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

CRYPTANALYSIS OF VIGGY'S EXTENDED FAMILY

DECIMATION IN DETAIL

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]

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.

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 eSetting 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.

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 ZWe 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 UFigure 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.

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.

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

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.

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]

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 10Consider 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 26Try 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 EReads:

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.

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 Pwithout 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 SThis 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 LWithout 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.

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 * * OkBy 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 ASliding 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 AWe 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:

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

* *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 ASince 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

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

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 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 6From 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 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.

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 CWith 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 ZWe set up two key word mixed alphabets and slide against eachother. With some trial and error we find:

NABLYCDFGHJKMPRVWXZQUESTIO QUESTIONABLYCDFGHJKMPRVWXZThe 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 BJNOQThe 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 QWe 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 Qare 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 FactorsOnly the factors 2 and 4 are common. We discard 2 asimprobable. We break up the message into groups of four.

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

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) XZIt 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 ZThe 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* UQRNRNZFXXWe 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 IFrom 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

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 61234What 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 YOk. 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 MVAJThe 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 YWe 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'sI 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.

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.

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.

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 DIEXZFactoring 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 AWe 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!

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.]

- 56123456123456123456123456123456123456123456123456123456123EBGSGGFIZTUAMFRFEQXPEPPOPCNBPQPOTXVNAIHHVRXCNHVGMFRFSIESQMVRKFUSVAGXLJKCNDVPROWBRHYFJMSHRFVSBAHWGZFAJOJMFAVCNDVDORZPHA

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.

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.

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]

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.

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.

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.)

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F., "War Secrets in the Ether - Volume I," Aegean Park Press, Laguna Hills, CA, 1977.[FLIC] Flicke, W. F., "War Secrets in the Ether - Volume II," Aegean Park Press, Laguna Hills, CA, 1977.[FLIC] Flicke, W. F., "War Secrets in the Ether," Aegean Park Press, Laguna Hills, CA, 1994.[FORE] DELAC, "Solving a Foreign Periodic by Lining Up the Alphabets," JJ46, The Cryptogram, published by the American Cryptogram Association, 1946.[FOWL] Fowler, Mark and Radhi Parekh, " Codes and Ciphers, - Advanced Level," EDC Publishing, Tulsa OK, 1994. (clever and work)[FRAA] Friedman, William F. , "American Army Field Codes in The American Expeditionary Forces During the First World War, USA 1939.[FRAB] Friedman, W. F., Field Codes used by the German Army During World War. 1919.[FRAN] Franks, Peter, "Calculator Ciphers," Information Associates, Champaign, Il. 1980.[FRE] Friedman, William F. , "Elements of Cryptanalysis," Aegean Park Press, Laguna Hills, CA, 1976.[FREA] Friedman, William F. , "Advanced Military Cryptography," Aegean Park Press, Laguna Hills, CA, 1976.[FREB] Friedman, William F. , "Elementary Military Cryptography," Aegean Park Press, Laguna Hills, CA, 1976.[FREC] Friedman, William F., "Cryptology," The Encyclopedia Britannica, all editions since 1929. A classic article by the greatest cryptanalyst.[FRSG] Friedman, William F., "Solving German Codes in World War I, " Aegean Park Press, Laguna Hills, CA, 1977.[FR1] Friedman, William F. and Callimahos, Lambros D., Military Cryptanalytics Part I - Volume 1, Aegean Park Press, Laguna Hills, CA, 1985.[FR2] Friedman, William F. and Callimahos, Lambros D., Military Cryptanalytics Part I - Volume 2, Aegean Park Press, Laguna Hills, CA, 1985.[FR3] Friedman, William F. and Callimahos, Lambros D., Military Cryptanalytics Part III, Aegean Park Press, Laguna Hills, CA, 1995.[FR4] Friedman, William F. and Callimahos, Lambros D., Military Cryptanalytics Part IV, Aegean Park Press, Laguna Hills, CA, 1995.[FR5] Friedman, William F. Military Cryptanalysis - Part I, Aegean Park Press, Laguna Hills, CA, 1980.[FR6] Friedman, William F. Military Cryptanalysis - Part II, Aegean Park Press, Laguna Hills, CA, 1980.[FR7] Friedman, William F. and Callimahos, Lambros D., Military Cryptanalytics Part II - Volume 1, Aegean Park Press, Laguna Hills, CA, 1985.[FR8] Friedman, William F. and Callimahos, Lambros D., Military Cryptanalytics Part II - Volume 2, Aegean Park Press, Laguna Hills, CA, 1985.[FR22] Friedman, William F., The Index of Coincidence and Its Applications In Cryptography, Publication 22, The Riverbank Publications, Aegean Park Press, Laguna Hills, CA, 1979.[FRS6] Friedman, W. F., "Six Lectures On Cryptology," National Archives, SRH-004.[FR8] Friedman, W. F., "Cryptography and Cryptanalysis Articles," Aegean Park Press, Laguna Hills, CA, 1976.[FR9] Friedman, W. F., "History of the Use of Codes," Aegean Park Press, Laguna Hills, CA, 1977.[FRZM] Friedman, William F.,and Charles J. Mendelsohn, "The Zimmerman Telegram of January 16, 1917 and its Cryptographic Background," Aegean Park Press, Laguna Hills, CA, 1976.[FROM] Fromkin, V and Rodman, R., "Introduction to Language," 4th ed.,Holt Reinhart & Winston, New York, 1988.[FRS] Friedman, William F. and Elizabeth S., "The Shakespearean Ciphers Examined," Cambridge University Press, London, 1957.[FUMI] Fumio Nakamura, Rikugun ni okeru COMINT no hoga to hatten," The Journal of National Defense, 16-1 (June 1988) pp85 - 87.[GAJ] Gaj, Krzysztof, "Szyfr Enigmy: Metody zlamania," Warsaw Wydawnictwa Komunikacji i Lacznosci, 1989.[GAR1] Gardner, Martin, "536 Puzzles and Curious Problems," Scribners, 1967.[GAR2] Gardner, Martin, "Mathematics, Magic, and Mystery ," Dover, 1956.[GAR3] Gardner, Martin, "New Mathematical Diversions from Scientific American," Simon and Schuster, 1966.[GAR4] Gardner, Martin, "Sixth Book of Mathematical Games from Scientific American," Simon and Schuster, 1971.[GARL] Garlinski, Jozef, 'The Swiss Corridor', Dent, London 1981.[GAR1] Garlinski, Jozef, 'Hitler's Last Weapons', Methuen, London 1978.[GAR2] Garlinski, Jozef, 'The Enigma War', New York, Scribner, 1979.[GE] "Security," General Electric, Reference manual Rev. B., 3503.01, Mark III Service, 1977.[GERH] Gerhard, William D., "Attack on the U.S., Liberty," SRH-256, Aegean Park Press, 1981.[GERM] "German Dictionary," Hippocrene Books, Inc., New York, 1983.[GILE] Giles, Herbert A., "Chinese Self-Taught," Padell Book Co., New York, 1936?[GIVI] Givierge, General Marcel, " Course In Cryptography," Aegean Park Press, Laguna Hills, CA, 1978. Also, M. Givierge, "Cours de Cryptographie," Berger-Levrault, Paris, 1925.[GLEN] Gleason, Norma, "Fun With Codes and Ciphers Workbook," Dover, New York, 1988.[GLE1] Gleason, Norma, "Cryptograms and Spygrams," Dover, New York, 1981.[GLEA] Gleason, A. M., "Elementary Course in Probability for the Cryptanalyst," Aegean Park Press, Laguna Hills, CA, 1985.[GLOV] Glover, D. Beaird, "Secret Ciphers of the 1876 Presidential Election," Aegean Park Press, Laguna Hills, CA, 1991.[GODD] Goddard, Eldridge and Thelma, "Cryptodyct," Marion, Iowa, 1976[GORD] Gordon, Cyrus H., " Forgotten Scripts: Their Ongoing Discovery and Decipherment," Basic Books, New York, 1982.[GRA1] Grandpre: "Grandpre, A. de--Cryptologist. Part 1 'Cryptographie Pratique - The Origin of the Grandpre', ISHCABIBEL, The Cryptogram, SO60, American Cryptogram Association, 1960.[GRA2] Grandpre: "Grandpre Ciphers", ROGUE, The Cryptogram, SO63, American Cryptogram Association, 1963.[GRA3] Grandpre: "Grandpre", Novice Notes, LEDGE, The Cryptogram, MJ75, American Cryptogram Association,1975[GRAH] Graham, L. A., "Ingenious Mathematical Problems and Methods," Dover, 1959.[GRAN] Grant, E. A., "Kids Book of Secret Codes, Signals and Ciphers, Running Press, 1989.[GRAP] DR. 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." Paris:Plon, 1973.[GYLD] Gylden, Yves, "The Contribution of the Cryptographic Bureaus in the World War," Aegean Park Press, 1978.[HA] Hahn, Karl, " Frequency of Letters", English Letter Usage Statistics using as a sample, "A Tale of Two Cities" by Charles Dickens, Usenet SCI.Crypt, 4 Aug 1994.[HAFT] Haftner, Katie and John Markoff, "Cyberpunk," Touchstine, 1991.[HAGA] Hagamen,W. D. et. al., "Encoding Verbal Information as Unique Numbers," IBM Systems Journal, Vol 11, No. 4, 1972.[HAWA] Hitchcock, H. R., "Hawaiian," Charles E. Tuttle, Co., Toyko, 1968.[HAWC] Hawcock, David and MacAllister, Patrick, "Puzzle Power! 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. Cryptography in an Algebraic Alphabet. American Mathematical Monthly. 36:306-312.[HIL2] Hill, L. S. 1931. Concerning the Linear Transformation Apparatus in Cryptography. American Mathematical Monthly. 38:135-154.[HINS] Hinsley, F. H., "History of British Intelligence in the Second World War", Cambridge University Press, Cambridge, 1979-1988.[HIN2] Hinsley, F. H. and Alan Strip in "Codebreakers -Story of Bletchley Park", Oxford University Press, 1994.[HIN3] Hinsley, F. H., et. al., "British Intelligence in The Second World War: Its Influence on Strategy and Operations," London, HMSO vol I, 1979, vol II 1981, vol III, 1984 and 1988.[HISA] Hisashi Takahashi, "Military Friction, Diplomatic Suasion in China, 1937 - 1938," The Journal of International Studies, Sophia Univ, Vol 19, July, 1987.[HIS1] Barker, Wayne G., "History of Codes and Ciphers in the U.S. Prior to World War I," Aegean Park Press, Laguna Hills, CA, 1978.[HITT] Hitt, Parker, Col. " Manual for the Solution of Military Ciphers," Aegean Park Press, Laguna Hills, CA, 1976.[HODG] Hodges, Andrew, "Alan Turing: The Enigma," New York, Simon and Schuster, 1983.[HOFF] Hoffman, Lance J., editor, "Building In Big Brother: The Cryptographic Policy Debate," Springer-Verlag, N.Y.C., 1995. ( A useful and well balanced book of cryptographic resource materials. )[HOF1] Hoffman, Lance. J., et. al.," Cryptography Policy," Communications of the ACM 37, 1994, pp. 109-17.[HOLM Holmes, W. J., "Double-Edged Secrets: U.S. Naval Intelligence Operations in the Pacific During WWII", Annapolis, MD: Naval Institute Press, 1979.[HOM1] Homophonic: A Multiple Substitution Number Cipher", S- TUCK, The Cryptogram, DJ45, American Cryptogram Association, 1945.[HOM2] Homophonic: Bilinear Substitution Cipher, Straddling," ISHCABIBEL, The Cryptogram, AS48, American Cryptogram Association, 1948.[HOM3] Homophonic: Computer Column:"Homophonic Solving," PHOENIX, The Cryptogram, MA84, American Cryptogram Association, 1984.[HOM4] Homophonic: Hocheck Cipher,", SI SI, The Cryptogram, JA90, American Cryptogram Association, 1990.[HOM5] Homophonic: "Homophonic Checkerboard," GEMINATOR, The Cryptogram, MA90, American Cryptogram Association, 1990.[HOM6] Homophonic: "Homophonic Number Cipher," (Novice Notes) LEDGE, The Cryptogram, SO71, American Cryptogram Association, 1971.[HYDE] H. Montgomery Hyde, "Room 3603, The Story of British Intelligence Center in New York During World War II", New York, Farrar, Straus, 1963.[IBM1] IBM Research Reports, Vol 7., No 4, IBM Research, Yorktown Heights, N.Y., 1971.[IC1 ] GIZMO, "Bifid Period Determination Using a Digraphic Index of Coincidence, JF79, The Cryptogram, American Cryptogram Association, 1979.[IC2 ] PHOENIX, "Computer Column: Applications of the Index of Coincidence, JA90, The Cryptogram, American Cryptogram Association, 1990.[IC3 ] PHOENIX, "Computer Column: Digraphic Index of Coincidence, ND90, The Cryptogram, American Cryptogram Association, 1990.[IC4 ] PHOENIX, "Computer Column: Index of Coincidence (IC), JA82, The Cryptogram, American Cryptogram Association, 1982.[IC5 ] PHOENIX, "Computer Column: Index of Coincidence, (correction) MA83, The Cryptogram, American Cryptogram Association, 1983.[IMPE] D'Imperio, M. E, " The Voynich Manuscript - An Elegant Enigma," Aegean Park Press, Laguna Hills, CA, 1976.[INDE] PHOENIX, Index to the Cryptogram: 1932-1993, ACA, 1994.[ITAL] Italian - English Dictionary, compiled by Vittore E. Bocchetta, Fawcett Premier, New York, 1965.[JAPA] Martin, S.E., "Basic Japanese Conversation Dictionary," Charles E. Tuttle Co., Toyko, 1981.[JAPH] "Operational History of Japanese Naval Communications, December 1941- August 1945, Monograph by Japanese General Staff and War Ministry, Aegean Park Press, 1985.[JOHN] Johnson, Brian, 'The Secret War', Arrow Books, London 1979.[KADI] al-Kadi, Ibrahim A., Cryptography and Data Security: Cryptographic Properties of Arabic, Proceedings of the Third Saudi Engineering Conference. Riyadh, Saudi Arabia: Nov 24-27, Vol 2:910-921., 1991.[KAHN] Kahn, David, "The Codebreakers", Macmillian Publishing Co. , 1967.[KAH1] Kahn, David, "Kahn On Codes - Secrets of the New Cryptology," MacMillan Co., New York, 1983.[KAH2] Kahn, David, "An Enigma Chronology", Cryptologia Vol XVII,Number 3, July 1993.[KAH3] Kahn, David, "Seizing The Enigma: The Race to Break the German U-Boat Codes 1939-1943 ", Houghton Mifflin, New York, 1991.[KARA] Karalekas, Anne, "History of the Central Intelligence Agency," Aegean Park Press, Laguna Hills, CA, 1977.[KASI] Kasiski, Major F. W. , "Die Geheimschriften und die Dechiffrir-kunst," Schriften der Naturforschenden Gesellschaft in Danzig, 1872.[KAS1] Bowers, M. W., {ZEMBIE} "Major F. W. Kasiski - Cryptologist," The Cryptogram, XXXI, JF, 1964.[KAS2] ----, "Kasiski Method," JF64,MA64, The Cryptogram, American Cryptogram Association, 1964.[KAS3] PICCOLA, "Kasiski Method for Periodics," JJ35,AS35, The Cryptogram, American Cryptogram Association, 1935, 1935.[KAS4] AB STRUSE, "Who was Kasiski?" SO76, The Cryptogram, American Cryptogram Association, 1976.[KATZ] Katzen, Harry, Jr., "Computer Data Security,"Van Nostrand Reinhold, 1973.[KERC] Kerckhoffs, "la Cryptographie Militaire, " Journel des Sciences militaires, 9th series, IX, (January and February, 1883, Libraire Militaire de L. Baudoin &Co., Paris. English trans. by Warren T, McCready of the University of Toronto, 1964[KOBL] Koblitz, Neal, " A Course in Number Theory and Cryptography, 2nd Ed, Springer-Verlag, New York, 1994.[KONH] Konheim, Alan G., "Cryptography -A Primer" , John Wiley, 1981, pp 212 ff.[KORD] Kordemsky, B., "The Moscow Puzzles," Schribners, 1972.[KOTT] Kottack, Phillip Conrad, "Anthropology: The Exploration Of Human Diversity," 6th ed., McGraw-Hill, Inc., New York, N.Y. 1994.[KOZA] Kozaczuk, Dr. Wladyslaw, "Enigma: How the German Machine Cipher was Broken and How it Was Read by the Allies in WWI", University Pub, 1984.[KRAI] Kraitchek, "Mathematical Recreations," Norton, 1942, and Dover, 1963.[KULL] Kullback, Solomon, Statistical Methods in Cryptanalysis, Aegean Park Press, Laguna Hills, Ca. 1976.[LAFF] Laffin, John, "Codes and Ciphers: Secret Writing Through The Ages," Abelard-Schuman, London, 1973.[LAI] Lai, Xuejia, "On the Design and Security of Block Ciphers," ETH Series in Information Processing 1, 1992. (Article defines the IDEA Cipher)[LAIM] Lai, Xuejia, and James L. Massey, "A Proposal for a New Block Encryption Standard," Advances in Cryptology -Eurocrypt 90 Proceedings, 1992, pp. 55-70.[LAKE] Lakoff, R., "Language and the Women's Place," Harper & Row, New York, 1975.[LANG] Langie, Andre, "Cryptography," translated from French by J.C.H. Macbeth, Constable and Co., London, 1922.[LAN1] Langie, Andre, "Cryptography - A Study on Secret Writings", Aegean Park Press, Laguna Hills, CA. 1989.[LAN2] Langie, Andre, and E. A. Soudart, "Treatise on Cryptography, " Aegean Park Press, Laguna Hills, CA. 1991.[LATI] BRASSPOUNDER, "Latin Language Data, "The Cryptogram," July-August 1993.[LAUE] Lauer, Rudolph F., "Computer Simulation of Classical Substitution Cryptographic Systems" Aegean Park Press, 1981, p72 ff.[LEAR] Leary, Penn, " The Second Cryptographic Shakespeare," Omaha, NE [from author] 1994.[LEA1] Leary, Penn, " Supplement to The Second Cryptographic Shakespeare," Omaha, NE [from author] 1994.[LEAU] Leaute, H., "Sur les Mecanismes Cryptographiques de M. de Viaris," Le Genie Civil, XIII, Sept 1, 1888.[LEDG] LEDGE, "NOVICE NOTES," American Cryptogram Association, 1994. [ One of the best introductory texts on ciphers written by an expert in the field. Not only well written, clear to understand but as authoritative as they come! ][LENS] Lenstra, A.K. et. al. "The Number Field Sieve," Proceedings of the 22 ACM Symposium on the Theory of Computing," Baltimore, ACM Press, 1990, pp 564-72.[LEN1] Lenstra, A.K. et. al. "The Factorization of the Ninth Fermat Number," Mathematics of Computation 61 1993, pp. 319-50.[LEWF] Lewis, Frank, "Problem Solving with Particular Reference to the Cryptic (or British) Crossword and other 'American Puzzles', Part One," by Frank Lewis, Montserrat, January 1989.[LEW1] Lewis, Frank, "The Nations Best Puzzles, Book Six," by Frank Lewis, Montserrat, January 1990.[LEWI] Lewin, Ronald, 'Ultra goes to War', Hutchinson, London 1978.[LEW1] Lewin, Ronald, 'The American Magic - Codes, ciphers and The Defeat of Japan', Farrar Straus Giroux, 1982.[LEWY] Lewy, Guenter, "America In Vietnam", Oxford University Press, New York, 1978.[LEVI] Levine, J., U.S. Cryptographic Patents 1861-1981, Cryptologia, Terre Haute, In 1983.[LEV1] Levine, J. 1961. Some Elementary Cryptanalysis of Algebraic Cryptography. American Mathematical Monthly. 68:411-418[LEV2] Levine, J. 1961. Some Applications of High- Speed Computers to the Case n =2 of Algebraic Cryptography. Mathematics of Computation. 15:254-260[LEV3] Levine, J. 1963. Analysis of the Case n =3 in Algebraic Cryptography With Involuntary Key Matrix With Known Alphabet. Journal fuer die Reine und Angewante Mathematik. 213:1-30.[LISI] Lisicki, Tadeusz, 'Dzialania Enigmy', Orzet Biaty, London July-August, 1975; 'Enigma i Lacida', Przeglad lacznosci, London 1974- 4; 'Pogromcy Enigmy we Francji', Orzet Biaty, London, Sept. 1975.'[LYNC] Lynch, Frederick D., "Pattern Word List, Vol 1.," Aegean Park Press, Laguna Hills, CA, 1977.[LYN1] Lynch, Frederick D., "An Approach To Cryptarithms," ACA, 1976.[LYSI] Lysing, Henry, aka John Leonard Nanovic, "Secret Writing," David Kemp Co., NY 1936.[MACI] Macintyre, D., "The Battle of the Atlantic," New York, Macmillan, 1961.[MADA] Madachy, J. S., "Mathematics on Vacation," Scribners, 1972.[MAGN] Magne, Emile, Le plaisant Abbe de Boisrobert, Paris, Mecure de France, 1909.[MANN] Mann, B.,"Cryptography with Matrices," The Pentagon, Vol 21, Fall 1961.[MANS] Mansfield, Louis C. S., "The Solution of Codes and Ciphers", Alexander Maclehose & Co., London, 1936.[MARO] Marotta, Michael, E. "The Code Book - All About Unbreakable Codes and How To Use Them," Loompanics Unlimited, 1979. [This is a terrible book. Badly written, without proper authority, unprofessional, and prejudicial to boot. 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. Hellman, "Hiding Information and Signatures in Trap Door Knapsacks," IEEE Transactions on Information Theory 24, 1978, pp. 525-30.[MILL] Millikin, Donald, " Elementary Cryptography ", NYU Bookstore, NY, 1943.[MM] Meyer, C. H., and Matyas, S. M., " CRYPTOGRAPHY - A New Dimension in Computer Data Security, " Wiley Interscience, New York, 1982.[MODE] Modelski, Tadeusz, 'The Polish Contribution to the Ultimate Allied Victory in the Second World War', Worthing (Sussex) 1986.[MRAY] Mrayati, Mohammad, Yahya Meer Alam and Hassan al- Tayyan., Ilm at-Ta'miyah wa Istikhraj al-Mu,amma Ind al-Arab. Vol 1. Damascus: The Arab Academy of Damascus., 1987.[MULL] Mulligan, Timothy," The German Navy Examines its Cryptographic Security, Oct. 1941, Military affairs, vol 49, no 2, April 1985.[MYER] Myer, Albert, "Manual of Signals," Washington, D.C., USGPO, 1879.[NBS] National Bureau of Standards, "Data Encryption Standard," FIPS PUB 46-1, 1987.[NIBL] Niblack, A. 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