Lesson 10:
Polyalphabetic Substitution Systems I
Viggy's Family And Quagmires I - IV
Applications Of The Principals Of Symmetry
CLASSICAL CRYPTOGRAPHY COURSE
BY LANAKI
April 6, 1996
Revision 0
COPYRIGHT 1996
ALL RIGHTS RESERVED
LECTURE 10
POLYALPHABETIC SUBSTITUTION SYSTEMS I
VIGGY'S FAMILY AND QUAGMIRES I - IV
APPLICATIONS OF THE PRINCIPALS OF SYMMETRY
SUMMARY
In Lecture 10, we return to our course schedule with a study of
fascinating cipher systems based on multiple alphabets
-Polyalphabetic Substitution systems. What is amazing about
these systems is how long they remained secure. The Viggy
systems (my name for Vigenere) was considered unbreakable for
over 200 years. Along comes Major Kasiski, and poof, we have
recreational cryptography.
I think the best way to introduce the subject is via an
overview based on the Op-20-GYT course notes (Office of Chief
Of Naval Operations, Washington) [OP20]. From there, I will
bring in MASTERTON's dissolution of QUAGMIRES I-IV. [MAST]
In Lecture 11, we will revisit polyalphabetic cipher systems
and the polygraphic cases using Friedman's detailed analysis.
We will cover the PORTA system and other family members. I
will cover decimation processes in detail. [FRE4], [FRE5],
FRE6], [FRE7], [FRE8]
In Lecture 12, we will describe the aperiodic polyalphabetic
case and give a diagram of topics considered in Lectures 10 -
12. [FR3]
I have updated our Resources Section with many references on
these systems - focusing on the cryptanalytic attack and those
of historical interest. Kahn has some interesting stories about
the Viggy family. [KAHN]
POLYALPHABETIC SUBSTITUTION
A cipher system which employs two or more cipher alphabets and
includes a method for designating which cipher alphabet is to
be used for the encipherment of each plain-text letter, is
called a polyalphabetic substitution system. Cipher systems
employing variant values may appear to use more than one
alphabet, but they have characteristics of mono-alphabetic
substitution and are properly classified as such.
Polyalphabetic substitution systems consists of two general
types; periodic and non-periodic.
(a) In the periodic type the text of a message is divided
into definite, regular groups or cycles of letters which are
enciphered with identical portions of the key. Periodic
systems are further subdivided as follows:
(1) Multiple Alphabet Ciphers in which any number of
cipher alphabets are used in order designated by a
prearranged key.
(2) Progressive Alphabet Ciphers in which a primary
cipher alphabet and its 25 secondary alphabets are
used either in regular succession, sliding the
components one letter at a time, or in irregular
order according to a prearranged shift.
(b) In the non-periodic type there are no cyclic repetitions
of the key.
The cipher alphabets employed in multiple alphabet substitution
systems may be constructed by any number of methods. As an
example, the QUAGMIRE IV uses both vertical and horizontal
keywords.
Example:
Plain A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
Cipher 1 R T U V W X Y Z P E N C I L S A B D F G H J K M O Q
" 2 E N C I L S A B D F G H J K M O Q R T U V W X Y Z P
" 3 D F G H J K M O Q R T U V W X Y Z P E N C I L S A B
Here the plain component is a normal sequence, and the cipher
component are identical keyword sequences. The same keyword
sequences may be used in both the plain cipher components, or
different sequences may be used. The key which determines the
setting of the cipher alphabets against the plain component
(RED) may be any prearranged word or phrase. Also, each cipher
alphabet may be assigned a number and the alphabets used in
accordance with a prearranged numerical key.
The process of enciphering a message with the multiple alphabet
system above would appear as follows:
Cipher Alphabet No.
1-2-3-1-2-3-1-2-3-1-2-3-1-2-3-1-2-3-1-2-3-1-2-3
Plain - M Y C O U R S E Z E R O T H R E E Z E R O A T T
Cipher - I Z G S V P F L B W R X G B P W L B W R X R U N
1-2-3-1-2-3-1-2-3-1-2-3-1-2-3-1-2-3
Plain - H I R T E E N T H I R T Y T H R E E
Cipher - Z D P G L J L U O P R N O U O D L J
In order to reduce the chances of encipherment by the wrong
alphabet, the plain text is often written so that the letters
designated by the key for encipherment by each alphabet are
placed in the same vertical column.
Note the repetitions in the plain text which begin at the same
point in the key produce repetitions in the cipher text, while
others [may not] do not. Friedman discusses accidental
repetitions in [FR7].
PRINCIPLES OF FACTORING
Major Friedrich W. Kasiski (1805-1881) was a career officer in
East Prussia's 33 Infantry Regiment. He is credited with a
revolutionary insight regarding polyalphabetic repeating key
systems - that the conjuction of a repeated portion of the key
with the repetition in the plaintext produces a repetition in
the ciphertext. Like causes produce like effects. The
interval between plaintext or ciphertext repetitions is noted
throughout the cryptogram, factored and the commonality of the
factor is a good indication of the key and number of alphabets
used to encipher the original methods. The fall of the
Vigenere family is attributed to Kasiski's examination. [KASI]
[KAS1], [KAHN]
If there are several long repetitions in the cipher text of an
unknown system, the intervals between the initial letters of
these repetition have a common factor, this factor represents
the number of alphabets used to encipher the message and the
exact number of repetitions of the key.
A simple example:
Given the cryptogram:
IZGSV PFLBW RXGBP WLBWR XRUNZ
DPGLJ LUOPR NOUOD LJ
Factoring:
Repetition Interval Factors Common Factor(s)
LBWRX 9 3,3 3
LJ 12 2,2,3 3
UO 6 2,3 3
The "period" or common factor is three and this is the number
of alphabets employed.
Digraph and trigraph repetitions may be the result of chance
instead of plain text repetitions. [FR7] discusses in detail.
When factoring results in more than one common factor we shall
use the highest common factor and check with frequencies of the
expected alphabets to see how close to normal they are. Only
short messages fail to lead to the correct determination of the
number of cipher alphabets employed in the system. When
factoring fails on a longer message, an aperiodic cipher may
have been employed.
SOLUTION OF A MULTIPLE ALPHABET CIPHER
Phamplet Number 7, Office of Operations Cryptanalysis, Office
of the Chief of Naval Operations, Washington, 1930 [OP20]
prepared this problem for discussion.
From: A B (Black Force Commander)
To: CD, EF, GH, IJ (Black Ships)
Time Groups: 0013-2300 April 1930
Remarks: Cruiser transmitter.
Cryptogram written out in worksheet format:
Alpha. - 1 2 3 4 5 6 7 8 9 10 Alpha. - 1 2 3 4 5 6 7 8 9 10
1 K P T X S L I C T M 16 M V H A W A D G G Z
2 I A M C B B N M S Z 17 Y F A R Q V K M M Q
3 M J K A Q J B F Z A 18 K F M P S L G X A H
4 J G M B S L N P H H 19 E F W K G C B F T H
5 E E J Z W N C L O W 20 S V C B B U A H S S
6 Z F S A A S Z D E P 21 K P K D E C G O H Z
7 Z X C D J D D H A J 22 L V O D S C O C H A
8 O D B K A H P L G H 23 G V W B Z C A M O Z
9 A J M K T V A M K H 24 M J K A Q J B F J H
10 M B C A A C N W S Z 25 X B H A A V A K O S
11 Z D W I J K G M C X 26 K P K G U L T J O Q
12 M V X X U N B W Z T 27 D F Q Q J K K M H Z
13 I Y N C P O G H H W 28 H V H A E P Z W Q R
14 L G T B W P L V T T 29 O P L A U L B M O Z
15 O B O X J L R M H Z 30 M J K A Q J B F
Collateral Information:
The Black and Blue Fleets are engaged in war maneuvers in the
Caribbean Sea. The Fleets are not in contact. The location of
the enemy (the Black Fleet) is unknown. The message in
question was intercepted by the Blue Flagship at 0015 on 14
April 1930. The operator had reason to believe that a cruiser
sent the message.
The composition of the Black Fleet is as follows:
Battleships Cruisers
West Virginia (flag) Trenton (flag)
Maryland Marblehead
Tennessee Richmond
New Mexico Memphis
Mississippi
California
Destroyers Air Force
Litchfield (flag) Saratoga (flag)
Preble Langley
Pruitt Gannet
Noa
Decatur Submarine Force
Sicard
Hulbert Argonne (flag and tender)
V-1, V-2, V-3
William B. Preston
Factoring:
Repetition Interval Factors
ZMJKAQJBF 210 2,3,5,7,10
ZMJKAQJBF 270 2,3,3,5,10
ZMJKAQJBF 60 2,2,3,5,10
MHZMVHA 120 2,2,2,3,5,10
ZMV 40 2,2,2,5,10
ZMV 160 2,2,2,2,2,5,10
KPK 50 2,5,5,10
The highest common factor is 10; the period and number of
alphabets used is 10, so the sequence repeats itself after
each 10 letters.
"Lining-up" is one of the basic operations of solution.
We group the message in lines of ten letters. The letters in
each column are enciphered by the same alphabet. Checking the
frequency tables, each alphabet resembles a single alphabet.
Frequency Tables
#1 #2 #3 #4 #5 #6 #7 #8 #9 #10
A 1 A 1 A 1 A 9 A 4 A 1 A 4 A A 2 A 2
B B 3 B 1 B 4 B 2 B 1 B 6 B B B
C C C 3 C 2 C C 5 C 1 C 2 C 1 C
D 1 D 2 D D 3 D D 1 D 2 D 1 D D
E 2 E 1 E E E 2 E E E E 1 E
F F 5 F F F F F F 4 F F
G 1 G 2 G G 1 G 1 G G 4 G 1 G 2 G
H H H 3 H H H 1 H H 3 H 6 H 6
I 2 I I I 1 I I I 1 I I I
J 1 J 4 J 1 J J 4 J 3 J J 1 J 1 J 1
K 4 K K 5 K 1 K K 2 K 2 K 1 K 1 K
L 2 L L 1 L 1 L L 6 L 1 L 2 L L
M 7 M M 4 M M M M M 8 M 1 M 1
N N N 1 N N N 2 N 3 N N N
O 3 O O 2 O O O 1 O 1 O 1 O 5 O
P P 4 P P 1 P 1 P 2 P 1 P 1 P P 1
Q Q Q 1 Q 1 Q 4 Q Q Q Q 1 Q 2
R R R R 1 R R R 1 R R R 1
S 1 S S 1 S S 4 S 1 S S S 3 S 2
T T T 2 T T 1 T T 1 T T 3 T 2
U U U U U 3 U 1 U U U U
V V 6 V V V V 3 V V 1 V V
W W W 3 W W 3 W W W 3 W W 2
X 1 X 1 X 1 X 3 X X X X 1 X X
Y 1 Y 1 Y Y Y Y Y Y Y Y
Z 3 Z Z Z 1 Z 1 Z Z 2 Z Z 2 Z 9
30 30 30 30 30 30 30 30 29 29
SOLUTION BY KNOWN-WORD METHOD
When ample collateral information is available, the known-word
attack is the easiest and potentially the quickest method of
solution. From the given data, the message is presumably from
the Commander of a cruiser division to his four cruisers,
giving orders for scouting operations of the cruiser division.
The words most likely to appear are:
Scouting Scouting line Trenton Latitude
Course Scouting course Marblehead Longitude
Speed Scouting speed Richmond Hundred
Distance Scouting distance Memphis Numbers
Position Commence scouting Enemy Times/Dates
Our concern is not with guessing words but standardizing the
solution.
The Known-Word" method applied in two ways:
(1) Start at a particular point in the cryptogram indicated by
the repetitions, symmetrical sequences, and try to fit the
known-word at this point. This is called the "Obvious
Location Method."
(2) Start with a "Known-word" and find a place where it will
fit. This may be called the "Obvious Word Method."
The best method to use depends on the circumstances. In this
problem both methods apply.
OBVIOUS LOCATION
The long repetitions are words or phrases, important to the
subject of the message, and may be known-words. They are
excellent points of attack. The beginning of the message or
the end of the message are usually good points of attack.
The second longest repetition is the right length for Trenton,
Memphis, or Hundred; furthermore it links in the letters of the
longest repetition.
Original Assumptions -
MHZ MVHA lines 15-27 TRENTON is best assumption.
TRE NTON
MEM PHIS
HUN DRED
Check
MOZ MJKAQJBF lines 24, 30 MOZ MJKAQJBF could be
T E N N Excellent TEE NHUNDRED excellent
M M P S Poor THE E--N --- poor
H N D D Poor
Check
MCZ MVX lines 1-12
TWE NTY excellent
M M PH poor
H V DP poor
Check the values of TEEN HUNDRED and TRENTON
Line 2-3 12345678910 12345678910
IAMCBBNMSZ MJKAQJBFZA
T E NHUNDRED
suggests ATTE NHUNDRED
Line 23-24 GVWBZCAMOZ MJKAQJBFDI
T TEE NHUNDRED
suggests THIR
FOUR
FIF
SIX
ATSEVEN
EIGH
Lines 29-30 OPLAULBMOZ MJKAQJBF--
N ETEE NHUNDRED
suggests NINETEE NHUNDRED
It is possible that all the above assumptions are incorrect but
they are too good to ignore. We enter the above values into
the cryptogram to see if skeletons of words appear.
Possibilities are indicated below:
Lines 19-20 12345678910 12345678910
EFWKGCBFTH SVCBBUAHSS
ED T T
SPEEDFI FTEENKNOTS
SI X
Line 19 ED could be Speed.. building on that we have other
possibilities.
Lines 21-22 KPKDECGOHZ LVODSCOCHA
U RE T R
COURSETHRE ETHREEZERO
Lines 11-12 ZEWIJKGMCZ MVXXUNBWZT
T E NT E
TWE NTYMILES
T THREE
FIVE
TRENTON is the most obvious break. Check letter-combinations
of frequencies to see which of the three chosen words fitted
best.
HZ =1 ZMV=1 ZM =4 HA=1
RE ENT EN ON Trenton is only assumption
EM MPH MP IS
UN NDR ND ED
Frequency 869 7639
Cipher MHZ MVHA
Frequency XXX XXXX X = high frequency
Plain TRE NTON
- = intermediate frequency
Frequency -X- --XX
Plain MEM PHIS O + low frequency
Frequency --X -XX-
Plain HUN DRED
OBVIOUS WORD METHOD - LOCATION BY FREQUENCIES
One method of fixing the location of an obvious word is by
frequencies, provided the obvious word has one or more letters
of very low frequency. The word should be 10 or more letters
to be practical.
The possibilities are RENDEZVOUS and MARBLEHEAD.
First, frequencies are written over each letter of the
cryptogram. The Known-word is put on a card and slid over the
cryptogram until it fits with the very low frequency letters
and neighbors. This method is rather tedious and painful, but
good in a pinch.
OBVIOUS WORD METHOD - LOCATION BY SYMMETRY OR REPETITIONS
Location of words by symmetry is commonly employed when dealing
with single key ciphers. With double key ciphers its
application depends much on chance. If the alphabets are
repeated in the key or the key is short, we employ a limited
form of symmetry.
With a non repeating key or very long key, this method fails.
With a fairly short key we employ this method provided:
(1) We assume a word or phrase longer than the key, and
(2) This word or phrase happens to contain a letter repeated
at an interval equal to the length of the key.
For our sample problem, one of our choices might be
10 letter key - SCOUTINGDISTANCE
Therefore, any place in the cryptogram where two successive
lines have common letters in the same column is a possible
location of our word. Failure to find this location,
eliminates the possibility of this word.
Table one partially shows the ciphertext where repeated letters
are ten spaces apart. Of the twelve possibilities for the word
"SCOUTINGDISTANCE" some are eliminated by frequencies of the
letters C,G,C, others by letter combinations and the balance by
test. All fail.
Our Navy students would try the scouting line of cruisers as:
4 3 1 2
MEMPHIS RICHMOND TRENTON MARBLEHEAD
2 1 OR 3 4
MARBLEHEAD TRENTON RICHMOND MEMPHIS
(flag)
These names might appear as follows:
MEMPHISRIC MARBLEHEAD
HMONDTRENT OR TRENTONRIC
ONMARBLEHE HMONDMEMPH
AD IS
These can be checked against Table I and cross checked by
frequency or digram analysis.
We have a little luck at Line 14 - 15 - 16
Line 14 LGTBWPLVTT
--MEMPHISR
Line 15 OBOXJLRMHZ
ICHMONDTRE
Line 16 MVHAWADGGZ
NTONMARBLE
check
Line 29 OPLAULDMOZ Line 11 MOZ
I N N T E I E
NINETEE TWE
Line 30 MJKAQJBF Line 12 MVX
NHUNDRED NT
NTY
OBVIOUS LOCATION METHOD
Table I gives a list of obvious locations. We suspect
the word COURSE followed by a ZERO and ONE TWO or THREE.
Some possibilities are:
COURSEZERO COURSETHRE
FOUR EZERO
COURSEONET COURSETHRE
WO EONE
COURSEZERO (promising but no check)
FOUR
COURSETHRE
ETHREE (checks with #9 in Table I)
Assumption
Line 21 KPKDECGOHZ Line 26 S KPKGULT
COU
S COUTING
Line 22 LVODSCOCHA
ETHREEZERO
Both assumptions are entered into the cryptogram.
TABLE I
Lines Reference
6-7 ZFSAASZDEPZXCDJD 1
8-9 KAHPLGHAJMKTVAMK 2
8-9 HAJMKTVAMKHMBCAA 3
10-11 ZZDWIJKGMCZMVXXU 4
15-16 ZMVHAWADGGZYFARQ 5
17-18 FARQVKMMQKFMPSLG 6
18-19 FPMSLGXAHEFWKGCB 7
18-19 HEFWKGCBFTHSVCBB 8
21-22 DECGOHZLVODSCOCH 9
21-22 CGOHZLVODSCOCHAG 10
21-22 HZLVODSCOCHAGVWB 11
22-23 VCDSCOCHAGVWBZCA 12
22-23 COCHAGVWBZCAMOZM 13
24-25 AQJBFJHXBHAAVAKO 14
25-26 OSKPKGULTJOQDFQQ 15
28-29 AEPZWQROPLAULBMO 16
29-30 AVLBMOZMJKAQJBF 17
TABLE II
12345678910 12345678910 12345678910 12345678910
COURSEZERO COURSETHRE COURSEONE COURSETWO
ZERO EZERO ERO Z ERO Z
ONE ONE NE O NE O
TWO TWO WO T WO T
THREE THREE HREE T HREE T
FOUR FOUR OUR F OUR F
FIVE FIVE IVE F IVE F
SIX SIX IX S IX S
SEVEN SEVEN EVEN S EVEN S
EIGHT EIGHT IGHT E IGHT E
NINE NINE INE N INE N
COURSEZERO COURSETHRE COURSEONET COURSETWOT
FOUR EZER WO WO
EONE
ETHREE
DISCOVERY OF THE SYSTEM
We study the values assumed previously:
Value Alphabets Value AlphabetS
C=E 3,6,8 H=O, O=H 3,6,8
O=H 3,8 N=L,L=N 3,6,8
H=O 3,8 K=U, U=K 3,6,8
B=E 4,7 N=A,A=N 4,7
A=N 4,7 S=E,E=S 5
The common values indicate that alphabets 3,6, and 8 are
identical and similarly so are 4 and 7. Five reciprocal
values are noted without inconsistencies. Seven different
alphabets are used. The alphabets are probably reciprocal.
If the seven alphabets are Secondary (derived from the same
cipher component set against the same plaintext but in
different alignments) a short cut solution is possible. We can
next combine the alphabets into one system.
We have enough clear text to solve the cryptogram - I leave the
balance to the student.
Alpha. - 1 2 3 4 5 6 7 8 9 10 Alpha. - 1 2 3 4 5 6 7 8 9 10
1 K P T X S L I C T M 16 M V H A W A D G G Z
C O M E N E N T O N R E
2 I A M C B B N M S Z 17 Y F A R Q V K M M Q
T N A T T E D S T
3 M J K A Q J B F Z A 18 K F M P S L G X A H
N H U N D R E D O C T E N T Y I
4 J G M B S L N P H H 19 E F W K G C B F T H
T E E N A R I S S P E E D I
5 E E J Z W N C L O W 20 S V C B B U A H S S
R L N E T E E N K N O T S
6 Z F S A A S Z D E P 21 K P K D E C G O H Z
N C O U R S E T H R E
7 Z X C D J D D H A J 22 L V O D S C O C H A
E R R O E T H R E E Z E R O
8 O D B K A H P L G H 23 G V W B Z C A M O Z
S O N I A T S E V E N T E E
9 A J M K T V A M K H 24 M J K A Q J B F J H
H T S N T I N H U N D R E D I
10 M B C A A C N W S Z 25 X B H A A V A K O S
N E N E A S T E O N N U E S
11 Z D W I J K G M C X 26 K P K G U L T J O Q
S U T T W E C O U T I N R E
12 M V X X U N B W Z T 27 D F Q Q J K K M H Z
N T Y M I L E S U S T R E
13 I Y N C P O G H H W 28 H V H A E P Z W Q R
I H T O R N T O N S S
14 L G T B W P L V T T 29 O P L A U L B M O Z
E E O N N I N E T E E
15 O B O X J L R M H Z 30 M J K A Q J B F
H M N T R E N H U N D R E D
TABLE III
DECIPHERING TABLE
PLAIN- A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
1 G K L M
2 J P V
3 C O H J W K X
4 B X A D K
5 Q S U B G E Z
6 C U N L
7 N B A G O
8 F C O H W M
9 O H S C
10 Z H A S
TABLE IV
ENCIPHERING TABLE
PLAIN- A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
1 G K L M
2 J P V
3-6-8 F C O U N L H J W M K X
4-7 N B X A D K G O
5 Q S U B G E Z
9 O H S C
10 Z H A S
Op-20-G gives us the quick and dirty of the problem. We need
to understand what equivalent cipher alphabets are and how the
multiple alphabet system lends itself to reconstruction.
EQUIVALENT CIPHER ALPHABETS
Any sequence containing 26 letters may be rearranged so that
all the letters which are originally separated by equal
intervals will also be spaced at equal intervals in the new
related sequences. Including the original sequence, a total of
of six related sequences may be constructed. [Friedman expands
on this principle in FR7.]
Example:
1 3 5 7 9 11
1 A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
2 A D G J M P S V Y B E H K N Q T W Z C F I L O R U X
3 A F K P U Z E J O T Y D I N S X C H M R W B G L Q V
4 A H O V C J Q X E L S Z G N U B I P W D K R Y F M T
5 A J S B K T C L U D M V E N W F O X G P Y H Q Z I R
6 A L W H S D O Z K V G R C N Y J U F Q B M X I T E P
In this example, a normal alphabet sequence has been re-spaced
to form five related sequences. In constructing them, the
original sequence is regarded as a circle and the letters are
counted off in equal intervals, then written in adjacent
positions to form a related sequence.
Only the odd intervals from 3 - 11 can be used in re-spacing a
26 letter sequence to form different related sequences.
{primes} Even intervals will produce only 13 letter sequences,
and the interval 13 can not be used. Odd intervals from 15-25
will produce identical sequences with those from 1-11 but in
reversed direction. (like the Porta)
Cipher alphabets may be re-spaced to form equivalent cipher
alphabets by the same process as that applied to construct
related sequences.
Example:
Original Cipher Alphabet
Plain - D I P L O M A C Y B E F G H J K N Q R S T U V W X Z
Cipher - V W X Z T H U R S D A Y B C E F G I J K L M N O P Q
Equivalent Cipher Alphabet
Plain - D L A B G K R U X I O C E H N S V Z P M Y F J Q T W
Cipher - V Z U D B F J M P W T R A C G K N Q X H S Y E I L O
An equivalent cipher alphabet can not be distinguished from the
original cipher alphabet unless a systematic construction or
some outside information is available to identify the original
one. The secondary alphabets generated by shifting the points
of coincidence of the plain and cipher components are the same
alphabets regardless of which equivalent cipher alphabet has
been shifted.
Example:
Original Cipher Alphabet
Plain - D I P L O M A C Y B E F G H J K N Q R S T U V W X Z
Cipher - X Z T H U R S D A Y B C E F G I J K L M N O P Q V W
Equivalent Cipher Alphabet
Plain - D L A B G K R U X I O C E H N S V Z P M Y F J Q T W
Cipher - X H S Y E I L O V Z U D B F J M P W T R A C G K N Q
The secondary alphabet of this example has been derived by
shifting the cipher component of the original alphabet of the
previous paragraph, and the equivalent secondary cipher
alphabet by shifting the cipher component of the equivalent
alphabet of the previous paragraph.
The number of spaces each cipher component has been shifted is
not the same in each case, yet the plain and cipher values
correspond exactly. This illustrates the most important
principle of symmetry in the secondary alphabets.
RECONSTRUCTION OF MULTIPLE ALPHABET SYSTEMS
When the same sequence has been used for each of the cipher
components of a multiple alphabet system, there are definite
relationships between the individual cipher values which may be
used in recovering other cipher values after a few have been
identified through analysis.
(a) When the plain component is originally a normal sequence
the cipher sequences will be recovered in their original order
and new values may be placed in the various cipher components
as soon as their relative positions have been established.
(b) When the plain and cipher components are originally the
same mixed sequence, the plain component enters into the
reconstruction in the same manner as the other cipher
component.
(c) The reconstruction of a multiple alphabet system in
which the plain component is a different mixed sequence from
that used in the cipher components, requires a relatively large
number of values for analysis.
The principles are explained by another example in which the
plain and cipher components are different mixed sequences:
Plain 0 - D I P L O M A C Y B E F G H J K N Q R S T U V W X Z
Cipher 1 - O P Q V W X Z T H U R S D A Y B C D F G I J K L M N
2 - N O P Q V W X Z T H U R S D A Y B C E F G I J K L M
3 - E F G I J K L M N O P Q V W X Z T H U R S D A Y B C
The interval between letters of two cipher components, letters
which occur in the same vertical column, is equal to the amount
of displacement of one component from the other.
O (1) To N(2) is an interval of one, the amount of shift
between the cipher components (1) and (2).
E (3) to O (1) is the same interval as O (3) to U (1), and is
the same interval as U (3) to F (1), etc.
Thus a chain of letters, EOUF with current relative spacings
could be made from the vertical relationship alone, when the
order of plain component sequence is unknown. A set of
equivalent alphabets might be the result of construction by
this means, but the original in this case would be recognized
when the proper spacing is found.
If the vertical relationship is used between components which
are displaced an even number of letters, such as ciphers (2)
and (3), a chain of 13 letters will result, and if the
components were originally displaced 13 letters, they would
show only reciprocal relationships.
APPLICATION OF SYMMETRY PRINCIPLES
Suppose the Enciphering table obtained during the solution of a
cryptogram appeared as follows:
Plain 0 - A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
Cipher 1 - Z U T R D A P V C W G I H
2 - X H Z N U D O W B V E F G T
3 - L E P W F I K T J U R S
Since the interval between R and P in the cipher sequence is
the same as that between P and F, we may arbitrarily assume
this interval to be one and build up a cipher sequence
accordingly.
The vertical columns remain unchanged. We write:
0 E I R in the third cipher S E I
1 R P F component appears under G R P F U O
2 U O S plain, so we continue G R P F U O
3 R P F G R P F U O
The progress of adding values to the plain and cipher sequences
progresses through the various stages:
0 T S E I R B Y
1 I S G R P F U O E H T
2 I S G R P F U O E H T
3 I S G R P F U O E H T
0 O L T S E I R B Y N C
1 W J V I S G R P F U O E H C T B Z
2 W J V I S G R P F U O E H C T B Z
3 W J V I S G R P F U O E H C T B Z
0 M H O G L T S E I R B Y N C A
1 L X K A W J D V I S G R P F U O E H C T B Z
2 K A W J D V I S G R P F U O E H C T B Z L X
3 X K A W J D V I S G R P F U O E H C T B Z L
The intervals between E, F, G and between V, W, X in the cipher
sequence obtained above, indicate the equivalent alphabets have
been recovered which should be re-spaced by counting off every
third letter in the reverse direction.
0 I L O M A C Y B E G H N R S T
1 O P V W X Z T H U R S D A B C E F G I J K L
2 O P V W X Z T H U R S D A B C E F G I J K L
3 E F G I J K L O P V W X Z T H U R S D A B C
CONTINUATION OF BLACK FORCE CRYPTOGRAM
A few more values are necessary in Table IV in order to
completely reconstruct the system used.
Line 1 Line 18
Alpha 1 2 3 4 5 6 7 8 9 10 Alpha 1 2 3 4 5 6 7 8 9 10
Cipher K P T X S L I C Cipher K F M P S L G X A H
Plain C O M E N E Plain C T E N T Y I
New M C New W
Line 3 to 5
Alpha 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1
Cipher M J K A Q J B F Z A J G M B S L N P H H E
Plain N H U N D R E D O T E E N A R I
New F U R P L
Adding these new values to Table IV gives the following table
for use in reconstruction of the system:
TABLE IV
Revised
ENCIPHERING TABLE
PLAIN- A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
1 G K L E M J
2 J P V
3-6-8 F C O U N T L H P J W M K X
4-7 N I B X A D K G P O
5 Q S U B G E Z
9 O Z H S C
10 Z H A S
The reciprocal relationship will be ignored.
On account of L and B being found in two vertical columns, a
good starting point is to assume that L and B are adjacent in
the cipher component. Then we would have the following in the
cipher component: GN, KI, MA, FQ, CS, PQ, AND WE.
Using the PGN sequence in the first three cipher components,
partial reconstruction can be made:
PLAIN- W T A O R P L
1 P G N W E
2 V P G N
3-6-8 M A H J P G N
4-7 P G N D
5 P G N
9 C S H J
10 M A
Since HJ appears with the same interval as LB, then OC and SM
are also adjacent in the cipher sequence being constructed.
PLAIN- H E W T A S O R Z N P L U
1 L B P G N O C S M A W E H J
2 H J V L B P G N
3-6-8 O C S M A W E H J V L B P G N K
4-7 L B P G N K I D O C S M A
5 O C S M A W E H J V L B P G N
9 O G S M A W E H J V
10 O C S M A
We combine the three partials:
PLAIN- H E W T A S O R Z N P L U
1 L B P G N O C S M A W E H J
2 H J V L B P G N
3-6-8 O C S M A W E H J V L B P G N K I D
4-7 L B P G N K I D O C S M A
5 O C S M A W E H J V L B P G N
9 O G S M A W E H J V
10 Z O C S M A
I think you can see that most of the cipher sequence could be
obtained without considering the fact that the plain component
is the same sequence reversed. The important point is that the
complete system may be reconstructed from relatively few values
obtained through analysis of the cryptogram.
The sequence used in this problem is randomly mixed, therefore
the original one can not be distinguished from a related one
which may be reconstructed. The ten cipher components are set
with the key GUANTANAMO under the A plain.
FURTHER REMARKS
The same method used in determining which cipher values
probably represent vowels or consonants may be applied to poly-
alphabetic substitution ciphers as described in Lectures 1 and
2. However, the values in each alphabet must be considered
with their respective prefixes and suffixes in adjacent
alphabets, in studying the frequencies of their combinations.
After the original sequences of a poly-alphabetic substitution
system are recovered, subsequent messages using these sequences
may be solved by a modified method. The "generatrix frequency"
method was developed by W. F. Friedman and is described in FR7.
SOLVING CIPHER SECRETS
MASTERTON (Frank W. Lewis) was a personal 'pick' of William F.
Friedman. His experience and book [MAST] is as insightful as
it is brilliant. He takes us through the QUAGMIRE family. The
American Cryptogram Association calls the class of periodic
polyalphabetic substitution QUAGMIRES I, II, II, IV after the
terminology used for keying Aristocrats. QUAGMIRES have a
mixed alphabet in at least one of the components. QUAGMIRE I
uses a keyword-mixed plain component with a determined number
of normal cipher alphabets at different settings; QUAGMIRE II
uses a normal plain and various settings of the same mixed
cipher component; QUAGMIRE III employs the same mixed alphabet
for plain and cipher (juxtaposition repeated on a cycle); and
QUAGMIRE IV which has one mixed alphabet for plain and a series
of slides of another mixed alphabet for the cipher components.
[MAST] The use of normal alphabets on a cycle, either direct
or reverse, is a weakness because the components are known and
are more vulnerable to solution.
QUAGMIRE I
We will take the QUAGMIRES in turn, making sure we understand
the method of encipherment and tricks of unraveling the text.
Lets build an alphabet on the Keyword ENCIPHERMENT:
E N C I P H R M T A B D F G J K L O Q S U V W X Y Z
Let us take a NORMAL alphabet, with C under the first letter of
plain sequence. This is cipher setting No 1. Slide the normal
alphabet to I, under E, P, H, E, R to get:
Plain 0 E N C I P H R M T A B D F G J K L O Q S U V W X Y Z
Cipher 1 C D E F G H I J K L M N O P Q R S T U V W X Y Z A B
2 I J K L M N O P Q R S T U V W X Y Z A B C D E F G H
3 P Q R S T U V W X Y Z A B C D E F G H I J K L M N O
4 H I J K L M N O P Q R S T U V W X Y Z A B C D E F G
5 E F G H I J K L M N O P Q R S T U V W X Y Z A B C D
6 R S T U V W X Y Z A B C D E F G H I J K L M N O P Q
I have numbered the alphabets for ease of use. The initial
column keyword is standard practice.
To encipher the word regarding: The first R is found in the
plain sequence, and the letter under it in alphabet 1 is I, we
use the cipher alphabets sequentially and return to alphabet 1
after using the sixth alphabet.
QUAGMIRE I ATTACK
Given:
WBFWX LWVPY WICQJ HJYDL LNABF JCQFB BHMPA XGKIU CRHVK
YNEJO VMDEJ SPQPT GLFFB YOEYD MIHYY JJCPY YDVIE TOFXX
LWPSC YTBKJ ORCYZ DBYDH YHR.
The Cryptogram usually provides a tip: "ILEANDTHENREPLIED. "
This will appear in the text someplace.
The repeat method of factoring doesn't work to well on this
example. So assume 6, 7 or 8. Write the crib based on those
cycles.
awh awh awh
ILEAND ILEANDT ILEANDTH
THENRE HENREPL ENREPLIE
PLIED IED D
We have added a possible text of awh to the crib. The middle
crib has the I over an I 13 letters apart and the E's interval
of 6. The stretch of cipher we want will have a repeat as:
----X------Y-----XY---.
The stretch "glffbYoeydmihYyjjcpYYdvie" fits the bill. We
rewrite the cryptogram into a cycle of seven letters either in
columns or rows. We fill in the tip and number the alphabets:
1234567 1234567 1234567 1234567 1234567 1234567 1234567
WBFWXLW VPYWICQ JHJYDLL NABFJCQ FBBHMPA XGKIUCR HVKYNEJ
1234567 1234567 1234567 1234567 1234567 1234567 1234567
OVMDEJS PQPTGLF FBYOEYD MIHYYJJ CPYYDVI ETOFXXL WPSCYTB
a whILEAN DTHENRE PLIED
1234567 1234567 1
KJORCYZ DBYDHYH R.
We prepare a deciphering tableux, putting the plain values
above the normal cipher strip and using the plain E to start.
Plain 0 E
-----------------------------------------------------
Cipher 1
2
3
4 U V W X Y Z A B C D E F G H I J K L M N O P Q R S T 5
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
6
7 F G H I J K L M N O P Q R S T U V W X Y Z A B C D E
Since the fourth alphabet also has a plain L, we enter it on
the top line, and similarly place a plain N from the fifth
alphabet. The N is confirmed by its appearance in the 7th
alphabet, so we know we are on the right track.
Since we have the plain L, the second alphabet comes in too and
hence the plain H and T. This gives us the third alphabet and
the plain I. There is more help. Looking down the various
columns we find the Keyword COUNTRY which must have been placed
under the first letter of the plain sequence. Snowballs.
Plain 0 A B C D E H R T P L W I N G
-----------------------------------------------------
Cipher 1 J K L M N O P Q R S T U V W X Y Z A B C D E F G H I
2 V W X Y Z A B C D E F G H I J K L M N O P Q R S T U
3 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
4 U V W X Y Z A B C D E F G H I J K L M N O P Q R S T
5 A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
6 Y Z A B C D E F G H I J K L M N O P Q R S T U V W X
7 F G H I J K L M N O P Q R S T U V W X Y Z A B C D E
The clues add up. The Keywords are PLOWING and COUNTRY.
The RST sequence is obvious. The message reads: The city
slicker asked the farmer what's your mules name? The farmer
thought awhile and replied I don't rightly know but I call him
JACK.
QUAGMIRE II
This polyalphabetic substitution uses a Normal plain and a
keyword mixed cipher alphabet. Lets tackle a problem with the
tip of 20 letters TAPHORICORTABOONATUR and also the tip
"usage." Sometimes we have hunches. Assume the period is 10,
and write out the tip on this basis. Nice pattern with a
digraphic hit TT, OO, RR
TAPHORICOR
TABOONATURe I have added the e
possibility.
and the cipher is:
12345678910 12345678910 12345678910 12345678910 12345678910
GJGQHJLELW SZGGETGMQS YVAHUOLFYN NIRJHVKJDS XMZVUEPETG
12345678910 12345678910 1
HIAHWZOTFN HIHVWQUQDN UENAEQMFQA YXIOVUIVYG NYLUJMOCVL
TAPHORICOR TABOONATUR e
RXSOTVSSMT CIIFHVEFYA VJLEUVDQFX OZJHNNUHQY EOGQDYGHEG
RXVVVOBVYY SR
Now we develop the deciphering tableaux.
Plain 0 A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
-----------------------------------------------------
Cipher 1 U H
2 I
3 H A
4 H N
5 W
6 Q Z
7 U O
8 T Q
9 F D
10 N
QUAGMIRE II ATTACK
We know that the plain sequence is normal. it is in the right
order and we can base our interval analysis on the plain. We
introduce Mr. Friedman's principle of symmetry to discover the
relationships in the cipher alphabets.
We know that the cipher text reads from left to right just as
we see it. The skeleton sequence is:
H------V------A, Q---Z----T, U-------O, and F-----D,
We can fill in a few letters. The Q---Z is either QVW-Z or Q-
VWZ. In No 1 Q cipher is either Y or Z and Z cipher is either C
or D. [MASTERTON jumps in with a NIO combination and VW but I
didn't see this until after the solution.] Alpha 4 puts V +6
from H, transposing that to alpha 1, puts a V under the A
plain, and suggests Q V W X Z sequence with Y in the Keyword.
X is pretty unpopular in keywords so we will go with this
assumption.
INTERMEDIATE DECIPHERING TABLEUX
Plain 0 A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
-----------------------------------------------------
Cipher 1 V W X Z U ? ? A T O H Q
2 I
3 H Q V W X Z U ? ? A T O
4 H Q V W X Z U A T
5 H Q V W X Z U A T O
6 O H Q V W X Z U A T
7 U T O H Q V W X Z
8 A T O H Q V W X Z U
9 F D
10 N
So we build up alpha's 1, 3, 5, 6, 8. We can place the H's
back in them from the Q by -6. in alpha 8 and 5. We see that
U +8 = O in alpha 7. The sequence ---A starts the keyword from
alpha three. Look at the T behind the Q by -17 offset in
alpha 8. Remember my assumed 'e' = U in alpha 1. We place this
hunch and let it play through.
We have U - - AT ........Y. I see the prefix UN and digram SA.
The word "unsatisfactory" comes to mind but I haven't got
enough hard evidence yet. We have a U +8 to O in the 7th
alpha. Fill in the alphas.
FINAL DECIPHERING TABLEUX
Plain 0 A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
-----------------------------------------------------
Cipher 1 V W X Z U n s A T i f c O r Y b d e g H j k l m p Q
2 I
3 H Q V W X Z U ? ? A T O
4 H Q V W X Z U A T
5 H Q V W X Z U A T O
6 O H Q V W X Z U A T
7 U T O H Q V W X Z
8 A T O H Q V W X Z U
9 F D
10 N
I know that Y is in the keyword and could be the last letter of
it. Look at the F-----D sequence. F is in the keyword and the
O-------H is the only area than can fit the F and the Y.
Plug in my UNSATifcOrY guess. The lower letters require
checking. Alphabet 1 fits the key as UNSATISFACTORY adjusted
for duplicate letters.
The message reads in part: Slang is language or phrases of a
vigorous colorful metaphoric or taboo nature invented to ...
QUAGMIRE III
The QUAGMIRE III is a very important class of ciphers because
they introduce the one of the most important tools invented by
Mr. Friedman, as explained in his Riverbank papers, called
"Direct and Indirect Symmetry."
The title of this problem is "Inertia in the British Labor
Market" and has the tip "ANDTHREECALLINGFORAMANTOSTANDON."
IBWVU PLTPJ TKPPM YCTDV XYGNY QYNTW NFSUI XNACX CFTGV
AIKPS RTCOJ JWPRR VOLAA ZRURJ NUIXM XPQBV UIBWO GPCDP
LNNRD FPSLI BUGOC DOTWK CPIRQ RVQGY GCXLV MNOBE QFVOL
GBWGP ATNJL YWRMW EKLAA VICVE AQBKU VFJUR DVIOZ MPTZO
VSLIH QBQXF LLLWH PUSGV XP.
QUAGMIRE III ATTACK
Note the repeat of the first three letters IBW at interval 81.
If the message starts with THE and the period turns out to be 9
we have found a wedge. Next place the tip in columnar line for
a cycle of nine.
A N D T H R E E C A I K P S R T C O
A L L I N G F O R J J W P R R V O L
A M A N T O S T A A A A R U R J N U
N D O N t w o f e e t ? I X M X P Q B V U
t h e ------- ? I B W O G P C D P
(also first three IBW)
The three A's in the first column followed by the two N's
prove the period of 9. This is not accidental. My guesses
of additional plain text are partially right - 'the' as you
will see later. Note the triple R's, two U's and Two I's in
the ciphertext lined up by columns in a period of 9.
Break the ciphertext into groups of nine.
123456789 123456789 123456789 123456789 123456789
IBWVUPLTP JTKPPMYCT DVXYGNYQY NTWNFSUIX NACXCFTGV
AIKPSRTCO JJWPRRVOL AAARURJNU IXMXPQBVU IBWOGPCDP
ANDT HREECALLI NGFORAMAN TOSTANDON THE
LNNRDFPSL IBUGOCDOT WKCPIRQRV QGYGCXLVM NOBEQFVOL
GBWGPATNJ LYWRMWEKL AAVICVEAQ BKUVFJURD VIOZMPTZO
VSLIHQBQX FLLLWHPUS GVXP.
Place the extended tip. In a QUAGMIRE III, or in any case
where the cipher component is the same as the plain component,
if one cipher -plain matches E for E, all pairs must match,
for the sequence is set A to A, B to B, etc. When this
happens, we get a column of our write-out as "free plain text,"
which is of considerable help.
I can not overemphasize the next step. Because of the K3
nature of the keying, the Plain component and the Cipher 1
alphabet represents pairs that are the same distance removed -
H to J, N to A, T to I, in this case. Similarly G to A, H to
B, O to X, and R to J are equally separated - though not at the
same interval as the first pairs obtained from line 1.
(Obviously, if H to J is "x" distance, H to B cannot be the
same distance.) Check this observation of Symmetry on the
decipher tableaux.
INITIAL DECIPHERING TABLEUX
Plain 0 A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
-----------------------------------------------------
Cipher1 J A I
2 A B X J
3 W A M
4 P R X
5 P R U
6 R Q
7 B V J T
8 N C O V
9 L U O
Let us write down all the pairs we get by going from plain to
cipher in each of the alphabets in turn. We can also write
down the from the sidewise relationships. For instance, A to C
on the plain sequence is the same distance P to R on Row 5. In
addition, Row 7 to Row 8 tells us that BC is the same distance
apart as VO.
This is a most powerful tool in solution of a sequence against
itself. You can imagine a little "square" and go up, or down,
or across, to find relationships within and between both plain
and cipher components.
Plain sequence to Row 1 HJ NA TI
2 GA HB OX RJ
3 EW FA SM
4 EP OR TX
5 AP CRU (CR-RU)
6 AR NQ
7 DB LV MJ NT
8 AN DC LOV (LO-OV)
9 IL NU TO
>From Plain A to C AC PR
>From Row 7 to 8 BC VO
There are a lot of relationships. I have not listed the
sidewise ones like Plain to Row 1 - H to N and J to A.
MASTERTON points out that Row 1 is the reverse of Row 8.
[MAST] I didn't see this "little" jump.
But I did make sense of the three letter chains; if L-O is the
same as O-V we have a three letter segment. Do you see that
the pairs in the listing above are separated by one letter in a
sequence obtained from the next set, as evidenced by LV in 7
and LOV in 8? We can add the two together:
DCB LOV M-J AN-T
Look at the fragments, and realize that we have found some good
information about the sequence. First of all the sequences are
reversed alphabets. The sequence has BCD, VOL, JKM since we
have used L and T-NA in it? [We can also look at a process
called decimination to bring the sequence to bear. We will do
that in the Friedman section.] Remember the very important
part of the tool of symmetry - that because the plain and all
the cipher alphabets are the same, we can associated pairs in
the straight, sideways, down etc as we find them, using the
plain or all nine cipher alphabets. In a QUAGMIRE IV, we
cannot use the plain sequence in this way because of a
different key.
We continue our recovery with A to N plain as the same
distance as R to Q in alpha 6. We add QR to our line.
VOL TINA BCD HJKM QR
Notice the H to B and G to A in the plain to alphabet 2
relationship. This tells us to put G ahead of H, then A goes
behind B as we expect. Since O is in VOL and N is in TINA
VOL/TINABCD/GHIJM/QR
the only missing element is P which we place as follows:
ku VOL/?/TINABCD (f)GHJMPQR swxyz
missing elements at this stage are e, k, u, w , x , y , z which
likely the E and U are in the Keyword.
INTERMEDIATE DECIPHERING TABLEUX - PARTIALS
Plain 0 V O L T I N A B C D F G H J M P Q R S
-----------------------------------------------------
Cipher1 V O L T I N A B C D F G H J M P Q R S w
2 X T I N A B C D F G H J M P Q
3 T I N A B C D F G H J M P
4 Q R S W? X
5
6
7
8 V O L T I N A B C F G H J M P Q R S
9
The line ups are not correct. We can find where alphabets 1,
2 and three start by putting the low frequency X in the right
spot. I leave this part of the work to the you all. [ Hint:
compress the V O L -----T I N A space and what keyword will fit
into - V O L u? T I (O)N. and place the E in the beginning.]
The answer is with Keywords EVOLUTION and BLUEPRINT:
FINAL DECIPHERING TABLEUX
Plain 0 E V O L U T I N A B C D F G H J K M P Q R S W X Y Z
-----------------------------------------------------
Cipher1 V O L U T I N A B C D F G H J K M P Q R S W X Y Z E
2 S W X Y Z E V O L U T I N A B C D F G H J K M P Q R
3 W X Y Z E V O L U T I N A B C D F G H J K M P Q R S
4 P Q R S W X Y Z E V O L U T I N A B C D F G H J K M
5 C D F G H J K M P Q R S W X Y Z E V O L U T I N A B
6 F G H J K M P Q R S W X Y Z E V O L U T I N A B C D
7 Y Z E V O L U T I N A B C D F G H J K M P Q R S W X
8 Z E V O L U T I N A B C D F G H J K M P Q R S W X Y
9 X Y Z E V O L U T I N A B C D F G H J K M P Q R S W
The message reads: The British created a civil service job in
eighteen hundred and three calling for a man to stand on the
cliffs of Dover with a spyglass.....
QUAGMIRE IV
The QUAGMIRE IV is probably the most difficult of the QUAGMIRES
because we need to recovery two keyworded alphabets and direct
symmetry will not work with the plain.
We are given:
MWQYD KMCAO KHSEE YULIH WYTEW YRLHG LMEJC ZHAKE NYWUP
thegr reat
QSQSO ESYEP BIZEW QYPKZ FHAAM GWPTR XNYWR LKSQE XHGRA
QCWAV JNCPM HDHZT BCBHR AMXUE OLTWR RIKNQ AKKDZ VJOYW
bet?
WHQJR FGYVP GILWV WGPTF MLYKX TAKOZ ATFGL AUT.
weenl atese ptemb erand decem berof thaty ear
QUAGMIRE IV ATTACK
The Title is "Lost Horsepower", the tips are starts with THE
GREAT and has WEENLATESEPTEMBERANDDECEMBEROFTHATYEAR in the
text. The letters bet?WEEN might be inferred.
Finding the cycle is our first challenge.
The WQY is +58, a discouraging number for factors. The cribs
are pretty generous, so looking at them we might find
something. Obviously, a plain hit at the correct interval of
the cycle would result in a cipher coincidence at the same
interval. Two occurrences of a plain letter at some interval
other than the period or multiple of the cycle, the ciphers
cannot be the same. MASTERTON describes a graphical technique
for knocking out intervals. [MAST]
OYWWHQJRFGYVPGILWVWGPTFMLYKXTAKOZATFGLAUT
betweenlateseptemberanddecemberofthatyear
* --9-- *
Thus the Y over E and H and Q over E "knock out" the intervals
3, 4 which are too short anyway, and also 11 because of the Y
over P. Note the +9 hit for Y over E. So we write out the
cipher in a period of nine:
123456789 123456789 123456789 123456789 123456789
MWQYDKMCA OKHSEEYUL IHWYTEWYR LHGLMEJCZ HAKENYWUP
thegreatE E GH EE E A
QSQSOESYE PBIZEWQYP KZFHAAMGW PTRXNYWRL KSQEXHGRA
E ?HE E T EA R RT ER E R E E
QCWAVJNCP MHDHZTBCB HRAMXUEOL TWRRIKNQA KKDZVJOYW
T A TE NH E E R bet
WHQJRFGYV PGILWVWGP TFMLYKXTA KOZATFGLA UT.
weenlates eptembera nddecembe rofthatye ar
Even with all the help and correct hits, the message is not a
give a way.
INITIAL DECIPHERING TABLEUX
Plain 0 A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
-----------------------------------------------------
Cipher1 U P T K M W
2 F H W O G T
3 M Q Z I
4 L Y J A
5 Y T R W D
6 F V K
7 M O W X G
8 T Y G C
9 P A C V W
Since the alphabets are different we can not chain from the
plain to cipher. However, WITHIN the cipher, the same rules
apply as before - except their isn't nearly as much
information. In Cipher 1 row we see that U to P is the same
distance as F to K , M to W and P to A. Ok. Remember that we
are dealing with unknown decimations, so the relationships
between UPA, PK and PT is unknown.
By decimation I mean the process of selection of elements from
a sequence according to some fixed interval. For example, the
sequence A E I M is derived, by decimation , from a normal
alphabet by selecting every fourth letter. It is the key to
Symmetry solutions because the latent relationships in a cipher
alphabet can be made patent by decimation. Lecture 11 will
give two methods of decimation in detail.
Table of Relationships in foregoing example:
UPA FK MW Plain A to E and Rows 1 to 9
PT LJ " E to N
PK HT YG " E to R and Rows 1 to 6 adding UF
PM QI LAWG YC " E to T and Rows 9 to 7 and 4 to 9
UMG PW " A to T and Rows 1 to 7
TM JA " N TO T
FH MQ " D to E
WTD " H to R and Rows 2 to 5
FV MO " A to B
VK OW TY " B to E
OG TC " B to T
PH KT Rows 1 to 2
PQ MI Rows 1 to 3
PL TJ MA Rows 1 to 4
PY KG MC Rows 1 to 8
FM HQ KW VO Rows 2 to 0
HY TG Rows 2 to 9
QL IA Rows 3 to 4
QW IG Rows 3 to 7
QY IC Rows 3 to 8
QA IW Rows 3 to 9
LW AG Rows 4 to 7
LY AC Rows 4 to 8 and Plain A to G adding
Cipher C under Plain G on Row
FP KA Rows 6 to 9 9
OT WY GC Rows 7 to 8
YA CW Rows 8 to 9
Row 2 to 3 and 6 to 7 are combined. S and T in plain are most
likely adjacent from VW in Cipher 9. Partials FH and MQ look
good without an intervening letter.
LAWG is our best bet for the wedge. It ties together E and T
in the same decimation. So:
Plain E T
Cipher P M
H
Q I
L A W G
K
L A W G
Y C
L A W G
If FH and MQ are the right order, P is in the keyword, since
the reverse bits of above (MP, IQ, GWAL) would not be
consistent with MPQ. Unfortunately, we have run out of gas and
must guess more plain. The plain E-gh-EE most likely is
Eighteen and since they are talking about years, why not
Seventy, since so many E's are fitting? The plain T of seventy
is confirmed. The plain V may not produce much but the cipher
G might be a bonanza. These new values add KE and JR to the
chain.
123456789 123456789 123456789 123456789 123456789
MWQYDKMCA OKHSEEYUL IHWYTEWYR LHGLMEJCZ HAKENYWUP
thegreatE T EIGHTEEN SEVENTY E A
QSQSOESYE PBIZEWQYP KZFHAAMGW PTRXNYWRL KSQEXHGRA
E THE E T EA R RT ER E R E E
QCWAVJNCP MHDHZTBCB HRAMXUEOL TWRRIKNQA KKDZVJOYW
T A TE NH E E R bet
WHQJRFGYV PGILWVWGP TFMLYKXTA KOZATFGLA UT.
weenlates eptembera nddecembe rofthatye ar
FINAL DECIPHERING TABLEUX
Plain 0 A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
-----------------------------------------------------
Cipher1 U P T K L M W
2 F H W O G T
3 M Q Z W I G
4 L Y J A
5 Y T R W M D
6 F V K J E
7 M O W X G
8 T Y G C
9 P A C V W
We look at VW and LM and KLM under the plain RST. We must
conclude that G-C is correct. Rows 7 and 8 have a G and C
under plain T, and WY under E and OT under B. This suggests
that WXY and O-T are part of the final chain. So push the
following chains:
KLM, G-C, VWXY, EA, O-T
The cipher sequence appears to go:
JKLMQVWXYZ
0 A N D E I C B F G H
---------------------------------------------
1 U T P R A
2 F H J K L M Q V W X Y Z
3 F H J K L M Q V W X Y Z
4 F H J K L M Q V W X Y Z
5 F H J K L M Q V W X Y Z
6 F H J K L M Q V W X Y Z
7 F H J K L M Q V W X Y Z
8F H J K L M Q V W X Y Z
9 P R A
The cipher keyword has this form O U T - P R - A I N G
with S, E, D candidates. The keyword is SPREADING.
The plain keyword can be derived as PANDEMIC and the cipher
setting key is HORSETAIL. The groundwork is left to the
student. Notice how resistant the QUAGMIRE IV was even with
loads of help.
LECTURE 10 HOMEWORK PROBLEMS
QQ-1 QUAGMIRE I Travelogue. (Ends:SINGOUTOFTHESEA) RHIZOME
KKQHPQR KTYOHTA TLGAWBM XORKTAT BSOOIYI CGICEJV UCYZRJP
ALNSFRZ UCQDXIS TDRBFYS YTFDZBD USQWKMT CPPDOAI CAAKEHK
UAYFHQA TLNIFSI SIGJHAS V.
QQ-2 QUAGMIRE III Tedious. (CRYPTANALYTIC METHODS)
DOPPELSCHACH
PNATV SJBAQ WGMTR BZYLU ACACR GBNTQ FGGCN APNID ULMVD
SCEPB AMCQF BBPVR EOBSL AFSAN HFYVV MCYTF LEMAO MFHVU
KBAAU ATTEA NGOHU GTQEX ISUGU SAKCC TLIRT TLSZM PBMGV
APYRV YIIGL WGNUF JFROG SNQGN HBOTU TACUO JUVQH HUGWW
WBIMT WNHVO GTLSZ MPYQZ BNCEN UWLC.
QQ-3 QUAGMIRE IV Economics Lesson. EDNASANDE
(BUSINESSACTIVITYDURINGAPERIOD)
TDNSE PMBSV FURMQ UFYSJ PAGGY FVIKT GYVLV FBTPH IIIAD
HVIUY QSAFA VQVFU HPIHE BIXNN HBSTN IRMQH IIIAD OVIXT
CTNOW EOJOZ BOWBU ONLFN GOBJS HBOQS VZMOU JSFQH SAHPS
JBBJT AAMIE XILRA TOTVL TUAML FLNEJ PPMNT XHVQV FCYSB
JODNF XJSFT UIUTM ONKDO UMMSB NWUL.
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[LYNC] Lynch, Frederick D., "Pattern Word List, Vol 1.,"
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written, without proper authority, unprofessional, and
prejudicial to boot. And, it has one of the better
illustrations of the Soviet one-time pad with example,
with three errors in cipher text, that I have corrected
for the author.]
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[SEBE] Seberry, Jennifer and Joseph Pieprzyk, "Cryptography: An
Introduction to Computer Security," Prentice Hall, 1989.
[CAREFUL! Lots of Errors - Basic research efforts may
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[SHAN] Shannon, C. E., "The Communication Theory of Secrecy
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