## Lesson 8: Introduction To Cryptarithms And Hill Cipher

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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 ACA-L 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 (0-9). 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 (9-0)      9 8 7 6 5 4 3 2 1 0

3. From 1 to 0 (1-0)      1 2 3 4 5 6 7 8 9 0

4. From 0 to 1 (0-1)      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 0-9 we would have
a key for 0-X or 0-A.     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, (0-9), 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 September-October, 1993, issue of The Cryptogram:

C-1. Square root.  (Two words, 1-0) 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

C-3. Division.  (Three words, 9-0) by LI'L GAMIN.
AUSSIE v SHEEP = UE; - SHEEP = SUMAIE; - SPIBHP = LUHE

____UE
Rewritten:  SHEEP/AUSSIE
SHEEP
SUMAIE
SPIBHP
LUHE

C-6.  Subtractions.  (Two words, (0-9) by CAGEY KIWI.
LADIES - GENTS = GNSDGS.     DAMES - MALES = NDGSS

-GENTS    -MALES
GNSDGS     NDGSS

rearranged the same way.

C-8. Multiplication.  (Three words, 0-9) 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, C-8, 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 C-3
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, 9-0. 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 in-formation, 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, 0-9.  First, notice
that N time OTTAWA results in a 7-digit number and that O time
OTTAWA results in a 6-digit 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 2-digit 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 6-digit number and
checking your outcome.  Solve C-1 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, C-6 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
Decatur, GA 30033-2729.

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]

S-1  Squares end only in 0, 1, 4, 5, 6, or 9.

S-2  If (...S)**2 ends in ...S, then S=0,1, 5 or 6.

S-3  If (...S)**2 ends in B n.e. S, then S= 2,3,4,7,8 or 9
B= 1, 4, 6 or 9

S-4  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

S-5 If N contains k digits, then N**2 contains either 2k-1 or
2k digits.

HILL CIPHER SYSTEM  (by NORTH DECODER)

There are two basic ways to prevent the tell-tale 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-
and-paper 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 n-graphs), 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 10-graphs
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.

The product is a 3 x 1 matrix, say C.  Its entries are
converted to letters,  and these give the ciphertext for the
first group.  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 n-grams into
different ciphertext n-grams. 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 n-gram using the
Hill Cipher, a nonsingular n x n matrix M is selected.
The n-gram is written as an n x 1 matrix P, and the ciphertext
is the n x 1 matrix C determined by the equation

C = MP.

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 trial-and-error. 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 accumulator
represents 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 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
non-singular 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

/lanaki.crypt.class/docs/hill-gifs

There is a freeware gif file viewer at the CDB in

/msdos/gif-viewers

HOMEWORK SOLUTIONS FROM LECTURE 7

FRE-2. 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.

FRE-3. 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.

FRE-4.  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

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.

ITA-2.  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.

ITA-3.  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
nessunodicaesattamente quello che pensalinyutang.

SPA-1.                                    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

SPA-2. 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

SPA-3.  (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

POR-2. 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.

POR-3. 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

C-1   Give two solutions to:  (BE)**2 = ARE

C-2   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:

Hill-1

Decipher the message:  YITJP  GWJOW  FAQTQ  XCSMA  ETSQU
SQAPU  SQGKC  PQTYJ

Use the deciphering matrix   | 5  1 |
| 2  7 |

Hill-2

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
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illustrations of the Soviet one-time pad with example,
with three errors in cipher text, that I have corrected
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[NIC1] Nichols, Randall K., "Xeno Data on 10 Different
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[ZEND] Callimahos, L. D.,  Traffic Analysis and the Zendian
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