## A Few Up Front Questions

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 Cipher System:

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.

### Example Encipherment

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.

### A Few Important Notes

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.

### Example Bifid Solution:

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

#### Bifid. HiPointer and Capiti Tribute Cache. MSCREP

VCUHKKS OAVEVTP ECMWSAF ERRBKTP RAAOAKP VIRFHPA ANTVTPO

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

### A Few Salient Notes:

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.”

### Key Square Recovery: Methods

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

### Key Square Recovery: Practice

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

### Key Square Recovery: Round 1

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:

### Key Square Recovery: Round 2

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.)

### Key Square Recovery: Round 3

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:

### Key Square Recovery: Round 4

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:

### Key Square Recovery: Round 5

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

## Substitution and Analysis: Method

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.

### Substitution: Round 1

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!

### Analysis: Round 1

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.

### Substitution: Round 2

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.

### Substitution: Round 3

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**

### Analysis: Round 3

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.

### Substitution: Round 4

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

### Analysis: Round 4

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.

### Substitution: Final

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.`

### Analysis: Final

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.”

# Variations on a Theme: The Twin Bifid

### The Twin Bifid Cipher System:

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.

#### Another Example (Twin Bifid):

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

## The Twin Bifid

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.

### The Twin Bifid: Solution

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.

### Example Twin Bifid Solution:

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

#### Twin Bifids. Ancient times. [First 21 letters same] OZ

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 Cipher Blocks:

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

#### Equate the Letter Coordinates:

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

#### Key Square Recovery

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 -

#### The Proto-Key Square

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?*

#### Key Reconstruction

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’.

#### Substitution

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…

# Variations on a Theme: The Conjugate Matrix Bifid

## The CM Bifid Cipher System:

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.

### Another Example (CM Bifid)

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

#### The CM Bifid

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.

#### The CM Bifid: Solution

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.

### Example CM Bifid Solution:

#### NN-2. Conjugated Matrix Bifid. LEDGE

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)

### Key Square Recovery

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

### Equate the Letter Coordinates:

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

### Key Square Recovery

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

#### The Proto-Key Squares

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 * * * * * * * *

#### Substitution

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**** ****

#### Analysis

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.