Wasn’t me. Glurk has a partial solution here:

http://www.zodiackillerfacts.com/forum/ … ?f=11&t=83

I think the full solution is still unknown. Been a while since I looked at this.

I had the same "solution" back in Aug 2009.
Got the "Oakland, CA" from, apparently Glurk (couldn’t remember who it was at the time I posted it)
And took it from there:

http://www.zodiackillerfacts.com/forum/ … =51&t=1199

Long time no see..bit busy over the last days, partying, working, decrypting. Some things I figured out, you might find them useful (or not):

On the first look, the post stamp on the celebrity cipher card reads as ’23 September 1980′. Most websites claim the card to be from 25th or from 1990 or something like that. However I think it quite well could be the 23rd of September, which in fact is autumn equinox, too.

Then something different regarding the decryption: In fact we got two approaches to solve any substitution cipher, no matter if homophones or cesar:

1. Bottom-up – We take the cipher, look into its structure, replace some parts of it with language and continue to add more of it until it is solved. This method may work, however it very well may lead into a dead end. Previous because if you choose one letter wrong, everything else you continue with is going into the dustbin. Most people use this cipher decryption method.

2. Top-down – This is a different way, the one any cryptologist is afraid of: We take all (!) possible solutions technically thinkable. Therefore, for the first symbol all alphabetical letters may be correct, same with the second etc.. The trouble is programmed in advance…a cipher with approximately 20 different symbols thus has 20^26 different possibilities which shall be reduced by all those which do not comply with ‘encryption rules’. The most important (and potentially only) encryption rule is most likely that one symbol would not stand for two different letters, although one letter could very well be represented by two different symbols. A typical homophone substitution cipher. In our 20^26 case this leads us to 6710886,4 billion billion billion varietes to look at. However, this decryption method, fully NSA style, is rock solid as it definitely enables us to get the solution.

And I tried the latter one. Looking at it, we find a second ‘encryption rule’ which may be assumed. The celebrity cipher is, opposite to the 340 and the 408, showing up with single words instead of one fluent text without spaces. We therefore can start to decrypt, for this reason let’s name the first four words of the cipher as A, B, C and D. The following is the word length of these words:

A – 5 letters
B – 8 letters
C – 10 letters
D – 7 letters

leading us to the conclusion that the first word has 5 letters, the second word has 8 letters etc. Now let’s start! We take a huge dictionary and go for it!

1. Let’s take all 10-letter words of which the 1st and the 7th letter is the same AND of which the 4th letter and the 10th letter is the same (e.g. ‘transition‘)
2. Let’s take all 7-letter words of which the 3rd and the 7th letter are – each – identical with the 3rd and the 5th letter of the words found in step one. No more than a theory of sets.
3. Let’s take all 8-letter words of which the 2nd and the 5th letter are – each – identical with the 1st and the 3rd letter of the 10-letter words found in step one.

And we should (or should not) get a readable text of three words.

In fact we can continue this ‘reduction’ of varietes to the degree that we additionally involve potential 2-letter words. The first letter of the first 2-letter word in the cipher is in fact the 6th letter of the 10-letter words again! And so on..continuing with the two 4-letter words, the reduction goes on.

Furthermore, we can (mainstream thinkingly) throw away all those varietes leading to a first letter of any 2-letter words in the cipher that however does not lead to a valid 2-letter word of our dictionary (such as ‘Y’ as not many good 2-letter words starting with ‘Y’ actually exist – it is more useful to look for words such as no, go, do, so, to, we etc.). Also we can look for double 2-letter words, such as ‘to do’ etc.

Three possible outcomes:
– No solution at all
– One solution
– Multiple solutions (spectrum of solvable varieties)

The celebrity cipher DOES deliver multiple solutions. There are approximately 700 solutions resulting from the three steps performed (theory of sets of the 8-/7-/10-letter words, considering their symbol structure). This mainly is the solutions spectrum of the celebrity cipher. It EXPANDS however by adding the rest of the cipher, the 4-letter words, the 2-letter words etc..

An example:

‘xxxxx FILMSETS INSENSIBLE DUSTBIN, Sx ‘

is definitely one valid variety of the first five words of the cipher, with x being a joker, of course. According to this, the fifth word most likely would be ‘SO’. It could continue with ‘SO IF’, ‘SO GO’, etc. However we do not want to guess but rather use ALL varieties and only reduce them if it is absolutely necessary. So we may continue with the two 4-letter words, as mentioned above. 4-letter words are frequent, however, so we may find some of them starting with the 3rd letter of the 10-letter word and ending with any letter that would start a 2-letter word. In our case, ‘SODA’ could be such a word if e.g. the first 2-letter word is ‘AM’. It could, however, be also the word ‘SOHO’ if the first 2-letter words is ‘ON’, ‘OR’ or ‘OF’.

So far so good. The second 4-letter word is ending with the first letter of a 2-letter word on position 3 and the last letter of our 8-letter word. If we use ‘AM’ and ‘FILMSETS’, the structure of the second 4-letter word would therefore look like this: ‘xxAS’ and could the word GLAS very well match into this position.

Considering the theory of sets above plus letters 1 and 6 of the 10-letter word being a first letter of any 2-letter word, we get our spectrum of solutions. This spectrum delivers a variety of technically valid solutions for the celebrity cipher, of which most likely at least one must be correct as a readable massage.

Now here comes the number:

This is the list of 10-letter words matching the procedure above:
acidulated
aftertaste
appearance
ascendable
assessable
attendance
capsaicins
cheesecake
coldcocked
condescend
conspectus
effervesce
excellence
expedience
experience
horseshoes
horsewhips
improviser
improvisor
ingestible
insensible
lateralize
liberalise
liberalize
literalize
melanomata
nonsigners
postimpact
procryptic
redescribe
regenerate
restaurant
roisterers
roisterous
stochastic
streamside
synthesist
tolerative
transition
trisectors
wickerwork

Each of these words has at least one 7-letter word matching the structure of cipher symbols. Also, each of these words has approximately 20-150 words of length 8 matching the structure (procedure above). The point is that actually there are no more 10-letter-words, except possibly names or any other language, that would match the structure according to this procedure. Z’s or the author’s solution therefore should, in any case, be amongst these.

The completed structure of these solutions currently does look like this:

..
__X_e finishes insensible…, s_ … s__i … if __is i_ __X_e finishes
__X_e fireside insensible…, s_ … s__i … if __ie i_ __X_e fireside
__X_e filister insensible…, s_ … s__o … of __or i_ __X_e filister
__X_e filmsets insensible…, s_ … s__o … of __os i_ __X_e filmsets
..

and so on with the two 4-letter words still to be placed in as well as the 5-letter word. And these to replace is – all top-down method – possible, too!

So for each structure we can add our two 4-letter words according to the method already described. For the structure ‘__X_e fireside insensible…, s_ … s__i … if __ie i_ __X_e fireside’ we can actually find exactly 11 words matching the first 4-letter words and exactly 8 words matching the second 4-letter word.

We are talking about five (!) words with length>3 being solved in the celebrity cipher.

However we do get multiple solutions. And the possibility of cross-referencing, as amongst the three words of length 7, 8 and 10, is limited. For example, if we define one 4-letter-word as being the ‘real’ one, we still do have 8 different 4-letter words for the second one.

So far, I did check only the groups of ‘INSENSIBLE’ and ‘AFTERTASTE’ as potential 10-letter words. These two words alone deliver (and due to lack of cross-referencing in fact DO deliver) the complete amount of 258 ‘structures’ with exactly (!) 95,466 solutions, depending on which 4-letter words you chose.

IMO only few of the derived 10-letter words are useful. Eg. ‘restaurant’ is not as useful as ‘transition’, just due to the fact that there is rather no single standing 2-letter word starting with an ‘R’. Nevertheless at least these do remain:

aftertaste
ascendable
assessable
attendance
horseshoes
horsewhips
melanomata
nonsigners
stochastic
streamside
tolerative
transition

all of them to be balanced with both, the 8- and the 7-letter words availabe.

Now what does this mean?

It simply leads us to the definite (!) conclusion that – technically – at least a few hundred thousand solutions do exist for the celebrity cipher. Now, after few hours of excel configuration, it is quite easy to list all those technically correct solutions. ‘APPROACH PROCRYPTIC SHOOTER’ nevertheless will not be the only solution, however, and could be a worse one, too.

IMO therefore we should continue to

a.) select all solutions possible, at least those containing the longer words B, C and D
b.) connect these phrases to e.g. Marianne Moore’s literature if Z has used her stamp intentionally (there is some reason for this…Marianne Moore had written letters to David Browner, chief of the SIERRA CLUB…)

So it could be quite worth looking into the connection of Marianne Moore and the Sierre Club itself..

One of many possible solutions:

‘ROSES THEORIES HORSEWHIPS SURVIVE, WE DO NOT .. ROMA AT GLAS HE ROSES THEORIES’
one of many technically correct solutions of the celebrity cipher..however without any sense in it..

QT
(on demand or when it’s ready I can place the excel sheet here, too, still some ‘exceling’ to do, however)

Catalogue of copyright entries, third series 1969 Jan-June
connection between Marianne Moore and the Sierra Club

https://books.google.at/books?id=mSwhAQ … 22&f=false

Part of the cipher is covered by a stamp. Some have speculated what might lie beneath, but I think it’s actually pretty simple. Any of you older farts remember these commercials?

https://www.youtube.com/watch?v=jxsayi0ZpVs

With that in mind, I think the body of the postcard reads, "Do you know me? My name is…"

Unfortunately, this adds little to no insight into the name that follows. Maybe an anagram. And you know how much I just love anagrams.

Part of the cipher is covered by a stamp. Some have speculated what might lie beneath, but I think it’s actually pretty simple. Any of you older farts remember these commercials?

https://www.youtube.com/watch?v=jxsayi0ZpVs

With that in mind, I think the body of the postcard reads, "Do you know me? My name is…"

Unfortunately, this adds little to no insight into the name that follows. Maybe an anagram. And you know how much I just love anagrams.

Thought about that one, however this would imply that the ‘M’ symbol would once represent the letter W (‘know’) as well as the letter ‘M’ (‘my’). If that was possible, you could place any text into the cipher. Exactly that, imo, is not the case..there are a lot of different possible solutions and we should look for a solution working out with both – sense and cipher structure. The two 2-letter words e.g. could also be:

AS HE
AS IF
AS IN
AS IS
AS IT
AS NO
AS OF
AS ON
AS TO
AS US
AS WE
AT AN
AT ME
AT NO
AT ON
AT UP
AT US
BE AN
BE AS
BE AT
BE IF
BE IN
BE IT
BE ME
BE NO
BE OF
BE ON
BE OR
BE SO
BE UP
BE US
BY AN
BY IF
BY IN
BY IT
BY ME
BY NO
BY OR
BY SO
BY TO
BY UP
BY US
DO GO
DO IF
DO IT
DO ME
DO NO
DO OR
DO SO
DO TO
DO US
GO AS
GO AT
GO BY
GO ON
GO TO
GO UP
HE IS
HE OR
HI AT
HI TO
IF AT
IF HE
IF IN
IF IT
IF NO
IF WE
IN AN
IN AS
IN AT
IN HE
IN IF
IN IT
IN NO
IN OR
IN SO
IN TO
IN US
IN WE
IS AN
IS AT
IS BY
IS HE
IS IN
IS IT
IS ME
IS NO
IS OF
IS ON
IS OR
IS SO
IS UP
IS US
IT AS
IT BY
IT IN
IT IS
IT ON
IT OR
IT SO
IT TO
IT UP
ME AS
ME AT
ME BY
ME IF
ME IN
ME NO
ME OF
ME ON
ME OR
ME SO
ME TO
ME UP
NO AD
NO IF
NO OR
NO SO
NO TO
NO US
OF IT
OF NO
OF US
ON AS
ON AT
ON HE
ON IF
ON IT
ON ME
ON NO
ON OR
ON TO
ON US
OR AM
OR AN
OR AS
OR AT
OR BE
OR BY
OR SO
OR GO
OR HE
OR IF
OR IN
OR IS
OR IT
OR ME
OR NO
OR SO
OR TO
OR UP
OR US
OR WE
SO AN
SO AS
SO AT
SO BE
SO DO
SO GO
SO HE
SO IF
SO IN
SO IS
SO IT
SO NO
SO OF
SO OR
SO TO
SO WE
TO AN
TO BE
TO DO
TO GO
TO IT
TO ME
TO NO
TO OR
TO SO
TO US
UP AT
UP BY
UP GO
UP HE
UP IF
UP IN
UP IT
UP ON
UP OR
UP SO
UP TO
US AT
US BY
US IF
US IN
US IT
US NO
US OF
US ON
US OR
US SO
US TO
US UP
US WE
WE AS
WE DO
WE GO
WE IF
WE IN
WE OF
WE OR
WE SO

The symbols under the stamp are not really required as there might be a short word only. This one, again, might be replaced with multiple different e.g. 3-letter-words. Too many varieties then. So we should find a combination of the three longer words (7, 8, 10) with sense in the text. This in combination with the list above would be a good find, especially if we then find the correct 4-letter words. There are about 10 to 50 possibilities for each after having selected the previous.

One interesting content find is e.g. the following combination:

8-letter word – SCOOPFUL
10-letter word – CAPSAICINS
7-letter word – PAPRICA

Capsaicin is actually an alkaloid that makes the paprica spicy/hot. And there can be a scoopful of it. So we got some sense in it somehow and do follow the cipher structure. However I doubt this one to be correct as it would lead to a 2-letter word starting with ‘C’ and also any verb is missing so far, too.

This is the cipher structure:

Find the Excel sheet below, however I deleted both 4-letter word definition parts as well as the 10-letter-word selection procedure as the sheet would otherwise be too big (25mb, slow internet..). There you can see the B, C, D word combinations possible according to the cipher structure.

Best is to start in line 517 with the C-word, choosing the 10-letter word. Other 10-letter words do not match the cipher structure. The procedure is described on the second sheet of the file. The yellow 10-letter-words comply with the start of the first 2-letter word (‘B’ symbol is on position 6 of the 10-letter word and on position 1 of the first 2-letter word).

Then you go down, looking for a matching D-word with 7 letters. Finally you go up to line 1 and you’ll find a variety of B-words with 8 letters. Some of them are grey as they do not comply with 2-letter words either.

Now you can use the three words, B-C-D to find some potential content in it. Everything else, imo, does not match a substitution or homophone encryption of the english (dictionary..) language.

QT

The 9 being at the end all the time may be a full stop.
I like the phrase "do you know me" from the suspected fake 78 letter. Maybe it isn’t a fake

Other solutions possible..

‘__D_N STEWARDS TRANSITION DIARIES, IT IS (or ‘IF IT’)..’
‘__V_A EMBALMED MELANOMATA VILLAIN, O…’
‘__P_S THWARTED HORSESHOES PURPOSE, SO…’

https://en.wikipedia.org/wiki/Horseshoe … California)

Lake horseshoe is near Redding, not far away from Paradise town.

QT

Another fit for the "IZY+FBI>c+" section is "OGLETHORPE".

Which means "O<YL>4F" becomes "__L_R_T". A word that fits that pattern is "CULPRIT".

Choosing that, "zRO1+" becomes "__C_E" which matches these: "CACHE, NICHE, UNCLE, CYCLE, LOCKE, ROCHE, LUCIE, EMCEE, SUCRE"

I like "UNCLE" so that means "dI^2Y3lp" becomes "_I__L___" which matches over a hundred words, but I like that it matches "SINCLAIR" to complete the name:

UNCLE SINCLAIR OGLETHORPE CULPRIT

And this is where I give up, because as you say, there are too many possible solutions with no clear winner.

When I read solutions that read like:

GRANULAR POKEMON RICE
SPECIOUS CONCRETE TRACTOR
OBJECTIVE FROG PERSONA
LUDICROUS PAPYRUS IDEA

These just don’t work, IMO. Maybe it’s just me.

-glurk

It’s fun to play, but an insane waste of time after 5 minutes.

NEXT!!!

Part of the cipher is covered by a stamp. Some have speculated what might lie beneath, but I think it’s actually pretty simple. Any of you older farts remember these commercials?

https://www.youtube.com/watch?v=jxsayi0ZpVs

With that in mind, I think the body of the postcard reads, "Do you know me? My name is…"

Unfortunately, this adds little to no insight into the name that follows. Maybe an anagram. And you know how much I just love anagrams.

Thought about that one, however this would imply that the ‘M’ symbol would once represent the letter W (‘know’) as well as the letter ‘M’ (‘my’).

A symbol representing more than one letter… you mean like he did with the 408? Yeah, probably not very likely.

When I read solutions that read like:

GRANULAR POKEMON RICE
SPECIOUS CONCRETE TRACTOR
OBJECTIVE FROG PERSONA
LUDICROUS PAPYRUS IDEA

These just don’t work, IMO. Maybe it’s just me.

-glurk

Glurk, didn’t you know that Z, besides collecting slaves for his afterlife,also collected pokemon cards

"Objective frog persona" might have appeared in one of the profiles of Z over the years…

"Objective frog persona" might have appeared in one of the profiles of Z over the years…

He felt it wasn’t easy being green?

Stuff like ‘objective frog persona’ doesn’t match the cipher structure. Also I do not believe that Z had used one symbol for two letters either.

The cipher structure is complicated but not complex. In fact it is solvable in multiple ways, possibly thousands. At least one of these solutions should lead to readable text with some content somehow. On first view I could not find a solution that is both, matching the cipher structure and some content.

"my home at" could be the cleartext for the last three short words, for example, as the ‘M’ is actually represented by the same symbol. Therefore the two longer repeating words might be an address or place.

The cipher is solvable, the task is rather to find the correct solution.

QT

QT-

My point was (and is) that the solution should actually be some sort of coherent English, not a series of random words stuck together, or "word salad" as some call it.

And it seems very likely that in this cipher, we are probably going to find a street name / surname / address / building name, etc., that will not be found in any dictionary.

-glurk