I don’t completely understand the method either but I love pi’s efforts to look at creative ways of reading the cipher. I’m almost convinced that Z’s apparent cipher genius is a single, quirky twist of the substitution method used in the 408 cipher. There are an infinite number of ways in which he could have changed the construction of the 340 and it could simply come down to guessing at a creative appproach like this. Best of luck, pi!

I had hoped that some measurement of language-like features in a cipher text would help indicate which manipulations would unravel the cipher text into a standard substitution cipher. Perhaps that is wishful thinking.

I had the same hope. Perhaps some metric other than the n-gram count might be more useful: IOC, presence of pivots like you suggested, etc. I will investigate that. Speaking of wishful thinking, I also have a bit of hope that the correctly re-arranged z340 (again, assuming it is a transposition of some kind) would display a hint of some kind that the proper transposition has been achieved (maybe a word or phrase is displayed in the transposed cipher text, maybe the "+" symbols are arranged in a non-random formation, etc.)

Duman’s approach, where he scores manipulations of the ciphertext by running ZKD trials on candidate manipulations, seems like a good approach, except for the difficulties due to the size of the search space. There are so many possible manipulations to choose from.

What he does is interesting for sure. However, I have 3 issues with using ZKD to score candidates:
1. It is an extremely expensive test to conduct when considering large search spaces.
2. I am quite uncertain that ZKD scores can be linearly compared in order to rank non-solution candidates.
3. Depending on the input params provided to ZKD, it might not necessarily converge on a correct solution.

To better explain point #3, I will use my test cipher (mentioned in a previous post) as an example. When I first created the cipher using the generator, it had 60 different symbols with a flat distribution and exhibited a few repeating n-grams. I ran this cipher in ZKD with the IoC Weight parameter set to the default 5. ZKD decrypted the cipher in a few seconds. I then made some modifications manually: increase total number of symbols to 63, have 1 symbol repeated 23 times, reduce the number of n-grams. ZKD could not find the solution anymore with the same IoC Weight param value, even after searching for a few minutes. By arbitrarily setting the IoC Weight param to 7, ZKD could again solve the cipher in a few seconds.

This makes me think that running ZKD automatically might miss a solution, unless multiple runs are executed on the same candidate using varying input params. This would make using ZKD an even more expensive scoring mechanism.

Thanks for the kind and encouraging words! Trying to find a solution to these ciphers sometimes feels like trying to extract the meaning of life from the writings on a box of cereal.

Perhaps it’s possible to implement a faster hillclimber. The algorithm described in the paper "Efficient Cryptanalysis of Homophonic Substitution Ciphers" has some interesting optimizations that might make for a faster solver. I wonder how it compares to ZKD.

Perhaps it’s possible to implement a faster hillclimber. The algorithm described in the paper "Efficient Cryptanalysis of Homophonic Substitution Ciphers" has some interesting optimizations that might make for a faster solver. I wonder how it compares to ZKD.

I had skimmed through this paper a while ago but I read it more thoroughly this time.

This algorithm has a hard time decrypting homophonic ciphers with qualities similar to the z340 (340 characters with 63 different symbols) with a success rate of only 10 to 20%. Figure #21 is quite eloquent in that regard.

However, and I find this very interesting, they tested their algorithm on a test cipher that essentially is a shortened version of the z408 with similar qualities as the z340 (340 characters and 65 different symbols). They obtained a success rate between 4 and 13%. They then tried to use the z408 statistics as the target instead of using normal English statistics. They obtained a success rate of 70 to 84%.

They then conclude, regarding using this algorithm to break the z340:

The results in this section strongly indicate that it should be worthwhile to consider specialized “language” models when trying to break the Zodiac 340. It might also be useful to employ the slow outer hill climb, but that appears to yield a more modest improvement. However, for the Zodiac 340, we have a brief ciphertext (only 340 symbols) and a relatively large alphabet (62 symbols). From Figure 21, we see that for these parameter values, we will need every possible advantage to have a reasonable chance of breaking the cipher.

Also:

These results indicate that the Zodiac 340 cipher—assuming it is a homophonic substitution—may be out of range for our attack. However, tests indicate that with a more accurate plaintext “language” model and a more thorough outer hill climb, we should have a realistic chance of breaking a cipher such as the Zodiac 340.

In other words, a good way to optimize a hill-climber to try to break the z340 would be to use the zodiac’s corpus (or a subset) to construct a target matrix instead of using standard English statistics. I think ZKD can be configured to use such statistics.

Based on these results, I also wonder if ZKD is not more efficient than this algorithm…

Meant to say. Love the look of the route diagrams. Very stylish.

I’ve just finished reading this thread and find it interesting.

Clearly, by rearranging the z340 as if each quadrant is a distinct route cipher, it is possible to significantly increase the repeating n-gram count of the resulting cipher; in some cases, the pattern count is more than quadrupled.

_pi, notice that in the image "quadroutes_3.gif" most of these patterns include the "+" symbol. Making me wonder about the validity. I just would like to see a comparison versus another solveable "340" cipher with the "+" symbols enforced to the same positions.

ePCC0Og1Z]T+M]Leb
B`oSk+PiTLj+V;P0
?iT;h^17?eB3SEH^
]+MekOK196IdJWQD_
hFE8bI6Jjab]B+LZj
1Z;Pk+pTfkm078+R`
L=?:-LXOkg++2fI@;
e+2V]IJ8QdKjc`Fp-
EJ1^PH5@LJ^+OBVPT
N6dmjT4A+jKefF?G1
DH]A+:dmX4EgBLX=X
PTK7I1]TXh0p6+9KZ
]0E[01fFJog+36iC=
_a+Oi120_Om?:S4U[
^:o+G4:5oVfkgeFXQ
;kmZJOiQLV=+X@id
X7U[T240+7+Xf30O6
BTW4K=54+PYea39KK
;H+R+V`1:02f0C6IS
]iF:^DbVhW:3o[NY

Thank you for this cipher, the symbol cycle over the letters is random right? In terms of the uniques (see viewtopic.php?f=81&t=2114) this cipher differs from the 340 in a few ways I tested, it is much more random and removing the "+" symbol does not improve this, but it does vastly so for the 340.

BTW, this cipher can be solved by ZKD in a few seconds.

I’m using ZKDecrypto 1.0 and your cipher does not solve within the standard settings, max failure: 2000, random swaps: 5, revert period: 400. I saw a readable result from my decrypting program as soon as it fired up (within a second) it starts with "iapproachedthewitnessstandwith". I find this highly disturbing as I assumed ZKDecrypto would be able to solve ciphers within these parameters meaning I have to redo some of my work, can someone confirm or disconfirm this?

Perhaps it’s possible to implement a faster hillclimber. The algorithm described in the paper "Efficient Cryptanalysis of Homophonic Substitution Ciphers" has some interesting optimizations that might make for a faster solver. I wonder how it compares to ZKD.

If I’m not mistaken the guy(s) or university that wrote the paper have 3 papers on the subject from different years, in all of them the same mistake is made, they interpreted the 408 as a 53 symbol cipher and the 340 as a 62 symbol cipher, subsequently, using wrong mapping of the ciphers. They also work with a "random initial key layer", in their last paper they off-load to a GPU 1.000.000 iterations of these to get a better start. I tested that, at least, with "random swapping" such a key layer is not worth its time, at any number of iterations. Let it be 100 or 1.000.000 or 1.000.000.000.000 random keys, it is still nothing in face of the searchspace. They present interesting ideas however.

_pi, notice that in the image "quadroutes_3.gif" most of these patterns include the "+" symbol. Making me wonder about the validity. I just would like to see a comparison versus another solveable "340" cipher with the "+" symbols enforced to the same positions.

The + symbol does appear frequently in these n-grams. I don’t think however that these n-grams reveal patterns of natural language; they are simply artifacts appearing as a result of scrambling the cipher.

I have executed this quad-route descrambling strategy on some other ciphers and have noticed that for some ciphers, certain descrambled candidates will exhibit much higher n-gram counts than the original cipher. For other ciphers, it appears that all descrambled variations will have a lower n-gram count than the original. This means that for certain unscrambled substitution ciphers, applying this strategy (and thus scrambling the cipher) increases the n-gram counts.

In the end, I just came to realize that looking at the n-gram count alone to score a transposition candidate was a bad idea…

Thank you for this cipher, the symbol cycle over the letters is random right?

Probably. I used a cipher generator that I mention early on in this thread; I don’t know how it is implemented.

I’m using ZKDecrypto 1.0 and your cipher does not solve within the standard settings

That’s correct. To quote myself from a previous post in this thread:

When I first created the cipher using the generator, it had 60 different symbols with a flat distribution and exhibited a few repeating n-grams. I ran this cipher in ZKD with the IoC Weight parameter set to the default 5. ZKD decrypted the cipher in a few seconds. I then made some modifications manually: increase total number of symbols to 63, have 1 symbol repeated 23 times, reduce the number of n-grams. ZKD could not find the solution anymore with the same IoC Weight param value, even after searching for a few minutes. By arbitrarily setting the IoC Weight param to 7, ZKD could again solve the cipher in a few seconds.

This makes me think that running ZKD automatically might miss a solution, unless multiple runs are executed on the same candidate using varying input params. This would make using ZKD an even more expensive scoring mechanism.

Thank you,

Your cipher solves easily with IoC 3. Good to know.

I wonder how the randomness of the homophonic cycle over the symbols affects the n-grams, with and without transposition. We can assume that n-grams from the following three sources will be different somehow: random text with flat frequencies, random text with english frequencies, english text.

In the thread http://zodiackillersite.com/viewtopic.p … 4&start=20 the Z340 quadrant hypothesis lives up again.

I ran a test to calculate the odds of having "By", BY", "yB" or "YB" appear just once in every quadrant from random shuffles (q1: 9 by 10, q2: 8 by 10, q3: 9 by 10, q4: 8 by 10). Only horizontal appearances just like in the Z340.

The odds are roughly 1 in 10,000.

In the thread viewtopic.php?f=63&t=4824&start=20 the Z340 quadrant hypothesis lives up again.

I ran a test to calculate the odds of having "By", BY", "yB" or "YB" appear just once in every quadrant from random shuffles (q1: 9 by 10, q2: 8 by 10, q3: 9 by 10, q4: 8 by 10). Only horizontal appearances just like in the Z340.

The odds are roughly 1 in 10,000.

So fairly rare then ? I know this is old ground but I still wonder if what I said a while ago could possibly be an idea to investigate.

"I have always taken it a different way. The crosses Paradice / Slaves is a clue the he was referring to the cipher. Also the "by" appears in each quadrant. Could he have been giving us a word that appears in each quadrant to try to help us along I.e. Gun, Fire, Knife & Rope. Also could the direction of the writing clue us in on the direction of the writing, could it be treated as 4 separate parts with 3 being horizontal and one being vertical. But at the end of the it could be anything." – In Relation to the Halloween Card.

Could dieby be an encipherment key? Or perhaps diebyfire for one quadrant, diebyknife for another, etc?

Could dieby be an encipherment key? Or perhaps diebyfire for one quadrant, diebyknife for another, etc?

great idea. split each quadrant and use keyword similar to how a vigenere cipher works.

"I have always taken it a different way. The crosses Paradice / Slaves is a clue the he was referring to the cipher. Also the "by" appears in each quadrant. Could he have been giving us a word that appears in each quadrant to try to help us along I.e. Gun, Fire, Knife & Rope. Also could the direction of the writing clue us in on the direction of the writing, could it be treated as 4 separate parts with 3 being horizontal and one being vertical. But at the end of the it could be anything." – In Relation to the Halloween Card.

Good points. I am thinking of a test where each quadrant could be 16 different directions (normal, reversed, mirrored, flipped, columnar variations and diagonal variations). That would make a total of 16^4 = 65,536. Thus, a very manageable amount of variations to process. From there also other quadrant dimensions could be considered but since the "BY" observation strictly defines the quadrants perhaps there is no reason? Also don’t know what to do with the middle column and (possibly) row. Though, from a statistical point of view that column and row seems to be part of the cipher so I’d say that it was just used in the Avery card to clearly divide the quadrants.

I made the Z340 test ciphers, 1,229,312 in total, download: https://drive.google.com/file/d/1ij5m8o … sp=sharing

Feel free to conduct your own tests on these ciphers.

The quadrants considered are 9 by 10, 8 by 10, 9 by 10 and 8 by 10. A strict interpretation. Each quadrant can individually have 28 distinct transpositions (normal, mirrored, flipped, reversed, columnar, diagonal variations). The cipher is then recombined into the 17 by 20 grid itself and also into an array while taking the quadrants (rectangles) off in LRTB order (perhaps hinted at by Zodiac listing "by fire, by gun, by knife and by rope" alphabetically in LRTB order on the Avery card).

The test will take a while so hold on. Here’s how a cipher appears in the file:

[Grid] means that the quadrants have been recombined in the 17 by 20 grid itself and [Array] means they were recombined into an array linearly LRTB. TP means transposition and UTP means untransposition (or wrap/unwrap). I’ve added cycle spectrum and bigram repeats.

cipher_information=[Grid] Diagonal 3(TP), Diagonal 3(TP), Diagonal 2(TP), Flip(TP), Cycle spectrum * 10000: 5712, Bigram repeats: 19
p#U#+2U^+>.zB47B-
_85(Ob-VJ&2z+cK<y
p9R+G++z++4DL@yOF
cpM^K2dZlzRk(|LAp
OM7+FqJ<R)HK/c_96
p%+^zl%fM.WJ2dXNF
^+DUltO;j2<(35CGY
>VBBZ8j-2Ld*)pFkV
EpP(yG*d*|TWKLOPF
HRlN#:SV|k1GYf#++
URt2J9l4CZO8A|K;+
+5DR6|29O_YOB*-Cc
/|lzN54(COSHT/()p
Ec5+f:K6z+FlWB|)L
<*#^*B1Ckp^.fMqG2
|yN1zLWFtE>VUZ5-+
SFX3+)|7Db.cV4t++
GB.T+P<Mp5FBc(;8R
ycc+cW>NST4M.+&BF
|R<WdBHkzYBpbTMKO

cipher_information=[Array] Diagonal 3(TP), Diagonal 3(TP), Diagonal 2(TP), Flip(TP), Cycle spectrum * 10000: 5339, Bigram repeats: 20
p#U#+2U^+_85(Ob-V
Jp9R+G++z+cpM^K2d
ZlOM7+FqJ<Rp%+^zl
%fM^+DUltO;j>VBBZ
8j-2EpP(yG*d*HRlN
#:SV|>.zB47B-&2z+
cK<y+4DL@yOFzRk(|
LAp)HK/c_96.WJ2dX
NF2<(35CGYLd*)pFk
V|TWKLOPFk1GYf#++
URt2J9l4C+5DR6|29
O/|lzN54(CEc5+f:K
6z<*#^*B1Ck|yN1zL
WFtSFX3+)|7DGB.T+
P<Mpycc+cW>NS|R<W
dBHkzZO8A|K;+_YOB
*-CcOSHT/()p+FlWB
|)Lp^.fMqG2E>VUZ5
-+b.cV4t++5FBc(;8
RT4M.+&BFYBpbTMKO

Best cycle spectrum score: cipher_information=[Grid] Normal(TP), Normal(TP), Columnar 4(UTP), Diagonal 6(UTP), Cycle spectrum * 10000: 6372, Bigram repeats: 25
Best bigram repeats: cipher_information=[Grid] Normal(TP), Normal(TP), Diagonal 5(TP), Diagonal 6(TP), Cycle spectrum * 10000: 5753, Bigram repeats: 39

cipher_information=[Grid] Normal(TP), Normal(TP), Columnar 4(UTP), Diagonal 6(UTP), Cycle spectrum * 10000: 6372, Bigram repeats: 25
HER>pl^VPk|1LTG2d
Np+B(#O%DWY.<*Kf)
By:cM+UZGW()L#zHJ
Spp7^l8*V3pO++RK2
_9M+ztjd|5FP+&4k/
p8R^FlO-*dCkF>2D(
#5+Kq%;2UcXGV.zL|
(G2Jfj#O+_NYz+@L9
d<M+b+ZR2FBcyA64K
-zlUV+^J+Op7<FBy-
zBP<OC4|5OFKRBM+8
DStcC(92N&T++;+b2
*|k7W6K45-t(.pLG5
+|Ep<z1B:4cMBp)qZ
f#JtNWCLzVB4Yc)|M
*^yR5HdW+UcFT+C(B
31NSl/Dk)fV.5;-/W
T.XF9cRMF.>bK*Tl^
+ccBG6<+>E|BHFpAO
|+R|ylz2US+8YOO_Z