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

• Lesson 1
• Lesson 2
• Lesson 3
• Lesson 4
• Lesson 5
• Lesson 6
• Lesson 7
• Lesson 8
• Lesson 9
• Lesson 10
• Lesson 11
• Lesson 12
• # CLASSICAL CRYPTOGRAPHY COURSE

BY LANAKI

October 23, 1995

LECTURE 2
SUBSTITUTION WITH VARIANTS
Part I

## SUMMARY

In Lecture 2, we expand our purview of substitution ciphers,drop the requirement for word divisions, solve a lengthyPatristocrat, add more tools for cryptanalysis, look at somehistorical variations and solve the assigned homework problems.

## IDENTIFYING SUBSTITUTION AND TRANSPOSITION CIPHERS

Recall from Lecture 1, that the fundamental difference betweensubstitution and transposition ciphers is that in the former,the normal or conventional values of the letters of the PT arechanged, without any change in the relative positions of theletters in their original sequences, whereas in the latter,only the relative positions of the letters of the PT in theoriginal sequences are changed, without any changes to theconventional values for the letters.

I used the term uniliteral frequency distribution [UFD] (I alsomisspelled uniliteral as unilateral) to identify the simplesubstitution cipher. Three properties can be discerned fromthe UFD applied to CT of average length composed of letters:(1) Whether the cipher belongs to the substitution ortransposition class; (2) If to the former, whether it ismonoalphabetic or non-monoalphabetic in character, (3) Ifmonoalphabetic, whether the cipher alphabetic is standard(direct or reversed) or mixed.

## CIPHER CLASS

Because a transposition cipher rearranges the PT, withoutchanging the identities of the PT, the corresponding number ofvowels (A,E,I,O,U,Y), high frequency consonants (D,N,R,S,T),medium-frequency consonants (B,C,F,G,H,L,M,P,V,W) andespecially, low-frequency consonants (J,Q,X,Y,Z) are exactlythe same in the CT as they are in the PT. In a substitutioncipher, the conventional percentage of vowels and consonantsin the CT have been altered. As messages decrease in lengththere is a greater probability of departure from the normalproportion of vowels and consonants. As messages increase inlength, there is lesser and lesser departure from normalproportions. At 1000 letters or more, there is practically nodifference at all between actual and theoretical proportions.Friedman presents charts showing the normal expectationof vowels and high, medium, low and blanks for messages ofvarious lengths. For example, for a message of 100 letters inplain English, there should be between 33 and 47 vowels(A,E,I,O,U,Y). Likewise, there will be between 28 and 42 high-frequency consonants (D,N,R,S,T); between 17 and 31 mediumfrequency consonants (B,C,F,G,H,L,M,P,V,W); between 0 and 3low-frequency consonants (J,Q,X,Y,Z); and between 1 and 6blanks theoretically expected in distribution of the PT. Cipherclass is considered transposition if the above limits bound theCT message and substitution if the above expected limits areoutside the chart limits for the message length in question. [FR1/ p32-39]

## UFD

The uniliteral frequency distribution (UFD) may be used toindicate monoalphabeticity. The normal distribution showsmarked crests and troughs by virtue of two circumstances.Elementary sounds which the symbols represent are used withgreater frequency. This is one of the striking characteristicsof every alphabetic language. With few exceptions, each soundis represented by a unique symbol. The one-to-one mappingcorrespondence between PT and CT will dictate a shifted UFDwith different absolute positions of the crests and troughsfrom normal. A marked crest-and-trough appearance in the UFDfor a given cryptogram indicates that a single cipher alphabetis involved and constitutes one of the tests for a mono-alphabetic substitution cipher.

The absence of marked crests and troughs in the UFD indicatesthat a complex form of substitution is involved. The flattenedout appearance of the distribution is one of the criteria forrejection of a hypothesis of monoalphabetic substitution.

## LAMBDA BLANK EXPECTATION TEST - LB^

Friedman presents a chart supporting the LB^ test for blanks inEnglish messages up to 200 letters. [FR1] Soloman Kullbackderives the Lambda test and presents extensive probability dataon English, French, German, Italian, Japanese, Portuguese,Russian and Spanish. [KULL] Statistical studies show that thenumber of blanks in a normal PT message is predictable.Friedman's chart shows that the plaintext limit, P and therandom expectation, R limits are a function of message size.On his chart, random assortment of letters correspond topolyalphabetic CT. The number of alphabets used is largeenough to approximate a UFD identical to a distribution ofletters picked randomly out of a hat.

## PHI TEST FOR MONOALPHABETICITY

This test compares the observed value PHI(o) for thedistribution being tested with the expected value PHI(r) randomand the expected value of PHI(p) plain text. For Englishmilitary text:

`          PHI(r) = .0385N(N-1)          PHI(p) = .0667N(N-1)`
where N is the number of elements in the distribution. Theconstant .0385 is 1/26 decimal equivalent and constant .0667is the sum of squares of the probabilities of occurrence of theindividual letters in English PT. [FR3]

Example 1 of the PHI test on the following cryptogram is:

`  O W Q W Z   A E D T D   Q H H O B   A W F T Z   W O D E Q  T U W R Q   B D Q R O   X H Q D A   G T B D H   P Z R D Kf:      3 3   7 2 1 1 4     1       4 1 6 3   4 1   5 1   3CT:     A B C D E F G H I J K L M N O P Q R S T U V W X Y Zf(f-1): 6 6  42 2 0 0 12    0      12 030 6  12 0  20 0   6N = number of letters = sum fi = 50PHI(o) =  sum [fi(fi-1)] = 154PHI(r) =  .0358N(N-1) = .0385x 50 x49 =94PHI(p) =  .0667N(N-1) = .0667x 50 x49 =163`
Since PHI(o), 154, more closely approximates PHI(p) than doesPHI(r), we have mathematical corroboration of the hypothesisthat the CT is monoalphabetic.

Example 2: Given the frequency distribution of CT as:

`f:      1     1 2 3 4 2     1       4 2   1           1   3CT:     A B C D E F G H I J K L M N O P Q R S T U V W X Y Zf(f-1): 0     0 2 612 2     0      12 2   0           0   6N = 25 lettersPHI(o) = 42PHI(r) = 0.0385x25x24 = 23PHI(p) = 0.0667x25x24 = 40`
Since PHI(o) observed is closer to PHI(p), then this letterdistribution is monoalphabetic. But compare to example 3 with25 letters:

`f:          1       1 1 2 1 1 1 3 1 1   1 2   1 1   1 1 2 3CT:     A B C D E F G H I J K L M N O P Q R S T U V W X Y Zf(f-1):     0       0 0 2 0 0 0 6 0 0   0 2   0 0   0 0 2 6N = 25 lettersPHI(r) = 0.0385x25x24 = 23PHI(p) = 0.0667x25x24 = 40`
Since PHI(o) observed is closer to PHI(r), then this letterdistribution is non-monoalphabetic.

Before we think this test is perfect, the student should trythe above PHI test on the phrase:

`" a quick brown fox jumps over the lazy dog"He will find that N=33, PHI(o)= 20 and PHI(r) = 41; PHI(p)=70.`
Since the observed value is less than half of PHI random, thiswould suggest that the letters of this phrase could not beplain text in any language. Think about the cause of thisresult. For a simplified derivation, see Sinkov [SINK]

Kullback gives the following tables for Monoalphabetic andDigraphic texts for eight languages:

`                  Monoalphabetic        Digraphic                     Text                 Text   English        0.0661N(N-1)          0.0069N(N-1)   French         0.0778N(N-1)          0.0093N(N-1)   German         0.0762N(N-1)          0.0112N(N-1)   Italian        0.0738N(N-1)          0.0081N(N-1)   Japanese       0.0819N(N-1)          0.0116N(N-1)   Portuguese     0.0791N(N-1)   Russian        0.0529N(N-1)          0.0058N(N-1)   Spanish        0.0775N(N-1)          0.0093N(N-1)                     Random Text   Monographic         Digraphic        Trigraphic   .038N(N-1)         .0015N(N-1)      .000057N(N-1)`
Note that the English plain text value is slightly less thanFriedman's. [KULL] [SINK]

## INDEX OF COINCIDENCE (I.C.)

Friedman made famous the Index of Coincidence. It is anothermethod of expressing the monoalphabeticity of a cryptogram. Wecompare the theoretical I.C. with the actual I.C. I.C. isdefined as the ratio of PHI(o)/PHI(r). Thus, in example onethe I.C. is 154/94 = 1.64. The theoretical I.C. for English is1.73 or (.0667/.0385). The I.C. of random text is 1.00 or(.0385/.0385). Friedman wrote a paper entitled "The Index ofCoincidence and Its Application in Cryptography", which isperhaps the most ground breaking treatise in the history ofcryptography. [FR22]

## CIPHER ALPHABETS - STANDARD OR MIXED

Assuming a UFD that is monoalphabetic in character, we observethe crests and troughs of the distribution. If they occupyrelative offset positions to the normal UFD, than the alphabetis most likely standard, (A, B, C,..). If not, the CT isprepared using a mixed alphabet. The direction the crests andtroughs progress left to right or right to left tell us whetherthe alphabet is standard or reversed in direction.

## LONG WORD RISTIES - SHERLAC METHOD

When an Aristocrat consists of all long words, it may beattacked by the SHERLAC Method. The object is to compare vowelpositions and word endings in a columnar display of the CT byindividual word. We mark all low frequency ( f <= 3 ) , thenthe 2nd column position (vowel favorite) and word endings areexamined. For example, from S-TUCK: [TUCK], [B201]

`    fi  14 13 12 12 10 10 8 5 5 4 4 4 3 3 3 3 3 3 2 2 2 2    CT   D  Q  I  N  O  P A L X E R V C F H M S Y J K W Z    F= 127 letters  = sum fi`
The CT presented in columnar form and marked for low frequencyletters is:

`      c  .  c  v  .  .  v  c  c  v1.    X  W  V  I  M  S  O  Q  P  N  V      s     c  o     h  a  n  t  i      c  v  .  c  c  .  v  .  v  c2.    Q  I  F  E  D  Y  I  H  O  Q,      n  o  b  l  e  w  o  m  a  n      .  v  .  c  v  .  v  c  c3.    Z  I  Y  P  I  Y  N  Q  L         o  w  t  o  w  i  n  g      v  .  v  c  c  v  c  v  c  v  c  c  .4.    D  K  O  L  L  D  A  O  P  D  R  E  W ,      e  x  a  g  g  e  r  a  t  e  d  l  y      c  v  c  .  c  v  v     v  c5.    R  N  X  M  E  D  O  X  D  R      d  i  s     l  e  a     e  d      c  v  .  v  c  v  v  c  c6.    X  I  C  D  A  D  N  L  Q .      s  o     e  r  e  i  g  n      .  c  v  .  v  v  v  c  v  v  c7.    M  A  I  C  I  V  O  P  N  I  Q         r  o     o  e  a  t  i  o  n      v  c  v  c  c  v  c  v  c  v8.    N  Q  I  A  R  N  Q  O  P  D ,      i  n  o  r  d  i  n  a  t  e      .  v  c  v  c  .  .  v  c  c9.    F  O  Q  N  X  S  H  D  Q  P         a  n  i  s  h  m  e  n  t      v  c  c  v  c  c  c  v  .  v  c  c  v  .  c  v10.   N  Q  V  I  Q  P  A  I  C  D  A  P  N  F  E  D.      i  n  c  o  n  t  r  o  v  e  r  t  i  b  l  e      v  .  c  v  c  .  v  c  .11.   O  J  P  D  A  H  O  P  S,      a  f  t  e  r  m  a  t  h      .  v  c  c  .  v  .  v  c12.   Z  N  Q  L -J  N  K  D  A  !!!         i  n  g  f  i  x  e  r`
We mark the cipher as we put forth the following thoughts.Analysis of Column 2 in the above CT, shows that I appearsthree times with 11 low frequency contacts. It is probablya vowel but not "i." N appears twice with 5 low frequencycontacts and also in third end position 3 times. Probablevowel, may be i as in ion. Q appears twice; no low frequencycontacts, follows probable vowels 7 times. Might be n.i and n are placed in the CT. Word 7 yields I = o. Word 3yields the L = g. [Word 6 may not fit though.] Word 7also suggest that P =t for tion. P precedes N and I 4 times,and follows O 3 times. D begins one word, ends 2, has highfrequency, and is scattered. Let D = e. O contacts 6 lowfrequency, and precedes n 3 times, t 4 times and the t isfollowed by e twice. We have the 'ate' trigram, so O=a. Notethat A reverses with D and contacts vowels 10 times. A=r.

Word 10 shows incontrovertible. Playing it thru with V=c,F=b,and E=l, word 2 becomes noblewoman, Y=w,h=m. Word 11 isaftermath giving two more PT equivalents of J=f, and S=h.Word 4 gives us the K=x, R=D, W=y. Word 6 is sovereign andyields the PT s. The balance of the CT can be found by checkoff and testing.

The message reads: Sycophantic noblewoman, kowtowingexaggeratedly, displeased sovereign. Provocation inordinate,banishment incontrovertible. Aftermath king-fixer!

I have experimented with the SHERLAC method. Even when the CTincludes small words <= 4 letters, it seems to yield valuabledata. Just line the CT (words 5 or more letters) in columnsand ignore the shorter words. Then work back and forth withthe shorter words for confirmations.

## PATRISTOCRATS

When we remove the crutch of word divisions in the Aristocratand present in standard telegraphic five letter groups, we havethe "Patristocrat" or "undivided." S-TUCK gives a solutionprocedure for "undivideds" as well as Friedman. ELCY alsodiscusses the Aristocrat without word divisions in her chapter11. [TUCK], [FR1], [ELCY] Friedman's presentation isexcellent and is summarized for the reader:

`Given  P-1:SFDZF  IOGHL  PZFGZ  DYSPF  HBZDS  GVHTF  UPLVD  FGYVJ  VFVHTGADZZ  AITYD  ZYFZJ  ZTGPT  VTZBD  VFHTZ  DFXSB  GIDZY  VTXOI        ---                    -------------YVTEF  VMGZZ  THLLV  XZDFM  HTZAI  TYDZY  BDVFH  TZDFK  ZDZZJ                      -------------------------------SXISG  ZYGAV  FSLGZ  DTHHT  CDZRS  VTYZD  OZFFH  TZAIT  YDZYG                                              --------------AVDGZ  ZTKHI  TYZYS  DZGHU  ZFZTG  UPGDI  XWGHX  ASRUZ  DFUID           ----                                      -----EGHTV  EAGXX`
There are two basic attacks on the Patristocrat. The firstmethod creates a triliteral frequency table and the second usesthe "probable word" as a wedge into the cryptogram. The firstattack follows many of the vowel - consonant splitting stepsthat we have looked at previously.

## METHOD A: Vowel - Consonant Splitting.

Step 1: Inspect/mark for long repetitions, many letters of normally low frequency, such as F, G, V, X, Z; and vowels and high frequency consonants N and R are relatively scarce.

Step 2: Prepare UFD and apply PHI tests.

` 8 4 1 23 3 19 19 15 10 3 2 5 2 0 3 5 0 2 10 22 5 16 1 8 14 35 A B C  D E  F  G  H  I J K L M N O P Q R  S  T U  V W X  Y  Z   PHI(p) = 3668   PHI(r) = 2117   PHI(o) =  3862   ft = 235`
The marked crests and troughs and the PHI test support themonoalphabetic hypothesis. Friedman advises that "the beginnermust repress the natural tendency to place too much confidencein the generalized principles of frequency and to rely too muchupon them. [i.e. setting Z=e, D=t ] It is far better to intoeffective use certain other data concerning normal plain text,such as digraphic and trigraphic frequencies."
Step 3: Prepare a special worksheet; mark reversible digraphs and trigraphs, inscribe the frequencies of the first and last 10 letters, because these positions often lend themselves more readily to attack, and note positions of low frequency CT letters.

Step 4: Prepare a Triliteral Frequency Distribution (TFD) showing One Prefix and One Suffix Letter. Examine the TFD for digraphs and trigraphs occurring two or more times in the cryptogram. Note repeated digraphs and trigraphs. For the above CT,

`  DZ  = 9x,  DF  = 5x, DV  = 2x  ZDF = 4x,  YDZ = 3x, BDV = 2x`

`     Condensed Table Of Repetitions For P-1.    Digraphs         Trigraphs          PolygraphsDZ - 9  TZ - 5      DZY - 4             HTZAITYDZY - 2ZD - 9  TY - 5      HTZ - 4             BDVFHTZDF - 2HT - 8  FH - 4      ITY - 4             ZAITYDZY - 3ZY - 6  GH - 4      ZDF - 4             FHTZ - 3DF - 5  IT - 4      AIT - 3GZ - 5  VF - 4      FHT - 3        VT - 4      TYD - 3        ZF - 4      YDZ - 3        ZT - 4      ZAI - 3        ZZ - 4          IE          ZF                                              HV          GI                                              ZG          SZ                                              IY          VG    DU AX                                     ZK          YZ    ZZ EH                                     IY          ZO    FH WH                                     HZ          CZ    ZF PD GT                                  VY          ZT    VS TU GX                                  HC          ZZ    DK ZH GU                                  DH          ZF    VH DZ KI                                  HZ          BV    DM YA FT                                  IY          YZ    EV LZ FT                                  HZ          ZF    DX YA HT UD                            AR ZH          IZ    VH SZ TH DX                            YD VE EG       ZF    YZ MZ FT HT                            RV VX XS       BV    VV BI MT AT                            FL HZ GV       YZ    DG TP TL XS                            IG VZ ZI       AZ    TU TA FT AT       SG          UG       JX PV GV YD    VF    PH FY VT OY       LV          GT       XB ZG ZI SG    ZS VA ZG SV VT GD ZS    HL       DZ UL       DG IY ZI ZD    ZY DG ZI FZ FB AT ZZ TH PV FH    XI SF    SU YP HG GD HZ TD FZ TF SD OH GL FO VV FZ HP VG    IG LZ    ZS -F HF  A  B  C  D  E  F  G  H  I  J  K  L  M  N  O  P  Q  R  S  T  8  4  1 23  3 19 19 15 10  3  2  5  2  0  3  5  0  2 10 22                   KD                   TD                   DY    TE             TA UD    AD             XD FT    ST       ZS    ZT UF    AF       TZ    GZ DG    DF       ZG    DY YY    LX       TD    TD ZT    FM       TZ    TB GZ    YT       ZG    JT DY    YT    X- ZB    FJ TA    DF    GX TD    DY OF    TT    HA IV    ZA YD FI FH    IW ZV    DZ DR RZ JF    SI ZF    BD GD GP YJ    VZ TD    GD GY HZ LD    TO GV    PF ZJ FP GH XG FS DS    DF DZ  U  V  W  X  Y     Z `
Step 5: Classify the cipher letters into vowels and consonants. As we did in the Aristocrats, again we separate high frequency letters into probable vowels and consonants. If we find A, E, I, O, and N, R, S, T, we have values for 2/3 of the cipher text letters that normally (most likely) occur in the cryptogram.

Friedman's Table 7-B in Appendix 2 confirms that vowel combine differently from consonants. The top 18 digraphs compose about 25 per cent of English text. The letter E enters into 9 of the 18 digraphs: [FRE1]

`        ED  EN  ER  ES  NE  RE  SE  TE  VE`
The remaining 9 digraphs are:

`         AN  ND  OR  ST  IN  NT  TH  ON  TO`
None of the 18 digraphs is a combination of vowels. So Ecombines with consonants more readily than with other vowels oreven itself. So if the letters of the highest frequency arelisted with the assume CT =e , those that show a high affinityare likely N R S T and those that do not show any affinity arelikely A I O U. In P-1., Let Z = e because it is highfrequency and combines with several other high frequencyletters, D, F, G. The nine next highest frequency letters andtheir combinatorial affinity with Z are:

`Z as prefix      8      4      4      1      0               D(23)  T (22)  F(19)  G(19)  V(16)Z as suffix      9      5      2      5      0Z as prefix     0        6     0     0               H(15)  Y(14)  S(10  I(10)Z as suffix     0        2     0     0`
Step 6: Analysis of Data:
CT D occurs 23 times, 18 times combined with Z, 9 times as areversal ZD, DZ. T shows 9 combinations Z, 4 in ZT and 5 in TZ. D and T must be consonants. Similarly, F, G, Y are guessed as consonants. An initial cut is:

`              Vowels              Consonants          Z=e, V, H, S, I       D, T, F, G, Y`
Friedman's Table 6 in Appendix 2 gives us 10 most frequently occurring diphthongs: [FRE1]

` Diphthong:   io  ou  ea  ei  ai  ie  au  eo  ay  ue Frequency:   41  37  35  27  17  13  13  12  12  11`
Also, O is usually the vowel of second highest CT frequency.Looking at V, H, S, I not = i, can we find the CT equilvalentof PT o?

List the combinations of V, H, S, I and Z=e in the message. Weexamine the combinations they make among themselves and with Z= e.

`     ZZ = 4   VH = 4  HH = 1  HI = 1 IS = 1  SV = 1`
Now, ZZ = ee. HH is oo, because aa, ii, uu are practicallynon-existent. oo is the second highest frequency double vowelnext to ee. If H=o, then V =i, where VH occurs twice and io isa high frequency diphthong in English. So our analysis results(unconfirmed) so far are:

`            Z = e,  H = o,  V = i`
So I and S should be a and u. Here we use another Friedmantool to look at the possibilities. We define the alternativePT diphthongs and add frequency values as a set.

`    1) either I = a and  S = u,      each digraph occurs 1x    2) or     I = u and  S = a.  HI = oa   value =  7           HI = ou  value = 37  SV = ui   value =  5           SV = ai  value = 17  IS = au   value = 13           IS = ua  value =  5                  ====                          ====  Total             25                            59`
Alternative two seems more likely. A more precise method forchoosing between alternative groups of Digraphs by consideringlogarithmic weights of their assigned probabilities, ratherthan PT frequency values. These weights are given by Friedmanin [FRE2] Appendix 2. The method is detailed on pp 259-260.Tables 8 and 9A - C give the data for 428 digraphs based on50,000 words of text. See also [KULL].

`  HI = oa   L224  = .48          HI = ou  L224  = .79  SV = ui   values= .42          SV = ai  values= .64  IS = au         = .59          IS = ua        = .42                  =====                         =====  Total Log base   1.49                          1.85             224`
Multiple occurrences of a digraph would be multiplied by itslog base 224 relative weight and added as a group.

`So we now have Z = e , H = o,  V =i,  S = A,  I = ufor vowel equivalents.`
The consonants may be viewed from their combination withsuspected vowels. Since VH = io might infer sion or tiontetragraphs we look at the CT and find
`           GVHT      and    FVHT`
T most likely is the n and G or F could be s or t. Note thatthe CT D is neither PT t or n or PT s. The reversal with Z=e, suggests the letter r.

As an alternative, the Consonant-Line approach would yield:

`                B C E J K M O R W             -----------------------                      Y ?                    D D ?D D D        vowel ?                    S S ?S S          vowel                        ?G G G G G                Z Z Z Z ?Z Z Z Z      vowel                      H ?H H          vowel                  T T T ?                  V V V ?V            vowel                        ?A                    F F ?F            vowel?                    X X ?                      I ?I            vowel?                        ?U`
The general principles are repeated. Vowels distinguishthemselves from consonants as they are represented by:

`1) high frequency letters2) high frequency letters that do not contact each other3) high frequency letters with great variety of contacts4) high frequency letters with am affinity for low frequency   PT consonants`
Step 7: Prepare the partial enciphering alphabet and substituteinto the cryptogram.

`PT: A B C D E F G H I J K L M N O P Q R S T U V W X Y ZCT: S       Z       V         T H     D G F I                                        F G`
P-1 revisited and rewritten:
`S F D Z F I O G H L P Z F G Z D Y S P F H B Z D Sa t r e t u   s o     e t s e r   a   t o   e r a  s     s     t         s t           sG V H T F U P L V D F G Y V J V F V H T G A D Z Zs i o n t       i r t s   i   i t i o n s   r e et       s           s t         s       tA I T Y D Z Y F Z J Z T G P T V T Z B D V F H T Z  u n   r e   t e   e n s   n i n e   r i t o n e              s         t                 sD F X S B G I D Z Y V T X O I Y V T E F V M G Z Zr t   a   s u r e   i n     u   i n   t     s e e  s       t                           s     tT H L L V X Z D F M H T Z A I T Y D Z Y B D V F Hn o     i   e r t   o n e   u n   r e     r i t o                sT Z D F K Z D Z Z J S X I S G Z Y G A V F S L G Zn e r t   e r e e   a   u a s e   s   i t a   s e      s                     t      t      s     tD T H H T C D Z R S V T Y Z D O Z F F H T Z A I Tr n o o n   r e   a i n   e r   e t t o n e   u n                                  s sY D Z Y G A V D G Z Z T K H I T Y Z Y S D Z G H U  r e   s   i r s e e n   o u n   e   a r e s o        t       t                           tZ F Z T G U P G D I X W G H X A S R U Z D F U I De t e n s     s r u     s o     a     e r t   u r  s     t     t         t                 sE G H T V E A G X X  s o n i     s  t           t`
I have left out the frequencies above the letters for editorialspace only.

We can see from a first reading that PT words operations, nineprisoners, and afternoon come thru. G = t, F = s, B = p, L =f.

Step 8: Complete the solution. Prepare the Ct/Pt Key Alphabets.

Message: As result of yesterdays operations by first divisionthree hundred seventy nine prisoners captured including sixteenofficers. One hundred prisoners were evacuated this afternoon,remainder less one hundred thirteen wounded are to be sent bytruck to chambersburg tonight.

`PT: A B C D E F G H I J K L M N O P Q R S T U V W X Y ZCT: S U X Y Z L E A V N W O R T H B C D F G I J K M P Q`

## METHOD B: Probable Word Attack.

"Patties" in the CM usually come with a tip and are generallyshorter than the above example. The tip constitutes a probablePT word or phrase and we search the CT for a pattern of CTletters that exactly match the probable word. The choice ofprobable words is aided or limited by the number and positionsof repeated letters. Repetitions may be patent (visibleexternally) or latent ( made patent as a result of theanalysis. For the example DIVISION with a repeated I orBATTALION with the reversible AT, TA help the cryptanalysiseven though word divisions were removed. Friedman named whatwe call patterns as idiomorphs. [FRE2] gives many patternlists for solution of substitution and Playfair ciphers. TEAcomputer program at the Crypto Drop Box is an automated patternlist to 20 words. [CAR1], [CAR2], [WAL1], [WAL2] giveidiomorphic data. The process of superimposing the plaintext word over the correct cipher text will effect the entryto the cryptogram. Other references include: [RAJ1], [RAJ2],[RAJ3], [RAJ4], [RAJ5], [HEMP], [LYNC].

## SOLUTION OF ADDITIONAL CRYPTOGRAMS PRODUCED BY SAME COMPONENTS.

Once the cryptogram has been solved and the keying alphabetreconstructed, subsequent messages which have been encipheredby the same means solve readily.

P- 2.

Suppose the following message is intercepted slightly later atthe same station.

`  I Y E W K   C E R N W   O F O S E   L F O O H   E A Z X X`
P - 1. reconstruction and arbitrarily set at L = a.

`PT  a b c d e f g h i j k l m n o p q r s t u v w x y zCT  L E A V N W O R T H B C D F G I J K M P Q S U X Y Z     Cryptogram     I Y E W K    C E R N W     Equivalents    P Y B F R    L B H E F`
running down the sequence yields CLOSE YOURS as a generatrix.
`                    I Y E W K    C E R N W                    P Y B F R    L B H E F                    Q Z C G S    M C I F G                    R A D H T    N D J G H                    S B E I U    O E K H I                    T C F J V    P F L I J                    U D G K W    Q G M J K                    V E H L X    R H N K L                    W F I M Y    S I O L M                    X G J N Z    T J P M N                    Y H K O A    U K Q N O                    Z I L P B    V L R O P                    A J M Q C    W M S P Q                    B K N R D    X N T Q R            ***     C L O S E    Y O U R S                    D M P T F    Z P V S T                    E N Q U G                    F O R V H                    G P S W I                    H Q T X                    I R U Y                    J S V Z                    K T W A                    L U X B                    M V Y C                    N W Z D                    O X A E`
Set the cipher component against the normal at C = i.
`PT  a b c d e f g h i j k l m n o p q r s t u v w x y zCT  F G I J K M P Q S U X Y Z L E A V N W O R T H B C D`
Solving: Close your station at two PM.

[FRE1] discusses keyword recovery processes on pp 85 -90.Also see [ACA].

## VOWEL CIPHER

Louis Mansfield introduced the concept of a vowel substitutioncipher in 1936. [MANS] In the vowel cipher the keyalphabet is written into a square with j=i, like this:

`                     COLUMN                 A   E   I   O   U                ?-----------------              A ?a   b   c   d   e                ?              E ?f   g   h   i/j k                ?              I ?l   m   n   o   p   ROW          ?              O ?q   r   s   t   u                ?              U ?v   w   x   y   z`
Enciphering is row by column or t = OO ; h = EI ;and e = AU.

The entire CT message enciphered is a succession of vowels.The CT will be exactly twice as long as the PT. This method isnot more difficult to crack than the standard Aristocrat butour focus is on the frequency of vowel combinations in the CT.The PT equivalents do not have to be in standard order. Trythis example.

VV-1.

`  1      2              3                 4         5OUOE    OUAE   OUIUIAAIUIUUOEOUAOAI    OEEOUUOE   OEEOAI     6                 7            8UOEUUEUEAIAEOE   OOAIOEUUOUUEAE   EEUO  9          10             11              12       13UUUEUE   OEUIEEEEIA   IUEEAOAIIUAIOAOEAE    IOAI   EIUUUI-                         14           15       16AIUOEUUEUEEA    EIEEIUIAOUUEAIOO   UUOAOO   AEAIOAOE17       18            19OEEE    EUAE     UIAIIIEUUEUUUIUEEA.`
To solve this cipher we list the various combinations of twovowels:
`    AA        EA  (2)    IA  (4)   OA  (3)    UA    AE  (6)   EE  (6)    IE        OE  (11)   UE  (10)    AI  (11)  EI  (2)    II  (1)   OI         UI  (5)    AO  (2)   EO  (2)    IO  (1)   OO  (3)    UO  (3)    AU        EU  (4)    IU  (4)   OU  (6)    UU  (7)`
AI and OE appear 11 times and often as finals. Either might bethe "e". OE appears as an initial. WE try OE as t and AI ase. Word 5 becomes 'the.' Word 4 confirms as 'that.' UU =a. UE = l. The normal procedure of test and confirm gives usthe message: It is imperative that the fullest details of alltroop movements be carefully compiled and sent to us regularly.

The partial keying square is:

`                  A    E    I    O    U              --------------------------              A        s    e    v              E   y    o    c    h    u              I   p         g    b    m              O   n    t         d    i              U        l    r    f    a`

## MIRABEAU'S CIPHER

Comte de Mirabeau (1749 - 1791) was one of the great oratorsin the National Assembly, the body that governed France duringthe early phases of the French Revolution. He was a politicalenemy of Robespierre. He developed a simple substitutionvariant to relay his court messages to Louis XVI (who rejectedhis moderate advise). His father, the Marquis de VictorRiqueti Mirabeau imprisoned his son for failure to pay debts.He devised this system during his stay in debtors prison.

The Mirabeau system of ciphering letters of the alphabet aredivided into five groups of five letters each. Each letter isnumbered according to its position in the group. The group isalso numbered. The key alphabet is arranged as follows:

`      1 2 3 4 5      1 2 3 4 5     1 2 3 4 5      I S U W B      K T D Q R     X L P A E          6              8             4                 1 2 3 4 5   1 2 3 4 5                 G O Y V F   Z M C H N                     7           5`
Encipherment of the phrase ' the boy ran' wouldbe 82.54.45, 65.72.73, 85.44.55 where the t isreferenced by group 8,letter2. Solution of messagesis clearly by frequency analysis, the key being reconstructedfrom the message. Mirabeau experimented by reversing numberorder in this positional number system and adding nulls toconfuse the interloper. One of the interesting complicationsadded by Mirabeau was to express the CT as a fraction withgroup number as numerator and position number as denominator.Other figures were added to foil decipherment. In such a casethe alphabet is grouped into fives as before, but the groupsand positions are each numbered with the same five figures.So:

`     1 2 3 4 5     1 2 3 4 5      1 2 3 4 5     O A G P U     T C N H Y      X M F I S         1             2              3     1 2 3 4 5     1 2 3 4 5     L Q W B V     R E K Z D         4             5`
Enciphering 'the boys run' :
`     2 2 5    4 1 2 3    5 1 2     - - -    - - - -    - - -     1 4 2    4 1 5 5    1 5 3`
Adding numerals 6, 7, 8, 9, 0 as non-values both above andbelow the line increase the security slightly. Or:
`     29  27  50     48  17  29  39     56  10  28     --  --  --     --  --  --  --     --  --  --     71  64  92     74  94  85  65     91  75  83`
The recipient reads the message by cancelling the non-valuesand using the others. A key to recognition of this cipher isthat the non-values (nulls) are never employed as group orposition numbers.

The complicated form of the Mirabeau is solved by preparing aFractional Bigram sheet and reducing out the non-values.Suppose we encipher the phrase 'we have been here':

`     2  1  4  5      4  5  5  2     1  5  5  5     -  -  -  -      -  -  -  -     -  -  -  -     4  2  5  2      4  2  2  3     4  2  1  2`
Using non-values (6,7,8,9,0) as:
`     62  10  40  65    47  57  75  62    27  58  57  85     --  --  --  --    --  --  --  --    --  --  --  --     48  62  95  20    84  27  92  30    49  62  19  29`
The five 'e' s which occur are different each time.
`                 65   57  75  58  85                 --   --  --  --  --                 20   27  92  62  29`
The fractional group sheet proceeds like a Bigram analysis.Instead of letters we use fractions.

The first fraction would be noted in four different ways, e.g.,

`              6  6  2  2              -  -  -  -              4  8  4  8              65                            6  6  5  5the group     --      would be catalogued   -  -  -  -              20                            2  0  2  0              5The fraction  -   (which is the real e) will eventually assume              2`
Its normal frequency and thus display its identity. Armed withthe fact that 5/2 represents e, we cancel out all thenon-values which occur with this fraction. Each time wecancel out a non-value, we do so for the entire cryptogram.Even if the 5/2 represents another letter, such as t, theuniliteral frequency distribution will be present in the CT.

## TELEPHONE CIPHER VARIATION - CHARLES SCHWAB

Hardly a cipher, but a modern substitution system effecting 10million brokerage customers is the Charles Schwab TelephoneAutomated Customer Service System. The telephone is used forthe enciphering of literal and numerical data to the Schwabcomputer system. So:

`       1  -       2  -  A B C       3  -  D E F       4  -  G H I       5  -  J K L       6  -  M N O       7  -  P R S          NO Q     use 99       8  -  T U V       9  -  W X Y          NO Z     use 98       *  -       0  -       #  -        J F M A M J JU A S O N DCALLS   A B C D E F G  H I J K LPUTS    M N O P Q R S  T U V W XSTRIKE PRICE CODES A   B    C    D    E    F    G    H    I    J    K    L 5   10   15   20   25   30   35   40   45   50   55   60105  110  115 120  125  130   135  140  145 150  155  160205  210  215 220  225  230   235  240  245 250  255  260305  310  315 320  325  330   335  340  345 350  355  360 M   N    O    P    Q    R    S    T    U    V    W    X65   70   75   80   85   90   95   100   7.5 12.5 17.5 22.5165  170  175 180  185  190   195  200265  270  275 280  285  290   295  300365  370  375 380  385  390   395  400`
Other combinations of the above indicate special actions.10 = accept. 90 = reject. * = return, end, # - accountterminator. Note that 1 number represents 3 equivalents.Schwab uses the position to indicate the letter similar to theancient Masonic Cipher. For example, Janus Enterprise Fundsymbol is JAENX = 51-21-32-62-92.

An order to buy 750 shares JAENX at a limit price of 23.5 plusa MAY call for 100 shares of BAX at strike price 25 mightinclude the following entries in the electronic order:

`18002724922, 61702554#, xxxx#, 1, 1, 750, 5121326292, 54,23*50, *, 1, 1,  100, 222192, 32, 32, 10, *`
Which represents the telephone number of Charles Schwab,account number and PIN, order codes, limit code, transfercodes, menu response items. Other codes would allow you tomove around your account and monitor the order.[Schw]

Note also that the basis telephone code is a 12 by 3 matrix.

`                       1 2 3                 1  -  b b b                 2  -  A B C                 3  -  D E F                 4  -  G H I                 5  -  J K L     b - represents available                 6  -  M N O         information slot                 7  -  P R S                 8  -  T U V     i.e. 9 = W X Y, but 93 = Y                 9  -  W X Y                           only            10   *  -  b b b            11   0  -  b b b            12   #  -  b b b`
It is easy to see how this process could be expanded to largerand larger keyspaces. See references [BOSW], [KOBL] and [WEL].for a fair discussions of the numerical requirements involved.A good discussion of the Information Theory is found inreference [RHEE]. A look at modern design criteria for bankfund transfer and similar PIN systems in found in Meyer andMatyas. [MM]

## MORE COMPUTER AIDS

Dr. Caxton C. Foster who wrote "Cryptanalysis for Micro-computers, while at the University of Massachusetts, hasgenerously donated his computer programs on substitution andtransposition to the class. I have sent an updated disk to ourCDB. [CCF] GWEGG has Cryptodyct on disk written in DbaseIV.Contact him for a copy.

A review of the entire field of applied cryptography ispresented in Bruce Schneier's book. Most of the material isbeyond the scope of this class, however a PC source / programdiskette is included with his book. There are ITAR limitationsassociated with his disk. We will cover some of the historicsymmetric algorithms such as Vigenere and Playfair ciphers.[SCHE]

## HOMEWORK ASSIGNMENTS

`Pd-1.                                            DanielH Z K L X   A L H X P   N C I N Z   X F L I X   G N W Q XP N Z K T   L N K X O   L X N I Z   X G I N X   P N E Z KX W Q X P   Z X L H X   P N C I N   Z X S N Q   N T X W QX P N W V   S N I K L   K H B L X   N W Q L X   H F Z I LN X A Z K   S B W E N   I.Pd-2.   Join the army.                             DanielF L B B A   O I A F Q   E A O M Z   U I L O N   R Z O Q AO P I L O   M O L S F   P F L I P   F L B B A   O E R I CA O Q E F   O P Q B L   O W A V H   Z O W E A   P X Z Q QG A P Z I   V V A Z Q   E G A Q E   F H T E L   G L S A PL R O W L   R I Q O U   F I E F P   E A Z O Q   Z I V I LQ T F Q E   E F P G F   M P L I G   U B L G G   L T H A.`

## SOLUTION TO HOMEWORK PROBLEMS FROM LECTURE 1

First assume that only English is involved in all 5 problems.(This may not be true third round.) My thanks to both WALRUSand SNAIL PACE for detailed solutions.

Problem A-1. can be solved by the Pattern Word method.

A-1. Bad design. K2 (91) AURION

`   1        2          3          4      5       6V G S   E U L Z K   W U F G Z   G O N   G M   V D G X Z A J U =                   7            8      9       10X U V B Z     H B U K N D W   V O N   D K   X D K U H H G D F =        11     12       13          14N Z X   U K   Y D K   V G U N   A J U X O U B B S       15           16     17        18X D K K G B P Z K   D F   N Y Z    B U L Z .`
Lots of two and three letter words. One of the three letterwords most likely 'the'. The GON GM combination is a possiblewedge. Note that N = 6/91 = 6.6% of CT, Z = 8/91 = 8.9% andV = 5/91 = 5.5% of CT. Word 10 pattern is (556) 291 on 12Lword = disappointed. Note the 'ted' ending fits with word 17t_e = the. so Y=h. Word 16 confirms DF = in.Plugging in A-1, we have:

`A-1.  Bad design.  K2 (91)                          AURION   1        2          3          4      5       6b o y     a   e s     a n o e   o u t   o     b i o d e g r aV G S   E U L Z K   W U F G Z   G O N   G M   V D G X Z A J U =                   7            8      9       10d a b l e     p l a   t i     b u t   i s   d i s a p p o i nX U V B Z     H B U K N D W   V O N   D K   X D K U H H G D F =        11     12       13          14t e d   a s   h i s   b o a t   g r a d u a l l yN Z X   U K   Y D K   V G U N   A J U X O U B B S       15           16     17        18d i s s o l   e s   i n   t h e    l a   eX D K K G B P Z K   D F   N Y Z    B U L Z .`
Words 11, 12 show: as his. Word 6 looks like biodegradable,and V confirms as a b in boat (word 13). Word 8 becomes but.Word 14 ends ally. The messages reads: Boy makes canoe out ofbiodegradable plastic but is disappointed as his boat graduallydissolves in the lake. The keyword recovered is MAYDAYCALL.Note that the tip was useless. One 12 letter pattern wordopens her up like a clam.

A-2. Not now. K1 (92) BRASSPOUNDER

`K D C Y   L Q Z K T L J Q X   C Y   M D B C Y J Q L :   " T RH Y D    F K X C ,     F Q   M K X   R L Q Q I Q   H Y D LM K L   D X C T W   R D C D L Q   J Q M N K X T M BP T B M Y E Q L   K   F K H   C Y   L Q Z K T L   T C . "`
Solve A-2 by eyball method. One letter word = a, look for you.. your combination, the word to in the first four wordsbefore quotation. The message reads: Auto repairmen tocustomer: " If you want, we can freeze your car until futuremechanics discover a way to repair it."

A-3. Ms. Packman really works! K4 (101) APEX DX

`* Z D D Y Y D Q T   Q M A R P A C ,   * Q A K C M K* T D V S V K .   B P   W V G   Q N V O M C M V B :   L D X VK Q A M S P D   L V Q U ,  L D B Z I   U V K Q F   P OW A M U X V ,   E M U V P   X Q N V ,  U A M O ZN Q K L M O V   ( S A P Z V O ) .`
APEX DX always sends an interesting con. Vowel splittingyields V, D, M, Q, P. Word 7 suggests that M=i. Words 15 and16 look like video game, the word fridge comes to bear.The message reads: Kuujjuaq airport, Arctic Quebec. So fewamenities: huge caribou head, husky decal on fridge, videogame, drink machine (broken). Kw = chimo; FORT.

`A-4.  Money value.  K4 (80)             PETROUSHKAD V T U W E F S Y Z   C V S H W B D X P   U Y T C Q P VE V Z F D A   E S T U W X   Q V S P F D B Y   P Q Y V D A F S ,H Y B P Q   P F Y V C D   Q S F I T X   P X B J D H W Y Z .`
Using the consonant-line method:
`               CEHZAUIXP               -----------               TTTT ?T        VOWEL                VV  ?VVV      VOWEL                YY  ?YY                Q   ?QQQ              DDDD  ?DD               WW   ?WWWW                SS  ?S               F    ?FFFFF     VOWEL                  BB?B                  P ?P                    ?`
J can be wrongly assumed to be a consonant. Digraph HW and rt/tr reversal fails but st/ts reversal gives information. Theword merchants can be found in a non-pattern word list. The chcombination fits CT HW. The message reads: Neighborlymerchants glimpse beyond bright personal splendor, clasp solemnprofit staunchly. Kw(s)= sprightly; BEHAVIOR.

A-5. Zoology lesson. K4 (78) MICROPOD

`A S P D G U L W ,   J Y C R   S K U Q   N B H Y Q I   X S P I NO C B Z A Y W N = O G S J Q   O S R Y U W ,   J N Y X UO B Z A   ( B C W S   D U R B C )   T B G A W   U Q E S L.* C B S W`
Note the entry B C W S... *C B S W. Try also.. LAOS. Theconsonant line yields S, U, Y, B as vowels. The message is;Koupreys, wild oxen having tough blackish=brown bodies, whiteback (also pedal) marks enjoy Laos. Kws = undomestic; BOVINE.

## REFERENCES / RESOURCES

`[ACA]  ACA and You, Handbook For Members of the American       Cryptogram Association, 1995.[BARK] Barker, Wayne G., "Cryptanalysis of The Simple       Substitution Cipher with Word Divisions," Aegean Park       Press, Laguna Hills, CA. 1973.[BAR1] Barker, Wayne G., "Course No 201, Cryptanalysis of The       Simple Substitution Cipher with Word Divisions," Aegean       Park Press, Laguna Hills, CA. 1975.[BOSW] Bosworth, Bruce, "Codes, Ciphers and Computers: An       Introduction to Information Security," Hayden Books,       Rochelle Park, NJ, 1990.[B201] Barker, Wayne G., "Cryptanalysis of The Simple       Substitution Cipher with Word Divisions," Course #201,       Aegean Park Press, Laguna Hills, CA. 1982.[BP82] Beker, H., and Piper, F., " Cipher Systems, The       Protection of Communications", John Wiley and Sons,       NY, 1982.[CAR1] Carlisle, Sheila. Pattern Words: Three to Eight Letters       in Length, Aegean Park Press, Laguna Hills, CA 92654,       1986.[CAR2] Carlisle, Sheila. Pattern Words: Nine Letters in Length,       Aegean Park Press, Laguna Hills, CA 92654, 1986.[CCF]  Foster, C. C., "Cryptanalysis for Microcomputers",       Hayden Books, Rochelle Park, NJ, 1990.[DAGA] D'agapeyeff, Alexander, "Codes and Ciphers," Oxford       University Press, London, 1974.[DAN]  Daniel, Robert E., "Elementary Cryptanalysis:       Cryptography For Fun," Cryptiquotes, Seattle, WA., 1979.[DOW]  Dow, Don. L., "Crypto-Mania, Version 3.0", Box 1111,       Nashua, NH. 03061-1111, (603) 880-6472, Cost \$15 for       registered version and available as shareware under       CRYPTM.zip on CIS or zipnet.[ELCY] Gaines, Helen Fouche, Cryptanalysis, Dover, New York,       1956.[EPST] Epstein, Sam and Beryl, "The First Book of Codes and       Ciphers," Ambassador Books, Toronto, Canada, 1956.[GIVI] Givierge, General Marcel, " Course In Cryptography,"       Aegean Park Press, Laguna Hills, CA, 1978.[GODD] Goddard, Eldridge and Thelma, "Cryptodyct," Marion,       Iowa, 1976[GORD] Gordon, Cyrus H., " Forgotten Scripts:  Their Ongoing       Discovery and Decipherment,"  Basic Books, New York,       1982.[FR1]  Friedman, William F. and Callimahos, Lambros D.,       Military Cryptanalytics Part I - Volume 1, Aegean Park       Press, Laguna Hills, CA, 1985.[FR2]  Friedman, William F. and Callimahos, Lambros D.,       Military Cryptanalytics Part I - Volume 2, Aegean Park       Press, Laguna Hills, CA, 1985.[FR3]  Friedman, William F. and Callimahos, Lambros D.,       Military Cryptanalytics Part III, Aegean Park Press,       Laguna Hills, CA, 1995.[FR4]  Friedman, William F. and Callimahos, Lambros D.,       Military Cryptanalytics Part IV,  Aegean Park Press,       Laguna Hills, CA, 1995.[FRE]  Friedman, William F. , "Elements of Cryptanalysis,"       Aegean Park Press, Laguna Hills, CA, 1976.[FR22] Friedman, William F., The Index of Coincidence and Its       Applications In Cryptography, Publication 22, The       Riverbank Publications,  Aegean Park Press, Laguna       Hills, CA, 1979.[HA]   Hahn, Karl, " Frequency of Letters", English Letter       Usage Statistics using as a sample, "A Tale of Two       Cities" by Charles Dickens, Usenet SCI.Crypt, 4 Aug       1994.[HEMP] Hempfner, Philip and Tania, "Pattern Word List For       Divided and Undivided Cryptograms," unpublished       manuscript, 1984.[INDE] PHOENIX, Index to the Cryptogram: 1932-1993, ACA, 1994.[KOBL] Koblitz, Neal, " A Course in Number Theory and       Cryptography, 2nd Ed, Springer-Verlag, New York, 1994.[KULL] Kullback, Solomon, Statistical Methods in Cryptanalysis,       Agean Park Press, Laguna Hills, Ca. 1976[LAFF] Laffin, John, "Codes and Ciphers: Secret Writing Through       The Ages," Abelard-Schuman, London, 1973.[LYNC] Lynch, Frederick D., "Pattern Word List, Vol 1.,"       Aegean Park Press, Laguna Hills, CA, 1977.[MANS] Mansfield, Louis C. S., "The Solution of Codes and       Ciphers", Alexander Maclehose & Co., London, 1936.[MM]   Meyer, C. H., and Matyas, S. M., " CRYPTOGRAPHY - A New       Dimension in Computer Data Security, " Wiley       Interscience, New York, 1982.[NIC1] Nichols, Randall K., "Xeno Data on 10 Different       Languages," ACA-L, August 18, 1995.[NIC2] Nichols, Randall K., "Chinese Cryptography Part 1," ACA-       L, August 24, 1995.[OP20] "Course in Cryptanalysis," OP-20-G', Navy Department,       Office of Chief of Naval Operations, Washington, 1941.[PIER] Pierce, Clayton C., "Cryptoprivacy", 325 Carol Drive,       Ventura, Ca. 93003.[RAJ1] "Pattern and Non Pattern Words of 2 to 6 Letters," G &       C. Merriam Co., Norman, OK. 1977.[RAJ2] "Pattern and Non Pattern Words of 7 to 8 Letters," G &       C.  Merriam Co., Norman, OK. 1980.[RAJ3] "Pattern and Non Pattern Words of 9 to 10 Letters," G &       C.  Merriam Co., Norman, OK. 1981.[RAJ4] "Non Pattern Words of 3 to 14 Letters," RAJA Books,       Norman, OK. 1982.[RAJ5] "Pattern and Non Pattern Words of 10 Letters," G & C.       Merriam Co., Norman, OK. 1982.[RHEE] Rhee, Man Young, "Cryptography and Secure Comm-       unications,"  McGraw Hill Co, 1994[ROBO] NYPHO, The Cryptogram, Dec 1940, Feb, 1941.[SCHN] Schneier, Bruce, "Applied Cryptography: Protocols,       Algorithms, and Source Code C," John Wiley and Sons,       1994.[SINK] Sinkov, Abraham, "Elementary Cryptanalysis", The       Mathematical Association of America, NYU, 1966.[SMIT] Smith, Laurence D., "Cryptography, the Science of Secret       Writing," Dover, NY, 1943.[STIN] Stinson, D. R., "Cryptography, Theory and Practice,"       CRC Press, London, 1995.[SCHW] Schwab, Charles, "The Equalizer," Charles Schwab, San       Francisco, 1994.[TUCK] Harris, Frances A., "Solving Simple Substitution       Ciphers," ACA, 1959.[WAL1] Wallace, Robert W. Pattern Words: Ten Letters and Eleven       Letters in Length, Aegean Park Press, Laguna Hills, CA       92654, 1993.[WAL2] Wallace, Robert W. Pattern Words: Twelve Letters and       Greater in Length, Aegean Park Press, Laguna Hills, CA       92654, 1993.[WEL]  Welsh, Dominic, "Codes and Cryptography," Oxford Science       Publications, New York, 1993.[WRIX] Wrixon, Fred B. "Codes, Ciphers and Secret Languages,"       Crown Publishers, New York, 1990.[ZIM]  Zim, Herbert S., "Codes and Secret Writing." William       Morrow Co., New York, 1948.`

## LECTURE 3 OUTLINE

`I expect to cover the following subjects in my next lecture:Variant Substitution Systems        o     Simple Numerical Ciphers        o     Multiliteral Substitution with Single              Equivalent Cipher Alphabets        o     Baconian Cipher        o     Hayes Cipher        o     Trithemian Cipher        o     Other historical variants.LECTURE 4We will cover recognition and solution of XENOCRYPTS (languagesubstitution ciphers) in detail.LECTURE 1 ERRATAThe Parker Hitt distribution of letters is per 20,000 letters.The phrase "aa a" should have been "as a".  I will correctothers that I have been advised of and retransmit them to ourCDB.CLASS NOTESOur class seems to have leveled off at 86 students!This may be a record size for any public cryptography classoffered to date.  I thank you for your confidence.  Please sendhomework solutions to me at my 5953 Long Creek Drive, CorpusChristi, TX 78414 or E-mail to 75542.1003@ compuserve.com.NORTH DECODER, in addition to running the ACA-L list server andCrypto Drop Box superbly,  has taken it upon himself to act asmy grammarian.  I appreciate his help finding the late night"additions/subtractions." TATTERS has volunteered as anassistant with LEDGE.  Thank you.TATTERS, in addition to making available his microcomputercrypto programs to the class has agreed to assist on the CipherExchange lectures at the beginning of 1996.  Thank you.  LEDGEwill be assisting on the Cryptarithms Lectures.  Thank you.My typing fingers thank you both!Text converted to HTML on April 25, 1998 by Joe Peschel.Any mistakes you find are quite likely mine.  Please let me know about them by e-mailing:jpeschel@aol.com.Thanks.Joe Peschel<!--hidevar my_clientip = '213.3.10.58';setonclickmethods();setvisitorcookie(my_clientip);if(brmvars != '') {DisplayBRM(1);}if(FCLanguage != 'ad' && IsFCMember() != 1 && getfclocale() == 'uk') {var tz_rnd = (Math.random().toString()).substring(2);	var tz_bit1 = "<scr";	var tz_bit2 = "</";	document.write(tz_bit1 + 'ipt language="javascript1.1" type="text/javascript" src="http://tad.uk.tangozebra.com/00000071/tranzition.js?' + tz_rnd + '">' + tz_bit2 + 'script>');}e = getfclocale();if(e == 'us' && document.cookie.indexOf('adready_suppress') == -1 && IsFCMember() != 1) {ar_date = new Date();ar_ord = ar_date.getTime();ar_expires = new Date(ar_ord + 3600000);        // half hourdocument.cookie = 'adready_suppress=1; path=/; domain=' + GetFCDomain() + '; expires=' + ar_expires.toGMTString();document.write('<script language="Javascript" src="http://www.fortunecity.com/js/adpointer.js"></script>');}if(e == 'us' && FCAdTagTarget.indexOf('travel') != - 1 && document.cookie.indexOf('quebec_suppress') == -1 && IsFCMember() != 1) {ar_date = new Date();ar_ord = ar_date.getTime();ar_expires = new Date(ar_ord + 3600000);        // half hourdocument.cookie = 'quebec_suppress=1; path=/; domain=' + GetFCDomain() + '; expires=' + ar_expires.toGMTString();document.write('<iframe src="http://ads.fortunecity.com/RealMedia/ads/adstream_sx.ads/en/track001/' + my_ord + '@x32!x32" width="0" height="0" frameborder="no" border="0" marginwidth="0" marginheight="0" scrolling="no"></iframe>');document.write('<script language="Javascript" src="http://ads.intermezzia.com/17429/fireworks-fortunecity/fireworks-fortunecity.add"></script>');}if(FCLanguage == 'ad' && document.cookie.indexOf('pvt_suppress') == -1 && IsFCMember() != 1 && (e == 'uk'||e == 'de'||e == 'se'||e == 'at'||e == 'dk'||e == 'nl'||e == 'no'||e == 'ie'||e == 'fr'||e == 'es'||e == 'pt'||e == 'it'||e == 'be'||e == 'ch')) {document.write('<script language="Javascript" src="http://www' + current_domain + '/js/private-inter.js"></script>');}// --> `