This is what I am guessing that he did; see if you can read it:

W H E N * O U A R E A N G R Y Y O
U * * * O T I O N S A R E L I K E
* U G E W * V E S D U R I N G A S
T * * M A N D Y O U * A * * O T H
E A R * O U R I N T * I * I O N T
* E S A * E W A Y T H A * Y O U C
A N * O T H E A R T H E H U M P B
A C K W H A L E * U N D E * W A T
E R S * D * E S S * * J E A L O U
S Y O R V * N G * F * L L * * S S
A * F E C T S H E A * * N G I N T
I U T I O N I T I S L I K * T * *
I N G T O H * A R A * * I E N D O
N T * E P H O N E W H E N P E * *
L * I N T H E R O O M A R E D A *
C I N G T * L O U * M U S I C O R
C H E * R I N G A * * L L * A M E
* * T H E T E L E V I S I O N Y *
U * F R I E N D * S S P * A K I N
G B U T Y * U C A N N O T H E A *

On the number of bigram repeats issue, I honestly don’t know. I guess what you are saying could be tested. The 340 is pretty darn cyclic. So I guess the answer to your thinking is to create a 63 symbol message with a lot of cyclic ciphertext groups. Include five or six symbols with count 9 to 10 that are in two to four respective cycles, one polyphonic symbol count of 24 randomly placed, and three other polyphonic symbols each with count 10-12 randomly placed. See what happens to the number of expected bigrams. If you are correct, then the number of bigrams will be higher than what you are finding with the 340.

With the wildcard hypothesis, Zodiac encoded the message in one of two ways. He either was encoding and just plopped down q’s, +’s, B’s and F’s (or maybe a different combination of symbols) instead of the symbols that were next in the cycle, or he made a first draft with cycles and then made a second draft with the polyphonic symbols plopped down. Result: difficult to solve. It would have been so easy to do.

By the way, I wonder if the pattern above is one that we could trace with ZKD. If we could take snapshots of the solving process and make a little movie that we could step through forward and backwards. Sometimes the solver finds little strings of words that appear and then disappear. I wonder if the little strings of words appear in certain locations more often than others. If there is a pattern. Has anybody ever tried to do that?

Just had to finish the above using the neck top computer.. Got this as a final with a few spell errors

WHEN yOU ARE ANGRY YOUr
emOTIONS ARE LIKE
HUGE WaVES DURING A STorM
AND YOU cAn nOT
HEAR YOUR INTuitION THE
SAmE WAY THAt YOU CAN
nOT HEAR THE HUMP BACK
WHALEs UNDEr WATER
SaDnESS or JEALOUSY
OR VeNGeFuLneSS
AfFECTS HEAriNG INTIUTION IT
IS LIKe TryING TO HeAR A frIEND
ON ThE PHONE WHEN PEopLe
IN THE ROOM ARE DAnCING
To LOUd MUSIC OR
CHEeRING A baLL gAME
On THE TELEVISION YoUr
FRIEND iS SPeAKING
BUT YoU CAN NOT HEAr

Then googled up the main lines and got Gary zukav… What an interesting writing style he has ..And an interesting "life style" he led.. Leads .. wiki. …. i had not heard of him.. The above was not verbatim.
interesting all the same

Mr. Lowe, that was a book that a friend gave to me. I had to modify the message a bit to fit into 340. That was my experiment when I thought that Zodiac may have randomly plopped wildcards onto the message. It sounds to me like you are very good with computers.

Mr. Lowe, that was a book that a friend gave to me. I had to modify the message a bit to fit into 340. That was my experiment when I thought that Zodiac may have randomly plopped wildcards onto the message. It sounds to me like you are very good with computers.

The end solve or completion on the above was done with the neck top computer. Just Honing ones skills on the tailings that you guys throw up..
Question did you throw in the spelling mistake deliberately.

I’m a bit of a spud with technology but a reasonable researcher
For me Trying to follow you guys is like trying to learn a new language ,it is an interesting ride I only wish I could understand it more.
I admire everyone’s work and really hope that you get a break soon..
Cheers

The spelling mistake was not intentional. And I am not particularly talented with technology either. I just like to work on the 340 because it is a fun and inexpensive hobby and I like puzzles and mysteries in general. Plus I want to be on T.V. some day and am hoping that solving it will get me there.

I am getting tired and should probably take a break if I can. Not a total break, but minimal involvement for a short period of time. I have other ideas that I would like to work on in the future as well. I think that if we keep exposing ourselves the the possibility of solving the 340, then it may happen.

That’s cool. I could use a little break from this thread as well.

That’s cool. I could use a little break from this thread as well.

It’s starting to bear an alarming resemblance to John Nash’s "crazy wall":

What people see when they stumble into this thread:

https://www.youtube.com/watch?v=hh84bamfe_A

As long as they also see Jennifer Connelly and hear (RIP) James Horner’s beautiful music it’s alright.

It seems to me that the best approach is to look at the least probable period 18 bigrams first. I still think that Zodiac may have encoded, and then after recognizing that he had created some recognizable patterns, made alterations to hide the patterns. I have other stuff going on right now, unfortunately, but am thinking about taking a look at the symbols close to or between the least probable to see if there are corresponding bigrams, high scoring cycles, or patterns. Also comparisons with test messages to see what "least probable" really means. I think that to calculate the probability of any repeating bigram:

[ ( Symbol A Count / 340 ) * ( Symbol B Count / 340 ) ] ^ Number of Occurrences

The top six symbol pairs in order of least probability are:

29 and 42 (three occurrences )
5 and 19 ( four occurrences )
15 and 19 ( three occurrences )
19 and 53 ( three repeats occurrences )
21 and 19 ( three repeats occurrences )
27 and 62 ( two occurrences )

I wonder if he slid some columns up or down one space and if looking at these will show any evidence of that.

EDIT:

Sliding columns up or down wouldn’t have to make all of the Period 18 bigrams into Period 1 bigrams. Some of the Period 18 bigrams may be "false." Assuming of course that some of them are "true."

For example, sliding columns K and L down one space would make some Period 1 bigrams. Maybe sliding some rows two spaces would make for a better vertical message. Something to that effect.

Period 19 corresponds to this reading rule (as an example):

– Start at top left of cipher (the "H"), and print the symbol there.
– Go down 1 row
– Go right 2 columns and print the symbol there.

(When you fall off the grid, wrap back around as if the text has no "edge")

Continue until you’ve read all 340 symbols. Here’s the result:

I had a version where I used a simple 19×18 columnar transposition (the resulting cipher is linked in the post). It has the same 67 bigram repeats, but only 2 trigram repeats, vs 4 in your rearrangement. I just realized that I haven’t tested mirrored and "snake" routes for this, only the forward and reverse directions. I think I need to put it in my solver that it always tests all of them automatically, so I don’t keep forgetting…

Smokie I’ve been experimenting a bit with your idea of column and row slides and have found a possible explanation for something that occurs in the first half of the 340 aside from the period 19 bigram find.

First half of the 340 with non-repeats measurement: normal: 4168, mirrored: 4500. Since this measurement correlates with the cyclicity (higher is better) it’s rather interesting to find the mirrored version scoring higher. Though my 2-symbol measurement (inspired by yours, normal: 155, mirrored: 146) seems to say the mirrored version is less cyclic.

Now with your idea of slides I noticed an interesting property. I encoded a 10 row message (p1) with 63 symbols: normal: 9228, mirrored: 7814. As it should be, normal>mirrored. Then I’ve added an incrementing row slide (shift), for example right shift the 1st row by 1 position, the 2nd row by 2, the 3rd row by 3, etc. normal: 8374, mirrored: 8596. And 2-symbol cycles: normal: 256, mirrored: 244. A bit of further testing shows that it doesn’t have to be an incremental row shift, a fixed value for all the rows will do as well. So such a scheme could cause this.

The message I made solves normally, but doesn’t with the added difficulty of the row shifts (with the already high multiplicity). So something like that could be actual for the first half of the 340 (after encoding). A full message with this scheme does solve. It seems interesting when considering the bigram difference for the halves. As the first part could have this scheme going on and the second part something a bit different.

Also, column slides randomize the cipher much more than row slides. So considering a thoroughly (uneven) columnary slided cipher (after encoding) is unrealistic as this would have left no cycle standing.

I had a version where I used a simple 19×18 columnar transposition (the resulting cipher is linked in the post). It has the same 67 bigram repeats, but only 2 trigram repeats, vs 4 in your rearrangement. I just realized that I haven’t tested mirrored and "snake" routes for this, only the forward and reverse directions. I think I need to put it in my solver that it always tests all of them automatically, so I don’t keep forgetting…

I discovered that my rule above wasn’t exactly the same as a real "19 period bigram" count, which accounts for the difference in trigram repeats. (It’s because my horizontal wrapping rule stayed on the same line rather than moving to the next line down).

But it led me to put another avenue of exploration on my todo list: Generating transposed ciphers that correspond to all possible sets of "reading rules" of length N. It may be possible to generate them by brute force for small N. Then for larger N maybe a hill climber could be used. But the idea would be to look for rules that produce interesting peaks in ngram repeats and candidate cycles.

Jarlve, I like your new avatar.

What about like this. There are a lot of places where adjacent columns can be switched, and then slide the new right column down one space and the new left column up one space, and then you have new bigrams that Zodiac could have been hiding. See columns B and C below.

A variation of daikon’s "strips of paper" hypothesis. You could also do this with columns G and H, H and I, N and O and get results. Probably others. Something like that anyway. B and C switched decreased cycle scores but N and O increased cycle scores.

Mr. Lowe, do you see any other column pairs that could be switched and slid to make multiple period 18 bigram repeats into period 0 bigrams?

Jarlve, I spent some time today updating my spreadsheets so that I can paste a message and it will tell me the period bigram repeats, cycle scores, score for the period bigram repeats, and top ten scores for the period bigram repeats. I think that bigram repeat count is important, but score is important too. I suggest that you look to see how many randomizations it would take to get a certain high score. I believe that the 5 – 19 cannot be a coincidence.

There’s another few ways to increase the bigrams,

Set the cipher in a 20 by 17 grid and reading SW-SE (diagonal) produces 38 bigrams and 35 mirrored.

HEBR(U>#Z3pOGp+l%WO&
D^D(+4(GVW)+k#2bPYLR
/5J+Jk.#Kp+fZ+B|<z28
KjROp+1*H_Rq#2pb&RLK
J9^%OF7TBlXTfSMF;+B<
MFG1BG)p+l2_cFKzF*K<
2BpzOUNyBO6N:(+Hdy7t
-cYAy29^4OFT-N:^j*Xz
6-<Sf9pl/CpclddG+4Uc
y5C^W(c+M8|CV@K+l#2E
.B)>+*5k.L-RR+4>f|pM
VFFz9z/JNbVM)|DP>Ldl
5||.UqLFH2|<Ut*5cZG+
kN(MVE5FV52+dp++|TB4
-R)Wk^D4ct+cW<SYM(+|
TC7z.;+c+ztZ8y.LWBOB
31c_8z6PYACOO|SBK*;+

And in 19 by 18 reading SN-WE (vertical) produces 39 bigrams and 37 mirrored (added one new symbol "?").

>)B.E2#l+K@VC|8M+HM
p|f>4+RR-L.k5*+BED|
)MVbNJ/z9zFFVU(RHFL
qU.||5ldL>P3Z#>Nk+G
Zc5*tU<|2+pGOppd+25
VF5EVM(D&OW%lkW)R-4
BT|++G(4+(D^S<Wc+tc
4D^b2#k+)WVz7CT|+(M
YJ+J5/RLYPZtz+c+;.B
+Zf+pK#.kOBWL.y8+pO
RjK82z<|8_c13BR&bp2
#qR_H*1AYP6zXlBT7FO
%^9JKL|OOCB1GFM<B+;
FMSfTKBS<K*FzKFc_2l
+p)G;*H+(:N6OByNUOz
pB2+-TFO4^92yAYc-t7
yd?C/lp9fS<-6zX*j^:
N?c(W^C5ycU4+Gddlcp

I suggest that you look to see how many randomizations it would take to get a certain high score.

Sure but what score do you mean?