Lesson 15: Statistical Attacks
CLASSICAL CRYPTOGRAPHY COURSE
BY LANAKI
July 01, 1996
COPYRIGHT 1996
ALL RIGHTS RESERVED
LECTURE 15
STATISTICAL ATTACKS
SUMMARY
Lecture 15 considers the role and influence that statistics
and probability theory exert on the cryptanalysis of unknown
ciphers. We develop our subject by the following references:
[FRE3], [SINK], [MAST], [ELCY], [GLEA], [KULL].
DISCUSSION
As you may know, William F. Friedman and Dr. Solomon Kullback
were the first Americans to apply Probability Theory and
Applied Statistics to the Science of Cryptanalysis. Their
achievements were so dynamic that American Crypee's were
able to read the secret messages of many of the Foreign
Governments that it dealt with. [YARD]
SCOPE
We shall look at three tests: Kappa test for coincidences,
Chi test or cross product test for superimposition, and Phi
test for monoalphabeticity. We will briefly touch on
Gleason's logarithmic weighting scheme for determination of
number of letters to differentiate a transposition. The
References and Resource section is substantially broadened
with nearly 150 more choice plums.
BASIC THEORY OF COINCIDENCES
We have already looked at a table of Phi Values For
Monoalphabetic and Digraphic Text By Kullback in Lecture 1.
We have also studied various Phi values for Xenocrypts
in Lecture 5. We found that the probability is related to
coincidences and that it is of significance when we
investigate repetitions of letters in a cipher.
We know that the probability of monographic coincidence (1)
of random text employing a 26 letter alphabet is 0.0385 , (2)
in English telegraphic plain text is 0.0667. We have defined
these values as Kr and Kp respectively.
One of the most important techniques in cryptanalysis is that
of applying the Kappa Test or Test of Coincidences. The most
important purpose for this test is to ascertain whether two
or more sequences are correctly superimposed. Correct means
the sequences are so arranged to facilitate or make possible
a solution. The Kappa test has the following theoretical
basis the following circumstances:
(1) If any two rather lengthy sequences of characters are
superimposed, it will be found that as successive pairs
of letters are brought into vertical juxtaposition, that
in a certain number of cases the two superimposed letters
will coincide,
(2) If we are dealing with random text (26 alphabet) there
will be 38 or 39 cases of coincidence per 1000 pairs of
letters examined because Kr = 0.0385.
(3) If we are dealing with plain text (English) there will be
66 or 67 cases of coincidence per 1000 pairs of letters
examined because Kp = 0.0667.
(4) If the superimposed sequences are wholly monoalphabetic
encipherments of plain text by the same cipher alphabet,
there will be 66 or 67 cases of coincidence per 1000
pairs of letters examined because in monoalphabetic
substitution there is a fixed or unvarying relation
between plain text and cipher text, so that for
statistical purposes the cipher text behaves just as if
it were normal plain text.
(5) Even if the two superimposed sequences are polyalphabetic
in character, there still will be 66 or 67 cases of
coincidence or identity per 1000 pairs of letters
examined provided the two sequences really belong to the
same cryptographic system and are superimposed at the
proper point with respect to the keying sequence.
(6) This last point may be seen in the two polyalphabetic
messages below: They have been enciphered poly-
alphabetically by the same two primary components sliding
against each other. The two messages begin at the same
point in the keying sequence. Consequently, they are
identically enciphered, letter for letter, the only
differences between them is due to differences in plain
text.
No. 1
Alpha 16 21 13 5 6 4 17 19 21 21 2 6 3 6 13 13 1 7 12 6
Plain W H E N I N T H E C O U R S E L O N G M
Cipher E Q N B T F Y R C X X L Q J N Z O Y A W
No. 2
Alpha 16 21 13 5 6 4 17 19 21 21 2 6 3 6 13 13 1 7 12 6
Plain T H E G E N E R A L A B S O L U T E L Y
Cipher P Q N T U F B W D J L Q H Y Z P T M Q I
Note, that (a) in every case in which two superimposed cipher
letters are the same, the plain text letters are identical
and (b) in every case in which two superimposed cipher
letters are the different, the plain text letters are
different. In such a system, even though the cipher alphabet
changes from letter to letter, the number cases of identity
or coincidence in the two members of a pair of superimposed
cipher letters will still be about 66 or 67 per thousand
cases examined, because the two members of each pair of
superimposed letters are in the same alphabet and it has been
seen in (4) that in monoalphabetic cipher text K is the same
as for plain text, viz, 0.667. The fact that in this case
each monoalphabet contains just two letters does not affect
the theoretical value of K (Kappa) and whether the actual
number of coincidences agrees closely with the expected
number based upon Kp = 0.0667 depends upon the lengths of the
two superimposed sequences. Messages No's 1 and 2 are said
to be superimposed correctly , that is brought into proper
juxtaposition with respect to the keying sequences.
(7) Now change the situation by changing the juxtaposition to
an incorrect superimposition with respect to the keying
sequence.
No. 1
Alpha 16 21 13 5 6 4 17 19 21 21 2 6 3 6 13 13 1 7 12 6
Plain W H E N I N T H E C O U R S E L O N G M
Cipher E Q N B T F Y R C X X L Q J N Z O Y A W
No. 2
Alpha 16 21 13 5 6 4 17 19 21 21 2 6 3 6 13 13 1 7
Plain T H E G E N E R A L A B S O L U T E
Cipher P Q N T U F B W D J L Q H Y Z P T M
It is evident that the two members of every pair are not in
the same cipher alphabets and any identical letters after
superimposition is strictly accidental. Actually the number
of repetitions will approximate Kr = 0.0385.
Note again, that in every case in which two superimposed
cipher letters are the same, the plain text letters are
not identical and in every case in which two superimposed
cipher letters are the different, the plain text letters are
no always different. Look at the superimposed T(cipher)'s
representing two different plain text letters and that the S
in "COURSE" gives the value J (cipher) and in the word
ABSOLUTELY gives H (cipher). It should be clear that an
incorrect superimposition by two different plain-text letters
enciphered by two different alphabets may "by chance" produce
identical cipher letters, which on superimposition yield
coincidence but have no external indications as to
dissimilarity in plain text equivalents. This incorrect
superimposition will coincide by a value of Kr = 0.0385.
(8) Note the two Z's and they represent the plain text L.
This occurred because the same cipher alphabet came into
play by chance twice to encipher the same plain text
letter both times. This may distort the Kr value for
some systems.
(9) In general, in the case of correct superimposition the
probability of identity or coincidence is Kp = 0.0667; in
the case of incorrect superimposition, the probability is
greater than or equal to Kr = 0.0385. The Kappa test,
aka coincidence test is defined by these values.
APPLYING THE KAPPA TEST
When we say Kp = 0.0667, this means that in a 1000 cases
where two letters are drawn at random from a large volume of
plain text, we should expect 66 or 67 cases of two letters to
coincide or be identical. Nothing is specified what these
letters shall be; they can be two Z's or two E's. Another
way is to consider that at random 6.67% of the comparisons
made will yield coincidences. So for 2000 examinations, we
expect 2000 x 6.67% = 133.4 coincidences [ use integers and
round down to 133]. Or 20,000 comparisons means 1,334
coincidences.
A more practical approach is to find the ratio of observed
number of coincidences to the total number of cases in
question that may occur, i.e. the total number of comparisons
of superimposed letters. When the ratio is closer to 0.0667
than 0.0385 the correct superimposition has been found. This
is true because both members of each pair of superimposed
letters belong to the same monoalphabet and therefore the
probability of their coinciding is 0.067; whereas, in the
case of incorrect superimposition, each pair belongs to
different monoalphabets and the probability of their
coinciding approaches 0.0385 rather than 0.0667.
To use the Kappa test requires calculating the total number
of comparisons in a given case and the actual number of
coincidences in the case under consideration. When two
messages are superimposed, the total number of comparisons
made equals the number of superimposed letters. When more
than two messages are superimposed in a superimposition
diagram (Lecture 13) it is necessary to calculate the number
of comparisons based on the number of letters in the column.
n letters = n(n-1)/2 pairs or comparisons,
in column
For a column of 3 letters , there are 3(2)/2 = 3 comparisons.
We compare the 1st with the 2nd, 2nd with 3rd and 1st with
3rd columns. The more general probability formula is
nCr = n!/r!(n-r)!
where we determine the number of combinations of n different
things taken r at a time. For two letters, r is always 2,
so n!/r!(n-r)! is the same as
n(n-1)(n-2)!/2(n-2)!
becomes n(n-1)/2
with the cancellation of terms using (n-2)!.
RULE
The number of comparisons per column times the number of
columns in the superimposition diagram of letters gives the
total number of comparisons. The extension to this reasoning
is where the superimposition diagram involves columns of
various lengths, then we add together the number of
comparisons for columns of different lengths to obtain a
grand total. Table 15-1 shows the number of letters in a
column versus the number of comparisons calculated. [FRE3]
Table 15 -1
Number of Number of Number of Number of
letters in comparisons letters in comparisons
column column
2 1 16 120
3 3 17 136
4 6 18 153
5 10 19 171
6 15 20 190
7 21 21 210
8 28 22 231
9 36 23 253
10 45 24 276
11 55 25 300
12 66 26 325
13 78 27 351
14 91 28 378
15 105 29 406
30 435
In ascertaining the number of coincidences in the case of a
column containing several letters, we still use the n(n-1)/2
formula, only in this case, n is the number of identical
letters in the column. The reasoning is essentially the same
as above. The total number of coincidences is the sum of the
number of coincidences for each case of identity.
Given the column:
C
K
B
K
Z
K
C
B
B
K
There are 10 letters with 3B's, 2C's 4K's and 1 Z. The 3B's
yield 3 coincidences, the 2 C's yield 1 coincidence, the 4
K's yield 6 coincidences. The sum is 3 + 1 + 6 = 10
coincidences in 45 comparisons = 0.2222
ENCIPHERMENT WITH SAME KEY BUT DIFFERENT INITIATION POINTS
In Lecture 13, I ended with the note that several messages
enciphered by the same keying sequence but each beginning at
a different point presented a challenge. The best attack is
that by superimposition and the Kappa test is used to
correctly line up the messages with respect to each other.
It is understood that the messages may be shifted relative to
each other at many points of superimposition but their is
only one point of superimposition for each message which
corresponds to monoalphabetic columnar superimposition of the
cipher text.
The method:
(1) Number the message according to their lengths.
(2) Fix message 1, message 2 is placed under it so that the
first pair of letters coincide.
(3) Examine, calculate total number of cases in which
superimposed letters are identical, thus the observed
number of coincidences. The total number of superimposed
pairs is calculated and multiplied by 0.0667 to find the
expected number of coincidences.
(4) If the observed number is considerably below the expected
number, or if the ratio of the observed number of
coincidences to the total is closer to 0.0385 than
0.0667, then the superimposition is wrong and we shift
message 2 one letter to the left.
(5) Repeat steps (3) - (4) until the correct superimposition
is found.
(6) Test message 3 against message 1 and then against message
2.
(7) Continue the process until all the messages are lined up
correctly.
Computers are a big help in this process.
EXAMINE OF KAPPA TEST
Given 4 messages of 30 intercepted using a long enciphered
keying sequence:
Message 1
PGLPN HUFRK SAUQQ AQYUO ZAKGA EOQCN
PRKOV HYEIU YNBON NFDMW ZLUKQ AQAHZ
MGCDS LEAGC JPIVJ WVAUD BAHMI HKORM
LTFYZ LGSOG K. [101]
Message 2
CWHPK KXFLU MKURY XCOPH WNJUW KWIHL
OKZTL AWRDF GDDEZ DLBOT FUZNA SRHHJ
NGUZK PRCDK YOOBV DDXCD OGRGI RMICN
HSGGO PYAOY X. [101]
Message 3
WFWTD NHTGM RAAZG PJDSQ AUPFR OXJRO
HRZWC ZSRTE EEVPX OATDQ LDOQZ HAWNX
THDXL HYIGK VYZWX BKOQO AZQND TNALT
CNYEH TSCT. [99]
Message 4
TULDH NQEZZ UTYGD UEDUP SDLIO LNNBO
NYLQQ VQGCD UTUBQ XSOSK NOXUV KCYJX
CNJKS ANGUI FTOWO MSNBQ DBAIV IKNWG
VSHIE P [96]
Superimpose messages 1 and 2.
* * *
No. 1 PGLPN HUFRK SAUQQ AQYUO ZAKGA EOQCN
No. 2 CWHPK KXFLU MKURY XCOPH WNJUW KWIHL
*
No. 1 PRKOV HYEIU YNBON NFDMW ZLUKQ AQAHZ
No. 2 OKZTL AWRDF GDDEZ DLBOT FUZNA SRHHJ
* * *
No. 1 MGCDS LEAGC JPIVJ WVAUD BAHMI HKORM
No. 2 NGUZK PRCDK YOOBV DDXCD OGRGI RMICN
*
No. 1 LTFYZ LGSOG K. [101]
No. 2 HSGGO PYAOY X. [101]
The number of comparisons is 101 x 0.0667 = 7 coincidences
which is less than the observed 8. Nice start but
suspicious. Shifting one letter to right the number of
coincidences is 4. One more shift = 3. Then:
* * *
No. 1 PGLPNHUFRKSAUQQAQYUOZAKGAEOQCN
No. 2 CWHPKKXFLUMKURYXCOPHWNJUWKW
* *
No. 1 PRKOVHYEIUYNBONNFDMWZLUKQAQAHZ
No. 2 IHLOKZTLAWRDFGDDEZDLBOTFUZNASR
* **
No. 1 MGCDSLEAGCJPIVJWVAUDBAHMIHKORM
No. 2 HHJNGUZKPRCDKYOOBVDDXCDOGRGIRM
*
No. 1 LTFYZLGSOGK.
No. 2 ICNHSGGOPYAOYX. [98]
Now 98 x 0.0667 = 6.5366 versus 9 coincidences or 30% more
than the first comparison. The first test was accidental.
The jump is normal from incorrect to correct. The correct
superimposition is either 100% correct or incorrect.
Friedman suggests that tests be made first to the right and
then to the left, one letter at a time for best efficiency.
[FRE3]
It is possible to systematize our investigation by testing
three or four messages at a time.
We make a diagram where the number of coincidences are
tallied with all three messages:
1 2 3
-----------------
1| x 9 3
|
2| x x 3
|
3| x x x
The number of tallies in cell 1-2 is 9 as examined. A column
which shows identical letters in messages 1 and 3 yields a
tally in 1-3, between 2 and 3 goes to 2-3 and so forth. Only
when a superimposition yields three identical letters in a
column is a tally to be recorded in 1-3 or 1-2 (3
coincidences.
So adding message 3 to the investigation:
*
No. 1 PGLPNHUFRKSAUQQAQYUOZAKGAEOQCN
No. 2 CWHPKKXFLUMKURYXCOPHWNJUWKW
No. 3 WFWTDNHTGMRAAZGPJDSQAUPFROXJRO
* * *
No. 1 PRKOVHYEIUYNBONNFDMWZLUKQAQAHZ
No. 2 IHLOKZTLAWRDFGDDEZDLBOTFUZNASR
No. 3 HRZWCZSRTEEEVPXOATDQLDOQZHAWNX
* *
No. 1 MGCDSLEAGCJPIVJWVAUDBAHMIHKORM
No. 2 HHJNGUZKPRCDKYOOBVDDXCDOGRGIRM
No. 3 THDXLHYIGKVYZWXBKOQOAZQNDTNALT
No. 1 LTFYZLGSOGK.
No. 2 ICNHSGGOPYAOYX.
No. 3 CNYEHTSCT.
so:
1 2 3
-----------------
1| x 9 3
|
2| x x 3
|
3| x x x
Successive number of columns are examined and coincidences
(of messages 1 and 3 and 2 and 3) are tabulated. We find:
Combination Total Number Number of Coincidences
of Delta
Comparisons Expected Observed %
1 - 3 99 ~ 7 3 -57
2 - 3 96 ~ 6 3 -50
1- 2- 3 293 ~ 20 15 -21
A correct superimposition for one of the three combinations
may yield such good results as to mask the bad results for
the other two combinations.
We shift message 3 one space to the right with the following
results:
*
No. 1 PGLPNHUFRKSAUQQAQYUOZAKGAEOQCN
No. 2 CWHPKKXFLUMKURYXCOPHWNJUWKW
No. 3 WFWTDNHTGMRAAZGPJDSQAUPFROXJR
* * * *
No. 1 PRKOVHYEIUYNBONNFDMWZLUKQAQAHZ
No. 2 IHLOKZTLAWRDFGDDEZDLBOTFUZNASR
No. 3 OHRZWCZSRTEEEVPXOATDQLDOQZHAWN
* *
No. 1 MGCDSLEAGCJPIVJWVAUDBAHMIHKORM
No. 2 HHJNGUZKPRCDKYOOBVDDXCDOGRGIRM
No. 3 XTHDXLHYIGKVYZWXBKOQOAZQNDTNAL
* *
No. 1 LTFYZLGSOGK.
No. 2 ICNHSGGOPYAOYX.
No. 3 TCNYEHTSCT.
1 2 3
-----------------
1| x 9 10
|
2| x x 7
|
3| x x x
Combination Total Number Number of Coincidences
of Delta
Comparisons Expected Observed %
1 - 3 99 ~ 7 10 +43
2 - 3 97 ~ 6 6 0
1- 2- 3 294 ~ 20 25 +25
The results are very good. We add the fourth message.
No. 1 PGLPNHUFRKSAUQQAQYUOZAKGAEOQCN
No. 2 CWHPKKXFLUMKURYXCOPHWNJUWKW
No. 3 WFWTDNHTGMRAAZGPJDSQAUPFROXJR
No. 4 TULDHNQEZZUTYGDUEDUPSDLIOLNN
No. 1 PRKOVHYEIUYNBONNFDMWZLUKQAQAHZ
No. 2 IHLOKZTLAWRDFGDDEZDLBOTFUZNASR
No. 3 OHRZWCZSRTEEEVPXOATDQLDOQZHAWN
No. 4 BONYLQQVQGCDUTUBQXSOSKNOXUVKCY
No. 1 MGCDSLEAGCJPIVJWVAUDBAHMIHKORM
No. 2 HHJNGUZKPRCDKYOOBVDDXCDOGRGIRM
No. 3 XTHDXLHYIGKVYZWXBKOQOAZQNDTNAL
No. 4 JXCNJKSANGUIFTOWOMSNBQDBAIVIKN
No. 1 LTFYZLGSOGK.
No. 2 ICNHSGGOPYAOYX.
No. 3 TCNYEHTSCT.
No. 4 WGVSHIEP.
1 2 3 4
----------------------
1| x 9 10 7
|
2| x x 7 7
|
3| x x x 5
|
4| x x x x
Combination Total Number Number of Coincidences
of Delta
Comparisons Expected Observed %
1 - 3 96 ~ 6 7 +16
2 - 3 95 ~ 6 7 +16
3 - 4 96 ~ 6 5 -16
1,2,3,4 581 ~39 44 +10
This is actually the correct group of superimpositions.
Testing another message 4 movement to right shows us the
picture.
No. 1 PGLPNHUFRKSAUQQAQYUOZAKGAEOQCN
No. 2 CWHPKKXFLUMKURYXCOPHWNJUWKW
No. 3 WFWTDNHTGMRAAZGPJDSQAUPFROXJR
No. 4 TULDHNQEZZUTYGDUEDUPSDLIOLN
No. 1 PRKOVHYEIUYNBONNFDMWZLUKQAQAHZ
No. 2 IHLOKZTLAWRDFGDDEZDLBOTFUZNASR
No. 3 OHRZWCZSRTEEEVPXOATDQLDOQZHAWN
No. 4 NBONYLQQVQGCDUTUBQXSOSKNOXUVKC
No. 1 MGCDSLEAGCJPIVJWVAUDBAHMIHKORM
No. 2 HHJNGUZKPRCDKYOOBVDDXCDOGRGIRM
No. 3 XTHDXLHYIGKVYZWXBKOQOAZQNDTNAL
No. 4 YJXCNJKSANGUIFTOWOMSNBQDBAIVIK
No. 1 LTFYZLGSOGK.
No. 2 ICNHSGGOPYAOYX.
No. 3 TCNYEHTSCT.
No. 4 NWGVSHIEP.
1 2 3 4
----------------------
1| x 9 10 3
|
2| x x 7 3
|
3| x x x 2
|
4| x x x x
Combination Total Number Number of Coincidences
of Delta
Comparisons Expected Observed %
1 - 3 96 ~ 6 3 -50
2 - 3 96 ~ 6 3 -50
3 - 4 96 ~ 6 2 -83
1,2,3,4 582 ~39 33 -18
SUBSEQUENT SOLUTION STEPS
These four messages were enciphered by a long keying
sequence. We now have found the correct superimposition of
the four messages. Therefore, the text has been reduced to
monoalphabetic columnar form and can be solved. What was not
given on this example was that the enciphering device was a
U. S. Army Cipher Disk and that the key was intelligent as
well as the alphabets are reversed standard.
It doesn't matter to the Kappa test what kind of cipher
alphabets were used or whether or not the key is random or
intelligent. We try our favorite technique - the probable
word on message 1 of DIVISION.
Ciphertext P G L P N H U F R K S A U Q Q
Assumed Plain D I V I S I O N
Resultant Key S O G X F
nope, shift one letter right.
Ciphertext P G L P N H U F R K S A U Q Q
Assumed Plain . D I V I S I O N
Resultant Key . J T K
nope, shift one more, and one and finally to the end with no
resultant intelligent key.
Ciphertext P G L P N H U F R K S A U Q Q
Assumed Plain R E G I M E N T N O
Resultant Key E L A N D O F T H E
which suggests LAND of T(HE) which yields REGIMENT NO. More
assumptions yield an E before LAND and the cipher text
yielding IS for the plain. The process continues one letter
at a time and checking the cipher versus the plain for
reconstructive clues.
We can use all four messages to gives us clues by multiple
superimposition.
Key E L A N D O F T
No 1 Ciphertext P G L P N H U F R K S A U Q Q
Plain R E G I M E N T
No 2 Ciphertext C W H P K K X F L U M K
Plain I E L D T R A I
No 3 Ciphertext W F W T D N H T G M R A A Z
Plain L I N G K I T C
No 4 Ciphertext T U L D H N Q E Z Z U T Y
Plain T I T A N K G U
We see No. 2 gives us FIELD TRAIN, No 3 has ROLLING KITCHEN,
and No 4 with ANTITANK GUN. These words yield additional
letters. If the key is unintelligent text we use the
messages against each rather than against the key.
UNKNOWN SEQUENCES
The previous example assumed a known cipher alphabet. When
it is not known, Data for solution by indirect symmetry by
detection of isomorphs cannot be expected, for isomorphs may
not be produced by the system. Solution can be reached only
if there is sufficient text to permit analysis of columns for
superimposition diagram. Large amount of text yields
repetitions and the basis for probable word assumption.
After establishment of a few values for cipher text letters
does indirect symmetry come into play. Each column requires
15 -20 letters minimum. These can be studied statistically
and if two columns have similar characteristics, they may be
combined using the cross product test.
RUNNING KEY PRINCIPLE
The running - key principle may be interesting in principle
but difficult in practice. Mistakes in encipherment or
transmission, essentially decrease the likely hood of the
correct decipherment. The running Key does improve
cryptographic security but the mechanical details involved in
the production, reproduction, and distribution of such keys
represents a formidable challenge - enough to destroy the
effectiveness of the system for practical purposes
(voluminous communication).
Suppose a basic unintelligible, random sequence of keying
characters which is not derived from the interaction of two
or more shorter keys and which NEVER repeats is employed only
ONCE as a key for encipherment. Can such a cryptogram be
solved. No. No method of attack will solve this because the
system is not uniquely solvable.
Two things are required for solution: the logical answer must
be offered and it must be unique. The Bacon-Shakespeare
"cryptographers tend to overlook the latter issue.
To attempt to solve a cryptogram enciphered as previously
described is like solving an equation in two unknowns with
absolutely no data available for solution but the solution
itself. The key is one unknown and the plain text is the
other. Any one quantity may be chosen and yield a viable
result without the required uniqueness constraint being
observed. There are an infinite number of solutions
possible.
The problem is better defined when the running key
constitutes intelligent test, or if it is used to encipher
more than one message, or if it is the secondary result of
the interaction of two or more short primary keys which go
thru cycles themselves. The additional information in these
cases are enough to meet the uniqueness constraint.
CROSS-PRODUCT TEST OR CHI [X]
The KAPPA test is used to prepare data for analysis. It
circumvents the polyalphabetic obstacle. It moves the
solution from polyalphabetic to monoalphabetic terms. The
solution can be reached if their is some cryptographic
relationship between the columns, or the letters can be
combined into a single frequency distribution.
The amount of data has to be sufficient for comparison
purposes and this depends on the type of cipher alphabets
involved. Although the superimposition diagram may be
composed of many columns, often only a relatively small
number of different cipher alphabets are put into play.
The number of times that a secondary alphabet is employed is
directly related to the key text or number of keying elements
in the sequence.
In the running-key cipher using a long phrase or book as a
key, the key is intelligible text and it follows that the
secondary alphabets will be employed with frequencies
directly related to the respective frequencies of occurrence
of letters of plain text. The key letter 'E' alphabet should
be most frequent, 'T' next and so forth. J, K, Q, X, Z are
improbable, so the cryptanalyst usually handles no more than
19-20 secondary alphabets.
It is possible to study the various distributions for the
columns of the superimposition diagram with the view of
assembling those distributions which belong to the same
cipher alphabet, say 'E', thus making the determination of
values easier in a combined distribution.
If the key is random text, and assuming sufficient text
within the columns, the columnar frequency distributions may
afford the opportunity to amalgamate a large number of small
distributions into a smaller number of larger distributions.
This is known as matching and we use the Cross-Product or Chi
Test, aka X test.
The Chi test is used to identify distributions which belong
to the same cipher alphabet. It is used when the amount of
data is not very large.
DERIVATION OF CHI TEST [KULL]
The theory of monographic coincidence in plain text was
originally developed by Friedman and applied in his technical
paper written in 1925 dealing with his solution of messages
enciphered by a cryptographic machine known as the "Herbern
Electric Super-Code." The paper is among the Riverbank
Publications in 1934.
The probability of coincidence of two A's in plain text is
the square of the probability of occurrence of the single
letter A in such text. Samething with B's through Z's.
The sum of these squares for all letters of the alphabet as
shown in Table 15-2, is found to be 0.0667. This is almost
double the combined probability of random text for hitting
two random text letters coincidentally or:
26 letters x 1/26 x 1/26 = 1/26 = 0.0385 = Kr
Table 15-2
Letter Frequency in Probability Square of
1000 Letters of Occurrence Probability
Separately of Separate
Occurrence
-----------------------------------------------------------
A 73.66 0.0737 0.0054
B 9.74 .0097 .0001
C 30.68 .0307 .0009
D 42.44 .0424 .0018
E 129.96 .1300 .0169
F 28.32 .0283 .0008
G 16.38 .0164 .0003
H 33.88 .0339 .0012
I 73.52 .0735 .0054
J 1.64 .0016 .0000
K 2.96 .0030 .0000
L 36.42 .0364 .0013
M 24.74 .0247 .0006
N 79.50 .0795 .0063
O 75.28 .0753 .0057
P 26.70 .0267 .0007
Q 3.50 .0035 .0000
R 75.76 .0758 .0057
S 61.16 .0612 .0037
T 91.90 .0919 .0084
U 26.00 .0260 .0007
V 15.32 .0153 .0002
W 15.60 .0156 .0002
X 4.62 .0046 .0000
Y 19.34 .0193 .0004
Z .98 .0010 .0000
---------------------------------------------------------
Total 1,000.00 1.0000 0.0667
We have seen this value before as Kp. It is the probability
that any two letters selected at random in a large volume of
normal English plain text will coincide.
Given a 50 letter plain-text distribution:
3 1 1 7 1 2 3 1 2 5 6 2 5 6 2 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
The number of pairings that can be made are n (n-1)/2 =
(50 x 49)/2 = 1,225 comparisons. According to the theory of
coincidences, there should be 1,225 x 0.0667 = 81.7065 or
approximately 82 coincidences of single letters. We look at
the distribution and finds there are 83 for a very close
agreement. [N(N-1)/2]
3 1 1 7 1 2 3 1 2 5 6 2 5 6 2 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
3+0+0+1+21+0+0+1+3+0+0+0+1+10+15+0+0+1+10+15+1+0+1+0+0+0=83
If N is the total number of letters in the distribution, then
the number of comparisons is N(N-1)/2 and the expected number
of coincidences may be written:
.0067N(N-1)/2
or (.0067N**2 - 0.0667N)/2 eq. I
If we let Fa = number of occurrences of A in the foregoing
distribution, the number of coincidences for letter A is
Fa(Fa-1)/2. Similarly for B, we have Fb(Fb-1)/2. The total
number of coincidences for the distribution is:
Fa(Fa-1)/2 +Fb(Fb-1)/2+...+Fz(Fz-1)/2.
Let Fa = any letter A..Z and d = the sum of all terms that
follow it. The distribution d(Fa**2-Fa)/2 represents the
actual coincidences.
Although derived from different sources we equate the terms.
d(Fa**2-Fa)/2 = (.0067N**2 - 0.0667N)/2
and dFa = N
d(Fa**2-Fa) = (.0067N**2 - 0.0667N)
dFa**2 - N = (.0067N**2 - 0.0667N)
dFa**2 = .0067N**2 + 0.9333N eg. II
Equation II tells us the sum of the squares of the absolute
frequencies of a distribution is equal to 0.0667 times the
square of the total number of letters in the distribution,
plus 0.933 times the total number of letters in the
distribution. We let S2 replace dFa**2.
Suppose two monoalphabetic distributions pertain to the same
cipher alphabet. If they are to be correctly combined into a
single distribution, the latter must still be monoalphabetic.
We use subscripts 1 and 2 to indicate the distributions in
question. So:
d(Fa1+Fa2)**2 = .0067(N1+N2)**2 + 0.9333(N1+N2)
expanding terms:
dFa1**2 +2dFa1Fa2 +dFa2**2 =0.0667(N1**2 +2N1N2 + N2**2) +
.9333N1 +.9333N2 eq. III
dFa1**2 = .0067N1**2 + 0.9333N1
dFa2**2 = .0067N2**2 + 0.9333N2
and rearranging:
.0667N1**2 +.9333N1 +2dFa1Fa2 + .0667 N2**2 + .9333N2 =
.0667(N1**2 +2N1N2 +N2**2) + .9333N1 +.9333N2
further reducing:
2dFa1Fa2 = 0.667 (2N1N2)
finally:
dFa1Fa2 = 0.667 eq. IV
-------
N1N2
This equation permits the establishment of an expectant value
for the sum of products of the corresponding frequencies of
the two distributions being considered for amalgamation. The
Chi test or Cross-product test is based on Equation IV.
Given two distributions to be matched:
1 4 3 1 1 1 1 3 2 2 1 1 3 2
F1 - 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 3 1 1 1 1 3 1 1 1 2
F2 - 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 juxtapose the frequencies for convenience.
N1 = 26
Fa1 1 4 3 1 1 1 1 3 2 2 1 1 3 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
Fa2 2 3 1 1 1 1 3 1 1 1 2
N2 = 17
Fa1Fa2 0 8 0 0 0 3 0 0 1 0 0 0 0 0 1 0 0 9 2 2 0 0 0 0 0 4
d=30
N1N2 = 26 x 17 = 442
dFa1Fa2 30
------- = -- = 0.0711
N1N2 442
or 442 x 0.0667 = 28.15 expected value versus 30. The two
distributions very probably belong together.
To point out the effectiveness of the correct Chi test
placement, we look at the example but juxtaposed one interval
to the left.
N1=26
1 4 3 1 1 1 1 3 2 2 1 1 3 2
F1 - 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
F2 - B C D E F G H I J K L M N O P Q R S T U V W X Y Z A
2 3 1 1 1 1 3 1 1 1 2
N2=17
Fa1Fa2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 2 0 0 0 0 3 0 0
dFa1Fa2=2+3+2+3= 10
dFa1Fa2 10
-------- = ---- = 0.226
N1N2 442
Thus, if the two distribution pertain to the same primary
components then they are not properly superimposed. The Chi
test may be applied also to cases where two or more frequency
distributions must be shifted relatively in order to find the
correct superimposition. The problem determines whether we
use direct superimposition or shifted superimposition of the
second distribution in question.
APPLYING THE CHI TEST TO PROGRESSIVE-ALPHABET SYSTEM
We assume for this example that the secondary alphabets were
derived from the interaction of two identical mixed primary
components. The cipher alphabet is based on HYDRAYLIC...Z
sequence shifted one letter to the right for each encipher-
ment. Based on Figure 15-1, the horizontal sequences are all
identical and shifted relatively. The letters inside the
square are plain-text letters.
Instead of letters in the cells of the square we tally the
normal frequencies of the letters occupying the respective
cells. For the first 3 rows we have:
1 . . . 5 . . . . 10 . . . . 15 . . . . 20 . . . . . 26
A 7 3 4 8 3 1 12 3 2 3 8 7 3 6 9 1 1 3 2 4 8
B 112 3 2 3 8 7 3 6 9 1 1 3 2 4 8 7 3 8 3 1
C 3 112 3 2 3 8 7 3 6 9 1 1 3 2 4 8 7 3 4 8
The shift required in this case is 5 to the right to match up
A and B. Note that amount of displacement, or number of
intervals, the B sequence must be shifted to make it match A
sequence corresponds exactly to the distance between the
letters A and B in the primary cipher component.
..... A U L I C B ......
0 1 2 3 4 5
The fact that the primary plain component is identical with
the primary cipher component is coincidental. The
displacement interval is being measured on the cipher
component.
The Given Cipher message is written into a 26 column (26
alphabets) square rather than the standard 5 letter groups.
FIGURE 15-1
ALPHABET NO
1 5 10 15 20 26
A | AULICBEFGJKMNOPQSTVWXZHYDR
B | BEFGJKMNOPQSTVWXZHYDRAULIC
C | CBEFGJKMNOPQSTVWXZHYDRAULI
D | DRAULICBEFGJKMNOPQSTVWXZHY
E | EFGJKMNOPQSTVWXZHYDRAULICB
F | FGJKMNOPQSTVWXZHYDRAULICBE
C H | HYDRAULICBEFGJKMNOPQSTVWXZ
I I | ICBEFGJKMNOPQSTVWXZHYDRAUL
P J | JKMNOPQSTVWXZHYDRAULICBEFG
H K | KMNOPQSTVWXZHYDRAULICBEFGJ
E L | LICBEFGJKMNOPQSTVWXZHYDRAU
R M | MNOPQSTVWXZHYDRAULICBEFGJK
N | NOPQSTVWXZHYDRAULICBEFGJKM
O | OPQSTVWXZHYDRAULICBEFGJKMN
L P | PQSTVWXZHYDRAULICBEFGJKMNO
E Q | QSTVWXZHYDRAULICBEFGJKMNOP
T R | RAULICBEFGJKMNOPQSTVWXZHYD
T S | STVWXZHYDRAULICBEFGJKMNOPQ
E T | TVWXZHYDRAULICBEFGJKMNOPQS
R U | ULICBEFGJKMNOPQSTVWXZHYDRA
V | VWXZHYDRAULICBEFGJKMNOPQST
W | WXZHYDRAULICBEFGJKMNOPQSTV
X | XZHYDRAULICBEFGJKMNOPQSTVW
Y | YDRAULICBEFGJKMNOPQSTVWXZH
Z | ZHYDRAULICBEFGJKMNOPQSTVWX
1 . . . 5 . . . .10 . . . .15 . . . .20 . . . . . 26
1 W G J J M M M J X E D G C O C F T R P B M I I I K Z
2 R Y N N B U F R W W W W Y O I H F J K O K H T T A Z
3 C L J E P P F R W C K O O F F F G E P Q R Y Y I W X
4 M X U D I P F E X M L L W F K G Y P B B X C H B F Y
5 I E T X H F B I V D I P N X I V R P W T M G I M P T
6 E C J B O K V B U Q G V G F F F K L Y Y C K B I W X
7 M X U D I P F F U Y N V S S I H R M H Y Z H A U Q W
8 G K T I U X Y J J A O W Z O C F T R P P O Q U S G Y
9 C X V C X U C J L M L L Y E K F F Z V Q J Q S I Y S
10 P D S B B J U A H Y N W L O C X S D Q V C Y V S I L
11 I W N J O O M A Q S L W Y J G T V P Q K P K T L H S
12 R O O N I C F E V M N V W N B N E H A M R C R O V S
13 T X E N H P V B T W K U Q I O C A V W B R Q N F J V
14 N R V D O P U Q R L K Q N F F F Z P H U R V W L X G
15 S H Q W H P J B C N N J Q S O Q O R C B M R R A O N
16 R K W U H Y Y C I W D G S J C T G P G R M I Q M P S
17 G C T N M F G J X E D G C O P T G P W Q Q V Q I W X
18 T T T C O J V A A A B W M X I H O W H D E Q U A I N
19 F K F W H P J A H Z I T W Z K F E X S R U Y Q I O V
20 R E R D J V D K H I R Q W E D G E B Y B M L A B J V
21 T G F F G X Y I V G R J Y E K F B E P B J O U A H C
22 U G Z L X I A J K W D V T Y B F R U C C C U Z Z I N
23 N D F R J F M B H Q L X H M H Q Y Y Y M W Q V C L I
24 P T W T J Y Q B Y R L I T U O U S R C D C V W D G I
25 G G U B H J V V P W A B U J K N F P F Y W V Q Z Q F
26 L H T W J P D R X Z O W U S S G A M H N C W H S W W
27 L Y R Q Q U S Z V D N X A N V N K H F U C V V S S S
28 P L Q U P C V V V W D G S J O G T C H D E V Q S I J
29 P H Q J A W F R I Z D W X X H C X Y C T M G U S E S
30 N D S B B K R L V W R V Z E E P P P A T O I A N E E
31 E E J N R C Z B T B L X P J J K A P P M J E G I K R
32 T G F F H P V V V Y K J E F H Q S X J Q D Y V Z G R
33 R H Z Q L Y X K X A Z O W R R X Y K Y G M G Z B Y N
34 V H Q B R V F E F Q L L W Z E Y L J E R O Q S O Q K
35 O M W I O G M B K F F L X D X T L W I L P Q S E D Y
36 I O E M O I B J M L N N S Y K X J Z J M L C Z B M S
37 D J W Q X T J V L F I R N R X H Y B D B J U F I R J
38 I C T U U U S K K W D V M F W T T J K C K C G C V S
39 A G Q B C J M E B Y N V S S J K S D C B D Y F P P V
40 F D W Z M T B P V T T C G B V T Z K H Q D D R M E Z
41 O O
A frequency distribution square is compiled, each column of
the text forming a separate distribution in columnar form in
the square. See Figure 15-2. Note the size of each
distribution on the right side of the square under N.
The Chi test is applied to the horizontal rows in the square.
Since the test is statistical, it is more reliable as the
size of the distribution increases. We choose the V and W
distributions because they have the greatest total number of
tallies at 53 and 52 occurrences, respectively.
Figure 15-2
1 . . . 5 . . . .10 . . . .15 . . . .20 . . . . . 26 N
A 1 1 1 4 1 3 1 1 3 2 3 3 1 25
B 6 3 3 7 1 1 1 1 2 1 2 1 8 1 4 43
C 2 3 2 1 3 1 1 1 1 1 2 4 2 1 5 2 6 4 2 1 45
D 1 4 4 2 2 7 1 1 2 1 3 3 1 1 1 34
E 2 3 2 1 4 2 1 4 2 3 2 1 2 1 1 3 1 35
F 2 4 2 3 7 1 1 2 1 6 3 9 3 2 2 1 1 1 51
G 3 6 1 1 1 1 1 4 2 1 4 3 1 1 3 2 3 2 39
H 5 7 4 1 3 4 2 6 2 2 2 1 38
I 4 2 3 2 2 2 1 3 1 1 4 1 3 2 8 4 2 45
J 1 4 3 4 4 3 6 1 3 4 2 1 3 2 4 2 2 50
K 3 2 3 3 4 6 2 2 2 2 1 2 2 2 1 37
L 2 2 1 1 1 2 2 7 4 1 2 1 1 1 1 3 1 1 33
M 2 1 1 3 1 5 1 3 2 1 2 4 7 3 1 37
N 3 2 5 1 7 1 3 2 3 1 1 1 4 34
O 2 3 1 6 1 2 2 1 5 4 2 1 3 1 2 2 38
P 4 2 9 1 1 1 1 1 1 1 9 5 1 2 1 3 43
Q 5 3 1 1 1 1 3 2 2 3 2 5 1 7 5 3 45
R 5 2 1 1 2 1 4 1 1 3 1 2 1 3 4 3 4 1 3 1 2 46
S 1 2 2 1 5 4 1 4 1 3 6 1 8 39
T 3 2 6 1 2 2 1 1 1 2 6 4 3 2 1 1 39
U 1 3 3 2 4 2 2 1 2 1 1 1 2 1 2 4 1 33
V 1 2 2 6 4 8 7 2 1 1 1 1 1 6 4 2 4 53
W 1 1 5 3 1 2 8 1 7 6 1 2 3 2 1 2 4 2 52
X 4 1 3 2 1 5 3 2 3 2 3 1 2 1 1 3 37
Y 1 1 3 3 1 4 4 2 1 4 2 4 3 5 1 2 3 44
Z 2 1 1 1 3 1 2 2 2 2 1 3 3 3 27
1 . . . 5 . . . .10 . . . .15 . . . .20 . . . . . 26
The results of three relative displacements are given.
Test 1
FV 1 2 2 6 4 8 7 2 1 1 1 1 1 6 4 2 4
1 . . . 5 . . . .10 . . . .15 . . . .20 . . . . . 26
FW 4 2 1 1 5 3 1 2 8 1 7 6 1 2 3 2 1 2
24. . 1 . . . 5 . . . .10 . . . .15 . . . .20 . . .
FVFW 4 1018 8 14 14 6 1 18 2 8
NV = 53, NW =52
dFVFW = 103
dFVFW = 103
----- --- = 0.037 nok.
NVNW 2756
Test 2
FV 1 2 2 6 4 8 7 2 1 1 1 1 1 6 4 2 4
1 . . . 5 . . . .10 . . . .15 . . . .20 . . . . . 26
FW 2 3 2 1 2 4 2 1 1 5 3 1 2 8 1 7 6 1
. .20 . . . 24. . 1 . . . 5 . . . .10 . . . .15 . .
FVFW 2 4 16 16 35 2 2 8 1 36
NV = 53, NW =52
dFVFW = 122
dFVFW = 122
----- --- = 0.044 nok.
NVNW 2756
Test 3
FV 1 2 2 6 4 8 7 2 1 1 1 1 1 6 4 2 4
1 . . . 5 . . . .10 . . . .15 . . . .20 . . . . . 26
FW 3 1 2 8 1 7 6 1 2 3 2 1 2 4 2 1 1 5
. 5 . . . .10 . . . .15 . . . .20 . . . . .26 1 . .
FVFW 3 2 4 48 4 56 7 4 3 2 1 2 24 8 2 20
NV = 53, NW =52
dFVFW = 190
dFVFW = 190
----- --- = 0.069 OK!
NVNW 2756
More tests would indicate that we have found the best
correlation for these two cipher alphabets. Therefore, the
primary cipher component has the letters V and W in these
positions. The 4th cell of the W distribution must be placed
under the 1 st cell of the V distribution per Test 3.
1 2 3 4
. . . V . . W . . .
The next best row is F with 51 occurrences. We must test
this row against V, W, and V+W. Test 4,5 and 6 show the
correct superimpositions for the F row. Note that the
computer can be a big time help in this evaluation.
Test 4
FV 1 2 2 6 4 8 7 2 1 1 1 1 1 6 4 2 4
1 . . . 5 . . . .10 . . . .15 . . . .20 . . . . . 26
FF 1 1 2 1 6 3 9 3 2 2 1 1 1 2 4 2 3 7
. .10 . . . .15 . . . .20 . . . . .26 1 . . . 5 . .
FVFF 1 4 36 12 72 14 2 1 1 1 2 24 8 6 28
NV = 53, NF =51
dFVFF = 212
dFVFW = 212
----- --- = 0.078
NVNF 2703
Test 5
FW 1 1 5 3 1 2 8 1 7 6 1 2 3 2 1 2 4 2
1 . . . 5 . . . .10 . . . .15 . . . .20 . . . . . 26
FF 3 7 1 1 2 1 6 3 9 3 2 2 1 1 1 2 4 2
5 . . . .10 . . . .15 . . . .20 . . . . .26 1 . . .
FVFF 3 35 2 48 3 63 18 2 6 2 1 4 16 4
NW = 52, NF =51
dFWFF = 210
dFWFF = 210
----- --- = 0.078
NWNF 2703
Test 6
FV+W 4 3 414 515 6 8 4 4 1 3 2 3 10 6 1 3 9
1 . . . 5 . . . .10 . . . .15 . . . .20 . . . . . 26
FF 1 1 2 1 6 3 9 3 2 2 1 1 1 2 4 2 3 7
. .10 . . . .15 . . . .20 . . . . . 26 1 . . . 5 . .
FV+W 4 6 84 15 35 18 16 8 1 3 21 6 40 12 9 63
*FF
N(V+W) = 105, NF = 51
dF(W+V)FF = 422
dF(W+V)FF = 422
-------- --- = 0.079
N(W+V)NF 5355
This test yield the sequence:
1 2 3 4 5 6 7 8 9
V . . W . . . F .
As the work progresses, we use smaller and smaller
distributions. This decrease in information is
counterbalanced by the number of superimpositions being
reduced as the primary cipher alphabet comes to the surface.
The completely reconstructed primary cipher component (both
plain and cipher were specified as identical) is:
1 . . . 5 . . . .10 . . . .15 . . . .20 . . . . . 26
V A L W N O X F B P Y R C Q Z I G S E H T D J U M K
In practice, the matching process would be interrupted after
a few letters of the primary component were retrieved and the
skeleton of a few words became apparent.
We ascertain the initial position for the primary cipher
component and decipher the cryptogram.
1 . . . 5 . . . .10 . . . .15 . . . .20 . . . . . 26
1 W G J J M M M J X E D G C O C F T R P B M I I I K Z
W I T H T H E I M P R O V E M E N T S I N T H E A I
2 R Y N N B U F R W W W W Y O I H F J K O K H T T A Z
R P L A I N A N D T H E M E A N S O F C O M M U N I
3 C L J E P P F R W C K O O F F F G E P Q R Y Y I W X
C A T I O N A N D W I T H T H E V A S T S I Z E O F
...... and so forth.
The interesting point is that all the tallies in the
frequency square were made of cipher letters occuring in the
cryptogram, and the tallies represented their actual
occurences. We compared cipher alphabet to cipher alphabet.
The plain text letters were held as unknown through out the
process.
CRACKING THE PROGRESSIVE CIPHER USING INDIRECT SYMMETRY
What happens when we do not have enough data to foster the
statistical attack? We can use indirect symmetry because of
certain phenomena arising from the mechanics of the
progressive cipher encipherment method itself.
Take:
Plain HYDRAULICBEFGJKMNOPQSTVWXZ
Cipher FBPYRCQZIGSEHTDJUMKVALWNOX
Encipher FIRST BATTALION by the progressive method sliding
the cipher component to the left one interval after each
encipherment.:
1 2 3 4 5 6 7 8 91011121314
Plain F I R S T B A T T A L I O N
Cipher E I C N X D S P Y T U K Y Y
Index F E B C I L U A R D Y H Z X
shift(-) 1 2 3 4 5 6 7 8 910111213
Repeated letters in the text are two I's, three T's and two
A's. Lets look at them:
F I R S T B A T T A L I O N
1 2 3 4 5 6 7 8 91011121314
Plain . I . . . . . . . . . I . .
Cipher . I . . . . . . . . . K . .
Plain . . . . T . . T T . . . . .
Cipher . . . . X . . P Y . . . . .
Plain . . . . . . A . . A . . . .
Cipher . . . . . . S . . T . . . .
The two I's are 10 letters apart in both the plain and cipher
components. Since the cipher component is displaced one step
after each encipherment, two identical letters n intervals
apart in the plain text must yield cipher equivalents which
are n intervals apart in the cipher component. This leads to
the probable word and indirect symmetry attack on the
progressive cipher.
A second flaw concerns the repeated cipher letters. Look at
the three Y's.
1 2 3 4 5 6 7 8 91011121314
Plain . . . . . . . . T . . . O N
Cipher . . . . . . . . Y . . . Y Y
Reference to the plain component shows that the N O . . . T
is reversed in order with respect to the plain text. The
intervals are correct. Since the cipher component is shifted
one to the left each encipherment, two identical letters n
intervals apart in the cipher text must yield plain text
equivalents which are n intervals apart in the cipher
component. If the cipher is displaced to the left than the
order of the plain is logically reversed.
Given the following message, which is assumed to start with
the military greeting COMMANDING GENERAL FIRST ARMY (probable
words) the data yielded by this assumption is:
IKMKI LIDOL WLPNM VWPXW DUFFT
FNIIG XGAMX CADUV AZVIS YNUNL ...
1.......................26
Plain (assumed) COMMANDINGGENERALFIRSTARMY
Cipher IKMKILIDOLWLPNMVWPXWDUFFTF
Set up the decryption square in Figure 15-3.
Figure 15-3
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 I
2 K
3 M
4 K
5 I
6 L
7 I
8 D
9 O
10 L
11 W
12 L
13 P
14 N
15 M
16 V
17 W
18 P
19 X
20 W
21 D
22 U
23 F
24 F
25 T
26 F
Applying indirect symmetry to the above square gives:
1 . . . 5 . . . .10 . . . .15 . . . .20 . . . . . 26
Plain A L I C E F G M N O S Y D R
Cipher M K V . L W N O . F . P . . . . . I . . . . T
D . . . . . . . . . . M
Setting C (plain) = I (cipher) for the first encipherment,
the 8th value, I (plain) = D (cipher) which yields D and
eventually X. We use the partial sequences to unlock other
letters. Using the word ARMY we open the gaps some more.
1 2 3 4 5 6 7 8 9 10 11 12
Plain N I I G X G A M X C A D
Cipher . I L . . . . E O . . R
The next word after ARMY might be WILL. We then insert the
W in the plain and G in the Cipher.
The presence of MMM, WWW, FFF in the cipher might be a short
word used several time.. hmm how about THE?? replacing any
one of the triplets with THE, applying indirect symmetry, we
may have a wedge.
MACHINE CRYPTOGRAPHY
The principles discussed in the previous paragraph may be
used with progressive systems in which the interval is > 1
and with modifications to those intervals which are irregular
but follow a pattern such as 1-2-3, 1-2-3, ... or 2-5-7-3-1,
2-5-7-3-1- and so on. The latter type of progression is
encountered in certain mechanical cryptographs. [FRE3]
THE PHI TEST h FOR MONOALPHABETICITY
The Chi test is based on the general theory of coincidences
and the probability constants Kp and Kr. Now two
monoalphabetic distributions when correctly combined will
yield a single distribution which still will be
monoalphabetic in character. The Phi (h) test is used to
confirm that a distribution is in fact alphabetic.
DERIVATION Of PHI h TEST
Start with a uniliteral frequency distribution, the total
number of pairs of letters for comparison purposes is:
N(N-1)/2 for N letters
from the discussion on the Chi (a) test we found that the
expected value of Fa(Fa-1)/2 +..+Fz(Fz-1) for A...Z is equal
to the theoretical number of coincidences of two letters to
be expected in N(N-1)/2 for N letters, which for normal
English plaintext is Kp x N(N-1)/2 and for random text is Kr
x N(N-1)/2.
d Fi (Fi-1) = E(hp) = Kp x N(N-1)
for i= A to Z for plain text
d Fi (Fi-1) = E(hr) = Kr x N(N-1)
for i= A to Z for random text
E(a) means the average or expected value of the expression in
parenthesis, Kp = 0.0667 for normal English plain text, Kr =
0.0385 for random English text (26 letters).
Example 1:
Is the following enciphered monoalphabetically:
1 1 2 3 4 2 1 4 2 1 1 3 N=25
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(ao) = 1x0+1x0+2x1+3x2+4x3+2x1+1x0+4x3+2x1+1x0+1x0+3x2=
2+6+12+2+12+2+6 = 42 o = observed
E(ap) = Kp x N(N-1) = 0.0667 x 25 x 24 = 40 plain
E(ar) = Kr x N(N-1) = 0.0385 x 25 x 24 = 23.1 random
Since the E(ao) =42 is closer to E(ap) = 40, the distribution
is most likely monoalphabetic.
Example 2:
Y O U I J Z M M Z Z M R N Q C X I Y T W R G K L H
The distribution is
1 1 1 2 1 1 1 3 1 0 2 1 2 1 1 1 1 2 3 N=25
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
d Fi(Fi-1) = 18
Since E(ar) is closer to E(ao) the enciphement is probabably
polyalphabetic to suppress the frequency distribution. The
message was enciphered actually by 25 alphabets used in
sequence.
LOGARITHMIC WEIGHT: CHI SQUARED TEST
Gleason discusses an important application of the theory of
testing hypothesis. Given a number of messages, some of which
are transposed English text and some are flat text. We want
to develop a test for picking out the transpositions, and to
accomplish this is possible to frame a statistical hypothesis
concerning each message. Gleason discusses a 5 step
procedure to 1) obtain probability information, 2) calculate
its critical region, 3) differentiate by weighted logs 4)
calculate the values of alpha and beta statistical inference
5) examine the normal distribution for given values of alpha
and beta. The answer tells us how many letters to examine at
some level of certainty to determine if we are dealing with a
transposition. Chapter 13 Problem 1 gives a reasonable look
at the process. [GLEA] Problems 2 and 3 look at the concept
of Bayesian probability applied to transposition problems and
should be of interest.
WITZEND'S TABLES TO AID CRYPTARITHM SOLUTION
WITZEND has graciously produced several cryptarithmic tables
to aid in solution for problems involving bases from ten to
sixteen. They are given as Tables 15 - 3 through 15 - 9 and
should ease the pain.
Table 15 - 3
DECIMAL - BASE 10
ADDITION
0 1 2 3 4 5 6 7 8 9
----------------------------
0 | 0 1 2 3 4 5 6 7 8 9
1 | 1 2 3 4 5 6 7 8 9 10
2 | 2 3 4 5 6 7 8 9 10 11
3 | 3 4 5 6 7 8 9 10 11 12
4 | 4 5 6 7 8 9 10 11 12 13
5 | 5 6 7 8 9 10 11 12 13 14
6 | 6 7 8 9 10 11 12 13 14 15
7 | 7 8 9 10 11 12 13 14 15 16
8 | 8 9 10 11 12 13 14 15 16 17
9 | 9 10 11 12 13 14 15 16 17 18
MULTIPLICATION
0 1 2 3 4 5 6 7 8 9
----------------------------
0 | 0 0 0 0 0 0 0 0 0 0
1 | 0 1 2 3 4 5 6 7 8 9
2 | 0 2 4 6 8 10 12 14 16 18
3 | 0 3 6 9 12 15 18 21 24 27
4 | 0 4 8 12 16 20 24 28 32 36
5 | 0 5 10 15 20 25 30 35 40 45
6 | 0 6 12 18 24 30 36 42 48 54
7 | 0 7 14 21 28 35 42 49 56 63
8 | 0 8 16 24 32 40 48 56 64 72
9 | 0 9 18 27 36 45 54 63 72 81
N 1 2 3 4 5 6 7 8 9
----------------------------------------------------
N Square 1 4 9 16 25 36 49 64 81
N Cube 1 8 27 64 125 216 343 512 729
N Fourth 1 16 81 256 625 1296 2401 4096 6561
N Fifth 1 32 243 1024 3125 7776 16807 32768 59049
N Sixth 1 64 729 4096 15625 46656 117649 262144 531441
N Sevnth 1 128 2187 16384 78125 279936 823543 2097152 4782969
X 2 4 5 5 5 5 6 8
Y 6 6 3 5 7 9 6 6
X * Y 12 24 15 25 35 45 36 48
Table 15 - 4
UNDECIMAL - BASE 11
ADDITION
1 2 3 4 5 6 7 8 9 A
----------------------------
1 | 2 3 4 5 6 7 8 9 A 10
2 | 3 4 5 6 7 8 9 A 10 11
3 | 4 5 6 7 8 9 A 10 11 12
4 | 5 6 7 8 9 A 10 11 12 13
5 | 6 7 8 9 A 10 11 12 13 14
6 | 7 8 9 A 10 11 12 13 14 15
7 | 8 9 A 10 11 12 13 14 15 16
8 | 9 A 10 11 12 13 14 15 16 17
9 | A 10 11 12 13 14 15 16 17 18
A |10 11 12 13 14 15 16 17 18 19
MULTIPLICATION
1 2 3 4 5 6 7 8 9 A
----------------------------
1 | 1 2 3 4 5 6 7 8 9 A
2 | 2 4 6 8 A 11 13 15 17 19
3 | 3 6 9 11 14 17 1A 22 25 28
4 | 4 8 11 15 19 22 26 2A 33 37
5 | 5 A 14 19 23 28 32 37 41 46
6 | 6 11 17 22 28 33 39 44 4A 55
7 | 7 13 1A 26 32 39 45 51 58 64
8 | 8 15 22 2A 37 44 51 59 66 73
9 | 9 17 25 33 41 4A 58 66 74 82
A | A 19 28 37 46 55 64 73 82 91
N 1 2 3 4 5 6 7 8 9 A
-------------------------------------------------
N Square 1 4 9 15 23 33 45 59 74 91
N Cube 1 8 25 59 104 187 292 427 603 82A
Table 15 - 5
DUODECIMAL - BASE 12
ADDITION
1 2 3 4 5 6 7 8 9 A B
-------------------------------
1 | 2 3 4 5 6 7 8 9 A B 10
2 | 3 4 5 6 7 8 9 A B 10 11
3 | 4 5 6 7 8 9 A B 10 11 12
4 | 5 6 7 8 9 A B 10 11 12 13
5 | 6 7 8 9 A B 10 11 12 13 14
6 | 7 8 9 A B 10 11 12 13 14 15
7 | 8 9 A B 10 11 12 13 14 15 16
8 | 9 A B 10 11 12 13 14 15 16 17
9 | A B 10 11 12 13 14 15 16 17 18
A | B 10 11 12 13 14 15 16 17 18 19
B |10 11 12 13 14 15 16 17 18 19 1A
MULTIPLICATION
1 2 3 4 5 6 7 8 9 A B
-------------------------------
1 | 1 2 3 4 5 6 7 8 9 A B
2 | 2 4 6 8 A 10 12 14 16 18 1A
3 | 3 6 9 10 13 16 19 20 23 26 29
4 | 4 8 10 14 18 20 24 28 30 34 38
5 | 5 A 13 18 21 26 2B 34 39 42 47
6 | 6 10 16 20 26 30 36 40 46 50 56
7 | 7 12 19 21 2B 36 41 48 53 5A 65
8 | 8 14 20 28 34 40 48 54 60 68 74
9 | 9 16 23 30 39 46 53 60 69 76 83
A | A 18 26 34 42 50 5A 68 76 84 92
B | B 1A 29 38 47 56 65 74 83 92 A1
N 1 2 3 4 5 6 7 8 9 A B
---------------------------------------------
N Square 1 4 9 14 21 30 41 54 69 84 A1
N Cube 1 8 23 54 A5 160 247 368 569 874 92B
X 2 3 3 4 4 6 6 6
Y 6 4 8 3 6 2 4 6
X * Y 10 10 20 10 20 10 20 30
X 6 6 8 8 8 9 9 2
Y 8 A 3 6 9 4 8 1
X * Y 40 50 20 40 60 30 60 2
X 2 3 3 3 4 4 4 4
Y 7 1 5 9 1 4 7 A
X * Y 12 3 13 23 4 14 24 34
Table 15 - 6
TERDECIMAL - BASE 13
ADDITION
1 2 3 4 5 6 7 8 9 A B C
----------------------------------
1 | 2 3 4 5 6 7 8 9 A B C 10
2 | 3 4 5 6 7 8 9 A B C 10 11
3 | 4 5 6 7 8 9 A B C 10 11 12
4 | 5 6 7 8 9 A B C 10 11 12 13
5 | 6 7 8 9 A B C 10 11 12 13 14
6 | 7 8 9 A B C 10 11 12 13 14 15
7 | 8 9 A B C 10 11 12 13 14 15 16
8 | 9 A B C 10 11 12 13 14 15 16 17
9 | A B C 10 11 12 13 14 15 16 17 18
A | B C 10 11 12 13 14 15 16 17 18 19
B | C 10 11 12 13 14 15 16 17 18 19 1A
C |10 11 12 13 14 15 16 17 18 19 1A 1B
MULTIPLICATION
1 2 3 4 5 6 7 8 9 A B C
----------------------------------
1 | 1 2 3 4 5 6 7 8 9 A B C
2 | 2 4 6 8 A C 11 13 15 17 19 1B
3 | 3 6 9 C 12 15 18 1B 21 24 27 2A
4 | 4 8 C 13 17 1B 22 26 2A 31 35 39
5 | 5 A 12 17 1C 24 29 31 36 3B 43 48
6 | 6 B 15 1B 24 2A 33 39 42 48 51 57
7 | 7 11 18 22 29 33 3A 44 4B 55 5C 66
8 | 8 13 1B 26 31 39 44 4C 57 62 6A 75
9 | 9 15 21 2A 36 42 4B 57 63 6C 78 84
A | A 17 24 31 3B 48 55 62 6C 79 86 93
B | B 19 27 35 43 51 5C 84 78 86 94 A2
C | C 1B 2A 39 48 57 66 75 84 93 A2 B1
N 1 2 3 4 5 6 7 8 9 A B C
-------------------------------------------------
N Square 1 4 9 13 1C 2A 3A 4C 63 79 94 B1
N Cube 1 8 21 4C 98 138 205 365 441 5BC 785 A2C
Table 15 - 7
QUADECIMAL - BASE 14
ADDITION
1 2 3 4 5 6 7 8 9 A B C D
-------------------------------------
1 | 2 3 4 5 6 7 8 9 A B C D 10
2 | 3 4 5 6 7 8 9 A B C D 10 11
3 | 4 5 6 7 8 9 A B C D 10 11 12
4 | 5 6 7 8 9 A B C D 10 11 12 13
5 | 6 7 8 9 A B C D 10 11 12 13 14
6 | 7 8 9 A B C D 10 11 12 13 14 15
7 | 8 9 A B C D 10 11 12 13 14 15 16
8 | 9 A B C D 10 11 12 13 14 15 16 17
9 | A B C D 10 11 12 13 14 15 16 17 18
A | B C D 10 11 12 13 14 15 16 17 18 19
B | C D 10 11 12 13 14 15 16 17 18 19 1A
C | D 10 11 12 13 14 15 16 17 18 19 1A 1B
D |10 11 12 13 14 15 16 17 18 19 1A 1B 1C
MULTIPLICATION
1 2 3 4 5 6 7 8 9 A B C D
-------------------------------------
1 | 1 2 3 4 5 6 7 8 9 A B C D
2 | 2 4 6 8 A C 10 12 14 16 18 1A 1C
3 | 3 6 9 C 11 14 17 1A 1D 22 25 28 2B
4 | 4 8 C 12 16 1A 20 24 28 2C 32 36 3A
5 | 5 A 11 16 1B 22 27 2C 33 38 3D 44 49
6 | 6 C 14 1A 22 28 30 36 3C 44 4A 52 58
7 | 7 10 17 20 27 30 37 40 47 50 57 60 67
8 | 8 12 1A 24 2C 36 40 48 52 5A 64 6C 76
9 | 9 14 1D 28 33 3D 47 52 5B 66 71 7A 85
A | A 16 22 2C 38 44 50 5A 66 72 7C 88 94
B | B 18 25 32 3D 4D 57 64 71 7C 89 96 A3
C | C 1A 28 36 44 52 60 6C 7A 88 96 A4 B2
D | D 1C 2B 3A 49 58 67 76 85 94 A3 B2 C1
N 1 2 3 4 5 6 7 8 9 A B C D
-----------------------------------------------
N **2 1 4 9 12 1B 28 37 48 5B 72 89 A4 C1
N **3 1 8 1D 48 8D 116 1A7 288 3A1 516 6B1 8B6 B2D
X 2 4 6 7 7 7 7 7
Y 7 7 7 2 4 6 8 A
X * Y 10 20 30 10 20 30 40 50
X 7 8 A C 2 4 6 7
Y C 7 7 7 8 8 8 3
X * Y 60 40 50 60 12 24 36 17
X 7 7 7 7 7 8 A C
Y 5 7 9 B D 8 8 8
X * Y 27 37 47 57 67 48 5A 6C
Table 15 - 8
QUINDECIMAL - BASE 15
ADDITION
1 2 3 4 5 6 7 8 9 A B C D E
----------------------------------------
1 | 2 3 4 5 6 7 8 9 A B C D E 10
2 | 3 4 5 6 7 8 9 A B C D E 10 11
3 | 4 5 6 7 8 9 A B C D E 10 11 12
4 | 5 6 7 8 9 A B C D E 10 11 12 13
5 | 6 7 8 9 A B C D E 10 11 12 13 14
6 | 7 8 9 A B C D E 10 11 12 13 14 15
7 | 8 9 A B C D E 10 11 12 13 14 15 16
8 | 9 A B C D E 10 11 12 13 14 15 16 17
9 | A B C D E 10 11 12 13 14 15 16 17 18
A | B C D E 10 11 12 13 14 15 16 17 18 19
B | C D E 10 11 12 13 14 15 16 17 18 19 1A
C | D E 10 11 12 13 14 15 16 17 18 19 1A 1B
D | E 10 11 12 13 14 15 16 17 18 19 1A 1B 1C
E |10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D
MULTIPLICATION
1 2 3 4 5 6 7 8 9 A B C D E
----------------------------------------
1 | 1 2 3 4 5 6 7 8 9 A B C D E
2 | 2 4 6 8 A C E 11 13 15 17 19 1B 1D
3 | 3 6 9 C 10 13 16 19 1C 2A 2E 33 37 3B
4 | 4 8 C 11 15 19 1D 22 26 2A 2E 40 45 4A
5 | 5 A 10 15 1A 20 25 2A 30 35 3A 40 45 4A
6 | 6 C 13 19 20 26 2C 33 39 40 46 4C 53 59
7 | 7 E 16 1D 25 2C 34 3B 43 4A 52 5E 61 68
8 | 8 11 19 22 2A 33 3B 44 4C 55 5D 66 6E 77
9 | 9 13 1C 26 30 39 43 4C 56 60 69 73 7C 86
A | A 15 20 2A 35 40 4A 55 60 6A 75 80 8A 95
B | B 17 23 2E 3A 46 52 5D 69 75 81 8C 98 A4
C | C 19 26 33 40 4C 5E 66 73 80 8C 99 A7 B3
D | D 1B 29 37 45 53 61 6E 7C 8A 98 A7 B4 C2
E | E 1D 2C 3B 4A 59 68 77 86 95 A4 B3 C2 D1
N 1 2 3 4 5 6 7 8 9 A B C D E
---------------------------------------------------
N **2 1 4 9 11 1A 26 34 44 56 6A 81 99 B4 D1
N **3 1 8 1C 44 85 E6 17D 242 339 46A 5DB 7A3 9B7 C2E
X 3 3 5 5 5 5 6 6
Y 5 A 3 6 9 C 5 A
X * Y 10 20 10 20 30 40 20 40
X 9 9 A A A A C C
Y 5 A 3 6 9 C 5 A
X * Y 30 60 20 40 60 80 80 40
X 3 3 5 5 5 5 6 9
Y 6 B 4 7 A D B 6
X * Y 40 80 13 23 10 25 35 45
Table 15 - 9
SEXDECIMAL - BASE 16
ADDITION
1 2 3 4 5 6 7 8 9 A B C D E F
-------------------------------------------
1 | 2 3 4 5 6 7 8 9 A B C D E F 10
2 | 3 4 5 6 7 8 9 A B C D E F 10 11
3 | 4 5 6 7 8 9 A B C D E F 10 11 12
4 | 5 6 7 8 9 A B C D E F 10 11 12 13
5 | 6 7 8 9 A B C D E F 10 11 12 13 14
6 | 7 8 9 A B C D E F 10 11 12 13 14 15
7 | 8 9 A B C D E F 10 11 12 13 14 15 16
8 | 9 A B C D E F 10 11 12 13 14 15 16 17
9 | A B C D E F 10 11 12 13 14 15 16 17 18
A | B C D E F 10 11 12 13 14 15 16 17 18 19
B | C D E F 10 11 12 13 14 15 16 17 18 19 1A
C | D E F 10 11 12 13 14 15 16 17 18 19 1A 1B
D | E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C
E | F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D
F |10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E
MULTIPLICATION
1 2 3 4 5 6 7 8 9 A B C D E F
-------------------------------------------
1 | 1 2 3 4 5 6 7 8 9 A B C D E F
2 | 2 4 6 8 A C E 10 12 14 16 18 1A 1C 1E
3 | 3 6 9 C F 12 15 18 1B 1E 21 24 27 2A 2D
4 | 4 8 C 10 14 18 1C 20 24 28 2C 30 34 38 3C
5 | 5 A F 14 19 1E 23 28 2D 32 37 3C 41 46 4B
6 | 6 C 12 18 1E 24 2A 30 36 3C 42 48 4E 54 5A
7 | 7 E 15 1C 23 2A 31 38 3F 46 4D 54 5B 62 69
8 | 8 10 18 20 28 30 38 40 48 50 58 60 68 70 78
9 | 9 12 1B 24 2D 36 3F 48 51 5A 63 6C 75 7E 87
A | A 14 1E 28 32 3C 46 50 5A 64 6E 78 82 8C 96
B | B 16 21 2C 37 42 4D 58 63 6E 79 84 8F 9A A5
C | C 18 24 30 3C 48 54 60 6C 78 84 90 9C A8 B4
D | D 1A 27 34 41 4E 5B 68 75 82 8F 9C A9 B6 C3
E | E 1C 2A 38 46 54 62 70 7E 8C 9A A8 B6 C4 D2
F | F 1E 2D 3C 4B 5A 69 78 87 96 A5 B4 C3 D2 E1
N 1 2 3 4 5 6 7 8 9 A B C D E F
-----------------------------------------------------
N **2 1 4 9 10 19 24 31 40 51 64 79 90 A9 C4 E1
N **3 1 8 1B 40 7D D8 157 200 2D9 3E8 533 6C0 895 AB8 D2F
LECTURE 14 SOLUTIONS
14-1. Multiplication (Two words, 0-1) original by EDNASANDE
WOMEN X MEN = UTNNLM + TIWENO + NWTWNN = NLSMTUWM
0123456789
SLOWMINUET
14-2. Division (Two words, 0 -9) MORDASHKA
ATOM / ASK = N; - GNC = IS
O123456789
TASKCOMING
14-3. Multiplication. (No word, 0-1) FOMALHAUT
ASAP X MAB = RITMT + TMPRY + PDBYD =PAYDIRT
0123456789
DBARTMPIYS
14-4. Unidecimal multiplication. (Two words 0-X) WALRUS
TOUGH X DIG = IDIGDN + NYYDNG + UIHDOU = DDCUUILN
0123456789X
CLOUDYNIGHT
LECTURE 15 PROBLEMS - Taken from OP- 20 -G course:
15-1. Naval Text. Recover Keys.
J Z S S W B P D Z Z L F O M E K Q P D J H C K U M C
A B C O O X M Y S I I G B S G G Y V D S W A J O Q E
K U P W K N J K C C H W O Z Q Q B P Y N V J J O Q E
K U C D S L R W C F Q I A V M S R S I X Y T P O P G
D H U V N K V K C Y Y A L R Q O O Q D N Z C G L R E
K F H Q R N J B.
15-2. Naval Text.
A U V Z I S Z F B F Y E I R B I O W A O Y J L B L D
D G K U I T T Z B D B E Q I O C J R F W X D Y H G M
S P P I S W Y P F V S Y G G S H Q K L A L Z A Q F N
U T C Q H D G Y L B Z P D V C S J N W G N T P T M S
H J T W C K O C M X Z P Z R R U Y I W H H M E Z F L
O C F I S W L P D N W T Z H H T I R L Y I P N Q F N
U T C Q H D G Y L B Z P D V C S J N W G N T P T M E
O S V B W J B L V X Z P Z R R U Y I W H H P L P F T
R B P G X B U L V N W J P R H I H F Q X L N B L P S
H J T W I J T T Q W E E Q F O I I Z P M B J Q P Y M
D U Q W A T Z O W D C L Z Q M P U K.
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