Whatever Zodiac did seems to have also suppressed repeated bigrams (without masking the repeated trigrams).
The Hardens said they guessed Zodiac might have used the word "kill" repeatedly in the 408, so they looked for "patterns" where the word would fit. The "square"-"B" bigram (standing for "LL") turned out to have the highest number of repeats, and "kill" fits for some of them.
All Zodiac had to do was read the papers to figure out what hints the Hardens used to crack the 408. He could have easily taken steps to hide those hints in the 340.
Here’s the excerpt from the Aug 9, 1969 article:
The patterns:
Guessing that square-B was "LL", forward slash was "K", and triangle was "I" would have been a very nice first step to unlocking the 408, because those symbols cover about 10% of the plaintext.
I keep coming back to the question: "Is it possible to attribute the 340 not cycling as well as the 408 (despite its higher symbol count) due to some transposition after encoding?".
Either,
– The cycles are randomized to some degree.
– Transposition is causing the randomization of the cycles.
– Anything that does not relate to the above, "superencipherment" or an extra step.Can we come up with tests to differentiate between any of these, even if just for specific schemes?
I probably wouldn’t be too great at making tests, but at some point will try to replicate the 340 by enciphering and then transposing to come up with a lot of period 19 bigram repeats, only a couple of apparent bigram repeats per half, and similar cycle stats. I will try to cook up a practical way to do that in a while if you want. I need to take another look at the odd-even phenomenon again and try to understand it better.
I can see why you were asking about the period 15 mirrored bigram repeats. I guess it could be a similar scheme with 15 rows and after lifting the letters from the original draft during transposition he placed them right to left instead of left to right. I guess that it is possible, but I think that working with 19 rows is good for the present because it is simple and closely matches the shape of the final message. Transposing like what we are working with now is going to be shown in a cryptography book. Except that a more complicated version of transposition rearranges the columns according to a key. He didn’t do that because of the period 19 bigram repeats.
I am going to try to think positive about what we are doing now, even though it is neither exciting nor exotic. Does your computer program assume that he placed the transposed letters from left to right top down, or does it also consider right to left bottom up, which I believe would cause a backwards period 19 bigram repeat? The weird Z O ^ A I K "signature" at the end maybe suggests not. I don’t know.
Another idea that i could work on is to look at the current model and see if I can replicate apparent trigram repeats. I did notice some when I was trying different versions of smokie13, but wasn’t focusing my attention on it.
doranchak, my thinking is that Zodiac may have randomized his symbol selection more and more as he encoded the 340. I did a message with a paper and pen key and used check marks to track the use of cycle symbols. As the check marks accumulate into little clusters, they start to get difficult to count. At that point, random symbol selection is just easier.
With smokie13, I randomized the symbol selection 0% in rows 1-5, 20% in rows 6-10, and 30% in rows 11-19. I didn’t randomize the symbol selection in row 20 because it is just the last part of the message not transposed.
Here is what I have for 340 trigram repeats:
I ran a few new trials for smokie13 by just clicking on "calculate now" to encode the cycles. No changes to the model whatsoever.
After two tries, I got this:
After one more try, I got this:
After another try, I got this:
They don’t line up with each other, but this shows that the current transposition model with cycles can easily result in a couple of trigram repeats.
If you have any requests or want to know more information let me know. I flattened out the key for smokie13. One of the letters has only one symbols. Most of the letters have two symbols. A few of the letters have three symbols and a few of the letters have four symbols.
@smokie, there may be something wrong with your trigram detection as the 340 has 2 repeating trigrams. I don’t know how likely or unlikely trigram repeats are in any case but your point may be valid.
The test I’m running is about 1/5th done for both the smokie13 and 340. The highest scoring variation for the 340 currently is 20683 and for the smokie13 it’s almost 22000. Both tests have not passed the first transposition starting corner (1, 2, 3, 4).
I have about 300 of these which are all very similar, I suppose even better results will come in later.
corner: 1, message length: 326, add: 44, sub: 204 Score: 21916 Ioc: 722 M: 196 C: 326 S: 64 osedhepeopleofthe tsofwhichitiscomp ndwelfareofthepar heunionthesafetya antheexistenceoft ncesnothinglessth dinginitsconseque sortancecomprehen ectspeaksitsownim sofamericathesuck fortheunitadstate nanewconstitution upontodeliberateo nmentyouarecalled stingfederallover iciencyofthesubsi erienceoftheineff ranunequivocalers tateofnewyorkafte tot /^$V%J%/JX%/&4C^ 65Z&QVW>()]W5>/GJ .?9U,A![%I&4C%J![ (Ua.DZYNCUM!A@]ST !.6C^@:_54%3>UI&N H#%M.Z6V_Y'FUM5]V $DH'W.)]#/.M^K7% 0I[6!H#@#I-J2%V^H %>NJU<+_N5I9YD- 5ZA<G^LW>T]VUO"* &/2NC^7.)6<?NT]@ Y<Y^9>/H5ND]O])Z. 7J/H]/?UFW=^L!N%/ 3-@Y4;ZO!2@>TXF@? M4W3'&U$@LTFBIP^2 )#_%H#SIA]VU57=MW @L)@Y>^ZA](@DHUA& 2T3O3@17)8I#TFUR0 N!]UIA.^QSZ[E<A4U ]I4
I can see why you were asking about the period 15 mirrored bigram repeats. I guess it could be a similar scheme with 15 rows and after lifting the letters from the original draft during transposition he placed them right to left instead of left to right.
I’ve tried that but I’m not getting anything like the 340. I can’t seem to reproduce a bigram repeat peak period 15 (mirrored) verus 19 (normal). Some speculation and ideas… I’m thinking that it may be the reason why the 340 is not solving under the bigram hypotheses we’ve explored so far. Possibly a different (from normal left-to-right, top-to-bottom) encoding path is causing it but does that makes sense?
I can see in your solution scoring 21916 where the misalignment lines are (because I know where they are). I assume that the high scoring 340 is gibberish?
I think that the encoding path so far as we have found is left right top bottom. But what about transferring plaintext right left bottom top, which would cause period 19 bigram repeats?
Excellent job by the way.
osed he people of the
ts of which it is comp
nd welfare of the par
he union the safety a
an the existence of t
nces nothing less th
ding in its conseque
portance comprehen
ect speaks its own im
s of america the subj
for the united state
n a new constitution
upon to deliberate o
nment you are called
sting federal gover
iciency of the subsi
erience of the ineffe
r an unequivocal exp
tate of new york afte
to t
The above is read from the bottom up from left to right
Edit ..just to note the first and last two lines are out of kilter..the rest is near perfect format.
To the people of the state of New York
After an unequivocal experience of the inefficiency of the subsisting federal government you are called upon to deliberate on a new Constitution for the United States of America The subject speaks its own importance comprehending in its consequences nothing less than the existence of the Union the safety and welfare of the parts of which it is composed
I had to google the answer
Thx Mr lowe, since the solve is upside down one of the other corners should produce an even better solve.
So far the test is over 1/4th and variations of the 2nd corner are being processed. Returns from the 1st corners for the 340 were really weak, 95 variations scored over 20500 with a max of 20683. Now with the start of the 2nd corners there are already (300+) variations over 20500 with the highest scoring 20856. I’ve been checking some of these results and I don’t see anything in them.
corner: 2, message length: 317, add: 53, sub: 185 Score: 20856 Ioc: 652 M: 201 C: 317 S: 64 sischingiseitdedf oredinersttingato ftherisiabuseyesa nissyousprrywithi didoworpreadforad heyinpayingfodesi mshouairsthinthey dowistatentlikert ometorsaychildhel icarhoodnonitsmel ltheandnifcuntcho ftuntinerrorstarc emayssnotehhlandc reateasotthadefen ginotheryounduswh entattheststobomu sechpotsteweaofsf backsandsawayedpr elingisodua OCB1GFT7FO%F+p:pH S<K*FzK<B+;lz7c+E H+(:N6OFc_2Ot^MBR TFO4^92ByNU-jl+(> /lp9fS<yAYc*d8U#p (W^C5yc-6zXd|*ZOl )B.E2#lU4+GC5VG%^ p|f>4+R+K@VkF3WDV |)MVbNJR-L.FPp(WP FLqU.||/z9z>+O)Yk k+GZc5*5ldL2&+L.| d+25VF5tU<|D4+#<1 W)R-4BTEVM((kRz*L <Wc+tc4|++G#/KHKT 7CT|+(MD^b25p2JfG tz+c+;.YJ+J+8_S)2 BWL.y8+B+ZfKR9aBd _c13BR&pORjq^MpyN YP6zXlBbp2#
I said the test would run about 24 hours but I miscalculated, it will be about 4 days at a rate of nearly 2200 cipher variations being processed per minute.
@smokie, there may be something wrong with your trigram detection as the 340 has 2 repeating trigrams. I don’t know how likely or unlikely trigram repeats are in any case but your point may be valid.
I shuffled the cipher text 1,000,000 times and measured the number of repeated trigrams for each one. Here are the results:
0 repeats: 665,689
1 repeat: 266,895
2 repeats: 57,335
3 repeats: 8,857
4 repeats: 1,073
5 repeats: 133
6 repeats: 15
7 repeats: 3
This suggests there’s about a 6.7% chance for a shuffle to have 2 or more repeating trigrams. (The original 340 has 2 repeating trigrams).
The expected number of repeating trigrams for any given shuffle is about 0.4.
Thanks doranchak.
After randomization of a 340 character plaintext, the trigram count is still plenty. With non-cyclic/random encoding your numbers should be correct. But with cyclic/non-random encoding these trigrams should have a much higher chance of persistance. I believe smokie is correct.
Yay, the smokie13 solved almost perfectly, there are 84 results which score over 23000. The test for the 340 is over 2/3 in, its highest return scoring 20857, it’s looking grim.
corner: 3, message length: 327, add: 44, sub: 204 Score: 23840 Ioc: 711 M: 195 C: 327 S: 64 tothepeopleofthes tateofnewyorkafte ranunequivocalexp erienceoftheineff iciencyofthesubsi stingfederallover nmentyouarecalled upontodeliberateo nanewconstitution fortheunitadstate sofamericathesubj ectspeaksitsownim portancecomprehen dinginitsconseque ncesnothinglessth antheexistenceoft heunionthesafetya ndwelfareofthepar tsofwhichitiscomp osed ]I4V%J%/JX%/&4C^ N!]UIA.^QSZ[E<A4U 2T3O3@17)8I#TFUR0 @L)@Y>^ZA](@DHUA& )#_%H#SIA]VU57=MW M4W3'&U$@LTFBIP^2 3-@Y4;ZO!2@>TXF@? 7J/H]/?UFW=^L!N%/ Y<Y^9>/H5ND]O])Z. &/2NC^7.)6<?NT]@ 5ZA<G^LW>T]VUO"* %>NJU<+_N5I9YD- 0I[6!H#@#I-J2%V^H $DH'W.)]#/.M^K7% H#%M.Z6V_Y'FUM5]V !.6C^@:_54%3>UI&N (Ua.DZYNCUM!A@]ST .?9U,A![%I&4C%J![ 65Z&QVW>()]W5>/GJ /^$
The test for the 340 finished. I personally don’t see much in these solves and they don’t score highly either. Though 1 scored over 21k and has the fragment "mything" in it but that’s not very unplausible. Corner 2 and especially corner 4 returned the best results.
All results scoring over 20500: https://drive.google.com/open?id=0B5r0r … 0xoZnNtVlE
Highest scoring result:
corner: 4, message length: 317, add: 179, sub: 299 Score: 21028 Ioc: 650 M: 201 C: 317 S: 64 eutraittoothoseth sportiseeinremeah ansuscraticeiscri maresomesofsisift opusmullisineiler drinlargetingendo psinyonesstofindt hstoundbymollsore forareemadeitsref ormythinghoustopl eefulsignsnoleply earbinginorforthe rpeantusebyanyotg ogesfcdalsthedane gotlesinanddhisou rsowhattheuncgmai ldedoneathatmeusb caattomousignowel landgardasc Yc13BR&b7FO%F+pyN _WL.y8+pp2#q^MpBd Bz+c+;.BORjKR9S)2 tCT|+(MY+Zf+8_JfG 7Wc+tc4DJ+J5p2HKT <)R-4BTa^b2#/Kz*L W+25VF5|++G(kR#<1 d+GZc5*EVM(D4+L.| kLqU.||tU<|2&+)Yk F)MVbNJ5ldL>+O(WP ||f>4+R/z9zFPpWDV pB.E2#lR-L.kF3G%^ )W^C5yc+K@VC5VZOl (lp9fS<U4+Gd|*U#p /FO4^92-6zX*d8+(> T+(:N6OyAYc-jlMBR H<K*FzKByNUOt^c+E SCB1GFMFc_2lz7:pH P6zXlBT<B+;
20523_635_201_3324782.txt has:
the numbede data ro makes mple do in try murder ooms
Murder rooms? 
I combined all the plaintexts together and scored the words found in the result. Here they are:
https://docs.google.com/spreadsheets/d/ … sp=sharing
Longer words appear first. Within each length, words are in descending order of the combined score which is the product of the word’s relative frequency in the English language, and the number of times the word appears among the plain texts.
