Polyalphabetic Substitution Systems III

Cryptanalysis Of Viggy's Extended Family

Decimation In Detail

CLASSICAL CRYPTOGRAPHY COURSE BY LANAKI May 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. We will cover decimation processes in detail and investigate special solutions for periodic ciphers. The important principle of Superimposition will be introduced. The Resources Section has been updated with more than 50 ACA published references on these and similar systems - focusing on 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 aperiodic polyalphabetic 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 further Cryptarithm 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 in Lecture 11. The PORTAX uses pairs of letters as a unit for encipherment and decipherment as apart from single letters. A special slide is required for its operation, and a keyword is 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 the A-indicator of the stationary alphabet.) To encipher the digraph RE, we take the R in the upper row of letters (stationary slide) and the E from the lower pair of letters (sliding), and use the opposite corners of the rectangle formed to obtain the ciphertext, or PI. However, if the digram ER is to be enciphered, we take the E from the stationary alphabet at the top, and the R from the sliding alphabet at the bottom to obtain FP. Note that if the first letter of a digraph is in the range of A-M, the equivalent ciphertext is dependent on where the slide is used for the key-letter; but, if the first letter of the digraph is in the range of N-Z, then it slides along with the paired rows of lower letters, and therefore all such digraphs having the first letter in the N-Z are constant, without dependent of the key. There is an exception when both letters in the plaintext digraph are in the same column, in which case the key letter has to be known, for letters appearing above the needed letters are used for the ciphertext. [BRYA] To encipher with keyword, the plaintext is written in two rows under it; continuing to the end of the message. When the final group is reached, if there are not enough letters to make it complete (an even number), add a single null. For example, encipher the word INNOVATION using the key OFTEN : * 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 V A T I O N g w e b --------- S A R E F O U N D x u i k e Setting the O of the sliding pairs under the 'A' indicator of the stationary alphabet, we enciphering IA as GE (opposite corners); then SO, continuing down the column we encipher the whole column. We then slide the strip until E-F (key) is under the A indicator and encipher that column. To find the period in the PORTAX is dependent on possible fragments of the plaintext which are known (through the N-Z combinations produced from the unchanged relationship of letters). Lets partially decipher the following PORTAX: SNPOW LBAMP ISCWU OOBXC WKMAT ZKTOW JCBLN CBJGB TAAJD IWUKW HHVZN MNUFM APBJW PCBSX JCJQX TMVUB MDCBJ 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 the final group, we find -NTON -HUN-R (hundred?) We next test the keyword by putting T in the final position and testing the precursor letter; A C E F H I L N O P R S and U, At the E setting, OM = TC, making -OYST/-SOCCU with R in the next group confirming OCCUR. The E substitution also gives us the HUNDRED. The rest of the analysis is left for the student for credit. THE NIHILIST SUBSTITUTION CIPHER One of my favorite ciphers is the Nihilist Substitution Cipher. Classified as a periodic, it employs numbers to represent letters. Numbers are derived from a 5 x 5 Polybius Square. 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 Z So A = 11, L = 31, T = 44. (Row by Column) The Polybius Square can be keyed. For example, using UNITED STATES OF AMERICA and eliminating the duplicate letters, 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 by columns 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 U The 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 applied first. Now the message COME AT ONCE enciphered with a keyword of TENT (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 to produce the ciphertext: COME: 57 49 65 59; ATON: 55 59 67 77; CE: 57 30. Each column represents a monoalphabetic substitution in itself, and the reading or value of these letters is dependent on the letters on either side of them. WEAKNESSES The lowest number of any key-letter which may be added to the lowest plaintext letter is 11, with a total of 22; the highest combination is two 55's or 10 (110). The numbers 6,7,8, or 9, are not involved in either the tens or the one's additions - but they may result in a sum. Cipher 22 must equal 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 find that a 6-7-8-9 is involved we can discard the period assumed as 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 sum and then the one's sum. FINDING THE PERIOD There are two ways to find the period - the short and the long way. SHORT METHOD The short way of finding the period is to look for two or more 30's. We treat them like a repeated digraph and factor the interval between them looking for a common factor. We may also try the same procedure with the lowest number versus the highest number, for example the distance between two 94's or two 26's. LONG METHOD The long way is to assume a 3 period and test the 1'st and 4'th, 2'nd and 5'th, 3'rd and 6'th in the same manner as the short method. When conflicts arise, discard the choice. We continue with an assumption of periods 4, 5, 6, etc. and increase the differentials between ciphertext numbers. [BRYA] CRYPTANALYSIS OF THE NIHILIST SUBSTITUTION Gaines [ELCY] suggests that cracking this cipher parallels the Viggy. The period is found through repeated sequences, or in their absence, through repeated single letters, yielding individual frequency counts on the several alphabets of the period. If the arrangement of the ciphertext follows the normal Polybius (aka Checkerboard) Square, the frequency counts will follow the graph of the normal alphabet less one letter. 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 any number to all of them does not change this fact. the maximum difference between any to sums is still 4. Now the number added during encipherment is also a number containing no digit other than 1-2-3-4-5; thus any number found in the cryptogram can be considered as carrying two separate additions, one for tens and one for ones. The two 5's added give us the revealing 0; the carried digit 1 can be mentally borrowed back, by decreasing the size of the digit preceding the zero. If we find a 40 , we look at it as 3 tens with ten units or finding 110, we may regard this as ten tens and ten units. If we find the numbers 29 and 87 in the cryptogram, we know they were not enciphered by the same key. This is because a difference greater than 4 in the respective tens units exists and no digit whatever added to any two digits of the original square can produce a difference greater than 4. Say we have 30 and 77, with no difference greater than 4, the presence of the zero needs to be accounted for. The number 30 has 2 tens and ten units; 7 - 2 >4, hence, we reject the same key hypothesis. Four giveaways are 22, 30, 102, and 110. The presence of any one of these numbers gives away the key to the whole cipher alphabet. [BRYA] presents a useful aid for the standard Polybius Square in Table 12-1. At the top is the key-number, at the left is the plaintext letter, and at ciphertext is found at the intersection. Any two of the three variables yields the unknown 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 M A 11 22 23 24 25 26 32 33 34 35 36 42 43 B 12 23 24 25 26 27 33 34 35 36 37 43 44 C 13 24 25 26 27 28 34 35 36 37 38 44 45 D 14 25 26 27 28 29 35 36 37 38 39 45 46 E 15 26 27 28 29 30 36 37 38 39 40 46 47 F 21 32 33 34 35 36 42 43 44 45 46 52 53 G 22 33 34 35 36 37 43 44 45 46 47 53 54 H 23 34 35 36 37 38 44 45 46 47 48 54 55 I 24 35 36 37 38 39 45 46 47 48 49 55 56 K 25 36 37 38 39 40 46 47 48 49 50 56 57 L 31 42 43 44 45 46 52 53 54 55 56 62 63 M 32 43 44 45 46 47 53 54 55 56 57 63 64 N 33 44 45 46 47 48 54 55 56 57 58 64 65 O 34 45 46 47 48 49 55 56 57 58 59 65 66 P 35 46 47 48 49 50 56 57 58 59 60 66 67 Q 41 52 53 54 55 56 62 63 64 65 66 72 73 R 42 53 54 55 56 57 63 64 65 66 67 73 74 S 43 54 55 56 57 58 64 65 66 67 68 74 75 T 44 55 56 57 58 59 65 66 67 68 69 75 76 U 45 56 57 58 59 60 66 67 68 69 70 76 77 V 51 62 63 64 65 66 72 73 74 75 76 82 83 W 52 63 64 65 66 67 73 74 75 76 77 83 84 X 53 64 65 66 67 68 74 75 76 77 78 84 85 Y 54 65 66 67 68 69 75 76 77 78 79 85 86 Z 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 Z A 11 44 45 46 52 53 54 55 56 62 63 64 65 66 B 12 45 46 47 53 54 55 56 57 63 64 65 66 67 C 13 46 47 48 54 55 56 57 58 64 65 66 67 68 D 14 47 48 49 55 56 57 58 59 65 66 67 68 69 E 15 48 49 50 56 57 58 59 60 66 67 68 69 70 F 21 54 55 56 62 63 64 65 66 72 73 74 75 76 G 22 55 56 57 63 64 65 66 67 73 74 75 76 77 H 23 56 57 58 64 65 66 67 68 74 75 76 77 78 I 24 57 58 59 65 66 67 68 69 75 76 77 78 79 K 25 58 59 60 66 67 68 69 70 76 77 78 79 80 L 31 64 65 66 72 73 74 75 76 82 83 84 85 86 M 32 65 66 67 73 74 75 76 77 83 84 85 86 87 N 33 66 67 68 74 75 76 77 78 84 85 86 87 88 O 34 67 68 69 75 76 77 78 79 85 86 87 88 89 P 35 68 69 70 76 77 78 79 80 86 87 88 89 90 Q 41 74 75 76 82 83 84 85 86 92 93 94 95 96 R 42 75 76 77 83 84 85 86 87 93 94 95 96 97 S 43 76 77 78 84 85 86 87 88 94 95 96 97 98 T 44 77 78 79 85 86 87 88 89 95 96 97 98 99 U 45 78 79 80 86 87 88 89 90 96 97 98 99 00 V 51 84 85 86 92 93 94 95 96 02 03 04 05 06 W 52 85 86 87 93 94 95 96 97 03 04 05 06 07 X 53 86 87 88 94 95 96 97 98 04 05 06 07 08 Y 54 87 88 89 95 96 97 98 99 05 06 07 08 09 Z 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 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 24 85 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 66 47 67 26 Try period = 2. Starting at the first number 24 constant we scan the line looking for differences greater than 4 using a constant 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 entire cryptogram, comparing numbers at an interval of 4, without find any difference in either tens or units greater than 4. We now must look at the numbers collectively in columns to verify the period is 4. We recopy the cryptogram into a block. 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 the digit 2 and since this can only come from the addition of 1 plus 1, the only possible key digit is 1. The units-half has a range of 4-5-6-7-8, maximum range possible. The smallest digit to result in 8 is 3, the largest digit to result in 4 is also 3, that is the only digit which can result in all of the digits 4-5-6-7-8 is 3, so that the cipher key for this column is 13. It cannot be anything else. Alphabet 2: The tens-half of the second column ranges over the full five digits 4-5-6-7-8 (key 3), and the units-half ranges over 5-6-7-8-9 (key 4). This suggests the key digit is 34. Alphabet 3: The tens-half of the third column contains the 'giveaway' digit of 2 and the units-half also contains the digit 2. The key digit to produce this situation is 11. Alphabet 4: The tens-half of the fourth column ranges only over the digits 5-6-7-8, with nothing to indicate whether the missing 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 normal square suggests COAO or COAT as the key word. We use Table 12-1 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: We have here only the shell, the nut has gone. For the most difficult case presenting multiple key possibilities, we line up the alphabets graphically against their frequency counts to eliminate the extra key digits. GROMARK MASTERTON describes a cipher called the GROMARK. The Gromark is akin to the GRONSFELD in that the components never change their position relative to each other and every plain text values has 10 possible cipher representatives. The GROMARK uses a different keying method; encipherment is effected by means of a normal alphabet plain set against a mixed cipher text alphabet. However, instead of cycles or predictable slides of the cipher component, one finds the plain value on the top (normal) component and counts a specified number of positions to the right, then takes the letter in the cipher alphabet immediately below. The choice of how far to count along the sequence is determined by the digital key. One essentially is adding 0 to 9 to the plain value, as in the Gronsfeld, but it is on the mixed sequence, set underneath a plain sequence. The key is derived from a Fibonacci series. On some cycle (frequently 5 wide) the key is derived from a starting group, by adding the first position to the second and placing the result in the sixth position. Similarly, positions 2 and 3 are added to make position number 7, 3, and 4 to make 8, and so forth. All additions are non carrying -a very 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 crib placement 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 7 J 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 crib SUBSTITUTES and runs somewhere from the J to the final P above. Since the plain sequence is normal, a repeated cipher letter, with different key letters on it, must stand for plain values removed from each other exactly by the difference of the two numbers. Thus C A C with keys 9 8 2 above it implies that the first cipher C is M for example, the second C is seven positions 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 a favorable case where the differences in the cipher repeats matches the plain differences, at the correct interval. To match these differences, we measure them in one direction for the plain and the reverse for the cipher. Table 12-1 shows subtraction of the left hand letter from the right, and we must look at the cipher in the other direction. Differences may be calculated modulo 26. Table 12-1 adjacent 19 21 2 19 20 9 20 21 20 5 19 diff's S U B S T I T U T E S xx 2 7 17 1 15 11 1 25 11 14 x-x 9 24 18 16 0 12 0 10 x--x 0 25 7 ... There is a difference of 7 with the C-C hit, but it doesn't appear on the second row of the table. The keyword must first 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 7 J 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 build the cipher component by writing the cipher component, and a normal alphabet, count along it from any given plain the number of steps given by the key, then write the cipher value. Find S on the top strip, count 8 to right, place an A. C is two spaces to the right of the position held by the U, and so on. Decipher other letters by counting backwards the number of steps given by the key. Cipher C ahead of thew crib 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 Z A J Y P Q W N C L Without a tip the system will fall to statistics. The numbers associated with any given cipher letter represent a stretch of 10 consecutive values along a normal alphabet such as C to L or X to G, we could prepare a table with A to Z as the rows and 9 to 0 as the columns. Frequencies can be combined and a stretch such as PQRST area will show as the normal. The backwards normal sequence yields a bar graph of the segment of the normal alphabetic frequencies. DECIMATION PROCESSES - FURTHER REMARKS In Lecture 11, we presented QUAGMIRES I-IV and solved them by a variety of methods. Inherent in their solution was Friedman's principle of indirect symmetry. [FRE7] Prima facie to this symmetry principle is a process of alphabet dissociation called Decimation. This same process effects all Viggy class ciphers and is important from a theoretical point of view. Decimation is especially effective in solving mixed alphabet systems like the Quagmire III & IV. Decimation is a process of selection and derivation of a sequence of equivalent components according to some fixed interval. For example, the sequence A E I M is derived by decimation of extracting every fourth letter from a normal alphabet. Consider the two mixed alphabets in a QUAGMIRE III: O1 * * Plain: QUESTIONABLYCDFGHJKMPRVWXZ Cipher: QUESTIONABLYCDFGHJKMPRVWXZQUESTIONABLYCDFGHJKMPRVWXZ * * Ok By setting the two sliding components against each other in the two positions shown: A in the first set and B in the second set we can derive two, we can derive two different sets of secondary alphabets based on the key letters. O1 * * Plain: QUESTIONABLYCDFGHJKMPRVWXZ Cipher: QUESTIONABLYCDFGHJKMPRVWXZQUESTIONABLYCDFGHJKMPRVWXZ * * Ok 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 Z Cipher: 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 N 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 Z Cipher: 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 type table based on the Keyword QUESTIONABLY (see a partial table in Table 11-2. 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 no resemblance of symmetry despite the fact that they were both created from the same primary alphabet. We do find a Latent Symmetry 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 Z Cipher: 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 ciphertext equivalents, beginning at any point and continuing until the chain is completed. We start Aplain = Jcipher, Jplain = Qcipher, Qplain = Bcipher...., dropping the common letters we 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 When slid against itself it will produce exactly the same secondary alphabets as do the primary components based upon the word QUESTIONABLY. For example, compare the secondary alphabets given by the two settings of the externally different components below: * * Plain: QUESTIONABLYCDFGHJKMPRVWXZ Cipher: 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 Z Cipher: 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: AJQBKULMEYPSCRTDVIFWOGXNHZ Cipher: 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 Z Cipher: 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 same equivalents in the secondary alphabets as does the sequence QUEST......XZ, the former is cryptographically equivalent to the latter sequence. For this reason the A J Q B K .. sequence is termed an equivalent primary component. If the real or original primary component is a keyword mixed sequence, it is hidden or latent within the equivalent primary sequence; it can also be made patent by the process of decimation of the equivalent primary component. Friedman in [FRE7] describes the process as follows: find three letters in the equivalent primary component that are a likely unbroken sequence in the original primary component, and see if the interval between the first and second is the same as that of the second and third. Try X, Y, Z in the equivalent primary component above. Note the sequence ..W O G X N H Z...; the distance or interval between W X Z is three letters. Continuing the chain by adding letters three intervals removed, the latent original primary component is made patent. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 W X Z Q U E S T I O N A B L Y C D F G H J K M 24 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 of the secondary alphabets, likely to be sequent in the keyword- mixed sequence, such as JK, the following chains of digraphs may be produced. Thus JK plain stand over QU cipher respectively, 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 equivalent primary component, as shown above. Even number of intervals for sliding the alphabets will yield half chains or 13 letter chains. Friedman [FRE7] describes several methods to combine the half chains into fully equivalent primary components. FRIEDMAN'S OBSERVATIONS Friedman observed that in the case of a 26-element component sliding against itself (both components proceeding in the same direction), it is only the secondary alphabets resulting from odd-interval displacements of the primary components which permit reconstructing a single 26-letter chain of equivalents. This is true except for the 13th interval displacement, which acts like an even number displacement, in that no complete chain of equivalents can be established from the secondary alphabet. Friedman states the general rule as: any displacement interval which has a factor in common with the number of letters in the primary sequence will yield a secondary alphabet from which no complete chain of 26 equivalents can be derived for the construction of a complete equivalent primary component. Components sliding in opposite directions act as a 13 interval displacement because of their reciprocal nature. Friedman concluded that whether or not a complete equivalent primary component is derivable by decimation from an original primary component (and if not, the lengths and numbers of chains of letters, or incomplete components, that can be constructed in attempts to derive such equivalent components) will depend upon the number of letters in the original primary component and the specific decimation interval selected. [FRE7] Friedman constructed a table relating the number of characters in the original primary component, decimation interval and total number of complete sequences that can be formed. See Table 12-3. TABLE 12-3 Number of Characters in Original Primary Component Decimation Interval 32 30 28 27 26 25 24 22 21 20 18 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 16 Total Number Of Sequences 14 6 10 16 10 18 6 8 10 6 4 6 >From Table 12-3, we see that in a 26-letter original primary component, decimation interval 5 will yield a complete equivalent primary component of 26 letters, whereas decimation intervals of 4 or 8 will yield 2 chains of 13 each. In a 24-letter component, decimation interval 5 will also yield a complete equivalent primary component of 24 letters, but decimation interval 4 will yield 6 chains of 4 letters each, and decimation interval 8 will yield 3 chains of 8 letters each. It follows that in the case of an original primary component in which the total number of characters is a prime number, all decimation intervals will yield complete equivalent primary components. Table 12-3 omits the prime number sequences from 16-32. [FRE7] SPECIAL SOLUTIONS FOR PERIODIC CIPHERS Special circumstances give rise atypical solutions of periodic 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 text message enciphered by two different keys in the same cryptosystem. Lets use two monoalphabetic substitution systems to illustrate the point. Assume two messages are intercepted going from station A to B. B had called for a retransmit because of some error in transmission. We suspect the messages are the same plaintext content and they both have the same length. We superimpose one message over the other: 1. NXGRV MPUOF ZQVCP VWERX QDZVX WXZQE TBDSP VVXJK RFZWH 2. EMLHJ FGVUB PRJNG JKWHM RAPJM KMPRW ZTAXG JJMCD HBPKY chaining from 1 to 2: NE>EW>WK>KD>DA ...... 1. ZUWLU IYVZQ FXOAR 2. PVKIV QOJPR BMUSH Next we initiate a chain of ciphertext equivalents (reducing the 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 N E 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 QUESTIONABLY and the decimation interval of +5 Modulo 26. The complete 26 letter chain was available for reconstruction, but this is not a requirement. Why is it possible to reconstruct the primary component and solve the above two messages without having any plain text at all? Since the plain text of both messages is the same, the relative displacement of the same primary components in the case of message 1 differs from the relative displacement of the same primary components in message 2 by a FIXED interval. Therefore, the distance between N and E (1st two cipher letters of the two messages) on the primary component, regardless of what plaintext letter these two cipher letters represent, is the same distance between E and W (18th letters), W and K (17th letters), and so forth. Thus this fixed interval permits the establishing of a complete chain of letters separated by constant intervals and this chain becomes an equivalent primary component. To solve, we take the frequency distributions of message 1 and 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 6 1: 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 1 2: 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 each other. With some trial and error we find: NABLYCDFGHJKMPRVWXZQUESTIO QUESTIONABLYCDFGHJKMPRVWXZ The plain text reads: Five squadrons must be in position by H plus six zero two at Jackson Ridge. The same procedure is applied on two repeating key ciphers suspected of being Isologs: Message 1 YHYEX UBUKA PVLLT ABUVV DYSAB PCQTU NGKFA ZEFIZ BDJEZ ALVID TROQS UHAFK Message 2 CGSLZ QUBMN CTYBV HLQFT FLRHL MTAIQ ZWMDQ NSDWN LCBLQ NETOC VSNZR BJNOQ The first step is to find the length of the period. The usual method fails for lack of long repetitions and the digraphs are not promising. We use the Principle of Superimposition to get a hold on the period for both cryptograms. 1 2 3 4 5 6 7 8 9101112131415161718192021222324252627282930 Y 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 U C 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 Q 313233343536373839404142434445464748495051525354555657585960 N 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 K Z 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 will be employed based upon the theory of factoring. We search for cases of identical superimposition. 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 if they were ordinary repetitions. The most frequent factor should correspond to the period. We are dealing with Isologs. The plain text is the same in both messages, so the principle of identity of superimposition can only be the result of identity of encipherments by identical cipher alphabets. The same relative position in the keying cycle has been reached in both cases of the identity. The distance between identical superimpositions must be equal to or a multiple of the length of the period. The following is the complete 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 as improbable. We break up the message into groups of four. 1234 1234 1234 1234 1234 1234 1234 1234 1. YHYE XUBU KAPV LLTA BUVV DYSA BPCQ TUNG 2. CGSL ZQUB MNCT YBVH LQFT FLRH LMTA IQZW * * * * 1234 1234 1234 1234 1234 1234 1234 1. KFAZ EFIZ BDJE ZALV IDTR OQSU HAFK 2. MDQN SDWN LCBL QNET OCVS NZRB JNOQ We 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 Q 2 N C D G B M Z Q L 3 Q U T O W B E Z C R V F S 4 H L W Q A S B T N ------------------------------------------------------ Using the meyhods previously described, we build up the equivalent 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) H JKM(P) (R) (V) XZ It is not a long jump to a key word QUESTIONABLY and the equivalent 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 was pure chance, it could have been an equivalent primary sequence alphabet. >From here we apply the completion of the plain-component sequence using the high frequency letter assortments. For the first message: Gen Alphabet 1 Alphabet 2 Alphabet 3 Alphabet 4 1 YXKLBDBTKE 1HUALUYPUFF 5YBPTVSCNAI EUVAVAQGZZ 2 2CZMYLFLIMS 4JEBYECREGG 5CLRIWTDABO SEWBWBUHQQ 3 2DQPCYGYOPT 3KSLCSDVSHH 3DYVOXIFBLN TSXLXLEJUU 4 4FURDCHCNRI MTYDTFWTJJ 3FCWNZOGLYA ITZYZYSKEE 5 3GEVFDJDAVO PICFIGXIKK GDXAQNHYCB OIQCQCTMSS 6 2HSWGFKFBWN 4RODGOHZOMM HFZBUAJCDL 5NOUDUDIPTT 7 JTXHGMGLXA VNFHNJQNPP JGQLEBKDFY 8ANEFEFORII* 8 KIZJHPHYZB WAGJAKUARR 1KHUYSLMFGC 6BASGSGNVOO 9 MOQKJRJCQL XBHKBMEBVV 2MJECTYPGHD 5LBTHTHAWNN 10 PNUMKVKDUY ZLJMLPSLWW PKSDICRHJF YLIJIJBXAA 11 4RAEPMWMFEC QYKPYRTYXX RMTFODVJKG CYOKOKLZBB 12 3VBSRPXPGSD UCMRCVICZZ 2VPIGNFWKMH 2DCNMNMYQLL 13 4WLTVRZRHTF EDPVDWODQQ WROHAGXMPJ 2FDAPAPCUYY 14 XYIWVQVJIG 3SFRWFXNFUU XVNJBHZPRK 3GFBRBRDECC 15 ZCOXWUWKOH TGVXGZAGEE ZWAKLJQRVM 1HGLVLVFSDD 16 QDNZXEXMNJ IHWZHQBHSS QXBMYKUVWP 1JHYWYWGTFF 17 UFAQZSZPAK OJXQJULJTT UZLPCMEWXR KJCXCXHIGG 18 EGBUQTQRBM NKZUKEYKII EQYRDPSXZV MKDZDZJOHH 19 3SHLEUIUVLP 5AMQEMSCMOO SUCVFRTZQW PMFQFQKNJJ 20 6TJYSEOEWYR? 4BPUSPTDPNN TEDWGVIQUX RPGUGUMAKK 21 IKCTSNSXCV 8LRETRIFRAA* ISFXHWOUEZ 3VRHEHEPBMM 22 5OMDITATZDW? 3YVSIVOGVBB OTGZJXNESQ WVJSJSRLPP 23 NPFOIBIQFX 3CWTOWNHWLL NIHQKZASTU XWKTKTVYRR 24 5ARGNOLOUGZ? DXINXAJXYY AOJUMQBTIE ZXMIMIWCVV 25 4BVHANYNEHQ FZOAZBKZCC 5BNKEPULIOS QZPOPOXDWW 26 LWJBACASJU GQNBQLMQDD 7LAMSREYONT* UQRNRNZFXX We choose generatrices 20/22/24; 21; 26; 7 because of the highest two category scores. it is not much of a jump to find 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 keyword QUESTIONABLY, we find the key words SOUP for message 1 and TIME 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 A A L L A R R A N G E M E N T S F O R R E L I E F B P C Q T U N G K F A Z E F I Z B D J E Z A L V O F Y O U R O R G A N I Z A T I O N H A V E B E I D T R O Q S U H A F K E N S U S P E N D E D X T 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 H A L L A R R A N G E M E N T S F O R R E L I E F L M T A I Q Z W M D Q N S D W N L C B L Q N E T O F Y O U R O R G A N I Z A T I O N H A V E B E O C V S N Z R B J N O Q E N S U S P E N D E D X Figure 12-1 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 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 Q E S T I O N A B L Y C D F G H J K M P R V W X Z Q U S T I O N A B L Y C D F G H J K M P R V W X Z Q U E T I O N A B L Y C D F G H J K M P R V W X Z Q U E S I O N A B L Y C D F G H J K M P R V W X Z Q U E S T O N A B L Y C D F G H J K M P R V W X Z Q U E S T I N A B L Y C D F G H J K M P R V W X Z Q U E S T I O A B L Y C D F G H J K M P R V W X Z Q U E S T I O N B L Y C D F G H J K M P R V W X Z Q U E S T I O N A L Y C D F G H J K M P R V W X Z Q U E S T I O N A B Y C D F G H J K M P R V W X Z Q U E S T I O N A B L C D F G H J K M P R V W X Z Q U E S T I O N A B L Y D F G H J K M P R V W X Z Q U E S T I O N A B L Y C F G H J K M P R V W X Z Q U E S T I O N A B L Y C D G H J K M P R V W X Z Q U E S T I O N A B L Y C D F H J K M P R V W X Z Q U E S T I O N A B L Y C D F G J K M P R V W X Z Q U E S T I O N A B L Y C D F G H K M P R V W X Z Q U E S T I O N A B L Y C D F G H J M P R V W X Z Q U E S T I O N A B L Y C D F G H J K P R V W X Z Q U E S T I O N A B L Y C D F G H J K M R V W X Z Q U E S T I O N A B L Y C D F G H J K M P V W X Z Q U E S T I O N A B L Y C D F G H J K M P R W X Z Q U E S T I O N A B L Y C D F G H J K M P R V X Z Q U E S T I O N A B L Y C D F G H J K M P R V W Z 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 PRIMARY COMPONENTS BUT WITH KEY WORDS OF DIFFERENT LENGTHS The example previous had two keywords the same lengths. The Method of Superimposition works with Keywords of different lengths. Friedman works an interesting example: Message 1 VMYZG EAUNT PKFAY JIZMB UMYKB VFIVV SEOAF SKXKR YWCAC ZORDO ZRDEF BLKFE SMKSF AFEKV QURCM YZVOX VABTA YYUOA YTDKF ENWNT DBQKU LAJLZ IOUMA BOAFS KXQPU YMJPW QTDBT OSIYS MIYKU ROGMW CTMZZ VMVAJ Message 2 ZGANW IOMOA CODHA CLRLP MOQOJ EMOQU DHXBY UQMGA UVGLQ DBSPU OABIR PWXYM OGGFT MRHVF GWKNI VAUPF ABRVI LAQEM ZDJXY MEDDY BOSVM PNLGX XDYDO PXBYU QMNKY FLUYY GVPVR DNCZE KJQOR WJXRV GDKDS XCEEC. Both messages permit factoring at periods of 4 and 6 letters, respectively. Superimposing the two messages and marking the position of each letter in the corresponding period, we have: 12341 23412 34123 41234 12341 23412 No. 1 VMYZG EAUNT PKFAY JIZMB UMYKB VFIVV No. 2 ZGANW IOMOA CODHA CLRLP MOQOJ EMOQU 12345 61234 56123 45612 34561 23456 34123 41234 12341 23412 34123 41234 No. 1 SEOAF SKXKR YWCAC ZORDO ZRDEF BLKFE No. 2 DHXBY UQMGA UVGLQ DBSPU OABIR PWXYM 12345 61234 56123 45612 34561 23456 12341 23412 34123 41234 12341 23412 No. 1 SMKSF AFEKV QURCM YZVOX VABTA YYUOA No. 2 OGGFT MRHVF GWKNI VAUPF ABRVI LAQEM 12345 61234 56123 45612 34561 23456 34123 41234 12341 23412 34123 41234 No. 1 YTDKF ENWNT DBQKU LAJLZ IOUMA BOAFS No. 2 ZDJXY MEDDY BOSVM PNLGX XDYDO PXBYU 12345 61234 56123 45612 34561 23456 12341 23412 34123 41234 12341 23412 No. 1 KXQPU YMJPW QTDBT OSIYS MIYKU ROGMW No. 2 QMNKY FLUYY GVPVR DNCZE KJQOR WJXRV 12345 61234 56123 45612 34561 23456 34123 41234 No. 1 CTMZZ VMVAJ. No. 2 GDKDS XCEEC. 12345 61234 What is neat about this superimposition is that we can establish secondary alphabets by distributing the letters from the 12 different superimposed pairs of numbers. The 1 - 1 superimposition is placed in the tableau at the 0 - 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 Z 2-2 H V N G U W E D M L X 3-3 E M X G I D J N R A O 4-4 X O C D K A F Y Q V N 1-5 B T W L R E M N Y U A 2-6 M O I C D U V F R 3-1 O G R L P S D Z 4-2 L P H U V E D M F 1-3 Q J V W K O X Y M A 2-4 B J X P O A F Y D 3-5 N R Y B C G Q S 4-6 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 direct sequence. We attempt to solve and fail. Normal reverse will also fail. We assume a K3 situation, i.e. the plain and cipher components are identical. Again the test fails. We assume that the plain is in reverse mode. Nope. So we have a K4 situation, both primary components are different mixed sequences. Message 1 transcribed into periods of four letters. Message 1 VMYZ GEAU NTPK FAYJ IZMB UMYK BVFI VVSE OAFS KXKR YWCA CZOR DOZR DEFB LKFE SMKS FAFE KVQU RCMY ZVOX VABT AYYU OAYT DKFE NWNT DBQK ULAJ LZIO UMAB OAFS KXQP UYMJ PWQT DBTO SIYS MIYK UROG MWCT MZZV MVAJ The Uniliteral frequency distributions for the four secondary alphabets are shown in 1A -4A. We have the reconstructed cipher 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 1 1A 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 4 2A 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 2 3A 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 1 4A 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 1 1B 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 2 2B 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 7 3B 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 1 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 Y We now shift 1B-4B for superimposition and combine the distributions. The latter distributions may be combined so as to yield a single monoalphabetic distribution for the entire message. In other words, the polyalphabetic message can be converted into monoalphabetic terms, and thereby simplifying the situation considerably. 1 3 2 1 1 4 1 5 2 2 1 2 1 1 1 5 3 3 1 1B 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 2 2B 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 7 3B 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 4 4B 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 1 1B-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 Y combinedH M L R S O A I Y N E T Plain Equiv's I have converted 2B-4B into terms of 1B. The 2 E's of 2B add to 1B I. The two K's of alphabet 3 becomes I's and the N becomes a T, and so forth. We solve the monoalphabetic cipher. 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 or equivalent plain component is straightforward. We may base the reconstruction upon any of the secondary alphabets, since the 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 V with Key words of STAR and OCEANS for messages 1 and 2. NECESSARY AND SUFFICIENT CONDITIONS FOR SUPERIMPOSITION AND CONVERSION TO MONOALPHABETIC TERMS This example shows the power of the method of superimposition and conversion of a polyalphabetic cipher to monoalphabetic terms. This conversion is possible because the sequence of letters forming the cipher component has been reconstructed and was known, and the uniliteral distributions for the respective secondary cipher alphabets could theoretically be shifted to correct superimpositions for monoalphabeticity. The data was sufficient to give proper indications for alignment of the alphabets and relative displacements. The chi test could also have been brought to bear to match columns. The above constitutes the necessary and sufficient conditions to convert theory to actuality. SOLUTION OF ISOLOGS INVOLVING DIFFERENT PAIRS OF UNKNOWN PRIMARY COMPONENTS The principle of superimposition continues to work for us even when the primary components are different, and the repeating keys are of different lengths. There are two general attacks. The first is a slight modification of the procedures previously discussed. We first factor the messages, then superimpose the messages on a width of the least common multiple, then create a reconstruction matrix based on the cipher values. We must limit our observations to within the matrix, because the given messages are different and therefore the indirect symmetry does not extend to the 0 or assumed plain line. The wrinkle in the fabric is we must restrict our observations to a homogeneous set of lines, like 1-1,1-2,1-3,1-4 etc. From this data, we reduce the reconstruction matrix to a smaller set and solve for the equivalent primary component. It is possible to invert the matrix so that values for the second message will yield its equivalent primary component. ARBITRARY REDUCTION METHOD It is not necessary to recognize the plain text to solve a problem involving Isologs. The next cryptanalytic attack is applicable for many types of ciphers. The procedure exposes latent letter relationships and reduces the imposed chaos of the 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 the least 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 first letter of the plain text. Since in message 1, B(cipher)= A(plain), then every B(cipher) in alphabet 1 must equal A(plain); these values are entered in the table above. Also the 65th and 73rd letter of message 1 are A(plain), this establishes that in message 2, G(cipher) in alphabet 5 and F(cipher) in alphabet 3 are also A(plain); we enter these values. Similarly, every J(cipher) in alphabet 1 of message 2 equals A(plain). We continue the process and recover all the A(plains) of the pseudo-plain text with the resulting worksheet 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) of message number 1 establishes the E(cipher) of message 2 also as a B(plain). This procedure of arbitrary assignments is continued until all the cipher letters of alphabet 1 of message 1, are placed. we are able to reduce most of the text 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 idiomorphic repetition ACHDIIFC representing Artillery becomes patent in the reduction process. This is rather exciting. From no patent 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 represents the pseudo-plain text and the values inside the matrix being cipher 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 E 2 L Q W K S E B Z O H C X 3 U P V Q B C X N S I W 4 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 C 2 S E H U W A Z I V N X 3 F U C A M L H K B G 4 I T K E S Z U N A J B Y Q 5 G F E C D B Y J A U M L ------------------------------------------------------ >From the above we chain out the equivalent primary components used for each message. Having reconstructed the cipher component for each message, the alphabets are aligned, combined and reduced to monoalphabetic terms. After solution of these messages, we find message 1 is a case of direct symmetry with the cipher component based on the keyword HYDRAULIC, and message 2 is a case of indirect symmetry with both components being keyword-mixed sequences based on our favorite keyword QUESTIONABLY. Friedman points out that the keywords are prime to each other (9 vs 11). Primality is not a necessary condition for solution based on this procedure. [FRE7] The method of Arbitrary Reduction is very powerful and works in other ares besides solving periodic polyalphabetic ciphers. It represents a workable approach where the cryptosystem involves nonrelated, random-mixed secondary alphabets 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 two isologous initial fragments which are identical except by one letter (called a 'stagger') we can solve the isologous portions of the messages and recover the primary cipher component by the process of indirect symmetry. Transmission garble usually creates stagger messages. Machine cipher systems sometimes produce these when a word separator is added. Staggers may be progressively larger as further word separators are omitted or added. Given: Message A * * ZFWAY ITBVX XWZQV PEBGS GGFIZ TUAMF RFEQX PEPPO PCNBP QPOTX VNAIH HVRXC NHVGM FRFSI ESQMV * Message B * * ZFWAY ITBVX XWZQV PDRKF USVAG XLJKC NDVPR OWBRH YFJMS HRFVS BAHWG ZFAJO JMFAV CNDVD ORZPH A * We note that both messages have the same 16 letter beginnings and that message B is 1 letter longer than message A. Note that the tetragraphs MFRF (29) and (65) are spaced 1 less letter than CNDV at (30) and (66). The D in position 17 of message 2 is the extra letter. Starting from the E in position 17 of message 1, we superimpose message one over message 2 starting at the R in position 18. [We use a period of 6 because the tetragraph delta equals 36 which factors into 3,4,6 and 9; 6 is confirmed via the message.] 56123456123456123456123456123456123456123456123456123456123 EBGSGGFIZTUAMFRFEQXPEPPOPCNBPQPOTXVNAIHHVRXCNHVGMFRFSIESQMV RKFUSVAGXLJKCNDVPROWBRHYFJMSHRFVSBAHWGZFAJOJMFAVCNDVDORZPHA 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 X 2-3 S V F H R U L B 3-4 P S H D J A 4-5 K V O H Y R J 5-6 W R A C F O 6-1 K J N G V W Z ------------------------------------------------------- It is fairly easy to align properly the cipher components after the primary cipher component or its equivalent have been recovered, thereby expediting the reduction of the cipher 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) alphabet 2;P(cipher) of alphabet 4 is under E(cipher) of alphabet 1. From this point on solution follows the normal path of reconstruction, keyword recovery and combination of alphabets, reduction to monoalphabetic terms and solution by frequency analysis. LONG LATENT REPETITIONS The stagger procedure applies to a periodic cryptogram which contains a long passage repeated in its plain text, the second occurrence occurring at a point in the keying cycle different from the first occurrence. If the passage is long enough, the equivalencies from the two corresponding sequences may be chained together to yield an equivalent primary component. In effect, we by-pass the solution by frequency analysis or making assumptions in the plain text of a 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 minor consideration. It paves the way for the second step, which consists of allocating the letters of the cryptogram into individual monoalphabetic distributions. The third step is to solve these distributions. The text is transcribed into its periods and written out in successive lines corresponding to the length of the period. The columns of letters as a series belong to the same monoalphabet. We also can see the letters as transcribed into superimposed periods; in such a case the letters in each column have undergone 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 short message ( few cycles in the text) we have a difficult problem. But supposing there were several short cryptograms enciphered by the same key, each message beginning at identical starting points in the key. We can superimpose these messages "in flush depth" or "head on" and know that 1) the letters in the columns belong to the same individual alphabets, 2) and that if there are enough messages (about 25-30 in English), then the frequency distributions applicable to the successive columns of text can be solved - without knowing the length of the key. Any difficulties that may have arisen because we were not able to factor the problem correctly are circumvented. The second step of the normal solution to the problem is by-passed. The assumption of probable initial words of messages and stereotyped beginnings is a powerful method of attack in such situations. Since the superimposed texts in these cases comprise only the beginnings of messages, assumptions of probable words are more easily made than when words are sought in the interior of the messages. Some common introductory words are REQUEST, REFER, ENEMY, WHAT, WHEN, and SEND. High frequency initial digraphs will manifest themselves in the first two columns of the superimposed diagram. The high frequency RE diagram manifests 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 the messages start at different initial points, providing the messages can be correctly superimposed, so that the letters which fall in one column really belong to one cipher alphabet. The superimposed messages are said to be "in depth." The chi test may be used to advantage in finding and combining columns of the superimposed diagram which were enciphered by identical keys, thus assisting in the analysis of frequencies of larger samples than were available before the amalgamation. [FRE7] CONCLUSION In summary, we have seen that the chaining process between cipher texts applies to the latent characteristics of the cipher components, regardless of the identity of the plain components and regardless whether direct or indirect symmetry is involved in the cryptosystems. The principle of super- imposition ranks as one of the most important principles of cryptanalysis. 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 Substitution 74 46 66 44 79 47 45 37 58 66 37 60 25 54 33 69 78 35 68 27 47 36 28 88 36 60 33 48 43 29 87 35 49 57 76 37 37 88 36 60 33 77 74 50 86 55 47 27 76 45 40 55 56 58 66 78 57 30 94 58 38 26 55 57 59 88 56 79 46 46 66 60 58 55 48 56. (DGGLWLRQ, ends WXEOIW) 12.2 Nihilist Substitution 38 76 54 76 64 76 76 54 74 55 35 76 77 76 47 58 76 85 74 44 65 88 63 74 47 36 95 74 63 44 37 58 57 96 65 36 66 85 74 63 55 79 53 67 57 56 58 64 67 67 56 67 57 74 55 55 57 86 03 43 46 67 73 96 67 39. (ETARVQITCO, ends HSMX) 12.3 PORTA QLAMU CHQGO FTESV XKEWC GMXPH UCLUS WSGXT EVURH TMTSU TKVSQ GCQCW LHMDX NUFUE EFXRF XPHUN RGPKC OXULB BBCUS IBBHW. (HAVE) 12.4 PORTA XFXYW ZJICZ IBUZN HJXEA ACWBE JOOCZ UPXFQ BXHFI CGMAZ KVQEG BBCAF KLLXF BVOUN TSAYZ KKXLR CWAJC LVVVI XNBFQ JVWBW BSWEY VUNGX ODFRZ PTEWO PJQNH WZPNA YRCLV YYWCQ ULOJB VK. (GSRWXERX) 12.5 PORTAX UXCUD ZMVBA FWWPV DIKDO JISMA WRBBA YLOYX AKUXR JGDCJ MYAPV RJWJA DMUKL KLUAM KAOEN YBFCC IQGFK QZAA. (PQXKEG) 12.6 PORTAX WWQPE JBDTM TMNWH CTJSW WKIAC BJKWL YHBYN OAKRZ PDYZM DIVGB QKNJP RNSRU FXWMU TKMJS KDNLW WFHKR JSCVF HTJIS JD. (UHDOLCH) 12.7 GROMARK HPMZU IBQHI SDHHH JKUNC OYJSC 24106 RBLOF REXTG EXAZA ILAXX XHFNH CDUYQ YUOMQ NVOIN XYMBR WAHNT FGPFB DOOMA CWHDH JXTTX CJIUR PVMZR EILDZ QJJTT ILNNP TREVL BQLL. ( tip: UCAUKYKUJK; ends tivpw.) REFERENCES / RESOURCES [updated 30 May 1996] [ACA] ACA and You, "Handbook For Members of the American Cryptogram Association," ACA publications, 1995. [ACA1] Anonymous, "The ACA and You - Handbook For Secure Communications", American Cryptogram Association, 1994. 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[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. 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[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. P., "Proposed Day, Night and Fog Signals for the Navy with Brief Description of the Ardois Hight System," In Proceedings of the United States Naval Institute, Annapolis: U. S. Naval Institute, 1891. [NIC1] Nichols, Randall K., "Xeno Data on 10 Different Languages," ACA-L, August 18, 1995. [NIC2] Nichols, Randall K., "Chinese Cryptography Parts 1-3," ACA-L, August 24, 1995. [NIC3] Nichols, Randall K., "German Reduction Ciphers Parts 1-4," ACA-L, September 15, 1995. [NIC4] Nichols, Randall K., "Russian Cryptography Parts 1-3," ACA-L, September 05, 1995. [NIC5] Nichols, Randall K., "A Tribute to William F. Friedman", NCSA FORUM, August 20, 1995. [NIC6] Nichols, Randall K., "Wallis and Rossignol," NCSA FORUM, September 25, 1995. [NIC7] Nichols, Randall K., "Arabic Contributions to Cryptography,", in The Cryptogram, ND95, ACA, 1995. [NIC8] Nichols, Randall K., "U.S. Coast Guard Shuts Down Morse Code System," The Cryptogram, SO95, ACA Publications, 1995. [NIC9] Nichols, Randall K., "PCP Cipher," NCSA FORUM, March 10, 1995. [NICX] Nichols, R. K., Keynote Speech to A.C.A. Convention, "Breaking Ciphers in Other Languages.," New Orleans, La., 1993. [NICK] Nickels, Hamilton, "Codemaster: Secrets of Making and Breaking Codes," Paladin Press, Boulder, CO., 1990. [NIHL] PHOENIX," Computer Column: Nihilist Substitution," MA88, The Cryptogram, American Cryptogram Association, 1988. [NIH1] PHOENIX," Computer Column: Nihilist Substitution," MJ88, The Cryptogram, American Cryptogram Association, 1988. [NIH2] PHOENIX," Computer Column: Nihilist Substitution," JA88, The Cryptogram, American Cryptogram Association, 1988. [NIH3] PHOENIX," Computer Column: Nihilist Substitution," JA89, The Cryptogram, American Cryptogram Association, 1989. [NIH4] FIDDLE and CLEAR SKYS," FIDDLE'S slide for Nihilist Number Substitution," ON48, The Cryptogram, American Cryptogram Association, 1948. [NIH5] RIG R. MORTIS," Mixed Square Nihilist," JA60, The Cryptogram, American Cryptogram Association, 1960. [NIH6] PICCOLA," Nihilist Number Cipher," AS37, The Cryptogram, American Cryptogram Association, 1937. 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Adleman, "A Method for Obtaining Digital Signatures and Public Key Cryptosystems," Communications of the ACM 21, 1978. [ROAC] Roach, T., "Hobbyist's Guide To COMINT Collection and Analysis," 1330 Copper Peak Lane, San Jose, Ca. 95120- 4271, 1994. [ROBO] NYPHO, The Cryptogram, Dec 1940, Feb, 1941. [ROHE] Jurgen Rohwer's Comparative Analysis of Allied and Axis Radio-Intelligence in the Battle of the Atlantic, Proceedings of the 13th Military History Symposium, USAF Academy, 1988, pp 77-109. [ROHW] Rohwer Jurgen, "Critical Convoy Battles of March 1943," London, Ian Allan, 1977. [ROH1] Rohwer Jurgen, "Nachwort: Die Schlacht im Atlantik in der Historischen Forschung, Munchen: Bernard and Graefe, 1980. [ROH2] Rohwer Jurgen, et. al. , "Chronology of the War at Sea, Vol I, 1939-1942, London, Ian Allan, 1972. [ROH3] Rohwer Jurgen, "U-Boote, Eine Chronik in Bildern, Oldenburs, Stalling, 1962. Skizzen der 8 Phasen. [ROOM] Hyde, H. Montgomery, "Room 3603, The Story of British Intelligence Center in New York During World War II", New York, Farrar, Straus, 1963. [ROSE] Budge, E. A. Wallis, "The Rosetta Stone," British Museum Press, London, 1927. [RSA] RSA Data Security, Inc., "Mailsafe: Public Key Encryption Software Users Manual, Version 5.0, Redwood City, CA, 1994 [RUNY] Runyan, T. J. and Jan M. Copes "To Die Gallently", Westview Press 1994, p85-86 ff. [RYP1] A B C, "Adventures in Cryptarithms (digital maze)," JA63, The Cryptogram, published by the American Cryptogram Association, 1963. [RYP2] CROTALUS "Analysis of the Classic Cryptarithm,"MA73, The Cryptogram, published by the American Cryptogram Association, 1973. [RYP3] CLEAR SKIES "Another Way To Solve Cryptarithms,"DJ44, The Cryptogram, published by the American Cryptogram Association, 1944. [RYP4] CROTALUS "Arithemetic in Other Bases (Duodecimal table),"JF74, The Cryptogram, published by the American Cryptogram Association, 1974. [RYP5] LEDGE, "Basic Patterns in Base Eleven and Twelve Arithmetic,"SO77, ND77, The Cryptogram, published by the American Cryptogram Association, 1977,1977. [RYP6] COMPUTER USER, "Computer Solution of Cryptarithms," JF72, The Cryptogram, published by the American Cryptogram Association, 1972. [RYP7] PIT, "Cryptarithm Crutch," JA80, The Cryptogram, published by the American Cryptogram Association, 1980. [RYP8] DENDAI, DICK, "Cryptarithm Ccub root," ND76, The Cryptogram, published by the American Cryptogram Association, 1976. [RYP9] S-TUCK, "Cryptarithm in Addition," AM44, The Cryptogram, published by the American Cryptogram Association, 1944. [RYPA] APEX DX, "Cryptarithm Line of Attack," ND91, The Cryptogram, published by the American Cryptogram Association, 1991. [RYPB] HUBBUBBER and CROTALUS, "Cryptarithm Observations," ND73, The Cryptogram, published by the American Cryptogram Association, 1973. [RYPC] CROTALUS, "Cryptarithms and Notation," JF73, The Cryptogram, published by the American Cryptogram Association, 1973. [RYPD] JUNKERL, "Cryptarithms: The digital root method," AS43, The Cryptogram, published by the American Cryptogram Association, 1943. [RYPE] CROTALUS, "Divisibility by Eleven," ND89, The Cryptogram, published by the American Cryptogram Association, 1989. [RYPF] S-TUCK, "Double Key Division," JJ43, The Cryptogram, published by the American Cryptogram Association, 1943. [RYPG] NEOTERIC, "Duo-Decimal Cryptarithms," AM40, The Cryptogram, published by the American Cryptogram Association, 1940. [RYPH] QUINTUPLEX, "Duo-Decimal Cryptarithms," JJ40, The Cryptogram, published by the American Cryptogram Association, 1940. [RYPI] FIDDLE, "Exhausitive for Three," JF59, The Cryptogram, published by the American Cryptogram Association, 1959. [RYPJ] ---, "Finding the Zero In Cryptarithms," DJ42, The Cryptogram, published by the American Cryptogram Association, 1942. [RYPK] FILM-D, "Greater than Less than Diagram for Cryptarithms," DJ51, The Cryptogram, published by the American Cryptogram Association, 1951. [RYPL] MI TI TI, "Introduction To Cryptarithms," SO63, The Cryptogram, published by the American Cryptogram Association, 1963. [RYPM] FORMALHUT, "Leading Digit Analysis in Cryptarithms," JA91, The Cryptogram, published by the American Cryptogram Association, 1991. [RYPN] CROTALUS, "Make Your Own Arithmetic Tables In Other Bases," MJ89, The Cryptogram, published by the American Cryptogram Association, 1989. [RYPO] BACEDI, "Method for Solving Cryptarithms," JF78, The Cryptogram, published by the American Cryptogram Association, 1978. [RYPP] SHERLAC, "More on Cryptarithms," DJ44, The Cryptogram, published by the American Cryptogram Association, 1944. [RYPQ] FIRE-O, "Multiplicative Structures," MJ70, The Cryptogram, published by the American Cryptogram Association, 1970. 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