I posted the new zzgrams file earlier, maybe you missed it.

Download zzgrams.zip

All ngrams in the zzgrams file have these:

ABCDE

ZBCDE
AZCDE
ABZDE
ABCZE
ABCDZ

ZZCDE
AZZDE
ABZZE
ABCZZ

Duplicates have been removed afterwards so it may seem some permutations are missing. Averaging won’t be accurate either. If you can’t get it to work this way then I think you should consider removing the wildcards as I did and adjust the ngrams to compensate. As I posted earlier this gave me a couple of amazing solves given the circumstances.

Jarlve,

I did use the zz ngrams on the last experiment. But I didn’t see where there were any with letters between two z’s.

For AAAH I have:

AAAH : 6 0
ZAAH : 6 0
AZAH : 6 0
AAZH : 6 0
AAAZ : 6 0
ZZAH : 6 0
AZZH : 6 0
AAZZ : 6 0

But what about ZAZH, ZAAZ and AZAZ?

When you have time. I’m just excited about the most recent experiment. I used a wildcard to represent a bunch of different letters and got a decent solve!

Smokie

I hope it works out for the two of you. I have been sitting back just watching you work. Haven’t the foggiest idea what you two are talking about but, hell, I’m excited too.

Soze

@Soze, we are trying to debunk/prove smokie’s wildcard hypothesis.

Jarlve,

I did use the zz ngrams on the last experiment. But I didn’t see where there were any with letters between two z’s.

For AAAH I have:

AAAH : 6 0
ZAAH : 6 0
AZAH : 6 0
AAZH : 6 0
AAAZ : 6 0
ZZAH : 6 0
AZZH : 6 0
AAZZ : 6 0

But what about ZAZH, ZAAZ and AZAZ?

When you have time. I’m just excited about the most recent experiment. I used a wildcard to represent a bunch of different letters and got a decent solve!

Smokie

Then we’ll have so many duplicates which we don’t know how to score! Also when a piece of text is scored with the ngrams it goes like this.

For 4-grams:

ilikeagoodmystery
ilikeagoodmystery
ilikeagoodmystery
ilikeagoodmystery
ilikeagoodmystery
ilikeagoodmystery
ilikeagoodmystery
ilikeagoodmystery
ilikeagoodmystery
etc…

But you may be right. Maybe I should recompile ngrams from a corpus after adding the same (upscaled) frequencies of "Z", "ZZ", "ZZZ".

Thanks Jarlve.

I was thinking about how Decrypto must compare short pieces of the message to the list of ngrams, and I understand what you are saying about duplicates and scoring issues.

I like the idea of upscaled z,zz and zzz ngrams in the future but don’t want to waste your efforts in the present. Let me look at why Decrypto preferred double SS instead of double ZZ in the last experiment. I can look at the scoring of those ngraphs to see how they compare and report back to you about that.

In the mean time, we can think about how to score the consolidated z-ngrams. The trick for using z-ngrams to find the plaintext that a polyalphabetic ciphertext represents is going to have limitations. I will think about some simple experiments to explore those limitations. If you have any ideas for experiments, let me know and I will try them.

I have a busy day today (hopefully) away from the computer. Will be back soon.

Smokie

Addendum Analysis Experiment 4 ZZ Wildcard n-graphs

I checked the first three instances of wildcard 63 63 for score.

Ngraphs for SS scored higher than the correct ngraphs and the zz ngraphs. Substituting 63 63 for the original symbols caused Decrypto to find a solution where we can see what the plaintext should be, but Decrypto DID NOT use the zz ngraphs when it solved.

So it seems that using the one z ngrams worked in a direct sense, but the zz ngrams worked indirectly because it made Decrypto look for ngrams with double letters that would work for a solution.

Jarlve, I agree that we would not know how to score ngrams with multiple wildcard symbols.

At this point I am going to have to re-group and think about the problem for a while. Like I said, I get sort of obsessed with the 340 for about a week or so each year. But I am going to have to rest for a while.

1. Zodiac used cycles but so far there is no way to determine which of the thousands of cycles are Zodiac made or random;
2. The +, q, B and F symbols (my 5, 19, 20 and 51) don’t cycle with the other fifty nine symbols well and may have a special purpose;
3. These symbols are either 1:1 substitutes, fillers or polyalphabetic;
4. Even the most repeated perfect cycles have missing symbols, and it is possible that Zodiac used the +, q, B, and F symbols as wildcards for the cycle symbols (the Wildcard Hypothesis);
5. There is a way to use ZKDecrypto to analyze at least one symbol as polyalphabetic with modified ngraph files where the letter Z is substituted for each letter in each ngraph.

For mental health reasons, among others, I am going to try to stop doing this now, at least for a while.

Thank you very much for your assistance, Jarlve and doranchak.

Smokie

EDIT: My last idea was an application that people could download and use to help solve the 340. Merge alleged wildcards into one symbol, or do whatever else you think should be done. It would probably take years for one computer to check all of the possibilities. But with many computers working together, it could go faster. A master program that delegates and manages the tasks of all of the application programs, and application programs that work in concert with the master program and each other to solve. Sort of like how people can download an application to help SETI look at tiny pictures of the universe for radio waves from other intelligent life in the universe. Everybody who wants to participate could and it would truly be a group effort. I think that if a solution could be found that way, it would provide a great deal of satisfaction to all of the participants.

I agree on your summation and understand very well why you need to take a break. What you call obsession I call drive, and you literally drive yourself to an-almost-breaking-point where, you know from previous experience, you really need to take a break, like or dislike! It has happened to me probably close to a 100 times. The best way to recover from this is to disconnect from the cause.

The dude abides!

Well, unfortunately, I fell off of the wagon a little bit today. I still think that some of the symbols are polyalphabetic. I did some testing to try to figure out what symbols may be wildcards, some of the results I will withhold for now because they don’t seem to show much and are not finished. But then I had an idea.

On June 3, 2015, Jarlve made some ngrams for me with one z. I made a coded message with a polyalphabetic symbol that ZKD1.2 solved but with Z’s in place of the correct letters. See "Decrypto seemed to have corresponded the n-grams with the Z’s in them and used them in the solution" at viewtopic.php?f=81&t=267&start=100.

So this time I just applied the 340. I lowered the value for E in INIT KEY from 8 to 7 and raised the value for Z in INIT KEY from 0 to 1.

Without the z n graphs, ZKD scored 32969 and the result was gibberish.

With the z ngraphs, the score was only 34249 in 15 minutes and the result was gibberish. But the program assigned symbol 19, or the "+" to letter Z:

O.k., so why did ZKD1.2 assign symbol 19, or the "+" to letter Z, but not assign any other symbol to letter Z? This was using the Z ngraph files, not the ZZ ngraph files.

I may do some more testing in the future, but is this a possible way to flush out the polyalphabetic symbol(s), if they are there? Is the + symbol the only polyalphabetic symbol? Are there other ways to test this hypothesis?

Smokie

EDIT: Are the scores for the z ngram files the only obstacle for finding a solution? If symbol 19, or the "+" symbol was the only polyalphabetic symbol, could this message be solved?

If symbol 19, or the "+" symbol was the only polyalphabetic symbol, could this message be solved?

I believe if you assign a new unique symbol for each occurrence of "+" in the ciphertext that would make it essentially a "wildcard". And I think that idea has already been tested and no solution was found. Either because "+" is not a wildcard, or because trying to solve a ciphertext of less than 340 characters long with 85 unique symbols is highly unlikely.

Ah, I remember where I saw it. Go to this page, and then in the "Available ciphers" list at the top select "Z340: pluses changed to unique symbols":
http://www.oranchak.com/zodiac/webtoy/stats.html

340.zodiac.uniplus.txt is that file. It has been included since the very first release of ZKD.

-glurk

It’s amazing that people are now thinking of ideas we already tried 10 years ago. That’s cool, I guess. No harm in doing it all over again.

You never know until you try it for yourself.

-glurk

One difference is that they are applying different algorithms to the already-explored ideas. I suppose that aspect is new.

It’s true, at least in this thread. I surely do not want to discourage anyone, ever. I think I just have spent too much time on the ciphers and have gotten frustrated!

-glurk

EDIT: I come across as "Negative Nellie" or something a lot, and I do not mean to. Carry on. Someone may find a new thing.

My latest idea is that if wildcards and cycles can be identified, we can expand wildcards into individual symbols and merge cycles into single symbols, incrementally so that multiplicity is kept at a minimum, until we can use ZKD1.2 to find a solution.

So I decided to make my own message and cipher, apply wildcards one at a time to find out how ZKD solves, and find a point where the scoring and readability is similar ZKD 340 results (at 15 minutes, score of 32969 with default values).

Using the unoriginal "I like killing" first 340 of the 408, I made this cipher and coded message:

A 1 2 3 4 5
B 6
C 7 8
D 9 10 11
E 12 13 14 15 16 17 18
F 19
G 20
H 21 22 23 24
I 25 26 27 28
J
K 29
L 30 31 32
M 33 34
N 35 36 37 38
O 39 40 41 42 43
P 44
Q
R 45 46 47 48
S 49 50 51 52
T 53 54 55 56 57
U 58
V 59
W 60
X 61
Y 62
Z

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

Then I substituted 63 where 19 appears in the 340 (not shown here) and got a somewhat readable solution score 36840 in 15 minutes:

I L I D E D I D L I N G P E O P L
E B S C A U S E I T I S S O R U C
H F U N I S I U M O R E F U N T H
A N D I D L I N G W I L S S A R E
I N T S E F O R R E S T S E C A U
S E M A N I S T H E R O U T E A N
G E S O U S A N I M A D O F A L L
T O D I D L S O S E T H I S G G I
V E S S E S H E R O U T T H R I L
D I N G E S P E S E N C E I S E V
E S B E T T E R T H A N G E T T I
N G Y O U T R O C D S O F S W I T
H A G I R L S H E B E S T P A R T
O F I T I U T H A T W H E N I S S
L D B E T E B O R N I N P A R A S
I C E A N D A L L T H E I H A V E
D I D S E E W I L S B E C O M E R
S S L A V E S I W I L D N O T G I
V E Y O U M Y N A R E B E C A U S
E Y O U W I L L T R Y T O U D O S

Then I expanded the 63 into 24 individual symbols (not shown here) and got a good readable solution score 42203 in 15 minutes:

I L I K E K I L L I N G P E O P L
E R E C A U S E I T I S S O M U C
H F U N I T I S M O R E F U N T H
A N K I L L I N G W I L E S A M E
I N T H E F O R R E S T B E C A U
S E M A N I S T H E M O S T D A N
G E Y O U S A N I M A L O F A L L
T O K I L L S O N E T H I N G G I
V E S E E T H E M O S T T H R I L
L I N G E O P E R E N C E I S E V
E A R E T T E R T H A N G E T T I
N G Y O U P R O C K S O F F W I T
H A G I R L T H E R E S T P A R T
O F I T I S T H A T W H E N I W I
L L R E P E R O R N I N P A R A G
I C E A N D A L L T H E I H A V E
K I L I E D W I L D R E C O M E M
E A L A V E S I W I L L N O T G I
V E Y O U M Y N A M E R E C A U S
E Y O U W I L L T R Y T O S L O U

Then I added 63 where 19 appears and 64 where 5 appears in the 340 and still got a somewhat readable solution score 36015 in 15 minutes:

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

I L I T A T A L L I N G P E O P L
E A S C A U S E I T I S S E D U C
H F U N A S I N M O R E F U N T H
A A A I L L I N G W A L S S A D E
I N T S E F O U R E S T S E C O U
A E M I N I S T H E D O N T D A R
G E S O U S A N I M A L E F O L L
T O T A L L S O S E T H I S G G I
V E S S E S H E D O N T T H R I L
L A N G E S P E S E A C E I S E V
E S B E T T E R T H I A G E T T I
N G Y O U R R E C T S O F S W I T
H A G A U L S H E B E S T P A R T
O F I T I N T H A T W H E R I S S
L L B E R E B O R N I N P O U I S
I C E A N D A L L A H E I H A V E
T A L S E D W I L S B E C O M E D
S S L O V E S I W I L L R E T G A
V E Y O U M Y N I D E B E C A U S
E Y O U W A L L T R Y T O N L O S

Then I expanded both 63 and 64 into individual symbols and again got a somewhat readable solution with multiplicity 0.279 and score 42247:

25 30 26 29 101 29 27 31 32 28 35 20 44 13 39 44 30
14 102 121 7 1 58 49 16 25 53 26 50 51 40 33 58 8
21 19 58 36 27 122 28 52 34 41 45 17 19 58 37 55 22
2 103 104 25 31 32 26 35 20 60 105 30 123 124 3 33 18
28 36 56 125 12 19 42 46 47 13 49 57 126 14 7 4 58
106 15 34 5 37 25 51 53 24 16 33 43 52 54 10 1 38
20 17 127 39 58 49 2 35 26 34 3 31 40 19 4 32 30
55 41 29 27 31 32 50 42 128 18 56 21 28 129 20 20 25
59 12 51 130 13 131 22 14 33 43 52 53 54 23 45 26 30
31 27 37 20 15 132 44 16 133 17 107 8 18 28 49 12 59
13 134 6 14 55 56 15 47 57 24 5 108 20 16 53 54 25
37 20 62 39 58 48 45 40 7 29 50 41 19 135 60 26 55
21 1 20 27 46 32 136 22 17 6 18 51 57 44 2 47 53
42 19 28 54 25 52 55 23 3 56 60 24 12 38 26 137 138
30 31 6 13 48 14 6 43 45 35 28 36 44 4 46 5 139
25 8 15 1 37 9 2 32 30 109 21 16 26 22 3 59 17
29 27 31 140 18 10 60 28 30 141 6 12 7 39 34 13 33
142 143 32 4 59 14 50 25 60 26 30 31 38 40 53 20 110
59 15 62 41 58 34 62 35 5 33 16 6 17 8 1 58 51
18 62 42 58 60 111 32 30 54 47 62 55 43 52 31 39 144

I L I T A T A L L I N G P E O P L
E B E C A U S E I T I S S O M U C
H F U N A S I S M O R E F U N T H
A D F I L L I N G W I L E C A M E
I N T H E F O R R E S T H E C A U
S E M A N I S T H E M O S T D A N
G E Y O U S A N I M A L O F A L L
T O T A L L S O G E T H I N G G I
V E S E E T H E M O S T T H R I L
L A N G E O P E R E A C E I S E V
E R R E T T E R T H A N G E T T I
N G Y O U P R O C T S O F E W I T
H A G A R L E H E R E S T P A R T
O F I T I S T H A T W H E N I W I
L L R E P E R O R N I N P A R A D
I C E A N D A L L T H E I H A V E
T A L K E D W I L L R E C O M E M
O U L A V E S I W I L L N O T G I
V E Y O U M Y N A M E R E C A U S
E Y O U W I L L T R Y T O S L O U

So even with two wildcards, ZKD1.2 was able to find a solution with sentence fragments of several words, and expanding the wildcards worked!

When I add a third wildcard, 65 where 20 appears in the 340 and all wildcards not yet expanded, I get unreadable gibberish score 33253:

25 30 26 29 64 29 27 31 32 28 35 20 44 13 39 44 30
14 64 63 65 1 58 49 16 25 53 26 50 51 40 33 58 8
65 19 58 36 27 63 28 52 34 41 45 17 19 58 37 55 22
2 64 64 25 31 32 26 35 20 60 64 30 63 63 3 33 18
28 36 56 63 12 19 42 46 47 13 49 57 63 14 7 4 58
64 15 34 5 37 25 51 53 24 16 33 43 52 54 10 1 38
20 17 63 39 58 49 2 35 26 34 3 31 40 19 4 32 30
55 41 29 27 31 32 50 42 63 18 56 21 28 63 20 20 25
59 12 51 63 13 63 22 14 33 43 65 53 54 23 45 26 30
31 27 37 20 15 63 44 16 63 17 64 8 18 28 65 12 59
13 63 6 14 55 56 15 47 57 24 65 64 20 16 53 54 25
37 20 62 39 58 48 45 40 7 29 50 41 19 63 60 65 55
21 1 20 27 46 32 63 22 17 6 18 65 57 44 2 47 53
42 19 28 54 25 52 55 23 3 56 60 24 12 38 26 63 63
30 65 6 13 48 14 6 43 45 35 28 36 44 4 46 5 63
25 8 15 1 37 65 2 32 30 64 21 16 26 22 3 59 17
29 27 31 63 18 10 60 28 30 63 6 12 7 65 34 13 33
63 63 32 4 59 14 50 25 60 26 30 31 38 40 53 20 64
59 15 62 41 58 34 62 35 65 33 16 6 65 8 1 58 51
18 62 42 58 60 64 32 30 54 47 62 55 43 52 31 39 63

A I A M R M A N D H C H H E T H I
E R S E F O R E A D A L R E R O S
E T O O A S H I N G S E T O W N B
E R R A N D A C H T R I S S O R T
H O U S A T I N L E R D S E P O O
R I N E W A R D H E R T I L O F T
H E S T O R E C A N O N E T O D I
N G M A N D L I S T U S H S H H A
V A R S E S B E R T E D L Y S A I
N A W H I S H E S E R S T H E A V
E S C E N U I L D H E R H E D L A
W H A T O U S E P M L G T S T E N
S F H A N D S B E C T E D H E L D
I T H L A I N Y O U T H A T A S S
I E C E U E C T S C H O H O N E S
A S I F W E E D I R S E A B O V E
M A N S T O T H I S C A P E N E R
S S D O V E L A T A I N T E D H R
V I A G O N A C E R E C E S F O R
T A I O T R D I L L A N T I N T S

When I expand wildcards 63, 64 and 65 into individual symbols for multiplicity of 0.312, I get a higher score 39652, but unreadable message:

25 30 26 29 101 29 27 31 32 28 35 20 44 13 39 44 30
14 102 121 151 1 58 49 16 25 53 26 50 51 40 33 58 8
152 19 58 36 27 122 28 52 34 41 45 17 19 58 37 55 22
2 103 104 25 31 32 26 35 20 60 105 30 123 124 3 33 18
28 36 56 125 12 19 42 46 47 13 49 57 126 14 7 4 58
106 15 34 5 37 25 51 53 24 16 33 43 52 54 10 1 38
20 17 127 39 58 49 2 35 26 34 3 31 40 19 4 32 30
55 41 29 27 31 32 50 42 128 18 56 21 28 129 20 20 25
59 12 51 130 13 131 22 14 33 43 153 53 54 23 45 26 30
31 27 37 20 15 132 44 16 133 17 107 8 18 28 154 12 59
13 134 6 14 55 56 15 47 57 24 155 108 20 16 53 54 25
37 20 62 39 58 48 45 40 7 29 50 41 19 135 60 156 55
21 1 20 27 46 32 136 22 17 6 18 157 57 44 2 47 53
42 19 28 54 25 52 55 23 3 56 60 24 12 38 26 137 138
30 158 6 13 48 14 6 43 45 35 28 36 44 4 46 5 139
25 8 15 1 37 159 2 32 30 109 21 16 26 22 3 59 17
29 27 31 140 18 10 60 28 30 141 6 12 7 160 34 13 33
142 143 32 4 59 14 50 25 60 26 30 31 38 40 53 20 110
59 15 62 41 58 34 62 35 161 33 16 6 162 8 1 58 51
18 62 42 58 60 111 32 30 54 47 62 55 43 52 31 39 144

H D N C E C O M E N A T E R R E D
H E W A Y S O W H I N A P L E S T
H I S G O I N E D T O H I S S I T
M U C H M E N A T A N D I S H E A
N G L E S I N M O R O F T H E R S
O L D I S H P I Y W E R E T L Y A
T H E R S O M A N D H M L I R E D
I T C O M E A N D A L O N A T T H
E S P A R E T H E R S I T C O N D
M O S T L U E W T H A T A N D S E
R V E H I L L O F Y O U T W I T H
S T I R S T O L E C A T I N A T I
O Y T O M E S T H E A F F E M O I
N I N T H E I C H L A Y S A N D E
D B E R T H E R O A N G E R M I G
H T L Y S O M E D T O W N T H E H
C O M P A L A N D B E S E N D R E
S W E R E H A H A N D M A L I T H
E L I T S D I A R E W E N T Y S P
A I N S A R E D T O I I R E M R K

My conclusion is that if the 340 is anything like the model, then it must have more than two wildcards. The good thing to know is that we don’t necessarily have to compensate for the high count of symbol 19, or "+", by merging a lot of cycles. Perhaps we can merge only enough cycles to compensate for some of the wildcards with lower counts, like 5 or 20 ("q" or "B").