Vallejo Police Chief Jack E Stiltz urged the killer to send in more details to prove the writer of the July 31st letters was the killer. He promptly did this by sending the August 4th Debut letter. If the killer had written his identity at the foot of 408, that would obviously include the word "Zodiac", then why would he reveal "Zodiac" only 4 days later in the August 4th letter. Is it conceivable that the killer wrote the August 4th communication to prove a link to the crimes, but also included the word "Zodiac" to prove his link to the July 31st letters and the unsolved 18 characters at the foot of the 408? Had the 18 unsolved characters been deciphered, and it had contained the word "Zodiac", the author of the August 4th letter would be vindicated. Unfortunately the 18 characters were never decoded. We know his identity or pseudonym was "Zodiac" and he did promise this in the 408. So, providing "Zodiac" after only 4 days must have been done for a reason. Was it to bind the two sets of communications together?

Don’t forget, the Debut letter was only written because of Vallejo Police Chief Jack E Stiltz urging the killer to send in more details. Had this not been the case, the "Zodiac" would have been plain old "killer" until at least the Paul Stine letter. In other words the Zodiac Killer would never have been known by "Zodiac" during the span of his 5 murders – if ever.

Correctly posted earlier ebeorietemsthhpiti can actually be ebsorietemsthhpiti

Just for a laugh I played around with the second version. Apart from the letter P, you can spell BETTIES THEOREM with the remaining letters.

Not a POI name but if we refer to ‘you can play chess with me’, the EBEOR… phrase could contain the word ‘bishop’:

ebeorietemsthhpiti

BISHOP..

The rest would then be

eeretemthiti

for example

TIME THREE TIE BISHOP

or

THREE TIE TIME – BISHOP

or similar.

QT

Z13 cipher solve attempt:

Posting it here to get your thoughts. I will be posting additional details on my website in the near future (I have minimal info there now) – https://zodiacrevealed.com

In a nutshell, I’ve tried thousands of combinations for the Z340 and Z13, both programmatically and manually. I keep coming back to this as the best solution, although it does force you to consider the circle 8’s as wildcards. (Perhaps to confuse the person attempting to resolve the cipher?)

What is interesting about this cipher solve is that I was a tad perplexed by the spelling of the last name. i.e. Did Lawrence Kane ever actually refer to himself as LAWERENCE with the extra E? Also, did he refer to himself as CANE? Turns out to be true in both cases.

I also have a partial solve for the Z340 but it is not ready for review yet due to the parts of the cipher which contain garbage text and many misspellings.

TheNegotiator

IMO 5 out of 13 symbols simply do not fit. Even the last name has to be written first..if you hang onto the Z13, it could be time to change your suspect. Just an idea.

JOHNEVERETHJR, except the ‘H’ is not such a bad match and still have I discarded him as a suspect.

QT

I think "John Everech Jr" works.

We speculate a lot about Robert as a good combo pick for RH and the Z13.

You know what else would be good? Russell.

The last name would be 5 or 6 letters. You would get the -er ending. That takes it to 4 repeating pairs. So you would only need 1 more repeat of a common letter: r, s, l or e.

Like Russell Z. Tyler would fit.

Let’s say the initials are RH. Next, let’s say the first name is Robert (the most common R name, and possibly in the Z408 leftovers).

Then we have Robert H——

Let’s assume a straight substitution key, and then an unscramble.

Next, let’s assume that 2 of the duplicates form an -er. That gets us to 2 e’s and 3 r’s. So we have the triplet accounted for. Let’s assume the -er at the end or the 3rd and 2nd to last letters.

Robert H—-er or Robert H—er-

The second letter must be a vowel, unless it’s Hungarian or Asian. And we would need 2 more repeat letters.

OK. Now what?

So in the above examples ….

First, you can’t use all of O, B, and T together. But you have to use two of them in the solution. But not all three.

Second, the letters T and B don’t combine easily. The only real way is with a -bert ending (in which case any additional vowels would be something other than -o

That leads us to, third, the H almost certainly needs a vowel following it.

So , then one of two things are likely to be true:
An -o follows the H and then
1. the t or b is in the 4th slot, or
2. the t is in the 7th slot

The alternative to this, as mentioned above, would be a -bert ending that would exclude the use of an -o.

R o b e r t H o – (t/b) er (t)

So some of the most likely possibilities, in this scenario with these assumptions, would be:

Holberg
Hoiberg
Haybert
Humbert
Halbert
Helborg

Holberg appears to be Danish/Norwegian in origin, or at least popularity. Edvard Grieg composed a Holberg Suite, for example.

Russell Harper

But then we are allowing anagrams and there will be thousands….

Russell Harper

But then we are allowing anagrams and there will be thousands….

I think if we choose one name and limit it to the likeliest configurations, then maybe or maybe not.

There seems to be an idea in the case that every possibility deserves similar weight. I’m not sure that’s true. Mathematics wasn’t enough to solve the 340. Dave O. and Jarl and Sam had to use some intuition, as well.

Another question: if it is a straightforward one-for-one substitution, doesn’t it seem extremely likely that it’s an anagram? I say that because I strongly doubt you would have that many repeating letters (9 out of 13 are duplicated) without having some doubled up next to each other. None of the repeating symbols are doubled up next to each other.

So it seems like it’s either an anagram, or homophonic, or some other gizmo to it.

I was playing with the name Richard and almost immediately I arrived at this:

Richard Ho-ard

Huh. I wonder what that could be?

This is almost certainly going to be the most common RH name that fits the Z13 pattern, probably by a good margin:

Richard Howard

That’s the #2 R first name of the Z birth era (and #5ish overall name of that era) and the #4 H last name at present (which is probably similar to whatit was then). Howard is #75 overall now, and it certainly has been overtaken by numerous Spanish names and other immigrant names.

And just as I say that ….. I totally left a cake out in the rain.

Because the most common H last name fits, too:

Richard Harris

FTR: I am definitely not saying that Richard Harris was the Zodiac.

CORRECTION:
Richard Harris DOES NOT FIT] the pattern. One too many Rs.

Did not know Richard Harris was a singer…cranky, sure, but not a singer.