It seems like some of the even rows score higher here.

Yes, and also in general there appears to be more randomness in the even rows.

From page 57:

Rows, row 14 again with a massive 67% of the randomizations being higher than the original. For comparison, in the 408, the last row scores 48%. Also notice a bias for randomization in even rows:

I started work on a cycle hill climber this morning named "Restore homophone cycles". Which will work just like any other solver module in AZdecrypt, though it uses perfect cycles as a measurement instead of ngrams. I am not sure yet how well this will work but it is an exciting project.

Here is a very early preview of it on the 408. Scroll down the code box until you see "Changes:", notice how it finds the distribution near the bottom (second half) of the cipher. Right now it seems to know where the randomness is but not exactly. From that information a heatmap could be built that hilights areas of probable encoding randomness. It may also tell us how well the 340 responds to this treatment.

Score: 22845.17 Ioc 1: 0.01829 Ioc 2: 0.88040
Pefect cycles: 85 Score: 13120
Match: 0.94852

9%P/Z(UB%kOR=pX=B
WV+eGYF69HP@K!qYe
MJY^UIk7qTtNQYD5)
S(/9#BPORAU%fRlqE
k^LMZJdrpFHVWe8Y
@+qGD9KI)6qX85zS(
RNtIYElO8qGBTQS#B
Ld/P#B@Xq=HMU^RRk
cZKqpI)Wq!85LMr9#
BPDR+z=6N(eEU/kF
ZcpOVWI5+tL)l^R6H
I9DR_TYrde/@XJQA
P5M8RUt%LSNVEKH=G
rI!Jk598LMlFA@Z(P
zUpkA9#BdW+VTtOP
^=SrlfUe67DzG%%IM
Nk)KcE/9%%ZfAP#B(
peXqHq_F#8c@9A9B
%OT5RUc+_dYq_^SqW
VNeGYKE_TYA9%#Lt_
HkFBl9zXAD)7L!=q
_ed##6e(PORXQF%/c
Z@JTtq_8JI+rBpQK6
V^Xr9WINqEHM)=UIk

Changes:

     (           
                 
                 
                 
                 
                 
                 
         =       
                 
     z        /  
                 
                 
         S       
           F @   
        d        
                 
   K            (
    H           
                 
 N               
 k  l     )      
       (       / 
             p K 
 ^     N         

That is a very interesting project. I am looking forward to seeing how it works. Let me know if I can make any messages for you, my encoder allows for random symbol selection 0-100% by position, so I can, say, make perfect cycles except for any certain cells which have 100% randomness. Like a row, column, or even a shape. Thanks for you work!

Okay smokie,

For some reason the 340 is responding different than other ciphers: the hill climber is creating allot of small perfect cycles that are very indentical in length.

Thanks for your offer. Do you think you could make a cipher or 5 with the following properties?

– Same plaintext. No transposition, wildcards or anything.
– Perfect cycles but with increasing or decreasing key efficiency throughout the ciphers.

If needed, I will add the randomization myself.

Four messages with the same plaintext, no transposition or polyphones. All perfect cycles, except that symbol selection is 100% random for certain positions which make up a large shape that you should recognize. Some of the symbols in the shape may be in cyclical sequence, but most should not be. From highest key efficiency to lowest key efficiency. What is the famous shape?

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

23 30 24 28 11 29 25 31 32 26 37 18 44 12 41 45 33
13 5 14 7 1 56 49 15 27 52 23 50 51 42 34 57 8
20 16 56 38 24 53 25 49 35 43 46 11 17 57 39 54 21
2 40 28 26 30 31 27 37 19 60 23 32 9 18 3 36 12
24 38 55 22 13 16 41 47 48 14 50 52 6 15 7 4 56
51 11 34 1 39 25 51 55 21 12 35 42 50 54 10 2 40
19 13 46 43 57 51 3 37 27 36 4 30 41 17 1 30 31
55 42 29 27 30 33 49 43 35 14 52 21 24 38 18 19 24
58 15 50 35 13 53 22 12 35 41 51 54 52 20 47 25 30
31 26 37 18 12 62 45 15 47 15 39 8 13 24 54 23 49
12 59 13 37 5 14 53 54 11 46 55 21 3 38 19 11 52
53 24 39 18 64 42 56 47 48 43 7 28 49 41 16 17 61
25 54 22 3 19 23 46 32 53 20 12 6 13 51 52 45 4
47 53 42 16 27 54 27 51 53 22 1 52 60 22 14 40 24
9 25 15 26 61 27 33 30 5 11 48 12 6 43 46 37 23
38 44 2 47 3 10 24 8 11 4 39 9 1 31 32 53 20
14 25 21 2 58 15 29 26 33 30 11 10 60 27 31 32 5
12 7 41 34 13 35 64 50 33 3 59 14 51 23 61 24 30
31 40 42 54 18 25 58 15 63 43 57 36 64 37 4 34 11
6 12 8 1 56 49 13 63 41 57 60 26 32 33 55 48 64

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

26 32 27 29 13 30 28 33 34 26 38 20 44 14 41 44 32
15 4 16 7 1 54 48 13 27 51 28 49 50 42 35 55 8
23 17 56 39 26 52 27 48 36 43 45 14 18 54 40 53 24
2 38 31 28 33 34 26 39 22 58 27 32 10 22 3 37 15
28 40 51 25 16 19 41 46 46 13 49 52 5 14 9 1 55
50 15 35 2 38 26 50 52 24 13 36 42 49 51 11 3 39
20 13 45 43 56 48 1 40 26 37 2 33 41 17 3 34 32
52 42 29 28 34 34 48 43 36 14 53 24 27 38 21 22 27
57 15 49 36 13 51 25 13 36 41 50 52 52 23 46 28 32
33 26 40 20 13 61 44 15 45 14 40 8 16 27 53 28 48
14 57 15 38 5 16 52 53 14 45 51 24 3 39 21 14 52
53 26 40 22 63 42 54 46 46 43 8 30 49 41 18 19 59
27 51 25 2 20 26 45 34 53 23 15 5 16 50 53 44 3
46 51 42 17 26 52 28 50 52 24 2 51 60 25 13 38 28
12 26 14 27 58 28 32 33 5 15 47 16 6 43 45 39 26
40 44 2 46 3 10 27 9 16 1 38 11 2 34 32 52 23
14 28 24 3 57 15 31 26 33 34 16 12 59 27 32 33 4
13 7 41 35 14 36 63 49 34 1 57 15 50 28 60 26 32
33 39 42 53 21 27 57 16 64 43 55 37 62 40 2 35 13
5 14 8 3 56 48 15 63 41 54 58 28 34 32 51 47 64

Four messages with the same plaintext, no transposition or polyphones. All perfect cycles, except that symbol selection is 100% random for certain positions which make up a large shape that you should recognize. Some of the symbols in the shape may be in cyclical sequence, but most should not be. From highest key efficiency to lowest key efficiency. What is the famous shape?

Thanks, though, the cycle hill climber is not yet restoring the cycles correctly and therefore it has little chance of drawing out the right shape with a heatmap. I have called your ciphers smokie59abcd if that is alright. In randomization tests column 9 comes up and possibly row 10. Perhaps your shape is a cross.

Follow up on:

For some reason the 340 is responding different than other ciphers: the hill climber is creating allot of small perfect cycles that are very indentical in length.

Please, bare with me for a moment while I try to explain something.

Here follows an example (1) of the p1 plaintext encoded using a highly efficient key. By unigram frequencies, it has a raw ioc of 1551, 0.01345 normalized and a flatness of 0.96384 (a flatness of 1 would denote that all symbols have the same frequency). Looking at perfect cycles, it has a raw ioc of 17780 (cycle score), 0.15517 normalized and a flatness of 0.99193 (this means that the perfect cycles are all very identical in length). There are 207 cycles with an average length of 9.74.

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

And the following example (2) is a less efficient key, matching that of the 340 ioc wise. By unigram frequencies, it has a raw ioc of 2237, 0.01940 normalized and a flatness of 0.66827. And by perfect cycles, it has a raw ioc of 9104, 0.07945 normalized and a flatness of 0.63193. There are 122 symbol cycles with an average length of 7.38.

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

Both example 1 and 2 cycle perfectly, but because of differences in key efficiency some stats change. As key efficiency goes down, the unigram ioc increases. And especially note the difference between cycle flatness of example 1 (0.99) and example 2 (0.63). This is a consistent behaviour with key efficiency, as the key efficiency goes down, both the unigram and cycle flatness decrease.

So with the 340 we could generally assume a less efficient key since it has a reasonably high ioc. While the cycle hill climber does not correctly restore cycles it does seem to latch on to the original cycle statistics. For the 340 – contrary to its ioc and assumed lower key efficiency – these are a high cycle flatness, and many short length cycles. It shows more or less that a highly efficient key was used, which is the observation or main point that I am suggesting.

But how do you cram an efficient key into a cipher with a higher ioc? It doesn’t make sense. I don’t exactly know but it seems reasonable to suggest that a higher than normal amount of the symbols are 1:1 substitutes or something similar. The following could be a hypothesis, instead of using all letters to encode a cipher, only use the top 10 (about) most frequent and let the unused letters blend in with the cipher. It does not explain the encoding randomness but that may be another thing entirely.

Follow up on:

I don’t exactly know but it seems reasonable to suggest that a higher than normal amount of the symbols are 1:1 substitutes or something similar. The following could be a hypothesis, instead of using all letters to encode a cipher, only use the top 10 (about) most frequent and let the unused letters blend in with the cipher. It does not explain the encoding randomness but that may be another thing entirely.

I made such a cipher, but without encoding randomization and it does look good in some regards. For one it matches the 340 unusually high unigram distance: viewtopic.php?f=81&t=3444. Though I swapped out 1:1 substitutes for null symbols. The cipher started out with a 17 by 14 plaintext (238 characters) then encoded with an efficient key to 54 symbols. Then, another 9 high frequency symbols were added randomly throughout the cipher. Bigrams are low (27) and the cipher doesn’t solve.

1  2  3  4  5  6  7  8  7  9  10 11 10 12 13 14 15
16 5  17 9  18 19 20 21 18 2  10 22 23 24 9  4  25
26 27 28 29 30 31 1  32 33 34 35 4  27 25 36 10 37
6  27 38 5  3  39 10 40 10 9  11 28 5  41 42 40 43
8  44 27 9  45 40 46 47 48 5  49 20 10 16 40 7  15
6  12 13 30 38 50 51 1  10 6  2  52 53 9  54 35 19
3  45 55 36 40 22 56 21 57 58 10 10 24 32 14 59 23
29 25 26 27 33 8  41 10 49 16 11 28 9  60 47 19 35
10 34 32 31 52 54 5  20 38 10 61 6  22 43 42 27 33
26 48 48 45 15 40 48 41 49 12 21 37 54 13 2  4  39
34 7  30 12 13 28 42 35 24 46 36 1  16 20 17 50 3
40 62 29 32 41 8  55 48 47 19 35 21 33 59 14 60 14
36 57 20 5  6  11 48 10 10 2  12 40 15 14 6  48 38
53 9  5  22 6  63 10 40 18 24 4  58 29 45 25 6  8
30 31 5  4  10 14 10 1  28 48 19 62 20 22 10 16 20
4  23 24 39 14 46 29 43 55 47 26 10 38 61 8  10 59
60 3  45 20 17 40 20 6  64 5  34 10 4  27 57 58 48
42 25 7  32 21 44 56 51 10 11 31 48 40 36 49 50 20
15 43 13 39 14 47 19 23 5  4  22 33 46 18 54 57 55
34 37 30 59 10 1  28 5  60 9  36 26 31 51 47 24 16

Since that became an interesting hypothesis all of a sudden I wrote a new hill climber for AZdecrypt and cracked my own cipher very easily:

Score: 25390.34 Ioc: 0.06515

Removed symbols: XFIPG_;&] (9)

ILIKEKILLINGPEOPL
EBECAUSEITISSOMUC
HFUNITISMOREFUNTH
ANKILLINGWILDCAME
INTHEFORRESTBECAU
SEMANISTHEMOSTDAN
GEROUSANIMALOFALL
TOKILLSOMETHINGGI
VESMETHEMOSTTHRIL
LINCEDPERENCEITIS
EVENBETTERTHANGET
TINGYOURROCKSOFFW
ITHAGIRLTHEBESTPA
RTOFITISTHATWHEN

I am glad that you were able to detect the cross. It is a big circle cross hairs symbol.

Smokie, is this the kind of thing you are talking about?
I found the attached image at: http://unsolved.cba.pl/index.php?topic=7.0

Smokie, is this the kind of thing you are talking about?
I found the attached image at: http://unsolved.cba.pl/index.php?topic=7.0

Yes, but I have never seen that particular image before. The 340 is cyclic, meaning that the symbols appear together in groups and cycle with each other, similar to the I LIKE KILLING message, but not quite as cyclic. Jarlve is trying to figure out why, I think, with a computer program that can detect cycle discrepancies. I made some messages for him where the symbols are perfectly cyclic, except for the positions that make a bid circle cross hairs. I just did it for fun to see if he could detect the shape, and he did detect the cross, but not the circle.

The Zodiac 340 has some very interesting statistics regarding the symbols and their positions with relation to each other, such that any random assortment of the same symbols would never be the same. However, that is about all we know. A great deal of work has been done to try to figure out what Zodiac did, but unfortunately, we are still stumped. Jarlve has made some very significant findings and computer programs to help with the effort. The cross hairs symbols in the messages that I made were just for fun.

I am glad that you were able to detect the cross. It is a big circle cross hairs symbol.

Early preview of smokie59d heatmap with my new encoding randomization test: https://drive.google.com/open?id=0B5r0r … 0hmQTJYRlU

I am glad that you were able to detect the cross. It is a big circle cross hairs symbol.

Early preview of smokie59d heatmap with my new encoding randomization test: https://drive.google.com/open?id=0B5r0r … 0hmQTJYRlU

thats pretty cool the cross really stands out and is it a big circle because you can just make one out if it is..

Here are some more encoding randomization maps, the algorithm scans through the cipher periodically by rows and columns and herein lies its bias.

340: https://drive.google.com/open?id=0B5r0r … 2YtWHRudVU
408: https://drive.google.com/open?id=0B5r0r … m5oWXp0Nnc
smokie59a: https://drive.google.com/open?id=0B5r0r … DVWaVNVaDg
smokie59b: https://drive.google.com/open?id=0B5r0r … U94NVBrVkk
smokie59c: https://drive.google.com/open?id=0B5r0r … EhocW5IdzA
smokie59d: https://drive.google.com/open?id=0B5r0r … XlhajZMR28

Smokie, could you make another cipher, perfect cycles but with a recurrent unchanging shape (such as a pivot) placed here and there throughout the cipher that is random to the encoding. Then, share the cipher and the exact shape and I will try to extract its positions. Thank you in advance.

Jarlve can you do that heat map in two halves just to see what it looks like, two rows of 10 please.

Oh and can you do the heat map without the last row in the 408 and also without row 14 in the 340

Jarlve can you do that heat map in two halves just to see what it looks like, two rows of 10 please.

Oh and can you do the heat map without the last row in the 408 and also without row 14 in the 340

Mr lowe,

I am glad you asked for these. Removing row 14 from the 340 now turns the spotlight exactly in the middle of the other horizontal pivot fragment. If you ask me it looks more and more credible that the horizontal parts of the pivots are made up structures by the Zodiac as he went through the encoding process.

340 part 1: https://drive.google.com/open?id=0B5r0r … Uowdi1mTDQ
340 part 2: https://drive.google.com/open?id=0B5r0r … XI4dkdBT0U
340 row 14 removed: https://drive.google.com/open?id=0B5r0r … jNzRTNSUnc

408 part 1: https://drive.google.com/open?id=0B5r0r … E5JNUFNeGc
408 part 2: https://drive.google.com/open?id=0B5r0r … UdJZGdHQ0U
408 part 3: https://drive.google.com/open?id=0B5r0r … ElWN1lzVnM
408 row 24 removed: https://drive.google.com/open?id=0B5r0r … TQ0bDNSLWs