This cipher is an extension of the Null Sequence cipher introduced by SCORPIUS (JA2012) to a transposition type of cipher. It was inspired by conversations between LIONEL and MSCREP concerning the sequence technique. The Sequence Transposition Cipher (STC) was introduced in the ND2015 issue of the Cryptogram (Cm).

Here is a simple example.

Plain Text: The early bird gets the worm.

Let us encipher this with the GROMARK Type Enciphering Sequence: 69315

First, count out the number of letters (N) in the plain text (In this case N = 23). Then write out the enciphering sequence to a length equal to the length of the plain text. Assign a single sequence digit to each plain text letter in order.

For example:

PT: T H E E A R L Y B I R D G E T S T H E W O R M

SS: 6 9 3 1 5 5 2 4 6 0 7 6 0 6 7 3 6 6 3 0 9 2 9

Now set up columns for the digits 0-9. These can be in any order (e.g., based on a 10 letter keyword or phrase). In this case we will use the key phrase ‘Gummy Bears’. Place the plain text letters in the columns based on there sequence numbers. For example:

G U M M Y B E A R S

4 9 5 6 0 2 3 1 7 8

Y H A T I L E E R -

O R B G R S T

M D W E

E

T

H

Now draw off the cipher text by taking off the columns in order. In this case:

CT: Y HOM AR TBDETH IGW LR ESE E RT

SS: 4 999 55 666666 000 22 333 1 77

In blocks of 5 with the GROMARK type sequence, this provides:

CT: 69315 YHOMA RTBDE THIGW LRESE ERT 9

Decipherment of a message with a known sequence and column order is the reverse of encipherment.

The use of a sequence and the columnar keyword has done nothing more than provided a simple and reversible method of scrambling the plain text.

This cipher does not require the use of a GROMARK type sequence. Other methods of providing a sequence (e.g., Pi or the SQR (2)) will work just as well. The GROMARK sequence is recommended since the Krewe is already familiar with it.

Since providing the enciphering sequence (as with the GROMARK) is a necessity for this type of CON, decrypting an STC will essentially require the cryptanalyst to recover the column order from the cipher text.

Consider the following example CON:

37639 EEOUE ADHRC AAAES OAOAN THYYT ARRLM NNOEI IEAML NTQOS IHUSE

NDBHI CTPBW TRTTT CTWPE CCANH NLIUY RL 3. (Abraham Lincoln)

(GROMARK type sequence)

Note that the crib is ‘Abraham Lincoln’. The CON uses a GROMARK type sequence with 37639 as the starting seed.

First, note that the text is 82 characters long. Let us first lay out the complete sequence. This provides:

37639 03929 32112 53237 85505 30558 35031 85349 38732 15053 65589 10370 13071 43785 70532 75859 23

A frequency chart of the sequence numbers provides:

1:7 2:7 3:18 4:2 5:16 6:2 7:8 8:7 9:6 0:9

If we knew the order of the columns in the cipher text, we could use the above information to decipher the CON. Unfortunately, we don’t know the order. That is what we must uncover by cryptanalysis.

Consider the crib ‘Abraham Lincoln’. Notice the low probability letter ‘b’. If we are lucky, that ‘b’ will be followed in the cipher text by other letters in the crib (i.e., the piece of the sequence that enciphers the crib may have used the sequence number that enciphered the ‘b’ multiple times).

Note that b occurs only at position 53 (BHI CTPBW) and position 59 (BW TRTTT).

Only position 53 provides a sequence that is consistent with the crib. We have BHIC. Now match up these letters with their position in the crib:

Pos: 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Letter: A B R A H A M L I N C O L N

To place the crib, we need to find a 14 digit long portion of the enciphering sequence that has the 2nd, 5th, 9th, and 11th numbers identical.

A quick search reveals that position 74 provides the needed match. This provides:

Seq: 43785 70532 75859 23 PT: ***ab raham linco ln

This indicates that the sequence BHIC belongs the column 5. If we substitute what we now know into the sequence we get:

Seq: 37639 03929 32112 53237 85505 30558 PT: ***** ***** ***** ***** ***** ***** Seq: 35031 85349 38732 15053 65589 10370 PT: ***** ***** ***** ***** ***** ***** Seq: 13071 43785 70532 75859 23 PT: ***** ***ab raham linco ln

We now know that the plain text ends in “Abraham Lincoln”. This suggests the message may be a quote from that individual.

Now we already know that column 5 must have 16 characters. So the 16 characters in the cipher text ending with BHIC must belong to column 5.

This indicates that:

Col: ***** ***** ***** ***** ***** ***** ***** ***** 55555 55555 CT : EEOUE ADHRC AAAES OAOAN THYYT ARRLM NNOEI IEAML NTQOS IHUSE COL: 55555 5**** ***** ***** ***** ***** ** CT : NDBHI CTPBW TRTTT CTWPE CCANH NLIUY RL

Substituting Col 5 and the crib into the sequence provides:

Seq: 37639 03929 32112 53237 85505 30558 PT: ***** ***** ***** n**** *tq*o **si* Seq: 35031 85349 38732 15053 65589 10370 PT: *h*** *u*** ***** *s*e* *nd** ***** Seq: 13071 43785 70532 75859 23 PT: ***** ***ab raham linco ln

Examining the placement of the crib again, we notice column 2 must end with the letters ML. This digraph only appears at position 39 in the cipher text (…IEAML). By our frequency count, we know that column 5 must have 7 letters. So the 7 characters in the cipher text ending with ML must belong to column 2.

This indicates that:

Col: ***** ***** ***** ***** ***** ***** ***22 22222 55555 55555 CT : EEOUE ADHRC AAAES OAOAN THYYT ARRLM NNOEI IEAML NTQOS IHUSE COL: 55555 5**** ***** ***** ***** ***** ** CT : NDBHI CTPBW TRTTT CTWPE CCANH NLIUY RL

Substituting Col 2 into the sequence provides:

Seq: 37639 03929 32112 53237 85505 30558 PT: ***** ***e* *i**i n*e** *tq*o **si* Seq: 35031 85349 38732 15053 65589 10370 PT: *h*** *u*** ****a *s*e* *nd** ***** Seq: 13071 43785 70532 75859 23 PT: ***** ***ab raham linco ln

Examining the placement of the crib again, we notice column 7 must end with the letters RL. This digraph only appears at positions 28 (THYYT ARRL) and 81 (NLIUY RL) in the cipher text. Position 28 can be ruled out since it would leave a 4 letter gap between columns 7 and 2 and there are no 4 letter columns in the frequency analysis. By our frequency count, we know that column 7 must have 8 letters. So the last 8 characters in the cipher text must belong to column 7.

This indicates that:

Col: ***** ***** ***** ***** ***** ***** ***22 22222 55555 55555 CT : EEOUE ADHRC AAAES OAOAN THYYT ARRLM NNOEI IEAML NTQOS IHUSE COL: 55555 5**** ***** ***** ****7 77777 77 CT : NDBHI CTPBW TRTTT CTWPE CCANH NLIUY RL

Substituting Col 7 into the sequence provides:

Seq: 37639 03929 32112 53237 85505 30558 PT: *h*** ***e* *i**i n*e*n *tq*o **si* Seq: 35031 85349 38732 15053 65589 10370 PT: *h*** *u*** **l*a *s*e* *nd** ***i* Seq: 13071 43785 70532 75859 23 PT: ***u* **yab raham linco ln

The final piece of information from the crib indicates that column 3 must end in AN. While the digraph AN appears twice in the cipher text, its natural position would seem to be at position 73. Here the termination of column 3 would meet the start of column 7. By our frequency count, we know that column 3 must have 18 letters.

This indicates that:

Col: ***** ***** ***** ***** ***** ***** ***22 22222 55555 55555 CT : EEOUE ADHRC AAAES OAOAN THYYT ARRLM NNOEI IEAML NTQOS IHUSE COL: 55555 53333 33333 33333 33337 77777 77 CT : NDBHI CTPBW TRTTT CTWPE CCANH NLIUY RL

Substituting Col 3 into the sequence provides:

Seq: 37639 03929 32112 53237 85505 30558 PT: th*p* *b*e* wi**i ntern *tq*o t*si* Seq: 35031 85349 38732 15053 65589 10370 PT: th*t* *uc** t*lpa *s*ep *nd** **ei* Seq: 13071 43785 70532 75859 23 PT: *c*u* *cyab raham linco ln

Now note that the bit of text at the end of line 1 and the beginning of line two suggests the words “quotes is that…” Note that for this to be correct, column 0 must have the sequence UEA. This sequence, in fact, is located at position 3 in the cipher text. By our frequency count, we know that column 0 must have 9 letters.

This indicates that:

Col: **000 00000 0**** ***** ***** ***** ***22 22222 55555 55555 CT : EEOUE ADHRC AAAES OAOAN THYYT ARRLM NNOEI IEAML NTQOS IHUSE COL: 55555 53333 33333 33333 33337 77777 77 CT : NDBHI CTPBW TRTTT CTWPE CCANH NLIUY RL

Substituting Col 0 into the sequence provides:

Seq: 37639 03929 32112 53237 85505 30558 PT: th*p* ob*e* wi**i ntern *tquo tesi* Seq: 35031 85349 38732 15053 65589 10370 PT: that* *uc** t*lwa *sdep *nd** *heir Seq: 13071 43785 70532 75859 23 PT: *ccu* *cyab raham linco ln

Note that the two initial letters of the cipher text (EE) must belong to either column 4 or column 6 based on the frequency count. Since the first word in the plaintext appears to be ‘the’, this suggests that column 6 is the initial column.

This indicates that:

Col: 66000 00000 0**** ***** ***** ***** ***22 22222 55555 55555 CT : EEOUE ADHRC AAAES OAOAN THYYT ARRLM NNOEI IEAML NTQOS IHUSE COL: 55555 53333 33333 33333 33337 77777 77 CT : NDBHI CTPBW TRTTT CTWPE CCANH NLIUY RL

Substituting Col 6 into the sequence provides:

Seq: 37639 03929 32112 53237 85505 30558 PT: thep* ob*e* wi**i ntern *tquo tesi* Seq: 35031 85349 38732 15053 65589 10370 PT: that* *uc** t*lwa *sdep end** *heir Seq: 13071 43785 70532 75859 23 PT: *ccu* *cyab raham linco ln

This leaves 4 columns (1, 4, 8, and 9) to locate in the cipher text. However, there is plentiful partial plaintext to aide in their location. The remainder of this solution is left as an exercise for the student.

]]>*p r ou d ly p ro u dl y pr o ud l y*

As you have already noted, 2 PRs and a UD exist in the cipher text. This points to the third possibility. With two possible PR segments, we can produce two possible candidates for partial plain text.

Candidate No. 1 (First PR):

D ** N TI * FI O ** E PR * UD T ** M ON * UR P ** N

Candidate No. 2 (Second PR):

R ** N TO * FI N ** E PR * UD L ** M EY * UR D ** N

(N.B., The asterisks represent the missing letters in the middle column)

There is nothing to really distinguish the two candidates (i.e., no obviously wrong letter combinations). However, the second set has the interesting combination ‘EY UR’. This suggests the word ‘YOUR’. The candidate word ‘proud’ (Which also requires an ‘o’) is only two rows above. This suggests that somewhere in the cipher text there must be a sequence that is of the form ‘o**o’.

There is in fact just one combination like this, at the very beginning of the the cipher text (OAIO in the second group). Lets plug the letters surrounding this segment into the blanks in No. 2 and see if it makes any sense.

We get:

** E ** R LI N TO F FI N IN E PR O UD L AI M EY O UR D SE N

Note two thing. First, there are lots of promising words or fragments here which suggest we have made a good choice (e.g., nine, off, and your). There are also lots of fragments to build on (e.g., LAIM suggest claim). Second, because the double ‘o’ sequence was so close to the front of the cipher, we have identified the top of the columns. We can extend these columns downward to try to find the column lengths.

Substituting and lowering the columns we get:

OP E NH R LI N TO F FI N IN E PR O UD L AI M EY O UR D SE N NA L LD

Note that the LD in the last column are the last letters in the cipher text. This MUST be the end of a column! Therefore, we have identified the beginnings of three columns and the end of at least one. Now all we need to do is try columns from the remaining, unused letters, on either side of our start to extend the partial words we already have.

I’ll leave it to you at this point. If you need some hints, drop me an E-Mail.

]]>The Portax cipher is a periodic, digraph based cipher similar to the Slidefair. Like the Slidefair, it is based on one of the classic periodic ciphers (The Porta in place of the Vigenère, Variant or Beaufort used by Slidefair). The major difference between the two ciphers is the method of choosing the digraphs the cipher is applied too. In Slidefair the digraphs are taken sequentially from the plain text. In Portax, the plain text is laid out in pairs of rows of period length and the the digraphs are formed from the vertical columns. Why this is so will become apparent in a little while.

The example we are going to study is ND03 E-04:

QVNRE WLAEP MMTIA NOJJA LWWML MBBUA LUDAK CFXCC UMXCK ISAJJ FFTBP UQIWU RYKCH UGMTH WHEMW CLFTI XJVMA PUGDW NIMPA VOCJI JRJDN BQZEM HNKDL SZMEM WUGDI ZWADA KMBNK WOVPS VMHLM JRWJR AQIUO AUQBJ VWHLM J

At this point, a discussion of the Portax slide is important since (In my humble opinion) it is this ciphers greatest weakness. The slide is formed by two alphabets. Each alphabet is written in two sets of rows. The upper alphabet is the classic Porta slide. The lower alphabet is the the normal alphabet written alternately into two rows. The lower half of the slide (A1/2, A2/1, and A2/2) is allowed to move freely with respect to upper portion of the first alphabet (A1/1). The slide can be set in 1 of 13 positions. When set in the A,B position, the slide looks like this:

A1/1 ABCDEFGHIJKLM Split -------------------------- A1/2 NOPQRSTUVWXYZNOPQRSTUVWXYZ A2/1 ACEGIKMOQSUWYACEGIKMOQSUWY A2/2 BDFHJLNPRTVXZBDFHJLNPRTVXZ

If the slide were set in the C,D position it would look like this:

A1/1 ABCDEFGHIJKLM Split -------------------------- A1/2 NOPQRSTUVWXYZNOPQRSTUVWXYZ A2/1 ACEGIKMOQSUWYACEGIKMOQSUWY A2/2 BDFHJLNPRTVXZBDFHJLNPRTVXZ

Digraphs are enciphered as follows. The first letter of the digraph is located in alphabet A1 and the second letter of the digraph is found in alphabet A2. These are considered to form the diagonal of a rectangle. The other corners of the rectangle are found and then used as the cipher text. If the two letters of the plain digraph fall in the same column, the other two letters in that column are used to form the cipher text. In any case, the letter from A1 is used as the first letter of the cipher digraph and the letter from A2 is used as the second letter of the cipher digraph. For example,

In the A,B setting:

- Plain text ‘at’ is enciphered as ‘JB’
- Plain text ‘ta’ is enciphered as ‘NM’

In the C,D setting:

- Plain text ‘at’ is enciphered as ‘ID’
- Plain text ‘ta’ is enciphered as ‘NM’

Note that the encipherment of ‘ta’ was not changed by the setting of the slide. This is because the first letter of the digraph is located in the A1/2 portion of the A1 alphabet. This portion of alphabet A1 is fixed with respect to alphabet A2. In fact, all the plain text digraphs starting with letters n-z can be encoded in only one way (Unless both plain text letters fall in the same column). This accounts for the method used by Portax to pick digraphs for encipherment. If the encipherment were done in the same manner as Slidefair, almost half of the message could immediately be read!

The simplest way to determine the period of a Portax is just brute forcing checking of the different possibilities. Essentially, we can just set the cipher up for a candidate period, find the digraphs that start with N-Z, decipher these digraphs, and see if the resulting partial plain text looks reasonable. Our problem cipher is good example of this. The first four cipher letters are in the N-Z range and will produce a reasonably long string of plain text which will let us quickly determine if the period being tested is correct. For example:

Period Cipher Text Plain Text Notes 4 QVNR pysn Reject EWLA gqbi 5 QVNRE ysap* Reject WLAEP grbi* 6 QVNREW snpu*t Reject LAEPMM hqaj*s 7 QVNREWL nput*j* Reject AEPMMTI gqbi*s* 8 QVNREWLA putt*r**Accept!EPMMTIAN grai*s** 9 QVNREWLAE uttw*n*** Reject PMMTIANOJ hqaj*s***

Of the periods tested, only period = 8 produces a string of reasonable characters. Based on this, our initial assumption will be that the period is 8.

Before beginning the work necessary to find the key word, arrange the cipher into groups of period length. This will greatly assist the determination of which digraph belongs to which keyword letter. This provides:

Pos. 12345678 Key ******** (Note two rows for each key letter since each ******** alphabet has two letters associated with it)

CT1 QVNREWLA OJJALWWM DAKCFXCC JJFFTBPU HUGMTHWH XJVMAPUG CT2 EPMMTIAN LMBBUALU UMXCKISA QIWURYKC EMWCLFTI DWNIMPAV pt1 putt*r** s****ns* *****r** ****v*so ****s*j* o*t**un* pt2 grai*s** d****st* *****u** ****n*eo ****n*s* v*r**fo* CT1 OCJIJRJD KDLSZMEM DAKMBNKW MJRWJRAQ VWH CT2 NBQZEMHN WUGDIZWA OVPSVMHL IUOAUQBJ LMJ pt1 t****t** ***or*** *****t*s **un*v*r st* pt2 d****i** ***ly*** *****a*t **is*i*h rs*

Looking at what our guess of period 8 has produced, we notice that none of the plain text strings seems unreasonable. This is confirmation that we are now on the correct path.

Now look through the cipher and see if you can guess at some of the unrecovered plain text. This will allow us to begin recovering the key word. One obvious candidate is the string ‘grai*s’ in the first block. This looks a lot like the word ‘grains’. Therefore, at period position 5, the digraph ‘ET’ must decipher as ‘*n’. A quick look at the cipher slide shows this is possible only if the alphabet at position 5 is E,F. This substitution provides:

Pos. 12345678 Key ****E*** (Note two rows for each key letter since each ****F*** alphabet has two letters associated with it)

CT1 QVNREWLA OJJALWWM DAKCFXCC JJFFTBPU HUGMTHWH XJVMAPUG CT2 EPMMTIAN LMBBUALU UMXCKISA QIWURYKC EMWCLFTI DWNIMPAV pt1 putthr** s***ins* ****dr** ****v*so ****s*j* o*t*eun* pt2 grains** d***ast* ****ou** ****n*eo ****n*s* v*r*efo* CT1 OCJIJRJD KDLSZMEM DAKMBNKW MJRWJRAQ VMH CT2 NBQZEMHN WUGDIZWA OVPSVMHL IUOAUQBJ LMJ pt1 t***at** ***or*** ****it*s **univ*r st* pt2 d***wi** ***ly*** ****ha*t **iswi*h rs*

We now have at least three more possible fill-ins. These are:

- The partial text ‘putthr**grains’ in the block 1 suggests the word ‘three’. The first ‘e’ will require that for period position 7 the digraph ‘LA’ decipher as ‘e*’. This will require the alphabet at position 7 to be S,T.
- The second ‘e’ in ‘three’ will require that for period position 8 the digraph ‘AN’ also decipher as ‘e*’. This requires alphabet 8 to be E,F.
- The partial text ‘t*e’ in block 6 suggests the word ‘the’. This will require that for period position 4 the digraph ‘MI’ decipher as ‘h*’. This requires alphabet 4 to be U,V.

These substitutions provides:

Pos. 12345678 Key ***UE*SE (Note two rows for each key letter since each ***VF*TF alphabet has two letters associated with it)

CT1 QVNREWLA OJJALWWM DAKCFXCC JJFFTBPU HUGMTHWH XJVMAPUG CT2 EPMMTIAN LMBBUALU UMXCKISA QIWURYKC EMWCLFTI DWNIMPAV pt1 putthree s**dins* ***edr** ***av*so ***es*ac o*theuni pt2 grainsof d**vast* ***you** ***en*eo ***sn*ss v*rsefor CT1 OCJIJRJD KDLSZMEM DAKMBNKW MJRWJRAQ VMH CT2 NBQZEMHN WUGDIZWA OVPSVMHL IUOAUQBJ LMJ pt1 t**cathe ***or*cl ***withs **univer st* pt2 d**lwill ***ly*ac ***thant **iswith rs*

At this point, you should be able to finish this decipherment without further help.

]]>The Ragbaby Cipher uses a 24 letter alphabet as both the plain text and the cipher text. The letters I/J and W/X are combined.

The cipher alphabet is keyed to a keyword. That word provides the initial letters in the cipher alphabet with the rest of the alphabet filling out the remaining letters in the usual fashion. For example:

Key Word: ROBIN KA: ROBINACDEFGHKLMPQSTUVWYZ Pos: 000000000011111111112222 012345678901234567890123

The cipher requires that word divisions be maintained because the encipherment process for each word depends on it’s position in the message. The first letter in each word is enciphered by locating the plaintext letter in the cipher alphabet and then shifting to the right a number of places equal to the words position in the text. The remaining letters in the word are enciphered by the same process. However, the shift is increased by one place for each additional letter. For example:

PT: The early bird gets the worm.

The message is encoded as follows:

PT: t h e e a r l y b i r d g e t s... Shift: 1 2 3 2 3 4 5 6 3 4 5 6 4 5 6 7 CT: U L H G E N T N A D A L M L R R

By keeping track of the letter positions, enciphering and deciphering can be done by simple addition and subtraction. The formula is given by:

Ct = Pt + Shift

For example, enciphering the ‘t’ (The 18th letter in the alphabet) in the first word;

Ct = 18 + 1 = 19 = 'U'

Deciphering the first A (The 5th letter in the alphabet) in the third word:

Pt = Ct - Shift = 5 - 3 = 2 = 'B'

Two important thing to remember about this system:

- The entire alphabet can be shifted as a whole either to the right or left without effecting the cipher text of a message.
- Pattern words do not maintain their patterns when enciphered. However, the cipher text for any word will have the same pattern no matter what shift is used to encipher them.

The solution of this type of cipher is normally started by using the method of the probable word (In ‘The Cryptogram’, a probable word is usually supplied with the cipher). Consider the following example:

QKAOAHAEV YHGOWCP LOEQ ZQQ GSKPN OCCFUE FSTVZU FH NZZ WMUZQUNFBI UGLLQNY HOZA GK SNN *CUUHRN *LENKDB.

Before using the provided crib, first notice that there are 3 words of identical length and pattern in the cryptogram. These are:

Ct: Z Q Q Ct: N Z Z Ct: S N N Shift: 4 5 6 Shift: 9 10 11 Shift: 14 15 16

Since the pattern of these words is the same, it is tempting to say that they are 3 examples of the same word. Since the most common 3 letter word is ‘the’, lets start with this as our first guess.

Initially we start with a cipher alphabet with no know letters:

Pos: 000000000011111111112222 012345678901234567890123 Ct: ************************ Pos: 222222333333333344444444 456789012345678901234567

(Notice that I have two values for the position of each letter shown. This is for ease of use during enciphering or deciphering. When enciphering, take the plain text letters value from the top position and add the shift. If the resulting value is greater than 23, find the value of the cipher text from the lower position table. When deciphering, use the values in the lower table and subtract the shift…)

Now, let’s start with the first candidate word, ‘ZQQ’. To seed the cipher alphabet, lets assume a position of the letter ‘E’ is 0 (Zero).

This provides:

Ct = Pt + shift Ct = 0 + 6 = 'Q'.

Having determined the position of ‘Q’ = 6, we can decipher it in the same word to find the position of the letter ‘H’ since it is enciphered as the letter ‘Q’.

This provides: Ct = 6 = Pt + shift = Pt + 5. Or Pt = 1 = ‘H’.

Using the same method, we can use the other two words to determine that ‘Z’ = 11 and ‘N’ =16.

We now have:

Pos: 000000000011111111112222 012345678901234567890123 Ct: EH****Q****Z****N******* Pos: 222222333333333344444444 456789012345678901234567

Now we can work on the enciphered ‘T’ in each of these words. Since we already know the value of ‘Z’ = 11, we can use ZQQ to determine the value of ‘T’. This gives:

Ct = 11 = Pt + shift = Pt + 4 or Pt = 11 - 4 = 7 = 'T'.

Note that if we had used the known value of N and the word NZZ we would have gotten:

Ct = 16 = Pt + shift = Pt + 9 or Pt = 16 - 9 = 7 = 'T'.

This is very good! We have obtained the same value for ‘T’ in two different ways. Our assumption that the repeated pattern word is ‘the’ is looking very good!

We can now recover the value of ‘S’ from SSN and this provides:

Ct = Pt + shift = 7 + 14 = 21 = 'S'.

We now have:

Pos: 000000000011111111112222 012345678901234567890123 Ct: EH****QT***Z****N****S** Pos: 222222333333333344444444 456789012345678901234567

Some notes on what we have so far. E and H are adjacent in the cipher alphabet but are separated by only 2 letters in the normal alphabet. The same is true of Q and T. T and Z are separated by 3 letters in the cipher alphabet and by 4 letters (Remember W/X is a single letter) in the normal alphabet. This suggests that the letters from position 0 to position 11 are not part of the key word. This implies that F and G (Between E and H) as well as R and S (Between Q and T) are in the key word. Note that this is confirmed by the position of S in the cipher alphabet. Further note that one of the letters U, V, W/Y or Y (Between T and Z) is also in the keyword. Locating one of these letters within the keyword will locate the rest in their proper places in the gap between T and Z.

By this time, you must be getting the idea of how to solve this cipher type. Now find the position of the crib and continue the process. You will be finished before you know it!

]]>The Baconian cipher is one of the earliest examples of what is now called a ‘fractionated’ cipher. Essentially, individual letters are enciphered as 5 character groups. Each letter is represented by a unique 5 character group of a/b characters (Note that this is essentially a binary system with each letter represented by a binary number). The ACA uses the following Baconian table:

A | aaaaa | E | aabaa | I/J | abaaa | N | abbaa | R | baaaa | W | babaa | |||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|

B | aaaab | F | aabab | K | abaab | O | abbab | S | baaab | X | babab | |||||

C | aaaba | G | aabba | L | ababa | P | abbba | T | baaba | Y | babba | |||||

D | aaabb | H | aabbb | M | ababb | Q | abbbb | U/V | baabb | Z | babbb |

PT: H E L P CT: aabbb aabaa ababa abbba

The a/b units can be further concealed by assigning an a/b value to each letter of the alphabet. For example, consider the following cipher ‘alphabet’ and the above message:

P: ABCDEFGHIJKLMNOPQRSTUVWXYZ V: aabbbbabbbbbaabbaaaaababab

PT: H E L P a/b: aabbb aabaa ababa abbba CT: QUICK BROWN TIGER SHEDS

In this case, a double encipherment has been used to hide the simple message ‘help’. The cipher text will read ‘QUICK BROWN TIGER SHEDS’. By replacing the cipher text letters with their a/b equivalents, the underlaying letters can be determined and the plain text recovered.

Note that a Baconian can be used in many other ways. For instance, only the first letters of each word may represent an a/b. Symbols or pictures can also be used. For this reason, Baconian ciphers can be popular for use with ornamental ciphers. However, this cipher is a bit impractical for everyday use since the cipher text is at least 5 times larger than the plain text.

We will be using JA05 E-05 as an example. This is a five alphabet Baconian, but we will only work with the first encoded message. The cipher text is:

QNGBU FAARA DSHBB NARKT NBXUM BYFTI CKBMK NELFH NWWID HCIVI BKPBK BFLGA BWKRH VBBUA NEZSK CATCO IHRLK NMCHE CBVBK BHVJY NAQBB KRJAB NRWNH PKWOG BBDZR BHXIG OFHAM

We are given the crib ‘Said’.

Start with the crib. The Baconian system would encipher this as follows:

PT: S A I D a/b: baaab aabaa abaaa aaabb

This a/b pattern must be hidden in the cipher text. The job of the cryptanalyst is to use this pattern to begin to uncover the a/b values assigned to the cipher text letters. This is easier than it first appears. The primary tool used here is to compare the a/b pattern of the crib with the cipher text and look for contradictions. For example, if the crib were placed at position 2, the following would result:

PT S A I D CT FAARA DSHBB NARKT NBXUM a/b baaab aaaaa abaaa aaabb

A quick look at the repeated letters in the cipher text shows that the A’s in FAARA have been assigned both the values of ‘a’ and ‘b’. This is a contradiction since A can have only one value. Therefore, the crib can not be placed at position 2.

In fact, attempting to fit the crib in all possible positions in the cipher shows that it fits only position 1 provides no contradictions. Therefore, the crib must be placed at the start of the cipher. This provides:

PT S A I D CT QNGBU FAARA DSHBB NARKT a/b baaab aaaaa abaaa aaabb

P: ABCDEFGHIJKLMNOPQRSTUVWXYZ V: aa*a*aaa**b**a**babbb*****

s a i d * * * * * * * QNGBU FAARA DSHBB NARKT NBXUM BYFTI CKBMK NELFH NWWID HCIVI BKPBK baaab aaaaa abaaa aaabb aa*b* a*ab* *ba*b a**aa a***a a**** ab*ab * * * * * * * * * e * BFLGA BWKRH VBBUA NEZSK CATCO IHRLK NMCHE CBVBK BHVJY NAQBB KRJAB aa*aa a*baa *aaba a**bb *ab** *aa*b a**a* *a*ab aa*** aabaa ba*aa * * * * * NRWNH PKWOG BBDZR BHXIG OFHAM aa*aa *b**a aaa*a aa**a *aaa*

Now note that CKBMK = *ba*b and that PKWOG = *b**a. Since no letter is encoded in the form bb***, we now can identify C = a and P = a. This provides:

P: ABCDEFGHIJKLMNOPQRSTUVWXYZ V: aaaa*aaa**b**a*ababbb*****

s a i d * * * * * * k QNGBU FAARA DSHBB NARKT NBXUM BYFTI CKBMK NELFH NWWID HCIVI BKPBK baaab aaaaa abaaa aaabb aa*b* a*ab* aba*b a**aa a***a aa*** abaab * * * * * * * * * e * BFLGA BWKRH VBBUA NEZSK CATCO IHRLK NMCHE CBVBK BHVJY NAQBB KRJAB aa*aa a*baa *aaba a**bb aaba* *aa*b a*aa* aa*ab aa*** aabaa ba*aa * * * * * NRWNH PKWOG BBDZR BHXIG OFHAM aa*aa ab**a aaa*a aa**a *aaa*

We now have to make some guesses to advance the analysis. Note the partial letters VBBUA = *aaba, and CBVBK = aa*ab. If V= a, these letters are ‘c’ and ‘b’ respectively. If V = b, these letters are ‘t’ and ‘f’ respectively. Since the second set of letters is more common than the first, we guess V = b. This provides:

P: ABCDEFGHIJKLMNOPQRSTUVWXYZ V: aaaa*aaa**b**a*ababbbb****

s a i d * * * * * * k QNGBU FAARA DSHBB NARKT NBXUM BYFTI CKBMK NELFH NWWID HCIVI BKPBK baaab aaaaa abaaa aaabb aa*b* a*ab* aba*b a**aa a***a aa*b* abaab * * t * * * * f * e * BFLGA BWKRH VBBUA NEZSK CATCO IHRLK NMCHE CBVBK BHVJY NAQBB KRJAB aa*aa a*baa baaba a**bb aab** *aa*b a*aa* aabab aab** aabaa ba*aa * * * * * NRWNH PKWOG BBDZR BHXIG OFHAM aa*aa ab**a aaa*a aa**a *aaa*

Now the fragment following the ‘t’ is very interesting. Since the most common word in English is ‘the’, it might be worth while trying to fit this word at this point. The cipher text now provides:

t * * VBBUA NEZSK CATCO baaba a**bb aaba*

To fit the word ‘the’ would require:

t h e VBBUA NEZSK CATCO baaba aabbb aabaa

Since this would produce no contradictions, we set E = a, Z = b, and O = a. This provides:

P: ABCDEFGHIJKLMNOPQRSTUVWXYZ V: aaaaaaaa**b**aaababbbb***b

s a i d * * * * * * k QNGBU FAARA DSHBB NARKT NBXUM BYFTI CKBMK NELFH NWWID HCIVI BKPBK baaab aaaaa abaaa aaabb aa*b* a*ab* aba*b aa*aa a***a aa*b* abaab * * t h e * * f * e * BFLGA BWKRH VBBUA NEZSK CATCO IHRLK NMCHE CBVBK BHVJY NAQBB KRJAB aa*aa a*baa baaba aabbb aabaa *aa*b a*aaa aabab aab** aabaa ba*aa * * c * * NRWNH PKWOG BBDZR BHXIG OFHAM aa*aa ab*aa aaaba aa**a aaaa*

Note the partial letter KRJAB = ba*aa. If J= a, this letters is ‘r’. If J = b, this letters is ‘w’. Since ‘r’ is a more common letter than ‘w’, we guess J = a. This provides:

P: ABCDEFGHIJKLMNOPQRSTUVWXYZ V: aaaaaaaa*ab**aaababbbb***b

s a i d * * * * * * k QNGBU FAARA DSHBB NARKT NBXUM BYFTI CKBMK NELFH NWWID HCIVI BKPBK baaab aaaaa abaaa aaabb aa*b* a*ab* aba*b aa*aa a***a aa*b* abaab * * t h e * * f * e r BFLGA BWKRH VBBUA NEZSK CATCO IHRLK NMCHE CBVBK BHVJY NAQBB KRJAB aa*aa a*baa baaba aabbb aabaa *aa*b a*aaa aabab aaba* aabaa baaaa * * c * * NRWNH PKWOG BBDZR BHXIG OFHAM aa*aa ab*aa aaaba aa**a aaaa*

Note the partial letter BHVJY = aaba*. If Y= a, this letters is ‘e’. If j = b, this letters is ‘f’. Since ‘ffer’ is a more common word fragment than ‘feer’, we guess Y = b. This provides:

P: ABCDEFGHIJKLMNOPQRSTUVWXYZ V: aaaaaaaa*ab**aaababbbb**bb

s a i d * * * * * * k QNGBU FAARA DSHBB NARKT NBXUM BYFTI CKBMK NELFH NWWID HCIVI BKPBK baaab aaaaa abaaa aaabb aa*b* abab* aba*b aa*aa a***a aa*b* abaab * * t h e * * f f e r BFLGA BWKRH VBBUA NEZSK CATCO IHRLK NMCHE CBVBK BHVJY NAQBB KRJAB aa*aa a*baa baaba aabbb aabaa *aa*b a*aaa aabab aabab aabaa baaaa * * c * * NRWNH PKWOG BBDZR BHXIG OFHAM aa*aa ab*aa aaaba aa**a aaaa*

Note the partial letters NRWNH = aa*aa, and PKWOG = ab*aa. If W = a, these letters are ‘a’ and ‘i/j’ respectively. If W = b, these letters are ‘e’ and ‘n’ respectively. Since the word fragment ‘fferenc’ is more common than the word fragment ‘fferaic’, we guess W = b. This provides:

P: ABCDEFGHIJKLMNOPQRSTUVWXYZ V: aaaaaaaa*ab**aaababbbbb*bb

s a i d * * * * * * k QNGBU FAARA DSHBB NARKT NBXUM BYFTI CKBMK NELFH NWWID HCIVI BKPBK baaab aaaaa abaaa aaabb aa*b* abab* aba*b aa*aa abb*a aa*b* abaab * n t h e * * f f e r BFLGA BWKRH VBBUA NEZSK CATCO IHRLK NMCHE CBVBK BHVJY NAQBB KRJAB aa*aa abbaa baaba aabbb aabaa *aa*b a*aaa aabab aabab aabaa baaaa e n c * * NRWNH PKWOG BBDZR BHXIG OFHAM aabaa abbaa aaaba aa**a aaaa*

Note the partial word ‘fference’ suggests ‘difference’. To match this, we set I=a, L=b, M=b, and X=b. This provides:

P: ABCDEFGHIJKLMNOPQRSTUVWXYZ V: aaaaaaaaaabbbaaababbbbbbbb

s a i d h l m e n c k QNGBU FAARA DSHBB NARKT NBXUM BYFTI CKBMK NELFH NWWID HCIVI BKPBK baaab aaaaa abaaa aaabb aabb* abab* ababb aabaa abbaa aaaba abaab e n t h e d i f f e r BFLGA BWKRH VBBUA NEZSK CATCO IHRLK NMCHE CBVBK BHVJY NAQBB KRJAB aabaa abbaa baaba aabbb aabaa aaabb a*aaa aabab aabab aabaa baaaa e n c e b NRWNH PKWOG BBDZR BHXIG OFHAM aabaa abbaa aaaba aabaa aaaab

The plain text therefore reads:

Said H.L. Mencken, the difference b

The remaining four parts of this message are decrypted in the same manner. For each part a separate crib is given. The five parts placed together should create one continuous message. The fragment ‘b’ at the end of message one suggests the word ‘between’ may be split by messages 1 and 2.

Good luck with the rest of the sections!

]]>What is the Bifid cipher? – The Bifid is an example of a periodic fractionated cipher. A fractionated cipher breaks up the letters of a plain text message into pieces and then recombines the pieces from different letters to form the cipher text.

Where did this cipher originate? – It was invented around 1901 by Felix Delastelle and was first described in the ON44 and JJ45 issues of the Cryptogram by S-TUCK and TONTO respectively. From reading these articles it is apparent that ACA members and ‘Friends Groups’ were well aware of the Bifid by that time.

What does the name mean? – The name means ‘two feet’. It is meant to describe the cipher system which fractionates each letter into two coordinates via the use of a Polybius Square.

The Bifid system performs it’s fractionation by writing the alphabet into a 5×5 keysquare (i.e., A typical Polybius Square). Each letter is assigned a unique set of coordinates based on its Row and Column position. In this system, a letter such as ‘T’ is represented by T = (T1, T2), were T1 and T2 are respectively the row and the column position of T.

The plain text is then broken up into period size group. The coordinates of each letter are then written in vertical columns below each letter. This creates a sets of numbers of period length by two deep.

The cipher text for each group is then created drawing out sets of two numbers horizontally from the group, starting at the left end of the top row and proceeding in a normal fashion through both rows. Each pair of numbers that is drawn off in this fashion is reconverted into a text character utilizing the key square. These letters are the cipher text.

Key Word: Example Plain Text: The end is near. Key Square: 1 2 3 4 5 1 E X A M P 2 L B C D F 3 G H I K N 4 O Q R S T 5 U V W Y Z Use Period = 3. THE END ISN EAR 431 132 343 114 521 154 345 133 (Note that first letter T =(4,5) is the first column.) Ct: RPL ALY KIT EOI (Note that first cipher text block RPL is taken horizontally out of the first block R = (4,3), P = (1,5) and L =(2,1).

Note that the ACA standard form requires the cipher text to remain in period size groups.

Now, let us assume that the above example is part of a larger message, and that we don’t know the key square. All we know (By some method) is the equivalence of a small portion of the plain text and the cipher text. For example, we know the following:

Pt: the end isn ear Ct: RPL ALY KIT EOI

Can we learn anything about the unknown key square from this material? The answer is yes. We can write out each block in its plain text and cipher text forms using the generalized coordinates (i.e., A = (A1, A2)) for each letter and relate the two sets of coordinates.

The fractionation block of the first text group in plain text/cipher text form are as follows:

t h e R P L T1 H1 E1 R1 R2 P1 T2 H2 E2 P2 L1 L2

Since both of these blocks are the fractionation of the same plain text, we can equate each letter coordinate in plain text group with the matching position in the cipher group. In other words: T1=R1, H1=R2, E1=P1, T2=P2, H2=L1, and E2=L2.

Note, we haven’t learn the value of T1 or R1 from this method. However, we have learned that T1 and R1 are equal (N.B., Examination of the keysquare confirms that T1 = R1 = 4).

In our example, there are 4 (groups) x 3 (period) x 2 (R/C) = 24 identities produced by the comparison of the plain text and cipher text.

Not all the information will be unique. Some of the equalities may be duplicates. Others may be simple identities (e.g., A1=A1).

In any case, if enough plain text and cipher text are equated, a great deal of information can be gained concerning the relationships of the letter coordinates and the key array may be partially or completely reconstructed.

This can best be seen by the example of the decryption of an actual Bifid.

The example we will be decrypting is from the log of a puzzle type Geocache in New England (GCX33R). The cipher is:

VCUHKKS OAVEVTP ECMWSAF ERRBKTP RAAOAKP VIRFHPA ANTVTPO

BAEDMKM XXIVDPD VLTEXQO AESTAOQ IEAUTIK ICVUZED RVXTSDK ste elcylin deris REUADFK RHDTRHT RCKNT

By the size of the individual text groups we know that the cipher period is 7. It should be noted that it is standard ACA practice to provide the period for Bifid or Trifid in this manner. This is necessary because the statistical methods necessary to determine a Bifid/Trifid period are extremely complicated and a bit more work than is normally possible for a ‘hand’ cipher.

We are also shown which cipher text group corresponds to the plain text phrase ‘steel cylinder is’. Being given the placement of the crib is a great advantage. The only real method available for crib placement in Bifid/Trifid ciphers is a brute force attempt to fit it at each location using the method shown below. To say the least, this can be a bit tiresome.

The latitude and longitude coordinates of an object (the ‘geocache’) are enciphered in this CON. Consequently, we should expect the plain text to primarily be long hand numbers.

The cache log states that the cache is within 0.5 miles of 43 degrees 15.979 minutes North and 070 degrees 1.872 minutes West. This point is within the State of Maine.

Most geocaches are hidden in stumps, under logs, in stonewalls, or under trees. Caches can contain a log book, ‘stash’, geocoins, and/or travelbugs.

Lastly, the cache owner cryptically remarks in the log: I hope this cache shows the wisdom of the old Klingon proverb which notes that, “A tribute cache is a dish best served cold.”

Before moving forward, lets quickly discuss the methods we will use to keep track of all the letter coordinate identities we will be dealing with. As we uncover new identities, (e.g., T1=U1), they will be assigned a numerical value (e.g., T1=U1=5). Although, in reality, only five values are possible (1-5) this device will allow us to track which coordinates have the same values.

Our eventual goal will be to properly assign a value to each of these identifiers. To do proper bookkeeping, I use two separate table. One is an array of the alphabet characters with their assigned identifiers. A second is a listing of letter coordinates by assigned identifier. A blank setup appears as follows:

A blank setup appears as follows:

P 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 1 - - - - - - - - - - - - - - - - - - - - - - - - - - 2 - - - - - - - - - - - - - - - - - - - - - - - - - - 1: 6: 2: 7: 3: 8: 4: 9: 5: 10:

For the sake of the brevity of this tutorial, I will initially update the above tables for every four identities we process. However, I recommend that when you first work with these you write a new set of tables every time you process an identity. It is easy to make errors with this type of work (Trust me on this!). Recording every move will both allow for easier error trapping and will remove the need to start from scratch should you find an error during your work.

Now, lets take the crib and write out it’s fractionation block:

s t e e l c y l i n d e r i s S1 T1 E1 E1 L1 C1 Y1 L1 I1 N1 D1 E1 R1 I1 S1 S2 T2 E2 E2 L2 C2 Y2 L2 I2 N2 D2 E2 R2 I2 S2

The fractionation of the three cipher text blocks containing the crib is:

* * * * s t e : e l c y l i n : d e r i s * * X X I V D P D : V L T E X Q O : A E S T A O Q X1 X2 X1 X2 I1 I2 V1 : V1 V2 L1 L2 T1 T2 E1 : A1 A2 E1 E2 S1 S2 T1 V2 D1 D2 P1 P2 D1 D2 : E2 X1 X2 Q1 Q2 O1 O2 : T2 A1 A2 O1 O2 Q1 Q2

Note, ‘:’ = Period Divisions

Lining up the first rows to clearly show the identities provides:

P: S1 T1 E1 E1 L1 C1 Y1 L1 I1 N1 D1 E1 R1 I1 S1 C: I1 I2 V1 V1 V2 L1 L2 T1 T2 E1 A1 A2 E1 E2 S1

Lining up the second rows shows the following identities:

P: S2 T2 E2 E2 L2 C2 Y2 L2 I2 N2 D2 E2 R2 I2 S2 C: P2 D1 D2 E2 X1 X2 Q1 Q2 O1 O2 T2 A1 A2 O1 O2

From the first rows first four identities we get:

P: S1 T1 E1 E1 L1 C1 Y1 L1 I1 N1 D1 E1 R1 I1 S1 C: I1 I2 V1 V1 V2 L1 L2 T1 T2 E1 A1 A2 E1 E2 S1

- S1 = I1: Since neither coordinate has an assigned value we set S1=I1=1.
- T1 = I2: As above, T1 = I2 = 2.
- E1 = V1: Again, E1 = V1 = 3.
- E1 = V1: Nothing new here. Boring so far!

This provides:

P 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 1 - - - - 3 - - - 1 - - - - - - - - - 1 2 - 3 - - - - 2 - - - - - - - - 2 - - - - - - - - - - - - - - - - - 1: S1, I1 6: 2: T1, I2 7: 3: E1, V1 8: 4: 9: 5: 10:

From the first rows second four identities we get:

P: S1 T1 E1 E1 L1 C1 Y1 L1 I1 N1 D1 E1 R1 I1 S1 C: I1 I2 V1 V1 V2 L1 L2 T1 T2 E1 A1 A2 E1 E2 S1

- L1 = V2: L1 = V2 = 4.
- C1 = L1: Now, L1 = 4 so C1 = 4.
- Y1 = L2: Again, Y1 = L2 = 5.
- L1 = T1: L1 = 4 and T1 = 2. So set L1 = V2 = C1 = 2.

This provides:

P 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 1 - - 2 - 3 - - - 1 - - 2 - - - - - - 1 2 - 3 - - 5 - 2 - - - - - - - - 2 - - 5 - - - - - - - - - 2 - - - - 1: S1, I1 6: 2: T1, I2, C1, L1, V2 7: 3: E1, V1 8: 5: L1, Y1 9: 10:

(Note I have removed identifier 4 from the above list. It could be reused. However, I find it makes for less confusion and easier error trapping should a mistake be made in updating the array if identifiers are not reused.)

From the third set of identities we get:

P: S1 T1 E1 E1 L1 C1 Y1 L1 I1 N1 D1 E1 R1 I1 S1 C: I1 I2 V1 V1 V2 L1 L2 T1 T2 E1 A1 A2 E1 E2 S1

- I1 = T2: Now, I1 = 1 so T2 = 1.
- N1 = E1: Now, E1 = 3 so N1 = 3.
- D1 = A1: Now, A1 = 6 so D1 = 6.
- E1 = A2: E1 = 3 so A2 = 3.

This provides:

P 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 1 6 - 2 6 3 - - - 1 - - 2 - 3 - - - - 1 2 - 3 - - 5 - 2 3 - - - - - - - 2 - - 5 - - - - - - - 1 - 2 - - - - 1: S1, I1 6: A1, D1 2: T1, I2, C1, L1, V2 7: 3: E1, V1, N1, A2 8: 5: L1, Y1 9: 10:

From the last set of identities from the first row we get:

P: S1 T1 E1 E1 L1 C1 Y1 L1 I1 N1 D1 E1 R1 I1 S1 C: I1 I2 V1 V1 V2 L1 L2 T1 T2 E1 A1 A2 E1 E2 S1

- R1=E1: Now, E1=3 so R1=3.
- I1=E2: Now, I1=1 so E2=1.
- S1=S1: An identity. So no new information.

This provides:

P 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 1 6 - 2 6 3 - - - 1 - - 2 - 3 - - - 3 1 2 - 3 - - 5 - 2 3 - - - 2 - - - 2 - - 5 - - - - - - - 1 - 2 - - - - 1: S1, I1, E2 6: A1, D1 2: T1, I2, C1, L1, V2 7: 3: E1, V1, N1, A2, R1 8: 5: L1, Y1 9: 10:

For the sake of brevity, I will skip the details for the second row of identifiers. Lining up the second rows provides the following identities:

P: S2 T2 E2 E2 L2 C2 Y2 L2 I2 N2 D2 E2 R2 I2 S2 C: P2 D1 D2 E2 X1 X2 Q1 Q2 O1 O2 T2 A1 A2 O1 O2

This provides:

P 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 1 1 - 2 1 3 - - - 1 - - 2 - 3 2 - 9 3 1 2 - 3 - 5 5 - 2 3 - 8 1 1 - - - 2 - - 5 - 7 7 7 5 3 7 1 - 2 - 8 9 – 1: A1, D1, D2, E2, I1, S1, T2 2: C1, I2, L1, O1, T1, V2 3: A2, E1, N1, R1, R2, V1 5: L2, Q2, X1, Y1 7: N2, O2, P2, S2 8: C2, X2 9: Q1, Y2

We have now extracted all the information about the keysquare we can from the crib. To proceed we must now enter several rounds of substitution and analysis.

- Form a key square with the information at hand. Do not attempt to maintain a 5×5 square, at least initially. Only combine rows and columns when the cryptanalysis indicates they are related.
- Substitute the partial key square into the cipher text to obtain a partial decryption.
- Analyze the plain text and look for partial words. Complete partial words to recover additional info on the key square with the methods already demonstrated.
- Update the key square and start the cycle again.

1 2 3 4 5 6 7 8 9 10 1 D I A * * * S * * * 2 T * * * L * O C * * 3 E V R * * * N * * * 4 * * * * * * * * * * 5 * * * * * * * X Y * 6 * * * * * * * * * * 7 * * * * * * * * * * 8 * * * * * * * * * * 9 * * * * Q * * * * * 10 * * * * * * P * * *

VCUHKKS OAVEVTP ECMWSAF ERRBKTP RAAOAKP VIRFHPA ANTVTPO BAEDMKM XXIVDPD ******* t*ive*n *do**** ***ve** nea***o ****ne* ive**in ******* ****ste VLTEXQO AESTAOQ IEAUTIK ICVUZED RVXTSDK REUADFK RHDTRHT RCKNT elcylin deris*l **edi** ****et* eent*** ree**** er***it *ro**

This all looks like English, which is very encouraging. However, note that with positions assign to over half the alphabet, only about a quarter of the plain text appears. How discouraging!

Note that the partial text “*ne* ive” at position 39 appears to be “onef ive”. comparing the plain text and cipher text at this position supplies:

V I R F H P A * * * o n e f V1 V2 I1 I2 R1 R2 F1 = * * * O1 N1 E1 F1 F2 H1 H2 P1 P2 A1 A2 * * * O2 N2 E2 F2

Now P1=O2=7 and P2=N2=7. So P=(7,7)

Also F1=F1, which is an identity. Further F2=A2= 3. So F=(*,3).

Note that the partial text “*ne* ive” at position 39 appears to be “onef ive”. comparing the plain text and cipher text at this position supplies:

V I R F H P A * * * o n e f V1 V2 I1 I2 R1 R2 F1 = * * * O1 N1 E1 F1 F2 H1 H2 P1 P2 A1 A2 * * * O2 N2 E2 F2

Now P1=O2=7 and P2=N2=7. So P=(7,7)

Also F1=F1, which is an identity. Further F2=A2= 3. So F=(*,3).

Note that the partial text “*et* een” at position 88 appears to be “betw een”. comparing the plain text and cipher text at this position supplies:

I C V U Z E D * * * b e t w I1 I2 C1 C2 V1 V2 U1 = * * * B1 E1 T1 W1 U2 Z1 Z2 E1 E2 D1 D2 * * * B2 E2 T2 W2

Now B1=C2=8 and B2=E1=3. So B=(8,3)

Also W1=U1=?, which relates two unknowns. Further W2=D2 = 1. So W = (*,1).

Finally, note that the partial text “*ive*n *do” at position 08 appears to be “fiveon edo”. comparing the plain text and cipher text at this position supplies:

O A V E V T P * f i v e o n O1 O2 A1 A2 V1 V2 E1 = * F1 I1 V1 E1 O1 N1 E2 V1 V2 T1 T2 P1 P2 * F2 I2 V2 E2 O2 N2

Now F1=O2=7 and F2=V1=3. So F=(7,3)

Also P1=O2=7 and P2=N2=7. So P=(7,7), confirming what we had recovered previously.

1 2 3 4 5 6 7 8 9 10 1 D I A * * * S * * * 2 T * * * L * O C * * 3 E V R * * * N * * * 4 * * * * * * * * * * 5 * * * * * * * X Y * 6 * * * * * * * * * * 7 * * F * * * P * * * 8 * * B * * * * * * * 9 * * * * Q * * * * * 10 * * * * * * * * * *

VCUHKKS OAVEVTP ECMWSAF ERRBKTP RAAOAKP VIRFHPA ANTVTPO BAEDMKM XXIVDPD ******* tfiveon *do**** r**ven* nea**no r**onef ivepoin ******* ****ste VLTEXQO AESTAOQ IEAUTIK ICVUZED RVXTSDK REUADFK RHDTRHT RCKNT elcylin deris*l **edi** ***bet* eent*** rees*** er***it *ro**

Note that the partial text “point” between groups 7 and 8. Comparing the plain text and cipher text at this position supplies:

B A E D M K M t * * * * * * B1 B2 A1 A2 E1 E2 D1 = T1 * * * * * * D2 M1 M2 K1 K2 M1 M2 T2 * * * * * *

Now B1=T1 but B1=8 and T1=2. This indicates that row 2 and 8 must be combined as must columns 2 and 8. Nothing in the present keysquare prevents this, since no contradictions or conflicts result.

1 2 3 4 5 6 7 8 9 10 1 D I A * * * S * * * 2 T C B * L * O * * * 3 E V R * * * N * * * 4 * * * * * * * * * * 5 * X * * * * * * Y * 6 * * * * * * * * * * 7 * * F * * * P * * * 8 * * * * * * * * * * 9 * * * * Q * * * * * 10 * * * * * * * * * *

VCUHKKS OAVEVTP ECMWSAF ERRBKTP RAAOAKP VIRFHPA ANTVTPO BAEDMKM XXIVDPD ******* tfiveon *dot*** r**veno nea**no r**onef ivepoin t****** xt*oste VLTEXQO AESTAOQ IEAUTIK ICVUZED RVXTSDK REUADFK RHDTRHT RCKNT elcylin deris*l *cedi** ***bet* eent*** rees*** er***it *ro**

Note that the partial text “**ven” in group 4 that suggests ‘seven’. Comparing the plain text and cipher text at this position supplies:

E R R B K T P r s e v e n o E1 E2 R1 R2 R1 R2 B1 = R1 S1 E1 V1 E1 N1 O1 B2 K1 K2 T1 T2 P1 P2 R2 S2 E2 V2 E2 N2 O2

Now K1=S2=7 and K2=E2=1. So K=(7,1).

Note that the partial text “t*o” in group 9 that suggests ‘two’. Comparing the plain text and cipher text at this position supplies:

X X I V D P D x t w o s t e X1 X2 X1 X2 I1 I2 V1 = X1 T1 W1 O1 S1 T1 E1 V2 D1 D2 P1 P2 D1 D2 X2 T2 W2 O2 S2 T2 E2

Now W1=X1=5 and W2=D2=1. So W=(5,1).

Note that the partial text “*l*ced” between groups 11 and 12 that suggests “placed”. Comparing the plain text and cipher text at this position supplies:

A E S T A O Q d e r i s p l A1 A2 E1 E2 S1 S2 T1 = D1 E1 R1 I1 S1 P1 L1 T2 A1 A2 O1 O2 Q1 Q2 D2 E2 R2 I2 S2 P2 L2

Now Q1=P2 but Q1=9 and P2=7. This indicates that row 9 and 7 must be combined as must columns 9 and 7. Nothing in the present key square prevents this, since no contradictions or conflicts result.

1 2 3 4 5 6 7 8 9 10 1 D I A * * * S * * * 2 T C B * L * O * * * 3 E V R * * * N * * * 4 * * * * * * * * * * 5 W X * * * * Y * * * 6 * * * * * * * * * * 7 K * F * Q * P * * * 8 * * * * * * * * * * 9 * * * * * * * * * * 10 * * * * * * * * * *

VCUHKKS OAVEVTP ECMWSAF ERRBKTP RAAOAKP VIRFHPA ANTVTPO BAEDMKM XXIVDPD *oto*** tfiveon edot*** rseveno neandno r**onef ivepoin t**ne** xtwoste VLTEXQO AESTAOQ IEAUTIK ICVUZED RVXTSDK REUADFK RHDTRHT RCKNT elcylin derispl *cedin* ***bet* eentwot rees**d er***it ero*k

Note that the partial text “***r” in group 3 that suggests ‘four’. Comparing the plain text and cipher text at this position supplies:

E C M W S A F e d o t f o u E1 E2 C1 C2 M1 M2 W1 = E1 D1 O1 T1 F1 O1 U1 W2 S1 S2 A1 A2 F1 F2 E2 D2 O2 T2 F2 O2 U2

- Now U1=W1=5 and U2=F2=3. So W=(5,3).
- Now M1=F1=7 and M2=O1=2. So M=(7,2).

This leaves letters Z, G, and H to be positioned in the key square. Z most likely belongs in row 5. A little experimentation shows H belongs in row 1 and G in row 3.

1 2 3 4 5 6 7 8 9 10 1 D I A * H * S * * * 2 T C B * L * O * * * 3 E V R * G * N * * * 4 * * * * * * * * * * 5 W X U * Z * Y * * * 6 * * * * * * * * * * 7 K M F * Q * P * * * 8 * * * * * * * * * * 9 * * * * * * * * * * 10 * * * * * * * * * *

VCUHKKS OAVEVTP ECMWSAF ERRBKTP RAAOAKP VIRFHPA ANTVTPO BAEDMKM XXIVDPD gotowes tfiveon edotfou rseveno neandno rthonef ivepoin tninesi xtwoste VLTEXQO AESTAOQ IEAUTIK ICVUZED RVXTSDK REUADFK RHDTRHT RCKNT elcylin derispl acedinw allbetw eentwot reesund erawhit erock

`Go to west five one dot four seven one and north one five point nine six two. Steel cylinder is placed in wall between two trees under a white rock.`

A view of row five quickly shows the correct order for the rows and columns to be 3, 1, 2, 7, 5.

The key square in correct order is:

1 2 3 4 5 1 R E V N G 2 A D I S H 3 B T C O L 4 F K M P Q 5 U W X Y Z

The key phrase is revealed to be the old Klingon saying: “Revenge, a dish best served cold.”

The method of enciphering and deciphering a Twin Bifid is identical to that used for the regular Bifid system.

That is because a Twin Bifid consists of two messages which share some plain text. These two messages have been enciphered by the normal Bifid method using the same key square but different periods.

Key Word: Example Plain Text: The end is near. (Message # 1, Period 3) The end is distant. (Messag2 # 2, Period 5) Key Square: 1 2 3 4 5 1 E X A M P 2 L B C D F 3 G H I K N 4 O Q R S T 5 U V W Y Z THE END ISN EAR THEEN DISDI STANT 431 132 343 114 43113 23423 44134 521 154 345 133 52115 43443 45355 Ct: RPL ALY KIT EOI RENLP CQKKR SASWZ

Now, let us assume that the above example is part of a larger set of messages, and that we don’t know the key square or the underlying plain text. All we know (By some method) is the equivalence of a small portion of the plain text of both messages. For example, we know the following:

Pt: t h e e n d i s * * * M1: R P L:A L Y:K I T: M2: R E N L P:C Q K K R:

: = Period Divisions

Can we learn anything about the unknown key square from this material? Again, the answer is yes. We can write out each block in its cipher text form using the generalized coordinates (i.e., A = (A1, A2)) for each letter and equate the first eight columns of coordinates. This can only be done because we know the first eight letters of plain text are identical in both messages.

Writing out the text coordinates in terms of the fractionation of the cipher text we get:

1 2 3 4 5 6 7 8 9 10 Pt: t h e e n d i s * * (In reality Unknown)

M1 R P L :A L Y :L I T : R1 R2 P1:A1 A2 L1:L1 L2 I1: P2 L1 L2:L2 Y1 Y2:I2 T1 T2: M2 R E N L P :C Q K K R : R1 R2 E1 E2 N1:C1 C2 Q1 Q2 K1: N2 L1 L2 Y1 Y2:K2 K1 K2 R1 R2:

Since the first 8 columns of each cipher are the fractionation of the same plain text, we can equate the letter coordinate in the first message with the matching position in the second message. In other words:

From 1st Row:

M1: R1 R2 P1 A1 A2 L1 L1 L2 (i.e., R1=R1, R2=R2, P1=E1,…) M2: R1 R2 E1 E2 N1 C1 C2 Q1

From 2nd Row:

M1: P2 L1 L2 L2 Y1 Y2 I2 T1

M2: N2 L1 L2 Y1 Y2 K2 K1 K2

As in the case of the regular Bifid, if enough cipher text blocks are equated, a great deal of information can be gained concerning the relationship of the letter coordinates to one another and the key array can be reconstructed.

Once these initial letter coordinate equivalences are obtained, solution is identical to that for the regular Bifid. Essentially:

- Use the identities to begin the reconstruction of the key square by the methods already demonstrated.
- With the proto-key square, substitute into both messages and examine the resulting plain text.
- From the plain text, identify partial words and further reconstruct the key based on the additional coordinate equivalences you obtain.
- Repeat the process until finished.

Note that a Twin Bifid is (IMHO) easier than regular Bifid since:

- You never have to place a crib!
- You have two sources of partial plain text to advance the solution versus only the one source available for a regular Bifid.

The example we will be decrypting is MA08 E-20:

HGTIPIYKF RULVOYRCG HMRSIWDWC LSMAFLRDM HMNSYDFOK ASYOHBAQT HADHXHYLU INYBFFTWD MIYMBONLR ERFHIKIIN NAYOTAWQT OHHUBKLUF NYHHNTVNA GWMKAHAVF TIEXSCIUT VBBEHSSDZ PERBIYOAT ILPEHFYMI BHHAIGC. HGTILYAW MOTOKYRC RHQRGCFF FBIRIDQI PLNBPIGL OIYSWGAE RDSEIAWA RTVFTOSV INRWFDDT KTYLYUWE ACLHVHPC LMNSKFEB IITDTUAM TBDLDFIF OSGOBMAN GIDWAMEY YETHQDVL GTHTAWBG OEAEWRMN LSELGUCI NE.

Write out the fractionation blocks for each message:

Message #1:H G T I P I Y K F :R U L V O Y R C GH1 H2 G1 G2 T1 T2 I1 I2 P1:R1 R2 U1 U2 L1 L2 V1 V2 O1P2 I1 I2 Y1 Y2 K1 K2 F1 F2:O2 Y1 Y2 R1 R2 C1 C2 G1 G2:H M RS I W D W C ::H1 H2 M1M2 R1 R2 S1 S2 I1::I2 W1 W2D1 D2 W1 W2 C1 C2: Message # 2:H G T I L Y A W :M O T O K Y R C :R H Q R GC F F :H1 H2 G1 G2 T1 T2 I1 I2:M1 M2 O1 O2 T1 T2 O1 O2:R1 R2 H1 H2 Q1Q2 R1 R2:L1 L2 Y1 Y2 A1 A2 W1 W2:K1 K2 Y1 Y2 R1 R2 C1 C2:G1 G2 C1 C2 F1F2 F1 F2:

- : = Period Divisions
- Bold Letters = First 21

From 1st Row:

M1: H1 H2 G1 G2 T1 T2 I1 I2 P1 R1 R2 U1 U2 L1 L2 V1 V2 O1 H1 H2 M1 M2: H1 H2 G1 G2 T1 T2 I1 I2 M1 M2 O1 O2 T1 T2 O1 O2 R1 R2 H1 H2 Q1

From 2nd Row:

M1: P2 I1 I2 Y1 Y2 K1 K2 F1 F2 O2 Y1 Y2 R1 R2 C1 C2 G1 G2 I2 W1 W2

M2: L1 L2 Y1 Y2 A1 A2 W1 W2 K1 K2 Y1 Y2 R1 R2 C1 C2 G1 G2 C1 C2 F1

After much work you should get:

P A B C D E F G H I J K L M 1 08 - 08 - - 15 03 01 07 - 13 06 09 2 13 - 12 - - 13 04 02 08 - 12 07 10 P N O P Q R S T U V W X Y Z 1 - 07 09 09 10 - 05 12 12 12 - 08 - 2 - 12 06 - 07 - 06 05 10 15 - 08 -

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 * H * * * * * * * * * * * * * 2 * * * * * * * * * * * * * * * 3 * * * G * * * * * * * * * * * 4 * * * * * * * * * * * * * * * 5 * * * * * T * * * * * * * * * 6 * * * * * * L * * * * * * * * 7 * * * * * * * I * * * O * * * 8 * * * * * * * Y * * * C A * * 9 * * * * * P * * * M * * * Q * 10 * * * * * * R * * * * * * * * 11 * * * * * * * * * * * * * * * 12 * * * * U * * * * V * * * * W 13 * * * * * * * * * * * K * * * 14 * * * * * * * * * * * * * * * 15 * * * * * * * * * * * * F * *

*Can you see the keyword in this mess?*

Note that rows 7 and 8 contain out of sequence letters (Row 7: I and O. Row 8: Y, A, and C.) suggesting they may contain the key word.

Further note that rows 9 and 12 contain letter more or less in sequence ( Row 9: M, P, and Q. Row 12: U, V, W.). This suggests these may be part of the final, non-key word rows.

The sequence in row 12 and the presence of Y in row 8 suggests the final row of the key square could be U, V, W, X, and Z.

Further, V and M fall in the same column suggesting the following sequence:

* M * P Q * * * * * Space for a potential intervening row. U V W X Z

Note that the three letters in row (R, S, and T) exist between Q and U. This suggests that the intervening row shown in the last slide does not exist. Further, since O is suspected of being in the keyword, N must be the letter between M and P. This provides:

* * * * * * * * * * * * * * * * M N P Q U V W X Z

Note that this suggests that letters R, S, and T are contained in the keyword.

Note that the three letters in row 8 (Y, A, and C) suggest that Y may be the last letter of the keyword followed by A-C as the beginning of the alphabetic non-key word sequence. Further, note that the letters I and O from row 7 are in the columns containing Y and C respectively. This suggests that one of the following arrangements exists in the key square:

I * * O * * I * * O Y A B C * * Y A B C

Recalling that the letters R, S, and T are contained in the keyword, we ‘quickly’ guess that the key word may be ‘HISTORY’. This certainly fits with the CONs subject ‘Ancient Times’.

1 2 3 4 5 1 H I S T O 2 R Y A B C 3 D E F G K 4 L M N P Q 5 U V W X Z

M1: HGTIPIYKF RULVOYRCG HMRSIWDWC LSMAFLRDM HMNSYDFOK ASYOHBAQT theminoan civilizat ionaroseo nthelarge islandofc reteabout… M2: HGTILYAW MOTOKYRC RHQRGCFF FBIRIDQI PLNBPIGL OIYSWGAE RDSEIAWA theminoa nciviliz ationwas dealtcri pplingbl owsbyase riesofea…

The CM Bifid system performs its fractionation by writing the plain text and the cipher text alphabets into two separate 5×5 key-squares.

As before, the plain text is broken up into period size group. The coordinates of each letter are found in the plain text key-square and written in vertical columns below each letter.

The cipher text for each group is then created drawing out sets of two numbers as before. In this case, each pair of numbers is converted into a cipher text character utilizing the cipher text key square.

Key Words: Example and Problem Plain Text: The end is near. (Period 3) Key Squares: 1 2 3 4 5 1 2 3 4 5 1 e x a m p 1 P R O B L 2 l b c d f 2 E M A C D 3 g h i k n 3 F G H I K 4 o q r s t 4 N Q S T U 5 u v w y z 5 V W X Y Z Pt: the end isn ear 431 132 343 114 521 154 345 133 Ct: SLE OEY IHU PNH

Now, once again assume that the above example is part of a larger message, and that we don’t know the key squares. All we know (By some method) is the equivalence of a small portion of the plain text of and cipher text. For example, we know the following:

PT: the end isn ear CT: SLE OEY IHU PNH

Can we learn anything about the unknown key squares from this material? By now you must sense that the answer is yes. We can write out each block in its cipher text and plain text forms using the generalized coordinates (i.e., A = (A1, A2), a=(a1,a2)) for the text and equate the appropriate columns of coordinates. This can only be done because we know the first eight letters of plain text are identical in both messages.

Once these initial letter coordinate equivalences are obtained, solution is similar to that for the regular Bifid. Essentially:

- Use the identities to begin the reconstruction of the plain and cipher text key squares by the methods already demonstrated.
- With the proto-key squares, substitute into the message and examine the resulting plain text.
- From the plain text, identify partial words and further reconstruct the key squares based on the additional coordinate equivalences you obtain.
- Repeat the process until finished.

Note that a CM Bifid is (IMHO) more difficult than regular Bifid since:

- You still have to place a crib! But now several crib location methods available for the Bifid don’t work for the CM Bifid.
- You have two two key squares to reconstruct versus only one key square for a regular Bifid.

RDMOOSD OMYCNHU SYVRRLL PCCCVNA CDHLTHR YMXYDBY BSVOOAC YVUDSDT DRKOLDR DRKOLDR YCSBMTO UYYYGAI KMAAWAO ASEEOUA OYEQAUA FLRYRZR butthem odernci YLWENYQ ICIRWVS FYDVYUR YGQEVVH FWKQYEP EBLXIED HECFVKD OLCBRPT tydoesn otnurtu LNOBDBR KCKOYGP BBRR (Ends: countryside)

Now, lets take the crib and write out it’s fractionation block:

b u t t h e m: o d e r n c i: t y d o e s n: b1 u1 t1 t1 h1 e1 m1:o1 d1 e1 r1 n1 c1 i1:t1 y1 d1 o1 e1 s1 n1: b2 u2 t2 t2 h2 e2 m2:o2 d2 e2 r2 n2 c2 i2:t2 y2 d2 o2 e2 s2 n2: o t n u r t u o1 t1 n1 u1 r1 t1 u1 o2 t2 n2 u2 r2 t2 u2

The fractionation of the cipher text blocks containing the crib is:

O Y E Q A U A: F L R Y R Z R: Y L W E N Y Q: O1 O2 Y1 Y2 E1 E2 Q1: F1 F2 L1 L2 R1 R2 Y1: Y1 Y2 L1 L2 W1 W2 E1: Q2 A1 A2 U1 U2 A1 A2: Y2 R1 R2 Z1 Z2 R1 R2: E2 N1 N2 Y1 Y1 Q1 Q2: I C I R W V S I1 I2 C1 C2 I1 I2 R1 R2 W1 W2 V1 V2 S1 S2

From 1st Row:

P1: b1 u1 t1 t1 h1 e1 m1 o1 d1 e1 r1 n1 c1 i1 t1 y1 d1 o1 e1 s1 n1 C1: O1 O2 Y1 Y2 E1 E2 Q1 F1 F2 L1 L2 R1 R2 Y1 Y1 Y2 L1 L2 W1 W2 E1 P1: o1 t1 n1 u1 r1 t1 u1 C1: I1 I2 C1 C2 I1 I2 R1

From 2nd Row:

P2: b2 u2 t2 t2 h2 e2 m2 o2 d2 e2 r2 n2 c2 i2 t2 y2 d2 o2 e2 s2 n2 c2: Q2 A1 A2 U1 U2 A1 A2 Y2 R1 R2 Z1 Z2 R1 R2 E2 N1 N2 Y1 Y1 Q1 Q2 P2: b2 u2 t2 t2 h2 e2 m2 c2: R2 W1 W2 V1 V2 S1 S2

After much work you should get:

P 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 1 - 01 03 05 05 - - 02 03 - - - 06 02 07 - - 07 12 03 02 - - - 03 - 2 - 12 02 02 03 - - 16 03 - - - 05 12 03 - - 17 06 05 03 - - - 19 - C 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 1 03 - 02 - 02 07 - - 07 - - 05 - 19 01 - 06 02 05 - 05 03 05 - 03 17 2 05 - 02 - 05 05 - - 03 - - 07 - 02 02 - 12 03 03 - 16 17 12 - 03 12

1 2 3 5 6 7 12 16 17 19 1 * * * * * * b * * * 2 * * u * * * n h * * 3 * c i t * * * * * y 5 * d e * * * * * * * 6 * * * m * * * * * * 7 * * o * * * * * r * 12 * * * * s * * * * * 16 * * * * * * * * * * 17 * * * * * * * * * * 19 * * * * * * * * * * 1 2 3 5 6 7 12 16 17 19 1 * O * * * * * * * * 2 * C R E * * * * * * 3 * * Y A * * * * V * 5 * * S * * L W U * * 6 * * * * * * Q * * * 7 * * I F * * * * * * 12 * * * * * * * * * * 16 * * * * * * * * * * 17 * * * * * * Z * * * 19 * N * * * * * * * *

RDMOOSD OMYCNHU SYVRRLL PCCCVNA CDHLTHR YMXYDBY BSVOOAC YVUDSDT ******* *****th ecit*** *****u* ******e i*****i **dit** *ti**** DRKOLDR DRKOLDR YCSBMTO UYYYGAI KMAAWAO ASEEOUA OYEQAUA FLRYRZR ******* ******* ******* e**it*i ****t*c t*dthe* butthem odernci YLWENYQ ICIRWVS FYDVYUR YGQEVVH FWKQYEP EBLXIED HECFVKD OLCBRPT tydoesn otnurtu reit**i ti***** *ee**** ******* ******* **e**** LNOBDBR KCKOYGP BBRR ******* **u**** ****

The proto-key provides a good start. Notice the apparent phrase “its citizens” which appears in line 3.

The remainder of the decryption is left as an exercise for the student.

]]>The Checkerboard cipher is a simple substitution cipher based on a 5 x 5 Polybius Square. It was invented by the Greek historian Polybius in the 2nd century BC.

The substitution alphabet(s) is(are) based on a 5×5 keyed Polybius square that is formed in any of the usual ways (I/J are considered equivalent). The rows and columns are labeled by either a single 5 letter word/sequence (Checkerboard) or by two five letter words/sequences (Double Checkerboard). In the case of the simple checkerboard, each plain text letter has a single cipher digraph replacement. In the case of the double checkerboard, each plain text letter has 4 possible cipher digraph replacements. This tutorial will consider only the simple checkerboard.

The plain text is enciphered one letter at a time by replacing it with the two letter digraph that corresponds to that letters Row and Column designators. Deciphering is the reverse of enciphering.

**Keyword:**SONGBIRD**Row Designator:**ROBIN**Column Designator:**VIREO**Plain Text:**The early bird gets the worm.

Form the Keysquare:

V I R E O ---------- R :s o n g b O :i r d a c B :e f h k l I :m p q t u N :v w x y z

P: t h e e a r l y b i r d g e t s t h e w o r m C: IE BR BV BV OE OI BO NE RO OV OI OR RE BV IE RV IE BR BV NI RI OI IV Message: IE BR BV BV OE OI BO NE RO OV OI OR RE BV IE RV IE BR BV NI RI OI IV

The student should study the above example to become familiar with the encipherment and decipherment methods since this knowledge is critical for the cryptanalysis of the cipher.

The cryptanalysis of this cipher primarily involves reconstructing the key square. This is done by matching pieces of the cipher text with plain text and reconstructing the square from the identified plain/cipher text equivalences.

The example we will be working on is from the log of a puzzle type Geocache in New England. The cipher is:

EI IU GQ TU TT EQ IU II IU GI GQ TU GQ GT TL EQ RQ GU TL GQ EQ IT TU EI IT EQ II IL TT TL RQ IU II TL RQ EQ TI TQ IU EQ II TT RQ TL RU TL II TT RI IU GI EQ RU TL II IU ET TT GT TU II GU GI EQ RU TL TT RI IU GU IU TT GI EQ RU TL GI EQ RU TL GI IU IL ET RI TL RQ TT GQ TU GQ GT TL EQ RQ TU TT TT GT TL IQ TU RQ TL IU GI TU EI TU ET GL TL TT ET TL TL RI GT EQ EI TL EQ II TT GT TL TU ET TL TU TQ TU RL RL IU IL ET ET TL RQ TQ TL GQ TT RQ TT IU TT GT TL RU TL TT TL ET TU II RQ (Location)

Note the following:

- The plain text ‘location’ has been given as a crib. Checkerboard ciphers published in the ‘Cryptogram’ are normally provided with a healthy crib.
- The latitude and longitude coordinates of an object (the ‘cache’) are enciphered in this CON. Consequently, we should expect the plain text to primarily be long hand numbers.
- The cache log states that the cache is within a few miles of 43 degrees 17.382 minutes North and 070 degrees 51.264 minutes West. This point is within the town of Berwick, Maine.
- Most geocaches are hidden in stumps, under logs, in stonewalls, or under trees. Caches can contain a log book, ‘stash’, geocoins, and/or travelbugs. Some geocachers who lives near this cache are named ‘TigQuilt’ (short for TiggerQuilt) and “Pax Et Bonum’.
- The notation GCRTRX is the waypoint designation for the cache. It has nothing to do with the CON.

The first order of business is to determine the letters that are being used to designate the rows and columns. From this, we may be able to determine the actual designator words by anagrammimg. Examination of the cipher text digraphs reveals the following:

Rows: E, I, G, T, R Columns: I, U, Q, T, L

Anagramming these letters provides only, TIGER and QUILT as possible words. We will start the decrypt assuming these are the row and column designators.

The crib ‘location’ has the 8 letters with ‘o’ appearing twice. We need to search through the cipher text to find a sequence of 8 digraphs which has the 2nd and 7th digraph the same. It is quickly evident that the start of the cipher text meets this condition. Specifically, if we place the crib at position 1, the first ten digraphs provide:

CT: EI IU GQ TU TT EQ IU II IU GI PT: l o c a t i o n o *

Note that the first letter following the crib would also be an ‘o’. This suggests the phrase “location of’. This extended crib would provide:

EI = l, IU = o, GQ = c, TU = a, TT = t, EQ = i, II = n, and GI = f.

Using these identities and substituting we obtain:

Q U I L T ---------- T :* a * * t I :* o n * * G :c * f * * E :i * l * * R :* * * * *

EI IU GQ TU TT EQ IU II IU GI GQ TU GQ GT TL EQ RQ GU TL GQ EQ IT l o c a t i o n o f c a c * * i * * * c i * TU EI IT EQ II IL TT TL RQ IU II TL RQ EQ TI TQ IU EQ II TT RQ TL a l * i n * t * * o n * * i * * o i n t * * RU TL II TT RI IU GI EQ RU TL II IU ET TT GT TU II GU GI EQ RU TL * * n t * o f i * * n o * t * a n * f i * * TT RI IU GU IU TT GI EQ RU TL GI EQ RU TL GI IU IL ET RI TL RQ TT t * o * o t f i * * f i * * f o * * * * * t GQ TU GQ GT TL EQ RQ TU TT TT GT TL IQ TU RQ TL IU GI TU EI TU ET c a c * * i * a t t * * * a * * o f a l a * GL TL TT ET TL TL RI GT EQ EI TL EQ II TT GT TL TU ET TL TU TQ TU * * t * * * * * i l * i n t * * a * * a * a RL RL IU IL ET ET TL RQ TQ TL GQ TT RQ TT IU TT GT TL RU TL TT TL * * o * * * * * * * c t * t o t * * * * t * ET TU II RQ * a n *

The first line of the plain text reads, “Location of the cac**…”. This suggest the fourth word is cache. This provides:

GT = h and TL =e

Using these identities and substituting we obtain:

Q U I L T ---------- T :* a * e t I :* o n * * G :c * f * h E :i * l * * R :* * * * *

EI IU GQ TU TT EQ IU II IU GI GQ TU GQ GT TL EQ RQ GU TL GQ EQ IT l o c a t i o n o f c a c h e i * * e c i * TU EI IT EQ II IL TT TL RQ IU II TL RQ EQ TI TQ IU EQ II TT RQ TL a l * i n * t e * o n e * i * * o i n t * e RU TL II TT RI IU GI EQ RU TL II IU ET TT GT TU II GU GI EQ RU TL * e n t * o f i * e n o * t h a n * f i * e TT RI IU GU IU TT GI EQ RU TL GI EQ RU TL GI IU IL ET RI TL RQ TT t * o * o t f i * e f i * e f o * * * e * t GQ TU GQ GT TL EQ RQ TU TT TT GT TL IQ TU RQ TL IU GI TU EI TU ET c a c h e i * a t t h e * a * e o f a l a * GL TL TT ET TL TL RI GT EQ EI TL EQ II TT GT TL TU ET TL TU TQ TU * e t * e e * h i l e i n t h e a * e a * a RL RL IU IL ET ET TL RQ TQ TL GQ TT RQ TT IU TT GT TL RU TL TT TL * * o * * * e * * e c t * t o t h e * e t e ET TU II RQ * a n *

Now, the end of line 3 and the beginning of line four reads “fi*et*o”. This suggests the phrase “five two” and provides RU = v and RI = w.

Using these identities and substituting we obtain:

Q U I L T ---------- T :* a * e t I :* o n * * G :c * f * h E :i * l * * R :* v w * *

EI IU GQ TU TT EQ IU II IU GI GQ TU GQ GT TL EQ RQ GU TL GQ EQ IT l o c a t i o n o f c a c h e i * * e c i * TU EI IT EQ II IL TT TL RQ IU II TL RQ EQ TI TQ IU EQ II TT RQ TL a l * i n * t e * o n e * i * * o i n t * e RU TL II TT RI IU GI EQ RU TL II IU ET TT GT TU II GU GI EQ RU TL v e n t w o f i v e n o * t h a n * f i v e TT RI IU GU IU TT GI EQ RU TL GI EQ RU TL GI IU IL ET RI TL RQ TT t w o * o t f i v e f i v e f o * * w e * t GQ TU GQ GT TL EQ RQ TU TT TT GT TL IQ TU RQ TL IU GI TU EI TU ET c a c h e i * a t t h e * a * e o f a l a * GL TL TT ET TL TL RI GT EQ EI TL EQ II TT GT TL TU ET TL TU TQ TU * e t * e e w h i l e i n t h e a * e a * a RL RL IU IL ET ET TL RQ TQ TL GQ TT RQ TT IU TT GT TL RU TL TT TL * * o * * * e * * e c t * t o t h e v e t e ET TU II RQ * a n *

An examination of the key square suggests that the key word occupies the first two rows and that the last three rows are the remainder of the alphabet. Since we have already located ‘e’, this suggests that GU = d. We can also determine that GL =g and EU = k since only those letters can fit the gaps in the plain text.

Using these identities and substituting we obtain:

Q U I L T ---------- T :* a * e t I :* o n * * G :c d f g h E :i k l * * R :* v w * *

EI IU GQ TU TT EQ IU II IU GI GQ TU GQ GT TL EQ RQ GU TL GQ EQ IT l o c a t i o n o f c a c h e i * d e c i * TU EI IT EQ II IL TT TL RQ IU II TL RQ EQ TI TQ IU EQ II TT RQ TL a l * i n * t e * o n e * i * * o i n t * e RU TL II TT RI IU GI EQ RU TL II IU ET TT GT TU II GU GI EQ RU TL v e n t w o f i v e n o * t h a n d f i v e TT RI IU GU IU TT GI EQ RU TL GI EQ RU TL GI IU IL ET RI TL RQ TT t w o d o t f i v e f i v e f o * * w e * t GQ TU GQ GT TL EQ RQ TU TT TT GT TL IQ TU RQ TL IU GI TU EI TU ET c a c h e i * a t t h e * a * e o f a l a * GL TL TT ET TL TL RI GT EQ EI TL EQ II TT GT TL TU ET TL TU TQ TU g e t * e e w h i l e i n t h e a * e a * a RL RL IU IL ET ET TL RQ TQ TL GQ TT RQ TT IU TT GT TL RU TL TT TL * * o * * * e * * e c t * t o t h e v e t e ET TU II RQ * a n *

Now, the end of line 2 and the beginning of line 3 reads “*oint*even”. This suggests the phrase “point seven” and provides TQ = p and RQ = s.

Further the end of line 4 reads “fo**we*t”. This suggests the phrase “four west”. This provides IL = u and ET = r.

Using these identities and substituting we obtain:

Q U I L T ---------- T :p a * e t I :* o n u * G :c d f g h E :i k l * r R :s v w * *

EI IU GQ TU TT EQ IU II IU GI GQ TU GQ GT TL EQ RQ GU TL GQ EQ IT l o c a t i o n o f c a c h e i s d e c i * TU EI IT EQ II IL TT TL RQ IU II TL RQ EQ TI TQ IU EQ II TT RQ TL a l * i n u t e s o n e s i * p o i n t s e RU TL II TT RI IU GI EQ RU TL II IU ET TT GT TU II GU GI EQ RU TL v e n t w o f i v e n o r t h a n d f i v e TT RI IU GU IU TT GI EQ RU TL GI EQ RU TL GI IU IL ET RI TL RQ TT t w o d o t f i v e f i v e f o u r w e s t GQ TU GQ GT TL EQ RQ TU TT TT GT TL IQ TU RQ TL IU GI TU EI TU ET c a c h e i s a t t h e * a s e o f a l a r GL TL TT ET TL TL RI GT EQ EI TL EQ II TT GT TL TU ET TL TU TQ TU g e t r e e w h i l e i n t h e a r e a p a RL RL IU IL ET ET TL RQ TQ TL GQ TT RQ TT IU TT GT TL RU TL TT TL * * o u r r e s p e c t s t o t h e v e t e ET TU II RQ r a n s

Finally, line two suggests the phrase “minutes one six” providing IT = m and TI = x. Line 5 suggests the phrase “the base of” providing IQ = b. By the process of elimination this leaves EL = q, RL = y, and RT = z.

Using these identities and substituting we obtain:

Q U I L T ---------- T :p a x e t I :b o n u m G :c d f g h E :i k l q r R :s v w y z

EI IU GQ TU TT EQ IU II IU GI GQ TU GQ GT TL EQ RQ GU TL GQ EQ IT l o c a t i o n o f c a c h e i s d e c i m TU EI IT EQ II IL TT TL RQ IU II TL RQ EQ TI TQ IU EQ II TT RQ TL a l m i n u t e s o n e s i x p o i n t s e RU TL II TT RI IU GI EQ RU TL II IU ET TT GT TU II GU GI EQ RU TL v e n t w o f i v e n o r t h a n d f i v e TT RI IU GU IU TT GI EQ RU TL GI EQ RU TL GI IU IL ET RI TL RQ TT t w o d o t f i v e f i v e f o u r w e s t GQ TU GQ GT TL EQ RQ TU TT TT GT TL IQ TU RQ TL IU GI TU EI TU ET c a c h e i s a t t h e b a s e o f a l a r GL TL TT ET TL TL RI GT EQ EI TL EQ II TT GT TL TU ET TL TU TQ TU g e t r e e w h i l e i n t h e a r e a p a RL RL IU IL ET ET TL RQ TQ TL GQ TT RQ TT IU TT GT TL RU TL TT TL y y o u r r e s p e c t s t o t h e v e t e ET TU II RQ r a n s

The plain text is now completely decrypted. Note that the key square is a homage to the local geocachers TigQuilt and Pax Et Bonum. The cache coordinates are adjacent to Knox Lane. Hence, the title of the cache “Hard Knocks” is a pun.

]]>The Foursquare cipher is a digraphic substitution cipher similar to the Playfair. Unlike the Playfair cipher, which is based on a single Polybius square, Foursquare uses 4 Polybius squares arranged into a single large square. The four squares are labeled from 1 to 4 starting in the upper left and proceeding clockwise. Squares 1 and 3 contain unkeyed plaintext alphabets placed in the normal order. Squares 2 and 4 are keyed ciphertext alphabets and may be placed in the squares in any of the usual ways.

Message encipherment proceeds much like Playfair. The plain text message is broken up into digraphs. Unlike Playfair, double letter digraphs are acceptable and the introduction of nulls is not necessary. To encipher each plain text digraph, the first and second letters are located in squares 1 and 3 respectively. These cells are considered to be the opposite corners of a rectangle. The remaining corners of this rectangle are located and read off from squares 2 and 4 to obtain the first and second letters respectively of the ciphertext digraph. Deciphering is the inverse of the encipherment process.

**Msg:**The early bird gets the worm.**Keywords:**ROBIN, BLUEBIRD

Key Square:

a b c d e:R A G P V f g h i k:O C H Q W l m n o p:B D K S X q r s t u:I E L T Y v w x y z:N F M U Z - - - - -:- - - - - B L U E I:a b c d e R D A C F:f g h i k G H K M N:l m n o p O P Q S T:q r s t u V W X Y Z:v w x y z

pt: th ee ar ly bi rd ge ts th ew or mx CT: LC VI AO SV PD TL WL LS LC AZ DS KW

Note that the keyed ciphertext in alphabets 2 and 4 were placed in the squares by two different methods. Further, the plaintext digraph ‘ee’ did not require a null separator as is necessary in Playfair. If you are not familiar with the Foursquare cipher, study this example. The cryptanalysis of this cipher type requires a good working knowledge of the substitution scheme.

The example we will be working on is JF04 E-23:

E-23. Foursquare. Funny money. (within our means) KOSTY

DK XP VF EO DK GT EW UI QN NV SB LQ NK VT GL ZQ UN PD AD NY LY WA CC GW TA PZ GA WS IU DZ OT OL PM EY VI GA XH YC PL BE GW WE NY CP CT QD FD KS NZ QS NS OT.

We are given the plain text crib ‘within our means’. As with Playfair we break this crib up into digraphs in the two possible ways and see if there are any repeat digraphs. If repeats exist, we would look in the ciphertext for a similar pattern to allow us to place the crib. In this case, the digraphic breakdowns provide:

1: wi th in ou rm ea ns 2: *w it hi no ur me an s*

Note that neither of the breakdowns contains repeated digraphs! What is a poor Cryptanalyst to do?

In cases like this, it becomes necessary to independently look for patterns within the first and second letters of the digraphs. This method was demonstrated by LIONEL at the 2002 East Coast mini-con and is called “The Method of Isomorphism”!

Essentially, this involves assuming any desired alphabet for squares 2 and 4 (Remember, squares 1 and 3 are always the same). This trial key square will be used to encipher the crib. The ciphertext, which results from the use of this trial key square, will not be correct. However, because the alphabets in squares 1 and 3 are fixed, any patterns we find in either the first letters or the second letters of these trial cipher digraphs must exist in the real CON. Let’s try this method with the following trial key square:

a b c d e:A B C D E f g h i k:F G H I K l m n o p:L M N O P q r s t u:Q R S T U v w x y z:V W X Y Z - - - - -:- - - - - A B C D E:a b c d e F G H I K:f g h i k L M N O P:l m n o p Q R S T U:q r s t u V W X Y Z:v w x y z

pt1: wi th in ou rm ea ns CT: YG SI HO PT RM AE NS pt2: it hi no ur me an CT: IT IH ON RU PB CL

Notice that the first trial encipherment contains no repeats of either the first or second letters of the cipher digraphs. However, the second breakdown does provide a pattern. The first letter of the first and second digraph is the same. We search through the CON and find this pattern at two locations:

Digraph 31: OT OL PM EY VI GA Digraph 44: CP CT QD FD KS NZ

Note that digraph 44 can not be the position of our plain string crib. The third and fourth digraphs at this position have identical second letters. This pattern did not show up in our trial encipherment of the crib. Therefore, our crib can not be placed there. This leaves only digraph 33 as a possible position for our crib. We will adopt this as our working hypothesis.

Our crib provides:

pt: it hi no ur me an CT: OT OL PM EY VI GA

Substitution of these equivalencies into the key square provides:

a b c d e:* * G * * f g h i k:* * * O * l m n o p:* * * P V q r s t u:* E * * * v w x y z:* * * * * - - - - -:- - - - - * I * * *:a b c d e * * L * *:f g h i k A * M * *:l m n o p * * * T Y:q r s t u * * * * *:v w x y z

DK XP VF EO DK GT EW UI QN NV SB LQ NK VT GL ZQ UN PD AD NY LY WA ** ** ** ** ** ds ** ** ** ** ** ** ** ou ch ** ** ** ** ** ** ** CC GW TA PZ GA WS IU DZ OT OL PM EY VI GA XH YC PL BE GW WE NY CP ** ** ** ** an ** ** ** it hi no ur me an ** ** ni ** ** ** ** ** CT QD FD KS NZ QS NS OT ** ** ** ** ** ** ** it

Note the positions of L, M, T, and Y in square 4. This suggests that the keyword is written into this square in vertical columns. The gap between M and T suggests only one of the intervening letters is in the keyword. The gap between T and Y is exact. This allows us to fill in the letters from T to Z in square 4. This provides:

a b c d e:* * G * * f g h i k:* * * O * l m n o p:* * * P V q r s t u:* E * * * v w x y z:* * * * * - - - - -:- - - - - * I * * V:a b c d e * * L * W:f g h i k A * M * X:l m n o p * * * T Y:q r s t u * * * U Z:v w x y z

DK XP VF EO DK GT EW UI QN NV SB LQ NK VT GL ZQ UN PD AD NY LY WA ** ** ** ** ** ds ug ** ** ** ** ** ** ou ch ** ** ** ** ** ** ** CC GW TA PZ GA WS IU DZ OT OL PM EY VI GA XH YC PL BE GW WE NY CP ** eh ** py an ** ** ** it hi no ur me an ** ** ni ** eh ** ** ** CT QD FD KS NZ QS NS OT ** ** ** ** ** ** ** it

Examining the partial plain text, the fragment ‘ds ug’ suggests the word ‘suggest’. Further, the fragment ‘eh ** py’ suggests ‘happy’. This will provide ge = UI, ap = TA, and st = QN. Substituting these equivalencies provides:

a b c d e:* * G * T f g h i k:* * * O U l m n o p:* * * P V q r s t u:* E * Q * v w x y z:* * * * * - - - - -:- - - - - * I * * V:a b c d e * * L * W:f g h i k A * M * X:l m n o p * * N T Y:q r s t u * * * U Z:v w x y z

DK XP VF EO DK GT EW UI QN NV SB LQ NK VT GL ZQ UN PD AD NY LY WA ** ** ** ** ** ds ug ge st ** ** ** ** ou ch ** hu ** ** ** ** ** CC GW TA PZ GA WS IU DZ OT OL PM EY VI GA XH YC PL BE GW WE NY CP ** eh ap py an ** ** ** it hi no ur me an ** ** ni ** eh ** ** ** CT QD FD KS NZ QS NS OT ** ** ** ** ** ** ** it

Note how well the last substitutions fit into the key square. This provides us with confidence we are on the right track. Examining our crib and the second line of the CON suggests the partial text ‘happy and live within our means’. This will provide dl = WS, iv = IU, and ew = DZ. Substituting these equivalencies provides:

a b c d e:W D G * T f g h i k:I * * O U l m n o p:* * * P V q r s t u:* E * Q * v w x y z:* * * * * - - - - -:- - - - - * I * * V:a b c d e * * L * W:f g h i k A * M S X:l m n o p * * N T Y:q r s t u * * * U Z:v w x y z

DK XP VF EO DK GT EW UI QN NV SB LQ NK VT GL ZQ UN PD AD NY LY WA ** ** ** ** ** ds ug ge st ** ** ** ** ou ch ** hu ** ** ** ** al CC GW TA PZ GA WS IU DZ OT OL PM EY VI GA XH YC PL BE GW WE NY CP ** eh ap py an dl iv ew it hi no ur me an ** ** ni ** eh ** ** ** CT QD FD KS NZ QS NS OT ** ** ** ** ** to ** it

Again, note how well the last substitutions fit into the key square. Now, note the positions of E, G, and O in square 2. As with square 4, this suggests that the keyword has been written into this square in vertical columns. Further, now that the letter I has been placed, the position of the remaining unplaced letters between E and O is apparent. Making this substitution provides:

a b c d e:W D G N T f g h i k:I * H O U l m n o p:* * K P V q r s t u:* E L Q * v w x y z:* F M * * - - - - -:- - - - - * I * * V:a b c d e * * L * W:f g h i k A * M S X:l m n o p * * N T Y:q r s t u * * * U Z:v w x y z

DK XP VF EO DK GT EW UI QN NV SB LQ NK VT GL ZQ UN PD AD NY LY WA ** ** ** ** ** ds ug ge st ed ** ** ** ou ch ** hu ** ** et us al CC GW TA PZ GA WS IU DZ OT OL PM EY VI GA XH YC PL BE GW WE NY CP ** eh ap py an dl iv ew it hi no ur me an ** ** ni ** eh ** et ** CT QD FD KS NZ QS NS OT ** ** ** on ey to do it

At this point, several more words and phrases become apparent. You should be able to finish this example on your own.

]]>- The Gromark shift for each letter is calculated from a 5 digit primer from which a running key is formed by addition of successive pairs of digits and dropping tens. The Ragbaby shift is found from the position of the letter in the plain text (Both word and letter position).
- The Ragbaby has a single alphabet for both plain and cipher text. The Gromark has independent plain and cipher alphabets positioned over one another. However, the Gromark plain alphabet is always in standard form (i.e., abcde…xyz).

As with the Ragbaby, the major problem with solution of a Gromark is placing the crib in the cipher text. This placement provides a start on determining the Cipher alphabet.

Using JA03 E-06 as an example (And assuming you can do the running key arithmetic), we start with:

Key: 64317

Plain: abcdefghijklmnopqrstuvwxyz Cipher: **************************

64317 07487 71254 83792 10613 16744 73181 04991 43805 71852 89370 JHOIB CKYIV HCRZO IOQTW WLJYB FMUTS CLSAL BCSRG OTNEJ RGQGW ADLAE ***** ***** ***** ***** ***** ***** ***** ***** ***** ***** ***** 72077 92746 19107 00177 01847 19218 01398 14279 56964 15505 6 JDZAM IEXNO WNWMV TYVCN AIYOP IGZJK IZBEV MSFHT SPEPO RLCYC P ***** ***** ***** ***** ***** ***** ***** ***** ***** ***** *

The crib is the word ‘college’.

Unlike the Ragbaby, the Gromark has no simple way to place the crib. All positions are equally likely and each needs to be evaluated. This is done for each position as follows:

- The plaintext letters in the crib each have a unique position in the plaintext alphabet. List out these values. For ‘college’, these values are: c=3, o=15, l=12, l=12, e=5, g=7, e=5.
- For each position in the cipher text, calculate the position of the cipher letter in the cipher alphabet by adding the plaintext letter position to the positions shift value. Consider the following examples:

Position 1 pt: c o l l e g e p#: 03 15 12 12 05 07 05 sh: 6 4 3 1 7 0 7 C#: 09 19 15 13 12 07 12 CT: J H O I B C K Position 32 pt: c o l l e g e p#: 03 15 12 12 05 07 05 sh: 3 1 8 1 0 4 9 C#: 06 16 20 13 05 11 14 CT: L S A L B C S Position 35 pt: c o l l e g e p#: 03 15 12 12 05 07 05 sh: 1 0 4 9 9 1 4 C#: 04 15 16 21 14 08 09 CT: L B C S R G O Position 62 pt: c o l l e g e p#: 03 15 12 12 05 07 05 sh: 2 7 4 6 1 9 1 C#: 05 22 16 18 06 16 06 CT: E X N O W N W

- Each position is evaluated and placed into one of the following three classifications:
- Inconsistent: The position is inconsistent with the placement of the plain text. This can happen for one of two reasons. In the first case, the same cipher alphabet position is shown to have two different cipher letters. Position 1 is an example of this category (i.e., CT position 12 has both B and K assigned to it). In the second case, the same cipher letter has two different cipher alphabet positions assigned to it. Position 32 is an example of this (i.e., CT letter S is assigned positions 14 and 16).
- Neutral: This position is not inconsistent with the placement of the plain text. This is due to the fact that all the cipher text letters at that position are different and so are all the calculated cipher letter positions. Position 35 is an example of this category (i.e., No identical cipher text letters and no identical calculated cipher alphabet positions). This position must remain a candidate for placement of the cipher text, but the evidence is weak.
- Consistent: This position is consistent with placement of the cipher text. Repeated cipher text letter all have the same calculated cipher alphabet positions with no other inconsistencies. Position 62 is an example of this category (i.e., The cipher text letters N and W are repeated and have the same positions values in both cases (16 and 6 respectively)). Because the probability of such coincidences is low, this type of position is a far more likely placement for the crib than a ‘not inconsistent’ position.

For the example here, the total number of positions in each category breaks down as follows:

Inconsistent: 82 Neutral: 16 Consistent: 02

Of the two consistent positions (62 and 99), No. 62 has two consistencies while No. 99 has only one. Therefore, the first guess for the position of ‘college’ is at position No. 62.

From the evaluation of position 62 we get the following cipher alphabet positions:

E = 05, X = 22, N = 16, O= 18, W = 06.

This provides us with:

Key: 64317

Plain: abcdefghijklmnopqrstuvwxyz Cipher: ****EW*********N*O***X****

64317 07487 71254 83792 10613 16744 73181 04991 43805 71852 89370 JHOIB CKYIV HCRZO IOQTW WLJYB FMUTS CLSAL BCSRG OTNEJ RGQGW ADLAE **o** ***** ****n *o**d e**** ***** ***** ***** n*he* ****d ****e 72077 92746 19107 00177 01847 19218 01398 14279 56964 15505 6 JDZAM IEXNO WNWMV TYVCN AIYOP IGZJK IZBEV MSFHT SPEPO RLCYC P ***** *coll ege** ****i ***n* ***** ***v* ***** **v*n ***** *

The string of cipher text around position 42 suggests the plain text ‘the’. Based on this, we can find the position of the cipher text letter T. The pt position of ‘t’ is 20. The Shift at position 42 is 3. Therefore, ‘T’ is at position CT = 20 + 3 = 23.

This provides:

Key: 64317

Plain: abcdefghijklmnopqrstuvwxyz Cipher: ****EW*********N*O***XT***

64317 07487 71254 83792 10613 16744 73181 04991 43805 71852 89370 JHOIB CKYIV HCRZO IOQTW WLJYB FMUTS CLSAL BCSRG OTNEJ RGQGW ADLAE **o** ***** ****n *o*nd e**** ***s* ***** ***** nthe* ****d ****e 72077 92746 19107 00177 01847 19218 01398 14279 56964 15505 6 JDZAM IEXNO WNWMV TYVCN AIYOP IGZJK IZBEV MSFHT SPEPO RLCYC P ***** *coll ege** w***i ***n* ***** ***v* ****n **v*n ***** *

The string we just modified now suggest the phrase ‘in the’ or ‘on the’. Trying the first option, we can find the position of cipher text letter G. The pt position of ‘i’ is 09. The Shift at position 40 is 1. Therefore, ‘G’ is at position CT = 09 + 1 = 10.

This provides:

Key: 64317

Plain: abcdefghijklmnopqrstuvwxyz Cipher: ****EW***G*****N*O***XT***

64317 07487 71254 83792 10613 16744 73181 04991 43805 71852 89370 JHOIB CKYIV HCRZO IOQTW WLJYB FMUTS CLSAL BCSRG OTNEJ RGQGW ADLAE **o** ***** ****n *o*nd e**** ***s* ***** ****i nthe* *i*ed ****e 72077 92746 19107 00177 01847 19218 01398 14279 56964 15505 6 JDZAM IEXNO WNWMV TYVCN AIYOP IGZJK IZBEV MSFHT SPEPO RLCYC P ***** *coll ege** w***i ***n* *a*** ***v* ****n **v*n ***** *

The string starting at position 16 suggests the word ‘founded’. This will provide:

The pt position of ‘f’ is 06. The Shift at position 16 is 8. Therefore, ‘I’ is at position CT = 06 + 8 = 14.

The pt position of ‘u’ is 21. The Shift at position 18 is 7. Therefore, ‘Q’ is at position CT = 21 + 7 = 28 = 2 (Mod 26).

The pt position of ‘d’ is 04. The Shift at position 22 is 0. Therefore, ‘L’ is at position CT = 04 + 0 = 4.

This provides:

- Key: 64317

Plain: abcdefghijklmnopqrstuvwxyz Cipher: *Q*LEW***G***I*N*O***XT***

64317 07487 71254 83792 10613 16744 73181 04991 43805 71852 89370 JHOIB CKYIV HCRZO IOQTW WLJYB FMUTS CLSAL BCSRG OTNEJ RGQGW ADLAE **om* ***f* ****n found ed*** ***s* *a**c ****i nthe* *ited **a*e 72077 92746 19107 00177 01847 19218 01398 14279 56964 15505 6 JDZAM IEXNO WNWMV TYVCN AIYOP IGZJK IZBEV MSFHT SPEPO RLCYC P ***** ecoll ege** w***i *m*n* ma*** n**v* ****n **v*n *y*** *

The string starting at position 45 suggests the words ‘United States’. This will provide:

- The pt position of ‘u’ is 21. The Shift at position 45 is 5. Therefore, ‘J’ is at position CT = 21 + 5 = 26.
- The pt position of ‘n’ is 14. The Shift at position 46 is 7. Therefore, ‘R’ is at position CT = 14 + 7 = 21.
- The pt position of ‘s’ is 19. The Shift at position 51 is 8. Therefore, ‘A’ is at position CT = 19 + 8 = 27 = 1 (Mod 26).
- The pt position of ‘t’ is 20. The Shift at position 52 is 9. Therefore, ‘D’ is at position CT = 20 + 9 = 29 = 3 (Mod 26).

This provides:

Key: 64317

Plain: abcdefghijklmnopqrstuvwxyz Cipher: AQDLEW***G***I*N*O**RXT**J

64317 07487 71254 83792 10613 16744 73181 04991 43805 71852 89370 JHOIB CKYIV HCRZO IOQTW WLJYB FMUTS CLSAL BCSRG OTNEJ RGQGW ADLAE t*om* ***f* **s*n found edt** ***s* *a*sc ***li ntheu nited state 72077 92746 19107 00177 01847 19218 01398 14279 56964 15505 6 JDZAM IEXNO WNWMV TYVCN AIYOP IGZJK IZBEV MSFHT SPEPO RLCYC P sa*t* ecoll ege** w***i am*n* ma*y* n**v* ****n **v*n ty*** *

The string starting at position 59 suggests the words ‘The College of William and Mary’. This will provide:

- The pt position of ‘h’ is 08. The Shift at position 60 is 7. Therefore, ‘M’ is at position CT = 08 + 7 = 15.
- The pt position of ‘f’ is 06. The Shift at position 70 is 7. Therefore, ‘V’ is at position CT = 06 + 7 = 13.
- The pt position of ‘i’ is 09. The Shift at position 72 is 0. Therefore, ‘Y’ is at position CT = 09 + 0 = 09.
- The pt position of ‘l’ is 12. The Shift at position 74 is 7. Therefore, ‘C’ is at position CT = 12 + 7 = 19.
- The pt position of ‘d’ is 04. The Shift at position 80 is 7. Therefore, ‘P’ is at position CT = 04 + 7 = 11.
- The pt position of ‘r’ is 18. The Shift at position 83 is 2. Therefore, ‘Z’ is at position CT = 18 + 2 = 20.

This provides:

Key: 64317

Plain: abcdefghijklmnopqrstuvwxyz Cipher: AQDLEW**YGP*VIMN*OCZRXT**J

64317 07487 71254 83792 10613 16744 73181 04991 43805 71852 89370 JHOIB CKYIV HCRZO IOQTW WLJYB FMUTS CLSAL BCSRG OTNEJ RGQGW ADLAE t*om* s*eff *rson found edth* *i*s* la*sc *o*li ntheu nited state 72077 92746 19107 00177 01847 19218 01398 14279 56964 15505 6 JDZAM IEXNO WNWMV TYVCN AIYOP IGZJK IZBEV MSFHT SPEPO RLCYC P satth ecoll egeof willi amand mary* ns*ve n***n *even tynin e

The cipher is now essentially solved. Only 6 cipher text letters remain unplaced. A perusal of the uncovered plain text quickly shows that the text begins with the words ‘Thomas Jefferson’ and ends with the words ‘in seventeen seventy nine’. With this info, the positions of the remaining cipher text letters can be uncovered.

This provides:

Key: 64317

Plain: abcdefghijklmnopqrstuvwxyz Cipher: AQDLEWFBYGPHVIMNKOCZRXTSUJ

64317 07487 71254 83792 10613 16744 73181 04991 43805 71852 89370 JHOIB CKYIV HCRZO IOQTW WLJYB FMUTS CLSAL BCSRG OTNEJ RGQGW ADLAE thoma sjeff erson found edthe first lawsc hooli ntheu nited state 72077 92746 19107 00177 01847 19218 01398 14279 56964 15505 6 JDZAM IEXNO WNWMV TYVCN AIYOP IGZJK IZBEV MSFHT SPEPO RLCYC P satth ecoll egeof willi amand maryi nseve nteen seven tynin e]]>

The Homophonic cipher uses the numbers from 01 to 100 to produce 4 independent cipher alphabets of 25 characters each (i/j are represented by single value). The four alphabets span from 01-25, 26-50, 51-75 and 76-100 respectively. Each alphabet is a simple shift of the standard alphabet. The keyword is given by the values of 01, 26, 51, and 76. For example, consider the keyword “BLUE”. The cipher alphabets would be given by:

Letter/Alphabet 01-25 26-50 51-75 76-100 A 25 41 47 97 B 01 42 58 98 C 02 43 59 99 D 03 44 60 100 E 04 45 61 76 F 05 46 62 77 G 06 47 63 78 H 07 48 64 79 I/J 08 49 65 80 K 09 50 66 81 L 10 26 67 82 M 11 27 68 83 N 12 28 69 84 O 13 28 70 85 P 14 30 71 86 Q 15 31 72 87 R 16 32 73 88 S 17 33 74 89 T 18 34 75 90 U 19 35 51 91 V 20 36 52 92 W 21 37 53 93 X 22 38 54 94 Y 23 39 55 95 Z 24 40 56 96

Each letter can be represented by one of four numbers. For example, “e” can be replaced with 04, 45, 61, and 76. This allows a word to be enciphered in many different ways. For example, the word ‘grey’ can be replaced with 06 32 61 95 as well as 78 73 45 23.

The advantage of this system is that the frequency of individual letters can be masked by a judicious use of the multiple ‘homophones’. The disadvantage of the system is that the recovery of the single plain text equivalent for a single number is all that is required for the recovery of an entire alphabet.

The example we will be working on is JA10 E-01:

53 73 39 84 02 71 30 36 17 30 65 74 80 21 17 84 02 91 80 63 53 22 84 27 91 02 34 04 30 15 58 44 97 14 91 57 57 07 14 76 30 19 34 38 32 90 33 70 90 85

Note that we are provided a Caesar shifted crib. We are told that the word XUS = “day” exists in the plain text.

Normally, this cipher would be attacked by a frequency analysis of the four cipher alphabets involved. Essentially, the high frequency cipher text numbers would be found and their plain text equivalents identified by analysis of their relative position within each alphabet. This method is covered by LEDGE in “Novice Notes”. In this case, however, we are given a crib. This can make the analysis much simpler.

The crib is only three letters long. A simple probabilistic analysis would indicate that the chances are about 2 in 3 that two or more of the letters in the crib will be enciphered in the CON by the same alphabet. This indicates that some pattern may exist in the cipher text that we can search for. Our job is to determine the possible patterns and then hunt for them.

For example the first two letters in the crib are ‘d’ and ‘a’. If these two letters were enciphered by the same cipher alphabet, this would mean that we would have two sequential cipher text numbers that differed by a value of either 3(if the cipher alphabet was keyed with e-z or a) or 21 (if the cipher alphabet was keyed with b-d). Similar relations ships exist for the d-y and a-y combinations being enciphered by the same cipher alphabet.

If we search through cipher using the d-a relationship we find only one matching candidate sequence at position 53:

73 70 7

This would provide the relationships d = 73, a = 70, and y = 7. This allows us to tentatively recover the 1-25 and 51-75 alphabets as follows:

Letter/Alphabet 01-25 26-50 51-75 76-100 A 09 * 70 * B 10 * 71 * C 11 * 72 * D 12 * 73 * E 13 * 74 * F 14 * 75 * G 15 * 51 * H 16 * 52 * I/J 17 * 53 * K 18 * 54 * L 19 * 55 * M 20 * 56 * N 21 * 57 * O 22 * 58 * P 23 * 58 * Q 24 * 60 * R 25 * 61 * S 01 * 62 * T 02 * 63 * U 03 * 64 * V 04 * 65 * W 05 * 66 * X 06 * 67 * Y 07 * 68 * Z 08 * 69 *

Note that this indicates that the key word would be of the form: S*G*.

This fragment seems like reasonable English, so we substitute the two candidate alphabets into the cipher and get:

53 73 39 84 02 71 30 36 17 30 65 74 80 21 17 84 02 91 80 63 53 22 84 27 91 I D * * T B * * I * V E * N I * T * * T I O * * * 02 34 04 30 15 58 44 97 14 91 57 57 07 14 76 30 19 34 38 32 90 33 70 90 85 T * V * G O * * F * N N Y F * * L * * * * * A * * 38 74 73 70 07 34 47 80 82 19 * E D A Y * * * * L

The recovered plain text looks a lot like English. We appear to be on the right track. Note that the word “funny” is now apparent in the second line. This provides the identity u = 91. We have now recovered three of the four alphabets. This provides:

Letter/Alphabet 01-25 26-50 51-75 76-100 A 09 * 70 97 B 10 * 71 98 C 11 * 72 99 D 12 * 73 100 E 13 * 74 76 F 14 * 75 77 G 15 * 51 78 H 16 * 52 79 I/J 17 * 53 80 K 18 * 54 81 L 19 * 55 82 M 20 * 56 83 N 21 * 57 84 O 22 * 58 85 P 23 * 58 86 Q 24 * 60 87 R 25 * 61 88 S 01 * 62 89 T 02 * 63 90 U 03 * 64 91 V 04 * 65 92 W 05 * 66 93 X 06 * 67 94 Y 07 * 68 95 Z 08 * 69 96

Note that this provides the partial key word: S*GE.

Substitution into the cipher provides:

53 73 39 84 02 71 30 36 17 30 65 74 80 21 17 84 02 91 80 63 53 22 84 27 91 I D * N T B * * I * V E I N I N T U I T I O N * U 02 34 04 30 15 58 44 97 14 91 57 57 07 14 76 30 19 34 38 32 90 33 70 90 85 T * V * G O * A F U N N Y F E * L * * * T * A T O 38 74 73 70 07 34 47 80 82 19 * E D A Y * * I L L

Several partial words are now apparent in the plain text. Any of these can be used to recover the remaining alphabet. The remainder of this decryption is left as an exercise for the student.

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