Zodiackillersite.com

I don't know John. is he a member here? I think his proposed solution deserves more credit and another look.

http://zodiackillersolved.blogspot.com/ ... ution.html

I think his solution the best I've seen so far. Unfortunately it breaks down at the end like all not completely correct solutions will,
but I saw something in his solution that points directly to my suspect and wouldn't make sense with any other known suspect.
So I actually mentioned this to LE with my proposed solution's to the Z13, Z32, EBEORIETEMETHHPITI (it's not just filler), and the Celebrity Cypher.

I think splitting into quadrants using either number keys like 0,3,6,9 from the Z32, or Keywords like Rope Fire Knife Gun will eventually give the answer.

Now do I want to scramble my brain trying to help solve it though.

Due to the fact that it doesn't match with the cipher homophones I personally wouldn't look into this self-claimed solution too much..

QT

Because of the structure of the first three lines of the 340 you can make lots of sensible sentences. Small challenge see who can make the longest sentence.s

The solution has an invisible character at the end of line 11, and one character less at line 17. The best option for the hypothesis considered was to remove the character at line 11 and insert an extra one at line 17. But possibly the plaintext is broken somewhere else.

Code: Select all

IWILLMAKETHISVERY
CLEARMYONEREASONI
AMKILLINGIFIDONTK
ILLIAMLIKELYTOOAB
ANDONMYCHANCETOCO
LLECTMYSLAVESFORT
HEAFTERLIFEEVENWH
ENIKNOWYOUCANTFIN
DMEIAMNOTSAFEWALK
INGIFIAMCOLLECTMY
SLAVESWHENSLAVESY
OUFOUNDSAVEDDEADI
NACARHECHANGEDHIS
MINDANDRAODFROMME
HEWALKEDBYWITHOUT
HIDINGFROMAENOTHI
NGISSAFEHEWASHIDL
INGAKNIFEYOUEVENL
ISTENEDTOOMYSLAVE
WASCALLINGFORHELP


Looking at the 340 as a set of the plaintext,

In a whole, the solution plaintext maps with 22.80% accuracy to the 340, by comparison another plaintext mapped with 5.45% accuracy to the 340. Top halves map with 27.79% accuracy and bottom halves map with 16.85% accuracy. Quadrant 1 maps with 30.76% accuracy. Quadrant 2 maps with 24.44% accuracy. Quadrant 3 maps with 16.66% accuracy. Quadrant 4 maps with 17.74% accuracy.

The plaintext may be broken somewhere, but you would expect to see close to 100% accuracies if the hypothesis is correct.

Didn't he simply use multiple keys? I.e., he squeezed as much forced plaintext out with the 1st key until it stopped working, then arbitrarily started a new key later in the ciphertext to regain the freedom to squeeze out more forced plaintext. I have a hazy memory of doing an analysis of his solution a long time ago. Maybe have been on Butterfield's forum but my posts there got lost.

If I'm right then anyone could come up with interesting readable plaintexts this way.

I found some old code I wrote long ago that extracted the symbol assignments from his solution. It shows that he assigned numerous plaintext letters to each cipher symbol. He did this so freely that he might as well have been decoding a one-way pad and inserting his own solution with an arbitrary key!

Code: Select all

Cipher symbol: Plaintext assignments, Cipher symbol count, Ratio of symbol count to number of plaintext assignments
U: I , 5, 2.5
z: NI, 9, 4.5
p: LSO, 11, 3.6666667
X: E, 2, 2.0
R: IOEL, 8, 2.0
O: YOSRF, 10, 2.0
B: ATNHSO, 12, 2.0
(: RFTE, 7, 1.75
+: ELTOAIMCSDHUNP, 24, 1.7142857
|: HWNCTI, 10, 1.6666666
y: MEA, 5, 1.6666666
): IGN, 5, 1.6666666
b: AL, 3, 1.5
^: ACDM, 6, 1.5
W: EIAN, 6, 1.5
V: KVFW, 6, 1.5
P: EC, 3, 1.5
L: SDWI, 6, 1.5
<: AMEO, 6, 1.5
6: AN, 3, 1.5
l: MGFSW, 7, 1.4
f: NAE, 4, 1.3333334
Z: NTG, 4, 1.3333334
S: ICO, 4, 1.3333334
J: KMU, 4, 1.3333334
9: NAE, 4, 1.3333334
8: LHO, 4, 1.3333334
k: TCEL, 5, 1.25
c: IFUGDHNE, 10, 1.25
T: VAGE, 5, 1.25
N: CENA, 5, 1.25
F: NTSCDAIE, 10, 1.25
C: VDFK, 5, 1.25
G: EGNMT, 6, 1.2
4: OLVRK, 6, 1.2
.: EDOIA, 6, 1.2
*: SILDA, 6, 1.2
M: LDEVNA, 7, 1.1666666
K: OAFKEG, 7, 1.1666666
5: AEVSHN, 7, 1.1666666
t: MEOT, 4, 1.0
q: TO, 2, 1.0
j: YO, 2, 1.0
d: YCADE, 5, 1.0
_: AUO, 3, 1.0
Y: RAEM, 4, 1.0
H: ITUC, 4, 1.0
E: WSB, 3, 1.0
D: NRHS, 4, 1.0
A: WR, 2, 1.0
@: F, 1, 1.0
>: LFYW, 4, 1.0
;: RDL, 3, 1.0
:: KL, 2, 1.0
7: ILD, 3, 1.0
3: ED, 2, 1.0
2: RBOLITYDH, 9, 1.0
1: IWS, 3, 1.0
/: OAV, 3, 1.0
-: SIYOA, 5, 1.0
&: TE, 2, 1.0
%: OE, 2, 1.0
#: MOHWR, 5, 1.0



The total number of symbol-to-plaintext assignments is 249. That is equivalent to decoding a cipher that has multiplicity of 249/340 = 0.73. It's very easy to make up fully readable messages that way.

I cannot even play this game of calling one incorrect "solution" better than any other. There are so many "self-proclaimed" solutions out there, and none of them have any verification at all. It's always one person, claiming a "solve."

Show me a solution that has a second (or more) person that verifies it. Then I MIGHT take a look. As far as I am aware - correct me if I am wrong - there still is no proposed 340 solution that two qualified people agree is correct.

It is still unsolved.

-glurk