I will take a look at it soon.

looking forward to seeing it, sounds interesting, I really like the simplistic pen and paper methods of transposition.

LRTB I see a small spike at P5, but nothing at P19. RLTB I see a small spike at P2, but not at P19. I transposed the message at 90 degrees, mirrored that, flipped it and mirrored and flipped it, but no spike at P19.

There is one pivot. Is that a naturally occurring product of the cipher? A vigenere with a period of 16 can cause a propensity to make pivots like the one in the 340, but also a lot more P16, P32, P48, P64, etc. unigram repeats than in the 340 or your message.

Top row is number of L=2 consecutive alternations ("CA"), bottom row is 340 LRTB.

06 07 08 09 10 11 12 13 14 15 16 17 18 19 20
72 37 21 03 04 03 01 01 00 00 00 00 00 00 00

Your message LRTB. Fewer of 6, 7, and 8 CA, no of 11, 12 or 13 CA, but one really long cycle at 18 CA.

06 07 08 09 10 11 12 13 14 15 16 17 18 19 20
54 18 12 05 04 00 00 00 00 00 00 00 01 00 00

I haven’t tried to solve it. A little later I would like to know how you made the pivot. Thanks.

Hi smokie,

thanks for checking out my test cipher. I have double-checked it again and I definitely see a bigram peak at P19. On P1 I have 19 repeated bigrams, on P19 it is 33.
Tomorrow or the day after tomorrow I will describe my approach (I need some time). I will also post some ideas about the "broken" cycles and bigrams in z340, I hope there is something new in it. Just now I am pretty euphoric as I feel I have found some possible explanations.

Mr lowe: I like pen & paper methods too. Some are easy to use but still very effective.

PS: I don’t think it is possible to solve my example cipher without knowing the method. Here at least some hints (just scroll down).

The + symbols form a line after untranspose, as they are fillers.
Transposition uses a keyword, but it is not Keyed Columnar. Not in the usual way, anyway.

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

I have been having a rest from the 340 for a few months but its interesting that a while ago I ended up having a (bad) partial solve but had five + signs at the end which intrigued me. now I have to go look for it in my saved collection. I think it was from 4X4 spiralling squares.

This could be entirely my error, not sure. The spreadsheet recognizes lower and upper case letters as the same, and does some weird things with math symbols and colons. But knowing that I converted all symbols to numbers with the code() formula to begin with. This is what I got:

100 97 58 43 61 61 65 120 51 43 97 104 66 35 84 117 114
81 110 99 111 88 68 99 61 49 70 69 106 53 104 110 59 48
99 104 80 43 113 66 108 88 118 105 81 90 61 110 43 98 76
102 53 50 72 85 88 116 58 73 89 104 73 99 112 106 122 100
78 56 75 43 59 117 43 108 89 115 121 59 84 61 43 68 110
83 61 84 70 97 43 115 108 77 111 84 104 85 76 49 71 59
73 61 68 72 88 106 35 66 61 68 75 75 88 84 66 119 57
81 43 51 100 121 35 53 58 68 74 61 43 105 117 55 48 80
58 73 80 115 35 53 59 76 78 84 56 50 70 43 43 119 120
88 59 99 82 122 101 86 90 100 65 43 67 56 61 68 114 113
105 88 90 111 121 89 84 107 51 70 89 50 118 43 49 61 122
112 88 72 43 100 68 55 59 103 89 53 106 82 61 52 88 65
72 115 35 78 113 55 61 43 75 116 77 70 89 72 112 66 72
82 43 108 112 83 110 65 107 113 106 85 108 81 105 48 55 116
76 119 66 90 117 88 104 50 59 86 97 116 56 85 43 84 104
59 102 67 72 43 82 50 90 103 115 97 81 110 61 84 81 43
53 72 112 120 68 59 43 110 100 90 78 76 100 119 84 78 68
121 84 65 101 58 113 78 88 70 121 104 55 105 59 66 116 35
108 86 48 43 61 106 101 82 98 89 99 111 114 108 43 58 67
113 85 88 50 115 90 110 81 50 75 100 80 84 61 102 43 103

VREHTRUFLELMITEVALROE····HIYO·····IWIK·····NINI·····EFTN·····HFETOTHISWURESTINGST

smokie:

When I convert my cipher to numbers using AZDecrypt, I get the same result as you. Seems fine to me.
I’ts end of the work day now and have a little time to write a generator for this kind of cipher. It’ll take a while, though. If you want, I’ll post the solution and my new ideas for z340 in advance.

Zresearch:
?

Edit: Translation error

Thanks for checking that. Do you get a spike at P19 with the numbers? Go ahead and post. I am interested in your idea.

smokie can you show it with a heat map and largo can you show the pivot please.. very keen to see your method largo, was it from a book or is it your own.
cheers

Do you get a spike at P19 with the numbers?

Yes. You can copy your numbers into AZDecrypt and click "Format -> Convert to characters" and then "Stats 1 -> Plaintext direction: ngrams". There is a spike at P19.

largo can you show the pivot please.. very keen to see your method largo, was it from a book or is it your own.

Of course:

I found the method in the following book (I have the German version of the book btw)
https://www.amazon.de/Codes-Ciphers-Sec … 579124852/

The method I use is called "triangle or trapezoid transposition". A description and an online tool can be found on the following pages:
http://kryptografie.de/kryptografie/chi … sition.htm
http://kryptografie.de/kryptografie/chi … sition.htm

By the way: This is a great page. They have lots of interesting cipher stuff!

Unfortunately, everything is in German, but the illustrations should help you to understand the procedure.
With the triangle-transposition there are problems with the length of z340, which is why I excluded it. More interesting is the trapezoidal transposition with a starting width of 5 characters. This produces a result of exactly 341 characters.

This is how my example cipher can be solved:
– Copy my cipher and remove all line breaks in a text editor. This is important because the online tool does not do it automatically. I prepared it for you:

da:+==Ax3+ahB#TurQncoXDc=1FEj5hn;0chP+qBlXviQZ=n+bLf52HUXt:IYhIcpjzdN8K+;u+lYsy;T=+DnS=TFa+slMoThUL1G;I=DHXj#B=DKKXTBw9Q+3dy#5:DJ=+iu70P:IPs#5;LNT82F++wxX;cRzeVZdA+C8=DrqiXZoyYTk3FY2v+1=zpXH+dD7;gY5jR=4XAHs#Nq7=+KtMFYHpBHR+lpSnAkqjUlQi07tLwBZuXh2;Vat8U+Th;fCH+R2ZgsaQn=TQ+5HpxD;+ndZNLdwTNDyTAe:qNXFyh7i;Bt#lV0+=jeRbYcorl+:CqUX2sZnQ2KdPT=f+g+

– Add a "+" so that the length is correct (already done in the prepared text above).
– Paste the cipher into the field "Eingabe".
– Enter the following in the field "Schlüssel / Parameter": "TITWILLOW,5"
– TITWILLOW" is the keyword, "5" is the start width.
– Set the "Operation" dropdown box to "Eingabe entschlüsseln" (Decrypt input).
– Press the "ausführen" button.
– Now the untransposed cipher with 24 "+" symbols at the end should be shown in the field "Ausgabe".
– Copy cipher into AZDecrypt, press "Solve". Should solve now.

Now my ideas about z340 in connection with the shown transposition:

Frequent occurrence of "+"-symbols in z340
The trapezoid transposition explains the many "+"-symbols in z340 when they are used as fillers at the end of the trapezoid. I created some test ciphers and couldn’t deduce a pattern of the "+"-symbols from the first attempt. Yet I have not checked the following, but at first glance the trapezoid produces exactly what Jarlve found:

The average of the position numbers for all occurrences of the ‘+’ symbol is 171, which is only one position from the midpoint of the cipher. This suggests the + symbols are very uniformly distributed throughout the ciphertext. (Source: Jarlve)

Pivots
According to my analyses, the trapezoid transposition has no particular tendency to produce pivots. It doesn’t rule them out either. For testing I have analyzed 5998 plaintexts with my C#-Library. These are the results:

Plaintext, not transposed:

1425 have 1 pivot.
276 have 2 pivots.
45 have 3 or more pivots.

After trapezoid transposition, no substitution:

1054 have 1 pivot.
117 have 2 pivots.
17 have 3 or more pivots.

Plaintext, random shuffle. Just for comparison:

1526 have 1 pivot.
309 have 2 pivots.
49 have 3 or more pivots.

After trapezoid transposition, 100% cyclic substitution (homophonic):

11 have 1 pivot.
0 have 2 pivots.
0 have 3 or more pivots.

After trapezoid transposition, 25% random cycles (homophonic):

25 have 1 pivot.
0 have 2 pivots.
0 have 3 or more pivots.

After trapezoid transposition, 100% random cycles (homophonic):

30 have 1 pivot.
0 have 2 pivots.
0 have 3 or more pivots.

Conclusion:
The more cyclical a homophonic substitution, the fewer pivots appear. But that would have to be checked again with more test data. Trapezoid transposition has no particular pivot tendency, but does not rule out pivots. At the end of this post I’ll give you an idea why z340 has two pivots anyway.

Periodic Bigram-Peaks
My test cipher has 19 repeated bigrams on P1 and 33 repeated bigrams on P19. To check if this was just a coincidence, I encoded all 5998 plaintexts using the method shown and created graphs. These show how many ciphers have a peak on which period (P1 to P29)

100% cyclic:

25% random:

Conclusion:
Most 100% cyclical ciphers have a bigram peak at period 1, but also some at period 19, which I find very interesting. Even more interesting is the behavior of ciphers that were randomly substituted by 25%. There are peaks in P11 and P19, but I yet have no idea where it comes from.
Period 1 often has very few bigrams. My example encryption shows this clearly.

TODO: Distance between P1 and period n must be taken into account. This is very high at P1/P19 in z340 and only therefore significant.

Possible explanation of the incomplete cycles in z340
Now, I’d like to drift a little bit further into conjecture and hypotheses. Here are my ideas:

Zodiac read in the newspaper that his first encryption was cracked very quickly. As far as I know, the newspaper article also mentioned how the Hardens managed to make it so quickly: they were looking for patterns. Let’s say Zodiac actually decided to use some kind of transposition in z340. Then he was perhaps aware that the resulting text will have fewer "patterns". I deliberately say "patterns", because he certainly didn’t worry about bigrams and trigrams. Especially not about period n or other things. Since he knew that analysts would certainly be looking for patterns again, he might have thought:"I’ll give you your patterns and lead you on the wrong track". I imagine that he had a finished transposed text in 17×20 before him. Before the substitution, he simply looked for "patterns" and found pivots by accident. These are substituted first to maintain the pattern. After that he searched for bi- and trigrams and substituted some of them first. He then replaced all remaining letters cyclically. As a result, you have the incomplete cycles that we are currently seeing.
If anyone’s already had this idea, then sorry for a repost. I don’t know the whole forum by memory;)

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

I like the idea but a trapezoid transposition with a starting width of 5 and keyword gives 51 columns = keyed columnar transposition on steroids.

I transposed a simple message using a "spiral cipher" in that last message but it got ruined, I think by my spell check.

So I made a New one:

Vt·evolerpic··rah··nIyne··o··srs·ifcpositfool·stu

The spiral cipher should be very simple to crack, but I made an easier one for fun.

TISIADHTIISSNITREETSIGNSUTFIFWLILOLOIKNOTIWTHNEIAHVTEIEM

I’m looking into the code of this topic now.

I like the idea but a trapezoid transposition with a starting width of 5 and keyword gives 51 columns = keyed columnar transposition on steroids.

But if the keyword is short it can be brute forced since the remaining columns should then just follow up normally?

Keyword:

ZODIAC
653412789111...
         012

Or does the keyword pattern repeat?

Keyword:

ZODIACZODIAC
653412119178...
      21 0