Even if the anagram is the correct solution, it’s practically impossible to verify, because there are so many other possible anagrams of those 18 letters.

Here are a few perfect anagrams of EBEORIETEMETHHPITI:

I’M HERE TO BITE THE PIE
THE EERIE POET BIT HIM
I HOE THE TIBET EMPIRE
I BETTER HOPE I TEE HIM.
BE HERE. TIPTOE. TIE HIM.

There are 741 billion different ways to rearrange the letters in EBEORIETEMETHHPITI. Even if only one in a billion of them spelled out a person’s name, you’d still have 741 names to pick from. The problem gets much worse when you allow for misspellings, since that drastically increases the number of names that can be found.

He still could have used e.g. six fillers in the last line. Or three. Or five.

QT

I’m investing a couple of hours into a gimmicky little thing that I want to include with a new version of my solver. An n-gram based anagram solver, it seems to work reasonably well for shorter strings. It only transposes, it does not substitute.

"ollieilkeigpilkpne"

AZdecrypt 0.993 (Reddit corpus 6-grams)

Score: 21481.67

ilikekillingpeople

"ebeorietemethhpiti"

AZdecrypt 0.993 (Reddit corpus 6-grams)

Score: 19972.66

timeihopethereitbe

The n-gram scoring wraps around the cipher and I corrected the offset for "ilikekillingpeople" manually. The "timeihopethereitbe" may still need such correction. I believe the last line of the 408 is probably filler though. There are indications, but you never know with such short strings.

Quicktrader: Another idea..in his cipher, Z had stated: ‘I will not give you my name’. It seems to be obvious that he did not. However, the opposite might be the true:

First of all, he talks about giving his name. ‘Not’ giving his name could have two reasons. Either he simply didn’t write it into the cipher OR he actually did write his name into the cipher but didn’t ‘give’ it. Latter would mean that Z had actually encrypted his name (which means that he doesn’t ‘give’ it, although it’s there). In the past, some people have assumed that those ‘fillers’ Z had used at the end of the cipher are simply what they appear to be, fillers. Others have assumed that those fillers, 18 symbols/letters, actually do represent his name (e.g. as an anagram)." (Emphasis added.)

AK Wilks: This is what I think the final 18 letters mean. He did not "give" us his name, but it is there – but we have to work to find it.

This is the "Unolved 18" from the First Zodiac Code as solved by the Harden’s, put into a Caesar Code Matrix with 0-3-6-9 shift values, numbers used for their absolute value of distance from zero, with the vertical letter shifts going + or – to produce the name in a consistent manner.

C…N…Y…R…D…R…X…Y…G…N…A…C…C…V…N…S…N…R
Z…K…B…O…G…O…U…B…J…K…X…Z…Z…Y….K…V…K…O
W…H…E…L…J…L…R…E…M…H…U…W..W..B….H…Y…H…L
T…E…H…I….M…I…O…H…P…E…R….T…T..E….E….B…E…I
Q…B…K…F…P…F…L…K…S….B…O…Q…Q..H….B…E….B..F
N…Y…N…C…S…C…I…N…V…Y….L…N…N..K….Y…H….Y..C
K…V…Q…Z…V…Z…F..Q…Y…V….I….K…K..N….V…K….V..Z

+…+…-….+…-…+..+(-…-…+…+)…+…+..-….+….-…..+..+
0…3…3….6…9…6…3..3..6…6…9….9…6…6….9….9….6…0

I’m surprised that you still consider that to be a valid appearance of an enciphered name, considering how many different names and words can come out of your shifting process.

I’m surprised that you still consider that to be a valid appearance of an enciphered name, considering how many different names and words can come out of your shifting process.

Well the probability study done by you and Aquiman did show the huge number of names that can be produced, but also showed the 340 anomaly (THEO KACZYNSKI) I discovered would happen by chance about 2/10 of 1% of the time. Aqui couldn’t do the 18 letter EBE sequence because he said it would take days of computer time to run, but estimated that an 18 letter name of a math professor would appear using my shift criteria and anagram rules about 1% of the time.

I agree with QT that a possible interpretation of Zodiac’s words is that he will not "give" us his name, but it is there in an encrypted way.

Anyway let’s not debate this here. At this new ZKS board I have never really done a thread devoted to my discovery. As part of my farewell tour I am going to do one grand thread on what I call "Zodiac 340 – The Graysmith and Kaczynski Anomalies." We can debate it further there. Basically I think there are some potential important discoveries in the 340 that have largely been ignored. I’ll post them, make my case, welcome good counterpoints and debate from you and others, and people can decide if there is anything valid to potentially follow up on or disregard it as chance formations.

BTW I totally agree with you it is very likely to be a waste of time to pursue wide anagrams of the EBE sequence. Your showing of 741 billion possible arrangements is convincing proof of that. And most EBE anagrams produce a garbled baby talk message that needs to be creatively filled in and helped to make anything remotely inderstandable. Nowhere else in his solved code or letters does Z communicate in such a fashion.

To create something like ROBERT EMET THE HIPPIE involves massaging the letters, adding letters that aren’t there and what do you end up with? A baby talk sentence that doesn’t really communicate a coherent idea or theme that Z used elsewhere.

A similar criticism applies to proposed wide anagram solutions like I BE OR I SEE ME, I HOPE IT BE TIME THERE or dozens of other possibilities. Most are poor grammar baby talk sentences and none really express a coherent meaningful idea that resonates with actual known Zodiac ideas or messages in the solved code or in letters.

QT and others, take it from someone who wasted a lot of time pondering wide anagrams, until Doranchak, Glurk, Aquiman and others showed me the billions of possible solutions and therefore the high probability that no one solution is valid. It’s fun to explore different anagram possibilities of EBE, but none produce an "Aha!" stunner of a valid message, and with the billions of possibilities it is very likely just a waste of time IMO and based on the facts and evidence.

Regarding to glurk’s observation:

http://zodiackillerciphers.com/wiki/index.php?title=Encyclopedia_of_observations

„Many symbols among the last 18 symbols are found directly above in the same columns. The "QEHM" sequence is particularly noteworthy.“

I was wondering how likely a pattern like this occurs when the last line is created by using random letters from the 408 key. So I wrote a small test which fills the last line by picking random letters from the key and checks if the letters are found in the corresponding columns. The first 23 lines are the same like in the 408 cipher.

Here are the results from 1000000 test ciphers:

Highest count of columns which have at least one occurence of the corresponding filler letter: 15

Test ciphers which have exactly 12 columns with matching filler: 3304 (out of 1000000). -> 0.3304%

Test ciphers which have 12 or more columns with matching filler: 4163 (out of 1000000). -> 0.4163%

The percentage is so low that I have decided to skip further tests („how likely is it that a sequence like „QEHM“ occurs by random“ etc…).

I am pretty sure that the formerly posted idea of "picking random letters from the column above to create a filler" is correct. So my opinion is that the filler section is just…uhm….a filler section and nothing else.

But to leave the door open: I think it is still possible that Z decided to use them as some kind of a key or a clue when he wrote z340. (Even this is not very likely)

This morning I had an idea. We all know that there is at least one missing word between part 2 and part 3 of z408 ("all the ..?people?.. I have killed"). Could it be that the original plaintext exactly fit into 408 characters? In this case there is missing more than the assumed word "people". Maybe something like "all the (whatever 18 letters) I have killed". May this be an explanation for the disrupted homophonic cycles in the last part of the letter? Maybe he realized the missing words when he reached the last line and decided to use fillers instead of writing it again (another hint for my "he was lazy" theory).

It does almost seem like he missed an entire line, which might account for at least 17 of the 18 filler characters at the end. Carefully drawing out all the cipher symbols is probably too tedious to repeat after discovering a mistake! (See the "K" symbol correction in Z340 for an example). Plus, maybe he didn’t want to waste the nice Saks Fifth Avenue stationery. The cycles, however, don’t seem to consistently fail in the third section of the cipher.

http://zodiackillerciphers.com/408/key.html#4

There you can see that many of the cycles fail in the 3rd section, but why don’t they all fail right at the boundary between sections 2 and 3? Some keep going perfectly for a little while in section 3 before failing. Some *do* fail right after the boundary, though.

Perhaps we should plot exactly where each cycle breakdown first occurs (with transcription errors fixed), to see if they tend to happen on certain lines or regions.

Perhaps we should plot exactly where each cycle breakdown first occurs (with transcription errors fixed), to see if they tend to happen on certain lines or regions.

I am working on it. At the moment I am a bit tired of z340 and it seems to be a good idea to do something else (e.g. z408 )

This is the spreadsheet which shows when the cycles are interrupted: https://www.dropbox.com/s/g93t6tgjqocxbz7/Homophonic%20Cycles%20z408.pdf?dl=0
Sometimes there is just one missing symbol, sometimes the symbols are getting totally random. Nothing new so far
Maybe he created a huge list of cycles for each plaintext letter and crossed out each occurence. When reaching part three of the plaintext he recognized that he had not prepared enough cycles and so he chose them random. If this sounds confusing:

Plaintext letter "A" maps to "X", "Y", and "Z". Letter "B" maps to "#". So he created a list of substitutions:

A = "XYZXYZXYZ"….this should be enough (it isn’t)
B = "#####"

Now he had the following plaintext:

AABAAABAABAAAABAAB

He started to encipher:

XY#ZXY#ZX#YZ…. ouch…not enough prepared letters for "A", all letters are crossed out….now make it random

Thanks for posting the cycle analysis, Largo!

Plaintext letter "A" maps to "X", "Y", and "Z". Letter "B" maps to "#". So he created a list of substitutions:

A = "XYZXYZXYZ"….this should be enough (it isn’t)

…

He started to encipher:

XY#ZXY#ZX#XZ…. ouch…not enough prepared letters for "A", all letters are crossed out….now make it random

I think I understand what you’re saying. But did you mean to write this instead: XY#ZXY#ZX#YZ

YZ at the end instead of XZ. Then the list for "A" is completely crossed out.

I suppose he may have done it that way. It would be hard to verify without discovering his notes!

But did you mean to write this instead: XY#ZXY#ZX#YZ

YZ at the end instead of XZ. Then the list for "A" is completely crossed out.

Oops! You are right, thanks! I’ve corrected the mistake.

I suppose he may have done it that way. It would be hard to verify without discovering his notes!

Yes, indeed. I was interested in the position of the cycle breaks because I had an idea. He maybe had a table with letter frequencies and created a list of prepared symbols based on the frequency of each letter. In this case the breaks should occur after a calculatable amount of letters.
Example:
Letter "I" has a frequency of 9% in english texts. Maybe he was naive enough to prepare exactly 37 symbols (9% of 408) but needed 44.
Does not seem to be the case. The analysis was just for fun and I think the positions of the breaks are not too important. If they are somehow related to the probably missing line at the beginning of part three then I don’t have any idea how to use the cycle breaks to squeeze out information we don’t have yet.

It does almost seem like he missed an entire line, which might account for at least 17 of the 18 filler characters at the end.
…
There you can see that many of the cycles fail in the 3rd section, but why don’t they all fail right at the boundary between sections 2 and 3? Some keep going perfectly for a little while in section 3 before failing. Some *do* fail right after the boundary, though.

I just dug out this thread and had an idea. The reason that at the beginning of part 3 some cycles break and some do not, may be easily explained. I once suggested a possibility of how Zodiac might have done the substitution:
viewtopic.php?f=81&t=3337
If this was the case, then only the cycles whose plain text letters appeared in the presumed missing line would be interrupted. A cycle can still be correct if there were as many plain text letters in the forgotten line as there are symbols for it.

Just an idea…

Translated with http://www.DeepL.com/Translator

I’m simply stunned that so many obviously intelligent people have labored so long attempting to solve a non-existent code. Zodiac scrambled some literary garbage and presented it as a "code" for no other reason than to divert said intelligent people. The more people there are attempting to solve a non-existent code, the fewer there are to pursue legitimate clues.
This is all great fun, folks, but it’s a waste of time.