**By
Randy Nichols (LANAKI)** President of the American
Cryptogram Association from 1994-1996. Executive Vice
President from 1992-1994
#
CLASSICAL CRYPTOGRAPHY COURSE
BY LANAKI
March 5, 1996
Revision 0
COPYRIGHT 1996
ALL RIGHTS RESERVED
LECTURE 11
POLYALPHABETIC SUBSTITUTION SYSTEMS II CRYPTANALYSIS
OF VIGGY'S FAMILY
#
##
SUMMARY
##
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]
##
ZEN CRYPTO
##
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.
##
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 this
procedure in [FR7].
##
THE LONG AND SHORT OF KASISKI
##
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.
A 6 45 83 89 92 115 B 16 17 40 41 80 86 104 C 21 35 D
--- E 19 74 91 93 F 26 32 52 60 87 G 9 12 77 84 98 100 109 H 5
38 57 66 108 116 I 13 75 106 J 4 20 85 94 111 K 8 39 63 67 70
71 105 112 118 124 L 29 34 54 81 88 103 117 M 27 36 55 62 90
121 N 2 44 51 82 O 15 31 56 59 P 30 69 Q 3 28 49 R 1 22 37 79 S
18 25 50 72 78 101 T 33 43 64 76 110 119 U 7 24 42 114 120 V 10
14 61 65 68 97 102 W 11 46 53 73 95 96 122 123 X 107 Y 23 47 58
113 Z 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 and total the
tabulations for each factor in each of the letters of the
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 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.
##
WHAT CIPHERS MAKE UP THE VIGGY
FAMILY?
##
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 VIGENERE CIPHER
##
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:
For some fixed k, regard blocks of k letters as vectors in
(Z/NZ)**k. Where N is the N-letter alphabet and a digraph
integer correspondence exists between modulo N**2 array and it
is a vector mapping. Choose some fixed vector b which exists in
the plane (Z/NZ)**k which can be remembered by a key word and
encipher by means of the vector translation C = P +b where C is
the ciphertext message unit and P is the plaintext message unit
which is a k-tuple of the integers modulo N.
The object is to guess N and k, break up the ciphertext in
blocks of k letters and perform a frequency analysis on the
first letter of each block to determine the first component of b
and then proceeds onto the second letter in the block, etc.
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:
fi(a) = [(n-1)-a +(ki +1) ] mod n [SEAB]
The latter four descriptions are boring - even to my engineering
background. They also do not hold water for randomized alphabets
or tableauxs with disruption areas in place. These represent
discontinuities in the mathematical function. They are
discontinuous and tractable. Or differentiable if the model is
such. SCYER's program may have solved the discontinuity integer
problem by area limits or module limits. When he publishes the
procedure, maybe he will tell us.
##
WHICH WAY?
##
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:
Send Supplies To Morley's Station
enciphered with the repeating key, BED under the original method
of encipherment as might be described by Blaise de Vigenere
would be: Key : BEDB EDBEDBED BE DBEDBED BEDBEDB
Plain : SEND SUPPLIES TO MORLEYS STATION
Cipher: 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
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]
##
DECIPHERMENT BY PROBABLE WORD
##
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.
##
DECIPHERMENT BY PROBABLE TRIGRAM
SEQUENCE
##
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.
##
BRYAN'S SAINT-CYR 'HITS' METHOD
##
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.
##
VIGENERE COMPUTER SOLUTION IS
QUICKER
##
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:
The period was found within 1 second. The trial keyword
was PLQMBER, which I assumed was PLUMBER. Using PLUMBER as my
keyword, it typed out the answer: "AMONG CERTAIN TRIBES OF
INDIANS IN ALASKA.. ends BUT ARE USED AS SLAVES." The
process took less than 3 seconds of compute time on my 486/50.
I then rearranged the ciphertext with five nulls strategically
added. The next pass gave me a period of nine and a gibberish
trial keyword. So for well defined problems the computer is less
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
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:
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 the
component, but by convention only. We will assume from now on
that 01 is the initial letter of the component in which it is
located. 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. 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]
##
FURTHER REMARKS ON REPETITIONS
##
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 .002
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
By 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.
##
STATISTICAL PROOF OF THE
MONOALPHABETICITY OF THE DISTRIBUTIONS
##
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.
##
SOLUTION BY FITTING THE
DISTRIBUTIONS TO THE NORMAL
##
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.
Alphabet 1 I.C. = 1.44 5 1 2 3 3 3 2 2 6 2 1 6 1 5 3 1
2 1 6 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
Alphabet 2 I.C. = 1.47 5 1 1 3 3 1 2 4 9 1 2 5 1 2 4 4 4 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 Alphabet 3 I.C.
= 1.71 2 3 1 8 2 2 4 8 1 1 2 3 4 5 1 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 Alphabet 4 I.C. = 1.36 3 1 1 3 4
4 4 2 2 3 3 1 1 1 2 2 4 9 2 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 Alphabet 5 I.C. = 1.91 6 2 4 8 1 3 7 1 2 1 4
3 5 2 2 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
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,E
cipher. A little experimentation gets us Aplain in alphabet 1=
Wcipher. The key word under Aplain is WHITE. The five complete
cipher alphabets are shown in matrix form in Figure 11-7.
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 T
Look 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.
##
SOLUTION BY COMPLETING THE
PLAIN-COMPONENT SEQUENCE
##
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:
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 a
number) as in the Gronsfeld Cipher.
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.
For Alphabet 1, generatrix 5.. E N D N R I D E M N For
Alphabet 2, generatrix 20.. N T R F Y M A R E D For Alphabet 3,
generatrix 19.. C E E A E A T E N M For Alphabet 4, generatrix
8.. O R D N S T O G T A For 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 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.
##
VARIANT CIPHER
##
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 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 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.
##
BEAUFORT CIPHER
##
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 --
##
BEAUFORT SOLUTION BY COMPUTER
NEEDS WORK
##
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:
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 the
Vigenere, 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
ciphertext 1-2=-1 +26 = 25 = Z.
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:
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 same
key 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 13
will give one number from 0-12 and another 13-25. Solution is by
inspection.
##
PORTA (aka NAPOLEON'S TABLE)
##
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.
##
MODIFIED PORTA
##
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
##
PROBABLE WORD METHOD OF SOLUTION
FOR PORTA
##
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 2
Use 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.
##
COMPUTER SOLUTION OF PORTA
##
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 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 the
actual keyword or require it.
##
GRONSFELD
##
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
##
LECTURE
11 PROBLEMS
##
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.
##
ANSWERS TO LECTURE 10 PROBLEMS
##
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
##
REFERENCES / RESOURCES
##
####
[updated 5 May 1996]
####
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written, without proper authority, unprofessional, and
prejudicial to boot. And, it has one of the better
illustrations of the Soviet one-time pad with example,
with three errors in cipher text, that I have corrected
for the author.]
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