Key is in NARA RG92E623

[Juskevicius]

Look, you're completely missing the reality of the situation. Do you even realize just how many ways numbers can be used in the creating of a cipher? “A lot of different ways!” This is your first problem right out of the gate.

Take c1, per example. It contains a much wider range of code and also 19 four-digit codes. It could “very-easily” be those 19four-digit codes represent words that form a single sentence, all the other remaining code simply being random ghost code.

The problem is that “you are assuming way too much” in the “practicing of your in-theory process that just by its very nature the practice contains many uncertainties and flaws when there exist, and here it comes, no known means of the applied coding process.” And there could be dozens, if not hundreds of different ways, to use numbers in the drafting of a cipher. And the two remaining ciphers already possess strong suggestive evidence that they have indeed been drafted differently.

Without the “means” you are simply left to take wild stab, or “desired stab” as to the applied coding process right from the very start.

And I might add, since you doubted my knowledge of these things, this was just pulled from online resources,

“The coefficient of variation (CV) is the ratio of the standard deviation to the mean. The higher the coefficient of variation, the greater the level of dispersion around the mean. It is generally expressed as a percentage.” Or, in the case of uncertain or unknown conclusion, such as the remaining two ciphers, “probability in theory.”

I see you are much smarter than me, so I have nothing to answer you. I'd rather wait until someone checks the book NARA RG92E623.

 

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[bigscoop]

I see you are much smarter than me, so I have nothing to answer you. I'd rather wait until someone checks the book NARA RG92E623.

Look, a few post ago you suggested that, “I see, I you do not understand The Coefficient of Variation. The aim of it is not about of probabilities.” To which I then replied with an “exact definition” of the purpose of Coefficient of Variation which clearly outlines its intended practice as being a process to better understand probabilities and percentages.

Now, if what you are saying is true, that, “The aim of Coefficient of Variation is to expose a distinction of one dataset from another dataset or group of datasets.” Then it MUST also be true that you are developing data to compare with another so you can arrive at, and here it comes, “probabilities.”

Probabilities are not exact in nature, hence, “Probably will” and “probably won't” or “probably can” and “probably can't” Probabilities are based on percentages that lead to conclusions of likelihoods. Probabilities are in no way, exacting.

Having said all the above, now let us set aside all of your prior work and let's take the “required time and effort” to examine and compare each of the three ciphers. One we have a provided clear text for, the other two we do not. So let's us first see what is possibly different about these two remaining rascals that still challenge and completely miff all those would-be hopeful decoders today.

Well, C3, a smaller range of code with a lot more repeated code. That's certainly a difference that sets it apart from the other two.

C1, a much broader range of code and even 19 four-digit codes, the only such cipher to contain these four-digit codes other than C2 which contained only one as representative of the letter “x” because the DOI contained no words beginning in “x”. But here we suddenly have 19 of these four-digit codes? If we apply the same accepted coding practice then we would what, have a cipher with 19 four-digit codes that all represent a letter that the key didn't have? Maybe 19 different codes representing the letter “x”? Well, neither is very likely, is it? So why the sudden and uncharacteristic urge to use 19 four-digit codes unless there was a change in the coding practice or entire coding method, which is pretty darn apparent by their sudden presence.

So my “uneducated” point is simply this, even if your dataset was exactly the same or exactly the opposite of the one being compared to, neither of those summations do anything whatsoever in helping to decode the two remaining ciphers. So what you are attempting to do is entirely pointless right out of the gate because we have no way of knowing exactly how the two remining ciphers have been constructed. All we know for “certain” is that they don't conform to the method applied in C2. “And that is an already established FACT!”

I don't understand what you're doing or hoping to ultimately achieve as you're just spinning your wheels and wasting your time in the most “improbable” way.

 

[Juskevicius]

I don't understand what you're doing or hoping to ultimately achieve as you're just spinning your wheels and wasting your time in the most “improbable” way.

I like your determination. Keep going, find your way. I give you a green light. Meanwhile, I do nothing and waiting until someone checks the book NARA RG92E623.

 

[bigscoop]

I like your determination. Keep going, find your way. I give you a green light. Meanwhile, I do nothing and waiting until someone checks the book NARA RG92E623.

Unlike you, I'm nowhere near a newbie.....lol