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Cryptology
Lecture Three

Dr. Richard Spillman
Pacific Lutheran University
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Last Lecture

    History

    Introduction to Polyalphabetic Ciphers
– Cylinder Ciphers
– Vigenere Cipher
– AutoKey Cipher
– Rotor Cipher
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Review – Vigenere
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Cipher
   Use CAP to break this Vigenere cipher:

mcgaw   pssep   ycuma   kmstm   zfdpf   bqdle   hfltf   fgxsg   guvty
msrfr   asdrr   lspia   hbgua   mcudc   owfeq   lcwhy   mzrne   dsbsy
grkey   omfok   iiwar   bcqcy   gphuq   xrzey   ksdca   ngwok   xrwor
asldc   thkar   vfbpr   tbdlw   lwvnc   xrvsr   kcqgc   kqrmn   nhlne
ksvos   kqhsr   aoqtf   hghek   izryc   wpbtf   xquyn   mcxsc   kpxtr
awviq   tfhcc   ghshc   gcpel   hbiop   gsdrj   roolr   asool   zvlsr
hfbod   vfbpr   hzrgw   iosep   tbgpc   gqllu   tgwhc   hboyr   xqknm
ecbat   twoaz   es

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Review – Nihilist Cipher

   This is a two step cipher polyalphabetic cipher

– first, letters are represented by numbers using a
Polybius Square

– second, like the Vigenere, the key word (in number
form) is written above the plaintext (also in number
form) and the ciphertext is the sum of each
plaintext character and its keyword character

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Review - Example
keyword)
1   2    3   4   5         Select another keyword: NEXT
1   e   x    a   m   p
35 11 12 45
2   l   b    c   d   f
3   g   h    i   k   n          Plaintext: STOP THAT
4   o   q    r   s   t                     44 45 41 15 45 32 13 45
5   u   v    w   y   z
35 11 12 45 35 11 12 45
44 45 41 15 45 32 13 45
Ciphertext:   79 56 53 60 80 43 25 90
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Outline
   History

   Polygraphic Ciphers

   Introduction to Transpositions

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History

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World War I

   August 5, 1914 - the first day of WWI

   At dawn, the British cable ship Telconia, located in
the North Sea drops grappling hooks overboard
and pulls up Germany’s transatlantic cables. They
are severed and then dropped back into the sea
– It is England’s first offensive action of the war

the outcome of WWI?
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British Exploitation

    England began to exploit this advantage
almost immediately
– On the first day of WWI, England formed a new
black chamber

    By mid November, this black chamber
moved to room 40 of the old Admiralty
building so it became known as room 40

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Early Failure
   One of the first messages received by room 40,
had they been able to solve it, might have
affected the entire course of world history
– it was message 51 sent by the German naval high command to the
commander of the German navy in the Mediterranean on August
4th at 1:35 am
– the message read (it was finally decoded after it was to late)
“Alliance with Turkey concluded August 3. Proceed at once to
Constantinople.”

   If this message had been decoded in time, what
would England have known and how might
they have responded?

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Result
   England missed its chance to destroy the German
navy

   Germany was able to firm up its relationship with
Turkey

   Turkey sealed off Russia from her allies

   Russia eventually capitulated and fell into civil war
How might the history of the world be different if message 51
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Lesson

 Knowledge   must be:
Accurate
Complete
Timely
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A Lucky Break

   Early in September 1914, only one month after the
start of the war, the German light cruiser
Magdeburg was wrecked in the Baltic
– the body of a drowned German officer was picked up by the
Russians a few hours later
– clasped in his arms were the cipher and signal books of the
German Navy
– In October, Russia turned them over to the British (unusual)

   PROBLEM: none of the 4 letter code words in the
book appeared in any German message
– after 3 weeks, ROOM 40 discovered that the code had been
superenciphered, that is the code words themselves had been
ciphered using a keyword substitution

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The Growth of Room 40

   During December 1914, ROOM 40 began
to have trouble reading German messages
– After an all-night effort they discovered the new key:
a b c d e f g h i j k l m n o p q r s t u v w x y z
I L D S A M X Z O   B N C V U G   T W F Y P   R E H

Notice that vowels represented vowels and consonants consonants so the code
words were still pronounceable

The Germans began to change their key every 3 months, but by then ROOM 40
was so proficient that the new key was found by 2 or 3 am and nearly always by
9 or 10 am

Overall, during WWI, it is estimated that ROOM 40 intercepted and solved
15,000 German secret communications                                        14
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Polygraphic Ciphers

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Definitions

    Review of cipher types
– Monoalphabetic: each plaintext character is
assigned to one ciphertext character
– Polyalphabetic: each plaintext character is assigned
to more than one ciphertext character

    A polygraphic cipher works on more than one
plaintext character at a time
– Groups of plaintext characters are replaced by
assigned groups of ciphertext characters

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Digraphic Ciphers

    For example, a digraphic cipher looks at pairs
of characters in the plaintext Note a is
– “at” may be replaced by “ui” assigned 2 values
– “ai” may be replaced by “nj”

    This is a strong cipher because there are more
digraphs than single characters
– In English there are 26 letters but 625 (25 x 25)
possible digraphs
– It makes simple frequency analysis useless

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Playfair

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Playfair Cipher

    Playfair is a polygraphic method designed to
invalidate single frequency analysis
– the key is a 5 x 5 matrix of letters (j is not used)
arranged by a keyword
– For example, if the keyword is harpsicord, the
matrix key is:

H   A   R   P   S
I   C   O   D   B
NOTE:   the second R in harpsicord is
E   F   G   K   L
not used
M   N   Q   T   U
V   W   X   Y   Z

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Playfair Rules
    All text is preprocessed
– j’s are replaced with i’s
– a null letter such as q is placed between any identical pairs of letters so
that “tt” becomes “tqt”
– if the plaintext has an odd number of letters, the null letter is added on
to the end

    The ciphertext is produced by looking at each pair of letters in the
plaintext (m1 and m2)
– if m1 and m2 are in the same row in the key matrix, then they are
replaced by the characters to their right
– if m1 and m2 are in the same column then they are replaced by the
letters below them
– if m1 and m2 are in different rows and columns then they are replaced
by the letters found on the other two corners of a rectangle formed by
m1 and m2

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Rule One
   Using the keyword array formed from
“software”
– the plaintext letter pair: “RE” becomes “EB”
– Note: the rows circle back on
themselves so the letter pair
“LQ” becomes “ML”               S   O   F   T   W
A   R   E   B   C
D   G   H   I   K
L   M   N   P   Q
U   V   X   Y   Z

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Rule Two
   Again using the keyword array formed from
“software”
– the plaintext letter pair “AL” becomes “DU”
– since the columns circle back
on themselves, the letter
pair “TY” becomes “BT”         S   O   F   T
T   W
A   R   E   B   C
D   G   H   I   K
L   M   N   P   Q
U   V   X   Y   Z

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Rule Three
   Using the keyword array formed from
“software”
– the letter pair “OP” forms two corners of an
imaginary rectangle
– so it is replaced by the
characters on the other
O       T
corners “TM”                     S
A   R
F
E   B
W
C
D   G   H   I   K
L   M   N   P   Q
U   V   X   Y   Z

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Observations
   Playfair was the first cipher in history to be digraphic - to encipher two
letters at a time so that the result depends on letter pairs
– As a result, Playfair obliterates single letter characteristics
   In addition, encipherment by digraphs halves the number of elements
available for frequency analysis (a 100 letter text has only 50 digraphs)

   The number of possible digraphs is far greater than the number of
single letters so their linguistic characteristics are spread over many
more elements

   Little is known about its use but Britain’s War Office apparently used it
during the Boer War

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Playfair in CAP

 CAP       will implement playfair:

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Identifying Playfair
   There are several characteristics of Playfair
ciphertext that can be useful in identifying it as a
Playfair system
– Since it is a substitution system, the rare consonants j, k,
q, x, and z will appear in higher frequencies than
plaintext and digraphs containing these consonants will
appear more frequently
– There are an even number of letters in the ciphertext
– When the ciphertext is broken up into digrams, doubled
letters such as SS, EE, MM, . . . will not appear.

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CAP Feature
   CAP will help in the process of identifying a Playfair (or other polygraphic)
system

Select
PolyID

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Playfair Features 1
   There are several features of Playfair ciphertext that
may be exploited to aid in breaking the cipher
– Digraphs will be repeated and their recurrences will generally follow
the normal frequencies for their plaintext equivalents
– No single letter in the plaintext can ever be represented by itself, i.e.
“a” can never be represented by “a”
– Two reversed digraphs in the plaintext such as ER and RE will
always be enciphered by two reversed digraphs in the ciphertext
– Each single letter in the plaintext can be enciphered by one of only 5
other letters – that immediately below it in the same column, and one
of the four other letters in the same row.
   As a result, the 3 or 4 letters having the highest frequency in the
ciphertext will usually be found in the same row as “e”

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Playfair Features 2

   Other features include:
– It is twice as probable that the two letters of any
pair are at the corners of a rectangle, than as in
the same row or column
– Any given letter cannot represent a letter that it
combines with diagonally
– When a cipher letter has been identified as a
substitute for a plaintext letter there is a 20%
chance that it represents the same plaintext
letter in each other appearance.
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Reconstructing a
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Playfair Key
   There are several features that may prove helpful in
reconstructing the Playfair square from partial
knowledge of the substitution patterns:
– An equivalence such as DE = EF implies the
arrangement of DEF in the same row or the same column
– Two equivalences such as BF=ND and BM=NI indicate
that B and N are in the same row
– Two equivalences such as BY=RU and BT=NU indicate
that B and U are in the same column
– Two equivalences such as UC=PY and UH=YD indicate
that U and Y are in the same row or column and H and D
are in the same row or column
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Breaking Playfair –
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Initial Steps
frequency analysis because:
– High frequency ciphertext letters tend to be in
the upper rows of the Playfair square
– Low frequency ciphertext letters tend to be in
the lower rows of the Playfair square
– The digraph pattern follows plaintext diagrams
   If TH = HM then HM will be a high frequency
digraph

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Probable Word Method

 Given         the Playfair ciphertext below:
pk km km ew dw qn bs hl uf gq zk zp tl fc ls fq tn ca
zw ae ns fq tn zw ps el kz kc xc rb ke tm wg co ab fk
vn cl uf ui df ch hq kc mp

Assume that it is known that the plaintext phase
“a sample of” is in the ciphertext

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Attack Process
   The general process of a known plaintext
attack on Playfair involves the following steps:
failed
Find a            Construct          Merge
Pattern Match      Possible Squares     Squares

failed    Construct
Valid Square
succeed

DONE                   33
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Step One
   After a single and digraph frequency analysis of the
ciphertext, run the known plaintext across the
ciphertext to find possible alignments
– A possible alignment is one that does not violate any of
the known characteristics of a Playfair square

Known: te st a a
te st
t es ta
Ciphertext: ah ty cv vg ui no pq

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Example
    For our example try the following:

as   am   pl     eo   f-
pk   km   km     ew   dw qn bs hl uf gq zk zp tl fc ls fq tn ca
zw   ae   ns     fq   tn zw ps el kz kc xc rb ke tm wg co ab fk
vn   cl   uf     ui   df ch hq kc mp

Will not work because
1. km can not map to two different pairs
2. Neither m or e can map to themselves

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Example (continued)
   After two slides, the matchings are:

as   am   pl   eo   f-
pk km   km   ew   dw   qn bs hl uf gq zk zp tl fc ls fq tn ca
zw ae   ns   fq   tn   zw ps el kz kc xc rb ke tm wg co ab fk
vn cl   uf   ui   df   ch hq kc mp

Will not work because
1. km can not map to two different pairs
2. m cannot map to itself

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Example (continued)
   After another 2 characters:

as         am   pl   eo   f-
pk km km         ew   dw   qn   bs hl uf gq zk zp tl fc ls fq tn ca
zw ae ns         fq   tn   zw   ps el kz kc xc rb ke tm wg co ab fk
vn cl uf         ui   df   ch   hq kc mp

This is a possible match since it does not violate any Playfair rules
RESULT: 4 possible pairings as             –   km
am             –   ew
pl             –   dw
eo             -   qn
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Step Two – Possible
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Squares
   Each of the possible pairings represents a state
of the Playfair square
1                   2                    3                   4
as = KM             am = EW             pl = DW             eo = QN
as    –   km     A   K   S   M       A   E   M   W       P   D   L   W       E   Q   O   N

am    –   ew     K
S
E
M
D                   Q
L                   O
pl    –   dw     M           A   K   W           A   E   W           P   D   N           E    Q

eo    -   qn                 M   S               W   M               W   L               N    O

Ultimately this will not work out, so go back to Step 1

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Step One (again)
   Return to step 1 and slide the known text to the
right by 2 characters:
as      am   pl   eo   f-
pk km km ew      dw   qn   bs   hl uf gq zk zp tl fc ls fq tn ca
zw ae ns fq      tn   zw   ps   el kz kc xc rb ke tm wg co ab fk
vn cl uf ui      df   ch   hq   kc mp

This is also a possible match
RESULT: 4 possible pairings           as   –   ew
am   –   dw
pl   –   qn
eo   -   bs
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Step Two (again)
   Each of the possible pairings represents a state of the Playfair square
1                    2                      3                      4
as    –    ew         as = EW              am = DW               pl = QN                eo = BS
am    –    dw         A   E   S   W        A   D   M   W        P   Q   L   N       E    B   O   S
E                    D
pl    –    gn         S                    M
Q
L
B
O
eo    -    bs         W           A    E   W           A   D    N           P   Q   S            E   B
W    S               M   W                N   L                S   O

What we can deduce:
A and E must be in the same column or row                  A and W must be in the same column or row
W and S must be in the same column or row                  E and S must be in the same column or row
A and D must be in the same column or row                  W and M must be in the same column or row
D and M must be in the same column or row
P and N must be in the same column or row                  P and Q must be in the same column or row
L and Q must be in the same column or row                  L and N must be in the same column or row
E and B must be in the same column or row                  E and S must be in the same column or row
O and S must be in the same column or row                  O and B must be in the same column or row
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Step Three – Reasoning 1
   From the prior slide we know the following:
– A,B,E,D must be in the same row or column
– Since the alphabet is entered in order after the key word it is
likely that the letters A,B,C,D,E are not part of the key word so
they form the second line of the square
   This implies that the key word is 5 letters long

A & W must
be in the same   W
A   B   C   D   E   row or column    A   B   C   D   E

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Step Three – Reasoning 2

    Continuing to apply the required conditions:

W                                    W                               S                        W            M   S
E & S must                                               D & M must
A   B   C   D   E                    A       B       C       D       E                        A   B   C    D   E
be in the same                                           be in the same
row or column                                            row or column

B & O must
Any guess on the key?                                                       be in the same
row or column
Since this satisfies                         W       O       R       M       S
W   O        M   S
A       B       C       D       E
all the other conditions                     F       G       H       I       K
A   B   C    D   E

as well, it is a good                        L       N       P       Q       T

candidate for the key square:                U       V       X       Y       Z
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Using CAP 1

    CAP provides some
basic tools for breaking
playfair
– Including a known word
search

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Using CAP 2

    CAP also provides a
simple Playfair
worksheet

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Hill Cipher

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Hill Cipher

    Another polygraphic cipher was
developed by the mathematician Lester
Hill in 1929 - called the Hill Cipher

    It takes m successive plaintext letters and
substitutes for them m ciphertext letters

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Hill Process

   Each character is assigned a numerical value
– a = 0, b = 1, . . ., z = 25

   for m = 3 the transformation of p1p2p3 to c1c2c3
is given by 3 equations:

c1 = (k11p1 + k12p2 + k13p3) mod 26

KEY   c2 = (k21p1 + k22p2 + k23p3) mod 26

c3 = (k31p1 + k32p2 + k33p3) mod 26

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Hill Matrix
   The Hill cipher is really a matrix multiplication
system
– The enciphering key is an n x n matrix, M
– The deciphering key is M-1
   For example, if n = 3 one possible key is:
17   17     5                            4    9   15
M=          (   21
2
18
2
21
19
)           M-1 =   (   15
24
17
0
6
17
)
Encrypt ‘n o w’
13 14 22   (17 17 5
21 18 21
2 2 19
13
)( ) = ( ) mod 26
14
22
23
20
4

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CAP Implementation
    CAP will create a Hill cipher for groups of 2 to 6
characters         CAP will generate
a matrix for you or
First, enter the
group size – which
is the same as the
matrix size

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Breaking Hill

    The Hill cipher is highly resistant to a ciphertext only
attack.
– In fact, the larger the matrix, the more resistant the cipher
becomes.

    However, it is easy to break using a known plaintext
attack.
– The process is much like the method used to break an affine
cipher in that the known plaintext/ciphertext group is used to set
up a system of equations which when solved will reveal the key.

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Theory
   We know the ciphertext matrix was created by
multiplying the plaintext matrix by the key
matrix or:
C = MK
   So, the key matrix can be recovered if we know
the inverse of the plaintext matrix:
Process: Find the inverse of
K=     M-1C            the known plaintext, multiply
it times a segment of ciphertext,
and check the resulting key
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Example

   Eve intercepted the following ciphertext:

209    37 59       255   110   196   247    90 52     34     1    68
11   130 43        23    20    26   219   119 93    164    12    63
110   202 124      137   112   158   232    23 127   118   128   123
115    89 62       224   199    10   199   142 104   242   120     4
142    26 230      159   129   164   133   153 31    256   210    62

She suspects that the ciphertext contains the words:
“she can not attack us”

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Plaintext Matrix
   Eve translates the known plaintext into a matrix (using
ASCII representation for the characters which assigns a
number between 0 and 255)

She calculates the inverse:

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Search
   Eve proceeds to multiply this inverse matrix times
matrices formed from the ciphertext.
– Each result is a possible key which she uses to decipher the
ciphertext.
– Most of the time the possible key does not work but the one
time the possible message matched the ciphertext, Eve
discovered a working key:

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Using CAP
    Breaking a Hill cipher by hand is very time consuming
however CAP will automate the process
Enter a possible                       Enter your best
plaintext segment                      guess for the size

Try the possible
keys until one
works

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Transposition

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Cipher Structure
...

Classical             Stream       Block

Transposition             Substitution

polyalphabetic                             monoalphabetic

Nihilist     Vigenere       Hill       MultiLiteral      Keyword       Affine   Shift

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Transpositions
   Transposition ciphers are like jigsaw puzzles in
that all the pieces are present but are merely
disarranged

   There are two types of transpositions
– monographic methods - deal with individual letters
– polygraphic methods - deal with units greater than single letters (words,
phases, etc)

   Most transpositions involve a geometric figure
(square, rectangle, . . .)
– the letters are inscribed in the figure is some agreed upon direction
– the letters are then transcribed or rewritten according to another
direction to form the ciphertext
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Transposition Key

   The specific key controls
– the geometric figure and its dimension
– the variation in the direction of inscription and transcription

   If the transposition cipher involves only a single
process of inscription followed by a single process
of transcription it is called a monophase
transposition, if additional methods of inscription
and transcription intervene between the original
inscription and the final transcription it is called
a polyphase transposition
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Rail Fence
 Process:   rather than substitute
letters, rearrange letters in the text

 RAIL-FENCE cipher
– PLAINTEXT: this is a test
– CIPHERTEXT: tiehsstsiat
– METHOD
Inscribe by a     t               i               e
zigzag pattern
h       s       s       t       s
extract by rows           i               a               t
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Other Figures

 Use   a triangle of a fixed size:
PlainText: You must do that now

Y
Inscribe by rows         o u m
u s t d o
t h a t n o w

Extract by columns     tuhosayuttmdnoow

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Permutation Cipher

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Permutation Method
   Break the plaintext up into groups of a fixed size, d
– define a permutation of the integers 1 to d called f
– within each block, permute the letters according to f
– the key is (d,f)

   For example, let d = 5 and let f be given by:
1        3
g e t t h        e b a l l
2        4
3        1        t h g e t        al e b l
4        5
5        2                                      63
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CAP Implementation

 Select     Permutation under the cipher

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Breaking a Permutation
   A known word attack can match the word with
the ciphertext to discover the permutation

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Column Transposition

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Column Transposition
   Write the plaintext into a matrix by rows, then
generate the ciphertext by selecting the columns in
a given order

   For example, given the plaintext “encryption
algorithms” and a 5x4 matrix         KEY

1 2 3 4
E   N
N   C   R    Select the columns in the order 4 - 2 - 1 - 3
Y   P
P   T   I
O   N
N   A   L
R   I   L   I   S   N   P   N   O   H   E   Y   O   G   T   C   T   A   R    M
G   O
O   R   I
T   H
H   M   S

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Column Key

 The key for a column transposition
is the number of columns and the
order in which they are selected
Both can be specified by a keyword:

the length of the keyword is the number of columns

the order of the letters in the keyword determines
the order in which the columns are selected

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Example Key

   For example, the keyword “general” defines a
column transposition with 7 columns

   To define the order of columns assign each
letter a number based on its order in the
alphabet
g e n e r a l
4 2 6 3 7 1 5
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Transpositions in CAP

   When you select
Column Transposition,
CAP presents a key
window:

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Transpositions
   We will look at the process of breaking a keyed
columnar transposition with completely filled
rectangles

   Given the following cipher, what is the first thing
you must determine?
NETEF   LTDSR   TSSTF   MDCET DRHXS WHOHO EEADU OUUFI RRRRS
NEROT   CFIEM   EDSHA   RTCPJ AOEGE WNLHO EPMWA WERUV AAINA
TSDDS   OEOAC   EHNTL   HFLAU RAEEN OTOTS SOSYS TNNCG EMETT
YDYRR   NEOOE   RESTH   INR

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involved in breaking a
column transposition cipher:
– Find possible rectangle sizes
– Select the correct rectangle
– Find the column order

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Possible Rectangles

 Thefirst step in analyzing a column
transposition is determining the
number of columns
– in a completely filled transposition, the number of
characters is the product of the number of rows
and the number of columns
– so, factor the number of characters to determine
possible row and column sizes

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Example
   First, factor the message length of the example
ciphertext

   our message has 153 letters and 153 has 3, 9, 17, and
51 as factors

   possible rectangle sizes (columns by row): 3 x 51, 51 x
3, 9 x 17, or 17 x 9

   9 x 17 and 17 x 9 are the most probable because the
other two have poor distributions of rows vs columns
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Which Rectangle?

   Since the factors only supply possible
column sizes - test each possibility by
doing a vowel count
– any line of plaintext should contain about 40%
vowels
– So, count the vowels in each row of each possible
rectangle
– the one with the best match to 40% is the best
choice for the actual rectangle

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Rectangle 1
   The 9 x 17 rectangle - 3.6 vowels per row expected (9 x .4)
1 2 3 4        5   6   7   8   9     vowels
difference
N C U F        G   A   N   S   E      2       1.6
E E O I        E   A   T   S   Y      7       3.4
T T U E        W   I   L   O   R      4        .4
E D U M        N   N   H   S   R      2       1.6
F R F E        L   A   F   Y   N      3        .6
L H I D        H   T   L   S   E      2       1.6
T X R S        O   S   A   T   O      3        .6
D S R H        E   D   U   N   O      3        .6   Total Difference
S W R A        P   D   R   N   E      2       1.6         20.6
R H R R        M   S   A   C   R      1       2.6
T O S T        W   O   E   G   E      4        .4
S H N C        A   E   E   E   S      4        .4
S O E P        W   O   N   M   T      3        .6
T E R J        E   A   O   E   H      5       1.4
F E O A        R   C   T   T   I      4        .4
M A T O        U   E   O   T   N      5       1.4
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D D C E        V   H   T   Y   R      2       1.6
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Rectangle 2
   The second possible rectangle is 17 x 9 (expected per
row vowel count = 6.8)
1   2   3   4   5   6   7   8   9 10 11 12 13 14 15 16 17   Vowels   Difference
N   R   E   O   U   N   M   P   L E T C A O N Y E             8         1.2
E   T   T   H   U   E   E   J   H R S E U T N D R             6          .8
T   S   D   O   F   R   D   A   O U D H R S C Y E             6          .8
E   S   R   E   I   O   S   O   E V D N A S G R S             7          .2
F   T   H   E   R   T   H   E   P A S T E O E R T             6          .8
L   F   X   A   R   C   A   G   M A O L E S M N H             5         1.8
T   M   S   D   R   F   R   E   W I E H N Y E E I             7          .2
D   D   W   U   R   I   T   W   A N O F O S T O N             6          .8
S   C   H   O   S   E   C   N   W A A L T T T O R             5         1.8
8.4

Which of the two is the most likely rectangle?
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Letter Affinity
   Once the rectangle size is determined, the column
order must be discovered
– Advantage is taken of all the characteristics of the plaintext
language

   First, in all languages there are certain letters
usually of medium or low frequency which
combine with other letters to form diagrams of
high frequency
– H (medium frequency) combines with T to form TH (highest
frequency)
– H combines with C (medium frequency) to form CH
– V (low frequency) combines with E to form VE (medium frequency
in military text)

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Rules

    When there is an H, attempts should be
made to combine it with a T or a C

    A V should be combined first with an E

    A K should be combined first with a C

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Pilot Letters
   Second, there is usually in every language at least
one letter which can be followed by only certain
other letters forming an obligatory sequence or
invariable digraph
– Q is always followed by U
– J can only be followed by a vowel
– X can be preceded only by a vowel and except at the end of a
word can only be succeeded by a vowel or C, H, P, T

   Letters such as these with limited affinity are
called pilot letters

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Anagramming
   Breaking a transposition cipher is a process of anagramming by
selecting a pilot letter and trying to form digrams with the other
letters in its row
for example, select the J in column 8 and match it with . . .

1   2     3   4   5    6   7    8   9 10 11 12 13 14 15 16 17
N   R     E   O   U    N   M    P   L E T C A O N Y E
E   T     T   H   U    E   E    J   H R S E U T N D R
T   S     D   O   F    R   D    A   O U D H R S C Y E
E   S     R   E   I    O   S    O   E V D N A S G R S
F   T     H   E   R    T   H    E   P A S T E O E R T
L   F     X   A   R    C   A    G   M A O L E S M N H
T   M     S   D   R    F   R    E   W I E H N Y E E I
D   D     W   U   R    I   T    W   A N O F O S T O N
S   C     H   O   S    E   C    N   W A A L T T T O R
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Centiban Weights

 TheUS government studied a set of
5,000 digrams and produced a table
of what are called centiban weights

 Thistable contains the logarithm of
twice the frequency of each digram

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Centiban Table
Second Letter
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   33   45   61   73   13   38   45   25   64   13   25   76   61   89   25   58   00   82   80   83   59   48   33   00   58   00
B   38   00   00   00   66   00   00   00   25   13   00   45   13   00   38   00   00   25   13   13   25   00   00   00   48   00
F   C   67   00   33   13   76   13   00   61   48   00   38   42   13   13   80   00   00   38   13   61   38   00   13   00   13   00
D   76   38   38   51   77   51   25   25   73   13   00   33   42   38   63   42   25   58   59   62   42   33   38   00   13   00
i   E   78   38   76   88   81   66   38   48   73   13   00   74   61   99   58   67   58   94   86   79   33   67   48   48   38   13
F   42   00   25   13   55   56   13   00   80   00   00   25   13   00   80   13   00   53   33   56   33   00   13   00   13   00
r   G   48   00   25   13   61   25   13   67   42   13   00   25   13   33   45   25   00   42   33   38   25   00   13   00   00   00
H   67   13   33   25   67   42   00   00   77   00   00   13   25   33   67   13   13   64   38   74   51   00   13   00   13   00
s   I   51   25   69   45   59   55   67   00   00   00   25   70   53   92   80   48   00   73   78   73   00   72   00   62   00   25
J   18   00   00   00   25   00   00   00   00   00   00   00   00   00   25   00   00   00   00   00   25   00   00   00   00   00
t   K   13   00   13   00   45   00   00   00   25   00   00   13   00   13   00   00   00   00   13   00   00   00   00   00   00   00
L   74   33   33   53   79   33   13   13   67   00   00   73   25   13   59   33   00   25   45   51   25   25   25   00   55   00
M   78   45   33   13   72   13   00   13   53   00   00   00   59   00   55   51   00   25   38   25   25   00   00   00   25   00
L   N   72   25   67   85   87   53   73   38   75   13   25   42   42   51   66   33   13   38   71   93   48   33   33   00   42   00
O   48   38   51   58   33   72   25   33   42   13   25   67   72   92   45   72   00   89   61   67   79   48   51   13   25   00
e   P   61   13   13   13   70   25   00   33   45   00   00   59   38   13   64   56   00   66   45   51   33   13   13   00   13   00
Q   00   00   00   00   00   00   00   00   00   00   00   00   13   00   00   00   00   13   00   00   62   00   00   00   00   00
t   R   80   25   53   64   96   45   48   33   75   13   13   42   53   48   74   59   00   56   75   81   42   42   38   00   53   00
S   71   33   59   42   84   58   25   72   77   00   13   25   33   38   62   55   00   42   67   88   56   13   38   00   13   00
t   T   74   33   45   45   91   48   13   92   82   00   00   42   45   48   84   25   13   64   67   67   42   00   78   00   80   13
e   U   42   33   33   33   56   13   51   00   42   00   00   45   42   68   13   25   00   75   58   58   00   13   00   00   00   00
V   45   00   00   00   87   00   00   00   58   00   00   00   00   00   13   00   00   00   00   13   00   00   00   00   00   00
r   W   58   00   00   00   69   00   00   38   59   00   00   13   00   25   67   00   00   13   13   00   00   00   00   00   13   00
X   25   00   25   13   13   13   00   13   25   00   00   00   00   13   13   25   00   13   13   48   00   00   00   00   00   00
Y   45   25   38   38   53   56   13   13   33   00   00   25   25   45   55   33   00   38   56   62   13   00   13   00   00   00
Z   13   00   00   00   25   00   00   00   13   00   00   00   00   00   00   00   00   00   00   00   00   00   00   00   00   00

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Possible Pairings
   Try both JU pairings in the example and rank each
digram by the centiban numbers
8   5 rank   8    13   rank
P   U 33     P     A    61
J   U 25     J     U    25
A   F 38     A     R    82
O   I 42     O     A    48
E   R 94     E     E    81 Result: pair columns 8 and 13
G   R 42     G     E    61
E   R 94     E     N    99
W   N 25     W     O    67
N   S 71     N     T    93
464              617
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Form Trigrams
   The digram JU should be followed by a consonant,
preferably N or S - columns 15 and 11 are
candidates
8 13 15 rank   8 13 11 rank
P A N     89   P A T     83
J U N     68   J U S     58
A R C     53   A R D     64
O A G     45   O A D     73
E E E     81   E E S     86
G E M     61   G E O     58
E N E     87   E N E     87
W 0 T     67   W O O     45
N T T     67   N T A     74
618            628
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Look for Words
 Continue this process looking for new digrams and for possible
words
 For example, are there any possible words in this text?
8 13 11        1   2   3   4   5   6   7   9 10 12 14 15 16 17
P A T          N   R   E   O   U   N   M   L E C O N Y E
J U S          E   T   T   H   U   E   E   H R E T N D R
A R D          T   S   D   O   F   R   D   O U H S C Y E
O A D          E   S   R   E   I   O   S   E V N S G R S
E E S          F   T   H   E   R   T   H   P A T O E R T
G E O          L   F   X   A   R   C   A   M A L S M N H
E N E          T   M   S   D   R   F   R   W I H Y E E I
W O O          D   D   W   U   R   I   T   A N F S T O N
N T A          S   C   H   O   S   E   C   W A L T T O R

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Solution

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

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Column Size in CAP
   CAP will determine both possible column sizes and run a
vowel analysis of each column:

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Anagramming in CAP
   CAP provides a tool to aid in the anagramming
process

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Summary
   History

   Polygraphic Ciphers
– Playfair
– Hill

   Introduction to Transpositions
– Permutation
– Column Transposition

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