## Lesson 16: Transposition

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CLASSICAL CRYPTOGRAPHY COURSE
BY LANAKI

August 22, 1996

LECTURE 16

TRANSPOSITION

SUMMARY

Lecture 16 considers a whole range of Transposition (or
displacement) ciphers.  We develop our subject using the
following references: [BAR2], [FRE4], [KULL], [OP20],
[MAST], [COUR], [LEDG], [BOW1], [ELCY].

SIMPLE ROUTE TRANSPOSITIONS (TRAMPS)

Transposition ciphers have been defined as that type of
cipher in which the elements or units of the plain text,
whether one is dealing with individual letters or groups
of letters, retain their original identities but undergo
some change in their relative positions or sequences so
that the message becomes unintelligible.  The majority
of transposition methods involve the use of a design or
geometric figure, such as a square, rectangle, triangle,
trapezoid, etc., in which the letters of the plain text
are first inscribed or written into the design according
to a previously agreed upon direction of writing and
then transcribed or taken off according to another and
different previously agreed-upon direction, to form the
text of the cryptogram.

In their simplest form, TRAMPS may take any of the
following routes when employing rectangles or squares
for transposing text of a message as illustrated below.
The plain-text message is assumed to be merely the
normal sequence from A to X, for ease in following the
route.

Any geometrical form can be used, but it must be full
block; if the letters of a message do not complete the
assigned block, nulls (arbitrary letters) must be added.

a) Simple Horizontal:

ABCDEF      FEDCBA       STUVWX      XWVUTS
GHIJKL      LKJIHG       MNOPQR      RQPONM
MNOPQR      RQPONM       GHIJKL      LKJIHG
STUVWX      XWVUTS       ABCDEF      FEDCBA

b) Simple Vertical:

AEIMQU      DHLPTX       UQMIEA      XTPLHD
BFJNRV      CGKOSW       VRNJFB      WSOKGC
CGKOSW      BFJNRV       WSOKGC      VRNJFB
DHLPTX      AEIMQU       XTPLHD      UQMIEA

c) Alternate Horizontal:

ABCDEF      FEDCBA       XWVUTS      STUVWX
LKJIHG      GHIJKL       MNOPQR      RQPONM
MNOPQR      RQPONM       LKJIHG      GHIJKL
STUVWX      STUVWX       ABCDEF      FEDCBA

d) Alternate Vertical:

AHIPQX      DELMTU       XQPIHA      UTMLED
BGJORW      CFKNSV       WROJGB      VSNKFC
CFKNSV      BGJORW       VSNKFC      WROJGB
DELMTU      AHIPQX       UTMLED      XQPIHA

e) Simple Diagonal:

ABDGKO      GKOSVX       OKGDBA      XVSOKG
CEHLPS      DHLPTW       SPLHEC      WTPLHD
FIMQTV      BEIMQU       VTQMIF      UQMIEB
JNRUWX      ACFJNR       XWURNJ      RNJFCA

ACFJNR      JNRUWX       RNJFCA      XWURNJ
BEIMQU      FIMQTV       UQMIEB      VTQMIF
DHLPTW      CEHLPS       WTPLHD      SPLHEC
GKOSVX      ABDGKO       XVSOKG      OKGDBA

f) Alternate Diagonal:

ABFGNO      GNOUVX       ONGFBA      XVUONG
CEHMPU      FHMPTW       UPMHEC      WTPMHF
DILQTV      BEILQS       VTQLID      SQLIEB
JKRSWX      ACDJKR       XWSRKJ      RKJDCA

ACDJKR      JKRSWX       RKJDCA      XWSRKJ
BEILQS      DILQTV       SQLIEB      VTQLID
FHMPTW      CEHMPU       WTPMHF      UPMHEC
GNOUVX      ABFGNO       XVUONG      ONGFBA

g) Spiral, Clockwise:

ABCDEF      LMNOPA       IJKLMN      DEFGHI
PQRSTG      KVWXQB       HUVWXO      CRSTUJ
OXWVUH      JUTSRC       GTSRQP      BQXWVK
NMLKJI      IHGFED       FEDCBA      APONML

h) Spiral, Counterclockwise:

APONML      NMLKJI       IHGFED      FEDCBA
BQXWVK      OXWVUH       JUTSRC      GTSRQP
CRSTUJ      PQRSTG       KVWXQB      HUVWXO
DEFGHI      ABCDEF       LMNOPA      IJKLMN

Example 1  - Let the message be (military text):

At fourteen hundred sighted submarine bearing two
three five degrees true.  (63)

Suppose we agree to use a completely filled square
of eight rows by eight columns, then we must add 1 null
to give us a multiple of eight (64). We agree that
alternate diagonals will be used for inscription.

1 2 3 4 5 6 7 8
1 A T R T R E M A
2 F U E D D B R O
3 O E N S U I W T
4 N U I S N T H E
5 H G D E G R D G
6 H E B N E E R R
7 T E I E V E T U
8 A R F I E S E N

Next the letters are taken off by simple vertical to
form the cryptogram:

AFONH  HTATU  EUGEE  RRENI  DBIFT  DSSEN
EIRDU  NGEVE  EBITR  EESMR  WHDRT  EAOTE
GRUN

To decipher the cryptogram, the process is reversed.
The total number of letters in the cipher text is used
to reconstruct the rectangle. Then the cryptogram is
inscribed by the agreed upon route and the plain text is
taken off by the other agreed upon route.

OTHER GEOMETRICAL FIGURES

We are not limited to the square or rectangle.  The
routes indicated above work for other geometrical
designs with minor modifications.

(a) Trapezoidal design:

A T F O U
R T E E N H
U N D R E D S
I G H T E D S U
B M A R I N E M P

(b) Triangular design:

A T F O U R T E
E D S I G H E
B M A R T N
C M I E H
P N D U
E S N
U D
B
The cryptograms resulting from figure (a) taken off
according to an alternate vertical route is:

BIURA  TTNGM  AHDEF  OERTR  IEENU  HDDNE  SSUMP

That resulting from figure (b) taken off according to a
diagonal route is:

AEBCP  EURTD  MMNSB  FSAID  NOIRE  UUGTH  RHNTE  E.

SOLUTION HINTS FOR TRAMPS

When but one cryptogram is available, the solution of a
tramp is largely trial and error.  There are some
shortcuts. Use:

a) The beginning and end of the cryptogram will follow
the most frequent initials (T A W O B I C S D H) and
finals (E T S D N R Y O F L ). Words may be assumed
which contain these initial or final letters
near the beginning or end of the cryptogram.

b) The interval between the letters of expected words,
high frequency digraphs, QU and vowel.

c) Long groups of vowels or consonants show up when
English is written horizontally and transcribed
vertically; these may be assumed to be adjacent.

d) The presence of parts of words are found with certain
routes such as spirals and helps to identify the
route.

e) Use the total number of letters to suggest the
geometric design and fill in the arbitrary figure
with the ciphertext to give further clues

NUMERICAL KEYS

A numerical key can be derived from a literal key as we
saw in substitution problems:

A M E R I C A N
1 6 4 8 5 3 2 7

or can be used as a guided for transposing letters like:

7 2 4 5 3 6 1     7 2 4 5 3 6 1     7 2 4 5
R E P O R T N     O O N P O S I     T I O N

The letters are take off the above groups and
transcribed into standard groups of five letters, all
letters marked 1 being taken first, then all those
marked 2, etc giving:

NIEOI   ROPNO   OPNTS   ROT

MISCELLANEOUS TRANSPOSITION METHODS

Transposition ciphers come in several simple varieties.

The oldest form may be reversed writing. The reversing
process may be applied to regular or irregular groups of
plain text letters:

Let the plain text be:  Bridge destroyed at eleven pm.

Words Reversed:

E G D I R B   D E Y O R T S E D   T A   N E V E L E  M P

Words Reversed and Regrouped into False Lengths:

E G D I R B   D E Y O R T   S E D T A   N E V E  L E M P

Text Reversed and Regrouped into Fives:

MPNEV  ELETA  DEYOR  TSEDE  GDIRB

Text Reversed and Regrouped into Fives With Nulls every
Fifth Position:

TRIMM  PNEVP  ELETA  ADEYR  ORTSL  EDEGU  DIRBM

Columnar by Bigraph:

or
B S          B R
R T          I D
I R          G E
D O          D E
G Y          S T
E E          R O
D D          Y E
E            D

Cipher Text:

B S R T I    R D O G Y   E E D D E , or
B I G D S    R Y D R D   E E T O E

or let the new plain be "Prepare to get underway":

Digraphs Reversed :

Plain     PREPA   RETOG   ETUND   ERWAY
Cipher    RPPER   ATEGO   TENUE   DWRYA

RAIL FENCE CIPHER

Just as the name implies, the Rail Fence Cipher
resembles an old rail fence found in many parts of New
England today; with its zig-zag appearance.

Plain:  Prepare to get underway.

P   E   A   E   O   E   U   D   R   A
R   P   R   T   G   T   N   E   W   Y

Ciphertext is taken off horizontally:

PEAEO   EUDRA   RPRTG   TNEWY

It may be composed of any number of rails ( or letters
in depth) which may be written up or down, coming from a
point and then reversing the direction to the end of the
message, either filling the final stroke or being short
a letter or more.

Any message may be written in with the normal sequence
up and down, or visa versa, or it may be written into
the points first, and then into successive horizontal
rows.  It is then taken out by the alternate process.

Table 16-1 shows the total length of a Rail Fence cipher
versus the various peaks plus extra letters from 2-10
rails.  There is no technical way to solve this cipher,
however Table 16-1 can help look at possibilities.

Example:

TAOET  NMFOA  TNEHM  NHWKS  POIDI  SLFMU  HSOBE  ALEEW
AUFHE  ASNES  P.  (51)

Scanning Table 16-1, for 2 rails there are 26 peaks; 3
rails, 13 peaks plus two extra letters (..); 4 rails, 9
peaks plus two extra letters; 5 rails, 7 peaks plus two
extra letters, and so forth. We use the digit which
falls directly under the message length; if no digits
are shown, take the digit to the left and add for the
extra letters the dots.

For a 3-depth, set up a pattern:

1      5      9      13
2   4  6  8  10  12  14
3      7     11      15

The cipher text looks like this:

T       T       O
A   E   N   F   A         ( improbable)
O       M       T

We try to write in the cipher text at the points and
follow through to the second row:

T       A      O
H   M   N  H              ( good plain text)
E       E

TABLE 16 - 1
Total Length of Cipher versus Various Peaks plus extra
Letters of Rails from 2-10

3  5  7  9  11  13  15  17  19  21  23  25  27  29
2  2  3  4  5  6   7   8   9   10  11  12  13  14  15
3     2 ..  3  ..  4   ..  5   ..  6   ..  7   ..   8
4        2  .. ..  3   ..  ..  4   ..  ..  5   ..  ..
5           2  ..  ..  ..   3  ..  ..  ..  4   ..  ..
6               2  ..  ..  ..  ..   3  ..  ..  ..  ..
7                   2  ..  ..  ..  ..  ..  3   ..  ..
8                       2  ..  ..  ..  ..  ..  ..   3
9                           2  ..  ..  ..  ..  ..  ..
10                              2  ..  ..  ..  ..  ..

31  33  35  37  39  41  43  45  47  49  51  53  55
2  16  17  18  19  20  21  22  23  24  25  26  27  28
3  ..  9   ..  10  ..  11  ..  12  ..  13  ..  14  ..
4  6   ..  ..   7  ..  ..   8  ..  ..   9  ..  ..  10
5  ..   5  ..  ..  ..   6  ..  ..  ..   7  ..  ..  ..
6   4  ..  ..  ..  ..   5  ..  ..  ..  ..   6  ..  ..
7  ..  ..  ..   4  ..  ..  ..  ..  ..   5  ..  ..  ..
8  ..  ..  ..  ..  ..  ..   4  ..  ..  ..  ..  ..  ..
9  ..   3  ..  ..  ..  ..  ..  ..  ..   4  ..  ..  ..
10 ..  ..  ..   3  ..  ..  ..  ..  ..  ..  ..  ..   4

57  59  61  63  65  67  69  71  73  75  77  79  81
2  29  30  31  32  33  34  35  36  37  38  39  40  41
3  15  ..  16  ..  17  ..  18  ..  19  ..  20  ..  21
4  ..  ..  11  ..  ..  12  ..  ..  13  ..  ..  14  ..
5   8  ..  ..  ..   9  ..  ..  ..  10  ..  ..  ..  11
6  ..  ..   7  ..  ..  ..  ..   8  ..  ..  ..  ..   9
7  ..  ..   6  ..  ..  ..  ..  ..   7  ..  ..  ..  ..
8   5  ..  ..  ..  ..  ..  ..   6  ..  ..  ..  ..  ..
9  ..  ..  ..  ..   5  ..  ..  ..  ..  ..  ..  ..   6
10 ..  ..  ..  ..  ..  ..  ..  ..   5  ..  ..  ..  ..

83  85  87  89  91  93  95  97  99  101  103  105
2  42  43  44  45  46  47  48  49  50  51   52   53
3  ..  22  ..  23  ..  24  ..  25  ..  26   ..   27
4  ..  15  ..  ..  16  ..  ..  17  ..  ..   18   ..
5  ..  ..  ..  12  ..  ..  ..  13  ..  ..   ..   14
6  ..  ..  ..  ..  10  ..  ..  ..  ..  11   ..   ..
7  ..   8  ..  ..  ..  ..  ..   9  ..  ..   ..   ..
8  ..   7  ..  ..  ..  ..  ..  ..   8  ..   ..   ..
9  ..  ..  ..  ..  ..  ..  ..   7  ..  ..   ..   ..
10 ..  ..  ..  ..   6  ..  ..  ..  ..  ..   ..   ..

107  109  111  113  115  117  119
2  54   55   56   57   58   59   60
3  ..   28   ..   29   ..   30   ..
4  ..   19   ..   ..   20   ..   ..
5  ..   ..   ..   15   ..   ..   ..
6  ..   ..   12   ..   ..   ..   ..
7  ..   10   ..   ..   ..   ..   ..
8  ..   ..   ..    9   ..   ..   ..
9  ..   ..   ..    8   ..   ..   ..
10 ..    7   ..   ..   ..   ..   ..

REDEFENCE

The railfence cipher may be made more secure when a
numerical key is used in addition to the initial
transposition.  For example:

2  T     L     G     E
4   H   R Y   D E   H W
1    E A   B R   T T   O M
3     E     I     S     R

Cipher: EABRT  TOMTL  GEEIS  RHRYD  EHW
Key = 2413

Solution is similar to railfence with the help of a tip.

FOUR WINDS CIPHER

R           R           G          N         W
P -|- E     A -|- E     O -|- E    U -|- D   R -|- A
P           T           T          E         Y

Taken off clockwise from left to right:

Cryptogram:

RRGNW   PEAEO   EUDRA   PTTEY

HEDGES

R   O   E   P   E   W   R   U   Y

P   T   D   E   G   R   A   T   A   E   N

(Base jumps over two letters)     Link P to R to E.

Cryptogram:

ROEPE   WRUYP   TDEGR   ATAEN

DIAMOND

Friedman in [FRE4] describes solution to an unusual
diamond design that looks like this:

1
2  3  4
5  6  7  8  9
10  11 12 13 14 15  16
17  18  19 20 21 22 23  24  25
26  27  28  29 30 31 32 33  34  35  36
37  38  39  40  41 42 43 44 45  46  47  48  49
50  51  52  53 54 55 56 57  58  59  60
61  62  63 64 65 66 67  68  69
70  71 72 73 74 75  76
77 78 79 80 81
82 83 84
85

The numbers indicate the method of encipherment. The
cipher is taken off vertically by column.

The foregoing examples would indicate that almost any
figure can be used for this type of transposition,
including stars, polygons and irregulars.  It is merely
necessary to agree on the figure and the starting points
for inscription and transcription processes.

CIVIL WAR MESSAGES

David Kahn gives us an interesting look at Civil War
Cryptography. If there was a reason why the North won,
it had to be superior cryptography.  Anson Stager
first superintendent of the Western Union Telegraph
Company, was charged by Major General B. McClellan with
drawing up a military cipher along the lines that he had
devised for Governor Dennison of Ohio.   [KAHN]

Stager complied.  Soon McClellan was relying on the
cipher to protect his communications during his
successful campaign in West Virginia. One of the first
users was Allan Pinkerton, founder of the agency that
bears his name and bodyguard to President Lincoln.  The
key was very short, it was dependable and was used by
the Union forces throughout the Civil War. It was used
extensively because the Civil War first employed the
telegraph on a large scale.  Communications from
Washington could take 10 days to their troops depending
on weather, health of the telegrapher operators and
availability of lines (which sometimes took a circuitous
route).  During Sherman's march to the sea, the Union
had to rely on Southern newspapers for accounts of his
slash and burn campaign.

So effective was the Stager cipher that those same
Southern newspapers advertised for help from anyone who
recognized or could break the Yankee cipher.

Stager's cipher was a word transposition. Stager's
telegraphic experience evidently led him to a system in
which the ciphertext consisted - as in the new telegraph
codes - of ordinary words, which are far less subject to
dangerous garbles than groups of incoherent letters.
[There is a funny story how one of the Rebel commanders
could not read the cipher message sent to him by one of
his forward patrols - prior to Gettysburg no less - so
he sent a messenger to the forward post to get a
clarification of the cryptogram received.  The messenger
returned to find his commander under arrest.  The
message was a warning of a Union trap. The lines were
effected by rain that particular day.]

The Stager cipher was appealing because of its
simplicity: the plaintext was written out in lines and
transcribed by columns, up some and down others in a
specified order.  His cipher was improved by adding
nulls, mazed routes of diagonals and interrupted columns
through larger rectangles and per Samuel H. Beckwith,
Grants cipher operator, important terms were represented
by codewords which were carefully chosen to minimize
telegraph error. The cipher expanded from one listed on
a single card to that by the end of the war, required 12
pages to list routes and 36 for the 1,608 codewords.
This was Cipher 4, the last of a series of 12 that the
North employed at various times.

A good example of the system is given by encipherment of
the message by President Lincoln on 1 June 1863: For
Colonel Ludlow. Richardson and Brown, correspondents of
the Tribune , captured at Vicksburg, are detained in
Richmond. Please ascertain why they are detained and get
them off if you can. The President.  Cipher No 9 was in
use and provided for the following codeword substit-
utions: VENUS for colonel, WAYLAND for captured, ODOR
for Vicksburg, NEPTUNE for Richmond, ADAM for President
of U.S. and NELLY for 4:30 pm time of dispatch.   The
keyword of GUARD set the size of the rectangle and
routes. Nulls were added to the end of each column.
The encipherer chose to write out the message in seven
lines of five words each with three nulls to complete
the rectangle. The plaintext was:

For       VENUS           Ludlow   Richardson  and
Brown     correspondents  of       the         Tribune
WAYLAND   at              ODOR     are         detained
they      are             detained and         get
them      off             if       you         can

Ciphertext: [up the first column,(kissing=null),down
second,(turning=null),up fifth,(times=null),down fourth,
(belly=null), up third column]

GUARD ADAM THEM THEY AT WAYLAND BROWN FOR KISSING VENUS
CORRESPONDENTS AT NEPTUNE ARE OFF NELLY TURNING UP CAN
GET WHY DETAINED TRIBUNE AND TIMES RICHARDSON THE ARE
ASCERTAIN AND YOU FILLS BELLY THIS IF DETAINED PLEASE
ODOR OF LUDLOW COMMISSIONER.

Confederate cryptography centered around the Vigenere
which we have previously studied.  The south employed
only three keywords: MANCHESTER BLUFF, COMPLETE VICTORY
and COME RETRIBUTION. Known also as the Vicksburg cipher
the team of Tinker, Chandler and Bates, very early
yuppies, were able to read a whopping 90% of the
Confederates messages and report them to Lincoln.
For example, Grant's troops intercepted a message on
eight captured rebels at Vicksburg trying to slip into
Vicksburg with 200,000 percussion caps.

Jackson, May 25, 1863
Lieutenant General Pemberton: My XAFV. USLX WAS VVUFLSJP
by the BRCYAJ. 200000 VEGT. SUAJ. NERP. ZIFM. It will be
GFOECSZOD as they NTYMNX. Bragg MJTPHINZG a QRCMKBSE.
When it DDZGJX. I will YOIG. AS. QHY. NITWM do you YTIAM
the IIKM. VFVEY. How and where is the JSQMLGUGSFTVE.
J. E. Johnston.

Note the flow of the message and hints along the way.
The word separators, the clear text leads you into the
next word.  The size of the words is known and might be
guessed.

The Plaintext based on the Keywords MANCHESTER BLUFF is:

Lieutenant General Pemberton: My last note was captured
by the picket. 200000 caps have been sent. It will be
increased as they arrive. Bragg is sending a division.
When it joins I will come to you. Which do you think the
best route? How and where is the enemy encamped? What is

J. E. Johnston.

CRYPTANALYSIS OF THE SINGULAR COLUMNAR TRANSPOSITION
CIPHER

Colonel W. Barker has perhaps the best description for
cracking the single columnar transposition problem.
[BAR3] (a tad better than the master himself.)

Encipherment

NEED SUPPLIES AT ONCE STOP REQUEST REPLY IMMEDIATELY.

with Literal key: SUMMER TIME

Step 1: Derive the numerical key from literal key.

810 4 5 1 7 9 3 6 2
S U M M E R T I M E

Step 2: Write the plain text beneath the numerical key:

810 4 5 1 7 9 3 6 2
S U M M E R T I M E
---------------------
| N E E D S U P P L I |
| E S A T O N C E S T |
| O P R E Q U E S T R |
| E P L Y I M M E D I |
| A T E L Y|----------|
----------

Note we are starting off with the more difficult
incomplete rectangle.  Technically this is called a
matrix and is addressed by its rows and columns.  There
are two kinds of columnar matrices to be considered, the
completely filled matrix and the in-completely filled
matrix. The length of the message in a completely filled
matrix is a multiple of the key- length which greatly
simplifies the solution.

Terminology

The size or dimensions of any matrix depends on two
things:

(1) The length of the key.
(2) The length of the message.

Given these two things, we can determine the type of
matrix we are dealing with, the number of long and short
columns, and the number of letters in each type of
column.  Graphically we have in our example:

Key length
*-------10--------*
810 4 5 1 7 9 3 6 2
S U M M E R T I M E
---------------------
Length of  * | N E E D S U P P L I | *
long       | | E S A T O N C E S T | | Length of short
column is  | | O P R E Q U E S T R | | column = 4
5         | | E P L Y I M M E D I | *
* | A T E L Y|----------|
---------- * - - - *
* - - - *     Number of short
Number of       columns is 5
long columns
is 5

Step 3: Take the columns out in numerical order. Thus
the first column out is 1 or S O Q I Y, then I T R I
etc.

The cipher text is:

SOQIY  ITRIP  ESEEA  RLEDT  EYLLS  TDUNU  MNEOE  APCEM
ESPPT

Note that the original plain text has not been changed
but merely rearranged or transposed by a numerical key.

Decipherment

Consider the decipherment of the following:

UNCKO  MNHTA  NSEOT  NMIEG  OFPER  NMAWO  OLTGA  SFHDO
OLLEN  YINRI  SIECY  COTOR  FETNN  TSGOR  IPTHT  NOETX
ISENW  ICXMI  NREUE  T.         (96)

With Keyword:  APPLE BLOSSOMS

Step 1: Derive the numerical key from the literal.

1 910 4 3 2 5 71112 8 613
A P P L E B L O S S O M S

Step 2: Determine the size of the matrix used in
encipherment. This step is the most important step in
decipherment.

Since the key is 13 and the message length is 96, we
divide the key-length into the message length to give:

_7_
13|96
91
--
5

where 13 is length of key, 96 is length of the message,
7 is the length of the short columns, and 5 is the
number of long columns.  Since the length of the short
column is 7, the length of the long column is 1 more or
8. And, the number long columns is 5, so the number of
short columns is the key-length minus the number of long
columns (13-5) = 8.  Now we have the size of the matrix.

Key length is 13

1 910 4 3 2 5 71112 8 613
A P P L E B L O S S O M S
----------------------------
| U         T               |
| N         A               | Length of
Length of    | C         N               | short columns
Long         | K         S               | is 7
column is 8  | O         E               |
| M         O               |
| N         T               |
| H          ---------------
------------|
Number of     Number of short
long columns  columns is 8
is 5

Step 3. Place the columns back into the matrix according
to the numerical key.  The plain text can now be read
horizontally from left to right, top to bottom within
the matrix.

1 910 4 3 2 5 71112 8 613
A P P L E B L O S S O M S
----------------------------
| U R G E N T L Y N E E D I |
| N F O R M A T I O N C O N |
| C E R N I N G N E W Y O R |
| K T I M E S A R T I C L E |
| O N P A G E S I X C O L U |
| M N T W O O S I X T E E N |
| N T H O F T H I S M O N T |
| H S T O P ----------------
------------|

URGENTLY NEED INFORMATION CONCERNING NEW YORK TIMES
ARTICLE ON PAGE SIX COLUMN TWO OF SIXTEENTH OF THIS
MONTH STOP.

Comparing the steps:

Step         Encipherment          Decipherment
1         Derive numerical key   Derive numerical key

2         Write plain text       Determine size of
beneath key            matrix from key-length
and message length.

3         Take columns out of    Put columns into matrix
matrix in numerical    in numerical order
order

special cases and work up to the general solution.

Case 1: Plain Text beginning of a Message Longer than
the Key-Length

Given the cipher text message known to be a single
columnar transposition:

TTDTI  TIIIH  NNOBT  ERNOO  IGSRY  SVIAA  XNAFN
ASMMR  IE.

We suspect that the words TRANSMIT INFORMATION is at the
beginning of this message.

We begin our solution by writing the ciphertext without
group divisions.  We then make a biliteral frequency
distribution. ( The bigram frequency distribution will
be 1 less than the total frequency because the last
letter does not have a partner.)

T T D T I T I I I H N N O B T E R N O O I G S R Y
S V I A A X N A F N A S M M R I E.

and,

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

Write out the first few letters of known plain-text
beginning horizontally and then horizontally written
letters are written the succeeding letters of the given
plain text as follows:

T R A N S M    HITS
-----------
R A N S M I     1
A N S M I T     2
N S M I T I     0
S M I T I N     0
M I T I N F     1
I T I N F O     2
T I N F O R     3
I N F O R M     6
N F O R M A     1
F O R M A T     0
O R M A T I     1
R M A T I O     1
M A T I O N     0

We use the biliteral distribution to determine the key
length as follows.  We start in column 'T' and note that
T occurs 5 times in the cipher message with 4 different
letters (T, D, I, and E). We specify as a hit (or
circle, or mark in someway) any of those letters in the
first column.     Column 1 has 3 hits - I T I. in column
2 we mark the three letters N, Y, and I as noted from
the biliteral frequency distribution.  We note the above
figure shows the final results.   The word inform has 6
hits and is below the word segment TRANSM. For this to
have happened the key-length has to be eight (count down
the rows) and the beginning within the matrix will look
as follows:

T R A N S M I T
-----------
I N F O R M A T
-----------
I O N

We now see why the cipher text letter T is followed by
I. They are both from the first column to the left in
the matrix. Since the ending is I O N falls in the third
horizontal row of the matrix; the first column is T I I
and that is found in group two of the cipher text.

Since the key is 8, and the number of letters in the
message is 42, we can easily determine the size of the
matrix used in encipherment.  Dividing 8 into 42 gives
40 with 5 as the length of the short, 6 the length of
the long and 2 (remainder) being the number of long
columns and (8-2) = 6 short columns after them.

We set up the enciphering matrix and write in the known
plaintext.

2 4
-----------------
| T R A N S M I T |
| I N F O R M A T |
| I O N           |
| I O             |
| H I             |
| N G ------------
-----|

2              4
TTDTI  TIIIH  NNOBT  ERNOO  IGSRY  SVIAA  XNAFN
ASMMR  IE.

The TII is a giveaway that this column was the second
column (length 6) to be taken out of the matrix during
encipherment.  The entire column is T I I I H N.  We
note that R N O O I G is the 4th column (long) with all
others being short.  The rest of the columns are easily
identified and the plain is read horizontally.

2 4 7 3 5 8 6 1
-----------------
| T R A N S M I T |
| I N F O R M A T |
| I O N B Y R A D |
| I O A T S I X T |
| H I S E V E N I |
| N G ------------
-----|

TRANSMIT INFORMATION BY RADIO AT SIX THIS EVENING.

Note that our solution has yielded the numerical key
(which may have been derived as a result of many literal
keys). The numerical key is used to read other messages
by the same source, collectively called traffic. The
trivial issue is that the solution was possible because
the known plain was longer than the keyword and hence,
set up adjacent vertical letters to identify in the
cipher text.

The Analytic Matrix or Hat Diagram

The Analytic matrix (aka Hat Diagram) is a fundamental
tool to solve all columnar transposition cipher systems.
You only need the keylength and the cipher text itself.

Given the cryptogram:

EIPEI  EUFSS  ETODE  ERTJR  OOSCL  NTPLH  EDGRF  TEEEE
SAOIT  SNULP  VONPT  ADAEL  YVLT.       (64)

with Key length = 10.

Step 1: Determine the size of the enciphering matrix.

Key of 10 into 64 = 6 as the length of the short, 7 as
the length of the long columns, 4 as the number of long
columns and 10 - 4 = 6 short columns.

or in shorthand      64 = 4 - 7's
6 - 6's

Check: 4 x 7 =28 + 6 x 6 = 36 = 64

Step 2: Divide ciphertext into long and short columns
starting with long columns at the head of the message
and the short columns at the tail of the message.

EIPEI  EU / FSS  ETOD / E  ERTJR  O / OSCL  NTP / L
H EDGR / F TEEEE / SAOIT  S / NULP  VO / NPT  ADA / EL
YVLT.  (64)

Step 3: Write down vertically from left to right these
divided columns, keeping the bottoms of the columns on
the same line.  Thus:

1 2 3 4 5 6 7 8 9 10
E F E O
I S E S L F S N N E
P S R C H T A U P L
E E T L E E O L T Y
I T J N D E I P A V
E O R T G E T V D L
U D O P R E S O A T
*-4-- * * ----6---*
long        short

These letters represent the foundation of the hat
diagram.

Step 4: Extend the tops of the columns from right to
left such that the long columns come at the tail of the
message and the short columns at the head of the
message.

We begin by drawing a line across the top of the of the
present columns:

_______
E F E O|__________A
I S E S L F S N N E
P S R C H T A U P L
E E T L E E O L T Y
I T J N D E I P A V
E O R T G E T V D_L
U D O P R E S O A T

this column and make it long by adding one letter from
the column previous to it.  We move to the left and make
column and place on top of the line.  We mark the bottom
of the letters for the 'borrowed letters' to help us
determine the length of the column extended over the top
border.

_______         V
E F E O|________O_A
I S E S L F S N N E
P S R C H T A U P L
E E T L E E O L T Y
I T J N D E I P_A V
E O R T G E T V D_L
U D O P R E S O A T

We extend two more short columns making a total of four
extended to long columns.

E
E I
_______     E T V
E F E O|____E_S_O_A
I S E S L F S N N E
P S R C H T_A U P L
E E T L E E O_L T Y
I T J N D E I P_A V
E O R T G E T V D_L
U D O P R E S O A T

At this point, still going from right to left, we extend
the balance of the columns as short columns (six in
all).  Above the line is extend only to the length of
the short column.

E E
D E I
_______   G E T V
E F E O|__R_E_S_O_A
I S E S L F S N N E
P S R C H_T_A U P L
E E T L E E O_L T Y
I T J N D E I P_A V
E O R T G E T V D_L
U D O P R E S O A T

The final hat diagram looks like this:

1 2 3 4 5 6 7 8 9 10
J L E E
O R N D E I
__U_D_O T G E T V
E F E O|P_R_E_S_O_A
I S E S L F S N N E
P S R C_H_T_A U P L
E E T_L E E O_L T Y
I T_J N D E I P_A V
E_O R T G E T V D_L
U D O P R E S O A T

Note the lengths of the columns -above the drawn lines
at the bottoms of the columns - are from right to left
the four long columns and the six short columns.

What does the hat matrix represent?

It represents the columns of the enciphering matrix in
numerical order from left to right. Because we do not at
this stage know which columns are actually long and
short, the hat matrix contains some superfluous letters
from the adjacent column. However, regardless of how
long and short columns are arranged in the actual
enciphering matrix, the columns of the hat matrix will
contain all the letters of the columns as found in the
actual enciphering matrix.

Case 2: Plain Text Longer than the Key-Length Anywhere
in the Message.

Given Cipher Text:

BIEEH  VHBSR  UAHEE  OREBE  ECOWV  NTETM  TAQZT  TDRNI
EESNE  ELOLO  EOERL  NINNF  R.    (61)

Key length unknown.

Step 1: Make a biliteral frequency distribution (from
left to right.) [ie. B has 3 contacts I, S, E in that
order]

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

Step 2: "Complete the Plain" text for first few letters
of known plaintext and apply biliteral frequency
distribution.  In the columns thus extended, note the
'hits' which follow the top column letter in the
biliteral frequency distribution.

Q U E E N     HITS
---------
1     U E E N E      2
2     E E N E L      1
3     E N E L I      3
4     N E L I Z      1
5     E L I Z A      0
6     L I Z A B      0
7     I Z A B E      2
8     Z A B E T      5   KEY LENGTH =8
9     A B E T H      2

For Z A B E T to have fallen directly under queen, the
key-length must be 8, so:

Q U E E N E L I
Z A B E T H

Step 2: Determine size of matrix, columns and construct
the hat diagram.

Key of 8 into 61 message length gives 7 as the short
length, 8 as the long length, 5 for the number of long
columns, and (8-5) 3 short columns.  The Hat diagram
looks like this:

1 2 3 4 5 6 7 8
C M
E O T R
_B_O_W_A_N E
* B S R V Q I_L_R
I R E N Z E O L *
long     | E U B T T E L N |
column   8 E A E E T S O I 7  short
= 8      | H H E_T_D_N E N |  column = 7
V E_C M R E_O N
H_E O T N E E_F
* B O W A I L R R *
*---5---* *-3-*
long      short

Step 3: Juxtapose the known plain text with the hat
diagram information and rearrange the columns.

Start with the Q  found in the 5th column of hat.  The U
Z                                       A
is relatively easy to find in column 2.

5 2
M B
T S
A R
Q U E E N E L I
Z A B E T H
T H
T E
D E
R O
N
I

In a similar manner the remaining columns are easily
identified and added to the juxtaposition:

5 2 3 6 4 1 7 8
C
M B E R O B E R
T S O N W I L L
A R R I V E O N
Q U E E N E L I
Z A B E T H O N
T H E S E V E N
T E E N T H O F
D E C E M B E R
R O O E T   R
N   W L A
I

Inspection tells us that the 61 letters lie within the
juxtaposition as follows:

5 2 3 6 4 1 7 8
C
M B E*R O B E R
T S O N W I L L
A R R I V E O N
Q U E E N E L I
Z A B E T H O N
T H E S E V E N
T E E N T H O F
D E C E M B E R*
R O O E T   R
N   W L A
I

We eliminate the superfluous letters, and shift column 6
to the beginning of the message. The result is the
original enciphering matrix and numerical key:

5 2 3 6 4 1 7 8
_______________
R O B E R T S O
N W I L L A R R
I V E O N Q U E
E N E L I Z A B
E T H O N T H E
S E V E N T E E
N T H O F_D_E_C
E_M_B_E_R

The plain text message is:

ROBERTSON WILL ARRIVE ON QUEEN ELIZABETH ON THE
SEVENTEENTH OF DECEMBER.

Proper juxtaposition of columns of the analytic matrix
depends not only upon the known plain text portion of
the enciphering matrix, but also upon plain text
appearing on the horizontal rows.  This solution is a
little more general as it does not depend on the
location of the known plain text.

COMPLETELY COLUMNAR TRANSPOSITION  - SOLUTION GIVEN KEY-
LENGTH AND A COMPLETELY FILLED MATRIX

The completely filled columnar matrix is a simpler
problem because we are dealing with only one column
length. There is no problem of determining the long and
short columns.

Column Matching:

Given the cipher text:

GLLEF  PLUOT  HERPI  RDEBC  NLGEE  NNBAR  SETHO  TEYWP
EHIAO  LIRMC  SERTS  VIIEH  EALPO  OEAFW  TX.  (72)

KEY LENGTH = 6

Step 1:  Determine size of rectangle.

72 = 6 KEY LENGTH X 12 ROWS

1 2 3 4 5 6
G R E E M E
L P N Y C A
L I N W S L
E R B P E P
F D A E R O
P E R H T O
L B S I S E
U C E A V A
O N T O I F
T L H L I W
H G O I E T
E E T R H X

This is the hat diagram with just one column length.
Solution depends now on merely rearrangement of these
six columns into numerical order.  There are two ways to
do this: 1) Anagraming and 2) column matching.

Anagraming

The cryptanalyst may anagram expected combinations of
letters.  Finding a Q, we look for a U and a vowel next
to it. These could be in the same row. We then
juxtaposition the columns or match them for QU_ or
combinations like THE or THAT, CK, ING, ION and so
forth.

Matching columns based on Validity Weighting

We select a single column at random from the analytic
matrix above:

1
G
L
L
E
F
P
L
U
O
T
H
E

Unless by chance that our choice (1/6 chance) is the
right hand column, one of the remaining columns will
stand to the left of our chosen column. There are 5
juxtapositions:

VW         VW         VW         VW        VW
12         13         14         15         16
GR   2     GE   2     GE   2     GM   0     GE  2
LP   0     LN   0     LY   3     LC   0     LA  2
LI   2     LN   0     LW   0     LS   1     LL  2
ER   2     EB   0     EP   1     EE   0     EP  1
FD   0     FA   1     FE   1     FR   1     FO  2
PE   2     PR   2     PH   1     PT   1     PO  2
LB   0     LS   1     LI   2     LS   1     LE  3
UC   1     UE   1     UA   1     UV   0     UA  1
ON   3     OT   2     OO   1     OI   1     OF  3
TL   0     TH   4     TL   0     TI   3     TW  1
HG   0     HO   2     HI   2     HE   4     HT  1
EE   0     ET   0     ER   2     EH   0     EX  3
---        ---        ---        ---       ---
12         15          16        12         23

where: VW = validity weight for individual bigrams and
total column validity weight.

To answer the question, which one of the 5 possibilities
is the best fit, we can use Barker's "Letter Contact
Weight Chart," Figure 16-2 to match and evaluate the
column interactions.  [BAR3]

To find the weight of a bigram in Figure 16-2, the first
letter of the bigraph is found in the vertical column
on the left side of the chart and the second letter is
found in the horizontal row of letters along the top of
the Figure. The intersection of these two letters is the
weight of the bigraph in question. For example QU has a
weight of 5, WH has a weight of three. The weights given
in Figure 16-3 are of a general nature for English but
are roughly dependent upon the expected frequency of
occurrence of the bigraphs plus the concept of 'good'
bigraphs like QU, LY, CK, TH, etc.  Since each of the 5
columns has 12 bigrams to evaluate, we look at the sum
of these individual weights as a column validity weight
to determine the best column match. The highest column
validity weight represents the best fit.

We can speed up the process by using the anagraming
approach in addition to the column matching attack.
We find the bigram LL in the 16 column combination.
LL is usually proceeded by a vowel. Looking at the
analytic matrix above at the same row as LL, only column
2 fits the bill.  We then place column 2 in front of
columns 1 and 6 and check the bigrams, trigrams for
impossibilities.

2 1 6
R G E
P L A
I L L
R E P
D F O
E P O
B L E
C U A
N O F
L T W
G H T
E E X

step:
3 5 4 2 1 6
E M E R G E
N C Y P L A
N S W I L L
B E P R E P
A R E D F O
R T H E P O
S S I B L E
E V A C U A
T I O N O F
H I L L T W
O E I G H T
T H R E E X

EMERGENCY PLANS WILL BE PREPARED FOR THE POSSIBLE
EVACUATION OF HILL TWO EIGHT THREE X

As a general rule, the longer the columns the more
reliable the use of Figure 16-2 for matching columns.
Generally speaking, a random bigram has a validity
weight of 1. Thus a mismatched pair of columns of length
10 will have a total validity weight of 10.  A validity
bigram has a weight of 1.5.  So the same example of 10
bigrams will have on average a validity weight of 15.

FIGURE 16-2
LETTER CONTACT WEIGHT CHART

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   2 2 1   1 1   1 1 1 2 1 3   2   2 2 2   2 1 1 1
B 1 1     4       1     2     1     1     2       1
C 2       2     2 1   2 1     2     1   1 1
D 2     1 2 1     2 1         1     1     1 1     1
E 2   1 2   1 1         1 1 2   1 2 2 2     1 1 3
F 1       1 1     1     1     2     1   1 1
G 1       2     2 1     1     2     2     1
H 3       4       2           2         1
I 1 1 2 1 1 2 1       2 2 1 3 2     1 2 2   2   2
J 1       1                   2           1
K 1       3       2         1 1       1           1
L 2     2 3       2   1 2     2       1 1 1       3
M 3 1     3       2       1   2 1         1
N 1   1 2 1   2   1 1 1     1 1       1 2 1 1   1 1
O 1 1 1 1   3 1   1   2 1 2 3 1 2   2 1 2 3 2 2 1
P 2       2     1 1     2     2     2   1 1
Q                                         5
R 1     1 2   1   1   1 1 1   1         1 1 1
S 2       2     2 2   1 1 1 1 2 2 1   1 3 2   1
T 2       2     4 3           3     1 1 2 2   1
U 1 1 1 1 1   1         1 1 2   1   2 2 2
V 2       4       2           2
W 2       2     3 2           1
X 1   1   1       1           1 1       1
Y 1       1                   2
Z         3                   2

IDENTIFICATION OF THE COMPLETELY FILLED MATRIX

There is a valuable test for determining if a complete
filled columnar matrix is in play.  Table 16-1 is an
extensive table of Key Lengths and Column-Lengths of
Completely Filled Matrices - Given their Message Length.
Table 16-1 covers message lengths from 15 to 300
letters.  The key lengths considered are from 3 to 25.
In reading the expressions to the right of the various
message lengths, the first number is the key length and
the second number is the column length. So for 22 X 9
following message length 198, the key length is 22 and
the column length is 9. Many message lengths may be the
result of several sizes of the completely filled
matrices.  These various sizes are listed within the
limits of the Table 16-1.  We use this table to
determine whether or not a particular message might have
been enciphered with a completely filled matrix. Table
16-1 shows that about 30% of all lengths cannot have
resulted from completely filled matrices.

Let's examine the following message:

URIRT  PEGRV  ATEPI  AFZSS  ITLFU  MAHKI  ECOLT  CWVAW
PEYLO  RAESL  ERETO  M.  (56)

Key-Length = unknown

>From Table 16-1, we find at 56 message length four
possibilities for a completely filled matrix exist. The
key lengths 4, 7, 8 and 14.

We test each of these possibilities without actually
reading the message and obtain reliability or validity
of each of the particular key-lengths. Normal plain-text
contains approximately 40% letters as vowels and that
these vowels are evenly distributed throughout the text.
We use the number of vowels in message X 100./number of
letters in the message = % of vowels in message. We
compute the per row number of vowels, mean difference
from normal, expected mean and validity value for the
matrix under consideration.

Number of      Mean
(7 x 8)        Vowels         Difference
1 2 3 4 5 6 7
U R F U O P S        3             0.2
R V Z M L E L        1             1.8
I A S A T Y E        5             2.2
R T S H C L R        0             2.8
T E I K W O E        4             1.2
P P T I V R T        1             1.8
E I L E A A O        6             3.2
G A F C W E M        2             0.8
---           ----
22             14.0 / 8 = 1.80

Expected Mean =22/8 =2.8

Number of   Mean
(14 X 4)               Vowels     Difference
1 2 3 4 5 6 7 8 9 1011121314
U T R E F I U K O W P O S E      7       1.5
R P V P Z T M I L V E R L T      2       3.5
I E A I S L A E T A Y A E O     11       5.5
R G T A S F H C C W L E R M      2       3.5
---     ----

22      14.0
Expected Mean = 22/4 = 5.5
Validity Value= 14/4 = 3.50

Four rows, and if this is a valid analytic matrix there
should be 24/4 = 5.5 vowels on each row.  This is the
expected mean for the each row. We tabulate the
individual row mean differences and sum them for the
matrix. The total of the mean differences is
proportional to the number of rows, by dividing this
total by the number of rows we obtain the Validity value
for this matrix.  Table 16-2 shows the interpretation of
the validity values.  It shows that less than or equal
to 0.75 (or closer to it in the indefinite range) the
more likely we are dealing with a completely filled
matrix with correct ley length.  Above 1.10 the more
likely it is that the matrix is incorrect.

Number of    Mean
(8 X 7)            Vowels       Difference
1 2 3 4 5 6 7 8
U G I T K C Y L         3           .1
R R A L I W L E         3           .1
I V F F E V O R         3           .1
R A Z U C A R E         4           .9
T T S M O W A T         2          1.1
P E S A L P E O         4           .9
E P I H T E S M         3           .1
---         ---
22          3.3

Expected Mean = 22/7 =3.1          3.3/7 = 0.47
Validity Value

The rows are too short (6 or less) for individual
analysis, so we combined them for analysis.

Number of      Mean Difference
Vowels
(4 X 14)
1 2 3 4
-------
U I K Y         5               1.9
R A I L
-------
I F E O
R Z C R         3                .1
-------
T S O A
P S L E         3                .1
------
E I T S
G T C L         2               1.1
-------
R L W E
V F V R         1                2.1
-------
A U A E
T M W T         4                 .9

-------
E A P O
P H E M         4                 .9
---               ---
22                7.1

Expected mean = 22/7 combined rows = 3.1
Validity value = 7.1 / 7 = 1.01  ( indefinite)

The results indicate that key length = 8 is the correct
length.

Key Length      Validity Value
of Matrix

14                 3.5
8                 0.47
7                 1.80
4                 1.01

Solution proceeds by anagraming and juxtaposition.  The
CK indicates a possible in the first row. Try the word
LUCKY and see how that fits.

(8 X 7)
*       * * * *
1 2 3 4 5 6 7 8
U G I T K C Y L
R R A L I W L E
I V F F E V O R
R A Z U C A R E
T T S M O W A T
P E S A L P E O
E P I H T E S M
becomes:
* * * * *
8 1 6 5 7 4 3 2
L U C K Y T I G
E R W I L L A R
R I V E O F F V
E R A C R U Z A
T T W O A M S T
O P P L E A S E
M E E T S H I P

Key = 81657432

Plaintext:  LUCKY TIGER WILL ARRIVE OFF VERA CRUZ AT TWO

TABLE 16-2
VALIDITY VALUES

<-------------> 0.75 <----------> 1.10 <------------>

Valid               Indefinite         Incorrect
Matrix                                  Matrix

TABLE 16-1
Table of Key Lengths and Column Lengths of Completely
Filled Matrices - Given The Message Length

15  3 X 5, 5 X 3
16  4 X 4, 8 X 2
17
18  3 X 6, 6 X 3, 9 X 2
19
20  4 X 5, 5 X 4 10 X 2
21  3 X 7, 7 X 3
22  11 X 2
23
24  3 X 8, 4 X 6, 6 X 4, 8 X 3, 12 X 2
25  5 X 5
26  13 X 2
27  3 X 9, 9 X 3
28  4 X 7, 7 X 4, 14 X 2
29
30  3 X 10, 5 X 6, 6 X 5, 10 X 3, 15 X 2
31
32  4 X 8, 8 X 4, 16 X 2
33  3 X 11, 11 X 3
34  17 X 2
35  5 X 7, 7 X 5
36  3 X 12, 4 X 9, 6 X 6, 9 X 4, 12 X 3, 18 X 2
37
38  19 X 2
39  3 X 13, 13 X 3
40  4 X 10, 5 X 8, 8 X 5, 10 X 4, 20 X 2
41
42  3 X 14, 6 X 7, 7 X 6, 14 X 3, 21 X 2
43
44  4 X 11, 11 X 4, 22 X 2
45  3 X 15, 5 X 9, 9 X 5, 15 X 3
46  23 X 2
47
48  3 X 16, 4 X 12, 6 X 8, 8 X 6, 12 X 4, 16 X 3, 24 X 2
49  7 X 7
50  5 X 10, 10 X 5, 25 X 2
51  3 X 17, 17 X 3
52  4 X 13, 13 X 4
53
54  3 X 18, 6 X 9, 9 X 6, 18 X 3
55  5 X 11, 11 X 5
56  4 X 14, 7 X 8, 8 X 7, 14 X 4
57  3 X 19, 19 X 3
58
59
60  3 X 20, 4 X 15, 5 X 12,, 6 X 10, 10 X 6,12 X 5,
15 X 4, 20 X 3
61
62
63  3 X 21, 7 X 9, 9 X 7, 21 X 3
64  4 X 16, 8 X 8, 16 X 4
65  5 X 13, 13 X 5
66  3 X 22, 6 X 11, 11 X 6, 22 X 3
67
68  4 X 17, 17 X 4
69  3 X 23, 23 X 3
70  5 X 14, 7 X 10, 10 X 7, 14 X 5
71
72  3 X 24, 4 X 18, 6 X 12, 8 X 9, 9 X 8,12 X 6, 18 X 4,
24 X 3
73
74
75  3 X 25, 5 X 15, 15 X 5, 25 X 3
76  4 X 19, 19 X 4
77  7 X 11, 11 X 7
78  3 X 26, 6 X 13, 13 X 6
79
80  4 X 20, 5 X 16, 8 X 10, 10 X 8, 16 X 5, 20 X 4
81  3 X 27, 9 X 9
82
83
84  3 X 28, 4 X 21, 6 X 14, 7 X 12, 12 X 7, 14 X 6,
21 X 4
85  5 X 17, 17 X 5
86
87  3 X 29
88  4 X 22, 8 X 11, 11 X 8, 22 X 4
89
90  3 X 30, 5 X 18, 6 X 15, 9 X 10, 10 X 9, 15 X 6,
18 X 5
91  7 X 13, 13 X 7
92  4 X 23, 23 X 4
93  3 X 31
94
95  5 X 19, 19 X 5
96  3 X 32, 4 X 24, 6 X 16, 8 X 12, 12 X 8, 16 X 6,
24 X 4
97
98  7 X 14, 14 X 7
99  3 X 33, 9 X 11, 11 X 9
100 4 X 25, 5 X 20, 10 X 10, 20 X 5, 25 X 4
101
102 3 X 34, 6 X 17, 17 X 6
103
104 4 X 26, 8 X 13, 13 X 8
105 3 X 35, 5 X 21, 7 X 15, 15 X 7, 21 X 5
106
107
108 3 X 36, 4 X 27, 6 X 18, 9 X 12, 12 X 9, 18 X 6
109
110 5 X 22, 10 X 11, 11 X 10, 22 X 5
111 3 X 37
112 4 X 28, 7 X 16, 8 X 14, 14 X 8, 16 X 7
113
114 3 X 38, 6 X 19, 19 X 6
115 5 X 23, 23 X 5
116 4 X 29
117 3 X 39, 9 X 13, 13 X 9
118
119 7 X 17, 17 X 7
120 3 X 40, 4 X 30, 5 X 24, 6 X 20, 8 X 15, 10 X 12,
15 X 8, 20 X 6 , 24 X 5
121 11 X 11
122
123 3 X 41
124 4 X 31
125 5 X 25, 25 X 5,
126 3 X 42, 6 X 21, 7 X 18, 9 X 14, 14 X 9, 18 X 7,
21 X 6
127
128 4 X 32, 8 X 16, 16 X 8
129 3 X 43
130 5 X 26, 10 X 13, 13 X 10
131
132 3 X 44, 4 X 33, 6 X 22, 11 X 12, 12 X 11, 22 X 6
133 7 X 19, 19 X 7
134
135 3 X 45, 5 X 27, 9 X 15, 15 X 9
136 4 X 34, 8 X 17, 17 X 8
137
138 3 X 46, 6 X 23, 23 X 6
139
140 4 X 35, 5 X 28, 7 X 20, 10 X 14, 14 X 10, 20 X 7
141 3 X 47
142
143 11 X 13, 13 X 11
144 3 X 48, 4 X 36, 6 X 24, 8 X 18, 9 X 16, 12 X 12,
16 X 9, 18 X 8, 24 X 6
145 5 X 29
146
147 3 X 49, 7 X 21, 21 X 7
148 4 X 37
149
150 3 X 50, 5 X 30, 6 X 25, 10 X 15, 15 X 10, 25 X 6
151
152 4 X 38, 8 X 19, 19 X 8
153 3 X 51, 9 X 17, 17 X 9
154 7 X 22, 11 X 14, 14 X 11, 22 X 7
155 5 X 31
156 3 X 52, 4 X 39, 6 X 26, 12 X 13, 13 X 12
157
158
159 3 X 53
160 4 X 40, 5 X 32, 8 X 20, 10 X 16, 16 X 10, 20 X 8
161 7 X 23, 23 X 7
162 3 X 54, 6 X 27, 9 X 18, 18 X 9
163
164 4 X 41
165 3 X 55, 5 X 33, 11 X 15, 15 X 11
166
167
168 3 X 56, 4 X 42, 6 X 28, 7 X 24, 8 X 21, 12 X 14,
14 X 12, 21 X 8, 24 X 7
169 13 X 13
170 5 X 34, 10 X 17, 17 X 10
171 3 X 57, 9 X 19, 19 X 9
172 4 X 43
173
174 3 X 58, 6 X 29
175 5 X 35, 7 X 25,25 X 7
176 4 X 44, 8 X 22, 11 X 16, 16 X 11, 22 X 8
177 3 X 59
178
179
180 3 X 60,4 X 45, 5 X 36, 6 X 30, 9 X 20, 10 X 18,
12 X 15, 15 X 12, 18 X 10, 20 X 9
181
182 7 X 26, 13 X 14, 14 X 13
183 3 X 61
184 4 X 46, 8 X 23, 23 X 8
185 5 X 37
186 3 X 62, 6 X 31
187 11 X 17, 17 X 11
188 4 X 47
189 3 X 63, 7 X 27, 9 X 21, 21 X 9
190 5 X 38, 10 X 19, 19 X 10
191
192 3 X 64, 4 X 48, 6 X 32, 8 X 24, 12 X 16, 16 X 12,
24 X 8
193
194
195 3 X 65, 5 X 39, 13 X 15, 15 X 13
196 4 X 49, 7 X 28, 14 X 14
197
198 3 X 66, 6 X 33, 9 X 22, 11 X 18, 18 X 11, 22 X 9
199
200 4 X 50, 5 X 40, 8 X 25, 10 X 20, 20 X 10, 25 X 8
201 3 X 67
202
203 7 X 29
204 3 X 68, 4 X 51, 6 X 34, 12 X 17, 17 X 12
205 5 X 41
206
207 3 X 69, 9 X 23, 23 X 9
208 4 X 52, 8 X 26, 13 X 16, 16 X 13
209 11 X 19, 19 X 11
210 3 X 70, 5 X 42, 6 X 35, 7 X 30, 10 X 21, 14 X 15,
15 X 14, 21 X 10
211
212 4 X 53
213 3 X 71
214
215 5 X 43
216 3 X 72, 4 X 54, 6 X 36, 8 X 27, 9 X 24, 12 X 18,
18 X 12, 24 X 9
217 7 X 31
218
219 3 X 73
220 4 X 55, 5 X 44, 10 X 22, 11 X 20, 20 X 11, 22 X 10
221 13 X 17, 17 X 13
222 3 X 74, 6 X 37
223
224 4 X 56, 7 X 32, 8 X 28, 14 X 16, 16 X 14
225 3 X 75, 5 X 45, 9 X 25, 15 X 15, 25 X 9
226
227
228 3 X 76, 4 X 57, 6 X 38, 12 X 19, 19 X 12
229
230 5 X 46, 10 X 23, 23 X 10
231 3 X 77, 7 X 33, 11 X 21, 21 X 11
232 4 X 58, 8 X 29
233
234 3 X 78, 6 X 39, 9 X 26, 13 X 18, 18 X 13
235 5 X 47
236 4 X 59
237 3 X 79
238 7 X 34, 14 X 17, 17 X 14
239
240 3 X 80, 4 X 60, 5 X 48, 6 X 40, 8 X 30, 10 X 24,
12 X 20, 15 X 16, 16 X 15, 20 X 12, 24 X 10
241
242 11 X 22, 22 X 11
243 3 X 81, 9 X 27
244 4 X 61
245 5 X 49, 7 X 35
246 3 X 82, 6 X 41
247 13 X 19, 19 X 13
248 4 X 62, 8 X 31
249 3 X 83
250 5 X 50, 10 X 25, 25 X 10
251
252 3 X 84, 4 X 63, 6 X 42, 7 X 36, 9 X 28, 12 X 21,
14 X 18, 18 X 14, 21 X 12
253 11 X 23, 23 X 11
254
255 3 X 85, 5 X 51, 15 X 17, 17 X 15
256 4 X 64, 8 X 32, 16 X 16
257
258 3 X 86, 6 X 43
259 7 X 37
260 4 X 65, 5 X 52, 10 X 26, 13 X 20, 20 X 13
261 3 X 87, 9 X 29
262
263
264 3 X 88, 4 X 66, 6 X 44, 8 X 33, 11 X 24, 12 X 22,
22 X 12, 24 X 11
265 5 X 53
266 7 X 38, 14 X 19, 19 X 14
267 3 X 89
268 4 X 67
269
270 3 X 90, 4 X 54, 6 X 45, 9 X 30, 10 X 27, 15 X 18,
18 X 15
271
272 4 X 68, 8 X 34, 16 X 17, 17 X 16
273 3 X 91, 7 X 39, 13 X 21, 21 X 13
274
275 5 X 55, 11 X 25, 25 X 11
276 3 X 92, 4 X 69, 6 X 46, 12 X 23, 23 X 12
277
278
279 3 X 93, 9 X 31
280 4 X 70, 5 X 56, 7 X 40, 8 X 35, 10 X 28, 14 X 20,
20 X 14
281
282 3 X 94, 6 X 47
283
284 4 X 71
285 3 X 95, 5 X 57, 15 X 19, 19 X 15
286 11 X 26, 13 X 22, 22 X 13
287 7 X 41
288 3 X 96, 4 X 72, 6 X 48, 8 X 36, 9 X 32, 12 X 24,
16 X 18, 18 X 16, 24 X 12
289 17 X 17
290 5 X 58, 10 X 29
291 3 X 97
292 4 X 73
293
294 3 X 98, 6 X 49, 7 X 42, 14 X 21, 21 X 14
295 5 X 59
296 4 X 74, 8 X 37
297 3 X 99, 9 X 33, 11 X 27
298
299 13 X 23, 23 X 13
300 3 X 100, 4 X 75, 5 X 60, 6 X 50, 10 X 30, 12 X 25
15 X 20, 20 X 15, 25 X 12

SOLUTION GIVEN KEY LENGTH PLUS A PROBABLE WORD IN THE
TEXT

Given the key-length, we are able to construct the hat
diagram; and the analytically juxtapositioning of the
matrix columns is facilitated greatly by a probable
word.

Given:

RCRKA  LPTNA  TALMO  IDFNV  TRTIN  FLEFR  IONOI  WOPIE
CGOAF  RDCUH  OIAIT  ELLPR  IRPSN  EYRRC  IHITI  OTWUO
IDSPF  SOIEK  GMN.   (93)

Probable word = NEW YORK   Key length = 9

Perform a monoalphabetic frequency distribution.

NEW YORK
652 1992

The Y is a gimme.  Draw up the hat diagram based on key
length = 9.

1 2 3 4 5 6 7 8 9
T I C O I
T R O G I R H
R A T N O A P I S
C L I O A I S T P
R M N I F T N I F
K O F W R E E O S
A I L O D L Y T O
L D E P C L R W I
P F F I U P R U E
T N R E H R C O K
N V I C O I I I G
A T O G I R H D M
T R N O A P I S N

>From column 7 with the Y and column 1 with the K:

7     1
I
R
P
S     R
N     C
E     R
N E W Y O R K
R     A
R     L
C     P
I     T
H     N
I     A
T

We add column 3 with the N, followed by 5 with the R, 6
matches the E, 4 the W, and 8 brings out RICHMOND.

2  3  6  4  7  9  5  1  8
I
O  I  R     C     H
T  I  O  P  S  G     I
T  R  A  N  S  P  O  R  T
A  T  I  O  N  F  A  C  I
L  I  T  I  E  S  F  R  O
M  N  E  W  Y  O  R  K  T
O  F  L  O  R  I  D  A  W
I  L  L  P  R  E  C  L  U
D  E  P  I  C  K  U  P  O
F  F  R  E  I  G  H  T  I
N  R  I  C  H  M  O  N  D
V  I  R  G  I  N  I  A  S
T  O  P

O     A T
R N

TRANSPORTATION FACILITIES FROM NEW YORK TO FLORIDA WILL
PRECLUDE PICKUP OF FREIGHT IN RICHMOND, VIRGINIA. STOP.

Barker presents two more special cases leading to the
General solution but the basic concepts have been
presented in this lecture. [BAR3]

DOUBLE COLUMNAR TRANSPOSITION CIPHER

Courville, Friedman and the Army Extension Course Text
No 166 discuss double transposition in copious detail.
Cryptanalysis of the double transposition is covered in
detail.  Essentially the encipherment is polyphase and
the decryption hinges on sizing the matrices correctly -
especially the first transposition matrix. [COUR],
[FRE4], [ARMY]

AMSCO

The AMSCO Cipher is another type of incomplete columnar
transposition.  Its column-letters are not limed to a
column of single letters, but rather alternating single,
double, single, double throughout the plain text length.
A numerical key is employed.  For example:

3  1  4  2  5             2  4  6  1  5  3
TH E  WE A  RI            T  HE W  EA R  IN
N  GO F  DE C             GO F  DE C  OR A
OR A  TI V  EM            T  IV E  M  ED AL
E  DA L  SW A             SW A  SC O  MM O
SC O  MM O  NI            N  IN E  NG L  AN
N  EN G  LA N             DD U  RI N  GT H
DD U  RI N  GT            E  RE I  GN O  FH
H  ER E  IG N             EN R  YT H  EE I
OF H  EN R  YT            G  HT H
H  EE I  GH T
HX                        1-2-1-2-1-2
2-1-2-1-2-1
2-1-2-1-2                 1-2-1-2-1-2
1-2-1-2-1                 2-1-2-1-2-1
2-1-2-1-2                 1-2-1-2-1-2
1-2-1-2-1                 2-1-2-1-2-1
2-1-2-1-2                 1-2-1-2-1-2
1-2-1-2-1                 2-1-2-1-2-1
2-1-2-1-2                 1-2-1
1-2-1-2-1
2-1-2-1-2
1-2-1-2-1
2                             (B)
(A)

In matrix (A) the alternating pattern of 2-1-2-1 follows
from one end of one line to the next line; but in matrix
(B) it is possible to have two 1's or two 2's in the
continuation from one line to the next.  This is a
pecularity of this cipher.  Solution is done similarly
to the incomplete columnar. Use of a probable word is
important for this cipher.  Columns are extracted in
numerical order.

Example:  Tip = PRECIOUS

NTTIN  OENOE  NTUSD  PRTTE  RIUUN  TOLIV  EDSIS  ORDEW
LLTIL  STSII  CRTOL  NKOOU  XHKIG  NALHE  ENEOL  ESERY
GSPDL  SRWIO  ANSWI  AAENS  LEIFS  RHPSA  FIHRR   (115)

Solution:

Divide tip into alternative patterns.

-P RE C IO U S-       PR E CI O US;

The ciphertext hits RE = none, IO at 89; PR at 16, CI at
33, US at 13.  Accept 2'nd pattern with three hits.
Write in the ciphertext on both sides of the known pairs
to the extent of 8 - 9 letters.

UN     IN
T      O
OL     EN
I      O
VE     EN
D      T
PR  E  CI  O  US
S      D
OR     PR
D      T
EW     TE
L      R
LT     IU

We lightly cross out the used letters as we go along.
The existence of PR here, shows that the PR of PRECIOUS
- appearing just once in the cipher can not be used here
so our original assumption is wrong.  Therefore the tip
is found on two lines.

We test the O's  using the alternate pattern 1-2-1-2-1,
whenevr an O occurs and see what is plausible.  O-54: O
LN K OO U XH gives CIOUS SLN*; O-58: O OU X HK gives
CIOUS SOUND OXPR*; O-59: O UX H KI gives CIOUS SUX*; O-
90: O AN S WI A gives CIOUS SAND; ORSPR DWIT EWATE,
which looks good so we write in the column. After EWATE,
the rows go bad indicating the bottom of the block.
Remember the first letter of the cipher of this type
will be found somewhere in the top row of the plaintext.
We extend our columns up to the first letter.

The final message is:

O  RI  E  NT  A  LL
UX U   RY  T  EN T
H  UN  G  IN  S  IL
KI T   SP O   L  ES
G  OL  D  EN  I  TS
NA I   LS O   FS I
L  VE  R  EN  R  IC
HE D   WI T   HP R
PR  E  CI  O  US  S  TO
NE S   AN D   AF L
O  OR  S  PR  I  NK
LE D   WI T   HR O
S  EW  A  TE  R  --

Note the PR was not a bigraph but broken up -P R in the
line above.

MYSZKOWSKI

(Named after the famous flying Myszkowski family circus
high-wire act)  is another incomplete columnar
transposition with an erratic method of taking out the
ciphertext letters.  Keywords with repeated letters are
allowed and taken out in left to right order for the
repeated letters. In ciphertext 2-, 3- and even 4-
decimations are evidenced.  A 3-decimation, 3 letters in
the keyword are repeated would give rise to every third
letter being at issue.

Solution is by period and probable word.

Keying examples:

F I C T I O N        P A P I L L A
2 3 1 6 3 5 4        4 1 4 2 3 3 1
A M O O S E I        A M O O S E I
S S O C A L L        S S O C A L L
E D A S T H E        E D A S T H E
W O R D I S S        W O R D I S S
A I D T O M E        A I D T O M E
A N C R O P P        A N C R O P P
E R O R T R I        E R O R T R I
M M E R F R O        M M E R F R O
M T H E A N I        M T H E A N I
M A L S H A B        M A L S H A B
I T O F F E E        I T O F F E E
D I N G O N T        D I N G O N T
R E E B R A N        R E E B R A N
C H E S - - -        C H E S - - -
(A)                 (B)

Cipher (A)

OOARD  COEHL  ONEEA  SEWAA  EMMMI  DRCMS  SADTO  IIONO
RTMFT  AAHTF  IOERH  ILESE  PIOIB  ETNEL  HSMPT  RNAEN
AOCSD  TRRRE  SFGBS  (95)

Cipher (B)

MISLD  EOSIE  NPRIM  OTIAB  TEITE  NHOCO  DTRRR  ESFGB
SSEAL  THISO  MOPTR  FRANH  AFEON  RAAOS  OEAWR  ADACE
OMEMH  MLIOD  NRECE   (95)

Compare the keyword mixings for both ciphers and pick up
the decimation intervals.

As a partial example of the process:

Given:  Keyword = ERMEDICAL, PERIOD = 6

UEIES  OCOSH  IEIDF  AIPLH  MLCAU  SSRTT  OTMUE  NRAAN
NROSA  XSREF  KPNEL  OINEN  OCMII  FOAGZ  NADEM  CLPRO
SITOM  RMCYS  NIIAA  AKEFT  OSINL  ATTSQ  ESHON  YLETD
RTNEF  TUESE  BEMGA  AICRT  PONHG  OEPAA  HOARD  RRAFR
NET  (163)

Block size is known and may be drawn up as 6 X 27, plus
1.

O R I
N T E
N T O
C O M
I T I            Trigraphs off these
F M A            possibilities:
O U G
Z E N               N T E (R)
A N D               (I)NTO
E R M E D I         C O M (M)
C A L               I T I (ION)
P A R               OUG (HT)
O N S               (A or I) - ZEN
I N T               PAR(T)
O R M
R O M
C S Y

The final key is:  3 5 4 1 3 2

Non repeated columns are removed exactly previously
discussed. Repeated letter columns give rise to the 2-
or 3- or 4- decimations, so look at adjacent letters for
plain text.

The Cadenus is a double transposition type, employing a
keyword, as in columnar transposition, to shift the
order of the columns and in addition, to shift the
starting point of each column using the same key. The
second shift is accomplished by attaching a letter of
the alphabet to each row during construction.  V and W
used together.  The block must be complete and 25
letters long in each column.

Example:

EASY        AESY
2134        1234
ASEV   A    SYST
EREL   B    RETO
IMIT   C    MTAT
ATIO   D    TLUS
NONT   E    OATL
HEUS   F    EEES
EFUL   G    FIYH
NESS   H    EASD
OFTH   I    FNMS
ENUS   K    NEUV
ISTH   L    SNPM
ATEV   M    TOFA
ERYM   N    RENU
ESSA   O    SEIE
GEMU   P    EIEL
STBE   Q    TARL
AMUL   R    MENT
TIPL   S    IEET
EOFT   T    OGEV
WENT   U    ESIT
YFIV   VW   FAIS
ELET   X    LTNG
TERS   Y    EEUV
LONG   Z    OWUL

Cipher:

SYSTR  ETOMT  ATTLU  SOATL  etc

Solution procedure:

1. Count the number of letters, divide by 25 = number of
columns.

2. Write the cipher into the block by horizontals.

3. Write the probable word and examine the cipher block
for correct letters. Anagram.

4. When the entire block has been constructed, find the
beginning of the plain text and rewrite the block.
Place the alphabet at one edge of the block and note
the keyword.

AUTO-TRANSPOSITION

The auto transposition cipher is a multiple trans-
position by groups with a keyword controlling the first
cipher group. The letters of each group in turn are
converted into a numerical sequence. In some cases,
anagraming is an aid, but not always.

To encipher, select a keyword of any length and write in
the plaintext under it. Skip a line and repeat the
plaintext with the first group under the keyword. Then
assign numbers to the keyword's letters in their order
of the normal alphabet. Using this resulting numerical
sequence apply it to the first group of the plaintext.
Continue in this manner through each consecutive group.

key:

F R A G I L E*W H E N M E M*B E R S O F A*N O R G A N I
3 7 1 4 5 6 2*7 3 1 6 4 2 5*2 3 6 7 5 4 1*4 6 7 2 1 5 3

Plain:
W H E N M E M*B E R S O F A*N O R G A N I*Z A T I O N G

Cipher:
E M W N M E H*A R B F S E O*O R N I A G N*I N G A Z O T
1 5 7 6 4 2 3 1 6 2 4 7 3 5  ...

The complete cipher may be written in either groups of
five or its true period length. A tip is essential.
Placing the tip is an easy exercise and recovering the
text after the tip is straightforward but not before it.
Anagramming is essential to the solution.

Given: Period = 6, Tip = EIGHTEENEIGHTYSEVE(N)

RHEPTE  SCDESE  ROOFTO  ACYDOS  UMREPT  WASASS  TTTIAS
LIMCAA  NICEIH  NENEVT  BTDOUA  GEIHET  EGEIHN  TEYVSE
RSONUF  IENCMO  ILNPGI  IHUENT  TETASD  ESECNT  GUFSSE
(150)

The tip is found in groups 12, 13 and part of 14.

Cipher

G E I H E T   E G E I H N   T E Y V S E
1 5 3 4 6 2   1 6 2 5 3 4
Plain
E I G H T E   E N E I G H   T Y S E V E
1 5 3 4 6 2   1 6 2 5 3 4   4 6 3 1 5 2

Cipher text is:

THE PREDESSOR OF TODAYS COMPUTERS WAS A MACHINE INVENTED
ABOUT EIGHTEEN EIGHTY SEVEN FOR USE IN COMPILING THE
UNITED STATES CENSUS.

Example of the mechanism is:
7 3 1 6 4 2 5
B E R S O F A    Plain
1 2 3 4 5 6 7

=     A R B F S E O    Cipher

GRILLE  / TURNING GRILLE

Friedman, Masterton, Bowers, LEDGE, Elcy as well as OP-
G-20 cover the Grille in detail.  [FRE4], [OP20],
[MAST], [LEDG], [BOW1], [ELCY].

The grille is an ingenious transposition which the
'stuff that spy used in the field' are made of.
Cryptographic grilles are stencils cut with holes for
the purpose of uncovering a small part of the paper that
the plain text is written on.  Generally, both
correspondents have identical grilles and know the
routes in and out of the grille to inscribe / transcribe
the plain /cipher text.

There are eight positions that the grille can be turned,
two sides and four 90 degree turns.

Lets illustrate with a 6 X ^ square and the message
SORTIE WILL COMMENCE AT MIDNIGHT FOUR JUNE

Let open apertures be shown as O and closed be shown by
X.

GRILLE                  1st Position
1 2 3 4 5 6               1 2 3 4 5 6
1   O   O                 1   S   O
2 O                       2 R
3       O   O             3       T   I
4         O               4         E
5   O   O                 5   W   I
6           O             6           L

2nd Position                3rd Position
1 2 3 4 5 6               1 2 3 4 5 6
1         L               1 A
2   C       O             2     T   M
3                         3   I
4   M   M   E             4 D   N
5     N                   5           I
6 C     E                 6     G   H

4th Position              Complete Inscription
1 2 3 4 5 6               1 2 3 4 5 6
1     T     F             1 A S T O L F
2       O                 2 R C T O M O
3 U   R   J               3 U I R T J I
4                         4 D M N M E E
5 U       N               5 U W N I N I
6   E                     6 C E G E H L

Cipher may be taken out by any route.

ASTOL  FRCTO  MOUIR  TJIDM  NMEEU  WNINI  CEGEH L.

To decipher we reverse this process.  We may anagram the
letters to form another sequence of letters that are
intelligible.  We assign numbers to the cipher text to
facilitate the process. We look for our invariable
combinations like QU and CK. These form a good starting
point.  Grille positions are 180 degrees reciprocals. We
can write the grille message out and then reverse the
message under it to have reciprocal positions in the
square line up vertically.

Problem:
ARUDU  CSCIM  WETTR  NNGOO  TMIEL  MJENH  FOIEI  L

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
A R U D U C S C I M  W  E  T  T  R  N  N  G  O  O  T
L I E I O F H N E J  M  L  E  I  M  T  O  O  G  N  N

22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
M  I  E  L  M  J  E  N  H  F  O  I  E  I  L
R  T  T  E  W  M  I  C  S  C  U  D  U  R  A

Each anagram formed with letters on one of these lines
corresponds with the reversal of the other plain text
formed on the other line.  The partial recovery of the
plain by anagraming might proceed as follows.
(only correct trials are shown)

34 17  22 34 17  22 34 17 6  25 8 32 10 22 34 17 6 24
E N    M E N     M E N C      L C O M M E R C E -->
U O    R U O     R U O F      E N U J R U O F T  <--

The plain text will break after just nine letters,
because the grille used was of that capacity.
When one fourth of the total number of letters in the
cipher text have been anagrammed without a break in the
plain text on either line, the letters which were
originally inscribed in reciprocal positions of the
grille have been found.

The grille used in this problem may be reconstructed by
numbering the cells of the square of 6 x 6 in the normal
manner and then cutting out the cells numbered according
to the series found 25-8-32-10-22-34-17-6-24.

SWAGMAN / LAZY SWAGMAN TRANSPOSITION

An Australian version of the Tramp, the Swagman uses a
keying square composed of somewhat random arrangement of
digits between 4 - 8 wide wherein the digits have no
repeats within either row or column.

3 2 1 4 5
1 5 3 2 4
2 4 5 3 1
5 3 4 1 2
4 1 2 5 3

The plain text is written horizontally to form a
rectangle commensurate with the width selected - here 5
X 5. If necessary, nulls are added to complete the last
letters of the rectangle, which must be a multiple of
the width of the square.

Suppose we have a short "message" and a corresponding 4-
width box:

T H I S   I S A N   E X        2 3 1 4
A M P L   E O F T   H E        3 2 4 1    4 X 10
S W A G   M A N T   R A        4 1 3 2
N S P O   S I T I   O N        1 4 2 3

The first column of each box is rearranged according to
the first column of the square - thus T is put to the
2nd row (as I and E of the other boxes will be) A is put
on the third, S on the fourth, and N on the first. Then
H(and S and X) get put on the third row, M on the
second, W on the first, and S on the fourth - since the
second column of our digital box is 3 2 4 1 .  And so on
to form the intermediate version:

N W I L   S A A T   O A
T M P G   I O T T   E E
A H A O   E S N I   H X
S S P S   M I F N   R N

The final cipher is taken off vertically:

NTASW  MHSIP  APLGO  SSIEM  AOSIA  TNFTT  INOEH  RAEXN

Lets look at a problem:

POGTC  VEEIO  AIROR  LLDLE  NOWGP  AIAAN  FNGTA  THATL
ICTPN  HUEAX  YGELA  DIDAN  EUNMB  ILANT  RRICM  EAMIG
LAMPA  RTASR  POOOA  LUPDO  BROAS  ESELA  NSNQL  ODUHC
EIAAS  CGDSO  ORREM  BTOWI  SOUS.

The crib is Caesar UXPFMP which is TWOELO.

POGTC  VEEIO  AIROR  LLDLE  NOWGP  AIAAN  FNGTA  THATL

o      ee           ll      o
ICTPN  HUEAX  YGELA  DIDAN  EUNMB  ILANT  RRICM  EAMIG

LAMPA  RTASR  POOOA  LUPDO  BROAS  ESELA  NSNQL  ODUHC

EIAAS  CGDSO  ORREM  BTOWI  SOUS.
t w    o

Since there are only a couple of W's, we look for all
the other letters of the crib in ORDER around that area,
find them.  We come up with a width which puts each
letter in consecutive columns.  The unique TW and final
O narrow things down.

The break is at position 138, or one position removed.
138 is a factor of 6.  We find as we put one letter in
each several consecutive boxes , we prepare a worksheet
of cipher text 6 deep.

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

One row of the key must be 2 1 5 4?? and the key row
above it must be ????51.

The message is about an elphant. GARGANTUAN MAMMALIAN
LOCOMOTION, A FLEXIBLE PROBOSCIS, TWO LONGATED INCISORS,
ALSO OSCILLATING AURAL APPENDAGES.

3 5 1 6 4 2
6 4 3 2 1 5
1 2 6 5 3 4
5 3 4 1 2 6        KEY
4 6 2 3 5 1
2 1 5 4 6 3

ZIMMERMAN AND CIPHER 0075

Arthur Zimmermann, the German foreign minister sent a
message to Mexico that put the US in a fury.  German
cryptographers used a cipher known to them as 0075. The
message sent was:

C:CTLTZ  EMRTH  IERSI  TNAII  WETXC  AAMOR  OXCEA  ATWOA
AONIZ  NEETN  MXASA  LDINF  ESZRC  ATEIO  GZFXA  LAEIR
AOMBI  OWEWW.   (90)

Unfortunately for Zimmermann and the Mexicans, British
intelligence cracked 0075 and wired U.S. President
Woodrow Wilson the following information (paraphrased
here):

CONTENTS OF ZIMMERMANN CABLE FOR YOUR INSPECTION:

P:" CONFIRM THAT MEXICO WILL BE AWARDED TITLE TO ARIZONA
TEXAS NEW MEXICO IF MEXICO ENTER WAR AGAINST USA AZ AZ
AZ"

Encipherment:

The key = 0075 was used in a simple equation to obtain a
control key.

K=19999 +Key
----------   = Control Key
97

K=19999 +0075
-----------  = 206.948536
97

All 10 digits are used and the period is ignored.

K= 2069484536

This series of digits is ranked according to the value
of each digit and its place in the series. Zero = 10.

K = 2 0  6 9 4 8 4 5 3 6       control key
K = 1 10 6 9 3 8 4 5 2 7       ranked control key

The plaintext is written below the ranked control key ,
ten letters to the line, but are written into  as
opposed to out according to the ranked control key.

K = 1 106 9 3 8 4 5 2 7
C A R H N T F I O M   (C O N F I R M T H A)
T L C I E W X I M O
L D A E E D A W B R
T I T R T A L E I O
Z N E S N A A T O X
E F I I M O E X W C
M E O T X N I C E E
R S G N A I R A W A
T Z Z A S Z A A U A
1 2 3 4 5 6 7 8 910     (RE-RANKED KEY)

Next we re-rank below the columns the control key, in
this case a straight series 1...10.  We take out the
columns by this new order and divide into groups of 5.

Ciphertext:

CTLTZ  EMRTH  IERSI  TNAII  WETXC  AAMOR  OXCEA  ATWOA
AONIZ  NEETN  MXASA  LDINF  ESZRC  ATEIO  GZFXA  LAEIR
AOMBI  OWEWW.   (90)

We have a double transposition. The primary drawback is
that the plain text is out in the open (albeit
scrambled). In the above cryptogram, the probable word
MEXICO with the X is a good start.

SOLUTIONS FOR LECTURE 15 PROBLEMS - Taken from OP- 20 -G
course:

15-1.   Naval Text.  Recover Keys.

J Z S S W B P D Z Z L F O M E K Q P D J H C K U M C

A B C O O X M Y S I I G B S G G Y V D S W A J O Q E

K U P W K N J K C C H W O Z Q Q B P Y N V J J O Q E

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

D H U V N K V K C Y Y A L R Q O O Q D N Z C G L R E

K F H Q R N J B.

The text appears in lines of 26 letters, which was
determined as the key length by factoring.  This is an
example of a regular progressive cipher. We reconstruct
the cipher component from symmetrical sequences. The
symmetrical sequences found, with their space
relationships in the cipher component are:

(K U) M C A B C O O X M Y S I I  - 5
(U P) W K N J K C C H W O Z Q Q

S S W B P D Z Z - 7
Y Y A L R Q O O

C D S L R W C (F Q I) - 22
R E K F H Q R (N J B)

The letters in parentheses are assumed to belong to the
symmetrical sequences but must be checked.

The reconstruction progresses through stages to give:

1234567891011121314151617181920212223242526  Interval
O                                  Z     7
Y    O      C         K       S         Z     5
H   R     C                                 22
H                                    X      5
P      R                                       7
-------------------------------------------
PY H  OR     C         K       S        XZ   (combined)
P                      K          U            5
B         J L                    (assumed)
NB       FIJ L   Q         W        22
A      N              M           W        5
D           Q                  7
E       D           Q                 22
-------------------------------------------
PY(T)HA(G)OREN BCDFIJKLMQSU(V)WXZ            (combined)

The cryptogram is then converted to the basis of one
cipher alphabet. This conversion process makes use of
the known shift between components of the cipher
alphabet, reducing each letter to is equivalent value
had the components not been shifted during encipherment.
The shift is done based on the square table.

Similar to a Viggy:

1 PYTHAGORENBCDFIJKLMQSUVWXZ
2 YTHAGORENBCDFIJKLMQSUVWXZP
3 THAGORENBCDFIJKLMQSUVWXZPY
. .....                              sequences
.
.
.

Line 1
J Z S S W B P D Z Z L F O M E K Q P D J H C K U M C
Converted
J X M L Q G S G L K R T S G S Y H N S V N K S X S D
Plain
M Y P O S I T I O N L A T I T U D E T W E N T Y T H

Line 2
A B C O O X M Y S I I G B S G G Y V D S W A J O Q E
Converted
A N N H T Q D S D G A S X R L K C G S Y H N Q N U N
Plain
R E E D A S H T H I R T Y L O N G I T U D E S E V E

The primary cipher alphabet for this problem is

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

These sequences are constructed from the words
quadricular and pythagorean, both names for the square
table used for Viggy and other encipherments.

15-2.  Naval Text.

A U V Z I S Z F B F Y E I R B I O W A O Y J L B L D
A T T W E N T Y T W O T W E N T Y F I V E O P E N E

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

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

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

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

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

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

O S V B W J B L V X Z P Z R R U Y I W H H P L P F T
O U R T Y S E V E N O P E N E D F I R E O N U P S H

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

H J T W I J T T Q W E E Q F O I I Z P M B J Q P Y M
H I R T E E N M I N U T E S A T S I X T H O U S A N

D U Q W A T Z O W D C L Z Q M P U K.
D T O T W O T H O U S A N D Y A E T

LECTURE 16 PROBLEMS

1. Complete columnar transposition.

WKAII  GLFGA  TEYHN  ONSOH  LGIRI  IAAIR  LGAMO
IMHSF  IDFGW  NNEYH  NEFNH  SLNSE  THS.  (63)

2. Nihilist transposition.

UCTEO  UAMAA  LTDMI  SUDDS  SISNU  OLNNH  AALTA
EYELB  NEANU  NRAPH  SNENX  ESTAE  ASJH.

3. Incomplete columnar

IENOR  RENHR  NAITI  ETTEC  FCOIP  TREYA  RCHTH
SPOAL  YONCW  SNARL  TEESN  TOYEL  ERSOL  UAIOE
VEPOR LNRTS  HIMIM E. (relatively)

4. Myszkowski.  Battlefield.

YIITU  HSATS  OIRLF  TSTFD  NCUAW  WGSUS  NYATO
EBEHR  GIPNP  OUSOM  ELEPO  YOONR  AYOIO  URTES

5. Amsco.

HENTI  DAHOS  CLOSN  PRNSA  FENTT  TIOAM  LROTE
RTLEI  ANCSC  RCISO  EMGRI  YOUIT  EMTAC  AIAME
ILIVI SPAEW  AMIFA. (propaganda)

6. Tramp.

CGHES  NOONE  NAETT  SHTIA  NEQCB  AWRSI  LTAOH
OAUEY  OCENA  TOMRT  HAEFO  ROEAU  PLNSD  STHIG.
(QCTCLYAPMQQ)

IRHRC  GRETR  ESDEE  OFOWN  ETLNS  EOTIG  IMNEI
TSONH  LTIID  DVLTS  NIADS  LSRAM  TSORU  HSCNE
DNIHU  EAGCD  IGIRS  WSLSH  BITNI  IHNNH  DNICD

8. Railfence.

TOEYC  SOEFO  MSAHH  RMOYU  LDTAC  LATYA  LFLME
EBGOP  VIPRV  IEEVS  ALUDO  WTGIG  THILL  CONT.

9. Redefence. Astronomic improbability.
tip = THE MOON TO

REOEN  IOFGS  AITWE  UMTBA  PITNP  ACOUH  OTICN
SAGFP  TRLEE  HTREN  MROOH  LEORN  SIVSE  ONTAC
SRSEL TUERS  HDTRO  AGYAH  TRAON  LE.

10. Turning grille.

TIP = the most serious and; NQEJPGUU

STTAH  IRNED  GSERL  GEOGM  AETON  ENBIE  DOTNH
EAEOS  MSTFI  LSOCI  OEHST  SNIER  CNTEN  SHTEC
SOIOS  LHOAU  SUSIS  EANWA  TMNER  BOECD  OSKRC
MSILT  EONMB  TLAEA  CTNID  DIEKD  OFNMF  AXVEF
ESEU.

11. Swagman. Agreeable toil.

NNWTI  HYORS  TEKKR  IENII  VNLSN  LOTOO  SLAVT
RETSI  ROSIM  KSCFR  SEEAO  OMTAC  HETTI  IWEVO
RHEII N.

REFERENCES AND RESOURCES

(I will append to a future lecture.)

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