Why and how the text could be Bavarian

[JoJo_Jost]

The weirdest line in the VMS solved..

I now have a name for my Solving-Theory, and its abbreviation works in both German and English: The Vowel Bridge Model (VBM) (das Vokal Brücken Modell) .

In another thread testing a solution, oshfdk pointed out the weird lines in the VMS and asked if this solver could handle them. Among them was this line:

<f75r.13,+P0>    pchedy.keedy.qokedy.<->qokedy.qokedy.qokedy.qokain.olshedy

Four consecutive qokedy! I was curious what this looked like—could I solve it using the VBM pattern, that is, using VCCV

The sequence starts with the word “keedy.qokedy”; the last y in keedy belongs to qo (Vocal Brigde). So the vowel bridge is: “y.qo”

I’ll assume here that y.qo is an “e.” (Although “y.qo” = e is extremely likely overall.) The last vowel is also a y.qo from qokedy.qokain. (If kaiin = "k" aiin, then the last letter of the chain would also have to be "k.")

The chain is therefore actually a bit longer and, according to my VBM method, can be broken down as follows:

y.qokedy.qokedy.qokedy.qokedy.qo (k)

y.qo = "e"
The “ked” in qokedy is:
k    = C1 = "n" 
ed = C2 = "d"

Let’s break this down into the VCCV sequence

y.qokedy.qokedy.qokedy.qokedy.qo (k)
 V  CC   V  CC    V  CC   V  CC  V    (C )
 e  nd    e  nd    e   nd   e  nd   e  (n)

Is there such a concatenation? Is that realy possible? I don’t think so in many languages, but in German it is:

written together, with correct spaces:

end  endenden  de (n)

Here’s a completely normal German sentence to illustrate:

"Ich denke oft an den im Elend endenden, den keiner retten konnte".

“I often think of the one who ended in misery, whom no one could save.”

There you have it—the complete chain in a normal sentence. This proves that such a hardcore repetition is indeed (theoretically) possible within my Vocal Bridge Model—and the other, less extreme repetitions, fit into the model accordingly. In fact, it’s more accurate to say that repetitions are a typical feature of the model.

Jost

PS: Unfortunately, the VMS is likely a bit more complicated, since the VV establish relationships with the CC, and C1 also establishes relationships with C2, making such a simple substitution unlikely. For now, I am assuming a homophonic, position-dependent cipher whose individual components depend on the corresponding ones. Nevertheless, the example impressively demonstrates how easily such repetitions can be solved using the VBM.