Lesson 12:
Polyalphabetic Substitution Systems III
Cryptanalysis Of Viggy's Extended Family
Decimation In Detail
CLASSICAL CRYPTOGRAPHY COURSE
BY
LANAKI
May 30, 1996
Revision 0
COPYRIGHT 1996
ALL RIGHTS RESERVED
LECTURE 12
POLYALPHABETIC SUBSTITUTION SYSTEMS III
CRYPTANALYSIS OF VIGGY'S EXTENDED FAMILY
DECIMATION IN DETAIL
SUMMARY
In Lectures 12 - 13, we continue our study of the "Viggy"
cipher family or Polyalphabetic Substitution systems. We
will cover decimation processes in detail and investigate
special solutions for periodic ciphers. The important
principle of Superimposition will be introduced.
The Resources Section has been updated with more than 50 ACA
published references on these and similar systems - focusing
on the cryptanalytic attack and areas of historical interest.
Thanks to PHOENIX for his help in compiling these sources.
[INDE]
"INCOMING"
In Lecture 13, we will tackle the difficult aperiodic
polyalphabetic case and introduce auto/running key systems.
We will diagram the topics covered in Lectures 10 - 13.
Lecture 14 will be presented by LEDGE. He will cover further
Cryptarithm topics.
Lectures 15-18 will discuss the various geometric,
transposition and fractionation ciphers.
PORTAX CIPHER
We start with a difficult cousin of the PORTA described in
Lecture 11. The PORTAX uses pairs of letters as a unit for
encipherment and decipherment as apart from single letters.
A special slide is required for its operation, and a keyword
is needed.
A B C D E F G H I J K L M (stationary)
. N O P Q R S T U V W X Y Z N O P Q R S T U V W X Y Z ...
. C E G I H M O Q S U W Y A C E G I K M O Q S .. (sliding
. D F H J L N P R T V X Z B D F H J L N P R T .. key)
(The above slide-setting is for G-H (key) directly under the
A-indicator of the stationary alphabet.)
To encipher the digraph RE, we take the R in the upper row of
letters (stationary slide) and the E from the lower pair of
letters (sliding), and use the opposite corners of the
rectangle formed to obtain the ciphertext, or PI. However,
if the digram ER is to be enciphered, we take the E from the
stationary alphabet at the top, and the R from the sliding
alphabet at the bottom to obtain FP.
Note that if the first letter of a digraph is in the range of
A-M, the equivalent ciphertext is dependent on where the
slide is used for the key-letter; but, if the first letter of
the digraph is in the range of N-Z, then it slides along with
the paired rows of lower letters, and therefore all such
digraphs having the first letter in the N-Z are constant,
without dependent of the key. There is an exception when
both letters in the plaintext digraph are in the same column,
in which case the key letter has to be known, for letters
appearing above the needed letters are used for the
ciphertext. [BRYA]
To encipher with keyword, the plaintext is written in two
rows under it; continuing to the end of the message. When
the final group is reached, if there are not enough letters
to make it complete (an even number), add a single null.
For example, encipher the word INNOVATION using the key
OFTEN :
*
A B C D E F G H I J K L M (stationary)
. N O P Q R S T U V W X Y Z N O P Q R S T U V W X Y Z ...
. C E G I K M O Q S U W Y A C E G I K M O Q S .. (sliding
. D F H J L N P R T V X Z B D F H J L N P R T .. key)
*
O F T E N (keyword)
---------
I N N O V
A T I O N
g w
e b
---------
S A R E F
O U N D x
u i
k e
Setting the O of the sliding pairs under the 'A' indicator
of the stationary alphabet, we enciphering IA as GE (opposite
corners); then SO, continuing down the column we encipher the
whole column. We then slide the strip until E-F (key) is
under the A indicator and encipher that column.
To find the period in the PORTAX is dependent on possible
fragments of the plaintext which are known (through the N-Z
combinations produced from the unchanged relationship of
letters). Lets partially decipher the following PORTAX:
SNPOW LBAMP ISCWU OOBXC WKMAT ZKTOW JCBLN CBJGB
TAAJD IWUKW HHVZN MNUFM APBJW PCBSX JCJQX TMVUB
MDCBJ CGUGR. (90)
Assuming a period of 6:
S N P O W L
B A M P I S
n t u r natural ?
l e d s good
-----------
C W U O O B
X C W K M A
o y s
s o c ok
-----------
T Z K T O W
J C B L N C
r o s t o
n y n d s better
-----------
B J G B T A
A J D I W U
y
m
-----------
K W H H V Z
N M N U F M
t p t
s r y
-----------
A P B J W P
C B S X J C
n r o
f t e
-----------
J Q X T M V
U B M D C B
n t o n
h u n r
-----------
J C R - -
U G R
-----------
Note the NY-NDS which could be NYaNDS or NYeNDS. Look at the
final group, we find -NTON -HUN-R (hundred?) We next test the
keyword by putting T in the final position and testing the
precursor letter; A C E F H I L N O P R S and U, At the E
setting, OM = TC, making -OYST/-SOCCU with R in the next
group confirming OCCUR. The E substitution also gives us the
HUNDRED. The rest of the analysis is left for the student
for credit.
THE NIHILIST SUBSTITUTION CIPHER
One of my favorite ciphers is the Nihilist Substitution
Cipher. Classified as a periodic, it employs numbers to
represent letters. Numbers are derived from a 5 x 5 Polybius
Square.
We set up a block of 25 letters and combine I/J in one cell.
Figure 12-1a
1 2 3 4 5
1 A B C D E
2 F G H I/J K
3 L M N O P
4 Q R S T U
5 V W X Y Z
So A = 11, L = 31, T = 44. (Row by Column)
The Polybius Square can be keyed. For example, using
UNITED STATES OF AMERICA and eliminating the duplicate
letters, we have:
Figure 12-1b
1 2 3 4 5
1 U N I T E
2 D S A O F
3 M R C B G
4 H K L P Q
5 V W X Y Z
We can also mix it up further with a little transposition.
Use BLACKSMITH, transpose and remove the ciphertext by
columns starting at 1:
B L A C K S M I T H
D E F G N O P Q R U
V W X Y Z
B D V L E W A F X C G Y K N Z S O M P I Q T R H U
The resulting square reads:
Figure 12-1c
1 2 3 4 5
1 B D V L E
2 W A X F C
3 G Y K N Z
4 S O M P I
5 Q T R H U
Figure 12-1c shows the effect of the transposition applied
first.
Now the message COME AT ONCE enciphered with a keyword of
TENT (period = 4) is:
T-44 E-15 N-35 T-44
----------------------
C-13 O-34 M-32 E-16
A-11 T-44 O-34 N-33
C-13 E-15 - -
We add the key and the plaintext equivalents together to
produce the ciphertext: COME: 57 49 65 59; ATON: 55 59 67
77; CE: 57 30. Each column represents a monoalphabetic
substitution in itself, and the reading or value of these
letters is dependent on the letters on either side of them.
WEAKNESSES
The lowest number of any key-letter which may be added to the
lowest plaintext letter is 11, with a total of 22; the
highest combination is two 55's or 10 (110). The numbers
6,7,8, or 9, are not involved in either the tens or the one's
additions - but they may result in a sum. Cipher 22 must
equal 11 plus 11; and 10 can only mean the sum of two 55's.
Zero in the one's column means that two 5's have been added.
This is also true in the ten's column. If at any time we find
that a 6-7-8-9 is involved we can discard the period assumed
as wrong. What we are looking for is a number in the 1-2-3-
4-5 range that may be added to produce first the ten's sum
and then the one's sum.
FINDING THE PERIOD
There are two ways to find the period - the short and the
long way.
SHORT METHOD
The short way of finding the period is to look for two or
more 30's. We treat them like a repeated digraph and factor
the interval between them looking for a common factor. We may
also try the same procedure with the lowest number versus the
highest number, for example the distance between two 94's or
two 26's.
LONG METHOD
The long way is to assume a 3 period and test the 1'st and
4'th, 2'nd and 5'th, 3'rd and 6'th in the same manner as the
short method. When conflicts arise, discard the choice.
We continue with an assumption of periods 4, 5, 6, etc. and
increase the differentials between ciphertext numbers. [BRYA]
CRYPTANALYSIS OF THE NIHILIST SUBSTITUTION
Gaines [ELCY] suggests that cracking this cipher parallels
the Viggy. The period is found through repeated sequences, or
in their absence, through repeated single letters, yielding
individual frequency counts on the several alphabets of the
period. If the arrangement of the ciphertext follows the
normal Polybius (aka Checkerboard) Square, the frequency
counts will follow the graph of the normal alphabet less one
letter. Even with the keyword mixed ciphertext alphabet,
no matter how badly mixed, the frequency counts are parallel,
the several alphabets combined follow one graph, and can be
"lined up."
Notice that the primary alphabet contains only the digits 1-
2-3-4-5. The maximum difference is 4 and addition of any
number to all of them does not change this fact. the maximum
difference between any to sums is still 4. Now the number
added during encipherment is also a number containing no
digit other than 1-2-3-4-5; thus any number found in the
cryptogram can be considered as carrying two separate
additions, one for tens and one for ones. The two 5's added
give us the revealing 0; the carried digit 1 can be mentally
borrowed back, by decreasing the size of the digit preceding
the zero. If we find a 40 , we look at it as 3 tens with ten
units or finding 110, we may regard this as ten tens and ten
units. If we find the numbers 29 and 87 in the cryptogram,
we know they were not enciphered by the same key. This is
because a difference greater than 4 in the respective tens
units exists and no digit whatever added to any two digits of
the original square can produce a difference greater than 4.
Say we have 30 and 77, with no difference greater than 4, the
presence of the zero needs to be accounted for. The number
30 has 2 tens and ten units; 7 - 2 >4, hence, we reject
the same key hypothesis.
Four giveaways are 22, 30, 102, and 110. The presence of any
one of these numbers gives away the key to the whole cipher
alphabet.
[BRYA] presents a useful aid for the standard Polybius
Square in Table 12-1. At the top is the key-number, at the
left is the plaintext letter, and at ciphertext is found at
the intersection. Any two of the three variables yields the
unknown letter/number.
Table 12-1
11 12 13 14 15 21 22 23 24 25 31 32
A B C D E F G H I/J K L M
A 11 22 23 24 25 26 32 33 34 35 36 42 43
B 12 23 24 25 26 27 33 34 35 36 37 43 44
C 13 24 25 26 27 28 34 35 36 37 38 44 45
D 14 25 26 27 28 29 35 36 37 38 39 45 46
E 15 26 27 28 29 30 36 37 38 39 40 46 47
F 21 32 33 34 35 36 42 43 44 45 46 52 53
G 22 33 34 35 36 37 43 44 45 46 47 53 54
H 23 34 35 36 37 38 44 45 46 47 48 54 55
I 24 35 36 37 38 39 45 46 47 48 49 55 56
K 25 36 37 38 39 40 46 47 48 49 50 56 57
L 31 42 43 44 45 46 52 53 54 55 56 62 63
M 32 43 44 45 46 47 53 54 55 56 57 63 64
N 33 44 45 46 47 48 54 55 56 57 58 64 65
O 34 45 46 47 48 49 55 56 57 58 59 65 66
P 35 46 47 48 49 50 56 57 58 59 60 66 67
Q 41 52 53 54 55 56 62 63 64 65 66 72 73
R 42 53 54 55 56 57 63 64 65 66 67 73 74
S 43 54 55 56 57 58 64 65 66 67 68 74 75
T 44 55 56 57 58 59 65 66 67 68 69 75 76
U 45 56 57 58 59 60 66 67 68 69 70 76 77
V 51 62 63 64 65 66 72 73 74 75 76 82 83
W 52 63 64 65 66 67 73 74 75 76 77 83 84
X 53 64 65 66 67 68 74 75 76 77 78 84 85
Y 54 65 66 67 68 69 75 76 77 78 79 85 86
Z 55 66 67 68 69 70 76 77 78 79 80 86 87
Table 12-1
continued
33 34 35 41 42 43 44 45 51 52 53 54 55
N O P Q R S T U V W X Y Z
A 11 44 45 46 52 53 54 55 56 62 63 64 65 66
B 12 45 46 47 53 54 55 56 57 63 64 65 66 67
C 13 46 47 48 54 55 56 57 58 64 65 66 67 68
D 14 47 48 49 55 56 57 58 59 65 66 67 68 69
E 15 48 49 50 56 57 58 59 60 66 67 68 69 70
F 21 54 55 56 62 63 64 65 66 72 73 74 75 76
G 22 55 56 57 63 64 65 66 67 73 74 75 76 77
H 23 56 57 58 64 65 66 67 68 74 75 76 77 78
I 24 57 58 59 65 66 67 68 69 75 76 77 78 79
K 25 58 59 60 66 67 68 69 70 76 77 78 79 80
L 31 64 65 66 72 73 74 75 76 82 83 84 85 86
M 32 65 66 67 73 74 75 76 77 83 84 85 86 87
N 33 66 67 68 74 75 76 77 78 84 85 86 87 88
O 34 67 68 69 75 76 77 78 79 85 86 87 88 89
P 35 68 69 70 76 77 78 79 80 86 87 88 89 90
Q 41 74 75 76 82 83 84 85 86 92 93 94 95 96
R 42 75 76 77 83 84 85 86 87 93 94 95 96 97
S 43 76 77 78 84 85 86 87 88 94 95 96 97 98
T 44 77 78 79 85 86 87 88 89 95 96 97 98 99
U 45 78 79 80 86 87 88 89 90 96 97 98 99 00
V 51 84 85 86 92 93 94 95 96 02 03 04 05 06
W 52 85 86 87 93 94 95 96 97 03 04 05 06 07
X 53 86 87 88 94 95 96 97 98 04 05 06 07 08
Y 54 87 88 89 95 96 97 98 99 05 06 07 08 09
Z 55 88 89 90 96 97 98 99 00 06 07 08 09 10
Consider Edwin Linquist's challenge:
24 66 35 77 37 77 55 59 55 45 55 88 28 66 46
88 37 67 33 59 58 65 45 66 67 58 44 55 34 79
44 59 55 45 42 87 28 76 43 78 46 86 26 67 24
85 26 67 28 76 26 78 46 65 65 88 36 49 54 67
28 65 42 88 36 49 44 89 57 58 54 66 47 67 26
Try period = 2. Starting at the first number 24 constant we
scan the line looking for differences greater than 4 using a
constant difference of 2. We come to 33 and 38 and stop.
Try period = 3. The first comparison fails at 24 and 77.
Try period = 4. We are able to go through the entire
cryptogram, comparing numbers at an interval of 4, without
find any difference in either tens or units greater than 4.
We now must look at the numbers collectively in columns to
verify the period is 4. We recopy the cryptogram into a
block.
Key = 4?
24 66 35 77
37 77 55 59
55 45 55 88
28 66 46 88
37 67 33 59
58 65 45 66
67 58 44 55
34 79 44 59
55 45 42 87
28 76 43 78
46 86 26 67
28 76 26 78
46 65 65 88
36 49 54 67
28 65 42 88
36 49 44 89
57 58 54 65
47 67 26 -
Alphabet 1: The tens-half of the first column contains the
digit 2 and since this can only come from the addition of 1
plus 1, the only possible key digit is 1. The units-half has
a range of 4-5-6-7-8, maximum range possible. The smallest
digit to result in 8 is 3, the largest digit to result in 4
is also 3, that is the only digit which can result in all of
the digits 4-5-6-7-8 is 3, so that the cipher key for this
column is 13. It cannot be anything else.
Alphabet 2: The tens-half of the second column ranges over
the full five digits 4-5-6-7-8 (key 3), and the units-half
ranges over 5-6-7-8-9 (key 4). This suggests the key digit
is 34.
Alphabet 3: The tens-half of the third column contains the
'giveaway' digit of 2 and the units-half also contains the
digit 2. The key digit to produce this situation is 11.
Alphabet 4: The tens-half of the fourth column ranges only
over the digits 5-6-7-8, with nothing to indicate whether the
missing digit is 4 or 9. The key might be either 3 or 4.
The units has the full range of digits 5-6-7-8-9, hence key =
4. So we have either 34 o 44 for our key digit. The normal
square suggests COAO or COAT as the key word. We use Table
12-1 to good advantage and decipher this cryptogram.
We decipher the whole cryptogram a column at a time:
'C' 'O' 'A' 'T'
-- -- -- --
A M I N
I S T E
R A T T
E M P T
I N G E
U L O G
Y I N A
F U N E
R A L S
E R M O
M W E H
A V E H
E R E O
N L Y T
H E S H
E L L T
H E N U
T I S G
O N E
Reads: A minister attempting eulogy in a funeral sermon: We
have here only the shell, the nut has gone.
For the most difficult case presenting multiple key
possibilities, we line up the alphabets graphically against
their frequency counts to eliminate the extra key digits.
GROMARK
MASTERTON describes a cipher called the GROMARK. The Gromark
is akin to the GRONSFELD in that the components never change
their position relative to each other and every plain text
values has 10 possible cipher representatives. The GROMARK
uses a different keying method; encipherment is effected by
means of a normal alphabet plain set against a mixed cipher
text alphabet. However, instead of cycles or predictable
slides of the cipher component, one finds the plain value on
the top (normal) component and counts a specified number of
positions to the right, then takes the letter in the cipher
alphabet immediately below. The choice of how far to count
along the sequence is determined by the digital key. One
essentially is adding 0 to 9 to the plain value, as in the
Gronsfeld, but it is on the mixed sequence, set underneath a
plain sequence. The key is derived from a Fibonacci series.
On some cycle (frequently 5 wide) the key is derived from a
starting group, by adding the first position to the second
and placing the result in the sixth position. Similarly,
positions 2 and 3 are added to make position number 7, 3, and
4 to make 8, and so forth. All additions are non carrying -a
very common cryptographic practice. [MAST]
Example:
Use the starter or "seed" of 48671, the key is:
48671 24383 67119 382021 ...
Solution follows the normal Viggy methods. The crib
placement can be interesting.
Example:
7 7 2 6 6 4 9 8 2 0 3 7 0 2 3 0 7 2 5 3 7 9 7
J C N W Z Y C A C J N A Y N L Q P W W S T W P
without knowing the cipher sequence, we are given the crib
SUBSTITUTES and runs somewhere from the J to the final P
above.
Since the plain sequence is normal, a repeated cipher letter,
with different key letters on it, must stand for plain values
removed from each other exactly by the difference of the two
numbers. Thus C A C with keys 9 8 2 above it implies that
the first cipher C is M for example, the second C is seven
positions to the right on the plain sequence, or T.
Or:
J K L M N O P Q R S T U V W X
C
*
We prepare a difference table. We are looking for a
favorable case where the differences in the cipher repeats
matches the plain differences, at the correct interval.
To match these differences, we measure them in one direction
for the plain and the reverse for the cipher. Table 12-1
shows subtraction of the left hand letter from the right, and
we must look at the cipher in the other direction.
Differences may be calculated modulo 26.
Table 12-1
adjacent 19 21 2 19 20 9 20 21 20 5 19
diff's S U B S T I T U T E S
xx 2 7 17 1 15 11 1 25 11 14
x-x 9 24 18 16 0 12 0 10
x--x 0 25 7 ...
There is a difference of 7 with the C-C hit, but it doesn't
appear on the second row of the table. The keyword must
first between A (between C's) and W.
7 7 2 6 6 4 9 8 2 0 3 7 0 2 3 0 7 2 5 3 7 9 7
J C N W Z Y C A C J N A Y N L Q P W W S T W P
S U B S T I T U T E S
This is a good tip placement and confirmed by the N-N hit.
The A---A in the cipher matches the S---T plain. We build
the cipher component by writing the cipher component, and a
normal alphabet, count along it from any given plain the
number of steps given by the key, then write the cipher
value. Find S on the top strip, count 8 to right, place an
A. C is two spaces to the right of the position held by the
U, and so on. Decipher other letters by counting backwards
the number of steps given by the key. Cipher C ahead of thew
crib translates to N.
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
A J Y P Q W N C L
Without a tip the system will fall to statistics. The numbers
associated with any given cipher letter represent a stretch
of 10 consecutive values along a normal alphabet such as C to
L or X to G, we could prepare a table with A to Z as the rows
and 9 to 0 as the columns. Frequencies can be combined
and a stretch such as PQRST area will show as the normal.
The backwards normal sequence yields a bar graph of the
segment of the normal alphabetic frequencies.
DECIMATION PROCESSES - FURTHER REMARKS
In Lecture 11, we presented QUAGMIRES I-IV and solved them by
a variety of methods. Inherent in their solution was
Friedman's principle of indirect symmetry. [FRE7] Prima
facie to this symmetry principle is a process of alphabet
dissociation called Decimation. This same process effects
all Viggy class ciphers and is important from a theoretical
point of view. Decimation is especially effective in solving
mixed alphabet systems like the Quagmire III & IV.
Decimation is a process of selection and derivation of a
sequence of equivalent components according to some fixed
interval. For example, the sequence A E I M is derived by
decimation of extracting every fourth letter from a normal
alphabet.
Consider the two mixed alphabets in a QUAGMIRE III:
O1
* *
Plain: QUESTIONABLYCDFGHJKMPRVWXZ
Cipher: QUESTIONABLYCDFGHJKMPRVWXZQUESTIONABLYCDFGHJKMPRVWXZ
* *
Ok
By setting the two sliding components against each other in
the two positions shown: A in the first set and B in the
second set we can derive two, we can derive two different
sets of secondary alphabets based on the key letters.
O1 * *
Plain: QUESTIONABLYCDFGHJKMPRVWXZ
Cipher: QUESTIONABLYCDFGHJKMPRVWXZQUESTIONABLYCDFGHJKMPRVWXZ
* *
Ok
Secondary Alphabet (1)
Plain: A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
Cipher: H J P R L V W X D Z Q K U G F E A S Y C B T I O M N
Secondary Alphabet (2)
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: J K R V Y W X Z F Q U M E H G S B T C D L I O N P A
Sliding strips will yield the same results as a Viggy type
table based on the Keyword QUESTIONABLY (see a partial table
in Table 11-2.
Table 12-2
Partial Reconstruction
QUESTIONABLYCDFGHJKMPRVWXZ
UESTIONABLYCDFGHJKMPRVWXZQ
ESTIONABLYCDFGHJKMPRVWXZQU
STIONABLYCDFGHJKMPRVWXZQUE
TIONABLYCDFGHJKMPRVWXZQUES
IONABLYCDFGHJKMPRVWXZQUEST
ONABLYCDFGHJKMPRVWXZQUESTI
NABLYCDFGHJKMPRVWXZQUESTIO
ABLYCDFGHJKMPRVWXZQUESTION
BLYCDFGHJKMPRVWXZQUESTIONA
LYCDFGHJKMPRVWXZQUESTIONAB
YCDFGHJKMPRVWXZQUESTIONABL
CDFGHJKMPRVWXZQUESTIONABLY
. .
Superficially secondary alphabets (1) and (2) show no
resemblance of symmetry despite the fact that they were both
created from the same primary alphabet. We do find a Latent
Symmetry Of Position (aka Indirect Symmetry of Position).
This phenomenon has widespread use in the Viggy family.
Consider alphabet (2):
Secondary Alphabet (2)
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: J K R V Y W X Z F Q U M E H G S B T C D L I O N P A
We construct a chain of alternating plaintext and ciphertext
equivalents, beginning at any point and continuing until the
chain is completed. We start Aplain = Jcipher, Jplain =
Qcipher, Qplain = Bcipher...., dropping the common letters
we have A J Q B. The complete sequence of letters is:
A J Q B K U L M E Y P S C R T D V I F W O G X N H Z
When slid against itself it will produce exactly the same
secondary alphabets as do the primary components based upon
the word QUESTIONABLY. For example, compare the secondary
alphabets given by the two settings of the externally
different components below:
* *
Plain: QUESTIONABLYCDFGHJKMPRVWXZ
Cipher: QUESTIONABLYCDFGHJKMPRVWXZQUESTIONABLYCDFGHJKMPRVWXZ
* *
Secondary Alphabet (1)
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: J K R V Y W X Z F Q U M E H G S B T C D L I O N P A
* *
Plain: AJQBKULMEYPSCRTDVIFWOGXNHZ
Cipher: AJQBKULMEYPSCRTDVIFWOGXNHZAJQBKULMEYPSCRTDVIFWOGXNHZ
* *
Secondary Alphabet (2)
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: J K R V Y W X Z F Q U M E H G S B T C D L I O N P A
Since the sequence A J Q B K ... gives exactly the same
equivalents in the secondary alphabets as does the sequence
QUEST......XZ, the former is cryptographically equivalent to
the latter sequence. For this reason the A J Q B K ..
sequence is termed an equivalent primary component. If the
real or original primary component is a keyword mixed
sequence, it is hidden or latent within the equivalent
primary sequence; it can also be made patent by the process
of decimation of the equivalent primary component.
Friedman in [FRE7] describes the process as follows: find
three letters in the equivalent primary component that are a
likely unbroken sequence in the original primary component,
and see if the interval between the first and second is the
same as that of the second and third. Try X, Y, Z in the
equivalent primary component above. Note the sequence ..W O
G X N H Z...; the distance or interval between W X Z is three
letters. Continuing the chain by adding letters three
intervals removed, the latent original primary component is
made patent.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 W
X Z Q U E S T I O N A B L Y C D F G H J K M
24 25 26
P R V
KEYWORD - MIXED SEQUENCE
We can combine the previous steps into one operation.
Starting with any pair of letters in the cipher component of
the secondary alphabets, likely to be sequent in the keyword-
mixed sequence, such as JK, the following chains of digraphs
may be produced. Thus JK plain stand over QU cipher
respectively, QU in the plain stand over BL in the cipher,
respectively, etc. Connecting the pairs:
JK>QU>BL>KM>UE>LY>MP>ES>YC>PR>ST>CD>RV>TI>DF>VW>IO>FG>WX>
ON>GH>XZ>NA>HJ>ZQ>AB>JK.....
We then unite by common letters:
JK>KM>MP>PR>RV>VW>WX>XZ>ZQ>QU>UE>ES>ST>TI>IO>ON>NA>
AB>BL>LY>YC>CD>DF>FG>GH>HJ>JK.....
or:
JKMPRVWXZ-QUESTIONABLY-CDFGH
HALF CHAINS
Only 12 /26 alphabets will yield a complete equivalent
primary component, as shown above. Even number of intervals
for sliding the alphabets will yield half chains or 13 letter
chains. Friedman [FRE7] describes several methods to combine
the half chains into fully equivalent primary components.
FRIEDMAN'S OBSERVATIONS
Friedman observed that in the case of a 26-element component
sliding against itself (both components proceeding in the
same direction), it is only the secondary alphabets resulting
from odd-interval displacements of the primary components
which permit reconstructing a single 26-letter chain of
equivalents. This is true except for the 13th interval
displacement, which acts like an even number displacement, in
that no complete chain of equivalents can be established from
the secondary alphabet. Friedman states the general rule as:
any displacement interval which has a factor in common with
the number of letters in the primary sequence will yield a
secondary alphabet from which no complete chain of 26
equivalents can be derived for the construction of a complete
equivalent primary component. Components sliding in opposite
directions act as a 13 interval displacement because of their
reciprocal nature.
Friedman concluded that whether or not a complete equivalent
primary component is derivable by decimation from an original
primary component (and if not, the lengths and numbers of
chains of letters, or incomplete components, that can be
constructed in attempts to derive such equivalent components)
will depend upon the number of letters in the original
primary component and the specific decimation interval
selected. [FRE7] Friedman constructed a table relating the
number of characters in the original primary component,
decimation interval and total number of complete sequences
that can be formed. See Table 12-3.
TABLE 12-3
Number of Characters in Original Primary Component
Decimation Interval 32 30 28 27 26 25 24 22 21 20
18 16
----------------------------------------------
2 16 15 14 27 13 25 12 11 21 10 9 8
3 32 10 28 9 26 25 8 22 7 20 6 16
4 8 15 7 27 13 25 6 11 21 5 9 4
5 32 6 28 27 26 5 24 22 21 4 18 16
6 16 5 14 9 13 25 4 11 7 10 3 8
7 32 30 4 27 26 25 24 22 3 20 18 16
8 4 15 7 27 13 25 3 11 21 5 9 2
9 32 10 28 3 26 25 8 22 7 20 2 16
10 16 3 14 27 13 5 12 11 21 2 9 8
11 32 30 28 27 26 25 24 2 21 20 18 16
12 8 5 7 9 13 25 2 11 7 5 3 4
13 32 30 28 27 2 25 24 22 21 20 18 16
14 16 15 2 27 13 25 12 11 3 10 9 8
15 32 2 28 9 26 5 8 22 7 4 6
16 2 15 7 27 13 25 3 11 21 5 9
17 32 30 28 27 26 25 24 22 21 20
18 16 5 14 3 13 25 4 11 7 10
19 32 30 28 27 26 25 24 22 21
20 8 3 7 27 13 5 6 11
21 32 10 4 9 26 25 8
22 16 15 14 27 13 25 12
23 32 30 28 27 26 25
24 4 5 7 9 13
25 32 6 28 27
26 16 15 14
27 32 10
28 8 15
29 32
30 16
Total Number
Of
Sequences 14 6 10 16 10 18 6 8 10 6 4 6
>From Table 12-3, we see that in a 26-letter original primary
component, decimation interval 5 will yield a complete
equivalent primary component of 26 letters, whereas
decimation intervals of 4 or 8 will yield 2 chains of 13
each. In a 24-letter component, decimation interval 5 will
also yield a complete equivalent primary component of 24
letters, but decimation interval 4 will yield 6 chains of 4
letters each, and decimation interval 8 will yield 3 chains
of 8 letters each.
It follows that in the case of an original primary component
in which the total number of characters is a prime number,
all decimation intervals will yield complete equivalent
primary components. Table 12-3 omits the prime number
sequences from 16-32. [FRE7]
SPECIAL SOLUTIONS FOR PERIODIC CIPHERS
Special circumstances give rise atypical solutions of
periodic ciphers. We shall look at four special cases:
1) isologs, 2) 'stagger', 3) long latent repetition and 4)
superimposition.
ISOLOGS
Recall that an Isolog is defined as the exact same plain text
message enciphered by two different keys in the same
cryptosystem. Lets use two monoalphabetic substitution
systems to illustrate the point. Assume two messages are
intercepted going from station A to B. B had called for a
retransmit because of some error in transmission. We suspect
the messages are the same plaintext content and they both
have the same length. We superimpose one message over the
other:
1. NXGRV MPUOF ZQVCP VWERX QDZVX WXZQE TBDSP VVXJK RFZWH 2.
EMLHJ FGVUB PRJNG JKWHM RAPJM KMPRW ZTAXG JJMCD HBPKY
chaining from 1 to 2: NE>EW>WK>KD>DA ......
1. ZUWLU IYVZQ FXOAR
2. PVKIV QOJPR BMUSH
Next we initiate a chain of ciphertext equivalents (reducing
the common letter) from message 1 to message 2, yielding:
*
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 N
E W K D A S X M F B T Z P G L I Q R H Y O U
* * * * *
24 25 26
V J C
With some experimentation, we find the Key word QUESTIONABLY
and the decimation interval of +5 Modulo 26. The complete 26
letter chain was available for reconstruction, but this is
not a requirement.
Why is it possible to reconstruct the primary component and
solve the above two messages without having any plain text at
all? Since the plain text of both messages is the same, the
relative displacement of the same primary components in the
case of message 1 differs from the relative displacement of
the same primary components in message 2 by a FIXED interval.
Therefore, the distance between N and E (1st two cipher
letters of the two messages) on the primary component,
regardless of what plaintext letter these two cipher letters
represent, is the same distance between E and W (18th
letters), W and K (17th letters), and so forth. Thus this
fixed interval permits the establishing of a complete chain
of letters separated by constant intervals and this chain
becomes an equivalent primary component.
To solve, we take the frequency distributions of message 1
and 2:
E S T I O
1 1 1 2 2 3 1 1 1 1 1 1 1 1 2 3 4 4 1 1 3 7 4 6 1 6
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
E S T I O
2 3 1 1 1 1 3 4 1 7 4 1 6 1 1 7 1 4 1 1 2 3 2 1 1 1
2: 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
We set up two key word mixed alphabets and slide against each
other. With some trial and error we find:
NABLYCDFGHJKMPRVWXZQUESTIO
QUESTIONABLYCDFGHJKMPRVWXZ
The plain text reads: Five squadrons must be in position by H
plus six zero two at Jackson Ridge.
The same procedure is applied on two repeating key ciphers
suspected of being Isologs:
Message 1
YHYEX UBUKA PVLLT ABUVV DYSAB PCQTU
NGKFA ZEFIZ BDJEZ ALVID TROQS UHAFK
Message 2
CGSLZ QUBMN CTYBV HLQFT FLRHL MTAIQ
ZWMDQ NSDWN LCBLQ NETOC VSNZR BJNOQ
The first step is to find the length of the period. The
usual method fails for lack of long repetitions and the
digraphs are not promising. We use the Principle of
Superimposition to get a hold on the period for both
cryptograms.
1 2 3 4 5 6 7 8 9101112131415161718192021222324252627282930
Y H Y E X U B U K A P V L L T A B U V V D Y S A B P C Q T U
C G S L Z Q U B M N C T Y B V H L Q F T F L R H L M T A I Q
313233343536373839404142434445464748495051525354555657585960
N G K F A Z E F I Z B D J E Z A L V I D T R O Q S U H A F K
Z W M D Q N S D W N L C B L Q N E T O C V S N Z R B J N O Q
We employ a subterfuge will be employed based upon the theory
of factoring. We search for cases of identical
superimposition. We have:
4 44 6 18 30
E and E are separated by 40 letters, U, U and U which
L L Q Q Q
are separated by 12 letters. We factor these intervals as if
they were ordinary repetitions. The most frequent factor
should correspond to the period. We are dealing with
Isologs. The plain text is the same in both messages, so the
principle of identity of superimposition can only be the
result of identity of encipherments by identical cipher
alphabets. The same relative position in the keying cycle
has been reached in both cases of the identity. The distance
between identical superimpositions must be equal to or a
multiple of the length of the period. The following is the
complete set of superimposed pairs:
Repetition Interval Factors
--------------------------------------------
EL - EL 40 2,4,5,8,10,20
UQ - UQ -UQ 12 2,3,4,6
UB - UB 48 2,3,4,6,,8,12,24
KM - KM 24 2,3,4,6,12
AN -AN -AN 36/12 2,3,4,6;9,12,18
VT -VT -VT 8/28 2,4; 2,4,7,14
TV - TV 36 2,3,4,6,9,12,18
AH - AH 8 2,4
BL -BL -BL 8/16 2,4,;8
SR - SR 32 2,4,8,16
FD - FD 4 2
ZN - ZN 4 2
DC - DC 8 2, 4
------------------------------------------------
Only the factors 2 and 4 are common. We discard 2 as
improbable. We break up the message into groups of four.
1234 1234 1234 1234 1234 1234 1234 1234
1. YHYE XUBU KAPV LLTA BUVV DYSA BPCQ TUNG 2. CGSL ZQUB
MNCT YBVH LQFT FLRH LMTA IQZW
* * * *
1234 1234 1234 1234 1234 1234 1234
1. KFAZ EFIZ BDJE ZALV IDTR OQSU HAFK
2. MDQN SDWN LCBL QNET OCVS NZRB JNOQ
We develop a decipherment Tableaux:
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
------------------------------------------------------
1 L F S J O M Y N I Z C Q
2 N C D G B M Z Q L
3 Q U T O W B E Z C R V F S
4 H L W Q A S B T N
------------------------------------------------------
Using the meyhods previously described, we build up the
equivalent primary component and combine our digrams.
BL, DF, ES, HJ, IO, KM, LY, ON,TI, XZ, YC, ZQ.
BLYC .DF TION XZQ(U) [ES]TION(A)BLY CDF (G) H
JKM(P) (R) (V) XZ
It is not a long jump to a key word QUESTIONABLY and the
equivalent primary component:
Q U E S T I O N A B L Y C D F G H J K M P R V W X Z
The fact that the original primary component was exposed was
pure chance, it could have been an equivalent primary
sequence alphabet.
>From here we apply the completion of the plain-component
sequence using the high frequency letter assortments.
For the first message:
Gen Alphabet 1 Alphabet 2 Alphabet 3 Alphabet 4
1 YXKLBDBTKE 1HUALUYPUFF 5YBPTVSCNAI EUVAVAQGZZ
2 2CZMYLFLIMS 4JEBYECREGG 5CLRIWTDABO SEWBWBUHQQ
3 2DQPCYGYOPT 3KSLCSDVSHH 3DYVOXIFBLN TSXLXLEJUU
4 4FURDCHCNRI MTYDTFWTJJ 3FCWNZOGLYA ITZYZYSKEE
5 3GEVFDJDAVO PICFIGXIKK GDXAQNHYCB OIQCQCTMSS
6 2HSWGFKFBWN 4RODGOHZOMM HFZBUAJCDL 5NOUDUDIPTT
7 JTXHGMGLXA VNFHNJQNPP JGQLEBKDFY 8ANEFEFORII*
8 KIZJHPHYZB WAGJAKUARR 1KHUYSLMFGC 6BASGSGNVOO
9 MOQKJRJCQL XBHKBMEBVV 2MJECTYPGHD 5LBTHTHAWNN
10 PNUMKVKDUY ZLJMLPSLWW PKSDICRHJF YLIJIJBXAA
11 4RAEPMWMFEC QYKPYRTYXX RMTFODVJKG CYOKOKLZBB
12 3VBSRPXPGSD UCMRCVICZZ 2VPIGNFWKMH 2DCNMNMYQLL
13 4WLTVRZRHTF EDPVDWODQQ WROHAGXMPJ 2FDAPAPCUYY
14 XYIWVQVJIG 3SFRWFXNFUU XVNJBHZPRK 3GFBRBRDECC
15 ZCOXWUWKOH TGVXGZAGEE ZWAKLJQRVM 1HGLVLVFSDD
16 QDNZXEXMNJ IHWZHQBHSS QXBMYKUVWP 1JHYWYWGTFF
17 UFAQZSZPAK OJXQJULJTT UZLPCMEWXR KJCXCXHIGG
18 EGBUQTQRBM NKZUKEYKII EQYRDPSXZV MKDZDZJOHH
19 3SHLEUIUVLP 5AMQEMSCMOO SUCVFRTZQW PMFQFQKNJJ
20 6TJYSEOEWYR? 4BPUSPTDPNN TEDWGVIQUX RPGUGUMAKK
21 IKCTSNSXCV 8LRETRIFRAA* ISFXHWOUEZ 3VRHEHEPBMM
22 5OMDITATZDW? 3YVSIVOGVBB OTGZJXNESQ WVJSJSRLPP
23 NPFOIBIQFX 3CWTOWNHWLL NIHQKZASTU XWKTKTVYRR
24 5ARGNOLOUGZ? DXINXAJXYY AOJUMQBTIE ZXMIMIWCVV
25 4BVHANYNEHQ FZOAZBKZCC 5BNKEPULIOS QZPOPOXDWW
26 LWJBACASJU GQNBQLMQDD 7LAMSREYONT* UQRNRNZFXX
We choose generatrices 20/22/24; 21; 26; 7 because of the
highest two category scores. it is not much of a jump to
find Alphabet 1 generatrix as alphabet 24:
1 2 3 4
A L L A
R R A N
G E M E
N T S F
O R R E
L I E F
O F Y O
U R O R
G A N I
Z A T I
>From a Vigenere Square (Figure 12-1) based on the keyword
QUESTIONABLY, we find the key words SOUP for message 1 and
TIME for message 2.
S O U P S O U P S O U P S O U P S O U P S O U P
----------------------------------------------------
Y H Y E X U B U K A P L L L T A B U V V D Y S A
A L L A R R A N G E M E N T S F O R R E L I E F
B P C Q T U N G K F A Z E F I Z B D J E Z A L V
O F Y O U R O R G A N I Z A T I O N H A V E B E
I D T R O Q S U H A F K
E N S U S P E N D E D X
T I M E T I M E T I M E T I M E T I M E T I M E
____________________________________________________
C G S L Z Q U B M N C T Y B V H L Q F T F L R H
A L L A R R A N G E M E N T S F O R R E L I E F
L M T A I Q Z W M D Q N S D W N L C B L Q N E T
O F Y O U R O R G A N I Z A T I O N H A V E B E
O C V S N Z R B J N O Q
E N S U S P E N D E D X
Figure 12-1
Q U E S T I O N A B L Y C D F G H J K M P R V W X Z
U E S T I O N A B L Y C D F G H J K M P R V W X Z Q
E S T I O N A B L Y C D F G H J K M P R V W X Z Q U
S T I O N A B L Y C D F G H J K M P R V W X Z Q U E
T I O N A B L Y C D F G H J K M P R V W X Z Q U E S
I O N A B L Y C D F G H J K M P R V W X Z Q U E S T
O N A B L Y C D F G H J K M P R V W X Z Q U E S T I
N A B L Y C D F G H J K M P R V W X Z Q U E S T I O
A B L Y C D F G H J K M P R V W X Z Q U E S T I O N
B L Y C D F G H J K M P R V W X Z Q U E S T I O N A
L Y C D F G H J K M P R V W X Z Q U E S T I O N A B
Y C D F G H J K M P R V W X Z Q U E S T I O N A B L
C D F G H J K M P R V W X Z Q U E S T I O N A B L Y
D F G H J K M P R V W X Z Q U E S T I O N A B L Y C
F G H J K M P R V W X Z Q U E S T I O N A B L Y C D
G H J K M P R V W X Z Q U E S T I O N A B L Y C D F
H J K M P R V W X Z Q U E S T I O N A B L Y C D F G
J K M P R V W X Z Q U E S T I O N A B L Y C D F G H
K M P R V W X Z Q U E S T I O N A B L Y C D F G H J
M P R V W X Z Q U E S T I O N A B L Y C D F G H J K
P R V W X Z Q U E S T I O N A B L Y C D F G H J K M
R V W X Z Q U E S T I O N A B L Y C D F G H J K M P
V W X Z Q U E S T I O N A B L Y C D F G H J K M P R
W X Z Q U E S T I O N A B L Y C D F G H J K M P R V
X Z Q U E S T I O N A B L Y C D F G H J K M P R V W
Z Q U E S T I O N A B L Y C D F G H J K M P R V W X
SOLUTION OF ISOLOGS INVOLVING THE SAME SET OF PRIMARY
COMPONENTS BUT WITH KEY WORDS OF DIFFERENT LENGTHS
The example previous had two keywords the same lengths.
The Method of Superimposition works with Keywords of
different lengths. Friedman works an interesting example:
Message 1
VMYZG EAUNT PKFAY JIZMB UMYKB VFIVV
SEOAF SKXKR YWCAC ZORDO ZRDEF BLKFE
SMKSF AFEKV QURCM YZVOX VABTA YYUOA
YTDKF ENWNT DBQKU LAJLZ IOUMA BOAFS
KXQPU YMJPW QTDBT OSIYS MIYKU ROGMW
CTMZZ VMVAJ
Message 2
ZGANW IOMOA CODHA CLRLP MOQOJ EMOQU
DHXBY UQMGA UVGLQ DBSPU OABIR PWXYM
OGGFT MRHVF GWKNI VAUPF ABRVI LAQEM
ZDJXY MEDDY BOSVM PNLGX XDYDO PXBYU
QMNKY FLUYY GVPVR DNCZE KJQOR WJXRV
GDKDS XCEEC.
Both messages permit factoring at periods of 4 and 6 letters,
respectively. Superimposing the two messages and marking the
position of each letter in the corresponding period, we have:
12341 23412 34123 41234 12341 23412
No. 1 VMYZG EAUNT PKFAY JIZMB UMYKB VFIVV
No. 2 ZGANW IOMOA CODHA CLRLP MOQOJ EMOQU
12345 61234 56123 45612 34561 23456
34123 41234 12341 23412 34123 41234
No. 1 SEOAF SKXKR YWCAC ZORDO ZRDEF BLKFE
No. 2 DHXBY UQMGA UVGLQ DBSPU OABIR PWXYM
12345 61234 56123 45612 34561 23456
12341 23412 34123 41234 12341 23412
No. 1 SMKSF AFEKV QURCM YZVOX VABTA YYUOA
No. 2 OGGFT MRHVF GWKNI VAUPF ABRVI LAQEM
12345 61234 56123 45612 34561 23456
34123 41234 12341 23412 34123 41234
No. 1 YTDKF ENWNT DBQKU LAJLZ IOUMA BOAFS
No. 2 ZDJXY MEDDY BOSVM PNLGX XDYDO PXBYU
12345 61234 56123 45612 34561 23456
12341 23412 34123 41234 12341 23412
No. 1 KXQPU YMJPW QTDBT OSIYS MIYKU ROGMW
No. 2 QMNKY FLUYY GVPVR DNCZE KJQOR WJXRV
12345 61234 56123 45612 34561 23456
34123 41234
No. 1 CTMZZ VMVAJ.
No. 2 GDKDS XCEEC.
12345 61234
What is neat about this superimposition is that we can
establish secondary alphabets by distributing the letters
from the 12 different superimposed pairs of numbers.
The 1 - 1 superimposition is placed in the tableau at the
0 - 1 row, column in the tableaux.
0 1 2 3 4 5 6 7 8 91011121314151617181920212223242526
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
---------------------------------------------------
1-1 I J P D Q G C E K O R Z
2-2 H V N G U W E D M L X
3-3 E M X G I D J N R A O
4-4 X O C D K A F Y Q V N
1-5 B T W L R E M N Y U A
2-6 M O I C D U V F R
3-1 O G R L P S D Z
4-2 L P H U V E D M F
1-3 Q J V W K O X Y M A
2-4 B J X P O A F Y D
3-5 N R Y B C G Q S
4-6 M L O S U V W X
---------------------------------------------------
We construct the complete equivalent primary component:
1 2 3 4 5 6 7 8 91011121314151617181920212223242526
I T K N P Z H M W B Q E U L F C S J A X R G D V O Y
Ok. We have the cipher component. Is it normal? reversed?
Mixed? Same questions for the plain component sequence.
We assume that the primary plain component is normal direct
sequence. We attempt to solve and fail. Normal reverse will
also fail. We assume a K3 situation, i.e. the plain and
cipher components are identical. Again the test fails. We
assume that the plain is in reverse mode. Nope. So we have a
K4 situation, both primary components are different mixed
sequences.
Message 1 transcribed into periods of four letters.
Message 1
VMYZ GEAU NTPK FAYJ IZMB UMYK BVFI VVSE
OAFS KXKR YWCA CZOR DOZR DEFB LKFE SMKS
FAFE KVQU RCMY ZVOX VABT AYYU OAYT DKFE
NWNT DBQK ULAJ LZIO UMAB OAFS KXQP UYMJ
PWQT DBTO SIYS MIYK UROG MWCT MZZV MVAJ
The Uniliteral frequency distributions for the four secondary
alphabets are shown in 1A -4A. We have the reconstructed
cipher alphabet, 1B-4b shows the sequences rearranged.
1 1 1 5 2 1 1 3 2 4 2 3 1 1 2 5 3 1 1
1A 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 2 1 2 2 2 1 4 1 1 1 5 4 2 2 4
2A 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
4 1 2 7 1 2 3 1 3 1 4 1 1 7 2
3A 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 3 4 1 4 4 2 1 3 4 5 3 1 1 1 1
4A 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 3 2 1 1 4 1 5 2 2 1 2 1 1 1 5 3 3 1
1B I T K N P Z H M W B Q E U L F C S J A X R G D V O Y
2 1 2 4 4 3 2 2 1 1 6 2 1 5 1 2
2B I T K N P Z H M W B Q E U L F C S J A X R G D V O Y
1 1 2 1 1 2 3 1 4 7 2 1 4 3 7
3B I T K N P Z H M W B Q E U L F C S J A X R G D V O Y
1 5 4 1 1 3 4 3 4 4 1 1 3 1 1 2 1
4B I T K N P Z H M W B Q E U L F C S J A X R G D V O Y
We now shift 1B-4B for superimposition and combine the
distributions. The latter distributions may be combined so
as to yield a single monoalphabetic distribution for the
entire message. In other words, the polyalphabetic message
can be converted into monoalphabetic terms, and thereby
simplifying the situation considerably.
1 3 2 1 1 4 1 5 2 2 1 2 1 1 1 5 3 3 1
1B I T K N P Z H M W B Q E U L F C S J A X R G D V O Y
2 1 1 6 2 1 5 1 2 2 1 2 4 3 2
2B E U L F C S J A X R G D V O Y I T K N P Z H M W B Q 2 1 1
2 3 1 4 7 2 1 4 3 7
3B K N P Z H M W B Q E U L F C S J A X R G D V O Y I T
1 1 3 4 3 4 4 1 1 3 1 1 2 1 1 5 4
4B P Z H M W B Q E U L F C S J A X R G D V O Y I T K N
6 2 5 4 2 7 15 9 2 21 9 6 410 3 1 1 7 2 918 9 1
1B-4B I T K N P Z H M W B Q E U L F C S J A X R G D V O Y
combinedH M L R S O A I Y N E T
Plain
Equiv's
I have converted 2B-4B into terms of 1B. The 2 E's of 2B add
to 1B I. The two K's of alphabet 3 becomes I's and the N
becomes a T, and so forth. We solve the monoalphabetic
cipher.
12341 23412 34123 41234 12341 23412
ENEMY HASCA PTURE DHILL ONETW OONEO
VDVTG ISWNZ KOFMV LIRZZ UDVOB UUDVU
URTRO OPSHA VEDUG INAND CANHO LDFOR
FMOMU UKWIS YVLFC RDSDL NSDIU ZLJUM
ANHOU RORPO SSIBL YLONG ERREQ UESTR
SDIUF MUMKU WWRPZ GZUDC VMMVA FVWOM
EINFO RCEME NTSTO PADDI TIONA LTROO
VVDJU MNVTV DOWOU KSLLR ORDUS ZOMUU
PSSHO ULDBE SENTV IAGEO RGETO WNFRE
KWWIU FZLPV WVDOY RSCVU MCVOU BDJMV
DERIC KROAD.
LVMRN XMUSL.
Having the plain text, the derivation of the plain or
equivalent plain component is straightforward. We may base
the reconstruction upon any of the secondary alphabets, since
the plaintext - ciphertext relationship is known directly,
and the primary cipher component is at hand. So:
1 2 3 4 5 6 7 8 9 1011121314151617181920212223242526
H M P C B L . R S W . . O D U G A F Q K I Y N E T V
with Key words of STAR and OCEANS for messages 1 and 2.
NECESSARY AND SUFFICIENT CONDITIONS FOR SUPERIMPOSITION AND
CONVERSION TO MONOALPHABETIC TERMS
This example shows the power of the method of superimposition
and conversion of a polyalphabetic cipher to monoalphabetic
terms. This conversion is possible because the sequence of
letters forming the cipher component has been reconstructed
and was known, and the uniliteral distributions for the
respective secondary cipher alphabets could theoretically be
shifted to correct superimpositions for monoalphabeticity.
The data was sufficient to give proper indications for
alignment of the alphabets and relative displacements. The
chi test could also have been brought to bear to match
columns. The above constitutes the necessary and sufficient
conditions to convert theory to actuality.
SOLUTION OF ISOLOGS INVOLVING DIFFERENT PAIRS OF UNKNOWN
PRIMARY COMPONENTS
The principle of superimposition continues to work for us
even when the primary components are different, and the
repeating keys are of different lengths.
There are two general attacks. The first is a slight
modification of the procedures previously discussed. We first
factor the messages, then superimpose the messages on a width
of the least common multiple, then create a reconstruction
matrix based on the cipher values. We must limit our
observations to within the matrix, because the given messages
are different and therefore the indirect symmetry does not
extend to the 0 or assumed plain line. The wrinkle in the
fabric is we must restrict our observations to a homogeneous
set of lines, like 1-1,1-2,1-3,1-4 etc. From this data, we
reduce the reconstruction matrix to a smaller set and solve
for the equivalent primary component. It is possible to
invert the matrix so that values for the second message will
yield its equivalent primary component.
ARBITRARY REDUCTION METHOD
It is not necessary to recognize the plain text to solve a
problem involving Isologs. The next cryptanalytic attack is
applicable for many types of ciphers. The procedure exposes
latent letter relationships and reduces the imposed chaos of
the cryptogram. Given:
Message 1
BWXPS OBYII UYHLF KFSOP VGEYW PBVXO
UGJPB WDXUG HSWDH KHKHC UAYKP NFSPD
OBBYB INKFL WABOX PJXUV WKFXR WXYWS
SDYZQ ZHETA JXXZW XJROS PDEEW OJONK
GIRXR WUYDK NTJWR EVBUR DLISJ BLCKK
FODEV DYZQZ SHCTW DIEXZ
Factoring gives us periods of 4 and 5 for messages 1 and 2,
respectively. We write out the messages on a width of the
least common multiple of 20.
Message 2
JNLEJ HWUAH JHUIV YNCHC HLPKD EWZJJ
JNAHB HZBIM TUBQE FJAKM JVBEF XNCTL
FAAKV KIABG CVFNY FWBIQ GERSA TZUSD
SXBUD SHAWA YXLJD CQLED HXGZL ZWHNB
VTJSA TSUUC MIAKK JEMIY DSKGB VTJYC
XYLZE CXLSU MVMND ONFJY
12341 23412 34123 41234 20
BWXPS OBYII UYHLF KFSOP
JNLEJ HWUAH JHUIV YNCHC
12345 12345 12345 12345
A A A
12341 23412 34123 41234 40
VGEYW PBVXO UGJPB WDXUG
HLPKD EWZJJ JNAHB HZBIM
12345 12345 12345 12345
A A
12341 23412 34123 41234 60
HSWDH KHKHC UAYKP NFSPD
TUBQE FJAKM JVBEF XNCTL
12345 12345 12345 12345
A
12341 23412 34123 41234 80
OBBYB INKFL WABOX PJXUV
FAAKG KIABG CVFNY FWBIQ
12345 12345 12345 12345
A A A A
12341 23412 34123 41234 100
WQFXR WXYWS SDYZQ ZHETA
GERSA TZUSD SXBUD SHAWA
12345 12345 12345 12345
12341 23412 34123 41234 120
JXXZW XJROS PDEEW OJONK
YXLJD CQLED HXGZL ZWHNB
12345 12345 12345 12345
12341 23412 34123 41234 140
GIRXR WUYDK NTJWR EVBUR
VTJSA TSUUC MIAKK JEMIY
12345 12345 12345 12345
A A A
12341 23412 34123 41234 160
DLISJ BLCKK FODEV DYZQZ
DSKGB VTJYC XYLZE CXLSU
12345 12345 12345 12345
A
12341 23412 170
SHCTW DIEXZ
MVMND ONFJY
12345 12345
A
We arbitrarily assign the value of A(plain) as the first
letter of the plain text. Since in message 1, B(cipher)=
A(plain), then every B(cipher) in alphabet 1 must equal
A(plain); these values are entered in the table above. Also
the 65th and 73rd letter of message 1 are A(plain), this
establishes that in message 2, G(cipher) in alphabet 5 and
F(cipher) in alphabet 3 are also A(plain); we enter these
values. Similarly, every J(cipher) in alphabet 1 of message
2 equals A(plain). We continue the process and recover all
the A(plains) of the pseudo-plain text with the resulting
worksheet shown above.
We arbitrarily assign the value of B(plain) to the V(cipher)
at the 21st position of message 1. The other V(cipher) of
message number 1 establishes the E(cipher) of message 2 also
as a B(plain). This procedure of arbitrary assignments is
continued until all the cipher letters of alphabet 1 of
message 1, are placed. we are able to reduce most of the
text to monoalphabetic terms. The worksheet is as follows:
12341 23412 34123 41234 20
BWXPS OBYII UYHLF KFSOP
JNLEJ HWUAH JHUIV YNCHC
12345 12345 12345 12345
ACHDIIFCK ACCA FME D
12341 23412 34123 41234 40
VGEYW PBVXO UGJPB WDXUG
HLPKD EWZJJ JNAHB HZBIM
12345 12345 12345 12345
B CE F LI AMF F BHOAM
12341 23412 34123 41234 60
HSWDH KHKHC UAYKP NFSPD
TUBQE FJAKM JVBEF XNCTL
12345 12345 12345 12345
CEOOC D FCM AJODB MEBO
12341 23412 34123 41234 80
OBBYB INKFL WABOX PJXUV
FAAKG KIABG CVFNY FWBIQ
12345 12345 12345 12345
DGFCA IFMA OJAIH DFOA
12341 23412 34123 41234 100
WQFXR WXYWS SDYZQ ZHETA
GERSA TZUSD SXBUD SHAWA
12345 12345 12345 12345
EB EJ CHCEE LOOHE LCF J
12341 23412 34123 41234 120
JXXZW XJROS PDEEW OJONK
YXLJD CQLED HXGZL ZWHNB
12345 12345 12345 12345
FOHLE O HDE BOPFO FIIF
12341 23412 34123 41234 140
GIRXR WUYDK NTJWR EVBUR
VTJSA TSUUC MIAKK JEMIY
12345 12345 12345 12345
G EJ CACHD IIFC ABGAH
12341 23412 34123 41234 160
DLISJ BLCKK FODEV DYZQZ
DSKGB VTJYC XYLZE CXLSU
12345 12345 12345 12345
HAM F G ND HFC OOHEL
12341 23412 170
SHCTW DIEXZ
MVMND ONFJY
12345 12345
IJGIE MALH
The above table is about 85% reduced and note the idiomorphic
repetition ACHDIIFC representing Artillery becomes patent in
the reduction process. This is rather exciting. From no
patent clues to reduction and latent clues exposed. Clever.
The solution is continued by setting up sequence recon-
struction matrices for both messages. The 0 line represents
the pseudo-plain text and the values inside the matrix being
cipher text.
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
------------------------------------------------------
1 B V H O W J G D S R I X F K Y E
2 L Q W K S E B Z O H C X
3 U P V Q B C X N S I W
4 E W Y P X K R T A Z G D
-------------------------------------------------------
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
------------------------------------------------------
1 J H T F G Y V D M S C
2 S E H U W A Z I V N X
3 F U C A M L H K B G
4 I T K E S Z U N A J B Y Q
5 G F E C D B Y J A U M L
------------------------------------------------------
>From the above we chain out the equivalent primary components
used for each message. Having reconstructed the cipher
component for each message, the alphabets are aligned,
combined and reduced to monoalphabetic terms. After solution
of these messages, we find message 1 is a case of direct
symmetry with the cipher component based on the keyword
HYDRAULIC, and message 2 is a case of indirect symmetry with
both components being keyword-mixed sequences based on our
favorite keyword QUESTIONABLY. Friedman points out that the
keywords are prime to each other (9 vs 11). Primality is not
a necessary condition for solution based on this procedure.
[FRE7]
The method of Arbitrary Reduction is very powerful and works
in other ares besides solving periodic polyalphabetic
ciphers. It represents a workable approach where the
cryptosystem involves nonrelated, random-mixed secondary
alphabets among which no symmetry of any sort exists!
SOLUTION BASED ON INDIRECT SYMMETRY OF A "STAGGER'
Given two messages with group counts nearly identical and two
isologous initial fragments which are identical except by one
letter (called a 'stagger') we can solve the isologous
portions of the messages and recover the primary cipher
component by the process of indirect symmetry. Transmission
garble usually creates stagger messages. Machine cipher
systems sometimes produce these when a word separator is
added. Staggers may be progressively larger as further word
separators are omitted or added.
Given:
Message A
* *
ZFWAY ITBVX XWZQV PEBGS GGFIZ TUAMF
RFEQX PEPPO PCNBP QPOTX VNAIH HVRXC
NHVGM FRFSI ESQMV
*
Message B
* *
ZFWAY ITBVX XWZQV PDRKF USVAG XLJKC
NDVPR OWBRH YFJMS HRFVS BAHWG ZFAJO
JMFAV CNDVD ORZPH A
*
We note that both messages have the same 16 letter beginnings
and that message B is 1 letter longer than message A. Note
that the tetragraphs MFRF (29) and (65) are spaced 1 less
letter than CNDV at (30) and (66). The D in position 17 of
message 2 is the extra letter.
Starting from the E in position 17 of message 1, we
superimpose message one over message 2 starting at the R in
position 18. [We use a period of 6 because the tetragraph
delta equals 36 which factors into 3,4,6 and 9; 6 is
confirmed via the message.]
56123456123456123456123456123456123456123456123456123456123
EBGSGGFIZTUAMFRFEQXPEPPOPCNBPQPOTXVNAIHHVRXCNHVGMFRFSIESQMV
RKFUSVAGXLJKCNDVPROWBRHYFJMSHRFVSBAHWGZFAJOJMFAVCNDVDORZPHA
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
-------------------------------------------------------
1-2 B F Z M P D S X
2-3 S V F H R U L B
3-4 P S H D J A
4-5 K V O H Y R J
5-6 W R A C F O
6-1 K J N G V W Z
-------------------------------------------------------
It is fairly easy to align properly the cipher components
after the primary cipher component or its equivalent have
been recovered, thereby expediting the reduction of the
cipher into monoalphabetic terms. Note that B(cipher) of
alphabet 2 is under E(cipher) of alphabet 1; V(cipher) of
alphabet 3 is under F(cipher) alphabet 2;P(cipher) of
alphabet 4 is under E(cipher) of alphabet 1. From this point
on solution follows the normal path of reconstruction,
keyword recovery and combination of alphabets, reduction to
monoalphabetic terms and solution by frequency analysis.
LONG LATENT REPETITIONS
The stagger procedure applies to a periodic cryptogram which
contains a long passage repeated in its plain text, the
second occurrence occurring at a point in the keying cycle
different from the first occurrence. If the passage is long
enough, the equivalencies from the two corresponding
sequences may be chained together to yield an equivalent
primary component. In effect, we by-pass the solution by
frequency analysis or making assumptions in the plain text of
a polygraphic cipher.
FINAL REMARKS REGARDING SOLUTION BY SUPERIMPOSITION
In solving an ordinary repeating-key cipher the first step,
ascertaining the length of the period, is a relatively minor
consideration. It paves the way for the second step, which
consists of allocating the letters of the cryptogram into
individual monoalphabetic distributions. The third step is to
solve these distributions. The text is transcribed into its
periods and written out in successive lines corresponding to
the length of the period. The columns of letters as a series
belong to the same monoalphabet.
We also can see the letters as transcribed into superimposed
periods; in such a case the letters in each column have
undergone the same kind of treatment by the same elements
(plain and cipher components of the cipher alphabet.)
If we have a case of a very long repeating key and a short
message ( few cycles in the text) we have a difficult
problem. But supposing there were several short cryptograms
enciphered by the same key, each message beginning at
identical starting points in the key. We can superimpose
these messages "in flush depth" or "head on" and know that 1)
the letters in the columns belong to the same individual
alphabets, 2) and that if there are enough messages (about
25-30 in English), then the frequency distributions
applicable to the successive columns of text can be solved -
without knowing the length of the key. Any difficulties that
may have arisen because we were not able to factor the
problem correctly are circumvented. The second step of the
normal solution to the problem is by-passed. The assumption
of probable initial words of messages and stereotyped
beginnings is a powerful method of attack in such situations.
Since the superimposed texts in these cases comprise only the
beginnings of messages, assumptions of probable words are
more easily made than when words are sought in the interior
of the messages. Some common introductory words are REQUEST,
REFER, ENEMY, WHAT, WHEN, and SEND. High frequency initial
digraphs will manifest themselves in the first two columns of
the superimposed diagram. The high frequency RE diagram
manifests itself in such words as REQUEST, REQUIRE,
REFERENCE, REFERRING, REQUISITIONS, REPEAT, RECOMMEND,
REPORT, RECONNAISSANCE, REINFORCEMENTS and perhaps REGIMENT.
(I assume the military text here.)
This same superimposition principle applies even if the
messages start at different initial points, providing the
messages can be correctly superimposed, so that the letters
which fall in one column really belong to one cipher
alphabet. The superimposed messages are said to be "in
depth." The chi test may be used to advantage in finding and
combining columns of the superimposed diagram which were
enciphered by identical keys, thus assisting in the analysis
of frequencies of larger samples than were available before
the amalgamation. [FRE7]
CONCLUSION
In summary, we have seen that the chaining process between
cipher texts applies to the latent characteristics of the
cipher components, regardless of the identity of the plain
components and regardless whether direct or indirect symmetry
is involved in the cryptosystems. The principle of super-
imposition ranks as one of the most important principles of
cryptanalysis. A pretty impressive tool.
LECTURE 11 SOLUTIONS
Thanks to BOZOL for the quick response and correct too!
11.1 Vigenere. Key= SLEEP. "Any reputable physician will
agree..
11.2 Beaufort. Key = SILENCE. "Although every one may not
subscribe to ..
11.3 Variant. Key = IMPSHGXW (HINSNOTI). Because of the
many pressures... [the correct key is SOLITUDE]
11.4 GRONSFELD. 6-3-8-4-0. "Too much discussion, especially..
11.5 BEAUFORT. Key = OCCUPATION. "Almost every man has a
job, many find..
BOZOL reports that the tip did not help him and that the
first pass at the key was ORCUPATMON which he mystically
came up with organization.
LECTURE 12 PROBLEMS
12.1 Nihilist Substitution
74 46 66 44 79 47 45 37 58 66 37 60 25 54 33 69 78 35 68 27
47 36 28 88 36 60 33 48 43 29 87 35 49 57 76 37 37 88 36 60
33 77 74 50 86 55 47 27 76 45 40 55 56 58 66 78 57 30 94 58
38 26 55 57 59 88 56 79 46 46 66 60 58 55 48 56. (DGGLWLRQ,
ends WXEOIW)
12.2 Nihilist Substitution
38 76 54 76 64 76 76 54 74 55 35 76 77 76 47 58 76 85 74 44
65 88 63 74 47 36 95 74 63 44 37 58 57 96 65 36 66 85 74 63
55 79 53 67 57 56 58 64 67 67 56 67 57 74 55 55 57 86 03 43
46 67 73 96 67 39. (ETARVQITCO, ends HSMX)
12.3 PORTA
QLAMU CHQGO FTESV XKEWC GMXPH
UCLUS WSGXT EVURH TMTSU TKVSQ GCQCW
LHMDX NUFUE EFXRF XPHUN RGPKC OXULB
BBCUS IBBHW. (HAVE)
12.4 PORTA
XFXYW ZJICZ IBUZN HJXEA ACWBE
JOOCZ UPXFQ BXHFI CGMAZ KVQEG BBCAF
KLLXF BVOUN TSAYZ KKXLR CWAJC LVVVI
XNBFQ JVWBW BSWEY VUNGX ODFRZ PTEWO
PJQNH WZPNA YRCLV YYWCQ ULOJB VK. (GSRWXERX)
12.5 PORTAX
UXCUD ZMVBA FWWPV DIKDO JISMA
WRBBA YLOYX AKUXR JGDCJ MYAPV RJWJA
DMUKL KLUAM KAOEN YBFCC IQGFK QZAA. (PQXKEG)
12.6 PORTAX
WWQPE JBDTM TMNWH CTJSW WKIAC
BJKWL YHBYN OAKRZ PDYZM DIVGB QKNJP
RNSRU FXWMU TKMJS KDNLW WFHKR JSCVF
HTJIS JD. (UHDOLCH)
12.7 GROMARK
HPMZU IBQHI SDHHH JKUNC OYJSC
24106
RBLOF REXTG EXAZA ILAXX XHFNH CDUYQ
YUOMQ NVOIN XYMBR WAHNT FGPFB DOOMA
CWHDH JXTTX CJIUR PVMZR EILDZ QJJTT
ILNNP TREVL BQLL. ( tip: UCAUKYKUJK; ends tivpw.)
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[GRI2] DUN SCOTUS,"Binary Number Grille," JA60, The
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[GRI3] S-TUCK,"Grille Solved By the Tableaux Method," DJ42,
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[GRI4] The SQUIRE,"More About Grilles," ON40,DJ40, The
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[GRI5] OMAR,"Rotating Grille Cipher," FM41, The Cryptogram,
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[GRI6] S-TUCK,"Solving The Grille. A New Tableaux Method,"
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[GRI7] LABRONICUS,"Solving The Turning Grille," JF88, The
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[GRI8] BERYL,"The Turning Grille," ND92, The Cryptogram,
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[GRI9] SHERLAC and S-TUCKP,"Triangular Grilles," ON45, The
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[GRIA] SHERLAC,"Turning Grille," ON49, The Cryptogram,
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[GRIB] DUN SCOTUS,"Turning (by the numbers)," SO61, The
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[GRIC] LEDGE,"Turning Grille (Novice Notes)," JA77, The
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[GRO3] MARSHEN," Checking the Numerical Key,"JF70, The
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[GRO8] DAN SURR," Gromark Club Solution," MA75, The
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[GRO9] B.NATURAL," Keyword Recovery in Periodic Gromark,"
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[GROA] D.STRASSE," Method For Determining Term of Key," MA75,
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[GROC] DUMBO," Periodic Gromark ," MA73, The Cryptogram,
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[GROD] ROGUE," Periodic Gromark ," SO73, The Cryptogram,
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[GROE] ROGUE," Theoretical Frequencies in the Gromark," MA74,
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[HOFF] Hoffman, Lance J., editor, "Building In Big Brother:
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[HOLM Holmes, W. J., "Double-Edged Secrets: U.S. Naval
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[HOM1] Homophonic: A Multiple Substitution Number Cipher", S-
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[HOM2] Homophonic: Bilinear Substitution Cipher, Straddling,"
ISHCABIBEL, The Cryptogram, AS48, American Cryptogram
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[HOM3] Homophonic: Computer Column:"Homophonic Solving,"
PHOENIX, The Cryptogram, MA84, American Cryptogram
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[HOM4] Homophonic: Hocheck Cipher,", SI SI, The Cryptogram,
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[HOM5] Homophonic: "Homophonic Checkerboard," GEMINATOR, The
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[HOM6] Homophonic: "Homophonic Number Cipher," (Novice Notes)
LEDGE, The Cryptogram, SO71, American Cryptogram
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