By
Randy Nichols (LANAKI) President of the American
Cryptogram Association from 19941996. Executive Vice
President from 19921994
CLASSICAL CRYPTOGRAPHY COURSE
BY LANAKI
February 22, 1996
Revision 0
LECTURE 8
INTRODUCTION TO CRYPTARITHMS AND HILL CIPHER
SUMMARY
In Lecture 8, we depart from the schedule for a real
treat. In the first part of this Lecture, we introduce
Cryptarithms by our guest lecturer LEDGE (Dr. Gerhard D. Linz).
LEDGE has already produced one of our better references on
beginning cryptography [LEDG], and I appreciate his assistance
in our course. The cryptarithms portion of this course will be
presented in three lectures and for the final book labelled
Lectures 20  23.
Following the Cryptarithms section we introduce the Hill
Cipher.
Our second guest lecturer is NORTH DECODER. Dr. Jerry
Metzger and his team are presenting you with the Crypto Drop
Box and the ACAL Listserver. The Hill cipher has six GIF files
associated with it and can be found at the CDB.
Waiting in wings patiently for my resource materials is
TATTERS to present Cipher Exchange problems.
INTRODUCTION TO CRYPTARITHMS (by
LEDGE)
Here's the first of the Cryptarithm lectures. It consists
of a general introduction to the genre including how to read
the problems. That's followed by an explanation of modulo
arithmetic. Then we look at how to identify the letters that
represent 0, 1 and 9, called digital characteristics. Then
there are two sections on making inferences, each demonstrating
a problem solution. Finally, there's a section on extracting
square roots.
Next lecture LEDGE will give some aids for solving
multiplication problems and then go into base 11 and base 12
arithmetic. Perhaps after that I can go to the more complicated
problems such as double key division.
PART I
DEFINITION:
A
cryptarithm is a mathematical problem, generally in
arithmetic, in which the numerical digits have been replaced
systematically by letters. The challenge of the problem is to
identify the digit for each letter and the key, if any.
Rules: 1. Each digit is replaced by one and only one
letter throughout the problem. 2. All digits appear at least
once in the problem. 3. No letter represents more than one
digit. 4. The numerical base, if other than ten (decimal), is
named. 5. The highest order digit of a number cannot be zero.
KEYS:
A table consisting of each of the
letters used in the cryptarithm paired with its numerical
equivalent constitutes the key to the cryptarithm. When the
digits are arranged in numerical order, either from smallest
to largest or largest to smallest or other logical order, the
letter portion of the key may spell out one or more words. The
word or words are then known as the keyword or keywords.
Generally, the constructor of the problem indicates the number
of words or the fact that the letters do not spell out words.
When the letter portion of the key consists of a word or
several words with no repeated letters (rule 3, above), the
digits are assigned in one of four ways:
1. From 0 to 9 (09). Ex. L O G A R I T H M S
0 1 2 3 4 5 6 7 8 9
2. From 9 to 0 (90) 9 8 7 6 5 4 3 2 1 0
3. From 1 to 0 (10) 1 2 3 4 5 6 7 8 9 0
4. From 0 to 1 (01) 0 9 8 7 6 5 4 3 2 1
In the first and fourth case L represents 0; in the second it
represents 9; and in the third it represents 1, etc. Higher
base arithmetic systems require additional digits according to
the size of the base. Undecimal is based on 11 digits rather
than 10. Generally the letter X or A is used to represent the
11th digit, or ten. Thus instead of a key for 09 we would have
a key for 0X or 0A.
Ex. B I G N U M E R A L S
0 1 2 3 4 5 6 7 8 9 X
Here the digit X (ten) will be replaced by S when it occurs. In
undecimal, 10 means eleven. If you do not understand the
concept of higher base arithmetic systems now, you will get an
extended treatment of this topic later in the course.
If no word is used, that fact will be stated as well as
the order in which letter equivalents are to be reported,
e.g., No word, (09), indicating that the letters for
reporting purposes are to be arranged starting with the letter
representing 0, followed by the letter for 1, then 2, etc.,
with the letter for 9 last. The letters will then appear in
random order, generally not alphabetical order.
ARITHMETIC:
Knowledge of addition, subtraction,
multiplication and division of whole numbers in base 10
(decimal) system will be assumed. Extraction of square and
cube roots will be explained later. While base 13 problems
sometimes appear among the numbered problems in the
Cryptarithms section, they and higher base problems are
generally offered as specials. More esoteric operations, such
as powers, magic squares, Pythagorean equations, etc. are also
offered as specials for those who like extra challenges.
PROBLEM STATEMENT:
In order to conserve space in the
journal, the problems in the Cryptarithms section are written
sequentially on one or more lines. I recommend rewriting the
problems in normal arithmetic format on every other line, so
as to have room for trial numbers. The process, without
skipping lines, will be illustrated with each of the normal
type of problems presented for solution. The following sample
problems to be rewritten are taken from the SeptemberOctober,
1993, issue of The Cryptogram:
C1. Square root. (Two words, 10) by EDNASANDE.
VO'TI'NG gives root VTO;  IN = NNTI;  NNNT = HONG; UIGG =
NUFE
_V__T__O
Rewritten: {VO'TI'NG
IN___
NN TI
NN_NT
HO NG
UI_GG
NU FE
C3. Division. (Three words, 90) by LI'L GAMIN.
AUSSIE v SHEEP = UE;  SHEEP = SUMAIE;  SPIBHP = LUHE
____UE
Rewritten: SHEEP/AUSSIE
SHEEP
SUMAIE
SPIBHP
LUHE
C6. Subtractions. (Two words, (09) by CAGEY KIWI.
LADIES  GENTS = GNSDGS. DAMES  MALES = NDGSS
Rewritten: LADIES DAMES
GENTS MALES
GNSDGS NDGSS
Additions and equations (mixed additions and subtractions) are
rearranged the same way.
C8. Multiplication. (Three words, 09) by APEX DX.
OTTAWA x ON = HNNTLIL + IIIEHE = TOOINRL
Rewritten: OTTAWA
___xON
HNNTLIL
IIIEHE_
TOOINRL
At this point you should understand the mechanics of the
presentation of the problems. You should also be ready to
construct cryptarithms of your own, although they may not be
suitable as yet for publication. To be suitable for
publication, the problem must conform to the rules listed on
page 1, and have a unique numerical solution. There must be one
and only one key that will solve the problem. If you have
understood the material thus far, you are ready to consider
ways of analyzing a problem to obtain the solution.
MODULO ARITHMETIC:
Since we will be dealing with the ten
digits, 0  9, but sometimes adding or subtracting them to get
numbers that are either greater than 9 or less than 0 (in
other words negative numbers), we need a way of reducing those
results back to the digits mathematically. Modulo arithmetic
is that way. If you add 8 + 5, you get 13. If you want to talk
about only the units digit of the result, you could subtract
10 from the 13 and get that units digit, 3. We say, then, that
13 = 3 (modulo 10). The 10 comes from the fact that there are
ten digits in the decimal system. When we learned addition, we
learned to carry the 10 to the next column on the left, thus
avoiding having to write a two digit number in a space where
there is room for only one: 28 +5
33 or 20 + 13 (8 + 5). In subtraction, 5  8 = 3, but 3 is
not in the range of the positive digits. Here we could add 10
to 3: 3 + 10 = 7, or 3 = 7 (modulo 10). In a subtraction
problem we get the 10 by borrowing it from the next highest
order digit in the subtrahend: 25
8 = 17 or 20  3 or 10 + (10  3) The way we learned to
subtract eliminates the negative numbers by borrowing 10 from
the 20 in 25. Modulo arithmetic is another way of talking about
the same process.
DIGITAL CHARACTERISTICS:Gaining an entry into a
problem is often expedited by being able to identify one or
more of the digits. Those most commonly identifiable with a
little bit of study of the problem are 0, 9, and 1. Zero in
particular has a number of recognizable characteristics. Add
zero to a number and the sum is that number, i.e., A + 0 = A.
Similarly, subtracting zero from a number yields that number,
i.e., A 0 = A. Multiply a number by zero and you get zero.
Subtract a number from itself and you get zero, i.e., A  A =
0. Once zero is identified, you will have the first or last
letter of a keyword, if any.
If 0 cannot be identified through any of the
characteristics enumerated above, it may yet be possible to
discover the candidates for it through a process of
elimination. Given a number, we know that the highest order
digit of that number cannot be zero. So if we have a number,
ABC, then A is not zero. Let's use that fact and any other
inferences we can make on the example multiplication problem,
C8, from page 3.
OTTAWA
___xON
HNNTLIL
IIIEHE_
TOOINRL
This problem has five different numbers with four different
beginning letters: O, H, I, and T. None of those can be zero.
The multiplier, ON, contains the digit N which, when
multiplying OTTAWA, produces a product not equal to zero.
Hence, N does not equal zero. When adding the two partial
products, E + I yields R not either E or I. Hence neither E nor
I = zero. We have eliminated seven letters as candidates for
zero. So far, at least, L, W, or R could be zero. It will take
more detailed analysis to determine which one is actually zero.
The number 9 has some interesting characteristics, one
of which mimics zero. When subtracting 9 from a number, you
must borrow from the next higher digit. The difference between
9 and the number is then one more than that number, i.e., 24 
9 = 15 contains 4 9 yielding 4 + 1 or 5. The 2 in the
original number has been reduced by 1 because of the
borrowing.
Let's look at another subtracting operation involving
9.:
247 48 = 199 or 247
48
= 199
That example includes a digit that is subtracted from itself.
That operation normally would produce zero. Here it produces 9
because of a borrowing necessitated by a previous subtraction,
namely 7  8. 4  4 becomes 3 4 yielding 9 and reducing the 2
to 1. That sort of effect is not possible in the units place of
a number because there is no previous borrowing when dealing
with whole numbers. Thus, when given a problem that includes:
ABCDE
DCFE
GHIJ
H could equal 0 or 9. More information is needed to resolve the
ambiguity. We have it here in the units place where E E = J.
There is no ambiguity in that fact since there cannot have been
any previous borrowing. So J = 0 and H = 9.
The number one can often be recognized as the highest
order digit of a number particularly when, in a subtraction
problem, it is not carried down to the answer line. Note that
in the previous example, A is the highest order of the
subtrahend, the number from which another number is to be
subtracted. It does not appear in GHIJ, the difference between
the two numbers. Clearly, it must have disappeared in the
process of borrowing. D must be greater than B, thus B  D
yields G, a number that is greater than B and necessitating
borrowing one from A, reducing it to zero. Notice than when
subtracting a larger digit from a smaller one, the resulting
difference is larger than the subtrahend digit, e.g., 5  8
yields 7 or 15  8 = 7 > 5. If you now understand
subtracting using modulo arithmetic, you will recognize that 5
 8 = 3 which = 7 (modulo 10). In modulo arithmetic we can
add or subtract the base, here 10, as many times as necessary
to produce a number in the desired range, here 0 to 9. (See
page 3, "Modulo arithmetic," 1st sentence.)
The number one can also be spotted in multiplication
since one times a number equals that number, i.e. A x 1 = A.
One times one also yields one, making it one of three digits
that when squared or multiplied by itself yields a number
whose unit digit is the same as the number squared: 1 x 1 = 1,
5 x 5 = 25, and 6 x 6 = 36. Once again, modulo arithmetic lets
us know that 25 = 5 (modulo 10) and 36 = 6 (modulo 10).
MAKING INFERENCES:(Example 1). Once you have done what
you can to spot 0, 1 and 9, you will have to rely on your
knowledge of arithmetic to determine the possibilities of the
other letters and to make decisions about their values. To see
how that works, let's work on a simple problem, the division
problem C3 at the bottom of page 2. It's reproduced below:
____UE
SHEEP/AUSSIE
SHEEP
SUMAIE
SPIBHP
LUHE
Before reading on, see what you can do with this problem.
Remember, the key is three words, 90. When you are ready, read
on for the solution.
In the above problem, we are helped by being able to
find all three of the digits, 0, 1, and 9. In the first
subtraction, I  P = I. In the second subtraction, E  P = E.
Both facts make 0 = P. Note also that U x P = P and E x P = P,
both consistent with P = 0, but not sufficient to prove that P
is zero, since both of those equations, modulo 10, could be
true for P = 5, e.g., 3 x 5 = 15 and 7 x 5 = 35, both ending
in 5. Next for the letter that represents one. U x SHEEP =
SHEEP. Hence, U must be 1. Note also in the second
subtraction, U P = blank or zero. Since we know P to be zero,
U must be 1. These chains of reasoning are typical in the
solution of cryptarithms.
Now let's find the letter for 9. In the first
subtraction, note that U  H = U. That could make H be zero or
nine. In the absence of other information, you could not be
sure which of those is true. Here you already know that zero
is represented by P. Thus, H = 9.
You now have a lot of useful information. Let's look at
the multiplications for more. U x SHEEP is 1 x SHEEP = SHEEP,
not much help there. E x SHEEP = SPIBHP. You can replace the
identified letters with their digital equivalents and get: E x
S9EE0 = S0IB90. E x 0 = 0, so far so good. E x E = 9 (modulo
10).
What are the possible values of E. E could be 3, as 3 x
3 = 9, or 7, since 7 x 7 = 49 or 9 (modulo 10). Let's try out
each possibility. 3 x S9330 = ??7990 or ??IHHP, not consistent
with SPIBHP. So E is not 3. E must then be 7. Let's check that
and see what else you can uncover. 7 x S9770 = ??8390 making I
= 8 and B = 3. Now SPIBHP is S08390. 8 is preceded by 0 so 7 x
S must end in 4 since we are carrying a 6 from the
multiplication of 7 x 9 and 6 + 4 = 0 (modulo 10).
Hence, S must be 2 as 7 x 2 = 14. SPIBHP becomes 208390.
To bring order out of all this information, we need to
reconstruct as much of the key as we can.
9 8 7 6 5 4 3 2 1 0
H I E B S U P
The missing letters are A, L, and M, all found in the second
subtraction. Entering what is known now makes that subtraction:
21MA87
208390
L197
Remember, you can check a subtraction by adding the subtracter
and the difference to get the subtrahend. Here, 0 + 7 = 7; 9 +
9 = 18, carrying 1 to the next addition; 1 + 3 + 1(carried) =
5, so A = 5. Since L and M are both less than 8, representing
as they do the two remaining unidentified digits, 6 and 4. L +
8 = M (modulo 10), or 6 + 8 = 14 or 4 (modulo 10). So L = 6 and
M = 4. The key becomes HIELAMBSUP.
You could also have worked with the first subtraction,
as it contains the letters M and A. Try that now using the
partially reconstructed key above. The results should be the
same.
MAKING INFERENCES:(Example 2). The multiplication
example, C 8, given on page 3 presents somewhat more
difficulties than the previous one, as none of 0, 1, or 9 can
be initially identified. There are enough other clues, however,
to make the solution come through a straightforward series of
inferences. Before reading on, see what you can recover from
that problem on your own. When you are in a thoroughly stuck
place, read on for some help, or the complete solution.
Here is the problem:
OTTAWA
___xON
HNNTLIL
IIIEHE_
TOOINRL
It was determined that zero is represented by L, W, or R. On
page 3 the key is stated to be three words, 09. First, notice
that N time OTTAWA results in a 7digit number and that O time
OTTAWA results in a 6digit number, the same length as OTTAWA.
Examine the second product carefully. O x OTTAWA = IIIEHE. The
highest order I (first digit of IIIEHE) results from the
product O x O. O cannot = 1 for 1 x OTTAWA = OTTAWA. O cannot
be as large as 4, for 4 x 4 = 16, which would add a seventh
digit to the product. So O = 2 or 3. 3 x 3 = 9, which would
make I at least 9. Looking at the problem again, the first I is
added to H giving T, a digit, but adding anything other than
zero to 9 produces a two digit number. So I cannot be 9 and O
cannot be 3. So O = 2.
With O = 2, I must be 4 or 5 since O x O is 4 or could
be 5 if a 1 is carried from the previous multiplication (2 x
T). So we have the following multiplication: 2 x 2TTAWA =
444EHE or 555EHE. We can divide each of those products by the
multiplier, 2, getting respectively 222??? and 277???. The
first quotient gives 222??? to represent OTTAWA  not possible
(it would be OOO???). The second quotient is consistent in
making T = 7 and I = 5. OTTAWA becomes 277AWA. IIIEHE =
555EHE.
Now let's look at the first product, N x 277AWA =
HNN7LIL. The product must be less than 10 x OTTAWA and that
makes its first digit less than O. There is only one such
digit, so H = 1. You could now divide 1NN7LIL by various
values of N to find a quotient that begins 277. It's easier,
however, to look at the addition of the two partial products
as they contain N's.
1NN7LIL
555EHE_
7225NRL
Since 1 + 5 = 7 (highest order pair), N + 5 must be > 10.
That would allow a carried 1 to be added to 1 + 5. N + 5 +
1(carried from the previous N + 5) = 2 (modulo 10). That makes
N = 6. Let's pause to construct a partial key using the
information so far identified. The key table becomes:
0 1 2 3 4 5 6 7 8 9
H O I N T
It's also possible to rewrite the problem substituting digits
for the identified letters:
277AWA
____x26
1667L5L
555E1E_
72256RL
The sums produce the following modulo 10 equation: E + 7 = 5; L
+ 1 = 6; E + 5 = R. The equations ignore possible carries of 1
which you may have to supply. Accepting that contingency, the
first equation produces 8 as the only possible value of E. The
third equation then makes R = 3 since there is no carry
possible. The second equation makes 4 and 5 possible values of
L, but 4 is the only available digit. Of the three letters that
could be zero only W is left unidentified. Only 9 is left for
A. As a check, 9 x 6 = 4 or L and 9 x 2 = 8 or E, checking out.
The key has become WHORLINTEA as the solution.
EXTRACTING SQUARE ROOTS:

Not understanding the following algebraic analysis of the
process of extracting a square root is no barrier to
understanding how to follow the method. It is included
here for those who are interested in understanding how
it is that the method works.
 When
squaring a number, one doubles the number of digits of the
original number. If you square 9, you get 81, 2 digits.
Squaring 3 you get 09. Square 35 and you get 1225, 4 digits.
Square 12 and you get 0144. When extracting the square root of
a number, you take cognizance of this fact by making a mark
after every two numbers beginning from the decimal point in
both directions. So 45678.96 becomes 4'56'78.96' with the
initial 4 being understood as 04. As many pairs 00 can be
added after the last mark without changing the value of the
number.
The
first trial root is the largest number whose square is equal to
or less than the initial pair of numbers. We'll call that trial
root x. (One could use the largest number whose square is equal
to or less than the initial two or more pairs of numbers. That
makes no theoretical difference although in practice that's
more difficult.) The square of x is then subtracted from the
first pair of numbers. The next pair of numbers is appended to
the difference as in a long division problem.
Now there is room for a 2digit root whose first digit is
x. If we call its second digit y, the root becomes 10x+y.
Multiplying that number by itself produces 100x} + 20xy +y}.
That can be factored to produce 100x} + y(20x + y). As x} has
already been subtracted from the highest order two digit number
of the original number it remains to subtract y(20x + y) from
the current remainder to make sure that y is not too large and
to determine a new remainder.
Now let's extract the square root of 45678. First mark
after every second number starting at the decimal point. _______
{4'56'78 The first pair of numbers is 04. The
square root of 4 is 2, a number we'll
place above the 4. We'll then square 2
getting 4 and placing it under the 4 in
the number and subtracting. Since the
remainder is zero we'll merely pull down
the next pair, 56, and produce our trial
divisor. The work looks like:
2______
{4'56'78
4___
56 The trial divisor is produced by
multiplying the root we have, 2, by 20
making 40. 40 divides into 56 one time
(trial y) which is added to 40 making
41. The trial y, 1, is placed over the
second digit of the new pair, 6. 41 is
multiplied by y (1) and subtracted from
56. Then 78 is pulled down at the end of
the difference. The work looks like:
2__1___
{4'56'78
4___
41 56
41___
15 78 Again the root, now 21 is multiplied by
20 giving 420. 1578 divided by 420
gives 3, our new y which is added to
420 giving 423. 3 x 423 or 1269 is
then subtracted from 1578 giving a
remaider of 309. The 3 is put above the
8 of 78 making the new root 213 with a
remainder. If it were desired to extend
the calculation to the right of the
decimal point, a pair of zeroes could
be appended to the remainder and the
process repeated with a decimal point
placed in the root after the 3. The
work without going past the decimal
point becomes:
2__1__3
{4'56'78
4___
41 56
41___
423 15 78
12_69
3 09
You can check that by squaring 213 (213
x 213) and adding 309. You should get
45,678. Practice by taking the square
root of another 5 or 6digit number and
checking your outcome. Solve C1 on
page 2 for homework. If you want more practice
work, find divisions, square roots, and multiplication problems
in the two current issues of The Cryptogram. Do the subtraction
problem, C6 on page 3 if you wish. Discuss problems you may
have with your mentor. If you have suggestions, questions, or
other reactions you wish to share with me, my address is:
Dr. Gerhard D. Linz
2649 Tanglewood Road
Decatur, GA 300332729.
My email address is fs01gdl@panther.gsu.edu.
I hope your pleasure in solving Cryptarithms is enhanced by this
presentation. Next time I'll respond to any concerns you have. I
also plan to give you some more tools for multiplication and
introduce counting systems based on 11 and above.
LEDGE January 2, 1996
OBSERVATIONS ON SQUARES (LANAKI)
Dr. Andree gives us some hints on squares and square
roots. [OKLA]
S1 Squares end only in 0, 1, 4, 5, 6, or 9. S2 If
(...S)**2 ends in ...S, then S=0,1, 5 or 6. S3 If (...S)**2
ends in B n.e. S, then S= 2,3,4,7,8 or 9 B= 1, 4, 6 or 9 S4 If
(..X)**2 ends in 0 1 4 5 6 9 then (...X) ends in 0 1,9 2,8 5
4,6 3,7 S5 If N contains k digits, then N**2 contains either
2k1 or 2k digits.
HILL CIPHER SYSTEM (by NORTH
DECODER)
There are two basic ways to prevent the telltale
behavior of plaintext letters from showing through in
ciphertext. One method is to vary the ciphertext letter that
replaces a given plaintext letter. That is the solution offered
by the Vigenere and other polyalphabetic systems. A second
technique is to encipher the plaintext in chunks of several
letters at a time. The Playfair system provides a compact
method for enciphering digraphs, that is, pairs of letters.
While the Playfair does disguise the behavior of individual
letters, even better would be a system that operated on letters
in groups of three letter (or four or five or ...). It seems
that no convenient pencil andpaper method for handling such
trigraphic (or quadgraphic or ...) encipherment has been
devised.
In 1929, Lester Hill [HIL1, HIL2] described an algebraic
procedure that allows encipherment of plaintext letters n at a
time (that is, in ngraphs), where n can be any positive
integer 1,2,3,.... Hill's Cipher could be carried out by hand
probably without too much hardship for groups of letters up to
five. After that, it would become a challenge to keep the
computations accurate. However, on a computer it would be
feasible to work with large groups of letters, and it seems
that plaintext enciphered in such a system using say 10graphs
would be difficult to crack.
The first step in using Hill's system is to assign
numerical values to the 26 letters of the alphabet. There is
nothing sacred about 26 in the system. The ideas work just as
well for alphabets of any size. So it would be possible to add
a few punctuation marks to the usual alphabet to get say 29
symbols, or to work entirely with data in binary form with an
alphabet of just two symbols. Also the numerical equivalents of
the letters of the alphabet could be assigned in some arbitrary
way, which would probably add to the security of the system.
For these notes, the 26 letter alphabet will be used, and
letters will be given their standard numerical equivalents,
namely a = 00, b = 01, ..., z = 26. The encipherment of plaintext is most neatly described using
matrix multiplication. A matrix is a rectangular array of numbers
such as:
 1 5 3 
M =  0 2 1 
That particular matrix has 2 rows and 3 columns. More briefly,
it is a 2 x 3 matrix. In certain cases, the product of two
matrices can be computed. The rule for multiplication requires
that the number of columns in the lefthand factor match the
number of rows in the righthand factor.
For example, if:
 2 5 0 1 
N =  7 6 1 3 
 3 3 0 1 
Then the product MN can be formed since M has three columns and
N has three rows. On the other hand, the product NM is not
defined.
For these two matrices the product is:
 1 5 3   2 5 0 1   46 44 5 19 
MN =  0 2 1  X  7 6 1 3  =  17 15 2 7 
 3 3 0 1 
The upper lefthand entry in the product matrix is produced by
multiplying each number in the first row of M by the
corresponding number in the first column of N, and adding the
results:(1)(2)+(5)(7)+(3)(3)=46. That explains why the number of
columns in M must match the number of rows in N.
The second number in the first row of the product is
produced in the same way by multiplying the first row of M
times the second column of N: (1)(5)+(5)(6)+(3)(3)=44.
And so on, the third number in the first row of the
product is produced by multiplying the first row of M by the
third column of N, and finally, the fourth number in the first
row of the product comes from multiplying the first row of M by
the fourth column of N.
To produce the second row of the product, the second row
of M is used in place of the first row of M in the preceding
computations. So, for example,(0)(2)+(2)(7)+(3)(3)=17 gives the
first number in the second row of the product matrix. If M had
more rows, each would be used in turn in the same way to
produce one more row in the product matrix.
If you think about the multiplication process described
above, you will see that the result of multiplying an r x s
matrix and an s x t matrix will be an r x t matrix.
In the application of matrix multiplication to the Hill
Cipher system, all arithmetic will be carried out modulo 26. In
other words, any time a number appears which is 26 or larger,
it is divided by 26, and the number is replaced by the
remainder of the division. In the example above, the
computation of the top left number in the product of M and N
could be written as (1)(2)+(5)(7)+(3)(3) = 2 +35 + 9 = 2 + 9 +
9 = 20 (mod 26). The symbol (mod 26) is added here just to
indicate there is funny arithmetic being used, namely that
arithmetic is being done modulo 26. If the alphabet had 29
symbols instead of 26, operations would be carried out modulo
29. Since all the examples here will be done modulo 26, the
indicator (mod 26) will be omitted from now on. So we will
write the example above as:
 1 5 3   2 5 0 1   20 18 5 19 
MN =  0 2 1  X  7 6 1 3  =  17 15 2 7 
 3 3 0 1 
To encipher the plaintext message "send more money",
first the message is rewritten in groups of letters of the
selected length. For this example, length three will be used, so
the message becomes "sen dmo rem one ykz", where two
nulls have been added to fill out the last group. Next an
enciphering matrix, or key, is selected. If letter groups of
size n are being used, an n x n enciphering matrix will be
needed.
For this example, the 3 x 3 matrix:
 1 7 22 
E =  4 9 2 
 1 2 5 
will be used. Notice that the numbers in the matrix might as well be
selected between 0 and 25 since all arithmetic will be done
modulo 26 anyway. To encipher the first three letter group of
plaintext, it is written as a 3 x 1 matrix, say P, the letters
are replaced by their numerical equivalents, and the matrix
product EP is computed.
Here are the details:
 1 7 22   s   1 7 22  18  20  U 
EP =  4 9 2  X  e  =  4 9 2  X  4 =  4=  E 
 1 2 5   n   1 2 5  13  13  N 
So the first three letters of ciphertext are UEN. The second
trigraph is enciphered as:
 1 7 22   d   1 7 22   3   5  F 
EP =  4 9 2  X  m  =  4 9 2  X 12 = 18=  S 
 1 2 5   o   1 2 5  17  19  T 
Continuing in this way, the ciphertext is found to be UEN FST
XYH LZI UCN, or, in traditional five letter groups, UENFS TXYHL
ZIUCN. Notice that in this example, repeated plaintext letters
are replaced by different ciphertext letters, and repeated
ciphertext letters represent different plaintext letters.
Deciphering requires a second matrix that undoes the
effects of the enciphering matrix. For the enciphering matrix
given above, the deciphering matrix, or deciphering key, is:
 21 23 18 
D =  6 23 6 
 9 7 15 
and operating on the first ciphertext trigram UEN gives:
 U   21 23 18   20   18   s 
D  E  =  6 23 6  X  4  =  4  =  e 
 N   9 7 15   13   13   n 
Operating on the remaining ciphertext trigram produces the rest
of the plaintext message.
The enciphering key matrix cannot be selected
arbitrarily.
For example, the matrix:
 0 0 0 
Z =  0 0 0 
 0 0 0 
would convert every plaintext message into the ciphertext
AAAAAAAA. To allow unique decipherment, an n x n enciphering key matrix
should convert different plaintext ngrams into different
ciphertext ngrams. An n x n matrix that behaves that way is
called nonsingular.
There are a number of more or less efficient tests for
nonsingularity. Here is one test that involves the determinant
of an n x n matrix. The determinant of a square matrix is a
number computed from the entries in the matrix. The definition
builds up from small matrices to larger ones. First the
determinant of any 1 x 1 matrix is defined to be the number
that is the entry in that matrix. Thus det 7 = 7. To compute
the determinant of a 2 x 2 matrix, step across the entries in
the first row of the matrix, multiply each entry by the
determinant of the 1 x 1 matrix that appears when the row and
the column the entry appears in are eliminated from the matrix.
The numbers produced in this way are alternately added and
subtracted to produce the determinant of the matrix.
Here's an example.
 4 3 
det  8 2  = (4) (det 2 ) (3) ( det 8 ) =
4 x 2  3 x 8 = 8  24 = 16.
The determinant of a 3 x 3 matrix is produced in the same way:
step across the first row, multiply each entry by the
determinant of the 2 x 2 matrix that appears when the entry's
row and column are crossed out, and alternately add and subtract
the resulting numbers.
For the matrix of the earlier example, the computations,
carried out modulo 26 this time, look like:
 1 7 22  9 2  4 2 
det  4 9 2  = 1 x det 2 5  7 x det  1 5 
 1 2 5 
 4 9 
+ 22 x det  1 2  = 1 x 41  7 x 18 + 22 x(1) = 23
The computation of the determinant is extended to larger square
matrices in the same pattern. More efficient ways to compute
determinants are discussed in textbooks on Linear Algebra.
The importance of the determinant is that a matrix is
nonsingular (and so usable as an enciphering key matrix in
Hill's cipher) if and only if its determinant is relatively
prime to 26. The matrix above has determinant 23 which is
relatively prime to 26, so it is a legal enciphering key. More
generally, if the alphabet used for the plaintext is made up of
m symbols, then the usable enciphering matrices are those with
determinant relatively prime to m. In the case of an alphabet
of 26 symbols, the determinant of a usable matrix must be odd
but not 13. Notice that if the size of the alphabet is
increased to 29 by adding a few punctuation symbols, many more
legal enciphering matrices will be available, both because
operations will now be carried out modulo 29, and also because
every number from 1 to 28 would be an acceptable value for the
determinant of an enciphering key matrix.
Once an enciphering key matrix has been selected, the
companion deciphering matrix needs to be computed. There are
some reasonably efficient methods for finding the deciphering
matrix. The method given here is easy to describe, but not very
efficient. Check out a Linear Algebra text for better methods
to handle matrices larger than say 4 x 4.
 The first step is the computation of the determinant
of the enciphering key E. If det E = e, then a number d is
needed such that ed= 1 (mod 26). For a relatively small
modulus such as 26, the d can be found by trial and error.
Simply compute e times 1,3,5,7 ,9,11,15,17,19,21,23, and 25
until a product equivalent to 1 modulo 26 appears.
For larger alphabets with say m letters, solving ed =1
(mod m) can be carried out in a more sophisticated way using
the Euclidean Algorithm, for example.
Check a Number Theory text for details. Set the number
d aside for a minute.
 Second, each number in the enciphering key matrix is
replaced by the determinant of the matrix obtained when the
element's row and column are erased from the matrix.
 Third, plus and minus signs are prefixed to each entry
in the new matrix in a checkerboard pattern starting with a
plus sign in the upper lefthand corner.
 Next, the matrix is flipped over the diagonal from the
upper left corner to the lower right corner so that the first
row be comes the first column, the second rows becomes the
second column, and so on.
 Finally, each entry in the matrix is multiplied by the
d computed in the first step.
 The resulting matrix is D, the deciphering key matrix.
Here are the computations that produce the deciphering key D of
the example above. The determinant of the enciphering key E has
already been computed: det E = 23. Since (17)(23) = 1 (mod 26),
it follows that d = 17. Next, the 1 in the upper left handcorner
of E is replaced by:
 9 2 
det  2 5  = (9)(5)(2)(2) = 45  4 = 41 = 15 (mod 26).
where, in the last step, 41 has been reduced modulo 26.
The replacement for the 7 in the first row and second column is:
 4 2 
det 1 5  = (4)(5)(2)(1) = 18 (mod 26).
The replacement for the 9 in the second row and second column
is:
 1 22 
det 1 5  = (1)(5)  (22)(1) =  17 = 9 (mod 26).
When all nine entries in E have been replaced, the matrix looks
like:
 15 18 25 
 17 9 21 
 24 18 7 
Adding the plus and minus signs in a checkerboard pattern
produces and replacing negative numbers by equivalent positive
numbers modulo 26 gives:
 15 18 25   15 8 25 
 17 9 21  =  9 9 5 
 24 18 7   24 8 7 
Flipping over the diagonal gives:
 15 9 24 
 8 9 8 
 25 5 7 
Finally, multiplying every entry of the last matrix by the
d=17 computed earlier, and reducing the entries modulo 26, the
result is:
 (17)(15) (17)(9) (17)(24)   21 23 18 
D =  (17)( 8) (17)(9) (17)( 8)  =  6 23 6 
 (17)(25) (17)(5) (17)( 7)   9 7 15 
Arithmetic done with matrices has a lot in common with
arithmetic done with ordinary numbers. The n x n matrix whose
entries are all 0 except for 1's down the diagonal from the
upper left to the lower right is called the identity matrix. It
plays a role in matrix multiplication similar to the role 1
plays in multiplication of numbers. That is, for any number m,
(1)(m) = m, while for any n x k matrix M, it is easily checked
that IM= M. Moreover, for each number r (provided r is not equal
to 0), it is possible to find a number s so that sr=1. The
number s is called the multiplicative inverse of r, and is
written as r^{(1)} (that is, r to the 1 power).
Likewise, for each n x n matrix M (provided it is nonsingular),
there is an n x n matrix N for which MN=I. The matrix N is
called the inverse of M, and is written as M^{(1)}.
The Hill Cipher system can be expressed compactly using some
algebraic notation. To encipher a plaintext ngram using the
Hill Cipher, a nonsingular n x n matrix M is selected. The
ngram is written as an n x 1 matrix P, and the ciphertext is
the n x 1 matrix C determined by the equation:
The deciphering matrix is the inverse of M. When the ciphertext
C is multiplied by M^{(1)}, the plaintext is recovered:
M^{(1)} C = M^{(1)} MP = IP = P.
Hill suggested that a good choice for an enciphering key matrix M
is one that turns out to be its own inverse. If M = M^{(1)},
Mi s called an involuntary matrix. The advantage gained is that
it is not necessary to compute the deciphering key. There are a
number of methods that will automatically produce involuntary
matrices, so the process of finding involuntary matrices does not
have to proceed by trialanderror. In any case, almost all
papers written about the Hill Cipher system following Hill's time
down to the present day assume the key is involuntary.
It seems that Hill and a partner (Weisner) filed a patent
(Message Protector, patent number 1,854,947) for a mechanical
version of the Hill Cipher in 1929, which, according to Kahn
[KAHN], used an involuntary matrix enciphering key so that the
same machine could be used to both encipher and decipher.
The Message Protector patented by Weisner and Hill
provides a mechanical means of doing matrix multiplication. The
device illustrated in the patent application is more accurately
described as authentication indicator rather than a
cryptographic mechanism. The principle of operation is very
simple. The active component consists of three gears on an axle
which are connected to three accumulator gears by chains. The
three accumulator gears all have the same number of teeth (101
in the patent), and they can rotate independently. The three
gears on the axle have 101, 202 and 303 teeth. As the axle is
turned through a certain amount, the accumulator gears turn one,
two and three times as far respectively. The teeth on the
accumulator gears are numbered from 0 to 100, and small gear on
the axle also has its teeth numbered from 0 to 100.
Now suppose the three accumulator gears start in position
0,0,0. If the axle turned through an amount that rotates its
small gear through 43 teeth, then accumulator gear one will read
43, accumulator two will show 86 and accumulator three will show
28. The last value occurs since the third accumulator wheel will
have made more than one revolution. If the starting position of
the accumulator wheels had been 11,91,4, then the axle rotation
through 43 teeth would leave the accumulators showing 53,76,32.
In essence, the accumulators are modulo 101.
On the actual devise, there are six axles, and their gears
can be moved to engage the accumulator drive chain one axle at a
time. The placement of the gears on the axles vary from one axle
to the next. On the illustrated machine in the patent, the
sequence is:
axle 1: 101,202,303
axle 2: 202,303,101
axle 3: 303,101,202
axle 4: 101,303,202
axle 5: 202,101,303
axle 6: 303,202,101
Suppose the accumulators begin showing 0,0,0. Keeping track for
now of only the total on the accumulator that connects to first
gear on each axle, here is what happens as the axles are turned
as follows:
axle 1: 23,
axle 2: 10,
axle 3: 88,
axle 4: 17,
axle 5: 41, and
axle 6: 51.
Initially, all the axles are disengaged from the accumulator
drive chain. (Keeping in mind the number of teeth on the first
gear on each axle.) axle 1 is engaged, turned 23, and the
accumulator shows 23. Axle 1 is disengaged, axle 2 is engaged,
turned 10, and the accumulator shows 43. Axle 2 is disengaged,
axle 3 is engaged, turned 88 , and the accumulator shows 4. Axle
4 is disengaged, axle 4 is engaged, turned 17, and the
accumulator shows 21. Axle 4 is disengaged, axle 5 is engaged,
turned 41, and the accumulator shows 18. Axle 5 is disengaged,
axle 6 is engaged, turned 51, and the accumulator shows 70.
The final total on the on that accumulatorrepresents the
computation:
(1)(23)+(2)(10)+(3)(88)+(1)(17)+(2)(41)+(3)(51) = 70 (mod 101).
Likewise the value on the accumulator connected to the second
gear on each axle shows the result of the operation:
(2)(23)+(3)(10)+(1)(88)+(3)(17)+(1)(41)+(2)(51) = 2 (mod 101).
Matrix notation can be used to express to whole operation
compactly as:
 23 
 10 
 1 2 3 1 2 3   88   59 
 2 3 1 3 1 2  .  17  =  55 
 3 1 2 2 3 1   41   54 
 51 
where arithmetic has been carried out modulo 101. To use the machine to
authenticate a check for example, six numbers, between 0 and 101,
are selected from the check. Perhaps the dollar amount of
$1230.45 could be split up as 12 and 30 and the cents could be
ignored. The check number of say 22131 might contribute three
more numbers, 2, 21, and 31. Finally, the date of the check,
maybe January 25, 1996 might contribute a sixth number, say 25.
Of course, people must agree on how these numbers are selected.
The check writer runs the six values through the Message
Protector as described above, and the resulting triple of values
is stamped on the check. The bank, before cashing the check,
operates on the same six numbers with its Message Protector, and
makes sure that the numbers produced on the accumulators matches
the ones stamped on the check, thus being sure that none of the
important figures on the check have been changed.
Although the Message Protector is a clever engineering
construction, there are certainly many obvious mechanical
shortcomings as well as weaknesses in the cryptographic system
which probably explains why the machine never became popular. In
fact, it's not clear if any were actually constructed. It would
take a good salesman to get people to spend money on a machine
to multiple 3 x 6 and 6 x 1 matrices. There did not seem to be
and reasonable way to change the gear sizes. If a key matrix
with entries besides 1, 2, and 3 were wanted, the number of
teeth on the gears would soon become so large that the structure
would have to be made pretty large, instead of the shoebox size
Weisner and Hill diagrammed.
Weisner and Hill also explain how the Message Protector
could be modified to act as a cryptographic devise. First of
all, the numbers on the various gears would be replaced by
letters, and the number of teeth on the accumulator gears would
be 26 so that the arithmetic operations would be carried out
modulo 26. Next, the axles would now carry six gears each, with
the number of teeth on each gear being a multiple of 26. There
would be six accumulators, so that six plaintext are converted
to six ciphertext letters. They say that the number of teeth on
the various gears "have to be selected according to certain
mathematical principles". What they mean, of course, is the
6 x 6 matrix, each entry of which gives the multiple of 26 that
gives the number of teeth on the corresponding gear, has to be
nonsingular modulo 26. It is suggested that the matrix may be,
but does not have to be, selected to be involuntary.
The gearing in the devise cannot be changed easily, and
certainly cannot be changed arbitrarily, so it seems the gearing
set was intended to be selected once and for all. Since that
pretty much makes the device cryptographically pointless, the
inventors proposed that a plaintext message first be converted
to a preliminary ciphertext according to so system left
unspecified, but they probably had something like a Playfair in
mind. The resulting ciphertext is then passed through the 6 x 6
Message Protector, to yield an intermediate ciphertext which is
then passed through a third and final encipherment using another
unspecified cipher system. The final ciphertext is transmitted,
and the authorized recipient reverses each of the three
encipherments to recover the original plaintext. It's not very
clear how much additional security has been introduced passing
the text through the Message Protector.
Nearly all discussions of cryptanalysis of Hill enciphered
messages begin with the fairly generous assumptions that the
cryptanalyst knows that an involuntary key matrix of known size
has been used, and also knows the numerical values assigned to
the alphabet letters. The only unknown is the particular key
matrix used to encipher the message. For a key matrix of size 2
x 2, a brute force attack is feasible since there are only 736 2
x 2 involuntary matrices. As the size of the key grows, a brute
force attack is no longer practical. For larger key sizes, no
specific cryptanalytic approaches have been published. But,
several authors given more or less detailed descriptions of
cryptanalysis, with examples for small key size (2 x 2, 3 x 3)
using the classic probable word or crib technique. That is, a
piece of plaintext is assumed to appear in the message, and it
is tried in each possible position. At each test location, a
number of equations must be true if the crib is to generate the
ciphertext at that spot. It turns out that even with a
relatively modest crib (3 letters for a 2 x 2 key, and 4 for a 3
x 3 key), most positions can be eliminated as impossible by
applying a few principles of linear algebra. Each possible crib
location will produce a candidate matrix key. A trial
decipherment of the ciphertext is made. If recognizable
plaintext results, the cryptogram is broken. If not, the crib is
moved along to the next possible spot, and the process is
repeated. For details on see on cryptanalysis of the Hill
Cipher, see [LEV1], [LEV2], [LEV3], [SINK], [MELL].
NORTH DECODER advises that the Hill cipher patent diagrams (GIF
format) scanned in reasonable well into the CDB. If you would
like to look at them, the files they are at the CDB in: Hill
Patents
There is a freeware gif file viewer at the CDB in:
/msdos/gifviewers
{well, it doesn't appear to be there; but you're browser
likely supports GIF viewing.JP}
HOMEWORK SOLUTIONS FROM LECTURE 7
FRE2. K2. (105) Another species. {sauvage,fp=ST] MELODE
P Q N X B M H Q I Q A B C I Q D K E X Q B Q O Q
P' W M R R Q; D K E X Q B Q O Q U Q I Q E Q Q M C
T E X R X B X D Q , X P Q A B K P' W M R R Q N Q
V C Q N W K B O Q U M C B B X Q E Q Q A B K C
N W K B A K C D K U Q.
Solution reads: (PRIMITIVE)
Le citoyen est une variete de l'homme;
vavariete degeneree ou primitive, il est a l'homme ce que chat
de gouttiere est au chat sauvage. FRE3. K2. (87) (jamais, A=b)
It's fun trying. GUNG HO D G X Z Q N
J D P M C J P U P L S U E' Z D Z D H U Q J S E J S N P U Q E Z H
Z D P M J H  K N D P: G Z K U D I Q S N U , G Z H S P D L S U,
U Q G U P O Z H U P . * R J I Q U I U G G U Solution reads:
(AMOUR)
Il y a trois choses que j'ai aime
toujours et jamais compris: La peinture, la musique, et les
dames. Fontenelle. FRE4. PAT from [GIVI] page 13.and ff. (130)
Solve and recover key(s). YJXMG
XBXUF JGECU JEBZD XAMNM ZDFLG FAFNJ OFNDJ GVJXE FNNME VRJZJ
KAFNB FNZAG NCUJE BNRUX OFNJG NNXKX FELGF BJRVF NOFUI FXAAF
GTFVR FAFKU FNBJE NADXN VMXUF PAT format reads: JAIOU IDIRE
AUNGR ANDPH ILOSO PHEQU ELESA MESHA UTAIN ESSON TCAPA BLESD ESPLU
SGRAN DSCRI MESAU SSIBI ENQUE DACTE SMERV EILLE UXETC ELEBR ESDAN
SLHIS TOIRE. ITA2. K2. (88) ( ne, han, con) Thirty days hath
September. LABRONICUS I D S A I K Q
W P L A I K A L B S C M D S P L A K E D W Z S, U W O U A L S R S
I I S C M D S . Q W B S A I L I I L P S A ' S O A L. I O I I W U
Z W K Z I D W A S V K A I D S A U I O A L. Solution reads:
(CALANDRIO)
Trenta di contra Novembre con Aprile,
giugno e Settembre. Di ventotto ce n'e uno. Tutti gli altri ne
han trentuno. ITA3. K2. (117) (sulla, f=I). La frode necessaria.
MICROPOD G Z Q K E A F S Z L T K F Q
A Q S F N F Q K G K Q T G G Z P Z Q F R A T J Z E F N S Z M T Z
J S A S Z R A P T D A F F Q K G K Z L Z S S K E O F J F Q Q T J
K R A E Z F Q Z S S Z H F J S F M T F G G K E O F L F J Q Z G A
J X T S Z J D. Solution reads: (PERFIDO)
La sicieta puoesister esolo sulla
basediunacerta quantitadibugie esolo apatto che
nessunodicaesattamente quello che pensalinyutang. SPA1. BARKER Z K E P C U K Y T C Y D M S R V C T P E R
A Z P Z N D Z K G C T Y R Z K R N T D G R Y C V K K S T P Q D P
E R M K T C Y G R Z Y P Q P M P E K E E C M K S C Z S K E R G R
T C M U R U C Z S R. Partial solution: no key. TADI MAS RESULTO
HERIDO Y SPA2. K2. (96) (deseo, f=R) Musica. D. STRASSE T I Z Q B J N A Z K J K T F Z N B P L T B
B F K N A G B N A G K T F P J G T P A O Z F M B F S J G H N B R
T B T I K T N Z G B I Q B B P K J I Q Z I B J M P B B J N A Q G
A O J M B M Z I Y Z N. Partial solution: (HOMBRES) UNA TEORIA
POPULAR ES QUE SPA3. (122) (ulado, MZ=qk) Flight? LIFER N S P Y K I X P U A K P Z D X P S P E X K
R L K O K A X T S P Q K D X R K R R S S I N K Y K R L A R S D K
T Q L D L P X K T A S Q X S P X P R S O S P R X J K R K T O A S
T S P Q X L S D O A X I S A E C S D L R S C P V D L N L B A X O
C D K R L. Partial solution: (DIPLOMAT) BENJAMIN FRANKLIN
ENVIADO POR2. K2 (96) (tenta; gj=NQ) Machine Age? YO TAMBIEN E P E J T X D U R T C J Z X G C V R J D J
X I N R S O C H C D T C V R P U C D V R J Z J U D C T J H J D G
X U M P C H J A X H X O X T J T V R J A J U A C M C B J S X. *O.
*T R T M X I H *Q X U J D Solution reads: (VIDAREL)
Vivemos numa epoca que se orgulha das
maquinas que pensam e desconfia de todo homen que tenta fazelo.
 H. Mumford Jones. POR3. K1. (nossos va) Letter to horseman?
ZYZZ U C U C G V C J F D E F W E O C
B G C V S I H C L I T I W F Y C V F U H F W F T L F R F B C H W
F C E S H I L F G I C D E G T I J H C V G R P C V C J F V D E F
W F H C V L F V F H J I S K I X J I Z U I G V T I V V I V B C D
E F G H I V V C I F Y K F R F T W F V. Partial Solution reads:
(VAQUEIRO): Papai sabe que tu.
NEW PROBLEMS
C1 Give two solutions to: (BE)**2 = ARE
C2 Square root: [OKLA] [OKLI]
R, A, T, S

Q UA RT ET
A

T UA
T SI

U RT
A UT

E AO ET
E ES UB

R AR
From Sinkov [SINK] two Hill system problems:
Hill1
Decipher the message:YITJP GWJOW FAQTQ XCSMA ETSQU
SQAPU SQGKC PQTYJ
Use the deciphering matrix  5 1 
 2 7 
Hill2
Decipher the message:MWALO LIAIW WTGBH JNTAK QZJKA ADAWS
SKQKU AYARN CSODN IIAES OQKJY B
Use the deciphering matrix  2 23 
 21 7 
REFERENCES / RESOURCES
[updated 22 February 1996]
[ACA] ACA and You, "Handbook For Members of the American
Cryptogram Association," ACA publications, 1995.
[ACA1] Anonymous, "The ACA and You  Handbook For Secure
Communications", American Cryptogram Association,
1994.
[ACM] Association For Computing Machinery, "Codes, Keys and
Conflicts: Issues in U.S. Crypto Policy," Report of a
Special Panel of ACM U. S. Public Policy Committee
(USACM), June 1994.
[AFM] AFM  10080, Traffic Analysis, Department of the Air
Force, 1946.
[ALAN] Turing, Alan, "The Enigma", by A. Hodges. Simon and
Schuster, 1983.
[ALBA] Alberti, "Treatise De Cifris," Meister Papstlichen,
Princeton University Press, Princeton, N.J., 1963.
[ALKA] alKadi, Ibrahim A., Origins of Cryptology: The Arab
Contributions, Cryptologia, Vol XVI, No. 2, April 1992,
pp 97127.
[AND1] Andree, Josephine, "Chips from the Math Log," Mu Alpha
Theta, 1966.
[AND2] Andree, Josephine, "More Chips from the Math Log," Mu
Alpha Theta, 1970.
[AND3] Andree, Josephine, "Lines from the O.U. Mathematics
Letter," Vols I,II,III, Mu Alpha Theta, 1971,1971,1971.
[AND4] Andree, Josephine and Richard V., "RAJA Books: a Puzzle
Potpourri," RAJA, 1976.
[ANDR] Andrew, Christopher, 'Secret Service', Heinemann,
London 1985.
[ANNA] Anonymous., "The History of the International Code.",
Proceedings of the United States Naval Institute, 1934.
[ANN1] Anonymous., " Speech and Facsimile Scrambling and
Decoding," Aegean Park Press, Laguna Hills, CA, 1981.
[ANTH] Anthony  Cave Brown, "Bodyguard of Lies", Harper and
Row, New York, 1975.
[ASIR] Anonymous, Enigma and Other Machines, Air Scientific
Institute Report, 1976.
[AUG1] D. A. August, "Cryptography and Exploitation of Chinese
Manual Cryptosystems  Part I:The Encoding Problem",
Cryptologia, Vol XIII, No. 4, October 1989.
[AUG2] D. A. August, "Cryptography and Exploitation of Chinese
Manual Cryptosystems  Part II:The Encrypting Problem",
Cryptologia, Vol XIV, No. 1, August 1990.
[BADE] Badeau, J. S. et. al., The Genius of Arab Civilization:
Source of Renaissance. Second Edition. Cambridge: MIT
Press. 1983.
[BAMF] Bamford, James, "The Puzzle Palace: A Report on
America's Most Secret Agency," Boston, Houghton Mifflin,
1982.
[BARB] Barber, F. J. W., "Archaeological Decipherment: A
Handbook," Princeton University Press, 1974.
[B201] Barker, Wayne G., "Cryptanalysis of The Simple
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