resultingfrom odd-interval displacements of the primary componentswhich permit reconstructing a single 26-letter chain ofequivalents. This is true except for the 13th intervaldisplacement, which acts like an even number displacement, inthat no complete chain of equivalents can be established fromthe secondary alphabet. Friedman states the general rule as:any displacement interval which has a factor in common withthe number of letters in the primary sequence will yield asecondary alphabet from which no complete chain of 26equivalents can be derived for the construction of a completeequivalent primary component. Components sliding in oppositedirections act as a 13 interval displacement because of theirreciprocal nature.

    Friedman concluded that whether or not a complete equivalentprimary component is derivable by decimation from an originalprimary component (and if not, the lengths and numbers ofchains of letters, or incomplete components, that can beconstructed in attempts to derive such equivalent components)will depend upon the number of letters in the originalprimary component and the specific decimation intervalselected. [FRE7] Friedman constructed a table relating thenumber of characters in the original primary component,decimation interval and total number of complete sequencesthat can be formed. See Table 12-3.

                              TABLE 12-3          Number of Characters in Original Primary ComponentDecimation Interval    32  30  28  27  26  25  24  22  21  2018  16            ----------------------------------------------    2       16  15  14  27  13  25  12  11  21  10   9   8    3       32  10  28   9  26  25   8  22   7  20   6  16    4        8  15   7  27  13  25   6  11  21   5   9   4    5       32   6  28  27  26   5  24  22  21   4  18  16    6       16   5  14   9  13  25   4  11   7  10   3   8    7       32  30   4  27  26  25  24  22   3  20  18  16    8        4  15   7  27  13  25   3  11  21   5   9   2    9       32  10  28   3  26  25   8  22   7  20   2  16    10      16   3  14  27  13   5  12  11  21   2   9   8    11      32  30  28  27  26  25  24   2  21  20  18  16    12       8   5   7   9  13  25   2  11   7   5   3   4    13      32  30  28  27   2  25  24  22  21  20  18  16    14      16  15   2  27  13  25  12  11   3  10   9   8    15      32   2  28   9  26   5   8  22   7   4   6    16       2  15   7  27  13  25   3  11  21   5   9    17      32  30  28  27  26  25  24  22  21  20    18      16   5  14   3  13  25   4  11   7  10    19      32  30  28  27  26  25  24  22  21    20       8   3   7  27  13   5   6  11    21      32  10   4   9  26  25   8    22      16  15  14  27  13  25  12    23      32  30  28  27  26  25    24       4   5   7   9  13    25      32   6  28  27    26      16  15  14    27      32  10    28       8  15    29      32    30      16Total NumberOfSequences   14   6  10  16  10  18   6   8  10   6   4   6
    From Table 12-3, we see that in a 26-letter original primarycomponent, decimation interval 5 will yield a completeequivalent primary component of 26 letters, whereasdecimation intervals of 4 or 8 will yield 2 chains of 13each. In a 24-letter component, decimation interval 5 willalso yield a complete equivalent primary component of 24letters, but decimation interval 4 will yield 6 chains of 4letters each, and decimation interval 8 will yield 3 chainsof 8 letters each.

    It follows that in the case of an original primary componentin which the total number of characters is a prime number,all decimation intervals will yield complete equivalentprimary components. Table 12-3 omits the prime numbersequences from 16-32. [FRE7]

    SPECIAL SOLUTIONS FOR PERIODIC CIPHERS

    Special circumstances give rise atypical solutions ofperiodic ciphers. We shall look at four special cases:1) isologs, 2) 'stagger', 3) long latent repetition and 4)superimposition.

    ISOLOGS

    Recall that an Isolog is defined as the exact same plain textmessage enciphered by two different keys in the samecryptosystem. Lets use two monoalphabetic substitutionsystems to illustrate the point. Assume two messages areintercepted going from station A to B. B had called for aretransmit because of some error in transmission. We suspectthe messages are the same plaintext content and they bothhave the same length. We superimpose one message over theother:

    1. NXGRV MPUOF ZQVCP VWERX QDZVX WXZQE TBDSP VVXJK RFZWH2. EMLHJ FGVUB PRJNG JKWHM RAPJM KMPRW ZTAXG JJMCD HBPKYchaining from 1 to 2:  NE>EW>WK>KD>DA ......1. ZUWLU IYVZQ FXOAR2. PVKIV QOJPR BMUSHNext we initiate a chain of ciphertext equivalents (reducingthe common letter) from message 1 to message 2, yielding:                                           *1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 NE W K D A S X M  F  B  T  Z  P  G  L  I  Q  R  H  Y  O  U  *         *            *                 *              *24 25 26 V  J  C
    With some experimentation, we find the Key word QUESTIONABLYand the decimation interval of +5 Modulo 26. The complete 26letter chain was available for reconstruction, but this isnot a requirement.

    Why is it possible to reconstruct the primary component andsolve the above two messages without having any plain text atall? Since the plain text of both messages is the same, therelative displacement of the same primary components in thecase of message 1 differs from the relative displacement ofthe same primary components in message 2 by a FIXED interval.Therefore, the distance between N and E (1st two cipherletters of the two messages) on the primary component,regardless of what plaintext letter these two cipher lettersrepresent, is the same distance between E and W (18thletters), W and K (17th letters), and so forth. Thus thisfixed interval permits the establishing of a complete chainof letters separated by constant intervals and this chainbecomes an equivalent primary component.

    To solve, we take the frequency distributions of message 1and 2:

                                           E       S T I   O     1 1 1 2 2 3 1 1 1 1 1 1 1 1 2 3 4 4 1 1 3 7 4 6 1 61:   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                   E   S T   I     O     2 3 1 1 1 1 3 4 1 7 4 1 6 1 1 7 1 4 1 1 2 3 2 1 1 12:   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
    We set up two key word mixed alphabets and slide against eachother. With some trial and error we find:
                  NABLYCDFGHJKMPRVWXZQUESTIO              QUESTIONABLYCDFGHJKMPRVWXZ
    The plain text reads:
    Five squadrons must be in position by Hplus six zero two at Jackson Ridge.
    The same procedure is applied on two repeating key cipherssuspected of being Isologs:
    Message 1YHYEX  UBUKA  PVLLT  ABUVV  DYSAB  PCQTUNGKFA  ZEFIZ  BDJEZ  ALVID  TROQS  UHAFKMessage 2CGSLZ  QUBMN  CTYBV  HLQFT  FLRHL  MTAIQZWMDQ  NSDWN  LCBLQ  NETOC  VSNZR  BJNOQ
    The first step is to find the length of the period. Theusual method fails for lack of long repetitions and thedigraphs are not promising. We use the Principle ofSuperimposition to get a hold on the period for bothcryptograms.

    1 2 3 4 5 6 7 8 9101112131415161718192021222324252627282930Y H Y E X U B U K A P V L L T A B U V V D Y S A B P C Q T UC G S L Z Q U B M N C T Y B V H L Q F T F L R H L M T A I Q313233343536373839404142434445464748495051525354555657585960N G K F A Z E F I Z B D J E Z A L V I D T R O Q S U H A F KZ W M D Q N S D W N L C B L Q N E T O C V S N Z R B J N O Q
    We employ a subterfuge based upon the theoryof factoring. We search for cases of identicalsuperimposition. We have:
        4      44                               6  18    30    E  and E   are separated by 40 letters, U, U and U  which    L      L                                Q  Q     Q
    are separated by 12 letters. We factor these intervals as ifthey were ordinary repetitions. The most frequent factorshould correspond to the period. We are dealing withIsologs. The plain text is the same in both messages, so theprinciple of identity of superimposition can only be theresult of identity of encipherments by identical cipheralphabets. The same relative position in the keying cyclehas been reached in both cases of the identity. The distancebetween identical superimpositions must be equal to or amultiple of the length of the period. The following is thecomplete set of superimposed pairs:
         Repetition         Interval          Factors
         EL - EL             40           2,4,5,8,10,20     UQ - UQ -UQ         12           2,3,4,6     UB - UB             48           2,3,4,6,,8,12,24     KM - KM             24           2,3,4,6,12     AN -AN -AN          36/12        2,3,4,6;9,12,18     VT -VT -VT          8/28         2,4; 2,4,7,14     TV - TV             36           2,3,4,6,9,12,18     AH - AH             8            2,4     BL -BL -BL          8/16         2,4,;8     SR - SR             32           2,4,8,16     FD - FD             4            2     ZN - ZN             4            2     DC - DC             8            2, 4
    Only the factors 2 and 4 are common. We discard 2 asimprobable. We break up the message into groups of four.
         1234 1234 1234 1234 1234 1234 1234 12341.   YHYE XUBU KAPV LLTA BUVV DYSA BPCQ TUNG 2.   CGSL ZQUBMNCT YBVH LQFT FLRH LMTA IQZW     *    *    *    *     1234 1234 1234 1234 1234 1234 12341.   KFAZ EFIZ BDJE ZALV IDTR OQSU HAFK2.   MDQN SDWN LCBL QNET OCVS NZRB JNOQWe develop a decipherment Tableaux: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------------------------------------------------------1   L   F S     J O   M Y     N         I       Z C Q2 N     C   D   G       B       M Z       Q       L3 Q U T     O     W B   E   Z   C     R V   F     S4 H       L   W       Q           A S     B T       N------------------------------------------------------
    Using the meyhods previously described, we build up theequivalent primary component and combine our digrams.
    BL, DF, ES, HJ, IO, KM, LY, ON,TI, XZ, YC, ZQ.BLYC .DF    TION    XZQ(U) [ES]TION(A)BLY CDF (G) HJKM(P) (R) (V) XZ
    It is not a long jump to a key word QUESTIONABLY and theequivalent primary component:
    Q U E S T I O N A B L Y C D F G H J K M P R V W X Z
    The fact that the original primary component was exposed waspure chance, it could have been an equivalent primarysequence alphabet.

    From here we apply the completion of the plain-componentsequence using the high frequency letter assortments.For the first message:

     Gen Alphabet 1    Alphabet 2    Alphabet 3    Alphabet 41   YXKLBDBTKE   1HUALUYPUFF   5YBPTVSCNAI    EUVAVAQGZZ2  2CZMYLFLIMS   4JEBYECREGG   5CLRIWTDABO    SEWBWBUHQQ3  2DQPCYGYOPT   3KSLCSDVSHH   3DYVOXIFBLN    TSXLXLEJUU4  4FURDCHCNRI    MTYDTFWTJJ   3FCWNZOGLYA    ITZYZYSKEE5  3GEVFDJDAVO    PICFIGXIKK    GDXAQNHYCB    OIQCQCTMSS6  2HSWGFKFBWN   4RODGOHZOMM    HFZBUAJCDL   5NOUDUDIPTT7   JTXHGMGLXA    VNFHNJQNPP    JGQLEBKDFY   8ANEFEFORII*8   KIZJHPHYZB    WAGJAKUARR   1KHUYSLMFGC   6BASGSGNVOO9   MOQKJRJCQL    XBHKBMEBVV   2MJECTYPGHD   5LBTHTHAWNN10  PNUMKVKDUY    ZLJMLPSLWW    PKSDICRHJF    YLIJIJBXAA11 4RAEPMWMFEC    QYKPYRTYXX    RMTFODVJKG    CYOKOKLZBB12 3VBSRPXPGSD    UCMRCVICZZ   2VPIGNFWKMH   2DCNMNMYQLL13 4WLTVRZRHTF    EDPVDWODQQ    WROHAGXMPJ   2FDAPAPCUYY14  XYIWVQVJIG   3SFRWFXNFUU    XVNJBHZPRK   3GFBRBRDECC15  ZCOXWUWKOH    TGVXGZAGEE    ZWAKLJQRVM   1HGLVLVFSDD16  QDNZXEXMNJ    IHWZHQBHSS    QXBMYKUVWP   1JHYWYWGTFF17  UFAQZSZPAK    OJXQJULJTT    UZLPCMEWXR    KJCXCXHIGG18  EGBUQTQRBM    NKZUKEYKII    EQYRDPSXZV    MKDZDZJOHH19 3SHLEUIUVLP   5AMQEMSCMOO    SUCVFRTZQW    PMFQFQKNJJ20 6TJYSEOEWYR?  4BPUSPTDPNN    TEDWGVIQUX    RPGUGUMAKK21  IKCTSNSXCV   8LRETRIFRAA*   ISFXHWOUEZ   3VRHEHEPBMM22 5OMDITATZDW?  3YVSIVOGVBB    OTGZJXNESQ    WVJSJSRLPP23  NPFOIBIQFX   3CWTOWNHWLL    NIHQKZASTU    XWKTKTVYRR24 5ARGNOLOUGZ?   DXINXAJXYY    AOJUMQBTIE    ZXMIMIWCVV25 4BVHANYNEHQ    FZOAZBKZCC   5BNKEPULIOS    QZPOPOXDWW26  LWJBACASJU    GQNBQLMQDD   7LAMSREYONT*   UQRNRNZFXX
    We choose generatrices 20/22/24; 21; 26; 7 because of thehighest two category scores. it is not much of a jump tofind Alphabet 1 generatrix as alphabet 24:
                     1 2 3 4                 A L L A                 R R A N                 G E M E                 N T S F                 O R R E                 L I E F                 O F Y O                 U R O R                 G A N I                 Z A T I
    From a Vigenere Square (Figure 12-1) based on the keywordQUESTIONABLY, we find the key words SOUP for message 1 andTIME for message 2.
    S O U P  S O U P  S O U P  S O U P  S O U P  S O U P----------------------------------------------------Y H Y E  X U B U  K A P L  L L T A  B U V V  D Y S AA L L A  R R A N  G E M E  N T S F  O R R E  L I E FB P C Q  T U N G  K F A Z  E F I Z  B D J E  Z A L VO F Y O  U R O R  G A N I  Z A T I  O N H A  V E B EI D T R   O Q S U   H A F KE N S U   S P E N   D E D XT I M E  T I M E  T I M E  T I M E  T I M E  T I M E____________________________________________________C G S L  Z Q U B  M N C T  Y B V H   L Q F T  F L R HA L L A  R R A N  G E M E  N T S F  O R R E  L I E FL M T A  I Q Z W  M D Q N  S D W N  L C B L  Q N E TO F Y O  U R O R  G A N I  Z A T I  O N H A  V E B EO C V S   N Z R B  J N O QE N S U   S P E N   D E D X                  Figure 12-1Q U E S T I O N A B L Y C D F G H J K M P R V W X ZU E S T I O N A B L Y C D F G H J K M P R V W X Z QE S T I O N A B L Y C D F G H J K M P R V W X Z Q US T I O N A B L Y C D F G H J K M P R V W X Z Q U ET I O N A B L Y C D F G H J K M P R V W X Z Q U E SI O N A B L Y C D F G H J K M P R V W X Z Q U E S TO N A B L Y C D F G H J K M P R V W X Z Q U E S T IN A B L Y C D F G H J K M P R V W X Z Q U E S T I OA B L Y C D F G H J K M P R V W X Z Q U E S T I O NB L Y C D F G H J K M P R V W X Z Q U E S T I O N AL Y C D F G H J K M P R V W X Z Q U E S T I O N A BY C D F G H J K M P R V W X Z Q U E S T I O N A B LC D F G H J K M P R V W X Z Q U E S T I O N A B L YD F G H J K M P R V W X Z Q U E S T I O N A B L Y CF G H J K M P R V W X Z Q U E S T I O N A B L Y C DG H J K M P R V W X Z Q U E S T I O N A B L Y C D FH J K M P R V W X Z Q U E S T I O N A B L Y C D F GJ K M P R V W X Z Q U E S T I O N A B L Y C D F G HK M P R V W X Z Q U E S T I O N A B L Y C D F G H JM P R V W X Z Q U E S T I O N A B L Y C D F G H J KP R V W X Z Q U E S T I O N A B L Y C D F G H J K MR V W X Z Q U E S T I O N A B L Y C D F G H J K M PV W X Z Q U E S T I O N A B L Y C D F G H J K M P RW X Z Q U E S T I O N A B L Y C D F G H J K M P R VX Z Q U E S T I O N A B L Y C D F G H J K M P R V WZ Q U E S T I O N A B L Y C D F G H J K M P R V W X

    SOLUTION OF ISOLOGS INVOLVING THE SAME SET OF PRIMARYCOMPONENTS BUT WITH KEY WORDS OF DIFFERENT LENGTHS

    The example previous had two keywords the same lengths.The Method of Superimposition works with Keywords ofdifferent lengths. Friedman works an interesting example:

                    Message 1VMYZG  EAUNT  PKFAY  JIZMB  UMYKB  VFIVVSEOAF  SKXKR  YWCAC  ZORDO  ZRDEF  BLKFESMKSF  AFEKV  QURCM  YZVOX  VABTA  YYUOAYTDKF  ENWNT  DBQKU  LAJLZ  IOUMA  BOAFSKXQPU  YMJPW  QTDBT  OSIYS  MIYKU  ROGMWCTMZZ  VMVAJ                Message 2ZGANW  IOMOA  CODHA  CLRLP  MOQOJ  EMOQUDHXBY  UQMGA  UVGLQ  DBSPU  OABIR  PWXYMOGGFT  MRHVF  GWKNI  VAUPF  ABRVI  LAQEMZDJXY  MEDDY  BOSVM  PNLGX  XDYDO  PXBYUQMNKY  FLUYY  GVPVR  DNCZE  KJQOR  WJXRVGDKDS  XCEEC.
    Both messages permit factoring at periods of 4 and 6 letters,respectively. Superimposing the two messages and marking theposition of each letter in the corresponding period, we have:
              12341  23412  34123  41234  12341  23412No. 1     VMYZG  EAUNT  PKFAY  JIZMB  UMYKB  VFIVVNo. 2     ZGANW  IOMOA  CODHA  CLRLP  MOQOJ  EMOQU          12345  61234  56123  45612  34561  23456          34123  41234  12341  23412  34123  41234No. 1     SEOAF  SKXKR  YWCAC  ZORDO  ZRDEF  BLKFENo. 2     DHXBY  UQMGA  UVGLQ  DBSPU  OABIR  PWXYM          12345  61234  56123  45612  34561  23456          12341  23412  34123  41234  12341  23412No. 1     SMKSF  AFEKV  QURCM  YZVOX  VABTA  YYUOANo. 2     OGGFT  MRHVF  GWKNI  VAUPF  ABRVI  LAQEM          12345  61234  56123  45612  34561  23456          34123  41234  12341  23412  34123  41234No. 1     YTDKF  ENWNT  DBQKU  LAJLZ  IOUMA  BOAFSNo. 2     ZDJXY  MEDDY  BOSVM  PNLGX  XDYDO  PXBYU          12345  61234  56123  45612  34561  23456          12341  23412  34123  41234  12341  23412No. 1     KXQPU  YMJPW  QTDBT  OSIYS  MIYKU  ROGMWNo. 2     QMNKY  FLUYY  GVPVR  DNCZE  KJQOR  WJXRV          12345  61234  56123  45612  34561  23456          34123  41234No. 1     CTMZZ  VMVAJ.No. 2     GDKDS  XCEEC.          12345  61234
    What is neat about this superimposition is that we canestablish secondary alphabets by distributing the lettersfrom the 12 different superimposed pairs of numbers.The 1 - 1 superimposition is placed in the tableau at the0 - 1 row, column in the tableaux.

    0     1 2 3 4 5 6 7 8 91011121314151617181920212223242526      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      ---------------------------------------------------1-1   I J   P   D         Q G C E       K O   R Z2-2   H V N                   G   U     W       E D M L X3-3   E         M     X   G   I D J   N     R         A O4-4               X   O C         D K   A F Y Q       V N1-5         B   T W   L       R   E     M N   Y       U A2-6   M O     I       C       D               U V     F R3-1   O   G     R             L   P   S   D           Z4-2   L P     H         U V               E D M      F1-3       Q J             V W K O X Y         M A2-4   B               J   X P O             A   F Y     D3-5   N R       Y                 B C G               Q S4-6           M         L O             S U V W X      ---------------------------------------------------We construct the complete equivalent primary component:   1 2 3 4 5 6 7 8 91011121314151617181920212223242526   I T K N P Z H M W B Q E U L F C S J A X R G D V O Y
    Ok. We have the cipher component. Is it normal? reversed?Mixed? Same questions for the plain component sequence.We assume that the primary plain component is normal directsequence. We attempt to solve and fail. Normal reverse willalso fail. We assume a K3 situation, i.e. the plain andcipher components are identical. Again the test fails. Weassume that the plain is in reverse mode. Nope. So we have aK4 situation, both primary components are different mixedsequences.

    Message 1 transcribed into periods of four letters.

                    Message 1VMYZ GEAU NTPK FAYJ IZMB UMYK BVFI VVSEOAFS KXKR YWCA CZOR DOZR DEFB LKFE SMKSFAFE KVQU RCMY ZVOX VABT AYYU OAYT DKFENWNT DBQK ULAJ LZIO UMAB OAFS KXQP UYMJPWQT DBTO SIYS MIYK UROG MWCT MZZV MVAJ
    The Uniliteral frequency distributions for the four secondaryalphabets are shown in 1A-4A. We have the reconstructedcipher alphabet, 1B-4b shows the sequences rearranged.
         1 1 1 5   2 1   1   3 2 4 2 3 1   1 2   5 3     1 11A   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 2 1   2       2   2 1 4   1     1   1   5 4 2 2 42A   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     4 1 2     7     1   2   3 1 3 1 4   1 1         7 23A   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     1 3     4       1 4 4       2 1   3 4 5 3 1   1 1 14A   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    1   3 2 1 1   4   1     5 2 2 1 2   1   1 1 5 3 3 11B  I T K N P Z H M W B Q E U L F C S J A X R G D V O Y    2 1 2     4   4 3 2   2   1   1     6 2 1     5 1 22B  I T K N P Z H M W B Q E U L F C S J A X R G D V O Y    1 1 2 1 1 2   3   1 4       7 2 1   4           3 73B  I T K N P Z H M W B Q E U L F C S J A X R G D V O Y    1 5 4   1 1       3   4 3       4 4 1 1 3 1   1 2 14B  I T K N P Z H M W B Q E U L F C S J A X R G D V O Y
    We now shift 1B-4B for superimposition and combine thedistributions. The latter distributions may be combined soas to yield a single monoalphabetic distribution for theentire message. In other words, the polyalphabetic messagecan be converted into monoalphabetic terms, and therebysimplifying the situation considerably.
        1   3 2 1 1   4   1     5 2 2 1 2   1   1 1 5 3 3 11B  I T K N P Z H M W B Q E U L F C S J A X R G D V O Y    2   1   1     6 2 1     5 1 2 2 1 2     4     3 22B  E U L F C S J A X R G D V O Y I T K N P Z H M W B Q 2 1 1    2   3   1 4       7 2 1   4           3 73B  K N P Z H M W B Q E U L F C S J A X R G D V O Y I T    1 1       3   4 3       4 4 1 1 3 1   1 2 1 1 5 44B  P Z H M W B Q E U L F C S J A X R G D V O Y I T K N        6 2 5 4 2 7  15 9 2    21 9 6 410 3 1 1 7 2 918 9 11B-4B   I T K N P Z H M W B Q E U L F C S J A X R G D V O YcombinedH M       L   R S       O       A       I Y N E TPlainEquiv's
    I have converted 2B-4B into terms of 1B. The 2 E's of 2B addto 1B I. The two K's of alphabet 3 becomes I's and the Nbecomes a T, and so forth. We solve the monoalphabeticcipher.
           12341  23412  34123  41234  12341  23412       ENEMY  HASCA  PTURE  DHILL  ONETW  OONEO       VDVTG  ISWNZ  KOFMV  LIRZZ  UDVOB  UUDVU       URTRO  OPSHA  VEDUG  INAND  CANHO  LDFOR       FMOMU  UKWIS  YVLFC  RDSDL  NSDIU  ZLJUM       ANHOU  RORPO  SSIBL  YLONG  ERREQ  UESTR       SDIUF  MUMKU  WWRPZ  GZUDC  VMMVA  FVWOM       EINFO  RCEME  NTSTO  PADDI  TIONA  LTROO       VVDJU  MNVTV  DOWOU  KSLLR  ORDUS  ZOMUU       PSSHO  ULDBE  SENTV  IAGEO  RGETO  WNFRE       KWWIU  FZLPV  WVDOY  RSCVU  MCVOU  BDJMV       DERIC  KROAD.       LVMRN  XMUSL.
    Having the plain text, the derivation of the plain orequivalent plain component is straightforward. We may basethe reconstruction upon any of the secondary alphabets, sincethe plaintext - ciphertext relationship is known directly,and the primary cipher component is at hand. So:
       1 2 3 4 5 6 7 8 9 1011121314151617181920212223242526   H M P C B L . R S  W . . O D U G A F Q K I Y N E T Vwith Key words of STAR and OCEANS for messages 1 and 2.

    NECESSARY AND SUFFICIENT CONDITIONS FOR SUPERIMPOSITION ANDCONVERSION TO MONOALPHABETIC TERMS

    This example shows the power of the method of superimpositionand conversion of a polyalphabetic cipher to monoalphabeticterms. This conversion is possible because the sequence ofletters forming the cipher component has been reconstructedand was known, and the uniliteral distributions for therespective secondary cipher alphabets could theoretically beshifted to correct superimpositions for monoalphabeticity.The data was sufficient to give proper indications foralignment of the alphabets and relative displacements. Thechi test could also have been brought to bear to matchcolumns. The above constitutes the necessary and sufficientconditions to convert theory to actuality.

    SOLUTION OF ISOLOGS INVOLVING DIFFERENT PAIRS OF UNKNOWNPRIMARY COMPONENTS

    The principle of superimposition continues to work for useven when the primary components are different, and therepeating keys are of different lengths.

    There are two general attacks. The first is a slightmodification of the procedures previously discussed. We firstfactor the messages, then superimpose the messages on a widthof the least common multiple, then create a reconstructionmatrix based on the cipher values. We must limit ourobservations to within the matrix, because the given messagesare different and therefore the indirect symmetry does notextend to the 0 or assumed plain line. The wrinkle in thefabric is we must restrict our observations to a homogeneousset of lines, like 1-1,1-2,1-3,1-4 etc. From this data, wereduce the reconstruction matrix to a smaller set and solvefor the equivalent primary component. It is possible toinvert the matrix so that values for the second message willyield its equivalent primary component.

    ARBITRARY REDUCTION METHOD

    It is not necessary to recognize the plain text to solve aproblem involving Isologs. The next cryptanalytic attack isapplicable for many types of ciphers. The procedure exposeslatent letter relationships and reduces the imposed chaos ofthe cryptogram. Given:

                              Message 1           BWXPS  OBYII  UYHLF  KFSOP  VGEYW  PBVXO           UGJPB  WDXUG  HSWDH  KHKHC  UAYKP  NFSPD           OBBYB  INKFL  WABOX  PJXUV  WKFXR  WXYWS           SDYZQ  ZHETA  JXXZW  XJROS  PDEEW  OJONK           GIRXR  WUYDK  NTJWR  EVBUR  DLISJ  BLCKK           FODEV  DYZQZ  SHCTW  DIEXZ
    Factoring gives us periods of 4 and 5 for messages 1 and 2,respectively. We write out the messages on a width of theleast common multiple of 20.

                             Message 2           JNLEJ  HWUAH  JHUIV  YNCHC  HLPKD  EWZJJ           JNAHB  HZBIM  TUBQE  FJAKM  JVBEF  XNCTL           FAAKV  KIABG  CVFNY  FWBIQ  GERSA  TZUSD           SXBUD  SHAWA  YXLJD  CQLED  HXGZL  ZWHNB           VTJSA  TSUUC  MIAKK  JEMIY  DSKGB  VTJYC           XYLZE  CXLSU  MVMND  ONFJY           12341  23412  34123  41234        20           BWXPS  OBYII  UYHLF  KFSOP           JNLEJ  HWUAH  JHUIV  YNCHC           12345  12345  12345  12345           A             A  A           12341  23412  34123  41234        40           VGEYW  PBVXO  UGJPB  WDXUG           HLPKD  EWZJJ  JNAHB  HZBIM           12345  12345  12345  12345                         A         A           12341  23412  34123  41234        60           HSWDH  KHKHC  UAYKP  NFSPD           TUBQE  FJAKM  JVBEF  XNCTL           12345  12345  12345  12345                         A           12341  23412  34123  41234        80           OBBYB  INKFL  WABOX  PJXUV           FAAKG  KIABG  CVFNY  FWBIQ           12345  12345  12345  12345               A      A    A       A           12341  23412  34123  41234       100           WQFXR  WXYWS  SDYZQ  ZHETA           GERSA  TZUSD  SXBUD  SHAWA           12345  12345  12345  12345           12341  23412  34123  41234       120           JXXZW  XJROS  PDEEW  OJONK           YXLJD  CQLED  HXGZL  ZWHNB           12345  12345  12345  12345           12341  23412  34123  41234       140           GIRXR  WUYDK  NTJWR  EVBUR           VTJSA  TSUUC  MIAKK  JEMIY           12345  12345  12345  12345                   A            A  A           12341  23412  34123  41234       160           DLISJ  BLCKK  FODEV  DYZQZ           DSKGB  VTJYC  XYLZE  CXLSU           12345  12345  12345  12345            A           12341  23412                     170           SHCTW  DIEXZ           MVMND  ONFJY           12345  12345                    A
    We arbitrarily assign the value of A(plain) as the firstletter of the plain text. Since in message 1, B(cipher)=A(plain), then every B(cipher) in alphabet 1 must equalA(plain); these values are entered in the table above. Alsothe 65th and 73rd letter of message 1 are A(plain), thisestablishes that in message 2, G(cipher) in alphabet 5 andF(cipher) in alphabet 3 are also A(plain); we enter thesevalues. Similarly, every J(cipher) in alphabet 1 of message2 equals A(plain). We continue the process and recover allthe A(plains) of the pseudo-plain text with the resultingworksheet shown above.

    We arbitrarily assign the value of B(plain) to the V(cipher)at the 21st position of message 1. The other V(cipher) ofmessage number 1 establishes the E(cipher) of message 2 alsoas a B(plain). This procedure of arbitrary assignments iscontinued until all the cipher letters of alphabet 1 ofmessage 1, are placed. we are able to reduce most of thetext to monoalphabetic terms. The worksheet is as follows:

               12341  23412  34123  41234        20           BWXPS  OBYII  UYHLF  KFSOP           JNLEJ  HWUAH  JHUIV  YNCHC           12345  12345  12345  12345           ACHDIIFCK     ACCA   FME D           12341  23412  34123  41234        40           VGEYW  PBVXO  UGJPB  WDXUG           HLPKD  EWZJJ  JNAHB  HZBIM           12345  12345  12345  12345           B  CE   F LI  AMF F  BHOAM           12341  23412  34123  41234        60           HSWDH  KHKHC  UAYKP  NFSPD           TUBQE  FJAKM  JVBEF  XNCTL           12345  12345  12345  12345           CEOOC  D FCM  AJODB   MEBO           12341  23412  34123  41234        80           OBBYB  INKFL  WABOX  PJXUV           FAAKG  KIABG  CVFNY  FWBIQ           12345  12345  12345  12345           DGFCA   IFMA  OJAIH  DFOA           12341  23412  34123  41234       100           WQFXR  WXYWS  SDYZQ  ZHETA           GERSA  TZUSD  SXBUD  SHAWA           12345  12345  12345  12345           EB EJ  CHCEE  LOOHE  LCF J           12341  23412  34123  41234       120           JXXZW  XJROS  PDEEW  OJONK           YXLJD  CQLED  HXGZL  ZWHNB           12345  12345  12345  12345           FOHLE  O HDE  BOPFO   FIIF           12341  23412  34123  41234       140           GIRXR  WUYDK  NTJWR  EVBUR           VTJSA  TSUUC  MIAKK  JEMIY           12345  12345  12345  12345           G  EJ  CACHD  IIFC   ABGAH           12341  23412  34123  41234       160           DLISJ  BLCKK  FODEV  DYZQZ           DSKGB  VTJYC  XYLZE  CXLSU           12345  12345  12345  12345           HAM F  G  ND    HFC  OOHEL           12341  23412                     170           SHCTW  DIEXZ           MVMND  ONFJY           12345  12345           IJGIE   MALH
    The above table is about 85% reduced and note the idiomorphicrepetition ACHDIIFC representing Artillery becomes patent inthe reduction process. This is rather exciting. From nopatent clues to reduction and latent clues exposed. Clever.

    The solution is continued by setting up sequence recon-struction matrices for both messages. The 0 line representsthe pseudo-plain text and the values inside the matrix beingcipher text.

    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------------------------------------------------------1  B V H O W J G D S R I X F K Y E2  L Q W K S E B Z O H     C   X3  U P V   Q B C X N     S I   W4  E W Y P X K   R T A   Z G   D-------------------------------------------------------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------------------------------------------------------1  J H T F G Y V D M     S     C2  S E H   U W A Z I V     N   X3  F   U   C A M L H       K   B G4  I T K E S Z   U N   A J B Y Q5  G F E C D B   Y J A   U M   L------------------------------------------------------
    From the above we chain out the equivalent primary componentsused for each message. Having reconstructed the ciphercomponent for each message, the alphabets are aligned,combined and reduced to monoalphabetic terms. After solutionof these messages, we find message 1 is a case of directsymmetry with the cipher component based on the keywordHYDRAULIC, and message 2 is a case of indirect symmetry withboth components being keyword-mixed sequences based on ourfavorite keyword QUESTIONABLY. Friedman points out that thekeywords are prime to each other (9 vs 11). Primality is nota necessary condition for solution based on this procedure.[FRE7]

    The method of Arbitrary Reduction is very powerful and worksin other ares besides solving periodic polyalphabeticciphers. It represents a workable approach where thecryptosystem involves nonrelated, random-mixed secondaryalphabets among which no symmetry of any sort exists!

    SOLUTION BASED ON INDIRECT SYMMETRY OF A "STAGGER"

    Given two messages with group counts nearly identical and twoisologous initial fragments which are identical except by oneletter (called a 'stagger') we can solve the isologousportions of the messages and recover the primary ciphercomponent by the process of indirect symmetry. Transmissiongarble usually creates stagger messages. Machine ciphersystems sometimes produce these when a word separator isadded. Staggers may be progressively larger as further wordseparators are omitted or added.

    Given:

                        Message A                     *                *ZFWAY  ITBVX  XWZQV  PEBGS  GGFIZ  TUAMFRFEQX  PEPPO  PCNBP  QPOTX  VNAIH  HVRXCNHVGM  FRFSI  ESQMV    *                    Message B                     *                 *ZFWAY  ITBVX  XWZQV  PDRKF  USVAG  XLJKCNDVPR  OWBRH  YFJMS  HRFVS  BAHWG  ZFAJOJMFAV  CNDVD  ORZPH  A       *
    We note that both messages have the same 16 letter beginningsand that message B is 1 letter longer than message A. Notethat the tetragraphs MFRF (29) and (65) are spaced 1 lessletter than CNDV at (30) and (66). The D in position 17 ofmessage 2 is the extra letter.

    Starting from the E in position 17 of message 1, wesuperimpose message one over message 2 starting at the R inposition 18. [We use a period of 6 because the tetragraphdelta equals 36 which factors into 3,4,6 and 9; 6 isconfirmed via the message.]

    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-------------------------------------------------------1-2         B   F Z           M     P D   S           X2-3   S       V   F         H     R     U L       B3-4         P   S                 H     D   J A4-5 K           V   O         H Y   R J5-6 W       R A             C     F               O6-1   K J     N     G           V W     Z-------------------------------------------------------
    It is fairly easy to align properly the cipher componentsafter the primary cipher component or its equivalent havebeen recovered, thereby expediting the reduction of thecipher into monoalphabetic terms:
    Note that B(cipher) of: alphabet 2 is under E(cipher) of alphabet 1;          V(cipher) of: alphabet 3 is under F(cipher) of alphabet 2;          P(cipher) of: alphabet 4 is under E(cipher) of alphabet 1.
    From this pointon solution follows the normal path of reconstruction,keyword recovery and combination of alphabets, reduction tomonoalphabetic terms and solution by frequency analysis.

    LONG LATENT REPETITIONS

    The stagger procedure applies to a periodic cryptogram whichcontains a long passage repeated in its plain text, thesecond occurrence occurring at a point in the keying cycledifferent from the first occurrence. If the passage is longenough, the equivalencies from the two correspondingsequences may be chained together to yield an equivalentprimary component. In effect, we by-pass the solution byfrequency analysis or making assumptions in the plain text ofa polygraphic cipher.

    FINAL REMARKS REGARDING SOLUTION BY SUPERIMPOSITION

    In solving an ordinary repeating-key cipher the first step,ascertaining the length of the period, is a relatively minorconsideration. It paves the way for the second step, whichconsists of allocating the letters of the cryptogram intoindividual monoalphabetic distributions. The third step is tosolve these distributions. The text is transcribed into itsperiods and written out in successive lines corresponding tothe length of the period. The columns of letters as a seriesbelong to the same monoalphabet.

    We also can see the letters as transcribed into superimposedperiods; in such a case the letters in each column haveundergone the same kind of treatment by the same elements(plain and cipher components of the cipher alphabet.)

    If we have a case of a very long repeating key and a shortmessage ( few cycles in the text) we have a difficultproblem. But supposing there were several short cryptogramsenciphered by the same key, each message beginning atidentical starting points in the key. We can superimposethese messages "in flush depth" or "head on" and know that 1)the letters in the columns belong to the same individualalphabets, 2) and that if there are enough messages (about25-30 in English), then the frequency distributionsapplicable to the successive columns of text can be solved -without knowing the length of the key. Any difficulties thatmay have arisen because we were not able to factor theproblem correctly are circumvented. The second step of thenormal solution to the problem is by-passed. The assumptionof probable initial words of messages and stereotypedbeginnings is a powerful method of attack in such situations.Since the superimposed texts in these cases comprise only thebeginnings of messages, assumptions of probable words aremore easily made than when words are sought in the interiorof the messages. Some common introductory words are REQUEST,REFER, ENEMY, WHAT, WHEN, and SEND. High frequency initialdigraphs will manifest themselves in the first two columns ofthe superimposed diagram. The high frequency RE diagrammanifests itself in such words as REQUEST, REQUIRE,REFERENCE, REFERRING, REQUISITIONS, REPEAT, RECOMMEND,REPORT, RECONNAISSANCE, REINFORCEMENTS and perhaps REGIMENT.(I assume the military text here.)

    This same superimposition principle applies even if themessages start at different initial points, providing themessages can be correctly superimposed, so that the letterswhich fall in one column really belong to one cipheralphabet. The superimposed messages are said to be "indepth." The chi test may be used to advantage in finding andcombining columns of the superimposed diagram which wereenciphered by identical keys, thus assisting in the analysisof frequencies of larger samples than were available beforethe amalgamation. [FRE7]

    CONCLUSION

    In summary, we have seen that the chaining process betweencipher texts applies to the latent characteristics of thecipher components, regardless of the identity of the plaincomponents and regardless whether direct or indirect symmetryis involved in the cryptosystems. The principle of super-imposition ranks as one of the most important principles ofcryptanalysis. A pretty impressive tool.

    LECTURE 11 SOLUTIONS

    Thanks to BOZOL for the quick response and correct too!11.1 Vigenere.  Key= SLEEP. "Any reputable physician will     agree..11.2 Beaufort.  Key = SILENCE. "Although every one may not     subscribe to ..11.3 Variant.  Key = IMPSHGXW (HINSNOTI).  Because of the     many pressures...   [the correct key is SOLITUDE]11.4 GRONSFELD. 6-3-8-4-0. "Too much discussion, especially..11.5 BEAUFORT.  Key = OCCUPATION.  "Almost every man has a     job, many find..     BOZOL reports that the tip did not help him and that the     first pass at the key was ORCUPATMON which he mystically     came up with organization.

    LECTURE 12 PROBLEMS

    12.1 Nihilist Substitution74 46 66 44 79 47 45 37 58 66 37 60 25 54 33 69 78 35 68 2747 36 28 88 36 60 33 48 43 29 87 35 49 57 76 37 37 88 36 6033 77 74 50 86 55 47 27 76 45 40 55 56 58 66 78 57 30 94 5838 26 55 57 59 88 56 79 46 46 66 60 58 55 48 56.  (DGGLWLRQ,ends WXEOIW)12.2 Nihilist Substitution38 76 54 76 64 76 76 54 74 55 35 76 77 76 47 58 76 85 74 4465 88 63 74 47 36 95 74 63 44 37 58 57 96 65 36 66 85 74 6355 79 53 67 57 56 58 64 67 67 56 67 57 74 55 55 57 86 03 4346 67 73 96 67 39. (ETARVQITCO, ends HSMX)12.3 PORTA       QLAMU  CHQGO  FTESV  XKEWC  GMXPHUCLUS  WSGXT  EVURH  TMTSU  TKVSQ  GCQCWLHMDX  NUFUE  EFXRF  XPHUN  RGPKC  OXULBBBCUS  IBBHW.  (HAVE)12.4 PORTA       XFXYW  ZJICZ  IBUZN  HJXEA  ACWBEJOOCZ  UPXFQ  BXHFI  CGMAZ  KVQEG  BBCAFKLLXF  BVOUN  TSAYZ  KKXLR  CWAJC  LVVVIXNBFQ  JVWBW  BSWEY  VUNGX  ODFRZ  PTEWOPJQNH  WZPNA  YRCLV  YYWCQ  ULOJB  VK.  (GSRWXERX)12.5 PORTAX       UXCUD  ZMVBA  FWWPV  DIKDO  JISMAWRBBA  YLOYX  AKUXR  JGDCJ  MYAPV  RJWJADMUKL  KLUAM  KAOEN  YBFCC  IQGFK  QZAA. (PQXKEG)12.6 PORTAX       WWQPE  JBDTM  TMNWH  CTJSW  WKIACBJKWL  YHBYN  OAKRZ  PDYZM  DIVGB  QKNJPRNSRU  FXWMU  TKMJS  KDNLW  WFHKR  JSCVFHTJIS  JD.  (UHDOLCH)12.7 GROMARK       HPMZU  IBQHI  SDHHH  JKUNC  OYJSC       24106RBLOF  REXTG  EXAZA  ILAXX  XHFNH  CDUYQYUOMQ  NVOIN  XYMBR  WAHNT  FGPFB  DOOMACWHDH  JXTTX  CJIUR  PVMZR  EILDZ  QJJTTILNNP  TREVL  BQLL. ( tip: UCAUKYKUJK; ends tivpw.)

    REFERENCES / RESOURCES

    [updated 30 May 1996]

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CRYPTOGRAM,"The Graphic Position Chart (On       Aristocrats)," JF59, The Cryptogram, American       Cryptogram Association, 1959.[GREU] Greulich, Helmut, "Spion in der Streichholzschachtel:       Raffinierte Methoden der Abhortechnik, Gutersloh:       Bertelsmann, 1969.[GRI1] ASAP,"An Aid For Grille Ciphers," SO93, The       Cryptogram, American Cryptogram Association, 1993.[GRI2] DUN SCOTUS,"Binary Number Grille," JA60, The       Cryptogram, American Cryptogram Association, 1960.[GRI3] S-TUCK,"Grille Solved By the Tableaux Method," DJ42,       The Cryptogram, American Cryptogram Association, 1942.[GRI4] The SQUIRE,"More About Grilles," ON40,DJ40, The       Cryptogram, American Cryptogram Association, 1940,       1940.[GRI5] OMAR,"Rotating Grille Cipher," FM41, The Cryptogram,       American Cryptogram Association, 1941.[GRI6] S-TUCK,"Solving The Grille. A New Tableaux Method,"       FM44, The Cryptogram, American Cryptogram Association,       1944.[GRI7] LABRONICUS,"Solving The Turning Grille," JF88, The       Cryptogram, American Cryptogram Association, 1988.[GRI8] BERYL,"The Turning Grille," ND92, The Cryptogram,       American Cryptogram Association, 1992.[GRI9] SHERLAC and S-TUCKP,"Triangular Grilles," ON45, The       Cryptogram, American Cryptogram Association, 1945.[GRIA] SHERLAC,"Turning Grille," ON49, The Cryptogram,       American Cryptogram Association, 1949.[GRIB] DUN SCOTUS,"Turning (by the numbers)," SO61, The       Cryptogram, American Cryptogram Association, 1961.[GRIC] LEDGE,"Turning Grille (Novice Notes)," JA77, The       Cryptogram, American Cryptogram Association, 1977.[GRO1] DENDAI, DICK," Analysis of Gromark Special,"ND74, The       Cryptogram, American Cryptogram Association, 1974.[GRO2] BERYL," BERYL'S Pearls: Gromark Primers by hand       calculator," ND91, The Cryptogram, American Cryptogram       Association, 1991.[GRO3] MARSHEN," Checking the Numerical Key,"JF70, The       Cryptogram, American Cryptogram Association, 1970.[GRO4] PHOENIX," Computer Column: Gronsfeld -> Gromark,"       "MJ90, The Cryptogram, American Cryptogram       Association, 1990.[GRO5] PHOENIX," Computer Column: Perodic Gromark," MJ90       The Cryptogram, American Cryptogram Association, 1990.[GRO6] ROGUE," Cycles for Gromark Running Key," JF75, The       Cryptogram, American Cryptogram Association, 1975.[GRO7] DUMBO," Gromark Cipher," MA69, JA69, The Cryptogram,       American Cryptogram Association, 1969.[GRO8] DAN SURR," Gromark Club Solution," MA75, The       Cryptogram, American Cryptogram Association, 1975.[GRO9] B.NATURAL," Keyword Recovery in Periodic Gromark,"       SO73, The Cryptogram, American Cryptogram Association,       1973.[GROA] D.STRASSE," Method For Determining Term of Key," MA75,       The Cryptogram, American Cryptogram Association, 1975.[GROB] CRUX," More On Gromark Keys," ND87, The Cryptogram,       American Cryptogram Association, 1987.[GROC] DUMBO," Periodic Gromark ," MA73, The Cryptogram,       American Cryptogram Association, 1973.[GROD] ROGUE," Periodic Gromark ," SO73, The Cryptogram,       American Cryptogram Association, 1973.[GROE] ROGUE," Theoretical Frequencies in the Gromark," MA74,       The Cryptogram, American Cryptogram Association, 1974.[GRON] R.L.H., "Condensed Analysis of a Gronsfeld," AM38,       ON38,The Cryptogram, American Cryptogram Association,       1938,1938.[GRN1] CHARMER, "Gronsfeld," AS44, The Cryptogram, American       Cryptogram Association, 1944.[GRN2] PICCOLA, "Gronsfeld Cipher," ON35, The Cryptogram,       American Cryptogram Association, 1935.[GRN3] S-TUCK, "Gronsfeld Cipher," AS44, The Cryptogram,       American Cryptogram Association, 1944.[GROU] Groueff, Stephane, "Manhattan Project: The Untold       Story of the Making of the Atom Bomb," Little, Brown       and Company,1967.[GUST] Gustave, B., "Enigma:ou, la plus grande 'enigme de la       guerre 1939-1945." 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Multidimensional Codes, Illusions, Numbers,       and Brainteasers," Little, Brown and Co., New York,       1994.[HEBR] COMET, "First Hebrew Book (of Cryptology)," JF72, The       Cryptogram, published by the American Cryptogram       Association, 1972.[HELD] , Gilbert, "Top Secret Data Encryption Techniques,"       Prentice Hall, 1993.  (great title..limited use)[HEMP] Hempfner, Philip and Tania, "Pattern Word List For       Divided and Undivided Cryptograms," unpublished       manuscript, 1984.[HEPP] Hepp, Leo, "Die Chiffriermaschine 'ENIGMA'", F-Flagge,       1978.[HIDE] Hideo Kubota, " Zai-shi dai-go kokugun tokushu joho       senshi."  unpublished manuscript, NIDS.[HIER] ISHCABIBEL, "Hieroglyphics: Cryptology Started Here,       MA71, The Cryptogram, American Cryptogram Association,       1971.[HILL] Hill, Lester, S., "Cryptography in an Algebraic       Alphabet", The American Mathematical Monthly, June-       July 1929.[HIL1] Hill, L. S. 1929. 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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. D., "Breaking The Maya Code," Thames and       Hudson, New York, 1992.[MAZU] Mazur, Barry, "Questions On Decidability and       Undecidability in Number Theory," Journal of Symbolic       Logic, Volume 54, Number 9, June, 1994.[MELL] Mellen G.  1981. Graphic Solution of a Linear       Transformation Cipher. Cryptologia. 5:1-19.[MEND] Mendelsohn, Capt. C. J.,  Studies in German Diplomatic       Codes Employed During World War, GPO, 1937.[MERK] Merkle, Ralph, "Secrecy, Authentication and Public Key       Systems," Ann Arbor, UMI Research Press, 1982.[MER1] Merkle, Ralph, "Secure Communications Over Insecure       Channels," Communications of the ACM 21, 1978, pp.       294-99.[MER2] Merkle, Ralph and Martin E. Hellman, "On the Security       of Multiple Encryption ," Communications of the ACM       24, 1981, pp. 465-67.[MER3] Merkle, Ralph and Martin E. 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