We will demonstrate the encryption
technique by the following example: Take the first 20 letters of the
key phrase and divide it into two halves. Within each
half, we assign 1 to the letter earliest in the alphabet,
and so on, treating 0 as the last number, and assigning
digits in order to identical letters. The two 10 digit results are then
added, digit by digit, ignoring carries. The result of the addition is
then expanded to another 50 pseudorandom digits through
chain addition. Add the first two digits in the group
together, take only the last digit of the result and
append it to the end of the group, then ignore the first
digit, and repeat the process. Take the last row, assign 1 to
the smallest digit, and so on, treating 0 as the last
number, and assigning digits in order to identical
digits. Those 10 digits are used as the top row of
numbers for a straddling checkerboard. The second row of the
checkerboard contains the highest frequency letters
ESTONIA, with a blank in the 3th , 6 th and 9 th square.
Write the digits, located above an empty square,
downwards in the first column. Complete the checkerboard
with the following letters and numbers: However, we start the filling of
the rows in the column, pointed to by the digit at the
left of that row. In our example, the start positions are
underlined. Complete until the end of that row and
proceed at the beginning of the row. We convert the plain text into
numbers according to the straddling checkerboard: In our example, several spaces
are used. However, it is recommended to omit all spaces
where legibility preserved. To determine the number of columns for
the two transpositions, we take the unequal digits (that
is, skipp if encounter an already used digit), one by
one, starting at the end of the last row of the 50
generated numbers, from right to left, and add them until
the result is more than 9. The result is the width of the
first columnar transposition. We continue with reading
off unequal digits, from right to left, and add them to
retrieve the width of the second columnar transposition: 1st transposition: 7 + 2 + 3 = 12
columns Take the 10 digits from the second half
of the key phrase, and the 10 digits used for the
straddling checkerboard and add them, digit by digit,
ignoring carries: The keys for the two
transpositions are taken by reading out the 50 numbers by
columns, using the 10 digits as a transposition key.
Again, 0 is last. The digits for the transpositions
after reading off the 23 required digits: The first transposition is a
simple columnar transposition. We use the first 12 of the
23 digits as transposition key, and fill the
transposition block with the numbers, obtained by the
checkerboard conversion. At this stage, null digits are
appended to the message, so that it will fill a whole
number of 5-digit groups. In our example, we add one null
digit. The message is then read off in
columns, using the top row digits as transposition key: The second transposition is a
disrupted columnar transposition. We use the last 11 of
the 23 digits as transposition key. The first triangular
area starts at the top of the column which will be read
out first, and extends to the end of the first row. It
continues in the next row, starting one column later, and
so on until it includes only the digit in the last
column. Then, if possible, after one full row, a second
triangular area starts, this time in the column which
will be read out second. A third, fourth, and more
triangular areas can be added, if enough rows available. Since we know that the message is 105
digits long, we know that we have to fill 9 rows with 11
digits, and 1 row with 6 digits. First, we fill the
transposition block row by row with the numbers from the
first transposition, first avoiding the triangular areas:
Next, we fill in the triangular
areas, row by row as well: Again, the message is read off in
columns, using the top row digits as transposition key: Finally, the digits are divided
in groups of 5 to get the fully encrypted message: Decrypting a message: To decrypt a message, we use the key
phrase to calculate the digits for the checkerboard and
the two transpositions. Next, we apply the transpositions
in reversed order. We create the block for the second
-disrupted - transposition, with the appropriate column
lengths and triangular areas. We fill in the encrypted
message column by column, according to the 2nd
transposition key. First, we read of the message row by
row, avoiding the triangular areas. Next, we read off the
triangular areas, also row by row. The result is filled in the first -
simple - transposition block, also created with the
appropriate column lengths, column by column according to
the 1st transposition key. Again, we read off the digits
row by row. The resulting sequence of digits is
converted to plain text, using the checkerboard. Note
that, at the end of the sequence, up to four null digits
could be added to complete a block of five, and should be
disregarded during conversion.
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