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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]
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:
(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.
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 BHere 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].
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.
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 FCollateral 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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
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/DatesOur concern is not with guessing words but standardizing the solution.
The Known-Word" method applied in two ways:
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
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.
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:
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
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.
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.
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.
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.
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 OThe 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 LThe 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
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 LAdding 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 ASince 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 AWe 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 AI 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.
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.
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, III, 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.
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:
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.
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.
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
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.
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.
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 ...
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.
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.
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:
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 QRNotice 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/QRthe only missing element is P which we place as follows:
ku VOL/?/TINABCD (f)GHJMPQR swxyzmissing elements at this stage are e, k, u, w , x , y , z which likely the E and U are in the Keyword.
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 3 start
by putting the low frequency X in the right spot. I leave this part of the work
to 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:
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...
The QUAGMIRE IV is probably the most difficult of the QUAGMIRES because we need to recover 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
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 arEven with all the help and correct hits, the message is not a give a way.
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
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.
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.
REFERENCES / RESOURCES [updated 6 April 1996]
[ACA] ACA and You, "Handbook For Members of the American
Cryptogram Association," ACA publications, 1995.
[ACA1] Anonymous, "The ACA and You - Handbook For Secure
Communications", American Cryptogram Association,
1994.
[ACM] Association For Computing Machinery, "Codes, Keys and
Conflicts: Issues in U.S. Crypto Policy," Report of a
Special Panel of ACM U. S. Public Policy Committee
(USACM), June 1994.
[AFM] AFM - 100-80, Traffic Analysis, Department of the Air
Force, 1946.
[ALAN] Turing, Alan, "The Enigma", by A. Hodges. Simon and
Schuster, 1983.
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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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[CAREFUL! Lots of Errors - Basic research efforts may
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Text converted to HTML on June 29, 1998 by Joe Peschel.
Any mistakes you find are quite likely mine. Please let me know about them by
e-mailing:
jpeschel@aol.com.
Thanks.
Joe Peschel
