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Unigram distance curiosity

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(@largo)
Posts: 454
Honorable Member
 

Convert the cipher to numbered by appearance. You have 63 symbols numbered from 1 to 63, numerically add 63 to the symbols of the middle 8 rows to make these distinct from the other rows/key. Then transpose. You can do it with AZdecrypt, under functions, manipulation, raise periodic, from: 103, to: 238. 6-grams should do the job.

I didn’t think of that! Clever!
Do you think it is possible to extend the solver for encryptions with two keys? The solver splits the numerical version of the encryption into several parts (rectangles) and increases the numbers within one of the rectangles. Basically the same as you do for region-based transpositions.

Again, the font is so awesome and thanks Largo.

You’re welcome! I’m glad it’s useful to you. :)

 
Posted : December 10, 2017 6:08 pm
Jarlve
(@jarlve)
Posts: 2547
Famed Member
Topic starter
 

It is plausible the center piece is using one encoding, and the top+bottom uses another.

This would definitably be a solvable hypothesis, something that we can check. Though, the question is, how many different arrangements/combinations are we looking at or do we want to consider.

I am working on this and wrote a hill climber that should solve a 2 keyed cipher among any distribution of rows. Example, key 1 occupies rows 1 to 6 and 15 to 20 and key 2 occupies rows 7 to 14. Another example, key 1 occupies the even rows and key 2 occupies the uneven rows (such as smokie5). It worked very well on smokie5 and and the 340 is running now. I would be happy if someone could make a few more test ciphers that are in line with borkky’s hypothesis:

He splits the clear text in two parts, and encodes them separately.
First half with one cipher, the other half with the other cipher, adding filler symbols at the end to produce a neat rectangle.
Then he makes two horizontal cuts, and rearranges the pieces.

AZdecrypt

 
Posted : December 10, 2017 8:48 pm
smokie treats
(@smokie-treats)
Posts: 1626
Noble Member
 

Smokie 66ABC are all P20. I just used a +/-53 symbol key for the middle rows, and the same key but about 10 extra symbols for the top and bottom rows. That’s so I could get a high count of P20 repeats. Cycling is 25% random. There is no homophone, or + symbol, so RAW IOC will be a bit lower than the 340.

A

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

EDIT: B

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

C

9 6 22 37 40 38 10 10 17 30 39 12 27 13 56 28 25
53 33 1 2 34 9 63 10 24 17 37 26 3 31 55 18 35
11 62 23 33 65 59 16 54 1 17 45 38 32 39 23 41 53
7 54 63 19 29 47 37 37 60 2 56 45 16 32 24 3 53
43 14 28 44 11 47 31 42 32 2 40 8 50 20 15 25 38
39 21 28 37 30 38 13 36 9 38 24 43 9 27 43 11 18
46 3 9 35 47 10 19 31 12 11 22 32 54 17 16 9 43
53 21 20 54 19 33 30 28 45 25 19 8 34 26 48 14 43
17 21 5 10 8 49 48 46 24 11 20 39 7 28 25 38 1
18 41 51 2 39 5 3 1 2 41 3 37 29 1 44 2 3
1 9 22 45 10 11 9 38 53 26 10 40 39 31 47 43 37
41 52 54 16 53 39 23 8 53 32 4 39 44 37 11 26 26
28 30 28 9 49 10 21 31 12 42 2 15 45 3 40 1 14
39 26 11 35 32 39 3 17 27 33 19 20 22 28 37 29 30
13 31 2 1 39 35 35 30 9 43 32 33 38 9 34 18 46
31 23 32 24 47 7 11 46 2 12 16 40 44 47 37 15 9
21 17 35 41 40 63 41 19 13 42 56 44 11 45 46 64 10
2 16 23 14 2 20 40 1 11 41 16 42 50 60 16 37 5
38 62 59 65 2 27 56 39 31 23 63 43 44 33 45 9 43
32 37 61 19 3 30 12 18 51 31 58 58 63 22 28 10 23

Let’s see if anyone can determine how many rows in the top and bottom, which are the same, have the key with extra symbols.

 
Posted : December 10, 2017 10:00 pm
(@largo)
Posts: 454
Honorable Member
 

I would be happy if someone could make a few more test ciphers that are in line with borkky’s hypothesis

Here you are:

Largo_Sing_Along

8z0Oap:sI7n;WGkRN
2XvArxKhPlzMS8QVO
nL9:sJ4t3iZNabfBo
uVUx;AF38isL14qyN
lwuMk2erx8o07sVLd
:=r2vpy8e+aCGxTY0
a5K-p7:qPaRdw0PHM
Vit1o3ZGyTFmSIsji
RmbX=uYvZwBp+T;Lj
Ohn:cNS3;yPFQWfA-
0EkZJWUePFxF9T34J
b;z+Q:VA0LFwCyNBP
0UhtDUePFOMonXlaa
zq2U0pmvaeuuVKDx9
awSVgLiucj:37QyCP
L0rd5pmGw8iVHFtp=
Fox3G1jvMWq9SLPY4
FbI+QwCh57OZVpR0:
VRveZTxkuV;3rIXz5
EB=SoF4WfYb1uuOmw
 
Posted : December 10, 2017 10:41 pm
smokie treats
(@smokie-treats)
Posts: 1626
Noble Member
 

Smokie 66ABC are all P20. I just used a +/-53 symbol key for the middle rows, and the same key but about 10 extra symbols for the top and bottom rows. That’s so I could get a high count of P20 repeats. Cycling is 25% random. There is no homophone, or + symbol, so RAW IOC will be a bit lower than the 340.

A

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

EDIT: B

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

C

9 6 22 37 40 38 10 10 17 30 39 12 27 13 56 28 25
53 33 1 2 34 9 63 10 24 17 37 26 3 31 55 18 35
11 62 23 33 65 59 16 54 1 17 45 38 32 39 23 41 53
7 54 63 19 29 47 37 37 60 2 56 45 16 32 24 3 53
43 14 28 44 11 47 31 42 32 2 40 8 50 20 15 25 38
39 21 28 37 30 38 13 36 9 38 24 43 9 27 43 11 18
46 3 9 35 47 10 19 31 12 11 22 32 54 17 16 9 43
53 21 20 54 19 33 30 28 45 25 19 8 34 26 48 14 43
17 21 5 10 8 49 48 46 24 11 20 39 7 28 25 38 1
18 41 51 2 39 5 3 1 2 41 3 37 29 1 44 2 3
1 9 22 45 10 11 9 38 53 26 10 40 39 31 47 43 37
41 52 54 16 53 39 23 8 53 32 4 39 44 37 11 26 26
28 30 28 9 49 10 21 31 12 42 2 15 45 3 40 1 14
39 26 11 35 32 39 3 17 27 33 19 20 22 28 37 29 30
13 31 2 1 39 35 35 30 9 43 32 33 38 9 34 18 46
31 23 32 24 47 7 11 46 2 12 16 40 44 47 37 15 9
21 17 35 41 40 63 41 19 13 42 56 44 11 45 46 64 10
2 16 23 14 2 20 40 1 11 41 16 42 50 60 16 37 5
38 62 59 65 2 27 56 39 31 23 63 43 44 33 45 9 43
32 37 61 19 3 30 12 18 51 31 58 58 63 22 28 10 23

Let’s see if anyone can determine how many rows in the top and bottom, which are the same, have the key with extra symbols.

Please forgive me, I made a mistake and use the wrong transposition matrix.

 
Posted : December 11, 2017 2:02 am
smokie treats
(@smokie-treats)
Posts: 1626
Noble Member
 

Here are the new smokie66ABC messages. The irony is that if I untranspose at period 20 they will solve easily because I only added some symbols to the keys for the top and bottom rows to get the regional biases and high P20 repeats. If you use different keys, then no P20 repeats as we have determined.

A

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

B

44 17 18 39 50 1 2 28 30 57 3 15 19 33 54 20 10
45 6 1 22 2 57 46 20 7 36 37 16 39 53 3 11 59
23 63 34 55 54 41 28 49 42 34 44 56 45 1 45 25 55
26 51 11 55 17 63 40 44 43 33 21 42 13 19 2 3 32
19 12 38 41 20 34 41 56 17 55 57 48 25 59 60 44 57
36 20 21 12 1 41 33 31 14 45 10 11 8 32 39 10 15
25 17 46 26 32 10 11 9 44 45 16 50 38 38 40 1 15
2 34 46 8 10 28 11 42 42 52 11 18 51 44 19 48 45
3 27 4 31 32 17 22 37 3 46 2 3 1 39 10 35 33
22 6 2 41 34 34 7 44 47 10 1 9 13 30 25 29 45
31 11 50 6 24 26 42 48 26 12 18 43 38 39 11 21 9
25 36 19 33 1 2 17 45 34 18 9 8 3 11 5 41 43
19 40 44 14 26 12 4 16 38 5 17 32 11 7 30 45 39
11 46 27 21 43 18 35 11 43 41 27 9 33 40 43 13 38
10 34 11 41 35 12 28 46 12 26 34 34 11 12 42 39 35
2 27 53 41 31 14 37 34 53 42 17 57 58 14 34 39 43
45 11 47 32 12 20 62 6 35 34 23 2 12 21 40 34 46
63 49 35 17 3 15 8 50 59 1 10 61 23 36 30 25 41
7 50 19 56 20 31 11 42 17 45 48 9 32 62 12 55 2
8 14 7 60 22 11 43 59 45 29 30 51 18 46 20 9 19

C

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

 
Posted : December 11, 2017 2:15 am
(@borkky)
Posts: 28
Eminent Member
 

Tonight I extended a little on the widget, added a info box about the "center" area.

https://martinlindhe.github.io/zodiac-widget/

My Z340 widget for the two-ciphers hypothesis: https://martinlindhe.github.io/zodiac-widget/

 
Posted : December 11, 2017 4:32 am
Jarlve
(@jarlve)
Posts: 2547
Famed Member
Topic starter
 

Tonight I extended a little on the widget, added a info box about the "center" area.

https://martinlindhe.github.io/zodiac-widget/

This widget investigates the observation that an unusual number of symbols are missing from the middle rows, as noted by smokie treats

Sorry to bring this up borkky, I can be a little pedantic. If memory serves well then I believe it was me who first made this observation. At the main post of this thread and almost a year earlier in the Route Transposition and Phenomenon thread:

by Jarlve » Thu Dec 29, 2016 6:02 am

Thanks doranchak,

I’ve been looking at the distances between symbols and that these may be related to the encoding. While I was trying to distinguish between palindromic and randomization I stumbled upon another weird thing in the 340. If you square all the distances between the symbols and total it you will find that the 340 scores very highly compared to other ciphers. Here is why.

On average, less frequent symbols will show a greater distance between each other than more frequent occurring symbols, but why is this such a default thing in the 340 with top-bottom symmetry to boot?

AZdecrypt

 
Posted : December 11, 2017 12:15 pm
Jarlve
(@jarlve)
Posts: 2547
Famed Member
Topic starter
 

I would be happy if someone could make a few more test ciphers that are in line with borkky’s hypothesis

Here you are:

Largo_Sing_Along

8z0Oap:sI7n;WGkRN
2XvArxKhPlzMS8QVO
nL9:sJ4t3iZNabfBo
uVUx;AF38isL14qyN
lwuMk2erx8o07sVLd
:=r2vpy8e+aCGxTY0
a5K-p7:qPaRdw0PHM
Vit1o3ZGyTFmSIsji
RmbX=uYvZwBp+T;Lj
Ohn:cNS3;yPFQWfA-
0EkZJWUePFxF9T34J
b;z+Q:VA0LFwCyNBP
0UhtDUePFOMonXlaa
zq2U0pmvaeuuVKDx9
awSVgLiucj:37QyCP
L0rd5pmGw8iVHFtp=
Fox3G1jvMWq9SLPY4
FbI+QwCh57OZVpR0:
VRveZTxkuV;3rIXz5
EB=SoF4WfYb1uuOmw

Thank you, no periodic transposition right? It is running now and may take a while.

AZdecrypt

 
Posted : December 11, 2017 12:18 pm
(@borkky)
Posts: 28
Eminent Member
 

Sorry to bring this up borkky, I can be a little pedantic. If memory serves well then I believe it was me who first made this observation. At the main post of this thread and almost a year earlier in the Route Transposition and Phenomenon thread:

by Jarlve » Thu Dec 29, 2016 6:02 am

Thanks doranchak,

I’ve been looking at the distances between symbols and that these may be related to the encoding. While I was trying to distinguish between palindromic and randomization I stumbled upon another weird thing in the 340. If you square all the distances between the symbols and total it you will find that the 340 scores very highly compared to other ciphers. Here is why.

On average, less frequent symbols will show a greater distance between each other than more frequent occurring symbols, but why is this such a default thing in the 340 with top-bottom symmetry to boot?

Of course credit should be right!

As I’m new here, I found the info from the wiki http://zodiackillerciphers.com/wiki/ind … tions#Z340 under "Unigrams":

Some symbols are exclusive to the first 6 and last 6 rows. An unusual number of symbols are missing from the middle 8 rows. (Source: smokie treats)

I won’t judge who was first with the idea. Are you ok if i write "credit Jarlve & smokie treats"?, since he detailed the separation more?

My Z340 widget for the two-ciphers hypothesis: https://martinlindhe.github.io/zodiac-widget/

 
Posted : December 11, 2017 12:30 pm
Jarlve
(@jarlve)
Posts: 2547
Famed Member
Topic starter
 

Here are the new smokie66ABC messages. The irony is that if I untranspose at period 20 they will solve easily because I only added some symbols to the keys for the top and bottom rows to get the regional biases and high P20 repeats. If you use different keys, then no P20 repeats as we have determined.

Thank you for your ciphers smokie. Are you sure that 2 keys sufficiently diffuse bigrams?

Plaintext 1 to 100, 2 individual keys top/bottom totalling 10000 samples. Ioc mean: 2240.99. Bigrams mean: 34.51.

Ioc:

- Mean: 2240.9998
- Variance: 8288.891199959988
- Standard deviation: 91.04334791713224
- Sigma of 2236: -0.05491669753347465


Bigrams:

- Mean: 34.5131
- Variance: 30.86042839000066
- Standard deviation: 5.555216322520722
- Sigma of 37: 0.4476693355609145


AZdecrypt solve substitution score (1000000 iterations, 10 restarts):

- Mean: 20323.37121536038
- Variance: 1782381.007860658
- Standard deviation: 1335.058428631743
- Sigma of 20351: 0.02069481308614695

AZdecrypt

 
Posted : December 11, 2017 12:37 pm
Jarlve
(@jarlve)
Posts: 2547
Famed Member
Topic starter
 

I won’t judge who was first with the idea. Are you ok if i write "credit Jarlve & smokie treats"?, since he detailed the separation more?

If Z340 ever gets solved, I am sure it will be a collective effort, built on all the work made by many users here & elsewhere.

Perhaps you could just credit the cipher section of this forum? That would embrace the collective effort spirit, in which I believe too.

There is a double entry in the wiki:

If you highlight all occurrences of certain symbols, they seem to avoid the middle of the cipher text. (Source: Jarlve)
Some symbols are exclusive to the first 6 and last 6 rows. An unusual number of symbols are missing from the middle 8 rows. (Source: smokie treats)

AZdecrypt

 
Posted : December 11, 2017 12:47 pm
(@largo)
Posts: 454
Honorable Member
 

I had some thoughts this morning. Sorry if any of this has already been mentioned. In this case simply ignore it =)

I am still irritated by the frequent occurrence of the "+"symbol. If more than one key was used, there are few explanations:
– The "+"symbol maps to the same plaintext letters in both keys.
– The "+"symbol maps in both keys to a frequently occurring plaintext letter (E, T, A, O, I, N etc.).
– It is actually a polyphone (but why should you do that in this case? That seems unrealistic to me)

z340 was written on a rather large sheet that did not fit in the envelope. It must have been folded. If you look at the high-resolution scan, you can see where it was folded. Eventually, Zodiac might have intended this to display regions. I’ll take a closer look at that on occasion.

If different keys were used alternately, this could be the cause of incomplete but not completely absent cycling. Has anyone ever done some tests before? For example, rows 1-6 and 15-20 with key A, then rows 7-14 encoded with key B and check the cycling? If not, I’d like to put that on my todo list. I now simply call this type of encryption "cake encryption" because, as with a cake, there are different "layers". So I don’t always have to refer to rows.

When I needed a new key for testing, I made it relatively easy and used a trick. It also works very well if you work with paper and pencil. That’s how it works:

As you can see, I left the order of the cipher symbols unchanged. Only the plain text letters were swapped in their order. Of course, this has an effect on the efficiency of the key, since the result is no longer "smooth". It would be interesting to know how this affects the cycles of a cake cipher. And the frequent occurrence of the "+"symbol could also be explained by this, since a correlation could take place depending on the key.

 
Posted : December 11, 2017 2:07 pm
(@borkky)
Posts: 28
Eminent Member
 

Perhaps you could just credit the cipher section of this forum? That would embrace the collective effort spirit, in which I believe too.

Sounds good, fixed now!

My Z340 widget for the two-ciphers hypothesis: https://martinlindhe.github.io/zodiac-widget/

 
Posted : December 11, 2017 2:15 pm
(@largo)
Posts: 454
Honorable Member
 

Okay…check this out:

Rows 1-6, followed by rows 15-20:

HERabcdVPeIfLTGgh
Nb+BjkOlDWYmnoKpq
BrstM+UZGWjqLkuHJ
SbbvdcwoVxbO++RKg
yzM+u12hI7FP+34e5
bwRdFcO-ohCeFagDj
rBXfos4zCEaVUZ7-+
ItmxuBKjObdmpMQGg
RtT+Lf#Cn+FcWBIqL
++qWCuWtPOSHT5jqb
IFehWnv1ByYOBo-Ct
aMDHNbeSuZOwAIK8+

Perfect 5-Cycles:

VPDrsVPDrsVPD (72)
PDrsUPDrsUPD (56)
PDrsxPDrsxPD (56)
PDrszPDrszPD (56)
PDrs7PDrs7PD (56)
PDrUxPDrUxPD (56)
PDrU7PDrU7PD (56)
PDrz7PDrz7PD (56)
PDsUxPDsUxPD (56)
PDsU7PDsU7PD (56)
PDsz7PDsz7PD (56)
PrsUxPrsUxP (42)
PrsU7PrsU7P (42)
Prsz7Prsz7P (42)
GkrsUGkrsUG (42)
GDrsUGDrsUGD (56)
DrsUvDrsUvD (42)
DrsUxDrsUxD (42)
DrsUyDrsUyD (42)
DrsU1DrsU1D (42)
DrsU7DrsU7D (42)
DrsU5DrsU5D (42)
DrsvyDrsvyD (42)
Drsv1Drsv1D (42)
DrsxyDrsxyD (42)
Drsx1Drsx1D (42)
Drsx5Drsx5D (42)
Drsz1Drsz1D (42)
Drsz7Drsz7D (42)
Drsz5Drsz5D (42)
Drs75Drs75D (42)
Drs45Drs45D (42)
DrUvyDrUvyD (42)
DrUv1DrUv1D (42)
DrUxyDrUxyD (42)
DrUx1DrUx1D (42)
DrUx5DrUx5D (42)
DrU75DrU75D (42)
Drz75Drz75D (42)
DsUvyDsUvyD (42)
DsUv1DsUv1D (42)
DsUxyDsUxyD (42)
DsUx1DsUx1D (42)
DsUx5DsUx5D (42)
DsU75DsU75D (42)
Dsz75Dsz75D (42)
rsUvyrsUvy (30)
rsUv1rsUv1 (30)
rsUxyrsUxy (30)
rsUx1rsUx1 (30)
rsUx5rsUx5 (30)
rsU75rsU75 (30)
rsz75rsz75 (30)

Perfect 6-Cycles:

PDrsUxPDrsUxPD (72)
PDrsU7PDrsU7PD (72)
PDrsz7PDrsz7PD (72)
DrsUvyDrsUvyD (56)
DrsUv1DrsUv1D (56)
DrsUxyDrsUxyD (56)
DrsUx1DrsUx1D (56)
DrsUx5DrsUx5D (56)
DrsU75DrsU75D (56)
Drsz75Drsz75D (56)

I haven’t quite understood the scoring yet, but compared to the unmodified z340 it seems amazing to me

 
Posted : December 11, 2017 2:16 pm
Page 7 / 9
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