Porta Cipher

The Cryptogram SO 2014


The MA 2013 Cm Porta provides an excellent example of Period Determination and crib placing procedures at our disposal.

MA 2013. Porta. Dubious honor.  TSIOLKOVSKY





Our ACA and You Guidelines prompts us to be aware that the Porta construction uses only thirteen alphabets and contains a Period which must be no less than ten or more than fifteen lines deep.

The first half of the alphabet (A-M) can only represent a plaintext letter from its second half.” A” cannot stand for B or C and N cannot stand for O or P. This allows for some relatively simple crib placements.

Porta Slide

We will illustrate the Porta slide usage described in Novice Notes by Gerhard D. Linz (LEDGE) to make the Porta construction process clearer.


“The horizontal line separates the two parts of the slide. The * on the upper half locates the two possible key letters on the lower part. The slide is set in the AB position. To use it, substitute for a letter on the upper part the letter standing opposite to it on the lower and vice versa. Thus if the key is A or B, A = N, B = O, etc. If the key were C or D, we would move the lower slide one space to the left, creating a new set of substitutes.”

Our ACA and You Period definition allows us to define the MA 2013 Porta as either a five or six Period length for this 66 letter construction and the thirteen alphabet limitation allows us to search for its crib (success) in the opposite half of the alphabet from each of the crib letters.

Locating the crib ciphertext by alphabet position

S – First half (1)

U – First half (1)

C – Second half (2)

C – Second half (2)

E – Second half (2)

S – First half (1)

S – First half (1)

We’re looking for a string of ciphertext letters appearing in these halves of the alphabet – 1, 1, 2, 2, 2, 1, 1. You will find this cipher construction unique to only the following ciphertext letters in this Porta:

s u c c e s s

Of course, these ciphertext construction letters will follow each other whether our Period length is five or six. It becomes our task to determine the proper Period by examining the plaintext for each Period length.

Note that the placement of success under ciphertext FDURSMF forces the period to be six, because the plaintext letter “s” has to line up under both ciphertext letters “F.”  If the period were five then F and M would both equal s in the same column

After posting the ciphertext letters horizontally into six columns, note the resulting key letters generated by the Porta slide placement of the ciphertext / plaintext letters to the top of each column. Two letters will appear for each column (AB in example to the left).   Anagramming one of the two letters of each column into the cipher’s keyword will generate a keyword of MAILED or NAILED.