340..partially solved ;)

Re: 340..partially solved ;)

Postby doranchak » Mon Apr 15, 2019 4:56 am

Here's what my word break guesser does with it:

THE REVENSE EAR THIS SHARE A E SEVEN THE ES EH STERN AT ONESS RESTS SE THERE EE US ANTAE HIS ENRRE POLE SHS REPULSES RPENT EITE TOO REST SEE A FS POSTS
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Re: 340..partially solved ;)

Postby Quicktrader » Mon Apr 15, 2019 11:30 am

The strings appear in a row but are from different areas of the cipher (see May 31st, 2019, posting), fwiw.

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Re: 340..partially solved ;)

Postby Quicksilver » Mon Apr 15, 2019 11:46 am

I have worked with many manufacturing computer systems, but you guys are way over my head. I don't write code. I was the guy that worked with IT to figure things out to get the information that the end user needed. I know you guys have to figure out how the code was compiled first and it is probably in English. This may not be the thread to post this question, but what if the 340 is in a foreign language? Can a program be written to look at that possibility as you go along or do your programs already look for that? Has it been attempted before? Is it even worth it?
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Re: 340..partially solved ;)

Postby Jarlve » Mon Apr 15, 2019 1:40 pm

Quicksilver wrote:This may not be the thread to post this question, but what if the 340 is in a foreign language? Can a program be written to look at that possibility as you go along or do your programs already look for that? Has it been attempted before? Is it even worth it?

I tested many languages in early 2017: viewtopic.php?f=81&t=3242
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Re: 340..partially solved ;)

Postby Jarlve » Mon Apr 15, 2019 2:07 pm

Quicktrader,

The IOC of the 408 is 0.0184 and the IOC of the 340 is 0.0193. What now?

Feel free to use AZdecrypt Lite @ http://jarlve.vdm-service.be/ to determine the IOC by putting a cipher in the left hand window and then clicking Statistics, Overall:

Code: Select all
Length: 408
Symbols: 54
Dimensions: 17 by 24
Multiplicity: 0.1323529411764706
Entropy: 5.659921738167404
Index of coincidence:
- Raw: 3070
- Normalized: 0.01848773907597437 <--------------- IOC
- Flatness: 0.8712269272529858
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Re: 340..partially solved ;)

Postby Quicktrader » Mon Apr 15, 2019 7:49 pm

I will, thanks...IoC might depend on if you use 23 (in use), 24, 26 letters for its calculation. Just entered it into the online tool (previous link). Both values show it'd be better multiplicated with the average amount of homophones per letter, to get English language value. Which is logical as well, imo, as the encrypted 'alphabet' of homophones has more 'letters'.

IoC value of 0.0193, based on English language, implied somehow 3.42 homophones per letter (average), I'd say (0.066 divided by 0.0193). Just show how way more difficult the Z340 is compared to the Z408..

But as I said, not very much into IoC stats..


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Re: 340..partially solved ;)

Postby Quicktrader » Tue Apr 23, 2019 9:31 am

It's getting interesting...

For the 100th time or so, the setting of the FCCP program has been changed. After finding way too many partial solutions with 10, 11 or 12 words, I went back to find rather one longer string instead of multiple short ones.

This time the focus is on line 16 to 18, filling all up with trigrams, bigrams, IoFBc, plus symbols etc. to get a string of 27 letters.

Still, way too many results. So I had to reduce the amount of data:

As soon as the program now finds any word, not only in the first, but also in the second and third string, I told my PC to skip the printing of all of its variations. As a result, I do receive only the initial partial cleartext variations from the first string. However only those, which are at least backed up with at least one result in (currently) two other strings / areas of the cipher (line 13 and line 1).

This step might be smart as we now get an overview of all potential solutions for line 16 to line 18 (4th row of interest only):

temp.jpg


As you can see, there still is a long way to go. The first rows of the diagram are acually supporting to understand how much has already been computed.

Nevertheless, we already get some nice cleartext candidates for this area of the cipher, e.g. A SEEMS I BROWSE NORTH IS SHOOK OE or A SEEMS I EXCUSE FOR THIS SHOCK OE (not shown in the diagram).

Of course this is not the correct solution, but it shows that the program can now go very much 'lingual', with meanwhile considering all relevant cipher structures: Repeating bigrams/trigrams; homophones occurring on specific locations of the strings analyzed etc.. We are now finding text.

We will see, how fast the program is doing on that one, which is obviously crucial. So far, the program does not print 99% of all variations, but it still computes them. Therefore, the next goal will be to only compute until in all strings words are found - but skip the rest of the computing (as we are only interested in string #1..).

This could actually safe a tremendous amount of computation time, I'd say like 99.999% or even more. The solving process itself would therefore be performed like this:

- Set variables (e.g. IoFBc or the + symbol)
- Complete strings #1,...with eg. trigrams or alphabet letters
- Search string #1 for minimum 4 words of length >4 (one solution)
- Search string #2 for minimum 1 word of length >4 (one solution)
- Search string #3, #4, #5,... " "
- If any solution is found: Stop computation, print words of string #1 (and any of the first solution found for the other strings)
- Print string#1 and rest of data
- Find logical content in string#1 results
- Compute rest of the cipher based on logical result found on string #1

Thus, the program hopefully will start with the longer string #1, then verifies its correctness on various other strings of the cipher, finally goes back to string #1 to print (without computing all the other variations).

The advantage of this forward-back-forward method is obvious:

We only receive a list of 'successful' strings of length 27 letters. The length of this list can then be reduced more and more by adding additional strings in which at least one word shall be found. And those remaining partial solutions can be looked at.

If there is one string with LOGICAL content, it might be our cleartext candidate (without computing the rest of the cipher at all). Based on this specific string, all other partial solutions can be discarded. Then finally completing the rest of the cipher by computation, all based on the logical string #1 that might then have been found.

Z being crackproof? Surely not.

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Re: 340..partially solved ;)

Postby jacob » Tue Apr 23, 2019 10:38 am

Wow!!
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Re: 340..partially solved ;)

Postby Quicktrader » Wed May 08, 2019 8:46 am

We have 'learned' to create a fictitiously created partial string out of the cipher, as a cleartext, using e.g. frequent bigrams, trigrams, alphabet letters etc.. Let all variations of such string, here of length = 27 symbols/letters, be an array of 'variations'.

In a first step, we cut off all variations which do not contain e.g. 3 or 4 words of length>4.

Only those remaining are used for further computation, searching for additional words but in other parts of the cipher, by adding letters, bigrams etc.. As a result, a huge Matrix occurs, involving all potential cipher parts, letters, bigrams, trigrams etc..for the rest of the cipher - which is way too much to compute.

BUT..what we can actually do is to BREAK the computation process as soon as at least one setting has been found with at least one word in each of the multiple partial cipher strings. Obvious advantage of this proess: If any potential solution does exist for our first (long) string, the rest of its sub-variations is not (yet) computed at all. This saves a lot of time.

Thus, the focus is only on the first string. But this string is shown only if - currently - a total of 9 words are found, 5 of them being found in different 'background' strings. If no word is found in one of the background strings, e.g. #5, the computation proceeds until either one word is found. Or, if none has been found, there is the elimination of the whole variation/array.

All successful variations, however, with at least one word in each string, are printed.

So far, this method works fine with an estimated 99.99999999% of all computable variations not to be computed at all. The results look - similar to my previous post - like this:

Code: Select all
THERE HI TH TIMESSIEXPOSEEARTHISSHAPEAE SLTHERETHEMETEL TEISTILL EOSANTATHI LOOSESNMIERE
THERE HI TH TIMESSIIMPOSEEARTHISSHAPEAE SLTHERETHEMETEL TIISTILL EOSANTATHI LOOSESNMIERE
THERE HI TH TIMESSIREPOSEEARTHISSHAPEAE SLTHERETHEMETEL TRISTILL EOSANTATHI LOOSESNMIERE
THERE HI TH TIMESSIOPPOSEEARTHISSHAPEAE SLTHERETHEMETEL TOISTILL EOSANTATHI LOOSESNMIERE
THERE HI TH TONESSIEXPOSEEARTHISSHAPEAE SMTHERETHINETEM TEISTORM EASANTATHI ROASTSNNOERE
THERE HI TH TONESSIIMPOSEEARTHISSHAPEAE SMTHERETHINETEM TIISTORM EASANTATHI ROASTSNNOERE
THERE HI TH TONESSIREPOSEEARTHISSHAPEAE SMTHERETHINETEM TRISTORM EASANTATHI ROASTSNNOERE
THERE HI TH TONESSIOPPOSEEARTHISSHAPEAE SMTHERETHINETEM TOISTORM EASANTATHI ROASTSNNOERE
THERE HI TH STEERSIHOURSENORTHISSHOUTOE STTHERETEDENSET THIRSTCT EASHESOTHI CRASHSEETERE


Thus, let us ignore everything around the string of length 27, we find potential solutions such as "...TONES. I IMPOSE EARTH IS SHAPED E...", which is quite a nice cleartext variations (with at least the words THERE, THINE [poet.], STORM, SANTA and ROASTS being found on different places of the cipher. Please note that all letters not covered by words are still fully iterable, e.g. the D of the word SHAPED as the program has actually found SHAPE (and not SHAPED); which is part of future computation.

Please not that the specific string of length 27 contains a total of 19 different homophones, thus the theoretical amount of all variations is:

26^19 = 766,467,265,200,361,890,474,622,976 or 766.5 Septillion

out of which we are now able to select dozens of potential cleartext solutions in a few minutes only. Computation, of course, will still take it's time. Then, all results can be analyzed in a second computation process - already having eliminated most of the 766.5 Septillion variations - for further results.

Considering a pre-setup of the + symbol as well as one 5-gram (IoFBc), each 5-gram takes a computation time of a few hours only. Thus, guessing those two pre-setup variablse correctly and one can get Z's cleartext on the same day (assuming homophone substitution; non-statistical outlier).

So far, the method chosen currently covers ~60% of all symbols of the Z340 cipher.

The BREAK function makes it possible..hopefully.

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Re: 340..partially solved ;)

Postby glurk » Thu May 09, 2019 2:58 am

QT-

Assuming you are still trying to solve the 340 as a normal homophonic cipher - it's not - I ran the string you named (TONESSIIMPOSEEARTHISSHAPEAE) through an old program of mine, and tried it in every possible position in the 340. It fits in seven positions, starting with 0. They are positions 5, 6, 7, 8, 11, 246, and 270. And in every one, the bulk of the cipher becomes gibberish. Here are the results:

Code: Select all
(Position 5)
HER>pl^VPkI1LTG2d   =   ----ETONESSIIMPOS
Np+B(#O%DWY.<!Kf)   =   EEARTHISSHAPEAE--
By:cM+UZGW()L#zHJ   =   R----A--PHT-IH---
Spp7^l8!V3pO++RK2   =   -EE-OT-AN-EIAA-EO
_9M+ztjdI5FP+&4k/   =   ---A---SS--EA--S-
p8R^FlO-!dCkF>2D(   =   E--O-TI-AS-S--OST
#5+Kq%;2UcXGV.zLI   =   H-AE-S-O---PNP-IS
(G2Jfj#O+_NYz+@L9   =   TPO---HIA-EA-A-I-
d<M+b+ZR2FBcyA64K   =   SE-A-A--O-R-----E
-zlUV+^J+Op7<FBy-   =   --T-NAO-AIE-E-R--
U+R/5tEIDYBpbTMKO   =   -A-----SSARE-M-EI
2<clRJI!5T4M.+&BF   =   OE-T--SA-M--PA-R-
z69Sy#+NI5FBc(;8R   =   -----HAES--R-T---
lGFN^f524b.cV4t++   =   TP-EO--O--P-N--AA
yBX1!:49CE>VUZ5-+   =   -R-IA------N----A
Ic.3zBK(Op^.fMqG2   =   S-P--RETIEOP---PO
RcT+L16C<+FlWBI)L   =   --MAII--EA-THRS-I
++)WCzWcPOSHT/()p   =   AA-H--H-EI--M-T-E
IFkdW<7tB_YOB!-Cc   =   S-SSHE--R-AIRA---
>MDHNpkSzZO8AIK;+   =   --S-EES---I--SE-A

(Position 6)
HER>pl^VPkI1LTG2d   =   ----E-TONESSIIMPO
Np+B(#O%DWY.<!Kf)   =   SEEARTHISSHAPEAE-
By:cM+UZGW()L#zHJ   =   A----E--MSR-IT---
Spp7^l8!V3pO++RK2   =   -EE-T--EO-EHEE-AP
_9M+ztjdI5FP+&4k/   =   ---E---OS--NE--E-
p8R^FlO-!dCkF>2D(   =   E--T--H-EO-E--PSR
#5+Kq%;2UcXGV.zLI   =   T-EA-I-P---MOA-IS
(G2Jfj#O+_NYz+@L9   =   RMP-E-THE-SH-E-I-
d<M+b+ZR2FBcyA64K   =   OP-E-E--P-A-----A
-zlUV+^J+Op7<FBy-   =   ----OET-EHE-P-A--
U+R/5tEIDYBpbTMKO   =   -E-----SSHAE-I-AH
2<clRJI!5T4M.+&BF   =   PP----SE-I--AE-A-
z69Sy#+NI5FBc(;8R   =   -----TESS--A-R---
lGFN^f524b.cV4t++   =   -M-STE-P--A-O--EE
yBX1!:49CE>VUZ5-+   =   -A-SE------O----E
Ic.3zBK(Op^.fMqG2   =   S-A--AARHETAE--MP
RcT+L16C<+FlWBI)L   =   --IEIS--PE--SAS-I
++)WCzWcPOSHT/()p   =   EE-S--S-NH--I-R-E
IFkdW<7tB_YOB!-Cc   =   S-EOSP--A-HHAE---
>MDHNpkSzZO8AIK;+   =   --S-SEE---H--SA-E

(Position 7)
HER>pl^VPkI1LTG2d   =   ----S--TONESSIIMP
Np+B(#O%DWY.<!Kf)   =   OSEEARTHISSHAPEAE
By:cM+UZGW()L#zHJ   =   E----E--ISAESR---
Spp7^l8!V3pO++RK2   =   -SS----PT-STEE-EM
_9M+ztjdI5FP+&4k/   =   ---E---PE--OE--N-
p8R^FlO-!dCkF>2D(   =   S-----T-PP-N--MIA
#5+Kq%;2UcXGV.zLI   =   R-EE-H-M---ITH-SE
(G2Jfj#O+_NYz+@L9   =   AIM-A-RTE-OS-E-S-
d<M+b+ZR2FBcyA64K   =   PA-E-E--M-E-----E
-zlUV+^J+Op7<FBy-   =   ----TE--ETS-A-E--
U+R/5tEIDYBpbTMKO   =   -E-----EISES-I-ET
2<clRJI!5T4M.+&BF   =   MA----EP-I--HE-E-
z69Sy#+NI5FBc(;8R   =   -----REOE--E-A---
lGFN^f524b.cV4t++   =   -I-O-A-M--H-T--EE
yBX1!:49CE>VUZ5-+   =   -E-SP------T----E
Ic.3zBK(Op^.fMqG2   =   E-H--EEATS-HA--IM
RcT+L16C<+FlWBI)L   =   --IESS--AE--SEEES
++)WCzWcPOSHT/()p   =   EEES--S-OT--I-AES
IFkdW<7tB_YOB!-Cc   =   E-NPSA--E-STEP---
>MDHNpkSzZO8AIK;+   =   --I-OSN---T--EE-E

(Position 8)
HER>pl^VPkI1LTG2d   =   ----O---TONESSIIM
Np+B(#O%DWY.<!Kf)   =   POSEEARTHISSHAPEA
By:cM+UZGW()L#zHJ   =   E----S--IIEASA---
Spp7^l8!V3pO++RK2   =   -OO----A--ORSS-PI
_9M+ztjdI5FP+&4k/   =   ---S---MN--TS--O-
p8R^FlO-!dCkF>2D(   =   O-----R-AM-O--IHE
#5+Kq%;2UcXGV.zLI   =   A-SP-T-I---I-S-SN
(G2Jfj#O+_NYz+@L9   =   EII-E-ARS-PS-S-S-
d<M+b+ZR2FBcyA64K   =   MH-S-S--I-E-----P
-zlUV+^J+Op7<FBy-   =   -----S--SRO-H-E--
U+R/5tEIDYBpbTMKO   =   -S-----NHSEO-S-PR
2<clRJI!5T4M.+&BF   =   IH----NA-S--SS-E-
z69Sy#+NI5FBc(;8R   =   -----ASPN--E-E---
lGFN^f524b.cV4t++   =   -I-P-E-I--S----SS
yBX1!:49CE>VUZ5-+   =   -E-EA-----------S
Ic.3zBK(Op^.fMqG2   =   N-S--EPERO-SE--II
RcT+L16C<+FlWBI)L   =   --SSSE--HS--IENAS
++)WCzWcPOSHT/()p   =   SSAI--I-TR--S-EAO
IFkdW<7tB_YOB!-Cc   =   N-OMIH--E-SREA---
>MDHNpkSzZO8AIK;+   =   --H-POO---R--NP-S

(Position 11)
HER>pl^VPkI1LTG2d   =   ----I------TONESS
Np+B(#O%DWY.<!Kf)   =   IIMPOSEEARTHISSHA
By:cM+UZGW()L#zHJ   =   PEAE-M--EROAOS---
Spp7^l8!V3pO++RK2   =   -II----S--IEMM-SS
_9M+ztjdI5FP+&4k/   =   ---M---S----M----
p8R^FlO-!dCkF>2D(   =   I-----E-SS----SAO
#5+Kq%;2UcXGV.zLI   =   S-MS-E-S-E-E-H-O-
(G2Jfj#O+_NYz+@L9   =   OES-H-SEM-IT-M-O-
d<M+b+ZR2FBcyA64K   =   SI-M-M--S-PEE---S
-zlUV+^J+Op7<FBy-   =   -----M--MEI-I-PE-
U+R/5tEIDYBpbTMKO   =   -M------ATPI-N-SE
2<clRJI!5T4M.+&BF   =   SIE----S-N--HM-P-
z69Sy#+NI5FBc(;8R   =   ----ESMI---PEO---
lGFN^f524b.cV4t++   =   -E-I-H-S--HE---MM
yBX1!:49CE>VUZ5-+   =   EP-TSA----------M
Ic.3zBK(Op^.fMqG2   =   -EH--PSOEI-HH--ES
RcT+L16C<+FlWBI)L   =   -ENMOT--IM--RP-AO
++)WCzWcPOSHT/()p   =   MMAR--RE-E--N-OAI
IFkdW<7tB_YOB!-Cc   =   ---SRI--P-TEPS--E
>MDHNpkSzZO8AIK;+   =   --A-II----E---S-M

(Position 246)
HER>pl^VPkI1LTG2d   =   -OENI-SE--P---EA-
Np+B(#O%DWY.<!Kf)   =   -IMAT-H----S--RH-
By:cM+UZGW()L#zHJ   =   A--OAMSSE-T---E--
Spp7^l8!V3pO++RK2   =   -II-S---EEIHMMERA
_9M+ztjdI5FP+&4k/   =   --AME---PI--M----
p8R^FlO-!dCkF>2D(   =   I-ES--HI--T--NA-T
#5+Kq%;2UcXGV.zLI   =   -IMRP--ASO-EESE-P
(G2Jfj#O+_NYz+@L9   =   TEA-H--HM---EM---
d<M+b+ZR2FBcyA64K   =   --AM-MSEA-AO----R
-zlUV+^J+Op7<FBy-   =   IE-SEMS-MHI---A-I
U+R/5tEIDYBpbTMKO   =   SME-I-OP--AI--ARH
2<clRJI!5T4M.+&BF   =   A-O-E-P-I--ASM-A-
z69Sy#+NI5FBc(;8R   =   E-----M-PI-AOT--E
lGFN^f524b.cV4t++   =   -E--SHIA--SOE--MM
yBX1!:49CE>VUZ5-+   =   -A------TONESSIIM
Ic.3zBK(Op^.fMqG2   =   POSEEARTHISSHAPEA
RcT+L16C<+FlWBI)L   =   EO-M---T-M---AP--
++)WCzWcPOSHT/()p   =   MM--TE-O-H----T-I
IFkdW<7tB_YOB!-Cc   =   P-------A--HA-ITO
>MDHNpkSzZO8AIK;+   =   NA---I--ESH--PR-M

(Position 270)
HER>pl^VPkI1LTG2d   =   --N--E----TIISTO-
Np+B(#O%DWY.<!Kf)   =   --SR-----A--O---H
By:cM+UZGW()L#zHJ   =   R--E-S--TA-HI-E--
Spp7^l8!V3pO++RK2   =   -----E------SSN-O
_9M+ztjdI5FP+&4k/   =   ---SE---T-E-S----
p8R^FlO-!dCkF>2D(   =   --N-EE----P-E-O--
#5+Kq%;2UcXGV.zLI   =   --S----O-E-T--EIT
(G2Jfj#O+_NYz+@L9   =   -TO-----S---ES-I-
d<M+b+ZR2FBcyA64K   =   -O-S-S-NOERE--M--
-zlUV+^J+Op7<FBy-   =   -EE--S--S---OER--
U+R/5tEIDYBpbTMKO   =   -SN----T--R--S---
2<clRJI!5T4M.+&BF   =   OOEEN-T--S---S-RE
z69Sy#+NI5FBc(;8R   =   EM----S-T-ERE---N
lGFN^f524b.cV4t++   =   ETE----O---E---SS
yBX1!:49CE>VUZ5-+   =   -R-I----P-------S
Ic.3zBK(Op^.fMqG2   =   TE--ER---------TO
RcT+L16C<+FlWBI)L   =   NESSIIMPOSEEARTHI
++)WCzWcPOSHT/()p   =   SSHAPEAE----S--H-
IFkdW<7tB_YOB!-Cc   =   TE--AO--R---R--PE
>MDHNpkSzZO8AIK;+   =   --------E----T--S

Unless I am missing something, there is NO WAY to make readable text around that in any placement. It's easy to fit strings in parts of the cipher, but one has to look at the entire thing, right? I just do not see how this is progress.

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