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

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

• Lesson 1
• Lesson 2
• Lesson 3
• Lesson 4
• Lesson 5
• Lesson 6
• Lesson 7
• Lesson 8
• Lesson 9
• Lesson 10
• Lesson 11
• Lesson 12
• # CLASSICAL CRYPTOGRAPHY COURSE

BY LANAKI

March 5, 1996

Revision 0
LECTURE 11
POLYALPHABETIC SUBSTITUTION SYSTEMS II

CRYPTANALYSIS OF VIGGY'S FAMILY

## SUMMARY

In Lectures 11-12, we continue our course schedule with a studyof fascinating cipher systems known as the "Viggy" based onmultiple alphabets - Polyalphabetic Substitution systems.

We will continue developing our subject via an overview basedon the Op-20-GYT course notes (Office of Chief Of NavalOperations, Washington) [OP20]. We will revisit polyalphabeticcipher systems using Friedman's detailed analysis. We willcover the Viggy, Variant, PORTA systems and other familymembers. [FRE4], [FRE5], FRE6], [FRE7], [FRE8]. We will takematerial from ACA's Practical Cryptanalysis Volume V by WilliamG. Bryan on "Periodic Ciphers - Miscellaneous: Volume II"[BRYA] and Sinkov's [SINK] text to discover Viggy's secrets.We will look at [ELCY's] treatment of these systems.

In Lecture 12, we will describe the difficult aperiodicpolyalphabetic case and give a diagram of topics considered inLectures 10 - 12. [FR3] We will complete the Viggy family.I will also cover decimation processes in detail.

I have again updated our Resources Section with many referenceson these systems - focusing on the cryptanalytic attack andareas of historical interest. Kahn has some wonderful storiesabout the Viggy family. [KAHN]

## ZEN CRYPTO

In Lectures 1- 10, I have purposely stayed away from theheavier mathematics of cryptography (subject to change).Everything I am presenting can and has been reduced tomathematical models and computerized for ease of work. For myreaders who can not live without the math diet, there areplenty of guru' s like [SCHN] and [SCH2] to have breakfastwith. There are plenty of computer aids at the Crypto Drop Boxto help you do the setup work.

BUT those who embark on a course of "only the computer" do thiswithout knowing the real effort -the brain power - theshortcuts - the tradecraft - the historical implications, inmy opinion, have lost the real heart of Cryptography. The "ahha's" of inspiration are what make the difference. First, thereis a fundamental problem in that computer models do not applyto all variant cases. Simple changes to the system can fooleven the most adept computer program. For example, placingclever nulls will defeat many a statistical based model.

Second, we lose the sense of urgency that was required forwartime cryptography. If President Kennedy's Playfair message[that's right it was not English as in the movie PT-109] onthe back of a coconut had been intercepted and deciphered bythe Japanese [which they very capable of doing], we might nothave had the graceful light of his Presidency or who knows themoon landings. As another case in point, the solution ofENIGMA during the mid - final Atlantic Campaigns of World WarII, reduced the operational effectiveness of the U-Boat to oneday and hence saved allied tonnage and warships supplingEurope. The American and British Crypee's 'thought' more liketheir German counterparts than their counterparts. Computersolutions were bulky, machine dependent [the solution "stops"]and not reliable until 1945. People made the difference.

## SOLVING A PERIODIC POLYALPHABETIC CIPHER

There are three fundamental steps to solve a Periodic cipher.

1) Determine the period. This sets up the correct geometrical positioning of ciphertext alphabets.
2) Identify the Cipher System and reduce or consolidate the multiple alphabet distribution into a series of monoalphabetic frequency distributions.
3) Solve the monoalphabetic distributions by known principles. We have covered this in Lectures 1-3 and Lecture 10.

Friedman presents a more detailed and eloquent version of thisprocedure in [FR7].

## THE LONG AND SHORT OF KASISKI

Step one is finding the period. Bryan reminds us that thereare at least two ways to find the period. The short approachmakes use of the distances between patent cipher textrepetitions and factors the differentials. The long approachis used when there are no patent repetitions to factor. Inthis case we set up a possibilities matrix and factor everycombination looking for the highest probable common factor.[BRYA]

As an example of the first case take:

`          10            20            30            40BGZEY  DKFWK  BZVRM  LUNYB  QNUKA  YCRYB  GWMKC  DDTSP          50            60            70            80OFIAK  OWWHM  RFBLJ  JQFRM  PNIQA  VQCUP  IFLAZ  HKATJ          90          100            110          120UVVQE  EKESZ  DUDWE  KKESL  IZQAT  SBYUZ  UUVAZ  IXYEZ         130          140JFTAJ  EMRAS  QKZSQ  FOPHM  W.`
We tabulate the repetitions and the cipher text letterdifferences between repetitions.
`Delta        FactorsBG 29          -RM 45         3,5,9KA 53          -MR 77         7,11QA 39         3,13VQ 17          -AZ 40         4,5,8,10AT 26         13UV 31         -EK 9          3,9KES 10        5,10      .... this trigraph more importantSQ 4          4              than QA or AT digraphs.                             Suggest that the period is                             either 5 or 10. Practice dictates                             that the larger number is the                             proper.`
But suppose there are no repeats or those that do exist do notestablish a period. What then?
`Given:         10             20           30            40RNQJH  AUKGV  WGIVO  BBSEJ  CRYUS  FMQLP  OFTLC  MRHKB         50            60            70             80BUTNA  WXZQS  NFWLM  OHYOF  VMKTV  HKVPK  KSWEI  TGSRB         90            100           110           120LNAGJ  BFLAM  EAEJW  WVGZG  SVLBK  IXHGT  JKYUC  HLKTUMWWK.`
We set up the following vertical tally. We note theactual position of every letter.
`A  6 45 83 89 92 115B  16 17 40 41 80 86 104C  21 35D  ---E  19 74 91 93F  26 32 52 60 87G  9 12 77 84 98 100 109H  5 38 57 66 108 116I  13 75 106J  4 20 85 94 111K  8 39 63 67 70 71 105 112 118 124L  29 34 54 81 88 103 117M  27 36 55 62 90 121N  2 44 51 82O  15 31 56 59P  30 69Q  3 28 49R  1 22 37 79S  18 25 50 72 78 101T  33 43 64 76 110 119U  7 24 42 114 120V  10 14 61 65 68 97 102W  11 46 53 73 95 96 122 123X  107Y  23 47 58 113Z  48 99`
Now we take each difference and every difference in each case.For example, A45-6, 83-6,89-6,92-6,115-6; and 83-45,89-45,92-45,115-45; and 89-83,92-83,115-83; and 92-89,115-89, and 115-92. Then we factor these differences, setting up a matrix(Table 11-1) of potential periods from 3-12 inclusive andtotal the tabulations for each factor in each of the letters ofthe alphabet. The highest column total represents the period.The number is correct more than 98 per cent of the time.

`             Table 11-1     3  4  5  6  7  8  9  10  11  12     -------------------------------A    3  1     1  1     1   1   2   1B    9  7  4  5  3  7  4   2   1   2C       1  1     1  1      1DE    1  1  1  1         1      1   1F    2  3  3  1  2  1   1   1  1G    5  5  4  1  4  3   2   1  3   1H    6  3  2  2  3  1   1   2  1I    1J    3  1  2  1  1  1   3   1K    13 10 4  9  8  5   3   1   2   3L    4  3  4  1  4  1   3   1   2M    4  2  3  2  6      3   1   1N    1  1  1  1  3  1       1O    1  3  1     1  1           1P    1Q    1     1     1R    5  1  1  3  2      1           1S    4  4  2  3  2  1   1   1   1T    4  3  1  1  2      1   1   2   2U    5  1  2  5  1  2   3   1   2   2V    5  6  2  2  1  2   3       1   1W    9  4  5  3  8  1   4   4   3   1XY    2  2  3  2  1  2       1   3   1Z    1     ---------------------------------     87  61 47  43  57  30  35  21  25  16    Columns totalX     3  4   5   6   7   8   9  10  1  112     times period     ----------------------------------    261 244 235 258 399 240 315 210 275 192    Total                    ===The period is 7.`

## WHAT CIPHERS MAKE UP THE VIGGY FAMILY?

The Viggy (or more correctly the Vigenere) Family is group ofciphers. Included in this group are: Vigenere, Variant,Beaufort, Gronsfeld, Porta, Portax, and Quagmires I-IV.Other ciphers may be included in the group. They are NihilistSubstitution, Auto - Key, Running Key and Interrupted ciphers.Bryan includes the Tri-square, the periodic FractionatedMorse, the Seriated Playfair and the Homophonic in the sameclass of ciphers.

These ciphers were invented at different times by differentauthors, sometimes with confusion of authorship, and indifferent countries. They are similar in that they representpermutations of the same cryptographic concept and can becracked with the same general methodology, albeit with slightvariations in procedure. What is also interesting is thatthese ciphers can be viewed in tableaux form, in slide form ormatrix form.

The theory of polyalphabetic substitution is simple. Theencipherer has at his disposal several simple substitutionalphabets, usually 26. He uses one such alphabet to encipheronly one letter, another alphabet for the second letter,and so forth, until some preconcerted plan has been followed.The earliest known ciphers of this kind, the Porta (1563), theVigenere (1586) used tableau's for encipherment, in which allthe alphabets were written out in full below each other. TheGronsfield (1655) had a mental key, and the Beaufort (1857)which came two hundred years later, again used the tableaux.The process was reduced to strips or slides in 1880 at theFrench military academy of Saint-Cyr. The polyalphabeticdeciphering slides now bear that name. [ELCY]

To know thoroughly any of these ciphers is to understand thefundamental principles of all. Lets look at the papa bear.

## THE VIGENERE CIPHER

The father of the Viggy family is the Vigenere Cipher. Likemost of the periodic ciphers, the 'Viggy' is actually a seriesof monoalphabetic substitutions such as Aristocrats, and sincea keyword is used, under each letter of the keyword, there is aseparate simple substitution cipher - each one different- using all the letters, in such a manner, that the resultingcipher is a combination of several such substitutions.

Attributed to Blaise de Vigenere, the cipher named for him wasinvented by him in 1586. In his "Traicte des Chiffres"he did invent an autokey system which used both a priming keyand did not recommence his plaintext key with each word, butkept it running continuously. He described a second autokeysystem which was more open but still secure. Both systems wereforgotten and were re-invented in the 19th century. Historianshave credited Vigenere with the simpler polyalphabeticsubstitution system. Legend grew around this cipher that itwas "impossible of translation" as late as 1917. [KAHN]

The original Viggy was composed of an enciphering anddeciphering tableaux. Letters were enciphered and decipheredone letter at a time. The modern Vigenere tableaux isshown in Figure 11-1.

`                   Figure 11-1   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 zA  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 ZB  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 AC  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 BD  D E F G H I J K L M N O P Q R S T U V W X Y Z A B CE  E F G H I J K L M N O P Q R S T U V W X Y Z A B C DF  F G H I J K L M N O P Q R S T U V W X Y Z A B C D EG  G H I J K L M N O P Q R S T U V W X Y Z A B C D E FH  H I J K L M N O P Q R S T U V W X Y Z A B C D E F GI  I J K L M N O P Q R S T U V W X Y Z A B C D E F G HJ  J K L M N O P Q R S T U V W X Y Z A B C D E F G H IK  K L M N O P Q R S T U V W X Y Z A B C D E F G H I JL  L M N O P Q R S T U V W X Y Z A B C D E F G H I J KM  M N O P Q R S T U V W X Y Z A B C D E F G H I J K LN  N O P Q R S T U V W X Y Z A B C D E F G H I J K L MO  O P Q R S T U V W X Y Z A B C D E F G H I J K L M NP  P Q R S T U V W X Y Z A B C D E F G H I J K L M N OQ  Q R S T U V W X Y Z A B C D E F G H I J K L M N O PR  R S T U V W X Y Z A B C D E F G H I J K L M N O P QS  S T U V W X Y Z A B C D E F G H I J K L M N O P Q RT  T U V W X Y Z A B C D E F G H I J K L M N O P Q R SU  U V W X Y Z A B C D E F G H I J K L M N O P Q R S TV  V W X Y Z A B C D E F G H I J K L M N O P Q R S T UW  W X Y Z A B C D E F G H I J K L M N O P Q R S T U VX  X Y Z A B C D E F G H I J K L M N O P Q R S T U V WY  Y Z A B C D E F G H I J K L M N O P Q R S T U V W XZ  Z A B C D E F G H I J K L M N O P Q R S T U V W X Y`
The normal alphabet at the top of the tableaux is for plaintextand the keyletters are shown at the extreme left under the 'A'of the top row. Where the two lines intersect in the body ofFigure 11-1, the ciphertext is found.

For example using the keyword TENT, we encipher "COME AT ONCE"

`We have: TENT       TENT         ----       ----         COME       VSZX   (ciphertext)         ATON       TXBG         CE         VI--`
The enciphering and deciphering problem are done as a group ofletters to improve speed and accuracy of the process.

Another way to look at this is that the Viggy is really a twodimensional slide problem. We can construct (or purchase forabout \$2.00 from ACA) a set of two Saint-Cyr slides thatoperate the same way as the tableaux shown in Figure 11-1.What is useful is that each slide bears the standard normalalphabet from A-Z with high frequency letters colored orshaded. Each slide is a double-alphabet to allow flexibility.

`                          Figure 11-2   ABCDEFGHIJKLMNOPQRSTUVWXYZABCDEFGHIJKLMNOPQRSTUVWXYZ   GHIJKLMNOPQRSTUVWXYZABCDEFGHIJKLMNOPQRSTUVWXYZABCDEF   *                         *`
Figure 2 shows the Saint- Cyr slide at a key of G. Check withFigure 11-1 to see that the results are the same for Nplain =Tcipher or Iplain = Ocipher.

The practical use of the Saint Cyr slide is that the wholecolumn of plaintext is enciphered as a unit. So C A C would beenciphered as V A V, plaintext O T E becomes S X I, etc.This eliminates mistakes. The cipher is taken off in 5 lettergroups by rows, so we would have VSZXT XBGVI for our previousexample.

Friedman points out that the sliding components produce thesame type of cipher with the circular disks like the old U. S.Army version. [FRE7]

Koblitz [KOBL] describes the Viggy as follows:

For some fixed k, regard blocks of k letters as vectors in(Z/NZ)**k. Where N is the N-letter alphabet and a digraphinteger correspondence exists between modulo N**2 arrayand it is a vector mapping. Choose some fixed vector bwhich exists in the plane (Z/NZ)**k which can be rememberedby a key word and encipher by means of the vector translationC = P +b where C is the ciphertext message unit and P is theplaintext message unit which is a k-tuple of the integersmodulo N.

The object is to guess N and k, break up the ciphertext inblocks of k letters and perform a frequency analysis on thefirst letter of each block to determine the first component ofb and then proceeds onto the second letterin the block, etc.

Konheim's description is worse than Koblitz's. [KONH]

Seberry and Pieprzyk describe the Viggy as made up ofkey sequence k= k1...kd where ki , (i=1,d) gives the amount ofshift in the ith alphabet, fi(a) = a+ki(mod n) and theciphertext is described as fi**(-1) = (ki -c) mod n so that:

fi(a) = [(n-1)-a +(ki +1) ] mod n [SEAB]
The latter four descriptions are boring - even to myengineering background. They also do not hold water forrandomized alphabets or tableauxs with disruption areas inplace. These represent discontinuities in the mathematicalfunction. They are discontinuous and tractable. Ordifferentiable if the model is such. SCYER's program may havesolved the discontinuity integer problem by area limits ormodule limits. When he publishes the procedure, maybe he willtell us.

## WHICH WAY?

Does it matter with the Viggy, that we encipher S by B (Balphabet or Key B) to find cipher T or encipher B by S (Salphabet or Key S) to find T? No. This is an interestingcharacteristic not shared by all in the Viggy family. It maybe its downfall.

For instance, the message:

Send Supplies To Morley's Station

enciphered with the repeating key, BED under the originalmethod of encipherment as might be described by Blaise deVigenere would be:
`Key   : BEDB  EDBEDBED  BE  DBEDBED  BEDBEDBPlain : SEND  SUPPLIES  TO  MORLEYS  STATIONCipher: TIQE  WXQTOJIV  US  PPVOFCV  TXDUMRO`
The modern Saint-Cyr slide encipherment of the above would be:
`  Key        B E D     B E D     B E D  Plain      S E N     D S U     P P L  Cipher     T I Q     E W X     Q T O             I E S     T O M     O R L             J I V     U S P     P V O             E Y S     S T A     T I O             F C V     T X D     U M R             N             Owhich gives:        5         10          15          20          25T I Q E W   X Q T O J   I V U S P   P V O F C   V T X D U      30M R O X X    (two ending nulls and a bad choice at that)`
With the Saint Cyr slide, we would encipher S, I, E, N; thenD, T, S, and finally P, O , T by setting the B key on thebottom slide under the A key of the top slide and reading offthe equivalents. [SINK], [ELCY]

## DECIPHERMENT BY PROBABLE WORD

Refer to Figure 11-3:

`                     Figure 11-3Deciphering with the Key:Key   :  B E D B E D B E D B E D  ........Cipher:  T I Q E W X Q T O J I V  ........Plain :  S E N D S U P P L I E S  ........Deciphering with the Message:Plain :  S E N D S U P P L I E S  ........  (trial key)Cipher:  T I Q E W X Q T O J I V  ........Key   :  B E D B E D B E D B E D  ........  (true  key)`
Figure 11-3 indicates a possible solution method. The messagefragment works well as a trial key, and if applied in thesame manner as the true key, the true original key will berevealed. The Vigenere Cipher works equally well in reverse.It is this peculiarity that portends the use of a probable wordattack.

Suppose we have the cryptogram:

`   U S Z H L    W D B P B    G G F S ...`
in which we suspect the presence of the word SUPPLIES.We decipher the first 8 letters using this probable word as atrial key, and obtain the jumbled series: C Y K S A O Z J,which is unsatisfactory. We next drop the first U, and obtaingroup : A F S W L V X X. We fail again on the third andfourth trials. The fifth decipherment obtains the seriesTCOMETCO. We see the TCO repeats and the key word COMET.[ELCY]

F. R. Carter of the ACA shows us a more organized approach inFigure 11-4:

`                   Figure 11-4Cryptogram Fragment: U S Z H L W D B P B G G F S ......Probable Word:                             *           S         C A H P T E L J X J O O N A           U           Y F N R C J H V H M M L Y           P             K S W H O M A M R R Q D           P               S W H O M A M R R Q D           L                 A L S Q E Q V V U H           I                   O V T H T Y Y X K           E                     Z X L X C C B O           S                               O                                            *`
Look down at an angle between the stars to find the key wordCOMET. The first letter S was used to decipher every possiblekey letter which can produce S. The entire row of equivalentswere produced at the same time. The resulting rows ofdecipherment indicate all the possible keyletters that couldproduce S, then U, then P, and so on. Carter actuallyshortened the procedure to three full rows and then partialsthereafter. He assumes that the keyword is readable anddiscards non readable text.

## DECIPHERMENT BY PROBABLE TRIGRAM SEQUENCE

For the case where we have no probable word or the sequence isvery short, we may use Ohaver's Trigram Method. We start witha list of usual trigrams THE, AND, THA, ENT, ION, TIO. The keyfragments deciphered by these will be short and numerous, somecorrect and some incorrect to bring out the repeating keysequence. A secondary worksheet is used to test the variousfragments as keys. If any one of them is a fragment of theoriginal key, it must bring out fragments of plaintext atregular intervals.

A scheme like Carters can be used with the trigrams THE, AND..replacing the word SUPPLIES. Refer to Figure 11-5.

`Given:                   10                    20              26  L N F V E  O L N V M  R N G Q F  H H R N H  I R V F E  B`
The cipher text is only 26 letters long. Every letter exceptthe final two might begin a cipher trigram. So we have 24cipher trigrams. Write them out in full on two worksheets.

`                    Figure 11-5ION                Trial 1    LNF NFV FVE VEO EOL OLN LNV NVM VMR MRN RNG NGQ    AZS FRI XHR NQB WAY GXA DZI FHZ NYE EDA JZT FSD                                        ---    GQF QFH FHH HHR HRN RNH NHI HIR IRV RVF VFE FEB    YCS IRU XTU ZTE ZDA ZJU FTV ZUE ADI JHS NRR XQOEDA                Trial 2    LNF NFV FVE VEO EOL OLN LNV NVM VMR MRN RNG NGQ    HKF JCV BSE RBO ALL KIN HKV JSM RJR ION NKC JDQ                                        ---    GQF QFH FHH HHR HRN RNH NHI HIR IRV RVF VFE FEB    CNF MCH BEH DER DON NKH JEI DFR EOV NSF RCE BBB`
Trial 1 tests for THA, THE, AND fail but ION gives us FRI andWAY. But anyone of these 24 decipherments on the second rowmight be a fragment of the original key. Trial 2 fails toconfirm FRI or WAY but test of key-fragment EDA yields ION.If this sequence is actually a portion of the original key,then the plaintext will be brought out at some constantdistance apart. The point we found the trigram is the tenthcryptogram letter; that is every trigram presents only one newletter so to find a completely different trigram in eitherdirection, we must count backwards or forwards a distance ofthree trigrams.

Beginning at the tenth trigram we examine every third trigramin both directions. The following is found: HKF, RBO, HKV,ION,CNF, DER,JEI, NSF. These are incoherent. This would beequivalent to a period of three - not likely. Try every fourthdecipherment: JCV,KIN,ION,MCH,NKH,NSF. Not usable for aconsecutive sequence, continuously written cryptogram.Trying the decipherments at a proposed period of 5, we get ALL,ION, BEH, DFR. This possibility is good. We try todecipher the T before ION and get the letter C. We now havefour letters in our key C E D A. With a little anagraming wehave the word D A * C E. A probable word FRIDAY comes to mind.

## BRYAN'S SAINT-CYR 'HITS' METHOD

William G. Bryan shows us how to use the high frequency letterson the Saint-Cyr slide to good use.

Given the Viggy with a known period of 7 based on a similareffort used in Table 11-1:

`PXIZH  GVGEU  UOXIX  MYEEJ  ZCOCM  OWZCL  FMTOR  ISIGH  LKWPSMSIDX  WCFBR  KPYXO  PRJIL  HFMCR  IHUDU  LVRLJ  FVVVS  HTYFRRGPHQ  WIIBL  XQXMM  TDVGU  EITFM  QEEJH  WUHFW.We reset the problem in groups of 7:              1234567              PXIZHGV              GEUUOXI              XMYEEJZ              COCMOWZ              CLFMTOR              ISIGHLK              WPSMSID              XWCFBRK              PYXOPRJ              ILHFMCR              IHUDULV              RLJFVVV              SHTYFRR              GPHQWII              BLXQXMM              TDVGUEI              TFMQEEJ              HWUHFW`
Now each column represents a separate simple substitutioncipher. They will not produce consecutive plaintext, butmerely show isolated letters in that particular substitution,to be coupled with those letters that fall on either side inother substitutions, to make a true plain text sequence. Here'swhere the underlined high-frequency letters on the slide comein:

We go down column 1 and tabulate all the letters whichappear more than once. P-2, G-2, X-2, C-2, I-3, T-2. Werearrange them in their normal sequence = C G I P T X.The lower slide is moved successively so that the first letterC is under the high frequency letters, in turn, A E H I N O R ST, and a reading is made of the number of 'hits' , the numberof other cipher text letters G I P T X that fall below the highfrequency letters. If they do then the letter under A of thetop slide is the key letter for that column. If they don'tfurther trials are necessary.

High frequency letters don't always show up. Some times mediumfrequency letters may be required. So with C under A: G-E, I-G,P-N, T-R, X-V; With C under E:G-I, I-K, P-R, T-V, X-Z; With Cunder the H: G-L, I-N, P-U, T-Y, X-C; with C under the I: G-M,I-O,P-V, T-Z, X-D; and with C under the N: G-R, I-t, P -A, T-E, X-I (six hits); and we have found the setting. So we set Punder the A in the top slide, and decipher the entire column AR I N N T H I A T T C D R, and write it into a blank column ascolumn 1.

Proceeding with Column 2, we have no results. Column shows 2passable results at P and U, Column 4 seems to go with Y,column 5, setting B has 4 hits, Column 6 has 5 hits indicatingan E, and Column 7, R gives six hits.

The keyword thus recovered is P P Y B E R. We choose todecipher the ending B E R as the ending of a keyword toproduce:

`                 B E R                 -----                 G C E                 N T R                 O F I                 N S I                 S K A                 G H T                 R E M                 A N T                 O N S                 L Y A                 T H E                 U R E                 E N A                 V E R                 W I V                 T A R                 D A S                 E S -`
These are almost all good fragments. The GHT must have an I orU before it. Since cipher letter G is involved, we place the Gunder the I which results in the Y we already had and putting Gunder the U gives us M under the A, we choose the latter.

Now we have MBER has a key fragment. Deciphering column 4with M adds N I I S A A U A T C T R T M E E U E V to theevidence.

There are several possibilities NGCE preceded by an O, UGHTpreceded by an O, TANT preceded by an OR; TLYA preceded by anN; UTAR preceded by an O or A; and EWIV preceded by R/H.

With the Viggy cipher, remember to read the setting for thekeyword letter below the A of the Stationary slide; and theplain text appears on the same slide as this A, while thecipher text is in the lower slide.

## VIGENERE COMPUTER SOLUTION IS QUICKER

At this juncture, I wondered how our Viggy solver at the CDBwould do on this problem. I brought up my faithful computerprogram and entered the cipher text into Vigenere.exe withouttelling it the period and found the following:

The period was found within 1 second. The trial keywordwas PLQMBER, which I assumed was PLUMBER. Using PLUMBER as mykeyword, it typed out the answer: "AMONG CERTAIN TRIBES OFINDIANS IN ALASKA.. ends BUT ARE USED AS SLAVES." The processtook less than 3 seconds of compute time on my 486/50.

I then rearranged the ciphertext with five nulls strategicallyadded. The next pass gave me a period of nine and a gibberishtrial keyword. So for well defined problems the computer isless fun but a clear winner. For the clever cryptographer,the computer can be defeated.

## PRIMARY COMPONENTS

We have seen that equivalents obtainable from use of squaretables may be duplicated by slides or revolving disks [FR2],[FR7] or computer models. Cryptographically, the results maybe quite diverse from different methods of using suchparaphenalia, since the specific equivalents obtained from onemethod may be altogether different from those obtained fromanother method. But from the cryptanalytic point of view thediversity referred to is of little significance.

There are, not two, but four letters involved in every case offinding equivalents by means of sliding components;furthermore, the determination of an equivalent for a givenplaintext letter is represented by two equations involving fourequally important elements, usually letters.

Consider this juxtaposition: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 Z2. F B P Y R C Q Z I G S E H T D J U M K V A L W N O XQuestion - what is the equivalent of Pplain when the Key letteris K? Answer - without further specification, the cipherequivalent can not be stated. Which letter do we set K againstand in which alphabet? We have previously assumed that the Kcipher would be put against A in the plain. But this is only aconvention.

`                       Figure 11-6                          Index            Plain                             *              * 1. Plain:                   ABCDEFGHIJKLMNOPQRSTUVWXYZ 2. Cipher:FBPYRCQZIGSEHTDJUMKVALWNOXFBPYRCQZIGSEHTDJUMKVALWNOX                             *              *                          Key              Cipher`
With this setting Pplain = Zcipher.

The four elements are:

1. The Key letter, 0k
2. The index letter, 01
3. The plaintext letter, 0p
4. The cipher letter. 0c

The index letter is commonly the initial letter of thecomponent, but by convention only. We will assume from now onthat 01 is the initial letter of the component in which it islocated. Refer to Figure 11-6 to confirm this assumption.The enciphering equations above are:
`                (I) Kk = A1 ; Pp = Zc     k=key, p=plain,                                          c=cipher, 1= initial`
There is nothing sacred about the sliding components. ConsiderFigure 11-6b.

`                      Figure 11-6b                         Index            Cipher                            *              *1. Plain:                   ABCDEFGHIJKLMNOPQRSTUVWXYZ2. Cipher:FBPYRCQZIGSEHTDJUMKVALWNOXFBPYRCQZIGSEHTDJUMKVALWNOX                            *              *                         Key              Plainthus           (II) Kk = A1; Pp = Kc`
Since equations (I) and (II) yield different results even withthe same index, key and plain text letters, it is obvious thata more precise formula is required. Adding locations to theseequations does the trick.

`(I)  Kk in component (2) =A1 in component (1); Pp in component     (1) = Zc in component (2).(II) Kk in component (2) =A1 in component (1); Pp in component     (2) = Zc in component (1).In shorthand notation:          (1)  Kk/2 = A1/1; Pp/1 + Zc/2          (2)  Kk/2 = A1/1; Pp/2 + Zc/1`
Employing two sliding components and four letters impliestwelve different resulting systems for the same set ofcomponents and twelve enciphering conditions. Theseconstitute the Viggy Family:
`                           Table 11-2(1) 0k/2=01/1; 0p/1=0c/2        (7)  0k/2=0p/1; 01/2=0c/1(2) 0k/2=01/1; 0p/2=0c/1        (8)  0k/2=0c/1; 01/2=0p/1(3) 0k/1=01/2; 0p/1=0c/2        (9)  0k/1=0p/2; 01/1=0c/2(4) 0k/1=01/2; 0p/2=0c/1        (10) 0k/1=0c/2; 01/1=0p/2(5) 0k/2=0p/1; 01/1=0c/2        (11) 0k/1=0p/2; 01/2=0c/1(6) 0k/2=0c/1; 0p/1=0p/2        (12) 0k/1=0c/2; 01/2=0p/1`
The first two equations (1) and (2) define the Vigenere type ofencipherment and are widely used. Equations (5) and (6) definethe Beauford type and Equations (9) and (10) define theDelastelle type of encipherment. [FR7]

## FURTHER REMARKS ON REPETITIONS

I have said that the three steps in the cryptanalysis ofrepeating key systems are : 1) Find the length of the period,2) Allocate or distribute the letters of the ciphertext intotheir respective alphabets, thereby reducing the polyalphabetictext to monoalphabetic terms, and 3) analysis of the individualmonoalphabetic distributions to determine the plain text valuesof their cipher equivalents in each distribution or alphabet.

As a direct result of using a repeating key (no matter howlong) certain phenomena are manifested externally to thecryptogram. Regardless of what system is used, identical plaintext letters enciphered by the same cipher alphabet with singleequivalents must yield identical cipher letters. This happenseach time the same key letter is used to encipher identicalplaintext letters.

Since the number of columns or positions with respect to thekey are limited, and there is a normal redundancy in thelanguage, it follows that there will be in a message of fairlength many cases where identical plain text letters must fallinto the same column. This will be enciphered by the samecipher alphabet, resulting in many repetitions. There are twotypes of repetitions: causal and accidental (random)repetitions. The former we can trace back to the key. Thelatter occurs when different plaintext letters fall indifferent columns and by chance produce identical cipher textletters.

Accidental repetitions will occur frequently with individualletters, less frequently with digraphs (because the accidentmust occur twice in succession, much less in the case oftrigraphs and very much less in the case of a tetragraph.The probability of chance repetition decreases significantly asthe repetition increases in length. Friedman has developedstatistical tables based on the binomial and Poissondistributions to determine the individual and cumulativeprobabilities for expected number of repetitions in n lettertext to occur x or more times in samples of random text.

The use of these tables is important. They tell us whenwe are dealing with cryptographically maneuvered text versusrandom noise designed to fool the listener. They indicatewhat may be a hoax (Beale or Bacon - Shakespeare controversies)versus valid enciphered text.

Tables 11-3 to 11-6 show the above theory.

`                           Table 11-3Number          Expected Number of Digraphs Occurringof                  Exactly x TimesLetters E(2)  E(3)  E(4)  E(5)  E(6)  E(7)  E(8)  E(9)  E(10)--------------------------------------------------------------100     6.21  .298  .011200     21.8  2.12  .154  .009300     42.5  6.23  .683  .060  .004400     65.3  12.8  1.87  .220  .022  .002500     88.1  21.6  3.97  .582  .071  .008600    110.   32.3  7.11  1.25  .184  .023  .003700    129.   44.3  11.4  2.35  .403  .059  .008  .001800    145.   57.1  16.8  3.96  .777  .130  .019  .003900    158.   70.1  23.2  6.16  1.36  .257  .043  .006   .0011000   169.   83.0  30.6  9.03  2.21  .466  .085  .014   .002                           Table 11-4Number    Expected Number of Trigraphs Occurringof                  Exactly x TimesLetters E(2)  E(3)  E(4)--------------------------100     .269  .001200     1.10  .004300     2.48  .014400     4.40  .033500     6.85  .064600     9.81  .111  .001700    13.3   .175  .002800    17.3   .261  .003900    21.8   .371  .0051000   26.8   .505  .008                           Table 11-5Number    Expected Number of Tetragraphs Occurringof                  Exactly x TimesLetters E(2)     E(3)--------------------------100     .010200     .043300     .096400     .171500     .270600     .389700     .530800     .693900     .8771000    1.08      0.001                           Table 11-6Number    Expected Number of Pentagraphs Occurringof                  Exactly x TimesLetters E(2)----------------100200     .002300     .004400     .007500     .011600     .015700     .021800     .027900     .0341000    .042`
By way of illustration, of the use of these tables, from Table11-3, we obseve that in a sample of 300 letters of random text,we may expect 43 digraphs to occur twice, 6 digraphs to occurthree times and 1 digraph to occur four times. If we sum thevalues under E(2) through E(6) we have the cumulativeprobability in the 300 letter sample. The sum is 49.477, whichindicates that in a sample of 300 letters or so, 49 digraphswill occur two or more times.

## STATISTICAL PROOF OF THE MONOALPHABETICITY OF THE DISTRIBUTIONS

The second step in the solution of periodic ciphers is todistribute the cipher text into the component monoalphabets.The period once established tells us the number of cipheralphabets. By rewriting the message in groups corresponding tothe length of the key (period) in columnar fashion, weautomatically have divided up the text so that lettersbelonging to the same cipher alphabet occupy similar positionsin the groups or in the same columns.

If we make separate uniliteral frequency distributions for theisolated alphabets, each of these resulting distributions istherefore, a monoalphabetic frequency distribution. Were thisnot so, if they did not have the characteristic crest andtrough appearance including the expected number of blanks,if the observed values of Phi are not sufficiently close to theexpected value of Phi plain, or do not yield I.C.'s in theclose vicinity of the expected value, then the entire analysisis fallacious.

The I.C. values of these individual distributions may beconsidered an index of correctness of the factoring process.Both theoretically and practically, the correct hypothesis withrespect to these distributions will tend to conform moreclosely to the expected I.C. of a monoalphabetic frequencydistribution.

Friedman demonstrates the above with an example: [FR7]

`Plaintext Message:The artillery battalion marching in the rear of the advanceguard keeps its combat train with it insofar as practical.Keyword BLUE using direct standard alphabets.Cipher AlphabetsPlain: 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 C D E F G H I J K L M N O P Q R S T U V W X Y Z A2.     L M N O P Q R S T U V W X Y Z A B C D E F G H I J K3.     U V W X Y Z A B C D E F G H I J K L M N O P Q R S T4.     E F G H I J K L M N O P Q R S T U V W X Y Z A B C D   B L U E B L U E B L U E B L U E B L U E B L U E ...   T H E A R T I L L E R Y B A T T I L I O N M A R ...Cipher TextUSYES  ECPMP  LCCLN  XBWCS  OXUVD  SCRHTHXIPL  IBCIJ  USYEE  GURDP  AYBCX  OFPJWJEMGP  XVEUE  LEJYQ  MUSCX  JYMSG  LLETALEDEC  GBMFI`
Friedman gives a useful formula for monographic I.C. of a 26character text:
`    I.C.  =  26 sum f(f-1)/N(N-1) = Phi(o) / Phi (r)`
and since Phi (p) for English is 0.0667N (N-1)and Phi (r) = 0.0385 N ( N-1) where N is the total number ofelements in the distribution. I.C. for English plain = 1.73and 1.0 for random text. We may apply the I.C. test to thedistributions of periodic polyalphabetic ciphers to confirmthe monoalphabeticity of their character. This also confirmsthe period length and correctness. If the correct period isassumed, then the Phi test applied to each of the alphabetsshould approximate closely and consistently the value of Phi(p)and conversely, if the incorrect period is assumed, then thePhi(o) should approximate the value of Phi(r). Deviation fromthis hypothesis must be statistically significant. [FR7]

So we break down the four alphabets:

`  4 1 4 1 1 1 1 1 3     1   1 1     1 1 4      Phi =42A 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  I.C.=1.681   2   4   1         2 1     4     4 1     1 2 2 Phi=44A 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  I.C.=1.911   5     1   1 1 1   5 2 1     1       2 1   1 2 Phi=46A 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  I.C.=1.99    1   6   2   2 1     1   1 1   2 2     1 1 3 1 Phi=44A 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  I.C.=1.91`
It is seen that all these distributions are monoalphabeticsince their observed Phi's are closer to the Phi (p) = 40.rather than Phi (r) = 23. Any other period assumed at fouror a multiple of four, will not yield monoalphabeticdistributions.

In light of the foregoing principles, we now look at twoadditional cryptanalytic techniques for the Viggy family.The first compares the distributions to the normal and thesecond is very important - completing the plain-component.

## SOLUTION BY FITTING THE DISTRIBUTIONS TO THE NORMAL

Given message text A:

`           5         10         15         20         25A.  A U K H Y  J A M K I  Z Y M W M  J M I G X  N F M L XB.  E T I M I  Z H B H R  A Y M Z M  I L V M E  J K U T GC.  D P V X K  Q U K H Q  L H V R M  J A Z N G  G Z V X ED.  N L U F M  P Z J N V  C H U A S  H K Q G K  I P L W PE.  A J Z X I  G U M T V  D P T E J  E C M Y S  Q Y B A VF.  A L A H Y  P O I X W  P V N Y E  E Y X E E  U D P X RG.  B V Z V I  Z I I V O  S P T E G  K U B B R  Q L L X PH.  W F Q G K  N L L L E  P T I K W  D J Z X I  G O I O IJ.  Z L A M V  K F M W F  N P L Z I  O V V F M  Z K T X GK.  N L M D F  A A E X I  J L U F M  P Z J N V  C A I G IL.  U A W P R  N V I W E  J K Z A S  Z L A F M  H S`
The period is 5 and the I.C. confirms this hypothesis.

We make uniliteral frequency distributions for the 5 alphabetsto determine if we have standard alphabets.

`Alphabet 1     I.C. = 1.445 1 2 3 3   3 2 2 6 2 1   6 1 5 3   1   2   1     6A B C D E F G H I J K L M N O P Q R S T U V W X Y ZAlphabet 2     I.C. = 1.475   1 1   3   3 1 2 4 9 1   2 5     1 2 4 4     4 3A B C D E F G H I J K L M N O P Q R S T U V W X Y ZAlphabet 3     I.C. = 1.712 3     1       8 2 2 4 8 1   1 2     3 4 5 1 1   5A B C D E F G H I J K L M N O P Q R S T U V W X Y ZAlphabet 4     I.C. = 1.363 1   1 3 4 4 4     2 2 3 3 1 1   1   2   2 4 9 2 2A B C D E F G H I J K L M N O P Q R S T U V W X Y ZAlphabet 5     I.C. = 1.91        6 2 4   8 1 3   7   1 2 1 4 3     5 2 2 2A 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`
Except for possibly Alphabet 1, all are standard distributions.It is clear that the Aplain for alphabets 2,3,4,5 are H,I,T,Ecipher. A little experimentation gets us Aplain in alphabet 1=Wcipher. The key word under Aplain is WHITE. The five completecipher alphabets are shown in matrix form in Figure 11-7.
`                          Figure 11-70 A B C D E F G H I J K L M N O P Q R S T U V W X Y Z1 W X Y Z A B C D E F G H I J K L M N O P Q R S T U V2 H I J K L M N O P Q R S T U V W X Y Z A B C D E F G3 I J K L M N O P Q R S T U V W X Y Z A B C D E F G H4 T U V W X Y Z A B C D E F G H I J K L M N O P Q R S5 E F G H I J K L M N O P Q R S T U V W X Y Z A B C DApplying these values to the first groups of our message:A U K H Y  J A M K I  Z Y M W M  J M I G X  N F M L XE N C O U  N T E R E  D R E D I  N F A N T  R Y E S T`
Look at the I.C.'s for these alphabets. The expected is1.73. The third alphabet is almost exact. Three alphabets seemlow and one is high or are they? Actually these deviations arewithin one sigma of the samples of these sizes 55 tallies, sothe deviations are not abnormal. The standard deviations maybe calculated with:
`For plain text:           Sigma (O) = Sqrt[ (0.0048)N**3 + (.1101)N**2-                              (.1149) N]           Sigma(I.C.)= 26/(N-1)sqrt(N) * sqrt[ (0.0048)N**2 +                           (.1101)N- (.1149) ]The more important deviation is from random rather thanobserved:           Sigma(Phi) = 0.2720 sqrt[ N (N-1)]           Sigma(I.C.)= 7.0711/sqrt[N(N-1)]   where: sqrt is the square root function`
The latter two equations apply to a 26 letter alphabet only.

Since simage is defined as a difference between the observedand the expected number, divided by the standard deviation, itmay be shown that the I.C. of Alphabet 1 is 1.44-1.00/.13 =3.38 sigma over random; for this type of distribution whichfollows the Chi squared distribution, this amounts to 1 chancein 300 of being random.

In the foregoing example, standard alphabets were used.We could easily of used reversed standard alphabets. The U.S.Army Cipher Disk produces just this type of cipher. It is knownas the Beaufort Cipher. The direction of the crests andtroughs is reversed when fitting the distributions to thenormal.

## SOLUTION BY COMPLETING THE PLAIN-COMPONENT SEQUENCE

When direct standard alphabets are used we can mechanicallysolve the cipher by completing the plain component. The plaintext reappears on only one generatrix and this generatrix isthe same for the whole message. It is the only generatrix thatyields intelligible text. This same process can be modified towork with the alphabets of a Viggy. In this case the correctgeneratrix should be distinguishable from the others because itshows a more favorable assortment of high frequency letters,and thus can be selected by eye from the whole set ofgeneratrixes.

Using the previous example, we let the first ten cipher lettersin each alphabet be set down in a horizontal line and theassumption is made that the alphabets are direct standard withnormal sequences. See Figure 11-8.

We use the following selection rules:

1. Circle all low frequency letters J, K, Q, X, Z and discard any row that has two or more of these letters in it.
2. We weight the eight highest frequency letters (ETANORISH) as 1 and the remaining letters as 0. The sum of the weights is recorded at the side of each row.
3. Select the highest score. This works 8 out of 10 times. The correct answer is 10 out of 10 if we examine the top three scores. Friedman presents the statistical proof for this method in FRE7].

This method works regardless of the key (which might be anumber) as in the Gronsfeld Cipher.

`                   Figure 11-8Gen./   Alphabet 1    Alphabet 2    Alphabet 3     Alphabet 4 1      AJZJNEZAIJ  2 UAYMFTHYLK  2 KMMIMIBMVU     HKWGLMHZMT 2      BKAKOFABJK    VBZNGUIZML    LNNJNJCNWV   5 ILXHMNIANU 3    0 CLBLPGBCKL  4 WCAOHVJANM    MOOKOKDOXW     JMYINOJBOV 4    0 DMCMQHCDLM    XDBPIWKBON  2 NPPLPLEPYX     KNZJOPKCPW 5  * 7 ENDNRIDEMN    YECQJXLCPO    OQQMQMFQZY     LOAKPQLDQX 6    7 FOEOSJEFNO    ZFDRKYMDQP  7 PRRNRNGRAZ   3 MPBLQRMERY 7    2 GPFPTKFGOP    AGESLZNERQ  7 QSSOSOHSBA     NQCMRSNFSZ 8      HQGQULGHPQ  5 BHFTMAOFSR  6 RTTPTPITCB  *8 ORDNSTOGTA 9    5 IRHRVMHIQR  4 CIGUNBPGTS    SUUQUQJUDC   4 PSEOTUPHUB 10     JSISWNIJRS    DJHVOCQHUT  4 TVVRVRKVED     QTFPUVQIVC 11     KTJTXOJKST  4 EKIWPDRIVU  3 UWWSWSLWFE     RUGQVWRJWD 12     LUKUYPKLTU    FLJXQESJWV    VXXTXTMXGF     SVHRWXSKXE 13     MVLVZQLMUV    GMKYRFTKXW  1 WYYUYUNYHG   3 TWISXYTLYF 14   4 NWMWARMNVW    HNLZSGULYX    XZZVZVOZIH     UXJTYZUMZG 15     OXNXBSNOWX  4 IOMATHVMZY  5 YAAWAWPAJI     VYKUZAVNAH 16   3 PYOYCTOPXY    JPNBUIWNAZ    ZBBXBXQBKJ   3 WZLVABWOBI 17     QZPZDUPQYZ    KQOCVJXOBA  2 ACCYCYRCLK     XAMWBCXPCJ 18     RAQAEVQRZA  1 LRPDWKYPCB    BDDZDZSDML     YBNXCDYQDK 19   5 SBRBFWRSAB    MSQEXLZQDC *8 CEEAEATENM     ZCOYDEZREL 20   4 TCSCGXSTBC *6 NTRFYMARED  2 DFFBFBUFON   4 ADPZEFASFM 21   2 UDTDHYTUCD  5 OUSGZNBSFE  2 EGGCGCVGPO   4 BEQAFGBTGN 22   4 VEUEIZUVDE  4 PVTHAOCTGF  0 FHHDHDWHQP   2 CFRBGHCUHO 23   2 WFVFJAVWEF  1 QWUIBPDUHG    GIIEIEXIRQ   3 DGSCHIDVIP 24     XGWGKBWXFG    RXVJCQEVIH    HJJFJFYJSR     EHTDIJEWJQ 25     YHXHLCXYGH    SYWKDRFWJI    IKKGKGZKTS     FIUEJKFXKR 26     ZIYIMDYZHI    TZXLESGXKJ  2 JLLHLHALUT     GJVFKLGYLS        Alphabet 5 1      YIMXXIRMEG 2      ZJNYYJSNFH 3      AKOZZKTOGI 4    2 BLPAALUPHJ 5      CMQBBMVQIK 6    4 DNRCCNWRJL 7      EOSDDOXSKM 8    5 FPTEEPYTLN 9      GQUFFQZUMO 10   4 HRVGGRAVNP 11   4 ISWHHSBWOQ 12     JTXIITCXPR 13     KUYJJUDYQS 14     LVZKKVEZRT 15   3 MWALLWFASU 16     NXBMMXGBTV 17   3 OYCNNYHCUW 18     PZDOOZIDVX 19     QAEPPAJEWY 20     RBFQQBKFXZ 21   4 SCGRRCLGYA 22   3 TDHSSDMHZB 23  *8 UEITTENIAC 24     VFJUUFOJBD 25     WGKVVGPKCE 26     XHLWWHQLDF`
The high frequency generatrixes are selected and their lettersare juxtaposed in columns, the consecutive letters ofintelligible plain text present themselves. If reversedstandard alphabets are used, we must convert the cipher lettersof each isolated alphabet into their normal, plain componentequivalents, and then proceed as in the case of direct standardalphabets.

`For Alphabet 1, generatrix 5..  E N D N R I D E M NFor Alphabet 2, generatrix 20.. N T R F Y M A R E DFor Alphabet 3, generatrix 19.. C E E A E A T E N MFor Alphabet 4, generatrix 8..  O R D N S T O G T AFor Alphabet 5, generatrix 23.. U E I T T E N I A C`
(Read down the columns for plain text.)

Friedman describes a graphical method for generatrixdevelopment in [FR7] and [FR8].

Time to move on to other family members. We shall identify thesystems and peculiarities of each, but remember that thesolution techniques presented for the papa bear apply equallywell to the children and cousins.

## VARIANT CIPHER

The Variant Cipher is just that, a variant of the Vigenere,except that if the Viggy procedure is followed through, apeculiar keyword appears, like JYUWFT. Going back to theslides, In the Variant, the plaintext appears in the oppositeslide from the one containing the key letter: Vigenere belowthe 'A' and Variant above the 'A'. The application of the highfrequency letters is the same. The keyword is obtained in adifferent fashion. For the simple encipherment of COME ATONCE with the keyword TENT:

`   T E N T     T E N T   -------     -------   C O M E     J K Z L   A T O N     H P B U   C E - -     J A - -`
The setting of the slides for say , the initial T of thekeyword is:
`   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   H I J K L M N O P Q R S T U V W X Y Z A B C D E F G`
The decipherment of a Variant is the same as a Vigenere.

## VARIANT SOLUTION BY COMPUTER

From our trusty CDB, I found Variant.exe and applied it to thefollowing cryptogram:

` UALOT SILKH RWEBN NRHNL THURD VPVCH DLSUC OABSM YMXFO QAUBR NFHFR IBAOH YTMWT ENJVQ UPZHF AQWGZ MVHTB OENJD IGIMF SULUA BPMLZ RNFNX SMJTG DJHAF EKKSZ QWDZQ CLVRN FZXBZ WISTJ LMRNH RZ.`
The solution was found in two steps with a period of 7, keyword"RABBVTS" which is RABBITS, and reads: Lamp black isextensively in the manufacture of printing inks, as a pigmentfor oil painting and also for waxing and lacquering of leatheras well as in darkening a furniture polish. Total time 2 or3 minutes.

## BEAUFORT CIPHER

A third member of the Viggy family, the Beaufort, and while thesame procedure is applied, the slides (or tables) aredifferent. One is a normal alphabet, extending double lengthA-Z; the other is reversed, double length Z-A. So if I = T atone setting, then T=I at the same setting. It does not matterwhat the index for the key is, the results are the same.

So:

` ABCDEFGHIJKLMNOBQRSTUVWXYZABCDEFGHIJKL TSRQPONMLKJIHGFEDCBAZYWXVUTSRQPONMLKJI *                         *Again the simple example.  T E N T     T E N T  -------     -------  C O M E     R Q B P  A T O N     T L Z G  C E - -     R A --`

## BEAUFORT SOLUTION BY COMPUTER NEEDS WORK

I found BEAUFORT.exe at the CDB and applied it two thefollowing message:

`LDYUP AKUPT LVDTO BXUFW SERZP QMQPD NITHA NXUHE UGZTG HMGSMSRCUF LBQPZ XRYOB FDMNZ TGCUP QQUFB PANAQ HBOON XOOQP DJCJKTPFDV TBRKL TTSZG ODUFB TETEL POIEB HRTSM DBGGA YUT.`
Not so successful this time. It croaked at period = 6.The best i could get was "light-" I then reran the program witha wider key range and found that the true period was 10. Aftersome trial and error,the keyword is LIGHTHOUSE and the message starts:
A fine head land of granite pierced by a natural arch on..Solution time 15 minutes with at least two wrong trails.

## RELATIONSHIPS

LEDGE points out some interesting relationships between theVigenere, Variant and Beaufort. Let A=0, B=1, C=2 .. Z=25,then:

`   O  Vigenere: Cipher Letter = Plaintext letter + keyletter                (modulo 26)   O  Variant:  Cipher letter = Plaintext letter - keyletter                (modulo 26)   O  Beaufort: Cipher letter = Keyletter - Plaintext letter                (modulo 26)`
Suppose plain text = B and Key = C. Since B=1 and C=2,Vigenere ciphertext = 1 + 2 = 3 or D; For Variant ciphertext1-2=-1 +26 = 25 = Z.

For Vigenere and Variant if key letter = A, since A=0,thecipher text = plain text. If we reconstruct a cipher assumingit is a Vigenere, but it is actually a Variant, we will get thetrue plain text but strange keyword. By subtracting theVariant equation from the Vigenere equation and setting ciphertext (Viggy) = ciphertext (Variant) and similarly plaintext(Viggy) = plaintext (Variant), we get the keyletter (Variant)= - keyletter(Vigenere) the same relationship as that betweenciphertext and plaintext when the keyletter is A in theBeaufort (since A=0). Hence, we encipher our strange keywordwith the A Beaufort alphabet to get the Variant key. The sameholds true if we have a Variant and assume it a Viggy.

If we have a Vigenere and a fragment of the same messageenciphered with the same key in Variant (or visa versa) then:

`a. Plaintext = (Ciphertext(Variant)) +   Ciphertext(Vigenere))/2(modulo 13)b. Key  = (Ciphertext(Vigenere) - Ciphertext(Variant))/2   (modulo 13)`
If we have a Vigenere and a fragment of a Beaufort for the samekey and plaintext or visa versa then:
`c. Plaintext = (Ciphertext(Vigenere)) -   Ciphertext(Beaufort))/2(modulo 13)d. Key  = (Ciphertext(Vigenere) +   Ciphertext(Beaufort))/2(modulo 13)`
In equations a-d, two answers are produced because modulo 13will give one number from 0-12 and another 13-25. Solution isby inspection.

## PORTA (aka NAPOLEON'S TABLE)

Table 11-7 defines the PORTA Cipher. In this table thealphabets are all reciprocal, for example Gplain(Wkey) =Rcipher, Rplain(Wkey)=Gcipher. They are called complementaryalphabets. Either of two letters may serve as a key letterindifferently: Gplain(Wkey) or Gplain(Xkey) = Rcipher.

`                  Table 11-7          A B C D E F G H I J K L M AB       N O P Q R S T U V W X Y Z          A B C D E F G H I J K L M CD       O P Q R S T U V W X Y Z M          A B C D E F G H I J K L M EF       P Q R S T U V W X Y Z N O          A B C D E F G H I J K L M GH       Q R S T U V W X Y Z N O P          A B C D E F G H I J K L M IJ       R S T U V W X Y Z N O P Q          A B C D E F G H I J K L M KL       S T U V W X Y Z N O P Q R          A B C D E F G H I J K L M MN       T U V W X Y Z N O P Q R S          A B C D E F G H I J K L M OP       U V W X Y Z N O P Q R S T          A B C D E F G H I J K L M QR       V W X Y Z N O P Q R S T U          A B C D E F G H I J K L M ST       W X Y Z N O P Q R S T U V          A B C D E F G H I J K L M UV       X Y Z N O P Q R S T U V W          A B C D E F G H I J K L M WX       Y Z N O P Q R S T U V W X          A B C D E F G H I J K L M YZ       Z N O P Q R S T U V W X Y`
The Porta Cipher permits 13 different ways to disguise a plainletter.

Again our simple encipherment:

`T E N T    T E N TC O M E    Y M S NA T O N    W E I EC E - -    Y T - -`
A peculiarity of this system is that since half the alphabetis represented by the half of the alphabet, there never will befound the letters A-M of the plaintext appearing as A-M in theciphertext; no N-Z plaintext appearing as the N-Z ciphertext.This helpful in placing a tip. THE shows up as a (A-M) (N-Z)(N-Z) combination. [BRYA]

Table 11-8 shows a different view of the PORTA Cipher

`                       Table 11-8                       Plain Text     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,B  N O P Q R S T U V W X Y Z A B C D E F G H I J K L MC,D  O P Q R S T U V W X Y Z N M A B C D E F G H I J K LE,F  P Q R S T U V W X Y Z N O L M A B C D E F G H I J KG,H  Q R S T U V W X Y Z N O P K L M A B C D E F G H I JI,J  R S T U V W X Y Z N O P Q J K L M A B C D E F G H IK,L  S T U V W X Y Z N O P Q R I J K L M A B C D E F G HM,N  T U V W X Y Z N O P Q R S H I J K L M A B C D E F GO,P  U V W X Y Z N O P Q R S T G H I J K L M A B C D E FQ,R  V W X Y Z N O P Q R S T U F G H I J K L M A B C D ES,T  W X Y Z N O P Q R S T U V E F G H I J K L M A B C DU,V  X Y Z N O P Q R S T U V W D E F G H I J K L M A B CW,X  Y Z N O P Q R S T U V W X C D E F G H I J K L M A BY,Z  Z N O P Q R S T U V W X Y B C D E F G H I J K L M A`
Using the message text A from page 20 as an example with keyword WHITE , the distribution of 5 alphabets is:
`     2   6 2 1   6 1 5 3   1     6 5 1 2 3 3   3 2 2 11.   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 2 5     1   3   4 4     1 2 3 1 2 4 9 1   2 52.   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     5 3 3     2       5 1 1       3 4 7 2 2 4 8 1   1 23.   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   4 4     2 2 3 3 1 9 2 2 3 1   1 3 3 2   2 44.   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         5 2 2 2           4 3 2 1 6 2 4   9 1 3   7   15.   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`
Now we can divide the M and N distributions, and each half maybe used to fit a normal distribution. In alphabet 1, thesequence CDEFGHIJ cipher may easily be recognized as NOPQRSTUplain; this would fix the keyletters as WX, and therefor theA...Mplain sequence should begin with Ycipher. In alphabets2,3, and 5 the RSTplain sequence may be spotted at BCDcipher,ABCcipher, and CDEcipher, respectively, whereas in alphabet 4,if Ncipher = Eplain, then Ecipher = Nplain; therefore theoriginal assumptions for the first halves will be confirmed bythe goodness of fit of the distributions for the second halves.The keys fore these 5 alphabets are derived as (W,X), (G,H)(I,J), (S,T), and (E,F); from these letters we get WHITE.

In completing the plain component sequence for the Portaencipherment, the cipher letters are first converted to theirPorta plain-component equivalents and then these letters areused for the decipherment. EXCEPT, cipher letters A-M arecompleted in a downward direction and cipher letters N-Z arecompleted in an upward direction.

Reference [FR7] gives the example:

`P K T F F  C D V I T   O B V Z X  C V R E E  G I V J ET P R K T  O Q C F L   P B V P X  ....`
The conversion process and plain component completion of thefirst three alphabets are shown below using the generatrixelimination and weighting scheme developed earlier:
`    Alphabet 1         Alphabet 2         Alphabet 3  P C O C G T O P    K D B V I P Q B    T V V R V R C V  ---------------    ---------------    ---------------1 C P B P T G B C    X Q O I V C D O  6 G I I E I E P I3 D O C O S H C D  3 W P N J U D E N    H J J F J F O J6 E N D N R I D E    V O Z K T E F Z    I K K G K G N K  F Z E Z Q J E F  2 U N Y L S F G Y    J L L H L H Z L0 G Y F Y P K F G    T Z X M R G H X  2 K M M I M I Y M  H X G X O L G H  3 S Y W A Q H I W    L A A J A J X A3 I W H W N M H I    R X V B P I J V    M B B K B K W B  J V I V Z A I J    Q W U C O J K U  1 A C C L C L V C  K U J U Y B J K  3 P V T D N K L T  0 B D D M D M U D  L T K T X C K L  3 O U S E Z L M S  7 C E E A E A T E2 M S L S W D L M  5 N T R F Y M A R  1 D F F B F B S F5 A R M R V E M A    Z S Q G X A B Q  2 E G G C G C R G  B Q A Q U F A B  1 Y R P H W B C P  0 F H H D H D Q H`
The generatrixes with the highest scores are the correct ones.

## MODIFIED PORTA

Just as the Vigenere table consisting of direct standardalphabets has its complementary table of reversed standardalphabets, a variant of the Porta table can be constructedwhere the lower halves of the sequences run in oppositedirection to the upper half. For example:

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

## PROBABLE WORD METHOD OF SOLUTION FOR PORTA

The probable word method is very easy way to attack a Portacipher. Let 1 = any letter in the A-M sequence, and 2 equalany letter in the N-Z sequence.

`P K T F F  C D V I T   O B V Z X  C V R E E  G I V J E2 1 2 1 1  1 1 2 1 2   2 1 2 2 2  1 2 2 1 1  1 1 2 1 1T P R K T  O Q C F L   P B V P X  ....2 2 2 1 2  2 2 1 1 1   2 1 2 2 2`
Use the probable word INFANTRY, which has the class notation of12112222, but in encipherment is reversed to 21221111 pattern.At position 15, X C V R E E G I, we find:
`    plain       I N F A N T R Y    cipher      X C V R E E G I    key         E W G I S E W G    derived     F X H J T F X H`
Read diagonally, we see WHITE repeated.

## COMPUTER SOLUTION OF PORTA

At the trusty CDB is a program called PORTA.exe. Using it onthe following cipher message found a period of 9 with apossible key of KL/IJ/CD/MN/AB/OP/OP/EF/QR. I came up with thekeyword: LIDNAOOER:

`  EYWRR  MOTJJ  QOHFA  LTYQV  SQFPG  EPWTG  RVGUC  DVVBT  EMLMN  BYSOE  OHFKW  YARQL  PEBSB  ETVXM  WVBCV  XRTIT  JJAMX  EHADX  VCAXN  MMWZR  WALFY  BTJSP  RTLLP  LZDVD  FZHGE  PBKQR  RUKWQ  AEAOP  Y`
and behold the message cracked to:
While the Romans used leeks in the culinary depart...
The process took less than two minutes but did not yield theactual keyword or require it.

## GRONSFELD

The GRONSFELD Cipher uses a numerical key and restricts theViggy table to just ten alphabets. We can construct a slidewith one normal alphabet and numbered one like this:

`     ... 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 ...`
One half the digits are used for encipherment and the otherhalf for decipherment. For example the key is derived asfollows:
`       C O N S T I T  U  T  I O N       1 6 4 8 9 2 10 12 11 3 7 5 `
The first duplicate letter carries the lower number.

So back to:

`    6 2 3 4      6 2 3 4    C O M E      I Q P I    A T O N      G V R R    C E - -      I G - -`
Slide method: put the 0 over the C, take the letter to theright in juxtaposition of the 6 = I, same for A which is Gand so on. We decipher by looking to the left.

A typical decipherment might look like this for the test word"YOUR":

`        0 2 4 7         0 2 4 7         0 2 4 7         0 2T S V H Y Q B V Y I G L M G U X A S R M F K C I A A O V I Z-----------------------------------------------------------S R U G Y O U R X F H K M E N T Z R Q L F I V E Z Z N U I X        -------         -------         -------         ---R Q T F         W G E J         Y Q P K         Y Y M TQ P S E         V F D I         X P O J         W W K RT S V H Y Q B V Y I G L M G U X A S R M F K C I A A O V I Z-----------------------------------------------------------Y     9 0   3   0   8     8     2       7   4   2 2O         2   7             6     4                 0U           7   4             3                       1R             4                 9`

## LECTURE 11 PROBLEMS

`11.1  Viggy.SYCVT  HFXEQ  DPTLN  KTGMP  FHMPA  SRVIT  LSEXH  DPITXKELIQ  WDXEC  VNLIP  HPWXD  XXIXH  UTRIH.11.2  Beaufort.SXSXZ  IYLEQ  AWEQF  EZEPP  QZQRD  VANKH  HLZJX  OQSEUYSOVS  SZKLE  DRMRU  THTUW  SCLOX  NEHLA  OPEEU  GAZIAUUOQG  OJX.11.3  Variant.JQRSB  YBKNF  WWTGK  UXDTK  ZAOAA  MCVJU  KBCEX  GUYLBUASWY  TIENQ  XLPYX  CWASU  VAKOM  XIGIK  XHWZT  SWGOPWRTSJ  NAWG.11.4 Gronsfeld.ZRWQU  IKLMS  IXAWI  UQMWP  KFQEL  RBWJG  XHIXT  NLVKS  ZHVHSZRUEK  KWPIM  GSXIA  XVUEL  RHZPI  SLBWT  NHU.11.5  Viggy or Beaufort; same message and key starts ONOIHT.ORQGX  HPNKW  QQCHI  ABIFZ  NQCHR  VLVLU  HYUDT  MCYJN  WAUHPHLVIN  BZCCB  GCGKZ  JNLMM  WTVLY  DYCCV  JPUVG  KLKQX  YTTKIXOQYB  JJMHJ  BYHQY  LFQWF  NRYUC  XCECN  GPCBW  TPAXE  ABKGCPVHKL  OIKQW  TPKOW  KNCMM  HFFAV  A.`

## ANSWERS TO LECTURE 10 PROBLEMS

Thanks to JOE O for a fine analysis of all three problems.

`QQ-1  QUAGMIRE I  Travelogue. (Ends:SINGOUTOFTHESEA) RHIZOME1234567  1234567  1234567  1234567  1234567  1234567  1234567THEFIRS  TIMEaVI  SITOREX  CLAIMSA  HROMANT  ICVENIC  ESINKINKKQHPQR  KTYOiTA  TLGAWBM  XORKTAT  BSOOIYI  CGICEJV  UCYZRJPALNSFRZ  UCQDXIS  TDRBFYS  YTFDZBD  USQWKMT  CPPDOAI  CAAKEHKUAYFHQA  TLNIFSI  SIGJHAS  V.QQ-1 Quagmire I Solution.     VERDICT/nose. Period =7.The first time visitor exclaims "Ah, romantic Venice sinkinginto the sea." The seasoned traveler exclaims,"Ah, stinkingVenice rising out of the sea.  0  A B C D F G H I J K L M P Q R T U V W X Y Z N O S E  1  V W X Y Z A B C D E F G H I J K L M N O P Q R S T U  2  E F G H I J K L M N O P Q R S T U V W X Y Z A B C D  3  R S T U V W X Y Z A B C D E F G H I J K L M N O P Q  4  D E F G H I J K L M N O P Q R S T U V W X Y Z A B C  5  I J K L M N O P Q R S T U V W X Y Z A B C D E F G H  6  C D E F G H I J K L M N O P Q R S T U V W X Y Z A B  7  T U V W X Y Z A B C D E F G H I J K L M N O P Q R SQQ-2  QUAGMIRE III Tedious.   (CRYPTANALYTIC METHODS)DOPPELSCHACHPeriod= 612345  61234  56123  45612  34561  23456  12345  61234  56123THETI  MEREQ  UIRED  BYS.......PNATV  SJBAQ  WGMTR  BZYLU  ACACR  GBNTQ  FGGCN  APNID  ULMVDSCEPB  AMCQF  BBPVR  EOBSL  AFSAN  HFYVV  MCYTF  LEMAO  MFHVUKBAAU  ATTEA  NGOHU  GTQEX  ISUGU  SAKCC  TLIRT  TLSZM  PBMGVAPYRV  YIIGL  WGNUF  JFROG  SNQGN  HBOTU  TACUO  JUVQH  HUGWWWBIMT  WNHVO  GTLSZ  MPYQZ  BNCEN  UWLC.HARDER/decorative. Period = 6. The time required by somecryptanalytic methods  grows extremely rapidly as key length ormessage length increases.  All possible keys for a columnartransposition instead of making an entry by building up afrom a pair of columns is an example.0  D E C O R A T I V B F G H J K L M N P Q S V W X Y Z1  H J K L M N P Q S V W X Y Z D E C O R A T I V B F G2  A T I V B F G H J K L M N P Q S V W X Y Z D E C O R3  R A T I V B F G H J K L M N P Q S V W X Y Z D E C O4  D E C O R A T I V B F G H J K L M N P Q S V W X Y Z5  E C O R A T I V B F G H J K L M N P Q S V W X Y Z DQQ-3  QUAGMIRE IV  Economics Lesson.     EDNASANDE      (BUSINESSACTIVITYDURINGAPERIOD)THEEC  ONOMY  OFTHE  NATIO ..........TDNSE  PMBSV  FURMQ  UFYSJ  PAGGY  FVIKT  GYVLV  FBTPH  IIIADHVIUY  QSAFA  VQVFU  HPIHE  BIXNN  HBSTN  IRMQH  IIIAD  OVIXTCTNOW  EOJOZ  BOWBU  ONLFN  GOBJS  HBOQS  VZMOU  JSFQH  SAHPSJBBJT  AAMIE  XILRA  TOTVL  TUAML  FLNEJ  PPMNT  XHVQV  FCYSBJODNF  XJSFT  UIUTM  ONKDO  UMMSB  NWUL.     EXCHANGE/stock/MARKET.  The economy of the Nation is built     on supply and demand, the result of inflation. Recession     is a temporary falling off of business activity during a     period when such activity has been generally increasing..0  S T O C K A B D E F G H I J L M N P Q R U V W X Y Z1  E T B C D F G H I J L N O P Q S U V W X Y Z M A R K2  X Y Z M A R K E T B C D F G H I J L N O P Q S U V W3  C D F G H I J L N O P Q S U V W X Y Z M A R K E T B4  H I J L N O P Q S U V W X Y Z M A R K E T B D E F G5  A R K E T B C D F G H I J L N O P Q S U V W X Y Z M6  N O P Q S U V W X Y Z M A R K E T B D E F G H I J L7  G H I J L N O P Q S U V W X Y Z M A R K E T B D E F8  E T B C D F G H I J L N O P Q S U V W X Y Z M A R K`

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