I stayed one week in Duisburg for work a few months ago, the food was really cheap!

I wanted to answer this post for ages. Are you in Germany more often? If so, let me know next time, maybe we can meet if you like.

I stayed one week in Duisburg for work a few months ago, the food was really cheap!

I wanted to answer this post for ages. Are you in Germany more often? If so, let me know next time, maybe we can meet if you like.

Thanks for asking Largo. I have stayed in Germany only twice, so very rarely. I kept asking the bartender at the hotel for "orangensaft" the whole week and nothing else. At the end of the week he replied "ich habe das gehabt mit deinen orangensaft!", hehehe.

Gutter Deuttscher Orrangensafft

(couldn’t resist)

QT
We lost the war

Gutter Deuttscher Orrangensafft

(couldn’t resist)

Ah, sieh an…ein Österreicher! Mir ist das im Profil irgendwie nie aufgefallen, hast Du das erst vor Kurzem aktualisiert?
Naja, Orangen sind Südfrüchte, also nichts wirklich authentisches aus Deutschland, wenngleich sehr lecker!
Vor einigen Jahren war ich mal in Salzburg und habe dort den Rosengarten besucht, wirklich sehr schön! Schöne Sache, dass Leute aus aller Welt zusammenarbeiten, um z340 vielleicht eines Tages zu lösen. Ich bin alle paar Monate mal in Neuburg an der Donau, also gar nicht sooooo weit weg von der österreichischen Grenze. Leider ist Wien dann trotzdem noch etwa fünf Stunden entfernt, ansonsten hätte man sich mal treffen können, schade.

PS: http://www.deepl.com/translator is the best translator I know so far! (gutter deuttscher transslattor)

Largo-

Ich bin total vom Thema abgewichen, aber dieser Sprachübersetzer ist fantastisch. Also vielen Dank dafür!

-glurk

FYI –

I posted Largo’s font in a new topic to make it easier to find:

viewtopic.php?f=81&t=3660&p=57441

(Thanks, glurk, for the suggestion)

One of the assumptions why z340 cannot be solved is a possible error in a transposition. I have used an example for my tests as a basis for which I proceeded as follows:

– Write the plaintext into a 20×17 grid
– Perform a diagonal transposition (starting at top left)
– Result: Many texts will have a P19 spike afterwards.
– Convert to 17×20 and perform homophonic encryption
– Result: In some cases, P19 is retained. Depends on the plain text and the key.

AZDecrypt quickly and reliably solves these encryptions with its transposition solver. The situation is different when you force an error. For example, if you forget one or two letters when transferring the diagonal transposition into a 17×20 grid, you get an encryption that is very difficult to solve. The following example shows how diced the plain text is after untransposition:

Plaintext

ASIWASGOINOVERTHE
CORKANDKERRYMOUNT
AINSISAWCAPTAINFA
RRELLANDHISMONEYH
EWASCOUNTINIFIRST
PRODUCEDMYPISTOLA
NDTHENPRODUCEDMYR
APIERISAIDSTANDAN
DDELIVEROHORTHEDE
VILHEMAYTAKEYAITO
OKALLOFHISMONEYAN
DITWASAPRETTYPENN
YITOOKALLOFHISMON
EYYEAHANDIBROUGHT
ITHOMETOMOLLYSHES
WORETHATSHEDLOVEM
ENONEVERWOULDSHEL
EAVEMEBUTTHEDEVIL
TAKETHATWOMANYEAH
FORYOUKNOWSHETRIC

After untransposition with error

ASIWASGOINEVERTNDCOR
KMOUKERRSISANTAIAINF
AWCAPANDHISRRELHEWAS
COMONEIFIRSTPRUNTINM
YPISTOLADUCERPRODUCE
DMNDTELIISAIDSTANYRA
ILHEVEROHORTHDANOKAL
LMAYTAKEYAEDEDITWAOF
HISMONEITOYITOOSAPRE
TTYPYANNEYYEKALLOFHI
SENNGHTITAHANDIBROMO
LYSHEHOMETOMOLUTSHED
SWORETHAOONEVELOVEME
NHSHELEAWOULDABUTTHE
VEMEYILTAKEDEVNATWOM
ATHTYEAHFONLRYOUKNYN
OWSEDHETHERIPICDDWVR

However, AZDecrypt is simply "uber" and in many cases it finds a solution though. However, in this case it really takes a long time. If three or four errors occur, AZDecrypt will no longer be able to find the solution (at least not automated). So I generated a lot of brute-force test cases. In the first round I had 340 test cases in which a "+" at position 1-340 was inserted to repair a possible error. The same I have done with 2 possible missing symbols. For all these test cases I then reversed the transposition described above. In a third test I have untransposed P2 to P170 instead of the diagonal transposition. This gave me several million test cases, which I solved step by step with AZDecrypt by batch file. I looked at all results manually that scored over 20900. However, there weren’t many of them, so the additional work was limited. This all happened overnight and the next morning there were only a few ciphers to sort out. The "diagonal" test was executed completely, but then I stopped the period test. I had hoped somehow that one of the results might at least give a hint to get closer to the solution. Unfortunately, that was not the case.

Even if there is an error in the size of 1-2 symbols in z340, you can’t go any further because you don’t know which (and if) transposition was used. So with brute force, you’re wasting your time here. Do one of you have any idea how to combine the missing symbol hypothesis with the transposition solver via hillclimber? Or is the idea itself doomed to failure?

For now, I am giving up the idea of "ordinary" transpositions. I’m working on a new little tool that might be helpful for everyone, but more about it later. I also have a few ideas on how to trick AZDecrypt with simple pen/paper methods, although very simple transpositions are used. More about this later, too, as soon as I’m sure I don’t tell any crap. Spoiler: Do not think in grids

Translated with http://www.DeepL.com/Translator

Largo:

I made a heatmap that sometimes detects the position of a skipped symbol in a message similar to the 340. Slide through the message, at each position add a null and push the other symbols over one position, then calculate the repeats. Then rank each position by the count of repeats. The idea is to use a combination of detection and brute force, starting with the positions more likely to be where a skipped symbol is.

Here is a message at P20. I typically use a 17×20 inscription rectangle for symplicity.

15 36 19 5 9 23 20 57 28 1 37 51 52 7 38 10 53
54 39 24 24 11 12 11 40 48 48 13 2 9 27 21 12 3
32 16 60 11 30 15 43 20 6 9 29 19 36 44 8 4 11
12 45 33 17 36 38 1 13 51 47 14 2 26 3 14 55 62
12 34 13 15 5 35 23 39 52 18 4 20 53 46 6 35 7
1 54 8 51 9 41 14 2 11 58 51 3 2 9 11 11 53
62 3 12 9 36 37 21 7 54 13 17 52 14 5 6 28 57
8 22 38 2 42 3 38 15 29 11 19 20 51 24 5 5 2
28 29 25 11 26 10 33 55 35 43 23 10 31 4 11 60 16
11 2 52 14 36 50 35 32 37 10 24 23 59 7 60 11 33
30 26 53 40 5 21 48 34 11 58 18 55 41 12 8 44 11
3 35 17 8 54 22 45 61 23 41 39 49 56 15 32 55 11
14 25 59 51 6 14 11 19 12 13 3 33 36 16 52 28 9
54 46 5 34 1 14 4 37 29 3 10 49 31 6 38 60 43
48 44 45 20 39 4 18 54 43 5 15 9 21 11 11 56 14
35 1 11 26 52 13 26 39 17 8 14 2 62 51 60 38 7
32 47 10 16 52 33 39 22 48 30 31 11 29 43 34 49 13
3 51 52 11 51 55 56 12 18 35 13 48 52 4 35 36 37
9 19 33 44 60 38 47 1 48 29 45 14 49 30 53 39 34
20 54 46 41 11 11 31 2 43 11 44 12 50 50 45 3 36

There is a skipped symbol, which I just put at the end. Red is the count of P20 repeats with a null inserted at that position. Purple is rank. Here, at P20, the rank for the correct position, which is yellow, is only 83. So in this case the idea isn’t perfect, but maybe it would cut down on the number of messages to try to solve.

But then I do that with P40, P60, P80, and P100 and add up all of the repeat counts and rank that. In this case, the idea worked dead on and found the correct position.

That’s as far as I have gone with this idea. Maybe more multiples of the period would help, but I am not sure. It is not always accurate, but could cut down on the number of messages to try to solve.

Thank you very much for the images and the explanation smokie!

Slide through the message, at each position add a null and push the other symbols over one position, then calculate the repeats.

Are you talking about counting bigram repeats?

My brute force approach did the same thing, except that I just sent all generated ciphers blindly through AZDecrypt without looking at the bigrams. If I understand it correctly, you’ll evaluate each step using the bigram repeats. Is that correct?
If so, then I am not sure that this procedure is sufficient. I think it generates a lot of false positives and false negatives.I have reviewed lots of books with my ACME statistics tool and found that even common plaintext, which has been substituted 100% cyclically, can have only 20 or less bigramms. Perhaps a combination of bigram count, IOC and cycles is an idea.
I am currently working on a new tool, but not more than one hour per day. Other things currently take absolute priority. But as soon as the tool is finished, I will try to build a filter based on your idea into peek-a-boo.

Translated with the help of http://www.DeepL.com/Translator

Yes, above I skipped a plaintext at row 13, column 17. Then I slid through the message adding a null at each position and counted the period 20, 40, 60, 80, and 100 repeats. The top picture shows that going by just period 20 is not that accurate in this example, as row 13 column 17 ranked only 83 out of 340. I left out the pictures for P40, P60, P80 and P100. But I added all repeats for P20-100 in the summary below, and row 13, column 17 ranked #1.

Like I said, it is not always accurate. Sometimes it will work well for some periods, while not for others, and sometimes it will work for the summary. I probably should do some testing to find out how much weight to give to each period, and how many periods to go. Unfortunately detecting such things is very difficult with this much diffusion.

By the way, what I’ve always wanted to ask: You say you work exclusively with Excel. Do you use VBA as a programming language or do you do it all using the "normal" Excel functions? I use Excel only occasionally and then only for very basic tables. So I’m pretty impressed what you’re doing with it.

I just use normal functions, and keep them as short as possible. I like to break out the calculations across a lot of cells and sheets in a logical direction to keep things easy to follow. No macros. It is limited to be sure, but I like the visual aspect. Back in the 1980’s when I was a kid I had a commodore 64. I didn’t have any money, so wrote a lot of my own video games in basic. I would like to learn Javascript, but time is limited.

However, AZDecrypt is simply "uber" and in many cases it finds a solution though. However, in this case it really takes a long time. If three or four errors occur, AZDecrypt will no longer be able to find the solution (at least not automated). So I generated a lot of brute-force test cases. In the first round I had 340 test cases in which a "+" at position 1-340 was inserted to repair a possible error. The same I have done with 2 possible missing symbols. For all these test cases I then reversed the transposition described above. In a third test I have untransposed P2 to P170 instead of the diagonal transposition. This gave me several million test cases, which I solved step by step with AZDecrypt by batch file. I looked at all results manually that scored over 20900. However, there weren’t many of them, so the additional work was limited. This all happened overnight and the next morning there were only a few ciphers to sort out. The "diagonal" test was executed completely, but then I stopped the period test. I had hoped somehow that one of the results might at least give a hint to get closer to the solution. Unfortunately, that was not the case.

Yeah, the transposition solver works really well in allot of cases. So well that I wonder wether the 340 could even be a simple unkeyed transposition without additional difficulties. About nulls, it is easy to solve up to 5 skips with a specific hill climber and did so a while ago, though I only considered 1-dimensional periodical transposition at period 15 and 19.

A continuation from the Cycle types thread here:

viewtopic.php?f=81&t=3616&start=50

z340:

top 10 l2 isomorphic sequences (7grams) (10000 shuffles):
sequence count sigma
ABABABA 222.0 5.844441900977058
ABAABAA 220.0 5.242584748143618
ABAAABA 206.0 4.686882212721511
ABAABAB 162.0 4.272889915964004
ABABABB 150.0 3.519110242921718
AABAABA 187.0 3.4713584344638413
ABBABAB 147.0 3.2571199850212813
ABABAAB 145.0 3.126041816538682
ABABBAB 143.0 3.1028297955864357
ABABAAA 174.0 2.7589370221075473

These were the isomorphic patterns found. I consolidated and sorted all of them, regardless of length. The picture is tiny because of all of the different patterns, but I will post more in upcoming days, zooming in on the spikes or groups of spikes that have the patterns and continuations of patterns shown in red above, the four 7 gram patterns with the highest count, except that you can continue the pattern into longer ones that make spikes or groups of spikes.

Then I will be going somewhere with it to compare all of the symbols considering several other interesting patterns. It should hopefully be interesting.

ABAB spikes

ABAABA spikes