While messing around with some geometric transposition and inversion ideas with the Z340, I have come across something odd that I hope somebody else can explain.
If you make a 9 x 9 square at the top left corner of the cipher, and start at the zodiac symbol, spiraling up and counterclockwise(ignoring the four 4×4 squares), the 17 characters of the selection create very strange statistics in AZD.
z^+FtMMqj(9fdp_b|
This string makes the IOC plummet (0.00735) while making the score jump up to above 41,000????
AZD stalls on these letters for the symbols:
"SFINALLYTHEPRODUC"
I have not been able to reproduce this kind of outcome with any different selections of the cipher, but I’m sure I’m just missing something. I am hoping that Jarlve or somebody else can explain why this is happening?
Hey Jelberg,
Multiplicity = symbols / length. The string you suppose has 16 symbols and a length of 17 and therefore a multiplicity of 0.9411764705882353. In the best case AZdecrypt can handle ciphers with multiplicities around 0.5 but not much higher. Furthermore, ciphers with a multiplicity of > 0.7 typically cannot have a unique solution because of the weak constraints imposed by the very few repeating characters. Since there are almost no constraints AZdecrypt will fit in the highest scoring n-grams it can find resulting in a very high score.
To get more feel for this open up the 408 in AZdecrypt and solve it. Now remove one row and solve it again. Continue this process until the program fails. Use 6-grams or higher if you want to solve a cipher with multiplicity > 0.3.
Another way to look at this is: You selected 17 symbols and there was only one repeat. That is why the IoC is so low, since it is measuring repeats.
We can ask this question: What are the odds of selecting 17 symbols that have only one repeat?
We can answer by running an experiment:
1) Shuffle the cipher
2) Pick the first 17 symbols
3) Calculate multiplicity
4) Track count for each value of multiplicity
Result for 1,000,000 shuffles:
1.0=63795
0.9411765=214596
0.88235295=304786
0.8235294=243716
0.7647059=122286
0.7058824=40382
0.64705884=8884
0.5882353=1384
0.5294118=161
0.47058824=10
The results show that 214,596 shuffles produced the same multiplicity as your extraction of symbols. That implies there is a 21% chance that random selection of symbols will behave the same way as your extraction of symbols. Or 28% if you include the 63,795 shuffles that had multiplicity of 1.0 (i.e., they have no repeating symbols at all).
This suggests there isn’t anything unusual about the behavior you’ve observed in your selection of symbols.
Jarlve & doranchak
Thank you both, that makes much more sense to me now. But your answers make me think of another question.
so words like:
(nameless & salesman)
(stakes & steaks)
will have the same multiplicity scores?
Is there any way of measuring whether such small segments are any more "likely" to be part of a correct transposition? Would you just measure the pieces against other random Tetris like selections from other areas in the cipher?
Or am I just missing it completely?
so words like:
(nameless & salesman)
(stakes & steaks)
will have the same multiplicity scores?
Yes, nameless & salesman have the same multiplicity. Also stakes & steaks. However, it is usually about the multiplicity of the cipher, not that of the plaintext.
Is there any way of measuring whether such small segments are any more "likely" to be part of a correct transposition?
Yes, there is. First you need some measurement for natural language. One carries out a scoring on the basis of ngrams. You take a lot (very, very much!) of plain text and divide it into nGrams. Example:
THISISATEST
nGrams of length 5:
THIS
HISI
ISIS
SISA
ISAT
SATE
ATES
TEST
Now it is counted how often which nGram occurs. For example, in English texts, "OFTHE" is pretty much the most common. "XKAHF", on the other hand, is probably less common. A presumed plain text can now also be broken down into such nGrams and evaluated according to the scoring carried out above. For a single word this is relatively meaningless, but in relation to long texts it is very easy to determine whether a text is nonsense or corresponds to a natural language. This is how our Auto-Solvers work.
In more detail:
viewtopic.php?f=81&t=3907#p62802
Would you just measure the pieces against other random Tetris like selections from other areas in the cipher?
If the transposition is only partially correct and the result is disrupted, the method shown above works only to a limited extent. The closer the transposition is to the original transposition, the higher the achieved points will be. However, this has limits. If only single words or sentences are correct, you won’t find out this way.
Translated with http://www.DeepL.com/Translator
Would you just measure the pieces against other random Tetris like selections from other areas in the cipher?
If the transposition is only partially correct and the result is disrupted, the method shown above works only to a limited extent. The closer the transposition is to the original transposition, the higher the achieved points will be. However, this has limits. If only single words or sentences are correct, you won’t find out this way.
I see, that makes sense. Would measuring the similarities of the symbols in the high scoring areas give any indication of homophones?
If there was another area with similar symbol combinations would you be able to reverse engineer the transposition in those areas?
for example:
enlarge
I took the first 340 of the Z408 and made a route to transpose it with.
