Sequence Transposition Tutorial

Background

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).

Cipher Description

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.

Notes

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.

Cryptanalysis

Consider the following example CON:

xx-01. Sequence Transposition. Check Your Sources. MSCREP

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.