340..partially solved ;)

[Jarlve]

408-cipher: 26^54 = 25618918401207524005916276251028099629855783828009607898193480161907468926976
340-cipher: 26^63 = 139098011710742195590974259094795403842655842142490330518716727403333474672708595090456576
Sure about that? Each digit more (or less) is 10 times more variations..a difference of 13 digits means a x1000,000,000,000 higher amount of combinations that have to be 'cracked'..

Jarlve-test-cipher: 26^100 = 3142930641582938830174357788501626427282669988762475256374173175398995908420104023465432599069702289330964075081611719197835869803511992549376 (142 decimal digits)

Not a problem to crack at all. The search space is not the issue here but the multiplicity is. Consider the Beale 2 cipher which has 182 symbols. Or some of my personal test ciphers with over 500 symbols. Say 26^500. Now that is 708 decimal digits and these still solve because the ciphers are also longer. It is about the ratio of symbols to the cipher length (multiplicity) and that is why 26^54, 26^63 or 26^100 does not mean anything. The 340 has a multiplicity of 0.185 (63/340) and that is not very difficult.

Code: Select all

1   2   49  3   4   35  56  39  70  86  64  80  7   25  8   7   75
33  9   47  10  11  12  13  59  96  68  50  13  65  46  15  12  28
16  17  79  66  60  74  87  61  24  8   18  94  41  12  71  84  76
26  90  3   54  57  70  78  5   6   19  62  52  20  93  27  51  58
1   48  95  97  92  42  83  69  85  55  98  100 31  4   10  11  79
72  25  43  26  64  49  13  53  82  33  15  46  65  14  36  27  66
80  47  89  8   12  65  37  71  56  24  40  67  83  44  45  2   39
68  77  35  86  75  57  61  73  29  59  74  91  96  90  6   80  96
21  99  98  29  94  84  16  58  51  46  72  95  81  76  69  50  52
67  56  5   6   99  22  7   94  85  99  48  28  94  60  100 87  13
58  34  92  64  38  92  53  14  55  69  53  97  11  66  93  4   14
68  54  71  80  23  8   79  85  89  83  10  3   65  77  17  41  88
78  74  82  37  6   62  69  2   84  76  25  9   33  61  84  7   40
85  95  73  44  1   81  1   98  100 97  45  53  19  82  47  90  78
36  60  59  87  88  54  52  67  31  4   89  25  38  46  32  5   78
48  7   37  18  26  20  62  28  33  27  64  36  37  2   39  14  91
25  49  16  40  63  33  35  56  70  70  47  20  88  86  75  57  9
59  10  8   43  47  15  30  98  52  45  21  59  72  96  19  50  67
39  66  83  68  93  60  34  99  23  77  12  24  30  64  11  29  94
31  58  28  26  79  13  92  23  73  12  88  87  70  75  81  18  30


[Quicktrader]

Well, two thoughts..Beale forget, it's a book cipher. Not comparable at all. Second, the cipher length, of course, matters. The ratio between what you call search space to the cipher length is critical.

If we look at the multiplicity

54/408 = 0.132 or an average of 7.55 letters per homophone

and

63/340 = 0.185 or an average of 5.40 letters per homophone

Thus, regardless of the cipher length, the 340 has a 40% different multiplicity than the 408. This, however, is only the RATIO between the homophones used and the length of the cipher. It doesn't say anything about the complexity of the cipher, which may increase overproportionally to its multiplicity and may also depend on e.g. words used, accidential distribution of n-grams etc..

When running ZDK, I indeed got a solution on the first try (numeric). When I shuffled the key (F5), however, I didn't get the result anymore. Are you able to always solve the test cipher? Indeed it delivers a great result.

Have the homophones in your test cipher been used in sequences?

QT

[Jarlve]

Are you able to always solve the test cipher? Indeed it delivers a great result.

Have the homophones in your test cipher been used in sequences?

Yes, always solves with AZdecrypt. 25% sequential homophone randomization.

Here's another with no sequential homophones. It does not matter.

Code: Select all

71  56  90  3   72  3   40  35  35  71  50  6   7   32  65  89  56
100 69  55  57  53  12  64  100 83  73  86  13  59  74  27  36  57
33  31  36  5   90  14  1   64  81  8   29  39  17  12  92  73  67
43  98  79  90  2   35  97  98  48  37  90  84  52  48  53  27  60
71  50  82  16  60  63  99  29  29  25  34  87  69  60  57  11  58
13  100 27  75  5   86  13  91  67  95  81  70  59  82  20  62  50
48  32  29  8   36  80  94  92  71  81  94  56  24  31  53  2   84
87  65  41  26  84  84  34  99  15  32  73  33  40  98  28  6   26
61  95  34  15  4   54  67  39  81  70  64  78  82  16  51  90  46
56  97  92  48  4   22  7   4   51  4   45  10  55  26  73  40  59
72  21  25  68  69  100 91  14  55  29  91  42  62  5   28  95  87
14  86  5   6   23  8   88  29  18  65  10  30  64  99  66  17  19
90  54  33  75  28  85  47  44  54  42  60  9   39  80  54  89  94
47  73  74  66  26  14  93  34  82  67  75  87  49  42  95  5   90
52  83  95  93  37  97  46  35  9   55  47  39  9   24  47  45  71
92  7   75  47  53  77  83  57  55  75  68  76  94  44  44  78  42
4   97  42  43  21  25  30  71  35  46  72  77  19  97  46  44  69
4   10  8   81  39  81  23  64  84  11  21  72  64  93  37  71  2
46  50  99  82  48  86  96  55  38  65  58  15  23  5   75  15  39
69  100 10  11  88  34  100 38  24  12  37  40  35  44  78  18  38


[Quicktrader]

Jarlve, you are probably right with what you say (otherwise I wouldn't get an at least partial solution of your cipher, either ).

I continued to try on the cipher 'as it is', because I still believe in a fair chance that it is solvable as a homophone substitution without transposition. Some progress, at least..I was able to expand the Python code similar to the following pseudo-code:

Code: Select all

SET VARIABLES
dictionary = (word1, word2,...word7000) [i][now words of length 5-10 letters][/i]
bigrams = (EN, ER,....LS) [i][frequent ones, as they are repeating at least once][/i]
iofbc = (ENTER, THERE,...) [i][overlapping repeating trigrams][/i]
- all above ~185,000 different sets of variables ('line') -

SEARCH FUNCTION
search for words in dictionary >0 [i][Aho-Corasick-Algorithm][/i]

CREATE FCCP-LINES
read line1 of 185,000, then go next
cipher phrase: chain #1 = e.g. M+pl+P+pl+Z+Q+H+F+B+c+Y
---if words in chain #1 found >0: add variables
---chain #2 = e.g. M+q+G+H+Q+c+t+pl+L
------if words in chain #2 found >0: add variables
------chain #3....
...
------------------if words in chain #9 found >0
------------------print chain#1, chain#2, chain#3...chain#9, word found in chain #1, word found in chain #2.....word found in chain #9.


The idea is, based on the variables set, to find any word in a specific string location of the cipher. The program then expands usually two to three variables with each step to create a second, third, fourth etc. string from the cipher. Only if the program finds any word in the previous string, the program continues to the next string. The program also covers all words found in a string.

What is quite nice is that each time I add another string, the more homophones are 'covered'.

Results, so far, was that the program was able to find eight (!) different words in the cipher (as I had shown previously finding approximately 5 words). Now I do a run with the program to find a minimum of nine words in various spots of the cipher.

Not many configurations would allow nine different words in the 340 cipher (one may want to try to enter nine words in Oranchak's webtoy..).

Below you find some illustration in which areas the program is looking for words. Meanwhile, the whole setting covers approximately 60% of the cipher (as the homophones not only appear in the strings underlined but also elsewhere).

lines.JPG

QT

[Quicktrader]

Excel is powerful..over the last days I created a small (=partial) CLEARTEXT GENERATOR for the 340 cipher. This is how it works:

First, we know about the following two sequences in the Z340:

seq1.JPG

seq1.JPG

We now use a dictionary (this time a large one). We then assume that in sequence no. 2 there is at least one word of length >4 letters. Although this could also not be the case, this possibility is still a realistic assumption (as long as Z had not only used short or extremely long words in that specific sequence).

However, we still have no idea, where such a word of length >4 would actually start.

Now the pattern matching method helps us out. The word 'boob', for example, can be placed into a cipher structure like 'CDDC' or, for homophone ciphers even 'CDJC' or 'CDJK', but not into a cipher structure like 'CDCJ' (try it, if you want). 'C' is 'C' and the letter 'b' is definitely not an 'o'.

Thus, not all words from the dictionary can be placed onto all positions of the sequence. This because some of the homophones ocurring in the sequence are repeating (eg. the double '++').

With the support of Cryptool, it was possible to place every single word (of different length, but at least >4) onto any position in sequence 2. All of the previous under consideration of the cipher pattern as well as the possibility of homophones being used (polyalphabet, e.g. 'CDJK'). Starting from homophone #1 of the sequence, checking the whole dictionary for the pattern, then homophone #2 etc.

This results in a list of approx. 150,000 words on specific positions in sequence no. 2.

After this pattern-matching (cracking) process has been done - which was, by the way, described in an early NSA paper by Friedman, 'Military Cryptanalysis', 1952 - there are still additional steps to be done:

After defining a letter for the '+' symbol, e.g. the letter 'L' or 'S', we can still add a list of 5-grams for sequence no. 1.

This because sequence no. 1, most likely, contains two frequent, overlapping trigrams. 'Frequent', because these overlapping trigrams actually occur twice in the cipher although homophones are used (shuffling the alphabet all the time..). Non-frequent trigrams rather would not occur twice or at least not in combination with a second also repeating trigram. This can be argued but one may expect two repating trigrams showing up in any homophone cipher anyway, as some frequent trigrams (e.g. 'AND' or 'THE') would indeed statistically occur around 6-8 times in a 340 letter text.

Now if we combine all of the above , we end up with a 20MB Excel file and millions of rather complicated formulas (e.g. 'first letter equal to '+' and second letter equal to the third letter of the 5-gram but length of the word =8' etc.). To cover all those combinations, millions of calculations are performed 'in the background' to find all combinations of '+' symbol, 5-gram, sequence pattern and of course words.

Astonishing were the results:

For example, if you select the letter 'S' in combination with the 5-gram 'THECO', there are only 79 values found (from the whole dictionary) that actually match the pattern structure of sequence no.2 (at any spot).

Based on a dictionary of 50,000 words, this represents a ratio of 79:50,000, which is equal to an exclusion of 99.84% of all words of the dictionary (with that specific setting). All of this is done based on NSA Friedman's millitary cryptanalisys method.

Following the previous example, it was possible to find out that if you use 'S' and 'THECO', you will be able to partially complete the sequence no. 2 with e.g. the word 'SEDUCTRESS' but 78 other words such as 'DOCTOR' or 'PASSPORT'. You can try this in Oranchak's webtoy, too. It is not possible, however, to enter the word 'CROSSROAD'. But to enter this word is possible, if you chose the 5-gram 'CTION' instead of 'THECO').

That only a very small amount of words is found under such pre-setting is quite satisfying. It opens at least a chance to crack that cipher.

Conclusion:
Considering the most likely 5-grams (e.g. 30) in sequence no. 1 results in approximately 2,400 potential cleartext phrases (per each letter chosen for '+'). This, of course, including at least one full word of length >4. It is possible, that one of the examples already found (like discussed above) was indeed once written by Zodiac.

In other words:
Enter one letter for the '+' symbol, add a frequent 5-gram for sequence no. 1 and with a good chance you get some potential cleartext (list of only 50-100 words). Based on such a large dictionary (+50,000), other cleartext is unlikely (assuming a standard homophone substitution).

Or described as sort of a promise:
If you give me the correct letter for the '+' symbol as well as the correct 5-gram for sequence no. 1, you will receive 23-26% cleartext of the 340 cipher in return (will be explained later). This, including a list of approximately 50-100 alternatives, which may match the pattern of sequence no. 2, too.

For Excel, imo, this is quite surprising. We may estimate:

Approximately a max. of 10 letters for the '+' symbol (I believe ins 'S' or 'L' but ok - at least it'd be a frequent one). An estimated 200 different 5-grams consisting of repeating, thus frequent trigrams (overlapping each other). Approximately 70-100 results per configuration. All this leads us to approx.

10 x 200 x 70 = 140,000

potential settings. Those may further be used for additional computation (e.g. cross-checking with another sequence of the cipher, continuing with pattern matching, trial & error etc.). As an optimist, however, I'd say only 2 letters for the '+' symbol are in question and a maximum of 60 different 5-grams may be needed; leading us to:

2 x 60 x 70 = 8,400

potential settings, with one of it leading to Z's cleartext. Optimism, however, is just a lack of information..

With the correct setting, all depending on the position and length of the word, approximately 80 to 90 homophones of the Z340 are then 'solved'.This is equal to a total of 23%-26% of the cipher!

As long as someone doesn't have an IQ of +170 or being the world champion in scrabble, additional computation is most likely needed to then convert the complete cipher into cleartext. Needless to say that, if you have choosen the wrong settings from the beginning (e.g. wrong 5-gram), Z would continue to be 'crackproof' until you have found the right setting.

For those interested, here is how many of the formulas used acutally look like (German languag, this specific one actually found the word 'seductress'..no need to understand the formula as it is here shown out of context)

Code: Select all

=WENN(UND(LÄNGE($Y19510)>11;$A19510=$I19510;$A19510=$AB$1;$B19510=RECHTS(LINKS(AY$1;3);1);$E19510=RECHTS(LINKS(AY$1;4);1);$F19510=LINKS(AY$1;1);$D19510=$L19510;$G19510=$K19510;$I19510=$J19510);$Y19510;WENN(UND(LÄNGE($Y19510)=11;$A19510=$I19510;$A19510=$AB$1;$B19510=RECHTS(LINKS(AY$1;3);1);$E19510=RECHTS(LINKS(AY$1;4);1);$F19510=LINKS(AY$1;1);$G19510=$K19510;$I19510=$J19510);$Y19510;WENN(UND(LÄNGE($Y19510)=10;$A19510=$I19510;$A19510=$AB$1;$B19510=RECHTS(LINKS(AY$1;3);1);$E19510=RECHTS(LINKS(AY$1;4);1);$F19510=LINKS(AY$1;1);$I19510=$J19510);$Y19510;WENN(UND(LÄNGE($Y19510)=9;$A19510=$I19510;$A19510=$AB$1;$B19510=RECHTS(LINKS(AY$1;3);1);$E19510=RECHTS(LINKS(AY$1;4);1);$F19510=LINKS(AY$1;1));$Y19510;WENN(UND(LÄNGE($Y19510)>5;LÄNGE($Y19510)<9;$A19510=$AB$1;$B19510=RECHTS(LINKS(AY$1;3);1);$E19510=RECHTS(LINKS(AY$1;4);1);$F19510=LINKS(AY$1;1));$Y19510;WENN(UND(LÄNGE($Y19510)=5;$A19510=$AB$1;$B19510=RECHTS(LINKS(AY$1;3);1);$E19510=RECHTS(LINKS(AY$1;4);1));$Y19510;""))))))

QT

[smokie treats]

Is this the Friedman book that you were talking about?

https://www.nsa.gov/news-features/decla ... 078762.pdf

[Quicktrader]

Uploaded it for you

https://docs.zoho.com/file/9j5vv3dbaa85 ... 52b691bd1c
(you may have to click download to view the pdf file)

Picked it up somewhere in 2016, look for the latter part where lists of patterns compared to words are listed on multiple pages. Looks like this (p. 3-29), left row is a patterns (A for the first unknown letter, B for the second etc.) and in the second line is an idea which word could match such pattern. They only used few words, of course - still a longer list - instead of a full dictionary.

patternmatching.JPG

What you see is actually an approach for monoalphabetic substitution. What was written above actually does the same, with the full dictionary all at once, on multiple positions and in addition to that considering the possibility of multiple homophones for each alphabetical letter. Then continue to combine it with the '+' symbol and a 5-gram of (almost) free choice..

QT

[smokie treats]

Thanks a lot. It is the same one, which I have not read before. This will give me something new to read during my work breaks. Thanks again.

[Quicktrader]

ABC DEF GFHIBJF.

QT

[Simplicity]

Good read going thru these posts,
Im very happy to see i came to some conclusions on my own that are much like yours.

A thought ive been playing with is that many symbols are actually determiners / adjectives.

For instance the zodiac symbol i suspected to be "T" or "H" then i thought it was "TH" which makes a bit of sense but it opens the door to the possibility of being "THE"

And in response to "+" being "L" i suggest that L20 #15 & #16 offers two clues,
A) theres no full stop, infering that the ending is maybe else where.
B)Its possibly a PS.
C)Or just filler

Goodluck and look forward to seeing more of your work.