Classical Cryptography Course,
Volumes I and II from Aegean Park Press

By Randy Nichols (LANAKI)
President of the American Cryptogram Association from 1994-1996.
Executive Vice President from 1992-1994

Table of Contents
  • Lesson 1
  • Lesson 2
  • Lesson 3
  • Lesson 4
  • Lesson 5
  • Lesson 6
  • Lesson 7
  • Lesson 8
  • Lesson 9
  • Lesson 10
  • Lesson 11
  • Lesson 12


    February 22, 1996

    Revision 0


    In Lecture 8, we depart from the schedule for a real treat.In the first part of this Lecture, we introduce Cryptarithms byour guest lecturer LEDGE (Dr. Gerhard D. Linz). LEDGE hasalready produced one of our better references on beginningcryptography [LEDG], and I appreciate his assistance in ourcourse. The cryptarithms portion of this course will bepresented in three lectures and for the final book labelledLectures 20 - 23.

    Following the Cryptarithms section we introduce the HillCipher.

    Our second guest lecturer is NORTH DECODER. Dr. Jerry Metzgerand his team are presenting you with the Crypto Drop Box andthe ACA-L Listserver. The Hill cipher has six GIF filesassociated with it and can be found at the CDB.

    Waiting in wings patiently for my resource materials is TATTERSto present Cipher Exchange problems.


    Here's the first of the Cryptarithm lectures. It consists of ageneral introduction to the genre including how to read theproblems. That's followed by an explanation of moduloarithmetic. Then we look at how to identify the letters thatrepresent 0, 1 and 9, called digital characteristics. Thenthere are two sections on making inferences, each demonstratinga problem solution. Finally, there's a section on extractingsquare roots.

    Next lecture LEDGE will give some aids for solvingmultiplication problems and then go into base 11 and base 12arithmetic. Perhaps after that I can go to the morecomplicated problems such as double key division.

    PART I


    A cryptarithm is a mathematical problem, generallyin arithmetic, in which the numerical digits have been replacedsystematically by letters. The challenge of the problem is toidentify the digit for each letter and the key, if any.


    A table consisting of each of the letters used in thecryptarithm paired with its numerical equivalent constitutesthe key to the cryptarithm. When the digits are arranged innumerical order, either from smallest to largest or largest tosmallest or other logical order, the letter portion of the keymay spell out one or more words. The word or words are thenknown as the keyword or keywords. Generally, the constructor ofthe problem indicates the number of words or the fact that theletters do not spell out words.
    When the letter portion of the key consists of a word orseveral words with no repeated letters (rule 3, above), thedigits 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 itrepresents 9; and in the third it represents 1, etc. Higherbase arithmetic systems require additional digits according tothe size of the base. Undecimal is based on 11 digits ratherthan 10. Generally the letter X or A is used to represent the11th digit, or ten. Thus instead of a key for 0-9 we would havea 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 theconcept of higher base arithmetic systems now, you will get anextended treatment of this topic later in the course.

    If no word is used, that fact will be stated as well as theorder in which letter equivalents are to be reported, e.g., Noword, (0-9), indicating that the letters for reporting purposesare to be arranged starting with the letter representing 0,followed by the letter for 1, then 2, etc., with the letter for9 last. The letters will then appear in random order, generallynot alphabetical order.


    Knowledge of addition, subtraction, multiplicationand division of whole numbers in base 10 (decimal) system willbe assumed. Extraction of square and cube roots will beexplained later. While base 13 problems sometimes appear amongthe numbered problems in the Cryptarithms section, they andhigher base problems are generally offered as specials.  Moreesoteric operations, such as powers, magic squares, Pythagoreanequations, etc. are also offered as specials for those who likeextra challenges.


    In order to conserve space in the journal,the problems in the Cryptarithms section are writtensequentially on one or more lines. I recommend rewriting theproblems in normal arithmetic format on every other line, so asto have room for trial numbers. The process, without skippinglines, will be illustrated with each of the normal type ofproblems presented for solution.
    The following sample problems to be rewritten are taken fromthe 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                             LUHEC-6.  Subtractions.  (Two words, (0-9) by CAGEY KIWI.LADIES - GENTS = GNSDGS.     DAMES - MALES = NDGSS         Rewritten:  LADIES     DAMES                     -GENTS    -MALES                     GNSDGS     NDGSS  
    Additions and equations (mixed additions and subtractions) arerearranged 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 thepresentation of the problems. You should also be ready toconstruct cryptarithms of your own, although they may not besuitable as yet for publication. To be suitable forpublication, the problem must conform to the rules listed onpage 1, and have a unique numerical solution. There must be oneand only one key that will solve the problem. If you haveunderstood the material thus far, you are ready to considerways of analyzing a problem to obtain the solution.


    Since we will be dealing with the tendigits, 0 - 9, but sometimes adding or subtracting them to getnumbers that are either greater than 9 or less than 0 (in otherwords negative numbers), we need a way of reducing thoseresults back to the digits mathematically. Modulo arithmetic isthat way. If you add 8 + 5, you get 13. If you want to talkabout only the units digit of the result, you could subtract 10from 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 tendigits in the decimal system.  When we learned addition, welearned to carry the 10 to the next column on the left, thusavoiding having to write a two digit number in a space wherethere 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 thepositive digits. Here we could add 10 to -3: -3 + 10 = 7, or -3= 7 (modulo 10). In a subtraction problem we get the 10 byborrowing it from the next highest order digit in thesubtrahend: 25 -8 = 17 or 20 - 3 or 10 + (10 - 3)The way we learned to subtract eliminates the negative numbersby borrowing 10 from the 20 in 25. Modulo arithmetic is anotherway of talking about the same process.


    Gaining an entry into a problem isoften expedited by being able to identify one or more of thedigits. Those most commonly identifiable with a little bit ofstudy of the problem are 0, 9, and 1. Zero in particular has anumber of recognizable characteristics. Add zero to a numberand 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 anumber from itself and you get zero, i.e., A - A = 0. Oncezero is identified, you will have the first or last letter of akeyword, if any.

    If 0 cannot be identified through any of the characteristicsenumerated above, it may yet be possible to discover thecandidates for it through a process of elimination. Given anumber, we know that the highest order digit of that numbercannot be zero. So if we have a number, ABC, then A is notzero. Let's use that fact and any other inferences we can makeon the example multiplication problem, C-8, from page 3.

                   OTTAWA               ___xON              HNNTLIL              IIIEHE_              TOOINRL 
    This problem has five different numbers with four differentbeginning letters: O, H, I, and T. None of those can be zero.The multiplier, ON, contains the digit N which, whenmultiplying OTTAWA, produces a product not equal to zero.Hence, N does not equal zero. When adding the two partialproducts, E + I yields R not either E or I. Hence neither Enor I = zero. We have eliminated seven letters as candidatesfor zero. So far, at least, L, W, or R could be zero. It willtake more detailed analysis to determine which one is actuallyzero.

    The number 9 has some interesting characteristics, one of whichmimics zero. When subtracting 9 from a number, you must borrowfrom the next higher digit. The difference between 9 and thenumber is then one more than that number, i.e., 24 - 9 = 15contains 4 -9 yielding 4 + 1 or 5. The 2 in the originalnumber 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 9because of a borrowing necessitated by a previous subtraction,namely 7 - 8. 4 - 4 becomes 3 -4 yielding 9 and reducing the 2to 1. That sort of effect is not possible in the units place ofa number because there is no previous borrowing when dealingwith whole numbers. Thus, when given a problem that includes:         ABCDE         -DCFE          GHIJ 
    H could equal 0 or 9. More information is needed to resolve theambiguity. We have it here in the units place where E -E = J.There is no ambiguity in that fact since there cannot have beenany previous borrowing. So J = 0 and H = 9.

    The number one can often be recognized as the highest orderdigit of a number particularly when, in a subtraction problem,it is not carried down to the answer line. Note that in theprevious example, A is the highest order of the subtrahend, thenumber from which another number is to be subtracted. It doesnot 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 isgreater than B and necessitating borrowing one from A, reducingit to zero. Notice than when subtracting a larger digit from asmaller one, the resulting difference is larger than thesubtrahend digit, e.g., 5 - 8 yields 7 or 15 - 8 = 7 > 5. Ifyou now understand subtracting using modulo arithmetic, youwill recognize that 5 - 8 = -3 which = 7 (modulo 10). Inmodulo arithmetic we can add or subtract the base, here 10, asmany times as necessary to produce a number in the desiredrange, here 0 to 9. (See page 3, "Modulo arithmetic," 1stsentence.)

    The number one can also be spotted in multiplication since onetimes a number equals that number, i.e. A x 1 = A. One timesone also yields one, making it one of three digits that whensquared or multiplied by itself yields a number whose unitdigit 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 that25 = 5 (modulo 10) and 36 = 6 (modulo 10).


    (Example 1). Once you have done what you canto spot 0, 1 and 9, you will have to rely on your knowledge ofarithmetic to determine the possibilities of the other lettersand to make decisions about their values. To see how thatworks, let's work on a simple problem, the division problem C-3at 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, readon for the solution.

    In the above problem, we are helped by being able to find allthree of the digits, 0, 1, and 9. In the first subtraction,I - P = I. In the second subtraction, E - P = E. Both factsmake 0 = P. Note also that U x P = P and E x P = P, bothconsistent with P = 0, but not sufficient to prove that P iszero, since both of those equations, modulo 10, could be truefor 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. Thesechains of reasoning are typical in the solution ofcryptarithms.

    Now let's find the letter for 9. In the first subtraction, notethat U - H = U. That could make H be zero or nine. In theabsence of other information, you could not be sure which ofthose is true. Here you already know that zero is representedby P. Thus, H = 9.

    You now have a lot of useful information. Let's look at themultiplications for more. U x SHEEP is 1 x SHEEP = SHEEP, notmuch help there. E x SHEEP = SPIBHP. You can replace theidentified 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'stry out each possibility. 3 x S9330 = ??7990 or ??IHHP, notconsistent with SPIBHP. So E is not 3. E must then be 7. Let'scheck that and see what else you can uncover. 7 x S9770 =??8390 making I = 8 and B = 3. Now SPIBHP is S08390. 8 ispreceded by 0 so 7 x S must end in 4 since we are carrying a 6from 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. Tobring order out of all this in-formation, we need toreconstruct 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 secondsubtraction. Entering what is known now makes that subtraction:          21MA87         -208390            L197 
    Remember, you can check a subtraction by adding the subtracterand 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, representingas 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 itcontains the letters M and A. Try that now using the partiallyreconstructed key above. The results should be the same.


    (Example 2). The multiplication example, C-8, given on page 3 presents somewhat more difficulties than theprevious one, as none of 0, 1, or 9 can be initiallyidentified. There are enough other clues, however, to make thesolution come through a straightforward series of inferences.Before reading on, see what you can recover from that problemon your own. When you are in a thoroughly stuck place, read onfor 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. Onpage 3 the key is stated to be three words, 0-9. First, noticethat N time OTTAWA results in a 7-digit number and that O timeOTTAWA results in a 6-digit number, the same length as OTTAWA.Examine the second product carefully. O x OTTAWA = IIIEHE. Thehighest order I (first digit of IIIEHE) results from theproduct O x O. O cannot = 1 for 1 x OTTAWA = OTTAWA. Ocannot be as large as 4, for 4 x 4 = 16, which would add aseventh digit to the product. So O = 2 or 3. 3 x 3 = 9, whichwould make I at least 9. Looking at the problem again, thefirst I is added to H giving T, a digit, but adding anythingother than zero to 9 produces a two digit number. So Icannot 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 ifa 1 is carried from the previous multiplication (2 x T).So we have the following multiplication: 2 x 2TTAWA = 444EHEor 555EHE. We can divide each of those products by themultiplier, 2, getting respectively 222??? and 277???. Thefirst quotient gives 222??? to represent OTTAWA - not possible(it would be OOO???). The second quotient is consistent inmaking T = 7 and I = 5. OTTAWA becomes 277AWA. IIIEHE = 555EHE.

    Now let's look at the first product, N x 277AWA = HNN7LIL. Theproduct must be less than 10 x OTTAWA and that makes its firstdigit less than O. There is only one such digit, so H = 1. Youcould now divide 1NN7LIL by various values of N to find aquotient that begins 277. It's easier, however, to look at theaddition 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 usingthe 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 digitsfor 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 of1 which you may have to supply. Accepting that contingency, thefirst equation produces 8 as the only possible value of E. Thethird equation then makes R = 3 since there is no carrypossible. The second equation makes 4 and 5 possible values ofL, but 4 is the only available digit. Of the three letters thatcould be zero only W is left unidentified. Only 9 is left forA. As a check, 9 x 6 = 4 or L and 9 x 2 = 8 or E, checkingout. The key has become WHORLINTEA as the solution.


    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 howit is that the method works.
    When squaring a number, one doubles the number of digits of theoriginal 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 ofa number, you take cognizance of this fact by making a markafter every two numbers beginning from the decimal point inboth directions. So 45678.96 becomes 4'56'78.96' with theinitial 4 being understood as 04. As many pairs 00 can be addedafter the last mark without changing the value of the number.

    The first trial root is the largest number whose square isequal to or less than the initial pair of numbers. We'll callthat trial root x. (One could use the largest number whosesquare is equal to or less than the initial two or more pairsof numbers. That makes no theoretical difference although inpractice that's more difficult.) The square of x is thensubtracted from the first pair of numbers. The next pair ofnumbers is appended to the difference as in a long divisionproblem.

    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} hasalready been subtracted from the highest order two digit numberof the original number it remains to subtract y(20x + y) fromthe current remainder to make sure that y is not too large andto determine a new remainder.

    Now let's extract the square root of 45678. First mark afterevery 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 TheCryptogram. Do the subtraction problem, C-6 on page 3 if youwish. Discuss problems you may have with your mentor. If youhave suggestions, questions, or other reactions you wish toshare with me, my address is:

      Dr. Gerhard D. Linz
      2649 Tanglewood Road
      Decatur, GA 30033-2729.
      My e-mail address is

    I hope your pleasure in solving Cryptarithms is enhanced bythis presentation. Next time I'll respond to any concerns youhave. I also plan to give you some more tools formultiplication and introduce counting systems based on 11 andabove.

    January 2, 1996


    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 9S-4 If (..X)**2 ends in 0 1 4 5 6 9 then (...X) ends in 0 1,9 2,8 5 4,6 3,7S-5 If N contains k digits, then N**2 contains either 2k-1 or 2k digits.


    There are two basic ways to prevent the tell-tale behaviorof plaintext letters from showing through in ciphertext. Onemethod is to vary the ciphertext letter that replaces a givenplaintext letter. That is the solution offered by the Vigenereand other polyalphabetic systems. A second technique is toencipher the plaintext in chunks of several letters at a time.The Playfair system provides a compact method for encipheringdigraphs, that is, pairs of letters. While the Playfair doesdisguise the behavior of individual letters, even better wouldbe 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 quadgraphicor ...) encipherment has been devised.

    In 1929, Lester Hill [HIL1, HIL2] described an algebraicprocedure that allows encipherment of plaintext letters n at atime (that is, in n-graphs), where n can be any positiveinteger 1,2,3,.... Hill's Cipher could be carried out by handprobably without too much hardship for groups of letters up tofive. After that, it would become a challenge to keep thecomputations accurate. However, on a computer it wouldbe feasible to work with large groups of letters, and it seemsthat plaintext enciphered in such a system using say 10-graphswould be difficult to crack.

    The first step in using Hill's system is to assign numericalvalues to the 26 letters of the alphabet. There is nothingsacred about 26 in the system. The ideas work just as well foralphabets of any size. So it would be possible to add a fewpunctuation marks to the usual alphabet to get say 29 symbols,or to work entirely with data in binary form with an alphabetof just two symbols. Also the numerical equivalents of theletters of the alphabet could be assigned in some arbitraryway, which would probably add to the security of the system.For these notes, the 26 letter alphabet will be used, andletters will be given their standard numerical equivalents,namely a = 00, b = 01, ..., z = 26.

    The encipherment of plaintext is most neatly described usingmatrix multiplication.  A matrix is a rectangular array of numberssuch 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 twomatrices can be computed. The rule for multiplication requiresthat the number of columns in the lefthand factor matchthe 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 andN has three rows. On the other hand, the product NM is notdefined.

    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 producedby multiplying each number in the first row of M by thecorresponding number in the first column of N, and adding theresults:(1)(2)+(5)(7)+(3)(3)=46. That explains why the numberof 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 sameway by multiplying the first row of M times the second columnof 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 thefirst row of M by the third column of N, and finally, thefourth number in the first row of the product comes frommultiplying the first row of M by the fourth column of N.

    To produce the second row of the product, the second row of M isused in place of the first row of M in the precedingcomputations. So, for example,(0)(2)+(2)(7)+(3)(3)=17 givesthe first number in the second row of the product matrix. If Mhad more rows, each would be used in turn in the same way toproduce 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 andan s x t matrix will be an r x t matrix.

    In the application of matrix multiplication to the Hill Ciphersystem, all arithmetic will be carried out modulo 26. In otherwords, any time a number appears which is 26 or larger, it isdivided by 26, and the number is replaced by the remainder ofthe division. In the example above, the computation of thetop 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). Thesymbol (mod 26) is added here just to indicate there isfunny arithmetic being used, namely that arithmetic is beingdone modulo 26. If the alphabet had 29 symbols instead of 26,operations would be carried out modulo 29. Since all theexamples here will be done modulo 26, the indicator (mod 26)will be omitted from now on. So we will write the exampleabove 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 themessage is rewritten in groups of letters of the selectedlength. For this example, length three will be used, so themessage becomes "sen dmo rem one ykz", where two nulls havebeen added to fill out the last group. Next an encipheringmatrix, or key, is selected. If letter groups of size n arebeing 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 aswell be selected between 0 and 25 since all arithmetic will bedone modulo 26 anyway. To encipher the first three lettergroup of plaintext, it is written as a 3 x 1 matrix, say P, theletters are replaced by their numerical equivalents, and thematrix 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 secondtrigraph 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 FSTXYH LZI UCN, or, in traditional five letter groups, UENFSTXYHL ZIUCN. Notice that in this example, repeated plaintextletters are replaced by different ciphertext letters, andrepeated ciphertext letters represent different plaintextletters.

    Deciphering requires a second matrix that undoes the effects ofthe 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 restof 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 ciphertextAAAAAAAA.
    To allow unique decipherment, an n x n encipheringkey matrix should convert different plaintext n-grams intodifferent ciphertext n-grams. An n x n matrix that behaves thatway is called nonsingular.

    There are a number of more or less efficient tests fornonsingularity. Here is one test that involves the determinantof an n x n matrix. The determinant of a square matrix is anumber computed from the entries in the matrix. The definitionbuilds up from small matrices to larger ones. First thedeterminant of any 1 x 1 matrix is defined to be the numberthat is the entry in that matrix. Thus det |7| = 7. Tocompute the determinant of a 2 x 2 matrix, step across theentries in the first row of the matrix, multiply each entry bythe determinant of the 1 x 1 matrix that appears when the rowand the column the entry appears in are eliminated fromthe matrix. The numbers produced in this way are alternatelyadded 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 thedeterminant of the 2 x 2 matrix that appears when the entry'srow and column are crossed out, and alternately add andsubtract 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 largersquare matrices in the same pattern. More efficient ways tocompute determinants are discussed in textbooks on LinearAlgebra.

    The importance of the determinant is that a matrix isnonsingular (and so usable as an enciphering key matrix inHill's cipher) if and only if its determinant is relativelyprime to 26. The matrix above has determinant 23 which isrelatively prime to 26, so it is a legal enciphering key. Moregenerally, if the alphabet used for the plaintext is made up ofm symbols, then the usable enciphering matrices are those withdeterminant relatively prime to m. In the case of an alphabetof 26 symbols, the determinant of a usable matrix must be oddbut not 13. Notice that if the size of the alphabet isincreased to 29 by adding a few punctuation symbols, many morelegal enciphering matrices will be available, both becauseoperations will now be carried out modulo 29, and also becauseevery number from 1 to 28 would be an acceptable value for thedeterminant of an enciphering key matrix.

    Once an enciphering key matrix has been selected, the companiondeciphering matrix needs to be computed. There are somereasonably efficient methods for finding the decipheringmatrix. The method given here is easy to describe, but notvery efficient. Check out a Linear Algebra text for bettermethods to handle matrices larger than say 4 x 4.

    • The first step is the computation of the determinant ofthe enciphering key E. If det E = e, then a number d is neededsuch that ed= 1 (mod 26). For a relatively small modulus suchas 26, the d can be found by trial and error. Simply compute etimes 1,3,5,7 ,9,11,15,17,19,21,23, and 25 until a productequivalent 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 theEuclidean Algorithm, for example.

      Check a Number Theory text for details. Set the number d asidefor a minute.

    • Second, each number in the enciphering key matrix is replacedby the determinant of the matrix obtained when the element'srow and column are erased from the matrix.

    • Third, plus and minus signs are prefixed to each entry in thenew matrix in a checkerboard pattern starting with a plus signin the upper lefthand corner.

    • Next, the matrix is flipped over the diagonal from the upperleft corner to the lower right corner so that the first row becomes the first column, the second rows becomes the secondcolumn, and so on.

    • Finally, each entry in the matrix is multiplied by the dcomputed in the first step.

    • The resulting matrix is D, the deciphering key matrix.

    Here are the computations that produce the deciphering key D ofthe example above. The determinant of the enciphering key Ehas already been computed: det E = 23. Since (17)(23) = 1 (mod26), it follows that d = 17. Next, the 1 in the upper lefthandcorner 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 columnis:    | 1 22 | det| 1  5 | = (1)(5) - (22)(1) = - 17 = 9 (mod 26). When all nine entries in E have been replaced, the matrix lookslike:     | 15 18 25 |     | 17  9 21 |     | 24 18  7 |Adding the plus and minus signs in a checkerboard patternproduces and replacing negative numbers by equivalent positivenumbers 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 thed=17 computed earlier, and reducing the entries modulo 26, theresult 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 witharithmetic done with ordinary numbers. The n x n matrix whoseentries are all 0 except for 1's down the diagonal from theupper left to the lower right is called the identity matrix.It plays a role in matrix multiplication similar to the role 1plays in multiplication of numbers. That is, for any number m,(1)(m) = m, while for any n x k matrix M, it is easily checkedthat IM= M. Moreover, for each number r (provided r is notequal 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 iswritten as r(-1) (that is, r to the -1 power). Likewise, foreach n x n matrix M (provided it is nonsingular), there is ann x n matrix N for which MN=I. The matrix N is calledthe inverse of M, and is written as M(-1).

    The Hill Cipher system can be expressed compactly using somealgebraic notation. To encipher a plaintext n-gram using theHill Cipher, a nonsingular n x n matrix M is selected.The n-gram is written as an n x 1 matrix P, and the ciphertextis the n x 1 matrix C determined by the equation:

      C = MP.
    The deciphering matrix is the inverse of M. When theciphertext C is multiplied by M(-1), the plaintext isrecovered:
              M(-1) C = M(-1) MP = IP = P.
    Hill suggested that a good choice for an enciphering key matrixM is one that turns out to be its own inverse. If M = M(-1),Mi s called an involuntary matrix. The advantage gained is thatit is not necessary to compute the deciphering key. There area number of methods that will automatically produce involuntarymatrices, so the process of finding involuntary matrices doesnot have to proceed by trial-and-error. In any case, almost allpapers written about the Hill Cipher system following Hill'stime 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 mechanicalversion of the Hill Cipher in 1929, which, according to Kahn[KAHN], used an involuntary matrix enciphering key so that thesame machine could be used to both encipher and decipher.

    The Message Protector patented by Weisner and Hill providesa mechanical means of doing matrix multiplication. The deviceillustrated in the patent application is more accuratelydescribed as authentication indicator rather than acryptographic mechanism. The principle of operation is verysimple. The active component consists of three gears on an axlewhich are connected to three accumulator gears by chains. Thethree accumulator gears all have the same number of teeth (101in the patent), and they can rotate independently. The threegears on the axle have 101, 202 and 303 teeth. As the axle isturned through a certain amount, the accumulator gears turnone, two and three times as far respectively. The teeth onthe accumulator gears are numbered from 0 to 100, and smallgear on the axle also has its teeth numbered from 0 to 100.

    Now suppose the three accumulator gears start in position0,0,0. If the axle turned through an amount that rotates itssmall gear through 43 teeth, then accumulator gear one willread 43, accumulator two will show 86 and accumulator threewill show 28. The last value occurs since the thirdaccumulator wheel will have made more than one revolution. Ifthe starting position of the accumulator wheels had been11,91,4, then the axle rotation through 43 teeth would leavethe accumulators showing 53,76,32. In essence, theaccumulators are modulo 101.

    On the actual devise, there are six axles, and their gears canbe moved to engage the accumulator drive chain one axle at atime. The placement of the gears on the axles vary from oneaxle to the next. On the illustrated machine in the patent,the sequence is:

    axle 1: 101,202,303axle 2: 202,303,101axle 3: 303,101,202axle 4: 101,303,202axle 5: 202,101,303axle 6: 303,202,101 
    Suppose the accumulators begin showing 0,0,0. Keeping trackfor now of only the total on the accumulator that connects tofirst gear on each axle, here is what happens as the axles areturned as follows:
    axle 1: 23,axle 2: 10,axle 3: 88,axle 4: 17,axle 5: 41,   andaxle 6: 51. 
    Initially, all the axles are disengaged from the accumulatordrive chain. (Keeping in mind the number of teeth on the firstgear on each axle.) axle 1 is engaged, turned 23, and theaccumulator 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 theaccumulator shows 21. Axle 4 is disengaged, axle 5 isengaged, turned 41, and the accumulator shows 18. Axle 5 isdisengaged, axle 6 is engaged, turned 51, and the accumulatorshows 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 secondgear 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 operationcompactly 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, sixnumbers, between 0 and 101, are selected from the check.Perhaps the dollar amount of $1230.45 could be split up as 12and 30 and the cents could be ignored. The check number of say22131 might contribute three more numbers, 2, 21, and 31.Finally, the date of the check, maybe January 25, 1996 mightcontribute a sixth number, say 25. Of course, people mustagree on how these numbers are selected. The check writer runsthe six values through the Message Protector as describedabove, and the resulting triple of values is stamped on thecheck. The bank, before cashing the check, operates on thesame six numbers with its Message Protector, and makes surethat the numbers produced on the accumulators matches the onesstamped on the check, thus being sure that none of theimportant figures on the check have been changed.

    Although the Message Protector is a clever engineeringconstruction, there are certainly many obvious mechanicalshortcomings as well as weaknesses in the cryptographic systemwhich probably explains why the machine never became popular.In fact, it's not clear if any were actually constructed. Itwould take a good salesman to get people to spend money on amachine to multiple 3 x 6 and 6 x 1 matrices. There did notseem to be and reasonable way to change the gear sizes. If akey matrix with entries besides 1, 2, and 3 were wanted, thenumber of teeth on the gears would soon become so large thatthe structure would have to be made pretty large, instead ofthe shoebox size Weisner and Hill diagrammed.

    Weisner and Hill also explain how the Message Protector couldbe 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 sothat the arithmetic operations would be carried out modulo 26.Next, the axles would now carry six gears each, with the numberof teeth on each gear being a multiple of 26. There would besix accumulators, so that six plaintext are converted to sixciphertext letters. They say that the number of teeth onthe various gears "have to be selected according to certainmathematical principles". What they mean, of course, is the6 x 6 matrix, each entry of which gives the multiple of 26 thatgives the number of teeth on the corresponding gear, has to benon-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, andcertainly cannot be changed arbitrarily, so it seems thegearing set was intended to be selected once and for all. Sincethat pretty much makes the device cryptographically pointless,the inventors proposed that a plaintext message first beconverted to a preliminary ciphertext according to so systemleft unspecified, but they probably had something like aPlayfair in mind. The resulting ciphertext is then passedthrough the 6 x 6 Message Protector, to yield an intermediateciphertext which is then passed through a third and finalencipherment using another unspecified cipher system. Thefinal ciphertext is transmitted, and the authorized recipientreverses each of the three encipherments to recover theoriginal plaintext. It's not very clear how much additionalsecurity has been introduced passing the text through theMessage Protector.

    Nearly all discussions of cryptanalysis of Hill encipheredmessages begin with the fairly generous assumptions that thecryptanalyst knows that an involuntary key matrix of known sizehas been used, and also knows the numerical values assigned tothe alphabet letters. The only unknown is the particular keymatrix used to encipher the message. For a key matrix of size2 x 2, a brute force attack is feasible since there are only736 2 x 2 involuntary matrices. As the size of the key grows,a brute force attack is no longer practical. For larger keysizes, no specific cryptanalytic approaches have beenpublished. But, several authors given more or less detaileddescriptions of cryptanalysis, with examples for small key size(2 x 2, 3 x 3) using the classic probable word or cribtechnique. That is, a piece of plaintext is assumed to appearin the message, and it is tried in each possible position.At each test location, a number of equations must be true ifthe crib is to generate the ciphertext at that spot. It turnsout that even with a relatively modest crib (3 letters for a2 x 2 key, and 4 for a 3 x 3 key), most positions can beeliminated as impossible by applying a few principles of linearalgebra. Each possible crib location will produce a candidatematrix key. A trial decipherment of the ciphertext is made.If recognizable plaintext results, the cryptogram is broken. Ifnot, the crib is moved along to the next possible spot, and theprocess is repeated. For details on see on cryptanalysis ofthe Hill Cipher, see [LEV1], [LEV2], [LEV3], [SINK], [MELL].

    NORTH DECODER advises that the Hill cipher patent diagrams (GIFformat) scanned in reasonable well into the CDB. If you wouldlike 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/gif-viewers

    {well, it doesn't appear to be there; but you're browser likely supports GIF viewing.--JP}


    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 QP' W M R R Q;  D K E X Q B Q   O Q U Q I Q E Q Q    M CT E X R X B X D Q ,   X P    Q A B    K   P' W M R R Q   N QV C Q   N W K B   O Q   U M C B B X Q E Q   Q A B    K CN 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 ceque chat de gouttiere est au chat sauvage.
    FRE-3. K2. (87) (jamais, A=b) It's fun trying. GUNG HOD G X Z Q N J D P M C J P U P L S U E' Z DZ 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.
    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 QT G G Z P Z Q F R A T J Z E F N S Z M T Z J S A SZ R A P T D A F F Q K G K Z L Z S S K E O F J F QQ T J K R A E Z F Q Z S S Z H F J S F M T F G G KE 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 nessunodicaesattamentequello 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 AZ P Z N D Z K G C T Y R Z K R N T D G R Y C V KK S T P Q D P E R M K T C Y G R Z Y P Q P M P E K EE 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 YSPA-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 FK N A G B N A G K T F P J G T P A O Z F M B FS J G H N B R T B T I K T N Z G B I Q BB P K J I Q Z I B J M P B B J N A Q G A O J M BM 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 OK A X T S P Q K D X R K R R S S I N K Y K R L A RS D K T Q L D L P X K T A S Q X S P X P R S O S PR X J K R K T O A S T S P Q X L S D O A X I S A EC 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 ENVIADOPOR-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 JX I N R S O C H C D T C V R P U C D V R JZ J U D C T J H J D G X U M P C H J A X H XO 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 DSolution reads: (VIDAREL)
    Vivemos numa epoca que se orgulha das maquinas quepensam 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 LI T I W F Y C V F U H F W F T L F R F B C H W F CE S H I L F G I C D E G T I J H C V G R P C V C JF V D E F W F H C V L F V F H J I S K I X J I Z UI G V T I V V I V B C D E F G H I V V C I F Y K FR F T W F V.Partial Solution reads: (VAQUEIRO): Papai sabe que tu.


    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  ARFrom Sinkov [SINK] two Hill system problems:Hill-1Decipher the message:YITJP  GWJOW  FAQTQ  XCSMA  ETSQUSQAPU  SQGKC  PQTYJUse the deciphering matrix   | 5  1 |                             | 2  7 |Hill-2Decipher the message:MWALO  LIAIW  WTGBH  JNTAK  QZJKA  ADAWSSKQKU  AYARN  CSODN IIAES  OQKJY  BUse the deciphering matrix   | 2  23 |                             | 21  7 |


    [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 - 100-80, 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] al-Kadi, Ibrahim A., Origins of Cryptology: The Arab Contributions, Cryptologia, Vol XVI, No. 2, April 1992, pp 97-127.[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.[AS] 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 Substitution Cipher with Word Divisions," Course #201, Aegean Park Press, Laguna Hills, CA. 1982.[BALL] Ball, W. W. R., Mathematical Recreations and Essays, London, 1928.[BAR1] Barker, Wayne G., "Course No 201, Cryptanalysis of The Simple Substitution Cipher with Word Divisions," Aegean Park Press, Laguna Hills, CA. 1975.[BAR2] Barker, W., ed., History of Codes and Ciphers in the U.S. During the Period between World Wars, Part II, 1930 - 1939., Aegean Park Press, 1990.[BAR3] Barker, Wayne G., "Cryptanalysis of the Hagelin Cryptograph, Aegean Park Press, 1977.[BARK] Barker, Wayne G., "Cryptanalysis of The Simple Substitution Cipher with Word Divisions," Aegean Park Press, Laguna Hills, CA. 1973.[BARR] Barron, John, '"KGB: The Secret Work Of Soviet Agents," Bantom Books, New York, 1981.[BAUD] Baudouin, Captain Roger, "Elements de Cryptographie," Paris, 1939.[BAZE] Bazeries, M. le Capitaine, " Cryptograph a 20 rondelles- alphabets," Compte rendu de la 20e session de l' Association Francaise pour l'Advancement des Scienses, Paris: Au secretariat de l' Association, 1892.[BEES] Beesley, P., "Very Special Intelligence", Doubleday, New York, 1977.[BLK] Blackstock, Paul W. and Frank L Schaf, Jr., "Intelligence, Espionage, Counterespionage and Covert Operations," Gale Research Co., Detroit, MI., 1978.[BLOC] Bloch, Gilbert and Ralph Erskine, "Exploit the Double Encipherment Flaw in Enigma", Cryptologia, vol 10, #3, July 1986, p134 ff. (29)[BLUE] Bearden, Bill, "The Bluejacket's Manual, 20th ed., Annapolis: U.S. Naval Institute, 1978.[BODY] Brown, Anthony - Cave, "Bodyguard of Lies", Harper and Row, New York, 1975.[BOLI] Bolinger, D. and Sears, D., "Aspects of Language," 3rd ed., Harcourt Brace Jovanovich,Inc., New York, 1981.[BOSW] Bosworth, Bruce, "Codes, Ciphers and Computers: An Introduction to Information Security," Hayden Books, Rochelle Park, NJ, 1990.[BOWE] Bowers, William Maxwell, "The Bifid Cipher, Practical Cryptanalysis, II, ACA, 1960.[BP82] Beker, H., and Piper, F., " Cipher Systems, The Protection of Communications", John Wiley and Sons, NY, 1982.[BRAS] Brasspounder, "Language Data - German," MA89, THe Cryptogram, American Cryptogram Association, 1989.[BROO] Brook, Maxey, "150 Puzzles in Cryptarithmetic," Dover, 1963.[BRIT] Anonymous, "British Army Manual of Cryptography", HMF, 1914.[BROG] Broglie, Duc de, Le Secret du roi: Correspondance secrete de Louis XV avec ses agents diplomatiques 1752-1774, 3rd ed. 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P., Geschichte der Mathematik im Mittelatter, Liepzig, Germany: Teubner, 1964.[YUKI] Yukio Nishihara, "Kantogan tai-So Sakusenshi," Vol 17., unpublished manuscript, National Institute for Defense Studies Military Archives, Tokyo.,(hereafter NIDS Archives)[ZIM] Zim, Herbert S., "Codes and Secret Writing." William Morrow Co., New York, 1948.[ZEND] Callimahos, L. D., Traffic Analysis and the Zendian Problem, Agean Park Press, 1984. (also available through NSA Center for Cryptologic History)Text converted to HTML on June 18, 1998 by Joe Peschel.

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