Welcome back from the Thanksgiving holiday break. The goodnews is that this lecture will come to you about Christmas,therefore, no homework. The not so good news is that thisconcluding Lecture 4 on Substitution with Variants covers somedifficult material of wide practically in the field.
In Lecture 4, we complete our look into English monoalphabeticsubstitution ciphers, by describing multiliteral substitutionwith difficult variants. The Homophonic and GrandPre Cipherswill be covered. The use of isologs is demonstrated. Asynoptic diagram of the substitution ciphers described inLectures 1-4 will be presented.
Each English letter in plain text has a characteristicfrequency which affords definite clues in the solution ofsimple monoalphabetic ciphers. Associations which individualletters form in combining to make up words, and thepeculiarities which certain of them manifest in plain text,afford further direct clues by means of which ordinarymonoalphabetic substitution encipherments of such plain textmay be readily solved. [FR1]
Cryptographers have devised methods for disguising,suppressing, or eliminating the foregoing characteristics inthe cryptograms produced by methods described in Lectures 1-3.One category of methods called "variants or variant values" isthat in which the letters of the plain component of a cipheralphabet are assigned two or more cipher equivalents.
Systems involving variants are generally multiliteral. In suchsystems, there are a large number of equivalents made availableby combinations and permutations of a limited number ofelements, each letter of the plain text may be represented byseveral multiliteral cipher equivalents which may be selectedat random. For example, if 3-letter combinations are employedas multiliteral equivalents, there are 263 or 17,576available equivalents for the 26 letters of the plain text.
They may be assigned in equal numbers of different equivalentsfor the 26 letters, in which case each letter would berepresentable by 676 different 3 letter equivalents or theybe assigned on some other basis, for example proportionately tothe relative frequencies of the plain text letters. [FR1]
The primary object of substitution with variants is again toprovide several values which may be employed at random in asimple substitution of cipher equivalents for the plain textletters.
As a slight diversion, the reader may ask about uniliteralsubstitution with variants. It is but not very practical.Note the following cipher alphabet constructed in French byCaptain Roger Baudouin in reference [BAUD]:
(Note that the Captain was not an ACA member. The H=Hcombination is not allowed.)
Baudouin proposed that the J and Y plain be replaced by I plainand K plain by C plain or Q plain and W plain by VV plain. Fourcipher letters would be available as variants for the high-frequency plain text letters in French.
Mixed alphabets formed by including all repeated letters of thekey word or key phrase in the cipher component were common inEdgar Allen Poe's day but are impractical because they areambiguous, making decipherment difficult; for example:
Enciphering Alphabet:
Inverse form for deciphering:
The average cipher clerk would have difficulty in decrypting acipher group such as TOOET, each letter having 3 or moreequivalents, from which plain text fragments (n)inth, ftthi(s), it thi, etc. can be formed on decipherment. [FR1]
In simple or single-equivalent monoalphabetic substitution withvariants, two points are evident:
2) The same character or cipher unit of the cryptogram invariably represents one and always the same letter of the plain text.
2) The same character or cipher unit of the cryptogram nevertheless invariably represents one and always the same letter of the plain text.
Figure 4-1
Figure 4-2
Figure 4-3
Figure 4-5
Figure 4-6
Figure 4-7
Figure 4-8
Figure 4-9
Figure 4-10
The matrices in Figures 4 -1 to 4-10 represent some of thesimpler means for accomplishing monoalphabetic substitutionwith variants. The matrices are extensions of the basic ideasof multiliteral substitution presented in Lecture 3.
The variant equivalents for any plain text letter may be chosenat will; thus, in Figure 4-1, e= 10, 15, 60, or 65; in Figure4-2, e= AU, AZ, FU, FZ, LU or LZ.
Encipherment by means of matrices shown in Figures 4-2, 4-3,4-6 is commutative. The coordinates may be read row by columnor visa versa. There is no cryptographic ambiguity. Theremaining matrices are noncommutative. The general conventionis to read row by column.
In Figures 4-5 and 4-6, the letters in the square have beeninscribed in such a manner that, coupled with the particulararrangement of the row and column coordinates, the number ofvariants available for each plain text letter is roughlyproportional to the frequencies of the letters in theplain text. Figure 35 incorporates a keyword on top of thisidea. [FR1]
The Homophonic Cipher is a simple variant system. It is a4-level (alphabets) dinome cipher. Consider Figure 4-11.
Figure 4-11
The keyword TRIP is found by inspecting dinomes 01, 26, 51, and76. (The lowest number in each of the four sequences.)[FR1] [FR5]
The Russians added an interesting gimmick called the DisruptionArea. Consider Figure 4-12 and note the slashes under U - Xfor the fourth level of dinomes. The famous VIC cipher usedthis feature very effectively. [NIC4]
Figure 4-12
The keyword NAVY is represented by dinomes 01, 27, 53, and 79.
Security for Homophonic systems is greatly improved if thedinomes and the four sequences are assigned randomly. However,the easy mnemonic feature of the keyworded four sequences islost.
The Mexican Cipher device is a Homophonic consisting of fiveconcentric disks, the outer disk bearing 26 letters and theother four bearing sequences 01-26, 27-52, 53-78, 79-00.The cipher disk enhances frequent key changes. Figure 4-12shows the matrix without the disruption area. [FR5] [NIC4]
Lets solve the following cryptogram.68321 09022 48057 65111 88648 42036 45235 0914405764 22684 00225 57003 97357 14074 82524 4076851058 93074 92188 47264 09328 04255 06186 7988285144 45886 32574 55136 56019 45722 76844 6835045219 71649 90528 65106 11886 44044 89669 7055318491 06985 48579 33684 50957 70612 09795 2914856109 08546 62062 65509 32800 32568 97216 4428234031 84989 68564 53789 12530 77401 68494 3854411368 87616 56905 20710 58864 67472 22490 0913662851 24551 35180 14230 50886 44084 06231 1287605579 58980 29503 99713 32720 36433 82689 0451652263 21175 06445 72255 68951 86957 76095 6721553049 08567 9730
Assuming we did not know that the above cryptogram was aHOMOPHONIC, we might make a preliminary analysis to see if weare dealing with a cipher or a code. We will cover codesystems later in the course, but a few introductory remarksmight be in order. The five letter groups could indicateeither a cipher or a code.
If the cryptogram contains an even number of digits, as forexample 494 in the previous message, this leaves open thepossibility that the message is a cipher containing 247 pairsof digits; were the number of digits an exact odd multiple offive, such as 125, 135, etc., the possibility that thecryptogram is in code of the 5-figure group type must beconsidered.
We next study the message repetitions and what theircharacteristics are. If the cipher text is of 5-figure codetype, then such repetitions as appear should generally be inwhole groups of five digits, and they should be visible in thetext just as the message stands, unless the code message hasbeen superenciphered. If the cryptogram is a cipher, thenrepetitions should extend beyond the 5-digit groupings; if theyconform to any definite at all they should for the most partcontain even numbers of digits since each letter is probablyrepresented by a pair (dinome) of digits.
We start with 4-part frequency distribution. We next assumea 25 character alphabet from 01-00. This is the common schemeof drawing up the alphabets. Breaking the text into dinomes(2-digit) pairs yields:
What we have before us are four simple, monoalphabeticfrequency distributions similar to those involved in amonoalphabetic substitution cipher using standard cipheralphabets. The next step is to fit the distribution to thenormal. Since I=J for the 25 letter alphabet, we find thatthe Keyword is JUNE and the following alphabets result:
The first groups of the cryptogram decipher as follows:01 I-J 26 U 51 N 76 E
02 K 27 V 52 O 77 F
03 L 28 W 53 P 78 G
04 M 29 X 54 Q 79 H
05 N 30 Y 55 R 80 IJ
06 O 31 Z 56 S 81 K
07 P 32 A 57 T 82 L
08 Q 33 B 58 U 83 M
09 R 34 C 59 V 84 N
10 S 35 D 60 W 85 O
11 T 36 E 61 X 86 P
12 U 37 F 62 Y 87 Q
13 V 38 G 63 Z 88 R
14 W 39 H 64 A 89 S
15 X 40 IJ 65 B 90 T
16 Y 41 K 66 C 91 U
17 Z 42 L 67 D 92 V
18 A 43 M 68 E 93 W
19 B 44 N 69 F 94 X
20 C 45 O 70 G 95 Y
21 D 46 P 71 H 96 Z
22 E 47 Q 72 IJ 97 A
23 F 48 R 73 K 98 B
24 G 49 S 74 L 99 C
25 H 50 T 75 M 00 D
If a 26-element alphabet were used only the distributionanalysis would have been changed to be on a basis of 26, theprocess of fitting the distribution to the normal would be thesame.
Suppose we know that two correspondents have been using thesame variant system as in the previous Homophonic.The message intercepted is:48226 88423 52099 93604 76059 05651 36683 5226797114 54466 76
A variation of the plain-component completion method can beused to crack the new message. We copy the message intodinomes and separate by levels.