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In Lectures 11-12, we continue our course schedule with a study of fascinating cipher systems known as the "Viggy" based on multiple alphabets - Polyalphabetic Substitution systems.
We will continue developing our subject via an overview based on the Op-20-GYT course notes (Office of Chief Of Naval Operations, Washington) [OP20]. We will revisit polyalphabetic cipher systems using Friedman's detailed analysis. We will cover the Viggy, Variant, PORTA systems and other family members. [FRE4], [FRE5], FRE6], [FRE7], [FRE8]. We will take material from ACA's Practical Cryptanalysis Volume V by William G. 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 aperiodic polyalphabetic case and give a diagram of topics considered in Lectures 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 references on these systems - focusing on the cryptanalytic attack and areas of historical interest. Kahn has some wonderful stories about the Viggy family. [KAHN]
In Lectures 1- 10, I have purposely stayed away from the heavier mathematics of cryptography (subject to change). Everything I am presenting can and has been reduced to mathematical models and computerized for ease of work. For my readers who can not live without the math diet, there are plenty of guru' s like [SCHN] and [SCH2] to have breakfast with. There are plenty of computer aids at the Crypto Drop Box to help you do the setup work.
BUT those who embark on a course of "only the computer" do this without knowing the real effort -the brain power - the shortcuts - the tradecraft - the historical implications, in my opinion, have lost the real heart of Cryptography. The "ah ha's" of inspiration are what make the difference. First, there is a fundamental problem in that computer models do not apply to all variant cases. Simple changes to the system can fool even the most adept computer program. For example, placing clever nulls will defeat many a statistical based model.
Second, we lose the sense of urgency that was required for wartime cryptography. If President Kennedy's Playfair message [that's right it was not English as in the movie PT-109] on the back of a coconut had been intercepted and deciphered by the Japanese [which they very capable of doing], we might not have had the graceful light of his Presidency or who knows the moon landings. As another case in point, the solution of ENIGMA during the mid - final Atlantic Campaigns of World War II, reduced the operational effectiveness of the U-Boat to one day and hence saved allied tonnage and warships suppling Europe. The American and British Crypee's 'thought' more like their German counterparts than their counterparts. Computer solutions were bulky, machine dependent [the solution "stops"] and not reliable until 1945. People made the difference.
There are three fundamental steps to solve a Periodic cipher.
Step one is finding the period. Bryan reminds us that there are at least two ways to find the period. The short approach makes use of the distances between patent cipher text repetitions and factors the differentials. The long approach is used when there are no patent repetitions to factor. In this case we set up a possibilities matrix and factor every combination looking for the highest probable common factor. [BRYA]
As an example of the first case take:
10 20 30 40
BGZEY DKFWK BZVRM LUNYB QNUKA YCRYB GWMKC DDTSP
50 60 70 80
OFIAK OWWHM RFBLJ JQFRM PNIQA VQCUP IFLAZ HKATJ
90 100 110 120
UVVQE EKESZ DUDWE KKESL IZQAT SBYUZ UUVAZ IXYEZ
130 140
JFTAJ EMRAS QKZSQ FOPHM W.
We tabulate the repetitions and the cipher text letter differences between
repetitions.
Delta Factors
BG 29 -
RM 45 3,5,9
KA 53 -
MR 77 7,11
QA 39 3,13
VQ 17 -
AZ 40 4,5,8,10
AT 26 13
UV 31 -
EK 9 3,9
KES 10 5,10 .... this trigraph more important
SQ 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 not establish a
period. What then?
Given:
10 20 30 40
RNQJH AUKGV WGIVO BBSEJ CRYUS FMQLP OFTLC MRHKB
50 60 70 80
BUTNA WXZQS NFWLM OHYOF VMKTV HKVPK KSWEI TGSRB
90 100 110 120
LNAGJ BFLAM EAEJW WVGZG SVLBK IXHGT JKYUC HLKTU
MWWK.
We set up the following vertical tally. We note the actual position of
every letter.
Table 11-1
3 4 5 6 7 8 9 10 11 12
-------------------------------
A 3 1 1 1 1 1 2 1
B 9 7 4 5 3 7 4 2 1 2
C 1 1 1 1 1
D
E 1 1 1 1 1 1 1
F 2 3 3 1 2 1 1 1 1
G 5 5 4 1 4 3 2 1 3 1
H 6 3 2 2 3 1 1 2 1
I 1
J 3 1 2 1 1 1 3 1
K 13 10 4 9 8 5 3 1 2 3
L 4 3 4 1 4 1 3 1 2
M 4 2 3 2 6 3 1 1
N 1 1 1 1 3 1 1
O 1 3 1 1 1 1
P 1
Q 1 1 1
R 5 1 1 3 2 1 1
S 4 4 2 3 2 1 1 1 1
T 4 3 1 1 2 1 1 2 2
U 5 1 2 5 1 2 3 1 2 2
V 5 6 2 2 1 2 3 1 1
W 9 4 5 3 8 1 4 4 3 1
X
Y 2 2 3 2 1 2 1 3 1
Z 1
---------------------------------
87 61 47 43 57 30 35 21 25 16 Columns total
X 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.
The Viggy (or more correctly the Vigenere) Family is group of ciphers. 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 Nihilist Substitution, Auto - Key, Running Key and Interrupted ciphers. Bryan includes the Tri-square, the periodic Fractionated Morse, the Seriated Playfair and the Homophonic in the same class of ciphers.
These ciphers were invented at different times by different authors, sometimes with confusion of authorship, and in different countries. They are similar in that they represent permutations of the same cryptographic concept and can be cracked with the same general methodology, albeit with slight variations in procedure. What is also interesting is that these ciphers can be viewed in tableaux form, in slide form or matrix form.
The theory of polyalphabetic substitution is simple. The encipherer has at his disposal several simple substitution alphabets, usually 26. He uses one such alphabet to encipher only 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), the Vigenere (1586) used tableau's for encipherment, in which all the alphabets were written out in full below each other. The Gronsfield (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 the French military academy of Saint-Cyr. The polyalphabetic deciphering slides now bear that name. [ELCY]
To know thoroughly any of these ciphers is to understand the fundamental principles of all. Lets look at the papa bear.
The father of the Viggy family is the Vigenere Cipher. Like most of the periodic ciphers, the 'Viggy' is actually a series of monoalphabetic substitutions such as Aristocrats, and since a keyword is used, under each letter of the keyword, there is a separate simple substitution cipher - each one different- using all the letters, in such a manner, that the resulting cipher is a combination of several such substitutions.
Attributed to Blaise de Vigenere, the cipher named for him was invented by him in 1586. In his "Traicte des Chiffres" he did invent an autokey system which used both a priming key and did not recommence his plaintext key with each word, but kept it running continuously. He described a second autokey system which was more open but still secure. Both systems were forgotten and were re-invented in the 19th century. Historians have credited Vigenere with the simpler polyalphabetic substitution system. Legend grew around this cipher that it was "impossible of translation" as late as 1917. [KAHN]
The original Viggy was composed of an enciphering and deciphering tableaux. Letters were enciphered and deciphered one letter at a time. The modern Vigenere tableaux is shown 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 z
A 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
B 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
C 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
D 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
E 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
F 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 E
G 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 F
H 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
I 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
J 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 I
K 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 J
L 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 K
M 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 L
N 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
O 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 N
P 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 O
Q 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 P
R 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
S 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 R
T 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 S
U 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 T
V 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
W 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 V
X 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 W
Y 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 X
Z 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 plaintext and the
keyletters are shown at the extreme left under the 'A' of the top row. Where the
two lines intersect in the body of Figure 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 of letters to
improve speed and accuracy of the process.
Another way to look at this is that the Viggy is really a two dimensional
slide problem. We can construct (or purchase for about $2.00 from ACA) a set of
two Saint-Cyr slides that operate the same way as the tableaux shown in Figure
11-1. What is useful is that each slide bears the standard normal alphabet from
A-Z with high frequency letters colored or shaded. 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 with Figure 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 whole column of plaintext is enciphered as a unit. So C A C would be enciphered as V A V, plaintext O T E becomes S X I, etc. This eliminates mistakes. The cipher is taken off in 5 letter groups by rows, so we would have VSZXT XBGVI for our previous example.
Friedman points out that the sliding components produce the same type of cipher with the circular disks like the old U. S. Army version. [FRE7]
Koblitz [KOBL] describes the Viggy as follows:
Konheim's description is worse than Koblitz's. [KONH]
Seberry and Pieprzyk describe the Viggy as made up of key sequence k= k1...kd
where ki , (i=1,d) gives the amount of shift in the ith alphabet, fi(a) =
a+ki(mod n) and the ciphertext is described as fi**(-1) = (ki -c) mod n so
that:
Does it matter with the Viggy, that we encipher S by B (B alphabet or Key B) to find cipher T or encipher B by S (S alphabet or Key S) to find T? No. This is an interesting characteristic not shared by all in the Viggy family. It may be its downfall.
For instance, the message:
Key : BEDB EDBEDBED BE DBEDBED BEDBEDB Plain : SEND SUPPLIES TO MORLEYS STATION Cipher: TIQE WXQTOJIV US PPVOFCV TXDUMROThe 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
O
which gives:
5 10 15 20 25
T I Q E W X Q T O J I V U S P P V O F C V T X D U
30
M 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; then D, T, S, and
finally P, O , T by setting the B key on the bottom slide under the A key of the
top slide and reading off the equivalents. [SINK], [ELCY]
Refer to Figure 11-3:
Figure 11-3
Deciphering 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 message fragment
works well as a trial key, and if applied in the same manner as the true key,
the true original key will be revealed. The Vigenere Cipher works equally well
in reverse. It is this peculiarity that portends the use of a probable word
attack.
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 a trial key, and obtain the jumbled series: C Y K S A O Z J, which is unsatisfactory. We next drop the first U, and obtain group : A F S W L V X X. We fail again on the third and fourth trials. The fifth decipherment obtains the series TCOMETCO. We see the TCO repeats and the key word COMET. [ELCY]
F. R. Carter of the ACA shows us a more organized approach in Figure
11-4:
Figure 11-4
Cryptogram 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 word COMET. The
first letter S was used to decipher every possible key letter which can produce
S. The entire row of equivalents were produced at the same time. The resulting
rows of decipherment indicate all the possible keyletters that could produce S,
then U, then P, and so on. Carter actually shortened the procedure to three full
rows and then partials thereafter. He assumes that the keyword is readable and
discards non readable text.
For the case where we have no probable word or the sequence is very short, we may use Ohaver's Trigram Method. We start with a list of usual trigrams THE, AND, THA, ENT, ION, TIO. The key fragments deciphered by these will be short and numerous, some correct and some incorrect to bring out the repeating key sequence. A secondary worksheet is used to test the various fragments as keys. If any one of them is a fragment of the original key, it must bring out fragments of plaintext at regular 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 except the final two
might begin a cipher trigram. So we have 24 cipher trigrams. Write them out in
full on two worksheets.
Figure 11-5
ION 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 XQO
EDA 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 and WAY. But
anyone of these 24 decipherments on the second row might be a fragment of the
original key. Trial 2 fails to confirm 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 constant distance apart. The point we
found the trigram is the tenth cryptogram letter; that is every trigram presents
only one new letter so to find a completely different trigram in either
direction, we must count backwards or forwards a distance of three trigrams.
Beginning at the tenth trigram we examine every third trigram in both directions. The following is found: HKF, RBO, HKV,ION, CNF, DER,JEI, NSF. These are incoherent. This would be equivalent to a period of three - not likely. Try every fourth decipherment: JCV,KIN,ION,MCH,NKH,NSF. Not usable for a consecutive 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 to decipher the T before ION and get the letter C. We now have four letters in our key C E D A. With a little anagraming we have the word D A * C E. A probable word FRIDAY comes to mind.
William G. Bryan shows us how to use the high frequency letters on the Saint-Cyr slide to good use.
Given the Viggy with a known period of 7 based on a similar effort used in
Table 11-1:
PXIZH GVGEU UOXIX MYEEJ ZCOCM OWZCL FMTOR ISIGH LKWPS
MSIDX WCFBR KPYXO PRJIL HFMCR IHUDU LVRLJ FVVVS HTYFR
RGPHQ 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 substitution cipher. They
will not produce consecutive plaintext, but merely show isolated letters in that
particular substitution, to be coupled with those letters that fall on either
side in other substitutions, to make a true plain text sequence. Here's where
the underlined high-frequency letters on the slide come in:
We go down column 1 and tabulate all the letters which appear more than once. P-2, G-2, X-2, C-2, I-3, T-2. We rearrange them in their normal sequence = C G I P T X. The lower slide is moved successively so that the first letter C is under the high frequency letters, in turn, A E H I N O R S T, and a reading is made of the number of 'hits' , the number of other cipher text letters G I P T X that fall below the high frequency letters. If they do then the letter under A of the top slide is the key letter for that column. If they don't further trials are necessary.
High frequency letters don't always show up. Some times medium frequency 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 C under 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 P under the A in the top slide, and decipher the entire column A R I N N T H I A T T C D R, and write it into a blank column as column 1.
Proceeding with Column 2, we have no results. Column shows 2 passable results at P and U, Column 4 seems to go with Y, column 5, setting B has 4 hits, Column 6 has 5 hits indicating an E, and Column 7, R gives six hits.
The keyword thus recovered is P P Y B E R. We choose to decipher the ending B
E R as the ending of a keyword to produce:
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 or U before
it. Since cipher letter G is involved, we place the G under the I which results
in the Y we already had and putting G under the U gives us M under the A, we
choose the latter.
Now we have MBER has a key fragment. Deciphering column 4 with M adds N I I S A A U A T C T R T M E E U E V to the evidence.
There are several possibilities NGCE preceded by an O, UGHT preceded by an O, TANT preceded by an OR; TLYA preceded by an N; UTAR preceded by an O or A; and EWIV preceded by R/H.
With the Viggy cipher, remember to read the setting for the keyword letter below the A of the Stationary slide; and the plain text appears on the same slide as this A, while the cipher text is in the lower slide.
At this juncture, I wondered how our Viggy solver at the CDB would do on this
problem. I brought up my faithful computer program and entered the cipher text
into Vigenere.exe without telling it the period and found the following:
We have seen that equivalents obtainable from use of square tables may be duplicated by slides or revolving disks [FR2], [FR7] or computer models. Cryptographically, the results may be quite diverse from different methods of using such paraphenalia, since the specific equivalents obtained from one method may be altogether different from those obtained from another method. But from the cryptanalytic point of view the diversity referred to is of little significance.
There are, not two, but four letters involved in every case of finding equivalents by means of sliding components; furthermore, the determination of an equivalent for a given plaintext letter is represented by two equations involving four equally 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 Z 2. 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 X Question - what is the equivalent of Pplain when the Key letter is K? Answer - without further specification, the cipher equivalent can not be stated. Which letter do we set K against and in which alphabet? We have previously assumed that the K cipher would be put against A in the plain. But this is only a convention.
Figure 11-6
Index Plain
* *
1. Plain: ABCDEFGHIJKLMNOPQRSTUVWXYZ
2. Cipher:FBPYRCQZIGSEHTDJUMKVALWNOXFBPYRCQZIGSEHTDJUMKVALWNOX
* *
Key Cipher
With this setting Pplain = Zcipher.
The four elements are:
(I) Kk = A1 ; Pp = Zc k=key, p=plain,
c=cipher, 1= initial
There is nothing sacred about the sliding components. Consider Figure
11-6b.
Figure 11-6b
Index Cipher
* *
1. Plain: ABCDEFGHIJKLMNOPQRSTUVWXYZ
2. Cipher:FBPYRCQZIGSEHTDJUMKVALWNOXFBPYRCQZIGSEHTDJUMKVALWNOX
* *
Key Plain
thus (II) Kk = A1; Pp = Kc
Since equations (I) and (II) yield different results even with the same
index, key and plain text letters, it is obvious that a more precise formula is
required. Adding locations to these equations 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 implies twelve different
resulting systems for the same set of components and twelve enciphering
conditions. These constitute 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 of
encipherment and are widely used. Equations (5) and (6) define the Beauford type
and Equations (9) and (10) define the Delastelle type of encipherment. [FR7]
I have said that the three steps in the cryptanalysis of repeating key systems are : 1) Find the length of the period, 2) Allocate or distribute the letters of the ciphertext into their respective alphabets, thereby reducing the polyalphabetic text to monoalphabetic terms, and 3) analysis of the individual monoalphabetic distributions to determine the plain text values of their cipher equivalents in each distribution or alphabet.
As a direct result of using a repeating key (no matter how long) certain phenomena are manifested externally to the cryptogram. Regardless of what system is used, identical plain text letters enciphered by the same cipher alphabet with single equivalents must yield identical cipher letters. This happens each time the same key letter is used to encipher identical plaintext letters.
Since the number of columns or positions with respect to the key are limited, and there is a normal redundancy in the language, it follows that there will be in a message of fair length many cases where identical plain text letters must fall into the same column. This will be enciphered by the same cipher alphabet, resulting in many repetitions. There are two types of repetitions: causal and accidental (random) repetitions. The former we can trace back to the key. The latter occurs when different plaintext letters fall in different columns and by chance produce identical cipher text letters.
Accidental repetitions will occur frequently with individual letters, less frequently with digraphs (because the accident must occur twice in succession, much less in the case of trigraphs and very much less in the case of a tetragraph. The probability of chance repetition decreases significantly as the repetition increases in length. Friedman has developed statistical tables based on the binomial and Poisson distributions to determine the individual and cumulative probabilities for expected number of repetitions in n letter text to occur x or more times in samples of random text.
The use of these tables is important. They tell us when we are dealing with cryptographically maneuvered text versus random noise designed to fool the listener. They indicate what may be a hoax (Beale or Bacon - Shakespeare controversies) versus valid enciphered text.
Tables 11-3 to 11-6 show the above theory.
Table 11-3 Number Expected Number of Digraphs Occurring of Exactly x Times Letters E(2) E(3) E(4) E(5) E(6) E(7) E(8) E(9) E(10) -------------------------------------------------------------- 100 6.21 .298 .011 200 21.8 2.12 .154 .009 300 42.5 6.23 .683 .060 .004 400 65.3 12.8 1.87 .220 .022 .002 500 88.1 21.6 3.97 .582 .071 .008 600 110. 32.3 7.11 1.25 .184 .023 .003 700 129. 44.3 11.4 2.35 .403 .059 .008 .001 800 145. 57.1 16.8 3.96 .777 .130 .019 .003 900 158. 70.1 23.2 6.16 1.36 .257 .043 .006 .001 1000 169. 83.0 30.6 9.03 2.21 .466 .085 .014 .002By way of illustration, of the use of these tables, from Table 11-3, we obseve that in a sample of 300 letters of random text, we may expect 43 digraphs to occur twice, 6 digraphs to occur three times and 1 digraph to occur four times. If we sum the values under E(2) through E(6) we have the cumulative probability in the 300 letter sample. The sum is 49.477, which indicates that in a sample of 300 letters or so, 49 digraphs will occur two or more times.
Table 11-4 Number Expected Number of Trigraphs Occurring of Exactly x Times Letters E(2) E(3) E(4) -------------------------- 100 .269 .001 200 1.10 .004 300 2.48 .014 400 4.40 .033 500 6.85 .064 600 9.81 .111 .001 700 13.3 .175 .002 800 17.3 .261 .003 900 21.8 .371 .005 1000 26.8 .505 .008
Table 11-5 Number Expected Number of Tetragraphs Occurring of Exactly x Times Letters E(2) E(3) -------------------------- 100 .010 200 .043 300 .096 400 .171 500 .270 600 .389 700 .530 800 .693 900 .877 1000 1.08 0.001
Table 11-6 Number Expected Number of Pentagraphs Occurring of Exactly x Times Letters E(2) ---------------- 100 200 .002 300 .004 400 .007 500 .011 600 .015 700 .021 800 .027 900 .034 1000 .042
The second step in the solution of periodic ciphers is to distribute the cipher text into the component monoalphabets. The period once established tells us the number of cipher alphabets. By rewriting the message in groups corresponding to the length of the key (period) in columnar fashion, we automatically have divided up the text so that letters belonging to the same cipher alphabet occupy similar positions in the groups or in the same columns.
If we make separate uniliteral frequency distributions for the isolated alphabets, each of these resulting distributions is therefore, a monoalphabetic frequency distribution. Were this not so, if they did not have the characteristic crest and trough appearance including the expected number of blanks, if the observed values of Phi are not sufficiently close to the expected value of Phi plain, or do not yield I.C.'s in the close vicinity of the expected value, then the entire analysis is fallacious.
The I.C. values of these individual distributions may be considered an index of correctness of the factoring process. Both theoretically and practically, the correct hypothesis with respect to these distributions will tend to conform more closely to the expected I.C. of a monoalphabetic frequency distribution.
Friedman demonstrates the above with an example: [FR7]
Plaintext Message:
The artillery battalion marching in the rear of the advance
guard keeps its combat train with it insofar as practical.
Keyword BLUE using direct standard alphabets.
Cipher Alphabets
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
---------------------------------------------------
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 A
2. 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 K
3. 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 T
4. 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 Text
USYES ECPMP LCCLN XBWCS OXUVD SCRHT
HXIPL IBCIJ USYEE GURDP AYBCX OFPJW
JEMGP XVEUE LEJYQ MUSCX JYMSG LLETA
LEDEC GBMFI
Friedman gives a useful formula for monographic I.C. of a 26 character
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 of elements in the distribution. I.C. for
English plain = 1.73 and 1.0 for random text. We may apply the I.C. test to the
distributions of periodic polyalphabetic ciphers to confirm the
monoalphabeticity of their character. This also confirms the period length and
correctness. If the correct period is assumed, then the Phi test applied to each
of the alphabets should approximate closely and consistently the value of Phi(p)
and conversely, if the incorrect period is assumed, then the Phi(o) should
approximate the value of Phi(r). Deviation from this 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 =42
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 I.C.=1.68
1 2 4 1 2 1 4 4 1 1 2 2 Phi=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 I.C.=1.91
1 5 1 1 1 1 5 2 1 1 2 1 1 2 Phi=46
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 I.C.=1.99
1 6 2 2 1 1 1 1 2 2 1 1 3 1 Phi=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 I.C.=1.91
It is seen that all these distributions are monoalphabetic since their
observed Phi's are closer to the Phi (p) = 40. rather than Phi (r) = 23. Any
other period assumed at four or a multiple of four, will not yield
monoalphabetic distributions.
In light of the foregoing principles, we now look at two additional cryptanalytic techniques for the Viggy family. The first compares the distributions to the normal and the second is very important - completing the plain-component.
Given message text A:
5 10 15 20 25
A. A U K H Y J A M K I Z Y M W M J M I G X N F M L X
B. E T I M I Z H B H R A Y M Z M I L V M E J K U T G
C. D P V X K Q U K H Q L H V R M J A Z N G G Z V X E
D. N L U F M P Z J N V C H U A S H K Q G K I P L W P
E. A J Z X I G U M T V D P T E J E C M Y S Q Y B A V
F. A L A H Y P O I X W P V N Y E E Y X E E U D P X R
G. B V Z V I Z I I V O S P T E G K U B B R Q L L X P
H. W F Q G K N L L L E P T I K W D J Z X I G O I O I
J. Z L A M V K F M W F N P L Z I O V V F M Z K T X G
K. N L M D F A A E X I J L U F M P Z J N V C A I G I
L. 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 alphabets to determine if we have standard alphabets.
Figure 11-7 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 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 V 2 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 3 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 4 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 S 5 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 Applying 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 X E N C O U N T E R E D R E D I N F A N T R Y E S TLook at the I.C.'s for these alphabets. The expected is 1.73. The third alphabet is almost exact. Three alphabets seem low and one is high or are they? Actually these deviations are within one sigma of the samples of these sizes 55 tallies, so the deviations are not abnormal. The standard deviations may be 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 than
observed:
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 observed and the expected number, divided by the standard deviation, it may 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 which follows the Chi squared distribution, this amounts to 1 chance in 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 known as the Beaufort Cipher. The direction of the crests and troughs is reversed when fitting the distributions to the normal.
When direct standard alphabets are used we can mechanically solve the cipher by completing the plain component. The plain text reappears on only one generatrix and this generatrix is the same for the whole message. It is the only generatrix that yields intelligible text. This same process can be modified to work with the alphabets of a Viggy. In this case the correct generatrix should be distinguishable from the others because it shows a more favorable assortment of high frequency letters, and thus can be selected by eye from the whole set of generatrixes.
Using the previous example, we let the first ten cipher letters in each alphabet be set down in a horizontal line and the assumption is made that the alphabets are direct standard with normal sequences. See Figure 11-8.
We use the following selection rules:
Figure 11-8
Gen./ 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 letters are
juxtaposed in columns, the consecutive letters of intelligible plain text
present themselves. If reversed standard alphabets are used, we must convert the
cipher letters of each isolated alphabet into their normal, plain component
equivalents, and then proceed as in the case of direct standard alphabets.
Friedman describes a graphical method for generatrix development in [FR7] and [FR8].
Time to move on to other family members. We shall identify the systems and peculiarities of each, but remember that the solution techniques presented for the papa bear apply equally well to the children and cousins.
The Variant Cipher is just that, a variant of the Vigenere, except that if the Viggy procedure is followed through, a peculiar keyword appears, like JYUWFT. Going back to the slides, In the Variant, the plaintext appears in the opposite slide from the one containing the key letter: Vigenere below the 'A' and Variant above the 'A'. The application of the high frequency letters is the same. The keyword is obtained in a different fashion. For the simple encipherment of COME AT ONCE 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 the keyword 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 GThe decipherment of a Variant is the same as a Vigenere.
From our trusty CDB, I found Variant.exe and applied it to the following
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 is extensively in the manufacture of printing inks, as a pigment for oil painting and also for waxing and lacquering of leather as well as in darkening a furniture polish. Total time 2 or 3 minutes.
A third member of the Viggy family, the Beaufort, and while the same procedure is applied, the slides (or tables) are different. One is a normal alphabet, extending double length A-Z; the other is reversed, double length Z-A. So if I = T at one setting, then T=I at the same setting. It does not matter what 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 --
I found BEAUFORT.exe at the CDB and applied it two the following message:
LDYUP AKUPT LVDTO BXUFW SERZP QMQPD NITHA NXUHE UGZTG HMGSM SRCUF LBQPZ XRYOB FDMNZ TGCUP QQUFB PANAQ HBOON XOOQP DJCJK TPFDV 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 with a wider key range and found that the true period was 10. After some trial and error, the keyword is LIGHTHOUSE and the message starts:
LEDGE points out some interesting relationships between the Vigenere, Variant
and Beaufort. Let A=0, B=1, C=2 .. Z=25, then:
For Vigenere and Variant if key letter = A, since A=0,the cipher text = plain text. If we reconstruct a cipher assuming it is a Vigenere, but it is actually a Variant, we will get the true plain text but strange keyword. By subtracting the Variant equation from the Vigenere equation and setting cipher text (Viggy) = ciphertext (Variant) and similarly plaintext (Viggy) = plaintext (Variant), we get the keyletter (Variant) = - keyletter(Vigenere) the same relationship as that between ciphertext and plaintext when the keyletter is A in the Beaufort (since A=0). Hence, we encipher our strange keyword with the A Beaufort alphabet to get the Variant key. The same holds true if we have a Variant and assume it a Viggy.
If we have a Vigenere and a fragment of the same message enciphered with the
same key in Variant (or visa versa) then:
Table 11-7 defines the PORTA Cipher. In this table the alphabets are all reciprocal, for example Gplain(Wkey) = Rcipher, Rplain(Wkey)=Gcipher. They are called complementary alphabets. Either of two letters may serve as a key letter indifferently: 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 plain letter.
Again our simple encipherment:
T E N T T E N T C O M E Y M S N A T O N W E I E C E - - Y T - -A peculiarity of this system is that since half the alphabet is represented by the half of the alphabet, there never will be found the letters A-M of the plaintext appearing as A-M in the ciphertext; 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 M
C,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 L
E,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 K
G,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 J
I,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 I
K,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 H
M,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 G
O,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 F
Q,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 E
S,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 D
U,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 C
W,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 B
Y,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 key word 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 1
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
4 2 5 1 3 4 4 1 2 3 1 2 4 9 1 2 5
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
5 3 3 2 5 1 1 3 4 7 2 2 4 8 1 1 2
3. 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 4
4. 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 1
5. 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 may be used to
fit a normal distribution. In alphabet 1, the sequence CDEFGHIJ cipher may
easily be recognized as NOPQRSTU plain; this would fix the keyletters as WX, and
therefor the A...Mplain sequence should begin with Ycipher. In alphabets 2,3,
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 the original assumptions for the first halves will
be confirmed by the 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 Porta encipherment, the cipher letters are first converted to their Porta plain-component equivalents and then these letters are used for the decipherment. EXCEPT, cipher letters A-M are completed in a downward direction and cipher letters N-Z are completed 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 E T P R K T O Q C F L P B V P X ....The conversion process and plain component completion of the first three alphabets are shown below using the generatrix elimination 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 I
3 D O C O S H C D 3 W P N J U D E N H J J F J F O J
6 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 L
0 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 A
3 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 E
2 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 F
5 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.
Just as the Vigenere table consisting of direct standard alphabets has its
complementary table of reversed standard alphabets, a variant of the Porta table
can be constructed where the lower halves of the sequences run in opposite
direction 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
The probable word method is very easy way to attack a Porta cipher. Let 1 =
any letter in the A-M sequence, and 2 equal any 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 E 2 1 2 1 1 1 1 2 1 2 2 1 2 2 2 1 2 2 1 1 1 1 2 1 1 T 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 2Use the probable word INFANTRY, which has the class notation of 12112222, 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.
At the trusty CDB is a program called PORTA.exe. Using it on the following
cipher message found a period of 9 with a possible key of
KL/IJ/CD/MN/AB/OP/OP/EF/QR. I came up with the keyword: 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 Yand behold the message cracked to:
The GRONSFELD Cipher uses a numerical key and restricts the Viggy table to
just ten alphabets. We can construct a slide with 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 other half for decipherment. For example the key
is derived as follows:
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 the right in
juxtaposition of the 6 = I, same for A which is G and 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 2
T 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 T
Q P S E V F D I X P O J W W K R
T 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 2
O 2 7 6 4 0
U 7 4 3 1
R 4 9
11.1 Viggy. SYCVT HFXEQ DPTLN KTGMP FHMPA SRVIT LSEXH DPITX KELIQ WDXEC VNLIP HPWXD XXIXH UTRIH. 11.2 Beaufort. SXSXZ IYLEQ AWEQF EZEPP QZQRD VANKH HLZJX OQSEU YSOVS SZKLE DRMRU THTUW SCLOX NEHLA OPEEU GAZIA UUOQG OJX. 11.3 Variant. JQRSB YBKNF WWTGK UXDTK ZAOAA MCVJU KBCEX GUYLB UASWY TIENQ XLPYX CWASU VAKOM XIGIK XHWZT SWGOP WRTSJ NAWG. 11.4 Gronsfeld. ZRWQU IKLMS IXAWI UQMWP KFQEL RBWJG XHIXT NLVKS ZHVHS ZRUEK 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 WAUHP HLVIN BZCCB GCGKZ JNLMM WTVLY DYCCV JPUVG KLKQX YTTKI XOQYB JJMHJ BYHQY LFQWF NRYUC XCECN GPCBW TPAXE ABKGC PVHKL OIKQW TPKOW KNCMM HFFAV A.
Thanks to JOE O for a fine analysis of all three problems.
QQ-1 QUAGMIRE I Travelogue. (Ends:SINGOUTOFTHESEA) RHIZOME
1234567 1234567 1234567 1234567 1234567 1234567 1234567
THEFIRS TIMEaVI SITOREX CLAIMSA HROMANT ICVENIC ESINKIN
KKQHPQR KTYOiTA TLGAWBM XORKTAT BSOOIYI CGICEJV UCYZRJP
ALNSFRZ UCQDXIS TDRBFYS YTFDZBD USQWKMT CPPDOAI CAAKEHK
UAYFHQA TLNIFSI SIGJHAS V.
QQ-1 Quagmire I Solution.
VERDICT/nose. Period =7.
The first time visitor exclaims "Ah, romantic Venice sinking
into the sea." The seasoned traveler exclaims,"Ah, stinking
Venice 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 S
QQ-2 QUAGMIRE III Tedious. (CRYPTANALYTIC METHODS)
DOPPELSCHACH
Period= 6
12345 61234 56123 45612 34561 23456 12345 61234 56123
THETI MEREQ UIRED BYS.......
PNATV SJBAQ WGMTR BZYLU ACACR GBNTQ FGGCN APNID ULMVD
SCEPB AMCQF BBPVR EOBSL AFSAN HFYVV MCYTF LEMAO MFHVU
KBAAU ATTEA NGOHU GTQEX ISUGU SAKCC TLIRT TLSZM PBMGV
APYRV YIIGL WGNUF JFROG SNQGN HBOTU TACUO JUVQH HUGWW
WBIMT WNHVO GTLSZ MPYQZ BNCEN UWLC.
HARDER/decorative. Period = 6. The time required by some
cryptanalytic methods grows extremely rapidly as key length or
message length increases. All possible keys for a columnar
transposition instead of making an entry by building up a
from 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 Z
1 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 G
2 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 R
3 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 O
4 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 Z
5 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 D
QQ-3 QUAGMIRE IV Economics Lesson. EDNASANDE
(BUSINESSACTIVITYDURINGAPERIOD)
THEEC ONOMY OFTHE NATIO ..........
TDNSE PMBSV FURMQ UFYSJ PAGGY FVIKT GYVLV FBTPH IIIAD
HVIUY QSAFA VQVFU HPIHE BIXNN HBSTN IRMQH IIIAD OVIXT
CTNOW EOJOZ BOWBU ONLFN GOBJS HBOQS VZMOU JSFQH SAHPS
JBBJT AAMIE XILRA TOTVL TUAML FLNEJ PPMNT XHVQV FCYSB
JODNF 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 Z
1 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
2 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 W
3 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 B
4 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 G
5 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 M
6 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 L
7 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 F
8 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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Text converted to HTML on June 29, 1998 by Joe Peschel.
Any mistakes you find are quite likely mine. Please let me know about them by
e-mailing:
jpeschel@aol.com.
Thanks.
Joe Peschel
