## Lesson 13: Aperiodic Systems Improving Cryptographic Security In Polyalphabetic Systems

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CLASSICAL CRYPTOGRAPHY COURSE
BY LANAKI

June 10, 1996
Revision 0

LECTURE 13

APERIODIC SYSTEMS

IMPROVING CRYPTOGRAPHIC SECURITY IN POLYALPHABETIC SYSTEMS

SUMMARY

Lecture 13 describes the difficult aperiodic polyalphabetic
case and reconsiders the Principle of Superimposition.  We
diagram the topics (I consider the heart) considered in
Lectures 10 - 13.  We develop our subject via the following
references [FRE3], [SACC], [BRYA], [SINK], [OP20] and [ELCY].

COURSE SCHEDULE CHANGES

In order to be more cost-efficient, I have been thinking how
to condense some of my future lecture material.  Here is how
the schedule looks for the balance of my course:

Lecture 14 - Cryptarithms by LEDGE
Lecture 15 - Statistical Methods (Sinkov, Kullback, Friedman)
Lecture 16 - Transposition
Lecture 17 - Transposition
Lecture 18 - Fractionation, Advanced Monome - Dinome Systems
Lecture 19 - Law and Politics of Cryptography
Lecture 20 - Cipher Exchange Systems
Lecture 21 - Cipher Exchange Systems
Lecture 22 - Modern Crypto-Systems, Double Key Cryptography,
Cipher Machines, PGP and PGPphone, Diamond
Cipher Family
Lecture 23 - Volume I and II References / Resources, Index to
Volume II Lectures 11 - 22. Table of Figures,
Table of Tables; Presentation of Certificates of
Achievement to my Students and Grateful Thanks
form LANAKI!

LEDGE has done a marvelous job on the Cryptarithms section. I
will leave open a slot for him if he consents to a third
Lecture.  Expect Lecture 23, which is devoted to Resources
Several of our class are helping out with an extra set of
eyes, to correct my atrocious typing and other errors caused
by crossing the different E-Mail gateways.  I thank them for
their valued help.

IMPROVING CRYPTOGRAPHIC SECURITY IN POLYALPHABETIC SYSTEMS

The last two chapters have explored the effects of repeating
key ciphers and the periodicity that occurs in them.
Establishing the period opens the wedge for solution of this
type of cipher system.  The difficulty of solution is related
to the number of cipher alphabets employed and their type.

Two procedures suggest themselves to improve the crypto-
graphic security of these systems.  First, we can increase
the length of the key.  This is akin to what we do with
modern public key systems.  Second, since the first step in
solution of a Viggy or other polyalphabetic cipher is to
establish the period, hence the number of alphabets employed,
we can eliminate the periodicity, and therefore eliminate the
cryptanalyst's attack.

APERIODIC CIPHER SYSTEMS

What is the real nature of periodicity in polyalphabetic
substitution systems?  How do we remove periodicity from
ciphers?

We understand the cyclic and repeating nature of a keyword
based periodic system.  However, we have taken for granted
that the keying element acts on constant-length plain text
groupings.  If this were not true, there would not be any
external manifestation of periodicity, despite the repetitive
or cyclic use of a constant-length key.  The key is of a
constant or fixed character.

Two approaches for eliminating or suppressing periodicity
come to bear: 1) by using constant-length keying units to
encipher variable-length plain-text groupings or 2) by using
variable-length keying units to encipher constant-length
plain-text groupings.

In cases of encipherment by constant-length groupings, the
apparent length of the period (found by factoring) is a
multiple of the real length and the multiple corresponds to
the length of the groupings, i.e. the number of plain-text
letters enciphered by the same key letter. Periodicity still
exists because in every system studied so far both the keying
units and the plain-text groupings are constant in length.

EFFECT OF VARYING THE LENGTH OF PLAIN-TEXT GROUPINGS

Lets assume that the keying units are kept constant and vary
the plain-text groupings.  The effect is to suppress
periodicity, even though the key may repeat itself many times
in the cryptogram.  This is true unless the law governing the
variation in plain-text groupings is itself cyclic in
character, and the length of the message is at least two or
more times that of the cycle applicable to this variable
grouping. [FRE3]

For example we encipher the following message using the
keyword SIGNAL, but divide up the plain-text into groups:

S I   G   N     A    L S   I   G    N     A  L S   I     G
1 12 123 1234 12345  1 12 123 1234 12345  1 12 123 1234 12345
C OM MAN DING GENER  A LF IRS TARM YHASI  S SU EDO RDER SEFFE
Q UW UGT KFAH UWNWJ  L HN ARQ NGPU PGNVF  I TR OPE RFER OCBBC

N A   L   S    I     G N   A   L    S     I G  N   A    L
1 12 123 1234 12345  1 12 123 1234 12345  1 12 123 1234 12345
C TI VET WENT YFIRS  T AT NOO NDIR ECTIN  G TH ATT ELEP HONES
L HS QHS WOFZ KDARQ  N NU NMM YIDU OQZKF  C NZ NUU WPWL EXYHT

S I   G   N     A    L S  I
1 12 123 1234 12345  1 12 123...
C OM MAS WITC HBOAR  D SC OMM...
Q UW UGO RFUL TZMAJ  I AQ UWW...

Cryptogram

QUWUG  TKFAH  UWNWJ  LHNAR  QNGPU  PGNVF  ITROP ERFER
OCBBC  LHSQH  SWOFZ  KDARQ  NNUNM  MYIDU  OQZKF CNZNU
UWPWL  EXYHT  QUWUG  ORFUL  TZMAJ  IAQUW  W...

The cipher text above shows a tetragraphic and a pentagraphic
repetition.  The two occurrences of QUWUG (COMMA) are
separated by an interval of 90 letters, the two occurrences
of ARQN (=IRST) by 39 letters.  The former is the true
periodic repetition measured in grouping cycle rather than
letters.  The interval is the product of the keying cycle of
6 by the grouping cycle of 15. The latter repetition are
produced by the same key letters I and G but do not have the
same enciphering points and is considered a partial periodic
as opposed to a completely periodic type.

Kasiski analysis focuses on the intervals between repetition
letters and developing factors which indicate the number of
cipher alphabets employed.  We also can study the interacting
cycles that produce the intervals directly.  If we look at
the above as counting according to groupings and not
according to single letters, the two pentagraphs QUWUG are
separated by an interval of 30 groupings. The separation of
30 key letters is made up of a key 6 letters in length and
has gone through 5 cycles. So 30 is the product of the number
of letters in the keying cycle (6) times the number of
different-length groupings in the grouping cycle (5).

Friedman describes a clever little cipher system based on a
lengthy grouping cycle which is guided by a key of its own.
We can use the number of dots and dashes contained in the
International Morse signals for the letters composing the
phrase DECLARATION OF INDEPENDENCE.  Thus, A(._) has 2,
B(_...) has 4, and so on.  So:

D E C L A R A T I O N O F I N D E P E N D E N C E
3 1 4 4 2 3 2 1 2 3 2 3 4 2 2 3 1 4 1 2 3 1 2 4 1

The grouping cycle is 3+1+4+4+.., or 60 letters long. If the
same phrase is used as the enciphering key (25 letters) the
complete period of the system would be the least common
multiple of 25 and 60 or 300 letters.  The length of the
complete period is the least common multiple of the two
component or interacting periods.

One drawback - the variable factor introduced above is
subject to a law which itself is periodic in character.

SOLUTION OF SYSTEMS USING CONSTANT-LENGTH KEYING UNITS TO
ENCIPHER VARIABLE-LENGTH PLAIN-TEXT GROUPINGS

APERIODIC GROUPINGS ACCORDING TO WORD LENGTHS

The simplest way to introduce aperiodicity is to encipher our
message by actual word lengths.  Although the average number
of letters composing words of any alphabetical language is
fairly constant, successive words comprising plain text vary
a great deal in this respect, and the variation is subject to
no law. In English, the mean length of words is 5.2 letters
but the words may contain from 1 - 15 or more letters;
successive words vary in length in an extremely irregular
manner, no matter how long the text is.

The use of word lengths for determining the number of letters
to be enciphered by each key letter of a repetitive key
seems more secure than it is.  The reasoning goes something
like this: if there is no periodicity in the cryptogram, how
can the letters of the cipher text, written in groups of five
letters be distributed into their respective monoalphabets.
If the first step is foiled, how can the cryptograms be
solved?  The answer: using a variation of the completion of
the plain component sequence method discussed under the
monoalphabetic cipher cracking.

SOLUTION WHEN DIRECT STANDARD CIPHER ALPHABETS ARE EMPLOYED

Since the individual separate words of a message are
enciphered by different key letters, these words will
reappear on different generatrices of the diagram.

Given:

T R E C S   Y G E T I   L U V W V   I K M Q I   R X S P J
S V A G R   X U X P W   V M T U C   S Y X G X   V H F F B
L L B H G.

First step: Run down the first 10 letters for a clue.

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

We place these over the first ten cipher letters to backout
the keying letters of R E A.  We can either set up the
remaining letters of the message on a sliding normal alphabet
scale or assume various keywords such as READ, REAL, REAM.
The completed solution is:

R      E      A        D             E             R

C A N  Y O U  G E T   F I R S T   R E G I M E N T   B Y
T R E  C S Y  G E T   I L U V W   V I K M Q I R X   S P

S         D         I         G      E      S

R A D I O   O U R   P H O N E   N O W   O U T   O F
J S V A G   R X U   X P W V M   T U C   S Y X   G X

T

C O M M I S S I O N
V H F F B L L B H G.    Key = READERS DIGEST

The slide is very quick.  We place the C(1) over T(2) and
back out the index A(1) over the Key letter R(2). For the
second group the Y(1) over C(2) will back out the A(1) over
E(2).  If reversed standard alphabets are employed we either
convert the cipher letter to normal alphabets or employ the
reverse alphabet slide. The slides, if not out of stock,
referred to are available from ACA for about \$3. It may be
used to aid solutions for the entire Viggy family.

SOLUTION WHEN ORIGINAL WORD LENGTHS ARE RETAINED IN THE
CRYPTOGRAM

Given the enciphered message:
DIVISION
12324256
XIXLP  EQVIB  VEFHAPFVT  RT  XWK  PWEWIWRD  XM  NTJCTYZL

BATTALIONS      ARTILLERY
1233245678      123455627
OAS  XYQ  ARVVRKFONT  BH  SFJDUUXFP  OUVIGJPF  ULBFZ

OCLOCK
123124
RV  DKUKW  ROHROZ.

We crack the above using Idiomorphs and "Probable Word"
analysis.

We note the Idiomorphs and use the Cryptodyct or TEA:

1) 12324256    = 32426 (8) = DIVISION
PWEWIWRD

2) 1233245678  = 3328  (10) = BATTALIONS
ARVVRKFONT
3) 123455627   = 55627 (9) = ARTILLERY
SFJDUUXFP

4) 123124      = 3124 (6) = O'CLOCK
ROHROZ

Using the assumed equivalents a reconstruction matrix is
established on the hypothesis that the cipher alphabets have
been derived from a mixed component against a normal
sequence.   Note that O(plain) = R(cipher) in both DIVISION
and OCLOCK, so the same cipher alphabet has been used.

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
DIVISION|---------------------------------------------------
OCLOCK  |  n O P   s t v W x Z H   D R a u   I     E f   j k
|
BATTA-  |---------------------------------------------------
LIONS   |R A             F     K   N O       T V
|
ARTILL- |---------------------------------------------------
ERY     |
|S       X       D     U           F   J         P
|---------------------------------------------------

The interval between letters O and R in the first and second
alphabets is the same at 12, therefore direct symmetry of
position is assumed.   We fill in the additional letters
(lower case).

It is a short stretch to find the keyword HYDRAULIC and to
decipher the equivalents based on the HYDRAULIC...Z sequence
against the normal alphabet at any point of coincidence and
completing the plain component sequence.  The words of the
message will then reappear on different generatrices. The key
letters may be ascertained and the solution completed. The
first three words are deciphered as follows:

XIXLP       EQVIB        VEFHAPFVT
----------------------------------
YHYGS       KTWHJ        WKLAESLW
ZIZHT       LUXIK        XLMBFTMX
AJAIU       MVYJL        YMNCGUNY
BKBJV       NWZKM        ZNODHVOZ
CLCKW       OXALN        AOPEIWPA
DMDLX       PYBMO        BPQFJXQB
ENEMY       QZCNP        CQRGKYRC
Ap = Sc      SBEPR        ESTIMATE
TCFQS
UDGRT        Ap = Pc
VEHSU
WFITV
XGJUW
YHKVX
ZILWY
AJMXZ
BKNYA
CLOZB
DMPAC
ENQBD
FORCE
Ap = Uc

The key for this message is SUPREME COURT and the complete
message is:

ENEMY  FORCE  ESTIMATED  AS  ONE  DIVISION  OF  INFANTRY
XIXLP  EQVIB  VEFHAPFVT  RT  XWK  PWEWIWRD  XM  NTJCTYZL

AND  TWO  BATTALIONS  OF  ARTILLERY  MARCHING  NORTH
OAS  XYQ  ARVVRKFONT  BH  SFJDUUXFP  OUVIGJPF  ULBFZ

AT  SEVEN   OCLOCK
RV  DKUKW  ROHROZ.

In the case of plain component in reverse normal alphabets.
the procedure is the same , except the completion tableaux
is created after the cipher letters are converted into their
plain-component equivalents.

ILLUSTRATION OF THE USE OF ISOMORPHISM

Consider the following cryptogram which has been enciphered
using the primary key word-mixed alphabet of (HYDRAULIC...XZ)
against a normal sequence.  I have retained word lengths for
simplicity:

VCLLKIDVSJDCI   ORKD   CFSTV   IXHMPPFXU   EVZZ

FK  NAKFORA   DKOMP   ISE   CSPPHQKCLZKSQ   LPRO

JZWBCX   HOQCFFAOX   ROYXANO   EMDMZMTS

We look at three sets of isomorphs:

1)  a VCLLKIDVSJDCI             2) a IXHMPPFXU
b CSPPHQKCLZKSQ                b HOQCFFAOX
3) a NAKFORA
b ROYXANO

Rather than identifying these from a TEA or Cryptodyct
database, we build up the partial sequences of equivalents.
[TEA], [CRYP]

>From 1a and 1b:
V = C, C = S, L = P, K = H, I = Q, D = K, S = L, J = Z

so:     VCSLP    DKH   IQ   JZ  are constructed.

>From 1b and 1c:

C=P, S=A, P=D, H=E, Q=R, K=X, L=N, Z=B

We find:

CPD SA HE QR KX LN ZB

>From 1a and 1c:

V=P, C=A, L=D,K=E,I=R,D=X,S=N,J=B

and:
LDX  VP  CA  KE   IR   SN   JB

Noting that the three isomorphs may be combined (VCSLP and
CPD make VCSLP..D; the latter and LDX make VCSLP..D...X),
the following sequences are established:

1 2 3 4 5 6 7 8 9 10 11 12 13
1. V C S L P A N D K H  .   X  E
2. I Q . . R
3. J Z . . B

Chain 1 contains exactly 13 letters and suggests a half-chain
is disclosed. the latter represents a decimation of the
original primary component at an even interval.

1 2 3 4 5 6 7 8 9 10
The placement of the letters V . S . P . N . K .  suggests a
reversed alphabet; we reverse the half-chain and extend to 26
places as follows:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
E     j K   N   P  q S     V     X  z  H      D r  A

23 24 25 26
L  i  C  b

We add the data from the two partial chains (JZ..B and
IQ..R).  [Small letters]

The keyword is HYDRAULIC. the full sequence is:

1234567891011121314151617181920212223242526
HYDRAULIC B E F G J K M N O P Q S T V W X Z

We confirm from 2a and 2b that the interval between H and I
is 7; same for O to X and Q and H, and C and M. From
idiomorphs 3a and 3b, the interval between R and N is 13;
which is the same for O and A and Y and K.

We now convert the ciphertext letters into plain-component
equivalents then complete the plain component sequences.

Solution Key: Strike While The Iron is (Hot?)

>From the slide we put A/S which confirms O/S,M/L C/V etc.,

S           T      R         I         K
COMMUNICATION   WITH   FIRST   ARTILLERY   WILL
VCLLKIDVSJDCI   ORKD   CFSTV   IXHMPPFXU   EVZZ

E     W        H       I        L            E
BE  THROUGH   CORPS   AND   COMMUNICATION   WITH
FK  NAKFORA   DKOMP   ISE   CSPPHQKCLZKSQ   LPRO

T         H           E         I
SECOND   ARTILLERY   THROUGH   DIVISION
JZWBCX   HOQCFFAOX   ROYXANO   EMDMZMTS

R        O          N          I       S
SWITCHBOARD   NO   COMMUNICATION   AFTER   TEN

Four assumptions were made in the above:

1.  The actual word lengths were known.
2.  The words were enciphered monoalphabetically by
different alphabets, producing isomorphs and lengths
of isomorphs that are known.

3.  Repetitions of plain-text words enciphered by
different alphabets, produce isomorphs and the
lengths of the isomorphs are definitely known as a
result of this action.

What if the cryptogram is put in the form of a Patristocrat
with 5-letter-groups?

Take the same problem as above and destroy the word lengths.
The problem is a little more difficult and requires more
trial and error.

VCLLK  IDVSJ  DCIOR  KDCFS  TVIXH  MPPFX

UEVZZ  FKNAK  FORAD  KOMPI  SECSP  PHQKC

LZKSQ  LPROJ  ZWBCX  HOQCF  FAOXR  OYXAN

OEMDM  ZMTST  ZFVUE  AORSL  AUPAD  DERXP

NBXAR  IGHFX  JXI.

The 13 letter isomorps are relatively easy to spot:

1. VCLLKIDVSJDCI
2. CSPPHQKCLZKSQ      Column ends IQR

Number 1 is the "header" and the left-hand boundary is known.
The right hand boundary marked by IQR is fortuitous. Not
knowing the exact length by one or two letters is not fatal
to the solution because we are interested in reconstructing
cipher equivalents not looking up the pattern words.

Isomorphism is not restricted to cases where secondary
alphabets are derived from a primary component sliding
against the normal.  It is useful in all cases of
interrelated alphabets no matter what the basis of their
derivation may be. It is second only to the importance of the
"Probable Word" method which has nearly universal
applicability.

SOLUTION OF SYSTEMS USING VARIABLE-LENGTH KEYING UNITS TO
ENCIPHER CONSTANT-LENGTH PLAIN-TEXT GROUPINGS

THE INTERRUPTED KEY CIPHER

Periodicity can also be suppressed by applying variable-
length key groupings to constant length plain-text groups.
One such method is the Interrupted Key Cipher which employs
an irregularly interrupted key sequence, the latter may be of
fixed or limited length and restarting it from its initial
point after the interruption, so that the keying sequence
becomes equivalent to a series of keys of different lengths.

Take the phrase BUSINESS MACHINES and expand it to a series
of irregular-length keying sequences, such as BUSI/BUSINE/
interruption prearrangement are given by Friedman [FRE3]:

(1) The keying sequence merely stops and begins again at
the initial point of the cycle.

(2) One or more of the elements in the keying sequence
may be omitted from time to time irregularly.

(3) The keying sequence irregularly alternates in the
direction of progression, with or without omission
of some of the elements.

Using an asterisk to indicate an interruption, a sequence of
10 elements might look like this:

Letter No     1 2 3 4  5 6 7 8 9 10  11 12 13  14 15 16
Key Element No1-2-3-4*-1-2-3-4-5-6-* -1- 2- 3-*-1- 2- 3
Letter No     17 18 19 20  21 22 23 24 25 26 27 28 29 30
(1) Key Element No-4 -5 -6 -7-* 1 -2 -3 -4 -5 -6 -7 -8 -9 -10
Letter No       31 32 33   34 35
Key Element No_*-1 -2 -3-*- 1 -2

Letter No     1 2 3  4 5 6 7   8 9  10 11 12  13 14 15 16
Key Element No1-2-3-*-7-8-9-10-1-2-*-4- 5- 6-*-3- 4- 5- 6
Letter No     17 18 19 20  21  22 23 24 25 26  27 28 29
(2) Key Element No-7 -8 -9-10- 1-*-8- 9- 10 -1-2-* -5 -6 -7-*
Letter No       30 31 32   33 34 35
Key Element No - 9-10 -1-*-5 -6 -7-

Letter No     1 2 3 4 5   6 7  8 9 10 11 12 13 14 15   16
Key Element No1-2-3-4-5-*-4-3-*4-5 -6 -7 -8 -9-10 -1-*-10
Letter No     17 18 19  20 21 22 23 24 25  26 27 28 29 30
(3) Key Element No 9- 8- 7-*-8- 9-10- 1- 2- 3-*-2- 1-10- 9-8-
Letter No       31 32 33 34 35
Key Element No *-9-10- 1- 2- 3

Method three is a key progression direction reversing method
and if their were no interruptions in the key, it could be
handled as a special form of the second method.  However,
combined with the second method, it represents a difficult
cryptographic variant.

RETURNING TO THE PRINCIPAL OF SUPERIMPOSITION

If one knows when the interruptions take place in each cycle,
then successive sections of the basic keying cycle in the
three cases may be superimposed.  Obviously, if one does not
know when or how the interruptions take place, the then the
successive sections of keying elements cannot be
superimposed.  See Table 13-1.

The interruption of the cycle keying sequence practically
takes place according to some prearranged plan, and the three
basic methods of interruption will be described in turn using
a short mnemonic key as an example.

Suppose we agree to interrupt the keying sequence after the
occurrence of a specified letter (called an interruptor-
fancy that). This may be a plain or cipher text letter,
agreed to in advance.  Then since in either case their is
nothing fixed about the time of interruption will occur - it
will take place at no fixed intervals - not only does the
interruption become quite irregular, following no pattern,
but also the method never reverts back to one having
periodicity. We have the LANAKI equivalent of a DOOSEY in the
polyalphabetic arena.

We will use the mnemonic key BUSINESS MACHINES and the cipher
alphabet HYDRAULIC...XZ sequence which slides:

1234567891011121314151617181920212223242526
HYDRAULIC B E F G J K M N O P Q S T V W X Z

The keying set is a Viggy, so A(1)/K(2) = P(1)/C(1),
where A is the index, K is the key letter, P is the plain
text letter, C is the ciphertext letter, (1) and (2)
subscripts refer to the top and the bottom slides.

Table 13-1

Method (1)
Keying Element No  1  2  3  4  5  6  7  8  9  10
------------------------------
Letter No          1  2  3  4|
Letter No          5  6  7  8  9  10|
Letter No          11 12 13|
Letter No          14 15 16 17 18 19 20|
Letter No          21 22 23 24 25 26 27 28 29 30|
Letter No          31 32 33
Letter No          34 35

Method (2)
Keying Element No  1  2  3  4  5  6  7  8  9  10
------------------------------
Letter No          1  2  3  -  -  -  4  5  6  7
Letter No          8  9| -  10 11 12|
Letter No          -  -  13 14 15 16 17 18 19 20
Letter No          21|-  -  -  -  -  -  22 23 24
Letter No          25 26|-  -  27 28 29| - 30 31
Letter No          32|-  -  -  33 34 35

Method (3)
Keying Element No  1  2  3  4  5  6  7  8  9  10
------------------------------
Letter No          1  2  3  4  5| -  -  -  -  -
Letter No          -  -  7  6  -  -  -  -  -  -
Letter No          -  -  -  8  9  10 11 12 13 14
Letter No          15|-  -  -  -  -| 19 18 17 16
Letter No          23 24 25|-  -  -  -  20 21 22
Letter No          27 26 -  -  -  -  - |30 29 28
Letter No          33 34 35                31 32

PLAIN TEXT LETTER INTERRUPTOR

Let the plain text letter R be designated an interruptor.
Interruption will occur immediately after and R occurs in the
plain text.

Index  A(1)
Key    K(2) B U S I N E S S M A C H I|B U S|B U S I|B U S I
Plain  P(1) A M M U N I T I O N F O R F I R S T A R T I L L
Cipher C(2) B O L Y R P J D R O J K X K J F Y X S X D J U P

Index  A(1)
Key    K(2) N E|B U S I N E S S M A C H I N E S B U|B U S I
Plain  P(1) E R Y W I L L B E L O A D E D A F T E R A M M U
Cipher C(2) S Y I Y D P Y F X U R A F A E N M J J V B O L Y

Index  A(1)
Key    K(2) N E S S M A C H I|B U S I|B U S|B U S I N E|
Plain  P(1) N I T I O N F O R T H I R D A R T I L L E R
Cipher C(2) R P J D R O J K X D G D X G U F D J U P S Y

Index  A(1)
Key    K(2) B U S I N|
Plain  P(1) Y . . . .
Cipher C(2) I

examples:   with Index = A

group 1, plain letter 1 = A ; key B ; cipher B

Plain           ABCDEFGHIJKLMNOPQRSTUVWXYZ
Cipher HYDRAULICBEFGJKMNOPQSTVWXZHYDRAULICBEFGJKMNOPQSTVWXZ
*

group 1, plain letter 2 = M ; key U ; cipher O

Plain       ABCDEFGHIJKLMNOPQRSTUVWXYZ
Cipher HYDRAULICBEFGJKMNOPQSTVWXZHYDRAULICBEFGJKMNOPQSTVWXZ
*           *
group 1, plain letter 3 = M ; key S ; cipher L

Plain                      ABCDEFGHIJKLMNOPQRSTUVWXYZ
Cipher HYDRAULICBEFGJKMNOPQSTVWXZHYDRAULICBEFGJKMNOPQSTVWXZ
*           *
group 2, plain letter 1 = F ; key B ; cipher K

Plain           ABCDEFGHIJKLMNOPQRSTUVWXYZ
Cipher HYDRAULICBEFGJKMNOPQSTVWXZHYDRAULICBEFGJKMNOPQSTVWXZ
*    *

group 2, plain letter 2 = I ; key U ; cipher J

Plain       ABCDEFGHIJKLMNOPQRSTUVWXYZ
Cipher HYDRAULICBEFGJKMNOPQSTVWXZHYDRAULICBEFGJKMNOPQSTVWXZ
*       *

group 3, plain letter 2 = S ; key B ; cipher Y

Plain           ABCDEFGHIJKLMNOPQRSTUVWXYZ
Cipher HYDRAULICBEFGJKMNOPQSTVWXZHYDRAULICBEFGJKMNOPQSTVWXZ
*                 *

Cryptogram

B O L Y R   P J D R O   J K X K J   F Y X S X   D J U P S
Y I Y D P   Y F X U R   A F A E N   M J J V B   O L Y R P
J D R O J   K X D G D   X G U F D   J U P S Y   I X X X X

Instead of employing an ordinary letter for interruptor, we
can use a low frequency letter like J. Actually any letter,
no matter what frequency level will produce plentiful
repetitions. The only advantage is that the intervals will be
random and therefor suppress (or reduce) periodicity.

INTERRUPTOR CASES

The interruptor problem presents two cases for investigation.
The first is when the system has been used several times and
the cipher alphabets are known.  The second case is when the
cipher alphabets are not known but several messages have been
intercepted.

Case 1 - Cipher Alphabets Known, Problem is to find specific
key.

Attack: Probable word. Using probable word ARTILLERY on
previous example, starting from first letter we have:

Cipher      B O L Y R P J D R
Plain       A R T I L L E R Y
'Key'       B H J Q P I B F U

failed.  We move one cipher letter to right with assumed
word.  Continued failure until the following:

Cipher      S X D J U P S Y I
Plain       A R T I L L E R Y
'Key'       S I B U S I N E B
*           *

We note the BUSINE suggesting BUSINESS. We also note the
interruptor letter R.  We use this key part on the first part
of the message with success.

Key         B U S I N E S S B U S
Cipher      B O L Y R P J D R O J
Plain       A M M U N I T I U M T

The last three letters suggest that there is more to the key.
Using Ammunition and back calculating the Key, we find
MA. We use the cipher and the plain back and forth to find
the total key, taking into account the interruptor letter R.

Case 2 - Cipher Alphabets Unknown, Problem is to find both
cipher alphabet and specific key.

Assume that the repetitive key is very long and that the
message is short. Solution is difficult because there are not
enough superimposable periods to help line up the alphabets
to yield monoalphabetic distributions that can be solved by
frequency principles.  This is the first step in the
cryptanalytic attack.  The superimposed periods essentially
line up the letters in the columns so that the same treatment
has been used to process both plain and cipher components.

Attack: Solution by Superimposition.

The second most important attack on cryptanalytic problems is
the Solution by Imposition. First we need a sufficient number
messages (25 - 30 for English) enciphered by the same key to
work with.  It is clear that if we superimpose these
messages, 1) the letters in the respective columns will all
belong to individual alphabets; and 2) a frequency
distribution of the columnar letters can be solved without
knowing the length of the key. In other words, any
difficulties that may have arisen on account of failure to
ascertain the length of the period have been circumvented.
The second step in the solution is by-passed. (3) For a very
long key employed, and a series of messages beginning at
different initial points are enciphered by the same key, this
method of attack can be employed after the messages are
superimposed at the same initial point [done with the help of
the Chi square test].  An example of this will be done in a
later lecture on statistical techniques.

CIPHER TEXT LETTER INTERRUPTOR

If we use a cipher text letter, say Q, as the interruptor, we
find a more difficult case with no significant repetitions
available for superimposition.

Key    K(2) B U S I N E S S M A C H I N E S B U S I N E S S
Plain  P(1) A M M U N I T I O N F O R F I R S T A R T I L L
Cipher C(2) B O L Y R P J D R O J K X T P F Y X S X B P U U

Key    K(2) M|B U S I N E S S M A C H I N|B U S I N E S S M
Plain  P(1) E R Y W I L L B E L O A D E D A F T E R A M M U
Cipher C(2) Q H R N M Y T T X H P C R F Q B E J F I E L L B

Key    K(2) A C H|B U|B U S I N E S S M A C H|B U S I N E
Plain  P(1) N I T I O N F O R T H I R D A R T I L L E R Y
Cipher C(2) O N Q O Q V E C X B O D F P A Z Q O N U F I C

Cryptogram

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

The attack is first to find the interruptor and then to
recover the plain by method of superimposition. To accomplish
superimposition a statistical test is essential and for this
a good many letters are required.

THE AUTO-KEY CIPHER or AUTOCLAVE CIPHER

The purpose of the Auto-key Cipher or Autoclave Cipher is to
eliminate periodicity and introduce a long key for the entire
message.  The Autoclave may be used with the Vigenere,
Variant, Beaufort, Gronsfeld, Porta or the Nihilist
Substitutions' basic principles. The overall picture is the
same; its handling, however, depends on the system involved.
>From a purely theoretical standpoint, we are approximating
the features of a One-Time Pad.

In practice, the Auto-Key is a nightmare.  MASTERTON points
out that the slightest difficulty in transmission of cipher
letters destroys the communication.  [MAST] Other authors
[ELCY] and [BRYA] and ACA KREWE find the Auto-Key and
Progressive Ciphers a real challenge.  There are two possible
sources for successive key letters: the plain text or the
cipher text of the message itself. In either case, the
initial key letter or key letters are supplied by pre-
agreement between the correspondents; after which the text
letters that are to serve as the key are displaced 1,2,3..
intervals to the right, depending upon the length of the
prearranged key.

Lets review the methods.

Plain-text keying using the single letter X:

Index  A(1) A1
Key    K(2) X N O T I F Y Q U A R T E R M A S T E R . .
Plain  P(1) N O T I F Y Q U A R T E R M A S T E R . . .
Cipher C(2) K B H B N D O K U R K X V D M S L X V . . .

Plain-text keying using long phrase TYPEWRITER as initial:

Index  A(1) A1
Key    K(2) T Y P E W R I T E R|N O T I F Y Q U A R . .
Plain  P(1) N O T I F Y Q U A R T E R M A S T E R . . .
Cipher C(2) G M I M B P Y N E I G S K U F Q J Y R . . .

Plain-text keying using divided text [aka Running Key]:

Type  PORTA
Key    K(2) OFFICERSANDDIRE
Plain  P(1) CTORSOFTHELOCAL
Cipher C(2) WEMAEMNKUXZATVN

Cipher text auto key with single letter X:

Index  A(1) A1
Key    K(2) X K Y R Z E C S M M D W A R D D V O S . . .
Plain  P(1) N O T I F Y Q U A R T E R M A S T E R . . .
Cipher C(2) K Y R Z E C S M M D W A R D D V O S J . . .

Cipher text auto key with key phrase TYPEWRITER:

Index  A(1) A1
Key    K(2) T Y P E W R I T E R|G M I M B P Y N E I . .
Plain  P(1) N O T I F Y Q U A R T E R M A S T E R . . .
Cipher C(2) G M I M B P Y N E I G S K U F Q J Y R . . .

Cipher text auto key with key phrase TYPEWRITER using only
the last letter of keyphrase to seed progression:

Index  A(1) A1
Key    K(2) T Y P E W R I T E R|I B F W I I A T X . . .
Plain  P(1) N O T I F Y Q U A R T E R M A S T E R . . .
Cipher C(2) G M I M B P Y N E I B F W I I A T X O . . .

SOLUTION OF CIPHER-TEXT AUTO-KEYED CRYPTOGRAMS WHEN KNOWN
CIPHER ALPHABETS ARE EMPLOYED

Attack:  Decipher the message beyond the key letter or key
word portion and then work backwards.

Cryptogram

W S G Q V   O H V M Q   W E Q U H   A A L N B   N Z Z M P
E S K D

Write the cipher text as key letters (displaced one interval
to the right) and decipher by direct standard alphabets
yields the following:

Key   W S G Q V O H V M Q W E Q U H A A L N B N Z Z M P E S K
Ct  W S G Q V O H V M Q W E Q U H A A L N B N Z Z M P E S K D
Plain W O K F T T O R E G I M E N T A L C O M M A N D P O S T

Try the probable word REPORT on the initial group:

Key          F O R C E V O H V M Q . .
Cipher       W S G Q V O H V M Q . . .
Plain        R E P O R T T O R E . . .

A semi-automatic method of solving such a message is to use
sliding normal alphabets and align the strips so that as one
progresses from left to right, each cipher letter set
opposite the letter A on the preceding strip.  Take the
letters VMQWEQUHA in the above example and note how the
successive plain text letters of the word REGIMENT reappear
to the left of the cipher letters MQWEQUHA.

SOLUTION OF AUTOCLAVE BY FREQUENCY ANALYSIS

REDUCED REPETITIONS

Repetitions are not as plentiful in the Autoclave Cipher Text
as they are in the Plain text because in this system, before
a repetition can occur, two things must happen simul-
taneously. First the plain-text sequence must be repeated and
second, one or more of the cipher-text letters immediately
before the second appearance of the plain text repetition
must be identical with one or more of the cipher-text letters
immediately before the first appearance of the group.  This
can only happen as a result of chance.

ex:   Use single key letter X:

Key    X C K B T M D H N V H L Y...K D K S J M D H N V H L Y
Plain  F I R S T R E G I M E N T   T H I R D R E G I M E N T
Cipher C K B T M D H N V H L Y R .KD K S J M D H N V H L Y R

The repeated word REGIMENT has 8 letters but the repeated
cipher text has 9 letters. The plain letter R must be M in
cipher both times. The chances of this are 1/26. In general,
an n-letter repetition in the cipher text, represents an
(n-k) -letter repetition in the plain text, where n is the
length of the cipher-text repetition and k is the length of
the introductory key.

DOUBLETS

Define the 'base letter' as the letter opposite which the key
letter is placed. We also know this as the index. For
convenience, we have chosen A or the initial letter in the
Viggy sequence. When the first key is a single letter, if the
base letter occurs as a plain-text letter its cipher
equivalent is identical with the immediately preceding cipher
letter; there is produced a double letter in the cipher text,
no matter what the cipher component is and no matter what the
key letter happens to be for encipherment.

ex. use HYDRAULIC..XZ sequence for both primary components,
with H, the initial letter of the plain component as a base
letter, and using introductory X as key letter:

Key    X J O I I F L Y U T T D K K Y C X G
Plain  M A N H A T T A N H I G H J I N K S
Cipher J O I I F L Y U T T D K K Y C X G L

Each time the doublet appears it means the second letter
represents H(plain), which is the base letter in this case
(initial letter of the plain component).  If the base letter
happens to be high frequency in normal plain text, say E, or
T, then the cipher text will show a high number of doublets.
The number of doublets is directly proportional to the
frequency of the base letter. If the cryptogram has 1000
letters, we should expect 72 occurrences of doublets, if the
letter was A, and visa-versa. This observation acts as a
check and a guess for new values in the cryptanalysis of the
problem.

When the introductory key is 2 letters, the same phenomenon
will produce groups of the formula ABA, where A and be may be
any letters but the first and the third letters must be
identical. Combine this phenomena with our use of idiomorphs
and we have a powerful wedge into the problem. If we take
BATTALION, it will be enciphered by AABCCDEFG formula.  If
the plain component is a mixed sequence and happens to start
with an E, the word ENEMY would be enciphered by AABBCD
formula.  Used together, we have a powerful tool to open this
cipher.

AMOUNT OF TEXT REQUIRED FOR FREQUENCY ANALYSIS

The Autoclave cipher essentially shifts the key text or
"offsets" the key by at least one letter to the right of the
cipher text. Every cipher letter which immediately follows
the key letter in the cryptogram is monoalphabetically
distributed. If 26 distributions are made, one for each
letter of the alphabet, showing the cipher letter immediately
succeeding each different letter of cipher text, then the
text will be allocated into 26 monoalphabetic distributions
which can be solved by frequency analysis. To do this
effectively requires at least 680 letters of text.  Friedman
details a 6 page long solution by frequency analysis of a
seven message problem which uses the above techniques to good
form.  [FRE3]

SOLUTION BY ANALYSIS OF ISOMORPHISMS

Of more interest to me, is when the message is short and does
not have enough letters to solve by frequency analysis.
Isomorphism is a frequent phenomena in the Autoclave cipher
and generally leads to a reasonable solution.

Given the following intercepted cryptograms:

1.
USYPD  TRXDI  MLEXR  KVDBD  DQGSU  NSFBO
BEKVB  MAMMO  TXXBW  ENAXM  QLZIX  DIXGZ
PMYUC  NEVVJ  LKZEK  URCNI  FQFNN  YGSIJ
TCVNI  XDDQQ  EKKLR  VRFRF  XROCS  SJTBV
EFAAG  ZRLFD  NDSCD  MPBBV  DEWRR  NQICH
ATNNB  OUPIT  JLXTC  VAOVE  YJJLK  DMLEG
NXQWH  UVEVY  PLQGW  UPVKU  BMMLB  OAEOT
TNKKU  XLODL  WTHCZ  R.

2.

BIIBF  GRXLG  HOUZO  LLZNA  MHCTY  SCAAT
XRSCT  KVBWK  OTGUQ  QFJOC  YYBVK  IXDMT
KTTCF  KVKRO  BOEPL  QIGNR  IQOVJ  YKIPH
JOEYM  RPEEW  HOTJO  CRIIX  OZETZ  NK.

3.

HALOZ  JRRVM  MHCVB  YUHAO  EOVAC  QVVJL
KZEKU  RFRFX  YBHAL  ZOFHM  RSYJL  APGRS
XAGXD  MCUNX  XLXGZ  JPWUI  FDBBY  PVFZN
BJNNB  ITMLJ  OOSEA  ATKPB  Y.

Frequency distributions are made (26 x 26 matrix), based on
the second letters of pairs.  The data is relatively scanty
and not promising.

Fortunately, there are several isomorphs available to work
with.

Message 1     (1)  D B D D Q G S U N S F B O B E K . . .
(2)  N E V V J L K Z E K U R C N I F . . .
(3)  T N K K U X O L D L W T H C Z R| end of
message

Message 2     (4)  C R I I X O Z E T Z N K| end of message

Message 3     (5)  C Q V V J L K Z E K U R F R F X ..

First, it is necessary to delimit the length of the
isomorphs.

We confirm the isomorphs begins with the doubled letters.
There is an E before the VV and within the isomorph. If E
were included, then the letter preceding the DD would be an N
to match its homolog E in the isomorph, which it is not.
The evidence suggests a 10 letter isomorph, because of the
tie in letter Z and the impossibility of 11 letters because
of the recurrence of the letter R in isomorph (5). It is not
matched with the recurrence of R in isomorph (2) nor by the
recurrence of T in isomorph 3.

Applying the principles of indirect symmetry to the
superimposed isomorphs, partial chains of equivalents may be
constructed and most of the primary component can be
established. So:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
T E Z K R . I V F  .  .  . Q   . W   G .  N  U  S  B  X  J

24 25 26
D  O  L

The only missing letters are A, C, H, M, P and Y. We apply
decimation on this partial reconstructed alphabet. The
seventh decimation yields results:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
T V W X Z . . D R  . U  L  I  .  B  E  F  G  J  K  .  N  O

24 25 26
.  Q  S

Our old friend HYDRAULIC...XZ returns to the surface.
The plain component turns out to be just a normal sequence.

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

We assume a single letter key.

Key    ? U S Y P W T R X D I M L E X R K V D B D D Q G S
Cipher U S Y P W T R X D I M L E X R K V D B D D Q G S U
Plain  ? P H R F Y I V E F I R E O F L I G H T A R T I L ...

Our one letter assumption is wrong.

Key    W I C K E R|T R X D I M L E X R K V D B D D Q G S
Cipher U S Y P W T R X D I M L E X R K V D B D D Q G S U
Plain  I N T E N S I V E F I R E O F L I G H T A R T I L ...

The first word suggest INTENSIVE coupled with FIRE.  Plug it
in and we have a key word of WICKER.  The other two messages
are recoverable in the same way.

Key    P R O M I S E R X L G H O U Z O ..
Cipher R E Q U E S T V I G O U R O U S
Plain  B I I B F G R X L G H O U Z O ....

Key    C H A R G E D R R V M M H C V B
Cipher S E C O N D B A T T A L I O N
Plain  H A L O Z J R R V M M H C V B

There are always several ways to skin a cat.

COMMANDANT BASSIERES

Both [ELCY] and General Givierge [GIVI] describe the two
processes designed by Commandant Bassieres for solving the
Autoclave cipher. He describes the preliminary process as one
similar to the Kasiski analysis for determining the correct
period.  A search is made of the repeated letters standing
exactly the group length interval apart. The single letter
separation upon tabulation will present itself as one of the
predominating periods.

Bassieres has two processes that follow the discovery of the
period. Process 1 for a group length of 7 for instance, would
take the 1st, 8th, 15th 22nd letters and consider them as a
series or columns. The cryptogram is written into seven
columns which permits decipherment straight down the column.
Starting with key letter A the complete first column is
deciphered and checked.  Then we use B, C,.. until we have a
good decipherment. Then on to series 2, etc. The Bassieres
process no. 1 sets up the entire 26 possible decipherments
for each series (column) and checks for "good" decipherments.
The form of decipherment reduces to alternating Vigenere and
Beauford groups.  Alternate rows in his matrix of solutions
reverse direction with respect to the keys.

Bassieres process no. 2 sets up a trial key of say 7 A's and
this has the effect of introducing periodicity into the
cryptogram at double the key length of the original key.
Solution is based on periodic methods.

Phillip D. Hurst put together some tables to help solve the
plain text keyed auto-key cipher.  Where message and key are
made up of ordinary text, both components will be subject to
the 70% high frequency letter consideration - therefor, high
frequency letters in the key and high frequency letters in
the message will be paired again and again as the coeff-
icients of cryptogram letters, so that cryptograms enciphered
with this kind of key must contain a great many letters
caused by this kind of coincidence.  Tables 13 -2a,b,c show
Hurst's observations for the Vigenere, Beauford and Porta
ciphers.  The alphabet across the top of any table is the
list of possible cryptogram letters, each with its own
column, each column containing only E,T,A,O,N,I,R,S,H and
which if enciphered by another letter from the same group,
would result in the cryptogram letter at the top of the
column. The key is found to the left. Attacks are made on the
second letter as discussed previously.  [ELCY]

With Hurst's method the Index A(1) over Key O(2) for plain
O(1) yields cipher C(2) for a Viggy. Hurst's Table 13-2a gets
the answer on the first try.

Table 13-2a
Tables of High Frequency Coefficients for Autoclave
Vigenere

Keys                     Cipher Letters

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

Table 13-2b
Tables of High Frequency Coefficients for Autoclave
True Beaufort/ Variant

Keys                     Cipher Letters

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

Table 13-2c
Tables of High Frequency Coefficients for Autoclave
Porta

Keys                     Cipher Letters

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
---------------------------------------------------
E  |     R S T             N O     A       E     H I
T  |         N O     R S T     E     H I         A
A  | N O     R S T             A       E     H I
O  |             N O     R S T   H I         A        E
N  | T             N O     R S H I         A        E
I  | R S T             N O             A        E     H I
S  |         N O     R S T     E     H I          A
H  |   R S T             N O         A       E      H I
R  |           N O     R S T       H I         A        E
3 3 3 2 4 4 3 2 3 4 6 5 3 4 2 3 4 4 -2 3 3 3 3 3 3 2

C. Stanley Lamb also put out a table of rough estimate of
rank and frequency of letters in cipher text for auto-key
cipher.  See Table 13- 3.

Table 13-3

Where Key is a Segment of Ordinary Plain Text
Estimated Rank of Cryptogram Letters and Their Frequency
per 10,000 Letters - from a Table of Ohaver

Vigenere

V   A   I   S  ERL WHB XGM FOZ  K  N  T  P  U  J  Y  C  Q
344 314 304 296   intermediate  150112 84 84 84 72 72 64 49

Beaufort and
Variant

A   N   E   W   O   M   Z BQKJRTHVFGUDX  P   L   S  I   YC
480 262 246 246 196 196 191 intermediate 121 121 104 104 57

Porta

K   N   L   E   RMF TWP UYQ XGC AVI BJZ   D   H   O
329 300 282 275    intermediate           132 113 97

SOLUTION OF PLAIN-TEXT AUTO-KEY SYSTEMS

Plain text auto-keying presents a different problem. So let
me stop dancing and get on with the challenge.  The mechanics
of this method disclose that a repetition of n letters
produces a repetition of (n-k) letters in the cipher text.
When an introductory key is k letters in length then an n-
letter repetition represents an (n+k) -letter repetition in
the plain text.  If key k equals 1, then there will be as
many repetitions in the plain as in the cipher text except
for true digraphic repetitions which disappear.

SINGLE KEY LETTER CASE

Look at the following plain-text encipherments of common
military terms: COMMANDING, BATTALION, DIVISION, CAPTAIN.

Key    . B A T T A L I O N
Plain  B A T T A L I O N .
Cipher . B T M T L T W B .

Key    . C O M M A N D I N G
Plain  C O M M A N D I N G .
Cipher . Q A Y M N Q L V T .

Key    . D I V I S I O N .
Plain  D I V I S I O N
Cipher . L D D A A W B

Key    . C A P T A I N
Plain  C A P T A I N .
Cipher . C P I T I V .

Five observations:

1. The cipher equivalent of A(plain) is the plain text
letter immediately preceding A(plain).

2. A plain-text sequence of the general formula ABA yields
a doublet as a(cipher) equivalent of the final two
letters; see IVI OR ISI in DIVISION.

3. Every plain-text trigraph having A(plain) as its central
letter yields a cipher equivalent the last two letters
of which are identical with the initial and final
letters of the plain-text trigraph; see MAN in
COMMANDING.

4. Every plain-text tetragraph having A(plain) as the
initial and the final letter yields a cipher equivalent
the second and fourth letters of which are identical
with the second and third letters of the plain-text
tetragraph; see APTA in CAPTAIN or ATTA in BATTALION.

5. For a single letter initial key, a repetition of n
plain-text yields an (n-k) sequence of cipher letters.
The simplest method of solving this type of cipher is by
means of the probable word.

Message 1

B E C J I   B T M T L   T W B P Q   A Y M N Q   H V N E T
B   A T T A L   I O N
W A A L C

The sequence BTMTLTWB fits the isomorph Battalion and we
insert on the cipher text.  We proceed backward and forward

B E C J I   B T M T L   T W B P Q   A Y M N Q   H V N E T
E A C H B   A T T A L   I O N C O   M M A N D   E R W I L

W A A L C
L P L A C

CRITICAL REVIEW

Masterton was right in his negative assessment of the Autokey
or Autoclave cipher.  Both cipher text and plain text
versions have serious weaknesses which exclude them from
practical or military use. They are slow to work with, prone
to serious/disabling error and they can be solved even when
unknown cipher alphabets are employed.

Recognition is not an issue.  In both systems there are
characteristics which permit of identifying a cryptogram as
belonging to this class of substitution.  Both cases show
repetitions in the cipher text. In cipher text autokeying
there will be far fewer repetitions than in the original
plain text, especially when introductory keys of more than
one letter in length are employed.  In plain text autokeying
there will be nearly as many repetitions in the cipher text
as in the original plain text unless long introductory keys
are employed.  In either system the repetitions will show no
constancy as regards intervals between them, and a uniliteral
frequency distribution will come up as a polyalphabetic.
Cipher text autokeying may be distinguished from its sister
by the appearance of the frequency distributions of the
second number of sets of two letters separated by the length
of the introductory key. In the case of cipher text auto-
keying these frequency distributions  will be monoalphabetic
in nature; its plain text keying sister will not show this
characteristic.

EXTENDING THE KEY

We have looked at ciphers that suppress/destroy the
periodicity, interrupt the key, and used variable lengths for
grouping of plain text. We can also lengthen the key to the
point where it provides insufficient text to decipher.

We can select a phrase from a book, a long mnemonic or long
numerical key. However, any method of transposition applied
to a single alphabetic sequence repeated several times will
yield a fairly long key which approaches randomness.  Another
method of developing a long key from a short mnemonic one is
shown below:

Mnemonic Key   C H R I S T M A S
Numerical Key  2-3-6-4-7-9-5-1-8

Extended key
1 2  3       4             5     6         7
C H R I S T M A|C|CH|C H R I|C H R I S T M|C H R|C H R I S|
8           9
C H R I S T M A S|C H R I S T|

The original key was 9 letters and the extended one is 45
letters.

Another popular method is to take the reciprocal like 1/49
which has many digits = .02040815... as the interruptor for
the key. 0 means use the first letter then use the next
numbers as seeds to how many letters are to be enciphered.

RUNNING KEY CIPHER

The Running Key, aka Continuous Key, or Non Repeating Key
systems in which the key consists of a sequence of elements
which never repeats no matter how long the message to be
enciphered happens to be. Once though indecipherable, this
cipher is subject to the probable word attack and
cryptanalysis when several messages with the same or
superimposable initial keys are intercepted.

PROGRESSIVE KEY CIPHER

The basic principle is quite reasonable. Two or more primary
elements are arranged or provided for according to a key
which may be varied; the interaction of the primary elements
results in a set of cipher alphabets; all the latter are
employed in a fixed sequence or progression.  If the number
of alphabets is small, the text relatively long, this reduces
to a periodic method.

The series of cipher alphabets in such a system constitutes
the keying sequence. Once set up, the only remaining element
in the key for a specific message is the initial cipher
alphabet employed. If the keying system is used by a large
group of correspondents, and employ the same starting point
in the message, the cipher will fall to superimposition.

The probable word method still remains the best attack on
this cipher.  Suppose a cipher message contains the sequence
HVGGLOWBESLTR.. and suppose we assume that the phrase THAT
THE is in the key text, and find the plain text MMUNITI..

Assumed Key Text      . T H A T T H E
Cipher Text           . H V G G L O W B E S L T R . . .
Resultant Plain Text  . M M U N I T I

This suggests the word AMMUNITION. The ON in the cipher text
then yields PR as the beginning of the word after THE in the
Key Text.

Assumed Key Text      . T H A T T H E P R
Cipher Text           . H V G G L O W B E S L T R . . .
Resultant Plain Text  . M M U N I T I O N

PR must be followed by a vowel, perhaps O. The O yields W
which may suggest WILL yielding OTEC.

Assumed Key Text      . T H A T T H E P R O T E C . . .
Cipher Text           . H V G G L O W B E S L T R . . .
Resultant Plain Text  . M M U N I T I O N W I L L . . .

This suggests the words PROTECTION, PROTECTIVE, PROTECTING,
etc. We coerce a few letters in each direction.

When we have multiple messages, we can superimpose them
assuming the correct reference point.  Correct super-
imposition with reference to the key text will provide the
addition of two to three letters to the key and assumptions
for words in several messages.  This leads to the assumption
of more letters, etc.

SOLUTION OF A PROGRESSIVE-ALPHABET CIPHER WHEN CIPHER
ALPHABETS ARE KNOWN

Use the cipher alphabet HYDRAULIC..XZ sequence sliding
against itself continuously producing secondary alphabets in
1 - 26. Starting with alphabet 1:

Plain   HYDRAULICBEFGJKMNOPQSTVWXZHYD ...
Cipher  HYDRAULICBEFGJKMNOPQSTVWXZ

Letter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
Alpha  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
Plain  E N E M Y H A S P  L  A  C  E  D  H  E  A  V  Y  I  N
Cipher E O G P U U E Y H  M  K  Q  V  M  K  Z  S  J  Q  H  E

Letter 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39
Alpha  22 23 24 25 26 1  2  3  4  5  6  7  8  9  10 11 12 13
Plain  T  E  R  D  I  C  T  I  O  N  F  I  R  E  U  P  O  N
Cipher N  L  H  H  L  C  V  B  S  S  N  J  E  P  K  D  D  D

Letter 40 41 42 43 44 45 46 47 48 49 50 51 52 53
Alpha  14 15 16 17 18 19 20 21 22 23 24 25 26  1
Plain   Z  A  N  E  S  V  I  L  L  E  R  O  A  D
Cipher  G  P  U  H  F  K  H  H  Y  L  H  M  R  D

This method reduces to a periodic system involving 26
secondary alphabets and used in simple progression.  The 1st,
27th, 53rd letters are in the 1st alphabet; the 2nd, 28th,
54th are in the 2nd alphabet and so on.

Solving the above, knowing the two primary components is not
too difficult. We lack only the starting point.  The solution
becomes evident by completing the plain component and
examining the diagonals of the diagram, the plain text
becomes evident.  Try:

Given:    H I D C T E H U X I ;  We complete the plain
component sequences initiated by successive cipher letters,
the plain text E N E M Y  M A C H I  is seen to come out in
successive steps upwards in Figure 13-1. Had the cipher
component been shifted in opposite directions during
encipherment, the plain text would be visible downward on the
diagonal.  If the sliding strips had been set up according to
the sequence of cipher letters on the diagonal, then the
plaintext would be seen as one generatrix.

GENERAL SOLUTION FOR CIPHERS INVOLVING A LONG-KEYING SEQUENCE
OF FIXED LENGTH AND COMPOSITION

No matter how the keying sequence is derived, if all the same
correspondents employ the same key, or if this key is used
many times by a single office, and if it always begins at the
same point, the various messages can be solved by
superimposition.  If there are sufficient messages than the
successive columns can be solved by frequency analysis. This
holds no matter how long the keying sequence and regardless
of whether the keying sequence is intelligible or random
text.

If the messages do not start at the same point, then we must
find a tie element to line up the columns. The tie element is
similar to the chemical engineering principle of material
balance and provides a stable point for the text to be
anchored for analysis.  Find one 5 letter polygraph that is
common to two messages and align the messages for super-
imposition based on their position.  Next we find a smaller
tie group of say 4 letters and tie the second message with a
third. We can extend the process to trigraphs or even
digraphs between the messages.  The first step then is to
examine the messages for repetitions. If there are enough
we can use the probable word method to set up plain - cipher
equivalencies and to reconstruct the primary components.
[FRE3]

Figure 13-1

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

ANALYTICAL KEY FOR LECTURES 10 - 13

--------------------------
. POLYALPHABETIC CIPHERS .
-------------------------
|              |
-------------------   -------------------
| PERIODIC SYSTEMS |  | APERIODIC SYSTEMS |
-------------------   -------------------
.                    .
.                    .
.                    II
.
.
.
............................
.              .           .
----------------       .      --------------
.   Repeating Key.      .    .  Progressive  .
.   Systems      .      .    .  Alphabet     .
.                .      .    .    Systems    .
----------------       .      --------------
. ............................
.     ---------------
.     .  Enciphered .
. .. ..  Numerical  .
.  Systems    .
--------------
|..........|
-----------     --------------
. Additive .    . Subtraction .
-----------      -------------

|----------------------|-----------------------|
|                      |                       |
-----------          ----------                ----------
.Monographic.        .Digraphic .              .N-Graphic .
-----------          ----------                ----------
.                        ...e
.
.................................
.                          .
----------------              ---------------
.   Interrelated .           .    Unrelated    .
.   cipher       .           .    cipher       .
.   alphabets    .           .    alphabets    .
----------------              ----------------
.
............................
.                          .
----------------              ---------------
.   Standard     .           .    Mixed        .
.   Alphabets    .           .    alphabets    .
.                .           .                 ...a
----------------              ----------------
..............................
.                          .
----------------              ---------------
.   Direct       .           .    Reversed     .
.   Standard     .           .    Standard     .
----------------              ----------------

a..........................................
.                          .
----------------              ---------------
.  One component .           . Both components .
.   mixed        .           .    mixed        .
----------------              ----------------
.                              .
.                              .
b                              c

b.......
.
.
............................
.                          .
----------------              ---------------
.   Normal       .           .    Normal       .
.   plain        .           .    cipher       .
.   component    .           .    component    .
----------------              ----------------

c....
......................................
.                          .
----------------              ---------------
.   Identical    .           .   Different     .
.   components   .           .   components    .
----------------              ----------------
.
.
.
.

............................
.                          .
----------------              ---------------
. Sequences      .           .  Sequences      .
. proceed in same.           .  proceed in     .
. direction      .           .  different      .
----------------            .  directions     .

e......
.
.... ............................
.                          .
----------------              ---------------
. Related        .           .   Unrelated    .
.   Digraphic    .           .   Digraphic    .
.  encipherment  .           .   encipherment .
----------------              ----------------

II
.
-------------------
| APERIODIC SYSTEMS |
---------.---------
.    .  .  .
--------------------------- .    .  .  .
. Constant Length Keying    ..    a  b  c
. Units on Variable Length  .
. Plain Text Groups         .
---------------------------
.            .
.            .
| ------------ |  | ----------|
| Word Lengths |  | Non-Word  |
| ------------ |  |  Lengths  |
.          |-----------|
.                .
. ....................................
.                              .
---------------------------       ------------------------
. Original Plain Text Groups .    . Original Plain Text    .
. retained in Cryptograms    .    . Groupings not retained .
----------------------------     . in Cryptograms         .
------------------------

a
.
.
-----------------------------------------
.  Variable-length keying units encipher  .
.  constant length plain-text             .
.  groupings.                             .
-----------------------------------------
.
............................
.                          .
----------------              ---------------
.   Interruptor  .           .    Interruptor  .
.   plain text   .           .    cipher text  .
.   letter       .           .    letter       .
----------------              ----------------

b
.
.
----------------------
.   Auto Key Systems   .
----------------------
.
.
............................
.                          .
----------------              ---------------
.   Cipher Text  .           .    Plain Text   .
.   auto keying  .           .    Auto Keying  .
.   systems      .           .    systems      .
---------------- ...........  ----------------
.
............................
.                          .
----------------              ---------------
.   Introductory .           .  Introductory   .
.   key is a     .           .  key is a word  .
.   single letter.           .   or phrase     .
----------------              ----------------

c
.
.
-------------------------------
. Systems Using Lengthy Keys   .
-------------------------------
.
.
............................
.                          .
----------------              --------------
.  Running Key   .           .  Progressive  .
.   systems      .           .  Key  systems .
----------------              ---------------

LECTURE 12 SOLUTIONS

Thanks to DAVLIN & DR FOX-G for their solutions for Lecture
12 problems:

12.1  Nihilist Substitution
carefully to avoid mistakes.

Keywords: CAREFUL, GUIDANCE

12.2  Nihilist Substitution
Perseverance is the mark of a true cryptogram addict.
Be patient and you can do it.

Keywords: SOLVE, CRYPTFANS

12.3  Porta
Conservative scientists have predicted the end of change at
various times but they have always been proved wrong.

Keyword: CHANGE

12.4  Porta
In a leisure society constant rebuilding of your own home to
your own taste, filling it with personal ingenuities and bold
signs might become the fashionable thing to do.

Keyword:  EQUALITY

12.5  Portax
Thank you for your letters and suggestions for the Novice
Notes series.  Keep them coming.  (signed) LEDGE

Keyword:  THANKS

12.6  Portax
Now, that the new year has come, might be the time to realize
that even the best resolutions are meant to be broken.

Keyword:  BROKEN

12.7  Gromark
The crazy person says "I am Abraham Lincoln" and the neurotic
says "I wish I were Abraham Lincoln" and the healthy person
says "I am I and you are you".  Frederick Perls

Keyword: Neurotics

LECTURE 13 PROBLEMS

13-1 Beaufort

ABRVJ  UTAMP  YPLHZ  OZYAP  YPJNP  KNXUG
QRDPC  ELPNC  BVCEF  NLLSJ  LGOWC  VYCGA
EVGIX  XNDKY  U.  (butter) (INWVQH)

13-2 Vigenere.

DWNIT  KGEWZ  ENJQZ  WXLLZ  WZOKC  ETOWI  NXVQS
DQGAK  MGGBH  NAMWE  OWVAM  UJDVQ  IMDSB  VCCTR
YUIQX.  (making, UHVW)

13-3 Vigenere Running Key
YPOSC  DWVWY  CCHZT  AKALF  I.  (tolls -2)

13-4 Vigenere Progressive key.  "Fungi"

IPGPUPX  GTIAKNP  AMEHLAW  SJSTROZ  TCGYUND  STNPJZM
OESWAXG  VLHSPZC  GNEIWHP  EKHNOWW  PMEQFVV  PDQAWCA
GGFRKSO  RCHZVKL  NBWHYBV  CUNBBBB  AVGCJFA  FLTMKUV  K.

REFERENCES / RESOURCES

I will issue additional references/resources at a later