Section I
Systems Using Standard Cipher Alphabets

Approaches to Solution
When standard alphabets are used with monoalphabetic systems, three approaches
are possible. The simplest occurs when text can be immediately identified. Identifica-
tion of only two or three letters in a standard unilateral alphabet is sufficient to
reconstruct and confirm the entire alphabet. The other two methods, where text is not
readily identifiable, are to match frequency patterns to the normal A through Z pat-
tern and to generate all possible solutions. All three of these methods also apply to
standard alphabet periodic polyalphabetics.

by Probable Word Method
9-2. Solution
When the alphabets in a periodic system are known or suspected to be standard, the
identification of one plaintext word is usually enough to recover the whole system. The
period must be identified first, as explained in the previous chapter, either by analysis
of repeat intervals or by the phi test. Then when a word is recognized from repeats or
stereotypes, the alphabets can be written and tried throughout the cryptogram. If they
produce good plaintext throughout, the problem is solved.

Factor analysis does not show us a clearcut period length, but if we select the four
letter repeat as the most likely causal repeat, 7 appears to be the correct period. If we
also try STOP as the four letter repeat, it gives us the following text and alphabets.

From the partial plaintext that this produces, STOP is clearly correct. Such words as
RECONNAISSANCE, HEAVY, and REINFORCED are apparent, any one of which
will complete the solution. For another type of probable word approach, applicable to
periodics or aperiodic, see paragraph 10-3c on crib dragging.

Solution by Frequency Matching
With monoalphabetic systems using standard alphabets, the solution was very easy
whenever a message was long enough to give a recognizable pattern. The characteristic
pattern of highs and lows of a standard sequence cannot be easily concealed. The same
technique applies to polyalphabetic systems, although messages necessarily must be
longer to produce a recognizable pattern for each separate alphabet.

a. Factor analysisshows common factors of three and six for all repeat intervals.
Based, on this, a frequency count for six alphabets is produced, as listed in
Figure 9-1. If the period were actually three, the first and fourth, the second and
fifth, and the third and sixth frequency counts would be similar. This is clearly not
the case, so the period is confirmed as six.

b. The easiest patterns to match are generally those with the highest ICs. The first,
second, and fifth alphabets have the highest ICs, and all can be matched fairly
easily. In the first, plaintext A equals ciphertext B. In the second, plaintext A
equals ciphertext A, and in the fifth, plaintext A equals ciphertext O. Other
alphabets can be matched, too, but using these as an example, the partially
reconstructed text is shown below.

c. The letter combinations produced by the three recovered alphabets are consistent
with good plaintext. Expanded plaintext can be recognized in many places. The
first word is ENEMY for example. Filling in added plaintext is a surer and quicker
means of completing the solution at this point than trying to match more alphabets.
Here is the complete solution.

Solution by the Generatrix Method
With standard alphabets or any known alphabets, the method of completing the plain
component can be used. This method, when applied to periodic systems, is commonly
called the generatrix method. The advantage of this method over frequency matching
is that it will work even with fairly short cryptograms. Just as with a monoalphabetic
system (see paragraph 4-11), the first step is a trial decryption at any alphabet align-
ment, followed by listing the plain component sequence vertically underneath each
letter of the trial decryption. Whenever the plain and cipher sequences are identical
and in the same direction, no trial decryption is necessary. The key difference with
periodic systems is that the process must be applied to the letters of each alphabet
separately. Plaintext will not be immediately obvious when you look at the generated
lines of letters from only a single alphabet, so selection must be initially based on letter
frequencies and probabilities rather than recognizable text. The process is illustrated
with the following cryptogram enciphered with direct standard alphabets.

a. The cryptogram has a period of five, which can be confirmed either through
periodic-phi tests or factor analysis of all the repeats, including two letter repeats,
which are not underlined.

b. The most obvious step to try is to substitute STOP for the four letter repeat. It does
not produce plaintext elsewhere, however. More powerful methods of solution are
c. The cryptogram can be readily solved by the generatrix method. The first step is to
separate the letters produced by each alphabet. The letters from each of the five
alphabets are listed separately below. Notice that if you read all the first letters, it
produces the first group of the cryptogram. The second letters produce the second
group and so on.

d. No trial decryption is required, because the same sequence is expected for both the
plain and cipher components. Therefore, the next step is to complete the plain com-
ponent sequence for each letter grouping. This is illustrated in Figure 9-2.

e. To aid in selection of the most likely generated letter sequences, numeric
probability data has been added to each line of the listing. The numbers listed
below each letter are assigned on the basis of logarithmic weights of the letter
probabilities. To the right of each group of logarithmic weights is the sum of the
weights for that group. Using this kind of weighting lets us determine the relative
probabilities of each line by adding the weights for each letter. The weights in
Figure 9-2 have been added according to the log weights shown in Table 9-1.

f. The listing in Figure 9-2 was computer generated. When this work must be done
manually, it is easier to generate the sequences without the probability data. Then
scan the generated rows for each alphabet to visually select those with the most high
frequency letters. Finally, if necessary, the probability data can be added only for
the selected rows.

g. Only rarely will the correct rows consist entirely of those with the highest totals.
Normally, you will have to try different combinations of the high probability rows
until you find the correct match. The best place to start is with those rows that
stand out the most from others in the same alphabet groups. In the illustrated
problem shown below, alphabets four and five provide the most likely starting
point. In each case, the sum of the log weights for one row are well above any others.
These are listed below, superimposed above each other with room for the other three
alphabets to be added.


h. As the rows are superimposed, the plaintext will appear vertically. The next step is
to see which high probability rows from other alphabets will fit well with the
starting pair. Trying both of the two highest probability rows for alphabet three
produces the next two possibilities.

i. Reading the plaintext vertically, the grouping on the right is better than the one on
the left. The DTS sequence in the left grouping is unlikely, and all the letter com-
binations on the right are acceptable. Furthermore, the EMY combination at the
beginning of the right grouping suggests ENEMY. The letter sequences for the first
two alphabets which begin with E and N respectively are both high probability
sequences. The complete solution is shown below.


Section II
Systems Using Mixed Alphabets
With Known Sequences

Approaches to Solution
When mixed sequences are used in periodic systems, a variety of different techniques
can be used to solve them. When the plain and cipher sequences are known, the same
techniques used with standard alphabets can be used, adapted to the known
sequences. When one or both of the sequences are unknown, new techniques must be
used. Each situation is a little different. The major paragraphs of this section deal with
each situation: both sequences are known, the ciphertext sequence is known, or the
plaintext sequence is known. Techniques for solving periodics when neither sequence is
known are covered in the next section.

Solving Periodics With Known Mixed
Exactly the same techniques that were used with standard alphabets can be used with
any known mixed sequences.

a. Successful assumption of plaintext allows you to directly reconstruct the cipher
alphabets, as before.

b. The generatrix method works, making sure that a trial decryption is first performed
with the sequences set at any alignment. All possible letter combinations are then
generated by completing the plain component sequence, as before. The key points
to remember are to perform the trial decryption and to use the plain component as
the generatrix sequence, not a standard sequence.

c. Frequency matching also works, but there are some differences in its application.
Frequency counts must be arranged in the cipher sequence order, not in standard
order. The pattern that the frequency counts are matched to must be adjusted to
the order of the known plain component. Rearrange the patterns of peaks and
troughs to fit the plain component. For example, shown below is the pattern for a
standard plain sequence and the pattern that results if a keyword mixed sequence
based on POLYALPHABETIC is used as the plain component.

The new pattern resulting from the mixed plaintext sequence is just as easy to
match frequency counts to as the more familiar standard pattern. If it should prove
difficult to match by eye alone, there is also a statistical test, called the chi test,
which can be used to aid the matching process. Paragraph 9-7 demonstrates the use
of the chi test.

9-7. Solving Periodics With Known Cipher
The technique of frequency matching can be used any time the cipher sequence is
known, whether or not the plain sequence is also known. When the plain sequence is
known, the frequency patterns of the cipher sequences are best matched to the ex-
pected plain pattern as explained in paragraph 9-6. When the plain sequence is un-
known, the frequency patterns of the cipher sequences can be matched to each other.
In either case, the key is that the known cipher sequence allows the frequency count to
be arranged in the order of the original cipher sequence. The following problem

demonstrates frequency matching with a known cipher component sequence. The
cipher component sequence in the problem in Figure 9-3 is a keyword mixed sequence
based on NORWAY.

a. Examination of the frequency patterns in Figure 9-3 shows that they do not match
the usual standard sequence-pattern. This means that the plain component
sequence was not a standard sequence.
b. If the cipher sequences can be correctly matched against each other, the crypto-
gram can then be reduced to monoalphabetic terms and solved easily.

c. Figure 9-4 is a portion of a computer listing that matches the frequency count of the
cipher letters of the first alphabet with the frequency count of second alphabet
letters at every possible alignment. The alignments are evaluated by the chi test. In
the chi test, each pair of frequencies for an alignment is multiplied. The products of
all the pairs are totaled to produce the chi value for that alignment. Figure 9-5
shows the computation carried out for the first alignment. The chi test is also called
the cross-product test.

d. Figure 9-6 shows the highest chi values for each match of the first alphabet with the
other four alphabets. For all matches except the fourth alphabet, the chi values
were clearly the highest. Two matches are shown for the fourth alphabet, because
the difference between the two values is not significant. Either match could be the
correct one.

e. To resolve which of the two matches with the fourth alphabet is correct, the highest
chi values for matches between the second and fourth and the third and fourth
alphabets have also been determined. These are shown in Figure 9-7.

f. The matches of alphabet four with alphabets two and three clarify which of the
matches with the first alphabet was correct. This becomes apparent when we set up
the other four alphabets.

g. The match of N of the first alphabet with P of the fourth alphabetic correct. The
second alphabet and third alphabet matches confirm this.

h. The next step in the solution is to reduce the cryptogram to monoalphabetic terms
using the matches just determined. An A through Z sequence is arbitrarily used for
the plain component, and the message is decrypted just as if it were the original.

i. Reduced to monoalphabetic terms, many more repeats in the text that were sup-
pressed by the multiple alphabets now appear. The solution is completed the same
as any other monoalphabetic system.

Solving Periodics With Known Plaintext
Sequences by Direct Symmetry
When the plaintext sequence is known, but not the ciphertext sequence, a solution
technique known as direct symmetry is possible. Direct symmetry depends on the
probable word method for the initial entry into the cryptogram. It makes use of the
fact that the columns can be reconstructed in their original order as recoveries are
made. Consider the next example, which uses a standard plaintext sequence.

a. The period is five. The 14 letter repeat is probably RECONNAISSANCE.

b. With recovered letters filled in, we can see that the beginning phrase is the

c. With a known plain component, the columns are in their original order. This means
that the partially reconstructed cipher sequences are also in the right order. Each
cipher sequence is the same sequence, and whatever one row reveals about the spac-
ing of letters can be transferred to other rows as well. For example, in the second
row, X follows immediately after W. X can then be placed after W in row three.
Similarly, all common letters can be placed by carefully counting the intervals and
placing the same letters at the same intervals in each row. Here is what the matrix
looks like after all such values are placed.

d. Filling all the new values into the text reveals many more possibilities. Completion
of the solution is routine from this point.

e. The direct symmetry technique can also be used as an alternate method when the
cipher sequence is the known sequence. The matrix can be inverted, placing the
cipher sequence on the top of the matrix and the plaintext equivalents inside in
separate rows for each alphabet. Each row will be the plaintext sequence in the
correct order. Horizontal intervals recovered in one row can then be duplicated in
each sequence just as was demonstrated above for cipher sequence recovery. Unlike
the technique of frequency matching, it depends on successful plaintext assump-
tions, however. It is not as powerful a method of solution, but if plaintext can be
readily identified, it may be the quickest way to solve a cryptogram.

Section III
Solving Periodics With Unknown Sequences

Solving Periodics by Indirect Symmetry
When neither the plaintext nor the ciphertext sequence is known, the matrix cannot be
initially recovered with sequences in the correct order. Frequency matching cannot be
used, either. However, some of the interval relationships are preserved even when the
columns are not placed in the correct order, and these interval relationships can be
exploited to aid in matrix recovery.
a. To illustrate how interval relationships are preserved, consider the following two
matrices. The first is the matrix in its original form. The second is the same matrix,
rearranged with the plain component in A through Z order. This is the form in
which you will normally recover a matrix with unknown sequences until enough is
known to rearrange the columns in the correct order.

b. The key principle to understand when working with ananalyst™s matrix, like the
second one above, is that every pair of columns and every pair of rows represents an
interval in the original matrix. To illustrate this, look at the plaintext A column
and the plaintext G column in the bottom matrix. The letters D and R appear in
the first cipher sequence. If you count the distance between the D and R in the
original (top) matrix, you see that the interval is nine. Similarly, the interval for the
other pairs in the two columns, R and X, U and P, and M and S, are also nine. For
any two columns that you compare, the horizontal interval between the letters in
each alphabet will be the same. The interval will not always be nine, of course. It
depends on which two columns you are comparing. The point is that between any
pairs in the same row in the same two columns, the interval will be the same.

c. Next compare the letters in the first cipher sequence and the second in the bottom
matrix. In the first column, the letters D and R appear, which we already noted are
nine letters apart horizontally in the original matrix. The letters R and X appear in

another column in the first and second sequences, as do U and P, and M and S. The
first and second cipher sequences are an interval of nine apart. Whichever pair of
letters you look at in the first and second cipher sequences, they are nine apart in
the original cipher sequence. Each pair of cipher sequences represents a different
interval. For example, the interval between the first and third cipher sequence is
eleven. The interval between the first and fourth is seven. The interval between the
second and third is two, and so on.

d. There are a number of ways in which we can use an understanding of these interval
relationships to help solve a polyalphabetic cryptogram. The use of interval
relationships where sequences are unknown and columns are out of order is called
indirect symmetry. This contrasts with the earlier situation with known sequences
and columns in the correct order, where we used direct symmetry to aid in the
e. To put indirect symmetry to use, consider the following example. Initial recoveries
in a polyalphabetic system have produced the following information.

f. In comparing the plaintext A and E columns, we see that the letters R and T and the
letters M and F are the same interval apart. We do not know what the interval is,
but we know it is the same in each case.

g. The same interval appears when we compare the first and third cipher sequences,
where R and T appear in the first column. Since we know the interval will be the
same for any pair of letters between the first and third sequences, and we know M
and F have the same interval as R and T, we can add the letter F in the plaintext I
column in the third sequence under the letter M.
h. Any time we can establish an interval relationship for two pairs in a rectangular
pattern as above, and can find three of the four letters, also in a rectangular pattern
elsewhere, we can add the fourth letter to complete the pattern. The pairs must be
read in the same direction in each case. Notice that we cannot add F in the plain-
text G column in the first sequence. The interval from the first to the third sequence
is not the same as the interval from the third to the first.

i. Matching pairs are usually found by reading horizontally in one case, and vertically
with one letter in common in the second case, as in the above example. Matching
relationships may be found anywhere in matrix, however, and are not restricted to

cases with one letter in common. You can find most such matching pairs by examin-
ing every column in which you have recovered at least three letters. For each letter
in the column, look for a match with letters on the same row that are the same as one
of the other letters in the column. When you find such letters, check for every possi-
ble complete rectangular relationship, and see if you can find the same relationship
with one letter missing elsewhere. Often the addition of one or two letters is all you
need to recognize more plaintext in the cryptogram and complete a solution.

j. If you have reason to believe that the plaintext sequence is the same as the cipher
sequences, you can use the plaintext sequence in establishing interval relationships,
too. All the techniques that apply to the ciphertext sequences apply to the plaintext
sequence as well, when it is the same sequence.

Extended Application of Indirect Symmetry
Indirect symmetry can be used in other ways, too. For example, when enough letters
have been recovered, you can list all the pairs of letters between each pair of sequences,
and develop partial decimated chains of letters for each, as was explained in paragraph
4-8 with monoalphabetic substitution. These partial chains from different alphabet
combinations can then be combined together geometrically to recover the original
sequence. This technique is illustrated in the following indirect symmetry problem.

a. Through recognition of the stereotyped beginnings and the use of many numbers,
the text shown has been recovered, and the recovered values filled into the matrix.

More values can be filled into the text, but we will first concentrate on the applica-
tion of indirect symmetry.

b. To recover additional values through indirect symmetry, examine each column
with more than two recovered letters in it. Beginning with the fifth column, take
each letter in turn, and scan the same row as the selected letter for letters that are
the same as those in the column. The first letter, Z, has no letters in common in its
row with the letters M, B, P, and N.
c. For the second letter, M, the common letter Z does appear in its row. Having found
a common letter, examine each rectangular relationship that exists between the two
columns. We first see that Z and W have the same interval as M and Z. Links with
this common letter will not add any more values, however.

d. The next rectangular relationship shows that P and L have the same interval as M
and Z. Reading M and Z vertically, we look for P or L on the same rows as the M
and Z to complete the relationship. We find neither P in the second row nor L in the
first row. If either occurred, we could fill in the other. The letters can be written in a
column off to the side for future use.
e. Having observed all relationships from the column with the common letter Z, we
look for another column with a common letter on the M row. B and P do not occur
except in our added column. The letter N does occur in the second row, however.
Examining relationships in the N column, we see that Z and J have the same inter-
val as M and N reading horizontally. With that established, we read M and N ver-
tically and look for Z in the second row or J in the last row. This time we find Z in
the second row. We can add J in the last row in the same column with Z to complete
the rectangular relationship.

f. Continuing this process, all the letters shown in bold print can be added to the
matrix without making any new plaintext recoveries.

g. It would be easy at this point to return to plaintext recovery to complete the solu-
tion, but another technique can be used to recover the original cipher sequences and
rebuild the matrix. This technique involves listing all links that result by matching
each cipher sequence with every other cipher sequence. Sequence 1 is matched with

sequences 2, 3, 4, and 5, in turn. Then sequence 2 is matched with 3, 4, and 5;
sequence 3 is matched with 4 and 5; and sequence 4 is matched with 5. If the plain-
text sequence were the same as the ciphertext sequence, it would only have been
necessary to match the plaintext with each cipher sequence to get all combinations.
When all links have been plotted and combined into partial chains wherever possi-
ble, the results are shown below.

h. Each set of partial chains represents a decimation of the original sequence.
Sometimes, you will be fortunate at this point to find that one of the partial chains
directly represents the original sequence (decimation one). When this happens, the
original sequence is the obvious starting point. It does not occur in this example, so
the best technique is usually to select a set with one of the longer chains as a
starting point and relate all other sequence combinations to it. Notice that the
chains produced by sequences 1-2 and by sequences 2-3 are obviously produced by
the same interval, since many of the partial chains are identical. They make a good
starting point for this problem. Begin by listing each chain fragment on paper,
horizontally. Write the separate chains in different rows so they will not run into
each other.

i. The next step is to relate other chains to the existing plot. By examining the inter-
vals or patterns that letters from other chains have in relation to the starting chains,
they can be added by following the same rule. For example, the 1-3 combination can

be added by observing that it will fit the starting chains by skipping every other
letter. This will also enable linking the fifth fragment, AS, with the fourth. After
adding all the 1-3 chains, the plot looks like this example.

j. Next, search for another combination that can be added to the plot. The 3-4 com-
bination links by counting backwards every fifth letter, as shown by the V and C of
the NZIVC chain. This ties all the chain fragments together into one longer chain.
When all combinations are added, each by their own rule, it results in almost com-
plete recovery.

k. This technique is known as linear chaining. Sometimes you will be unable to com-
bine the fragments into one long chain. When all intervals are even, you will always
end with two separate 13-letter chains, which may be combined by trial and error or
by figuring out the structure of the original matrix. A second technique, called
geometric chaining, which could have been applied here also, is explained in
paragraph 9-11.

l. Continuing, the chain above must be a decimation of the original sequence. Since V,
W, and X are spaced consistently nine apart, trying a decimation of 9 produces the
next sequence.

m. With G missing from alphabetical progression, the sequence is keyword mixed,
based on GAMES. We can now return to the polyalphabetic matrix and rearrange
the columns using the GAMES sequence on each cipher row.

n. The unused letters can be determined by returning to the plaintext and deciphering
the rest of the message. The plaintext sequence turns out to be a simple transposi-
tion mixed sequence based on OLYMPIC. The repeating key is KOREA.

o. The approach shown to solving this problem is not necessarily the way in which you
would solve it in actual practice. It would probably be more effective to return to
the plaintext earlier than was done in this example. This approach was selected to
show the variety of indirect symmetry techniques that can be used, not necessarily
because it would yield the quickest solution.

9-11. Solution of Isologs
Whenever isologs are encountered between periodic messages with different period
lengths, it is possible to recover the original cipher sequences without any initial plain-
text recovery. The cryptograms can then be reduced to monoalphabetic terms and
quickly solved. Two different techniques may be used, depending on whether the same
alphabets or different alphabets are used in the isologs.
a. When isologous cryptograms use the same alphabets with different repeating keys,
the cipher sequences can be recovered by the indirect symmetry process. Take the
following two messages, for example.

(1) To solve the isologs, the two messages are first superimposed with the alphabets
numbered for each.

(2) With periods of 3 and 4, there are 12 different ways in which the alphabets of the
first are matched to the alphabets of the second. These begin with the first
alphabet of message 1 matched with the first alphabet of message 2 and con-
tinue through alphabet 3 matched with alphabet 4. After these 12 matches, the
cycle of matches starts over again. For other periods, the number of different
alphabet matches is the least common multiple of the two period lengths. The
least common multiple of 6 and 4 is 12. The least common multiple of 6 and 9 is
18. For periods of 8 and 9, 72 different alphabet matches are required.

(3) Analysis continues by plotting the links for each alphabet pair. For example,
the first link is A1=D1, the second link is O2=C2, and the third link is P3=F3.
The next example shows all links plotted and combined into partial chains.

(4) The 1-3 plot shows that the same alphabets were used in both these positions.

(5) The partial chains can be combined into one long chain by a process of
geometric chaining. Geometric chaining will often produce results when linear
chaining is not effective. Geometric chaining is plotted horizontally and ver-
tically, instead of in one straight line. Relationships between alphabet matches
can be discovered more readily with this method.

(6) Geometric chaining begins, as with linear chaining, by selecting one alphabet
match to plot horizontally. We can select the 1-1 match for its 5-letter chain as a
starting point. Next, select a second alphabet match to intersect it plotted ver-
tically. For our example, we will use the 2-2 match, producing the following in-
itial plot.

(7) To this initial plot, we add as many other fragments from the 1-1 and 2-2
matches as we can at this time. We can also set up plots separated from these
for each one that cannot be linked to it.

(8) The next step is to find another alphabet match that can easily be added to the
plot. For example, the 1-2 match proceeds in the diagram along a lower left to
upper right diagonal, as shown by the NSC and XJ fragments. All the 1-2 frag-
ments can be added by the same diagonal rule. This ties in the separate plots
from above, also.

(9) Each additional alphabet combination can be added to the plot now. In many
cases, you may see different possibilities for rules. For example, the 3-4 match
can be seen to proceed by an up 3, left 1 rule, as shown by the TO link. A simpler
equivalent is to plot by the upper left to lower right diagonal, as shown by the
PK link. The simplest way to describe the 3-3 match is up 1, right 2, as shown
by the TK or BY links. This is similar to a knight™s move in chess. When all
matches are plotted, they produce this diagram.

(10) The rows can easily be extended into one 26-letter chain at this point, but if
alphabetic progression can be spotted by any other rule, it can be used instead.
For example, starting with the V in the upper left part of the diagram, VWXY
appears by a descending knight™s move. Continuing from the Y that repeats
near the left side, the sequence can be extended further. The complete
sequence appears below.

(11) Using the new recovered sequence and the relationships between the alphabets
of messages 1 and 2, the matrices for both messages can be set up. Using the
first cipher sequence for message 1, all the cipher sequences for message 2 can
be lined up with it using the links already plotted. Here is how the message 2
alphabets line up with alphabet one. The first 1-1, 1-2, 1-3, and 1-4 links from
the isologs are shown in bold print to demonstrate how they were lined up.

(12) Similarly, the alphabets in the first matrix can be completed by plotting the
relationships between the second message and the first. The solution then
becomes a matter of reducing them to monoalphabetic terms.

(13) In cases where the two periods have a common factor, the sequences can still
be recovered, but they cannot be fully aligned. In this case, the chi test can be
used to match the sequences by frequencies, if necessary, once the sequences
are known.

b. A different technique must be used if different alphabets are used between the
isologs, not just different repeating keys. For example, consider the next two

(1) The sequences are different in the two messages, and they cannot be directly
chained together. If you listed the links resulting from the two messages using
the previous technique, they would lead nowhere and contradictions would
quickly develop. The cipher sequences of each must be kept separate.
(2) The method of recovering the cipher sequences when they are different is to set
up periodic matrices one over the other, as shown below. Message 1 and message
2 equivalents are then plotted in the correct sequence for each in the same
columns. Initially, this will result in more than 26 columns, but as incomplete
columns are combined with each other, the matrices will collapse to the correct
width. This method could be used with more than two isologs also, by superim-
posing as many matrices as there are isologous messages.

(3) The first three groups of each message are plotted above. Each time a previously
used letter appears in the same sequence, the two columns can be combined.
For example, in message 2, the Zs in the third sequence allow those two columns
to be combined, and similarly, the Xs in the fourth sequence can be combined.
In the next example, the complete messages are plotted and all possible
columns are combined.

(4) These matrices can easily be completed by direct symmetry, remembering that
the sequence in each matrix is different.

(5) Either cryptogram can now be reduced to monoalphabetic terms and solved, as