Amsco Tutorial

I’m going to use JA03 E-11 as the basis for this tutorial. The crib is that the word ‘proudly’ is in the plain text. In AMSCOs, the plain text is broken up alternately into single and double letters. Therefore, the crib must have been broken up in one of the following ways:

*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.