SOLUTION OF NUMERICALLY-KEYED

COLUMNAR TRANSPOSITION CIPHERS

Completely Filled Matrices - Determining

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Matrix Size

When completely filled matrices are known or suspected, the first step in their solution

is to determine the matrix size. As discussed in Chapter 11 for simple columnar

transposition, the correct width must be an even divisor of the message length. With

simple columnar transposition, the correct width could be confirmed easily, because

plaintext will appear on the rows when the width is correctly selected. It is not as sim-

ple with numerically-keyed transposition. Although each row will contain the letters of

plaintext for that row when the width is correctly selected, the letters will be out of

order. The key to recognition is the vowel count on each row. Vowels should appear

consistently with fairly even counts on each row when the correct width is tried. In

plaintext, vowels appear about 40 percent of the time even in small samples of text.

This is necessary for text to be pronounceable. If some of the rows have too many or too

few vowels, you probably have the wrong width. Consider the next cryptogram.

a. The cryptogram has 56 letters, assuming the final Xs are all nulls. If a completely

filled matrix is suggested by past experience, then the matrix is probably either 7 or

8 letters wide. Write the cryptogram by columns into a trial matrix of each width

and count the vowels in each row.

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b. The first matrix, with a width of seven letters, has the more regular spacing of

vowels. The letter Q in the first matrix also has a U on the same row, whereas the

second matrix does not. The first matrix is clearly the better possibility.

Matrix Reconstruction by Anagramming

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Continuing the same problem, the object now is to rearrange the columns into the

original order. The rearrangement of letters to find the original plaintext order is called

anagramming. You may be able to see possibilities for complete words on some of the

rows, but the Q and the U on the seventh row provide the most obvious starting point.

To recover the numerical key at the same time, number the columns in numerical

order before starting reconstruction.

a. All the letter combinations produced by placing columns 7 and 5 together look

reasonable for plaintext. At this point, you can see that the last two rows should

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both be followed by vowels. Both the 1 and 6 columns end with two vowels. Here is

what each looks like when added to the initial two columns.

b. Both possibilities give good plaintext letter combinations, but at this point, several

words are suggested in the second match. REF.. ..CE could be part of

REFERENCE. XTW could be part of SIX TWO, and the UMB in that case would

suggest NUMBER. With these probable words, clearly column 3 follows 756.

Column 7 is the left-hand column, because the letters needed for REFERENCE,

SIX, and NUMBER are on the row above in column 4. Adding columns 3 and 4

produces the next matrix.

c. The remaining two columns are easily filled in to complete the solution.

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Incompletely Filled Matrices - Hat Diagrams

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Incompletely filled matrices are also solved by anagramming, but it is a more difficult

process because you cannot initially tell which letters are on the same row with each

other. If you know or can correctly assume the width of the matrix, you can limit the

possibilities. Consider the next cryptogram, which is expected to have a matrix width

of eight letters.

a. With a length of 76 letters and a suspected width of 8, there must be four columns

with 10 letters and four columns with 9 letters. We can show the range of letters that

could be placed in each column by trying the first four columns as the longer

columns and alternately, the last four columns as the long columns. The true

arrangement is probably neither, but it will serve to show the possible range of first

and last letters for each column.

b. These two extreme situations can be combined into a single diagram, called a hat

diagram. It is constructed by using the first diagram. Next, combine the letters that

the second diagram shows can precede the already listed letters by adding them to

the top of the first diagram. Similarly, draw a line across the bottom of the first

diagram to show the possible bottom letters. The altered first matrix is now the

completed hat diagram.

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c. The completed hat diagram can now be used as a guide to show how columns may

be aligned together. Its value can be seen if you try to place the Q in the text before

a U. There are two Us in the cryptogram. The Q is necessarily near the top of the

matrix. The U in column 2 can only be at the bottom of the matrix. The U in

column 3 can only be at or near the top of the matrix. The correct U to place with

the Q is now obvious. Lining up the Q in column 8 with the U from column 3

produces an initial reconstruction.

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d. Next, there is an X near the bottom of the matrix in column 2. It will combine well

with the SI of the first two columns to produce SIX.

e. SIX is not the only number near the bottom of the matrix. FOUR and TWO are

likely on the last two rows, and column 4 is available with RO near the bottom.

f. The E after SIX suggests EIGHT. The numbers themselves suggest the word

COORDINATES, which appears in the middle of the matrix. With these words

written in, the rest of the columns can be placed.

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g. All letters are now used, but several letters appear at both the top and bottom of the

matrix. The first word of the message is ALTERNATE, and the letters before it all

appear correctly at the bottom of columns. The L at the bottom after ONE correctly

appears as part of ALTERNATE at the top. Removing the duplicated letters and

shifting ALTERNATE to begin at the left-hand column completes the solution.

h. This solution depended on correctly identifying the width of the matrix and the

fortunate appearance of the Q and U. Without the Q and U and without any

indication of the width, a great deal more trial and error would be required for a

successful solution. Hat diagrams can be constructed for different possible widths,

for example, and probable words searched for within the structure of the diagram.

The solution is still possible in most cases, although it will often take longer than

the example did. When the same keys are reused for a period, special situations can

arise which make the solution much easier. The next chapter shows the techniques

that can be used in these special situations.

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