"IofC" is easy, it’s Index of Coincidence (IC or IoC). But I can’t quite understand the rest. Especially the part about "numbers" in Z340. Is he treating symbols as numbers? But how does he map symbols to numbers?

I have to admit that I do keep returning to the card with "Paradice" and "Slaves" forming a cross quite a bit, but no idea what it means. I’ve seen suggestions that it hints at Z340 being a word search type of puzzle. But personally I don’t think it’s likely, mostly because you can’t write a meaningful message that way. I’m more inclined to think that it just happens that the decoded Z340 would spell out "Paradice" and "Slaves" in one of the rows/columns (not consecutive letters of course). Z might have noticed it later on, or as he was writing the message, and offered it as a hint as to which symbols would map to some of the letters. Which row and column in particular though? There are several possibilities that would allow such mapping due to repeated letters in "Paradice" and "Slaves". Another idea I had is that this type of "crossed words" arrangement hints at Z340 being encoded as both vertical and horizontal. Some sort of lattice route, where you start reading horizontally and skip every other letter, and then you finish skipped letters by reading them vertically, by columns? Since there is an odd number of letters in each row, it will create a perfect checkerboard pattern. I’ve tested that idea and didn’t get a solve though.

I wonder if the Halloween card is a hint that the cipher is in 2 parts with 2 keywords per half. If its a hint, it has to make some level of sense.

I could be wrong but I think all 4 byfire. byknife… could all work as proper keys. No repeating letters.

I think I’ve reproduced his result concerning the distances between recurring doubles:

Only 12.5% of the numbers in the 340 set should be divisible by 8, so if the distances between recurring doubles is a "random" event, the numbers should be in this range. In a "random" set about 42.5% of the numbers should be divisible by 2 or 3, while in this set 36% are divisible by 2 and 25% are divisible by 3. These numbers are somewhat low, but tolerable. The real thrill comes when we count the numbers divisible by 8, which should be at or less than 12.5% – we find that 28.57% are divisible by 8, more than twice the expected amount!

So, I think he measured the distances between recurring bigrams, and then looked for factors in those distances that occur more often than expected. Here is the output of a program I just wrote to demonstrate. Each bigram is shown, followed by a list of its positions in the cipher text. Then each difference is shown, along with all of the numbers that evenly divide the difference:

By: [34, 167]
 - difference: 133. factors: 7 19 
p7: [53, 163]
 - difference: 110. factors: 2 5 10 11 22 55 
+&: [80, 200]
 - difference: 120. factors: 2 3 4 5 6 8 10 12 15 20 24 30 40 60 
8R: [86, 219]
 - difference: 133. factors: 7 19 
FB: [145, 166, 214]
 - difference: 21. factors: 3 7 
 - difference: 48. factors: 2 3 4 6 8 12 16 24 
#O: [22, 125]
 - difference: 103. factors: 
Np: [17, 327]
 - difference: 310. factors: 2 5 10 31 62 155 
+R: [64, 171]
 - difference: 107. factors: 
|5: [76, 212]
 - difference: 136. factors: 2 4 8 17 34 68 
O+: [62, 126]
 - difference: 64. factors: 2 4 8 16 32 
Op: [162, 263]
 - difference: 101. factors: 
G2: [14, 120, 270]
 - difference: 106. factors: 2 53 
 - difference: 150. factors: 2 3 5 6 10 15 25 30 50 75 
5F: [77, 213]
 - difference: 136. factors: 2 4 8 17 34 68 
Bc: [146, 215]
 - difference: 69. factors: 3 23 
Fl: [89, 282]
 - difference: 193. factors: 
UZ: [40, 250]
 - difference: 210. factors: 2 3 5 6 7 10 14 15 21 30 35 42 70 105 
M+: [38, 70, 138]
 - difference: 32. factors: 2 4 8 16 
 - difference: 68. factors: 2 4 17 34 
(): [44, 303]
 - difference: 259. factors: 7 37 
++: [63, 236, 289]
 - difference: 173. factors: 
 - difference: 53. factors: 
(#: [21, 101]
 - difference: 80. factors: 2 4 5 8 10 16 20 40 
)L: [45, 287]
 - difference: 242. factors: 2 11 22 121 

So I think Glen is saying that there are 25 measurements of the differences in positions. In a random set of numbers, you’d expect 1/8 (12.5%) of them to be divisible be 8, on average. But among the 25 measurements, 7 of them are divisible by 8. 7/25 = 28% (not sure where he got 28.57% though). And that’s more than twice the expected amount.

Also, check out this spreadsheet: https://docs.google.com/spreadsheets/d/ … sp=sharing

It shows each factor, how often it’s expected to be seen among 25 random samples, and how many times it is actually seen. The ratio shows how far off the actual count is from the expected count. The ones in boldface are the "unusually high" ones. Note that 17 also occurs more than expected, coinciding with cipher width.

Not sure if any of these are significant but it’s an interesting observation by Glen.

How can a phrase like "Paradice Slaves" be used as a key (within the context of the post)?

I can only guess that the phrase can "seed" the first stage of symbol assignment. Write all plaintext letters from A-Z. Then underneath, write the key phrase PARADICE SLAVES with duplicate letters removed, then fill the rest with the unused letters. You now have the first substitutions, and can start putting extra symbols in the key.

What is meant by:

There’s also this odd distance grouping that points in the same direction – recurring doubles group at distances, several in the 60’s, several in the low 100’s, and several in the 130’s.

I’m guessing this relates to the distance measurements – you can see them in my above results (differences in the 60s, 100s, and 130s.)

What specifically is a low-level polyalphabetic cipher?

I think he refers to the similar phenomenon in Z408. That is, some symbols ended up being assigned to multiple plaintext letters (which I think is due to transcription mistakes). It ends up making Z408 slightly (low-level) polyalphabetic.

Some similar peaks seem to appear for Z408.

Here’s the raw output again but for Z408:

Wq: [143, 293]
 - difference: 150. factors: 2 3 5 6 10 15 25 30 50 75 
BV: [245, 287]
 - difference: 42. factors: 2 3 6 7 14 21 
I): [92, 141]
 - difference: 49. factors: 7 
S(: [51, 100]
 - difference: 49. factors: 7 
Ik: [39, 406]
 - difference: 367. factors: 
HM: [129, 401]
 - difference: 272. factors: 2 4 8 16 17 34 68 136 
A9: [242, 303, 333]
 - difference: 61. factors: 
 - difference: 30. factors: 2 3 5 6 10 15 
@X: [125, 199]
 - difference: 74. factors: 2 37 
LM: [70, 148, 229]
 - difference: 78. factors: 2 3 6 13 26 39 
 - difference: 81. factors: 3 9 27 
cZ: [136, 373]
 - difference: 237. factors: 3 79 
VE: [215, 391]
 - difference: 176. factors: 2 4 8 11 16 22 44 88 
eG: [20, 325]
 - difference: 305. factors: 5 61 
)S: [50, 274]
 - difference: 224. factors: 2 4 7 8 14 16 28 32 56 112 
r: [75, 194]
 - difference: 119. factors: 7 17 
DR: [155, 189]
 - difference: 34. factors: 2 17 
^R: [132, 183]
 - difference: 51. factors: 3 17 
6q: [94, 398]
 - difference: 304. factors: 2 4 8 16 19 38 76 152 
L): [180, 212]
 - difference: 32. factors: 2 4 8 16 
BP: [56, 153, 386]
 - difference: 97. factors: 
 - difference: 233. factors: 
P#: [122, 285]
 - difference: 163. factors: 
Tt: [43, 251, 377]
 - difference: 208. factors: 2 4 8 13 16 26 52 104 
 - difference: 126. factors: 2 3 6 7 9 14 18 21 42 63 
85: [97, 146]
 - difference: 49. factors: 7 
%%: [268, 280]
 - difference: 12. factors: 2 3 4 6 
WV: [17, 322]
 - difference: 305. factors: 5 61 
UI: [38, 405]
 - difference: 367. factors: 
WI: [175, 396]
 - difference: 221. factors: 13 17 
GY: [21, 326]
 - difference: 305. factors: 5 61 
OR: [10, 58, 366]
 - difference: 48. factors: 2 3 4 6 8 12 16 24 
 - difference: 308. factors: 2 4 7 11 14 22 28 44 77 154 
qE: [66, 127, 399]
 - difference: 61. factors: 
 - difference: 272. factors: 2 4 8 16 17 34 68 136 
#B: [55, 117, 123, 152, 244, 286]
 - difference: 62. factors: 2 31 
 - difference: 6. factors: 2 3 
 - difference: 29. factors: 
 - difference: 92. factors: 2 4 23 46 
 - difference: 42. factors: 2 3 6 7 14 21 
B%: [7, 305]
 - difference: 298. factors: 2 149 
9%: [0, 279, 334]
 - difference: 279. factors: 3 9 31 93 
 - difference: 55. factors: 5 11 
qG: [87, 111]
 - difference: 24. factors: 2 3 4 6 8 12 
c+: [299, 312]
 - difference: 13. factors: 
PO: [57, 365]
 - difference: 308. factors: 2 4 7 11 14 22 28 44 77 154 
RU: [208, 310]
 - difference: 102. factors: 2 3 6 17 34 51 
9#: [54, 151, 243]
 - difference: 97. factors: 
 - difference: 92. factors: 2 4 23 46 
VW: [80, 174, 246]
 - difference: 94. factors: 2 47 
 - difference: 72. factors: 2 3 4 6 8 9 12 18 24 36 
TY: [192, 331]
 - difference: 139. factors: 
EH: [128, 400]
 - difference: 272. factors: 2 4 8 16 17 34 68 136 
qW: [292, 321]
 - difference: 29. factors: 
r9: [150, 394]
 - difference: 244. factors: 2 4 61 122 
AP: [203, 284]
 - difference: 81. factors: 3 9 27 
Xq: [126, 291]
 - difference: 165. factors: 3 5 11 15 33 55 
/9: [53, 278]
 - difference: 225. factors: 3 5 9 15 25 45 75 

And the breakdown for each factor is on the 2nd tab on the spreadsheet: https://docs.google.com/spreadsheets/d/ … sp=sharing (the tabs are at the bottom of the page)

The factor 8 occurs 1.5 times as much as expected for Z408, compared to 2.2 times for Z340. Z408 also has strong peaks for factors 16 and 17.

I have no idea what to make of it all. Seems like noise at first glance, though.

I think he measured the distances between recurring bigrams, and then looked for factors in those distances that occur more often than expected.

It’s similar to the same symbol distance variance metric that I worked on for a while, except that it uses bigrams instead of monograms (i.e. individual symbols). It can be used to measure the randomness of the homophone cycles in the cipher (and in parts of it). The idea behind it is that perfect cycles spread out same symbols more evenly (see the linked post for detailed explanation), which proved to be reliably measurable by a simple number. Not sure what distances between same bigrams can tell us though. Cracking bifid ciphers involves looking at IoC of bigrams at a distance (or with a period, or with a step, there are many names for it), but that’s different from looking at distances between same bigrams. I think you look at factors of distances of the same symbols to find vigenere period (i.e. key length), but again, that’s individual symbols, not bigrams. There might be something here of course, just needs further research.

In a random set of numbers, you’d expect 1/8 (12.5%) of them to be divisible be 8, on average. But among the 25 measurements, 7 of them are divisible by 8. 7/25 = 28% (not sure where he got 28.57% though). And that’s more than twice the expected amount.

Well, we are not looking at a random set of symbols. So it’s a good thing that Z340 doesn’t behave like a random set of symbols. That means that there is likely a meaningful (or at least non-random) plaintext behind it, and that Zodiac also didn’t use a very good encryption method (all good/modern encryption methods make the ciphertext behave just like random numbers).

The various non-repeat measurements are interesting to me. I’ve found that my cipher generator does a good job of producing ciphers that have the same non-repeat measurements as Z340. So it makes me wonder, how strong can the non-repeat measurements be on completely random cycles? Or put another way, how easily can non-repeat measurements be a "false positive" on the underlying cipher text?

Or perhaps a more interesting question is: What other symbol assignment methods produce the low variance needed to contribute to apparently high non-repeat scores, besides cyclic homophones?

It does seem to be a good thing that all these seemingly non-random features are present in the cipher text. It lends hope to the hypothesis that a real message exists in there somewhere.

Thanks allot for your explanation daikon and dorachak! I thought that IofC was perhaps a variant of IoC, f for factor?

From your tables I can derive that the average bigram distance for the 340 is about 120 which has factors: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120.

I suppose the simplest answer is that with enough samples factors of 120 (the average) will be higher. Or to the extent factors are shared, for example 16 shares 3 out of 4 (ignoring 1). 8 and 10 share all factors with 120, since they are a factor of 120.