Of course, you could argue—based only on what you have presented already—that all doubled letters in this sense must evaluate to zero, and hence to ‘A’, as:

Double D = 4 – 4 = 0 → A; and

Double J = 10 – 10 = 0 → A

So, it might seem that we are back to ‘AA’, and Arthur Allen, after all our efforts. 🤣 

I believe that Zodiac and the Monster of Florence are the same person; however, there is no clear or conclusive evidence that it was Mr. Bevilacqua, and therefore, until proven otherwise, he should be considered unrelated to the case. That said, my solution to the Z32 includes, among various hypotheses, the initials “JB” as a possible identifier of the author.

Among other things, I’ve recently identified a possible cryptographic link to Zodiac within an official message from the Monster of Florence. It involves a feature that is quite characteristic of Zodiac, which I might expand on in a new topic.

Posted by: @shaqmeister

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As well as both these things, you also include something to give the name of The Zodiac, but I haven’t looked at that in detail so I cannot give any comment.

OK, I have gone back and read what you have written about this part and can understand your thinking here. And yes, I can see that you do arrive at ‘JB’ as one of your 4 alternatives.

It shows great restraint on your part to have reached ‘JB’, whilst having in mind a belief that the Monster of Florence and the Zodiac Killer may be one and the same, and then not to simply jump at Bevilacqua and say “case closed.”

Well… over the past few months I’ve put together a lot of “strange coincidences,” more than you might imagine. Unfortunately, I’m not a cryptographer, so I don’t really have the authority to say, “no, these aren’t coincidences.” That’s also why I ended up on this forum!

I have been thinking about how your solution to ‘HERCULES’ would be very much more satisfactory, if it hadn’t been for the fact that the omega block (5 symbols) breaks the simple pattern having its symbol at the beginning and not the end.

As it stands, it gives us the problem that we have two non-alphabetic symbols together at positions 12 and 13 in the cipher text. From your starting idea, however, we would not expect to see this as the non-alpha symbols are meant to be just block separators, and two symbols together don’t separate anything.

So, suppose we did just take the ‘Ω’ at position 13 and the ‘G’ at 17 and swapped them around.

Firstly, doing this would make the pattern consistent—a series of alphabetic blocks separated by non-alpha symbols. Also, it would have no negative effect on your summing for each block using A=0, etc. Nor would it affect the parallel homophonic solution, the ‘O’ constraint in this section is not moved out of its position with this straight swap.

How, though, could you explain this?

Well, simple error would be a possibility, if you wanted to go down this road. The Zodiac has made errors in each of his solved ciphers, as we have seen. There are also instances elsewhere where The Zodiac shows impatience and writes something, having missed a step along the way, then has to go back and make a correction. 

In the postscript at the end of the Little List letter, he writes “+ inches along the radians” and then has to go back and insert the ‘#’ which he had hurried past and missed out.

You could argue, then, that he was doing the same with the first line of Z32. Having just set down the closed triangle at position 12, he was already thinking about the end of the line marker, ‘Ω’, and wrote it immediately, by mistake. Knowing, however, that this error would not break the A=0 summing for this block, and aware that the ‘O’ did have to come next, he carried on.

And maybe, just maybe, it is this error that he is wanting to point out to us when he writes “Claif” instead of “Calif” on the John’s envelope.

“Five letters, two need swapping around.”

Posted by: @lendor-77

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Ciao @shaqmeister! Regarding what I think I’ll be publishing in the coming days (hopefully) you might want to watch this video.

Ciao @lendor-77. Yes, I will certainly do so. I have been reading even more about Il Mostro since we last spoke. 

I will, however, have to give you my thoughts maybe tomorrow, as I’m am just starting a run of late shifts at work.

Maybe you would like to start that new thread with this, so that we don’t let this thread get distracted from your work on the cipher?

Ok, let me see if I got this right: you think Zodiac accidentally put the omega symbol at the beginning of the sequence instead of at the end.

As for the errors in the ciphers or texts, here’s how I see it: I find it hard to believe they’re truly random. It seems more likely that they were intentional, maybe to hide something we still haven’t figured out. Sure, one or two mistakes can happen, but especially in the writing there are quite a few obvious oddities… so I tend to think they were deliberate.

However, for the purpose of identifying the author’s initials, this would lead to different results, since “omega” would already be in the correct position and the letter S would not be mirrored, thus having a positive value.

So what result would that give? Different initials for the author, if I’m not mistaken, either TL or TS.

Yes, in your solution as it is you have the advantage that you have all the letters in ‘HERCULES’ (although the ‘E’ only occurs once—not a problem though). If a possible error is considered, however, the best that you could achieve would be ‘HERCULE’, which is more of a problem. And, as you say, it would alter the steps that follow to reach different initials.

I will have to give this some more thought, and will do when I get back later.

In my thinking, if the basic pattern (summing, with non-alpha as separators) is valid—which is what we are assuming here—then where it evidently appears to change, we have to be able to account for why it changed. What is it that we have to do differently with the omega block, and is this obvious to us? Why do we reverse just the ‘T’? Why don’t we reverse all the letters, and then apply modulo 26 through summation? (Just thinking out loud here.) 

Also, it would be important to think about whether the Zodiac could reach ‘S’ in the same manner as in all the other blocks, merely through assigning the free letters ‘R’, ‘T’, ‘G’ differently, keeping ‘O’ where it is and pushing ‘Ω’ to the end. If this would be easily possible, then we would have to consider that maybe he isn’t looking for ‘S’ at all. Or maybe—just maybe—we have some indication here that we’ve been going astray with our basic pattern all along. (Again, just something to consider.)

Again, I will have to look later and check if there is a straightforward way to get ‘S’ in this block. There is sufficient freedom here, so I would have to suppose it must be possible. Maybe you have time to check, @lendor-77?

Alright, let me try to answer your questions step by step by summarizing the decryption process used in the analysis and retracing the reasoning.

1. Sequence splitting: it can be observed that the Z32 cipher can be divided into sequences arranged so that each one ends with a special character on the right, preceded by alphabetic characters (either normal or mirrored). The exception is the omega sequence, which can still be isolated because it appears at the edge of the first line; this allows it to be separated, but with the omega positioned on the left rather than on the right like the other sequences.

2. What to do with the omega sequence?
2.1 – First hypothesis: adjust it by moving the omega directly to the final position. However, this opens up several possibilities, since the sequence as divided — “omega ORTG” — needs to be modified. But how?

I could simply move the omega to the end by shifting all the letters, obtaining “ORTG omega,” or I could swap the G with the omega, resulting in “GORT omega.”

This leads to multiple possibilities and interpretations. However, it’s important to keep one point in mind: changing the position of the letters does not alter their value unless they are made mirrored. Therefore, whether I write “GORT omega” or “ORTG omega,” the result of the sum in the A=0 system would remain the same, namely “E omega.”