Nick Pelling's thoughts on the Zodiac ciphers

[doranchak Profile Website]

Also, one thing I didn't include in the experiment is to check for repeated tetragrams, similar to how "qEHM" appears in the unmolested 408. I think its repetition in the 408 makes it even less likely that Z simply by accident stumbled upon a selection that looks "pulled down." I don't see any such nice groupings of trigrams or tetragram in the randomized selections.

I didn't look for Ted's name anywhere, though. I hope I've flushed AK out of my brain.

[glurk Profile Website]

Also, one thing I didn't include in the experiment is to check for repeated tetragrams, similar to how "qEHM" appears in the unmolested 408. I think its repetition in the 408 makes it even less likely that Z simply by accident stumbled upon a selection that looks "pulled down." I don't see any such nice groupings of trigrams or tetragram in the randomized selections.

Does that mean that the repetition of "qEHM" makes it more likely that Z intentionally pulled them down?

-glurk

[doranchak Profile Website]

It seems so. It would have taken a good bit of luck to accidentally achieve the same feat using a non-pulldown method.

[doranchak Profile Website]

As expected, experimentation shows the relative rarity of four columnar repeats that occur on the same line:

http://oranchak.com/zodiac/pulldowns-3.html

Only found 4 hits so far (one of them happened upon a 5-gram), out of 5,811,384 random trials. So, if you were using a non-pulldown method to construct the filler for the 408, you'd have about a 1 in 1.5 million chance of coincidentally repeating a tetragram from above while constructing the final 18 symbols of the cipher text.

So, if the cipher author used some other method to construct the filler, he had to be really lucky to get all those in-column repeats, plus the perfectly repeated tetragram. One wacky possibility remaining is that the pulled down symbols mark other plaintext letters in some way to construct a message. For instance, what if the matching symbols in the rows above the final row were selected because they appear underneath plaintext letters that form the "filler" message? Or perhaps some other positional rule is used to identify the corresponding plaintext.

I did say it was wacky.

[Zsearcher Profile]

I mentioned the idea that in addition to filler it was possibly some kind of clue years ago.

[Wrench Profile]

Great work Doranchak. Now what's your take on the drop down trigram in the 340?

I know, never satisfied...

[traveller1st Profile]

So, if the cipher author used some other method to construct the filler, he had to be really lucky to get all those in-column repeats, plus the perfectly repeated tetragram. One wacky possibility remaining is that the pulled down symbols mark other plaintext letters in some way to construct a message. For instance, what if the matching symbols in the rows above the final row were selected because they appear underneath plaintext letters that form the "filler" message? Or perhaps some other positional rule is used to identify the corresponding plaintext.

I did say it was wacky.

I second. Great work.

EDIT: removed stupid question - was getting my ciphers confused lol.

Trav

[Zsearcher Profile]

If you count the last letter on the 23rd row: E

And the twelve filler letters in the 24th row: B OR ETEMETH TI

You get 13 total letters: EBORETEMETHTI

And the Es line up with the 8s in the middle. Creepy.

Maybe someone can figure out another step.

[doranchak Profile Website]

It occurred to me that an exact proportion could be calculated.

How many different ways can you construct the final 18 symbols of the 408 cipher?

* Let's assume that for each of the 18 positions, you can select one of the 54 unique symbols of the cipher.
* Therefore, there are 54^18 = 1.5 x 10^31 possible constructions of the final 18 symbols. The full number is 15,243,604,656,924,933,407,477,640,462,336.

How many different ways can you construct the final 18 symbols so you end up with a repeated tetragram plus 8 additional in-column repeats?

* There are 14 different columns in which a repeated tetragram can begin.
* There are 23 different rows in which the tetragram can appear.
* When a tetragram is selected, that leaves 13 columns for us to choose from.
* We choose 8 of them. There are 1,287 ways to select 8 columns from 13 possibilities.
* For each of the 8 columns, we can select one row among 23.
* So, the total number of constructions is: 14*23*1287*(23^8) = 3.2 x 10^16, or about 32 quadrillion.

Thus, if you randomly select uniformly from all possible constructions, the odds of happening upon an arrangement where there's a repeated tetragram plus eight columnar repeats would be 470 trillion to one AGAINST you.

Somebody check my math. It's easy to make mistakes with this.

EDIT: The above counts exclude constructions that have more than 8 columnar repeats in addition to the repeated tetragram. So, let's see. There are 1,287 ways to select 8 columns from 13. And there are 715 ways to select 9 columns from 13. 286 ways to select 10 from 13. 78 ways to select 11 from 13. 13 ways to select 12 from 13. And finally, 1 way to select 13 from 13.

So the updated number of possible "nicely pulled down" constructions is: 14*23*(1287*(23^8) + 715*(23^9) + 286*(23^10) + 78*(23^11) + 13*(23^12) + 1*(23^13)) = 2.8 x 10^20, or 282 quintillion.

That brings down the odds to about 54 billion to one against you if my math is right.

EDIT #2: Also, I didn't account for symbols that repeat within a column. Thus, there aren't really 23 rows to pick from when selecting a symbol for a columnar repeat. So, the 282 quintillion is an upper bound on the true number of constructions.

[glurk Profile Website]

doranchak-

I'll check your math as best as I am able. I fear it will make my head hurt, but I'll give it a go.

How many different ways can you construct the final 18 symbols of the 408 cipher?
* Let's assume that for each of the 18 positions, you can select one of the 54 unique symbols of the cipher.
* Therefore, there are 54^18 = 1.5 x 10^31 possible constructions of the final 18 symbols. The full number is 15,243,604,656,924,933,407,477,640,462,336.

OK, this is valid. As a smaller example, if we look at 3 numbers from the set of {0...9} then 10^3 = 1000 and indeed 000 to 999 enumerates 1000. So 54^18 is correct. QED.

How many different ways can you construct the final 18 symbols so you end up with a repeated tetragram plus 8 additional in-column repeats?
* There are 14 different columns in which a repeated tetragram can begin.

Certainly, with a 17 column-wide cipher, a tetragram in row 24 can only begin in columns 1-14.

* There are 23 different rows in which the tetragram can appear.

Assuming a 24-row cipher in which the tetragram appears in row 24, that leaves 23 to compare against.

* When a tetragram is selected, that leaves 13 columns for us to choose from.

Given that a tetragram is of length 4, and the column width is 17, 17-4=13.

* We choose 8 of them. There are 1,287 ways to select 8 columns from 13 possibilities.

Yes, via combinatorics, (13!) / (8!(13-8)!) = 1287:
http://www.wolframalpha.com/input/?i=%2 ... 8%29%21%29

* For each of the 8 columns, we can select one row among 23.

Yes, the combinations here should be 23^8 (78310985281 combinations.) Or is this wrong?

* So, the total number of constructions is: 14*23*1287*(23^8) = 3.2 x 10^16, or about 32 quadrillion.

Looks right to me. Any mistakes I've made in this "double-check" are my own, but doranchak's calculations look correct to me.

I'd also point out that the "pulldown-theory" includes a pulled-down bigram as well as the pulled-down tetragram. But then the math would get really hairy....

-glurk