At the risk of quoting things from the book of the bleeding obvious, here are my thoughts on z340 using a grid transposition cipher (in addition to homophonic)

I always considered transposition ciphers to be the poor cousin of substitution, after all, there are all the plain text letters to see – you’ve just got to rearrange them. The interesting thing is when I came to actually try and use one, and it proved slightly harder than I thought.

Say for simple example your plain text message was “abcdefghijklmnopq”

You pick a secret grid size – say 7 x 3, and write in the message left to right, top to bottom.

The problem is, your message is a little shorter than the 21 available squares, so you leave them blank. An alternative is to fill them with gibberish or a fixed character like x, but lets not do that.

To encipher, you transpose this into a different size grid, say 4×6.

Form the source grid, you start in column 1 reading down, and fill the values into the output grid left to right. So starting with a,h,o then b,i,p and c,j,q. However, when you come to column 4 you skip the blanks, only writing in d,k followed by e,l,f,m and g,n.

The result looks like this.

This would fit into a 4×5 grid, but I’ve selected a 4×6 to allow for a whole additional line of null symbols to be included at the end.

This output has an interesting property similar to the skip-19 cycle for z340 bigrams.

If you read out every 3rd character left to right you get…

abcdlgXXhijkfnXXopqemXXX

This output has sections that are correctly sequenced (abcd, hijk, opq) but overall you don’t recover the plain text. If this were further enciphered with a homophonic cipher, skipping 3 would recover a lot of bigrams, but not all of them.

Is it possible the z340 is a simple vertical to horizontal grid transposition from a source grid with 19 rows, but due to nulls at the end of the plain text it is necessary to switch to skipping every 18 characters at some point because of the shorter source columns?

CK