340..partially solved ;)

[AK Wilks]

QT-

That's a good analysis. I haven't checked your math - WAY too much work for me right now - but I'm sure you are probably correct. The problem, as I see it, and I've mentioned this before, is that the 340 is probably NOT just a normal homophonic substitution cipher like the 408 was. So I'm not sure that all of this really helps... That said, I do enjoy reading your posts and I'm glad that people are still working on the 340. I'm a bit lost on what to do with it at this point, to be honest.

-glurk

I agree that the 340 is probably not a normal homophonic substitution cipher like the 408 and probably QT does as well. But I am thinking it is possible that the first stage of the 340 solves mostly as a normal homophonic substitution cipher, with a second stage employing a Caesar or something else. But before we get there we have to solve the first stage. Parts of this first stage solution may read like normal English language words and sentences and parts may read very strange until we discover the second stage to apply.

[AK Wilks]

Long time no news here around..had to fight with a broken laptop, last file sadly completely gone. But now there is some news.

Some of us may have thought about the chances that three double symbols occur in a 340 cipher. Me too. And it is not enough to simply go for some probabilities of how often a double letter would actually occur in a normal text.

Because his statistical observation fades away with the usage of homophones. Therefore we got to review this issue under consideration of the amount of homophones Z might have used. Compared to the 408, by using 63 homophones, Z has in average increased the amount of homophones per alphabetical letter in the 340. We do assume, however, that he had used a similar (or identical) amount of homophones, nevertheless has increased the homophones in low-frequency letters. I do so because otherwise the statistical significance (later to be focussed on) would even grow higher.

Three main criteria are essential in the following consideration:

The amount of letters expected (overall frequency)
The amount of homophones
The statistical expectation regarding double letters.

Latter I have received from this list (http://www.wordfrequency.info/free.asp?s=y) which appears to be more precise (450m words). Indeed, it also covers the main 5,000 words, many of them might have been used by Z.

Let's jump quickly into a preliminary example:

1. The letter Q, for example, is not frequent (overall frequency) nor frequent as a double letter (double letter frequency). Our double symbol which shows up three times is the '++'. Generally, '+' alone has a frequency of 24, therefore Q can be ruled out for two (later even three) reasons:
a.) Q is not frequent enough as it should show up approximately 0.37 times only
b.) Q as a double letter (QQ) doesn't occur in 5,000 words even once.
It therefore is most unlikely that '++' is representing 'QQ'.

2. The letter 'E', instead, is very frequent (overall frequency) and somehow frequent as a double letter (33 times in 5,000 words). It therefore could - on first view - represent the '+' symbol.

However I will now demonstrate why this is not possible at all.

Please would you allow me a short introduction?
The main reason is that the statistical expectation of three double letters showing up, in homophone ciphers, is in advance being influenced by the amount of homophones. Assume that 'EE' would show up, let's say, 200 times in 5,000 words. With an average length of 5 letters per word (assumed), we could now think about the 340 cipher consisting out of approximately 68 words.

This leads us to

200 / 5000 x 68

therefore an expectation of 2.72 times the 'EE' is present in the 340 (this amount is incorrect, in fact it's much lower). Nevertheless, in fact it could be represented by the three (and this is crucial!) '++' symbols.

BUT:
The letter 'E' is represented not only by one symbol. It is rather estimated that 'E' is represented by at least 5 or up to 8 different symbols! Even if 'EE' would show up three times in the 340, it is most unlikely that those three double letters - accidentially - are represented by always (3x2 times) the same out of those 5 to 8 different homophones! The very reverse, chances are even much greater that other homophones come into play to represent at least one or more symbols in those three expected double letters. Therefore even only one or two different double symbol would show show up in the cipher. This difference is significant as we do not only discuss the double symbols being present twice or three times in the cipher, no...instead we do understand that, assuming 'E' being represented by e.g. 8 symbols, the chances for 'EE' to be represented three times all by the same symbols is actually very low! It even is low for just the first double symbol!

Shortly described...the MORE homophones a letter is represented by, the LESS is the chance that a double letter (such as 'EE') would show up us a double symbol ('++') in the 340, too. Even less likely, that the following double symbols are existing of the same symbols.

So what to do? In fact, we do already talk about combinatory mathematics. Everybody of us does know the chance to throw a dice to a 6, which is, throwing once, 1/6. This is easy. But what we've got here is way more complicated:

Scenario 1:
Imagine 16 balls in a bowl...with 8 different colors (two each - representing the double letter - and the 8 colors representing some 8 different homophones). We are now allowed to draw 3 times (representing e.g. the expected 2.72 double letters ('EE')). Now guess how low is the chance that you draw three times after another ALWAYS the SAME COLOR for BOTH BALLS. In this scenario, not even one ball would be allowed to have a different color. This most likely won't work, even if you may want to try it at home with your kids.

Scenario 2:
Imagine 2 balls in a bowl only, both of the same color. This would be the case if only one homophone existed for the letter, therefore '+' representing a letter such as Z. Then we definitely get the double homophones all being the same, without any problems. But what is the issue now? First, the letter Z usually doesn't occur 24 times in a 340 cipher. Second, with only 9 times in 5,000 words it is a bad candidate for showing up three times as a ZZ in a 340. Although 'everything is possible', this scenario is even less realistic than our first one.

Scenario 3:
The favorite...what if the amount of homophones (colors of the balls in a bowl) is quite limited, e.g. to three different homophones, AND the expectation of the letter to occur as a double letter is quite high, e.g. 7 times? And in addition to that, let this letter be even a medium to frequent one, too?

All those scenarios above can be calculated.

We therefore are able to say which letter (based on a certain number of assumed homophones and certain assumed double letter frequencies) is the most likely to be represented by three double '++'. In the beginning I thought this would be easy to calculate, but I soon realized that it is actually not. Mainly because we got two criteria of 'balls' to be drawn, the first double letter symbol and the second one (thus drawing two balls at the same time with the goal for them having the same color). This is circumstantial as we can write down all those combinations even manually down on a piece of paper in form of a probability tree. To make that easier, however, there is the great Bernoulli method to evaluate those chances considered above. Those, however, are still depending on the amount of homophones.

An example:
Let's assume two homophones for one letter, the symbols A and B. Therefore a double letter can be represented by AA, AB, BA or BB. The chance to get an AA is 0.25 or 25%, same with BB and the others. To get a 3 times in a row 'AA' situation is therefore 0.25^3 or 1.5625%. This might be doubled as we are completely satisfied with three BB's, too (however we don't expect BAs and ABs hanging around..maybe additionally but that again is against our double letter frequency expectation).

Probabilities for getting an 'AA' situation are:

1 homophone: 100% - AA
2 homophones: 25% - AA, AB, BA, BB
3 homophones: 11.11% (1/9) - AA, AB, AC, BA, BB, BC, CA, CB, CC
4 homophones: 6.25% (1/16) - AA, ..., DD
..
7 homophones: 2.04% (1/49) - AA, ..., ..., GG
(every homophone may be combined with every other homophone, therefore e.g. 7x7 in the latter example)

If we now have look at the 'E', with - let's now say 7 - homophones, this is getting different: 1/7 = 0.14285^3 = 0.00291545 or approximately 0.3%. This is simplified, as we only deal with three draws to get those identical double letters in a row.

But what if out of 7 draws, no matter when, 3 identical pairs of homophones show up? What are the odds then?

Great Bernoulli developed a formula for shortening up this tree of probabilities. It can be tested on both, a low and a high numbered basis, and does go like this (German link: http://www.mathe-online.at/lernpfade/Ko ... ?kapitel=4):

QT Chart One.jpg

With

n=number of 'tries' (expected double letters of a specific alphabetical letter in a 340 cipher text)
i=number of 'successes' (the required three double symbols present in the 340 cipher text)
p=probability of success (the odds of e.g. getting an 'AA' instead of an 'AB' or 'AC',..)

Putting it into the formula, we simply get the probability of how likely it is to get three double symbols at all. This we can do for each alphabetical letter seperately. Beware: We now do assume a certain amount of homophones for each letter, too, so the chances ('p') is differing for each individual letter of the alphabet, fully depending on how many homophones are expected to be represented by.

I did make a tableau consisting of 'overall letter frequency' & 'amount', 'double letter frequency' & 'amount' and went into the possibilities of getting an 'AA' situation which shall depend on 2, 3, 4 etc. homophones. In my excel sheet the formula for each alphabetical letter looked somehow like this:

=FACULTY(G3)/FACULTY(3)*IF(H3=3;P$3^3;IF(H3=4;R$3^3;IF(H3=2;N$3^3;IF(H3=1;L$3;IF(H3=7;T$3;fehler)))))*((1-IF(H3=3;P$3;IF(H3=4;R$3;IF(H3=2;N$3;IF(H3=1;L$3;IF(H3=7;T$3;fehler))))))^(G3-3))

which should be no more and no less than the Bernoulli formula (the 'if' refers to the different amount of homophones expected).

I have to admit that I was suprised:

1. Three letters showed up with an 'error' as I had not yet defined the precondition of '0' homophones, e.g. Z, Q, J. Didn't take much care about those.
2. All letters with only one homophone expected showed up with a different error (because the square in Bernoullis formula gets into zero or negative), those were the letters PCMGBZUJKQVWXY. All those letters have an overall expected frequency of 0.09% to 3.06%, therefore would most of them not even be considered to show up more than 10 times (as double and single letterS) - please reconsider: The '+' is present 24 times. Therefore those letters didn't even reach 46% of what '+' is coming up with. Also none of those letters is expected to show up even twice (therefore not even three times) as a double letter.
3. The letters LSHDFEROTAIN, however, went into a closer Bernoulli-consideration: Those letters gave values on what the chances are that they would show up in the cipher with three identical double symbols.

And guess what:

QT Chart Two.jpg

This is the result. Both, depending on a certain amount of homophones as well as oncertain double letter expectations (latter one based on 450m words). It is the statistical expectation for a letter to appear three times as a double symbol in a 340 cipher.

EVEN IF we assume e.g. the letter 'O' to show up with four different types of its homophones (therefore satisfied not only with AA but also BB, CC, DD), the probability doesn't even increase above the benchmark of 0.1 percent (0.024% x 4 = 0.096%).

Why is that? Why is this so low?

Well that's explained quicker than what I had written above:
'O' is expected to show up only 3.11 times as a double letter ('OO'). And, due to it's overall letter frequency, it is expected to have 4 homophones. To draw three times two identical balls at the same time out of a bowl with 8 balls (with putting them back), all of those being colored colored the same, is simply extremely hard.

With 71% it is obviously way easier to draw 7 times, with three colors in the bowl only, to get three times a pair of two identical colored balls (or symbols). It is, imo, not possible to follow that thought 'immediately' as such a drawing of balls could look like this (or completely different..many ways, seven levels):

AC, BB, BC, CA, BA, CB, BB, AA, BC, AC, BB

As you can see, in this example it took us 11 draws to complete the three double homophones.

And this is exactly what the table above had told us:
It is not enough to draw 7.1 times and expect three double 'LL' symbols. Instead even 10-11 draws are expected to be needed to get those three identical symbol double patterns in the ciphertext.

Accidentially, in the 340, 7.1 draws (occurrances) were enough. Or had Z used the double letter 'LL' not only 7.1 times but 10 or more times in his cipher?

What is a fact, however, is that due to a higher amounts of homophones (caused by a higher overall frequency of the letter) and due to lower expected amounts of double letters (e.g. the 'DD', which is rather seldom compared to 'LL'), the other letters do not (significantly!) come into consideration to be represented by the three double '++' symbols.

The first solved symbol, therefore, imo based on valid, significant statistical data, is therefore the '+' representing the consonant 'L' (and only that one, as long as no error or absolutely extreme outliers exist).

It should be mentioned, however, that assuming different amounts of homophones (higher ones) would even reduce this probability (lower ones would increase, however the 340 has more, not less homophones). 'LL' to occur three times in the 340 is not even expected at all. This would actually requre a value of 100%. Instead it's 71%. With Z using 'LL' a bit more often or the symbols accidentially falling into a good pattern, the triple (definitely exist more in the cleartext) 'LL' represented by the triple '++' is, as far as I can say, the only correct solution.

QT

IMO this is excellent work QuickTrader! And not only is LL a combination that occurs often in normal English language usage, the Zodiac in particular used the LL combination more than any other. look at this past work for reference, with comments from myself, glurk and doranchak. In terms of solving the first stage of the 340, I agree with you very much that as a working hypothesis it makes very strong sense to solve ++ as LL.

AK Wilks:

In the first Zodiac Code he used "S" 24 times. Only once was there a "SS", and that happened in the phrase "it iS So much fun". Zodiac neved used a word that has "SS" in it, not once. Compare that to the number of words he used in the first code that had "LL" - SEVEN different words! And they are words we might think it is likely he also used in the 340 code. They were:

kill,

killing,

thrilling,

collecting,

shall,

will,

all.

Zodiac also used these words with a single "L". - like, people, wild, girl, slaves, slow, animal, afterlife.

By my rough quick count, the letter "L" appears 33 times in the 408 code. About 8.1% of the letters are "L", which is about double normal usage, as it happens that many words Z liked to use have either L or LL in them. So if Z used "L" 8.1% of the time in the 408, it seems to me that 9.7% use in the 340 code is within normal deviation, really only two more uses of a word like kill, killing, collecting, etc., would do it. And if we strike the untranslated last 18, leaving us 390 translated letters, the 33 uses of "L" amount to 8.5%, which is even closer to my proposed 9.7%. in the 340 code.

GLURK: I've done a small bit of work on this, and I'm going to have to say that as far as Zodiac's use of the doubled letter "L," he did in fact use it a lot.
More than would be expected in normal writing. Even in words that, properly spelled, would not have the letter doubled.

I did not do exact word counts, sorry, but in a quick study I've found:

ALL, ALLREADY
AWFULLY
BILLIARD
BILLOWY
BULLET
BULLSHIT
CALLED
CELLING
COLLECT, COLLECTING
CONTINUALLY
FILLING
FULL
HELL
HILL, HILLS
HOLLY
KILL, KILLED, KILLER, KILLING
PULLED
REALLY
ROLLED
SHALL
SMALL
SQUEALLING
TELL, TELLING
THRILLING
TITWILLO
UNTILL
VALLEJO
WACHAMACALLIT
WALL
WELL
WILL

This should be all of them, I think, unless I missed something... I probably did miss something...
I don't, however, believe that the 340 is simply a homophonic cipher like the 408. But I also don't have any doubt at all that Zodiac often used "LL" in his writings. He clearly used it more than would be expected.

DORANCHAK: Here are some others:

ALLEY (confession letter)
ALLEYS (confession letter)
ALLWAYS (1971-03-13-times)
BALL (confession letter)
BELLI (1969-12-20-melvin-envelope)
CALL (confession letter, 1969-11-09-chronicle, 1970-07-26-chronicle)
CELL (1970-07-26-chronicle)
FALL (1970-06-26-chronicle-cipher)
FELLOWS (1970-07-26-chronicle)
FOLLOWED (confession letter)
UNWILLING (desktop poem)
SPELL (1974-02-14-sla)
SPILLING (desktop poem)
WILLING (confession letter)
WILLINGLY (confession letter)
YOULL (1970-10-05)

[The411]

Is it possible that this cipher could be a Vingiere cipher with a Caeser shift?

If that were the case then it would negate the effectiveness of distribution frequency analysis.

Based on the card that the Cipher was enclosed with perhaps the Key Word was "Thing" and if we multiply Thing x4 we would have a good explanation of the 20 rows in the 340 with Thing repeated 4x then codes to be adjusted by the letter value of T=20 H = 8 I= 9 N =14 G = 7

We also see Thing underlined 6 times with 6 exclamation points and we know 6x6 =36 + 26 = 62. Is it possible that the shifts don't happen alphabetically but on a Typewriter keyboard?

For example in Line 1 H + 20 = B or N depending on the direction. if we add twenty to H we get B or N depending on which direction you go. Then 6 spaces on the Typewriter keyboard would yield a shift key or backspace each of which would carry a numerical value assigned by placement on the typewriter based on location. 54 keys on some old typewriters.

I know sounds a bit convoluted but it seems that given the struggles with frequency that Z must have chosen a system that would guard against frequency analysis and cribbing.

[AK Wilks]

Those are all good posibilities. I think Z left multiple clues of 3-6-9, including the ones you mention, plus the +09 that keeps showing up in the 340, the 0-3-6-9 on the map he sent in and others. I think that could be clues for a 0-3-6-9 Caesar shift in the second stage of the code. But before we get there we have to solve the first stage.

I think Quicktrader has shown an excellent analysis of why LL is likely to recur more than EE, OO or anything else. And the comments from me, glurk and doranchak show that Z specifically used LL more than any other combination. For all those reasons I agree with QT that it makes sense to provisionally solve the ++ as LL.

[AK Wilks]

Using the excellent analysis by Quicktrader showing the ++ very likely solves as LL, and from Glurk and the FBI files that normal F and normal B possibly solves as L, this is what we get:





Building from that, as Graysmith also solves + , F, and B all as L, I then add in strong logical possible word solves from the Raw Graysmith of SEE A NAME and THESE FOOLSHALL SEE this is what we get:
Note: I did not intentionally create the 1st line THE or the 2nd line HELLTOO. Those were cteated by applying the QT solves, the Glurk/FBI solves and the RG word solves of SEEANAME and THESEFOOLSHALLSEE.

[masootz]

good work mr wilks. it opens up the possibility of the simple answer being that he created a cipher which, when properly decoded, reveals a message hidden in within lines of garbage text. that would make a lot of sense to me as i've never believed zodiac was an expert cryptologist. if you wanted to make an encryption harder to crack, a la the progression from the 408 to the 340, having your statement mixed up with garbage text would certainly do so without requiring you to acquire additional encryption skills.

i realize this opens up even more implausible solutions but i think we're on the right track and personally maintain that the solution to the 340, much like the 408, will make perfect sense once it's revealed.

[AK Wilks]

good work mr wilks. it opens up the possibility of the simple answer being that he created a cipher which, when properly decoded, reveals a message hidden in within lines of garbage text. that would make a lot of sense to me as i've never believed zodiac was an expert cryptologist. if you wanted to make an encryption harder to crack, a la the progression from the 408 to the 340, having your statement mixed up with garbage text would certainly do so without requiring you to acquire additional encryption skills.

i realize this opens up even more implausible solutions but i think we're on the right track and personally maintain that the solution to the 340, much like the 408, will make perfect sense once it's revealed.

Thanks!

I think the first stage solution to the 340 would likely have several lines of clear English, like perhaps SEEANAME and THESEFOOLSHALL SEE and maybe HELLTOO. Several other lines may seem to be mostly or all garbled garbage, perhaps intentionally to increase difficulty of solution as you suggest, but perhaps those seeming garbage lines require a second stage solution of Cesar or something else.

Using the excellent analysis by Quicktrader showing the ++ very likely solves as LL, and from Glurk and the FBI files that normal F and normal B possibly solves as L, this is what we get:





Building from that, as Graysmith also solves + , F, and B all as L, I then add in strong logical possible word solves from the Raw Graysmith of SEE A NAME and THESE FOOLSHALL SEE this is what we get:
Note: I did not intentionally create the possible 1st line THE or the 2nd line HELLTOO. Those were created by applying the QT solves, the Glurk/FBI solves and the RG word solves of SEEANAME and THESEFOOLSHALLSEE.

[Quicktrader]

After six months, finally another update:

PhD Donald Burleson has written a brilliant abstract regarding vowel-consonant-analysis.
http://www.blackmesapress.com/Crypto.htm

The main idea behind his abstract is to not only look for the contact frequencies of each symbol but to also check the contact frequencies of each adjacent symbol of those 'root' symbols, too.

It should be mentioned that his theory is based on Russian development, however has fairly improved the identification process. Also, due to his division with the frequency of each root symbol, it should work with homophone ciphers quite well, too.

The basic idea is to divide the contact values of each adjacent symbol by the frequency of the root symbol, therefore achieving a 'MACC' value while the contact values of each root symbol (the symbol analyzed) is divided by its frequency, too (called 'VCC'). I had known about the VCC approach regarding the root symbol, even developing it further 'through the cipher', however not thought about analyzing its adjacent symbols. So the MACC is a good, at least complementary approach to analyze symbols with regard to their vowel/consonant status.

I did the Burleson-Analysis on parts of the 340. For this reason I had selected some groups of symbols of length approximately L=8 to reduce the overall cipher complexity (less cipher, less symbols). I took such groups that appear to be connected, sort of tied to each other by using as many similar symbols as possible. This allows counterchecking the groups to each other, therefore eliminating errors in cleartext or vowel-consonant analysis.

The results of the Burleson-Analysis were astonishing: Most of the symbols had a MACC value of either less than 9 or more than 11. This meaning that there is only a small group of symbols which identity (vowel or consonant) could not cleary be identified. Those symbols are potentially good 'binders' such as the letters 'D', 'T' or 'N', while symbols with a MACC value of less than 9 are potential vowels (including the two best binders 'L' and 'R') and, further, symbols with a MACC value of greater than 11 being consonants.

Identification of symbols being vowels/consonants thus got possible very clearly.

Additionally to that, the MACC values appear to be close to Burleson's experience regarding the MACC value, e.g. 9.8 for the letter 'N'. Based on this analysis, a symbol can be associated to a group of cleartext letters, such as 'D', 'T' or 'N', instead of requiring a 24 letter alphabet as being potentially correct (8 symbols usually quote with 24^8 varieties..now being e.g. 3^8 varieties, which is of course dramatically less, therefore easier to solve).

One letter showed to be distinctively a super-consonant: The letter 'Y' has a low MACC value of 4.5, exactly such value was found in the group I had selected, too. Looking closer to this specific symbol, it indeed had followed a double symbol. Both, the identical MACC value plus the low VCC value of the root symbol itself plus the connection to a double symbol leads me to the conclusion that this specific symbol is actually representing the cleartext letter 'Y'. In the selected groups this symbol appears only twice, so I would not exclude the letter 'I' to be an alternative, however the MACC value of approx. 4.5 is in fact compliant and is a 'Y' following two double letters linguistically quite ok, too.

Will add some more analysis on the cipher, trying to associate various cleartext letters to symbols according to their MACC value, hopefully getting a readable cleartext in the near future.

groups.jpg

The selected groups..the first four letters cover approx. 30% of this cipher selection.

macc.jpg

MACC values with 3 having a 4.5, similar to Burleson's abstract (letter 'Y')

QT

[masootz]

qt - is it possible, using this method, to decode the initial substitution part of the cipher if we assume he additionally used some other method? in other words, if he did scramble words or purposely make spelling mistakes does this method fall apart because it's initial assumption is that certain letters appear near certain other letters or in groups?

in either case this is fascinating and potentially a very lucrative avenue for figuring out what he did. if we use this method to make determinations about what a decrypted substitution cipher 340 should look like then i assume it'd be pretty apparent, given enough time for the variants, to say with some level of certainty that he did or didn't use this method.

[jroberson]

Burleson must be totally blind to the color red.

Wow.

It's like tripping on LSD.