It’s nice to see that it works to some degree. But yes, I also think the problem are doubles (and even a triple).

Do you want me to create ngrams also with double ZZ’s for 3, 4 and 5-grams? Would be amazing if that worked!

If this works, then we will be able to use Decrypto to find words that have polyalphabetic symbols in them!

But hang on for a bit and let me do some confirmation testing. I want to make sure that I am thinking and doing this right.

Wildcard n-gram Experiment 2

Same message with perfect cycles, except that I changed the symbol L values a little bit.

Symbol Count
5 A 1 2 3 4 5
1 B 6
2 C 7 8
3 D 9 10 11
7 E 12 13 14 15 16 17 18
1 F 19
1 G 20
4 H 21 22 23 24
4 I 25 26 27 28
0 J
1 K 29
3 L 30 31 32
2 M 33 34
4 N 35 36 37 38
5 O 39 40 41 42 43
1 P 44
0 Q
4 R 45 46 47 48
4 S 49 50 51 52
5 T 53 54 55 56 57
1 U 58
1 V 59
1 W 60
1 X 61
1 Y 62
0 Z 63

25 30 26 29 12 29 27 31 32 28 35 20 44 13 39 44 30
14 6 15 7 1 58 49 16 25 53 26 50 51 40 33 58 8
21 19 58 36 27 54 28 52 34 41 45 17 19 58 37 55 22
2 38 29 25 31 32 26 35 20 60 27 30 9 20 3 33 18
28 36 56 23 12 19 42 46 47 13 49 57 6 14 7 4 58
50 15 34 5 37 25 51 53 24 16 33 43 52 54 10 1 38
20 17 48 39 58 49 2 35 26 34 3 31 40 19 4 32 30
55 41 29 27 31 32 50 42 33 18 56 21 28 36 20 20 25
59 12 51 34 13 57 22 14 33 43 52 53 54 23 45 26 30
31 27 37 20 15 61 44 16 46 17 38 8 18 28 49 12 59
13 35 6 14 55 56 15 47 57 24 5 36 20 16 53 54 25
37 20 62 39 58 48 45 40 7 29 50 41 19 19 60 26 55
21 1 20 27 46 32 56 22 17 6 18 51 57 44 2 47 53
42 19 28 54 25 52 55 23 3 56 60 24 12 38 26 60 27
30 31 6 13 48 14 6 43 45 35 28 36 44 4 46 5 11
25 8 15 1 37 9 2 32 30 57 21 16 26 22 3 59 17
29 27 31 32 18 10 60 28 30 31 6 12 7 39 34 13 33
62 49 32 4 59 14 50 25 60 26 30 31 38 40 53 20 27
59 15 62 41 58 34 62 35 5 33 16 6 17 8 1 58 51
18 62 42 58 60 28 32 30 54 47 62 55 43 52 31 39 60

Perfect solve in 15 seconds with the wildcard n-gram files.

To make things easy, I made 63 the wildcard and letter Z. I substituted 63 where the q appears in the 340 (not that it really matters exactly where). L=1 for 63.

25 30 26 29 63 29 27 31 32 28 35 20 44 13 39 44 30
14 63 15 7 1 58 49 16 25 53 26 50 51 40 33 58 8
21 19 58 36 27 54 28 52 34 41 45 17 19 58 37 55 22
2 63 63 25 31 32 26 35 20 60 63 30 9 20 3 33 18
28 36 56 23 12 19 42 46 47 13 49 57 6 14 7 4 58
63 15 34 5 37 25 51 53 24 16 33 43 52 54 10 1 38
20 17 48 39 58 49 2 35 26 34 3 31 40 19 4 32 30
55 41 29 27 31 32 50 42 33 18 56 21 28 36 20 20 25
59 12 51 34 13 57 22 14 33 43 52 53 54 23 45 26 30
31 27 37 20 15 61 44 16 46 17 63 8 18 28 49 12 59
13 35 6 14 55 56 15 47 57 24 5 63 20 16 53 54 25
37 20 62 39 58 48 45 40 7 29 50 41 19 19 60 26 55
21 1 20 27 46 32 56 22 17 6 18 51 57 44 2 47 53
42 19 28 54 25 52 55 23 3 56 60 24 12 38 26 60 27
30 31 6 13 48 14 6 43 45 35 28 36 44 4 46 5 11
25 8 15 1 37 9 2 32 30 63 21 16 26 22 3 59 17
29 27 31 32 18 10 60 28 30 31 6 12 7 39 34 13 33
62 49 32 4 59 14 50 25 60 26 30 31 38 40 53 20 63
59 15 62 41 58 34 62 35 5 33 16 6 17 8 1 58 51
18 62 42 58 60 63 32 30 54 47 62 55 43 52 31 39 60

Solved in 4:45. Quite a bit longer, but it did solve with the Z’s where they should be. For a while, the program thought that the Z’s were S’s, maybe because of expected frequency. I don’t know.

Note that the two Z’s on line 4 where the two q’s are on the 340 are at the end of the word "than" and the beginning of the word "killing." So no two-wildcard n-grams were needed to solve.

ILIKZKILLINGPEOPL
EZECAUSEITISSOMUC
HFUNITISMOREFUNTH
AZZILLINGWZLDGAME
INTHEFORRESTBECAU
ZEMANISTHEMOSTDAN
GEROUSANIMALOFALL
TOKILLSOMETHINGGI
VESMETHEMOSTTHRIL
LINGEXPEREZCEISEV
ENBETTERTHAZGETTI
NGYOURROCKSOFFWIT
HAGIRLTHEBESTPART
OFITISTHATWHENIWI
LLBEREBORNINPARAD
ICEANDALLZHEIHAVE
KILLEDWILLBECOMEM
YSLAVESIWILLNOTGZ
VEYOUMYNAMEBECAUS
EYOUWZLLTRYTOSLOW

Symbol 63 (or Z) is E, B, N, K, I, S, N, N, T, I, and I in that order

So yeah, if we want to test whether one of the symbols is polyalphabetic, and there are not two of those symbols in the same n-graph(?), we can use the one-wildcard n-grams to solve. I think that is what I am finding. Assuming that we know the L counts, and I am not sure how differences between set L counts and actual L counts affect things.

I will conduct another experiment before we should make two-wildcard n-grams.

Smokie

Glad to see it’s working a bit, though wonder what the limits are for this approach.

Made them anyway. And haven’t tried them.

By the way, I didn’t include the bigraphs for obvious reasons but come to think of it you should probably add a line at the bottom "ZZ : 100 0". The numeric value being the score of the particular ngram, maybe you’ll need to put it a bit higher even.

And your not locking the suspected wildcards to the "Z" letter?

zzgrams

@glurk

Is it a problem if the ngram files for ZKDecrypto are a bit unsorted?

Thanks

@glurk

Is it a problem if the ngram files for ZKDecrypto are a bit unsorted?

Thanks

The sorting does not matter. I just PM’d Smokie about this whole thing stating that I am "retired" damn it! RETIRED I tell you! I’m just burnt out on it. I did look at what you did with the ngram files, and it seems to work. But I did not spend too much time on it.

I’m much more willing to follow along with the work that you all are doing than to participate. I hope you understand. I am happily following the progress (and it IS progress) but I have nothing much to add at this point.

And I surely do not mean this to be discouraging or baleful, I just don’t spend much time on the work any longer.

-glurk

EDIT: I almost hate to comment on ZKD any more, because I tend to sound angry about it. I’m not angry. I’m just tired and worn out. I’m glad to see others taking up the work, and there is certainly more that can be done. I’m just tired of it for now.

Jarlve: I did not lock 63 to Z and will have to edit the recent post. I am going to look in the files to see if the n-graphs with the Z’s in them are there. The ones that I think Decrypto found. I don’t know to what extent this trick can be used, but it needs to be explored further for sure. Hopefully it will turn out to be another tool for the toolbox.

glurk: I completely understand your standpoint on these matters. Thanks for stopping me from trying to do something that would have been a huge waste of time. And thanks for ZKDecrypto!

Smokie

N-graphs for Wildcard n-graph Experiment 2

LIKZ : 81 0
ZECA : 75 0
THAZ : 54 0
ZILLI : 21 0
WZLD : 16 0
CAUZE : 6 0
REZCE : 13 0
THAZ : 54 0

Could not find ZHE or ZHEI. We would have to have ZHE because the word "the" is Zodiac’s most favorite trigraph. Am I missing something or did the program not solve correctly? We would also need ZHEI for the word "their."

Could not find GZVE or WZLL but ZKDecrypto constructed them somehow. The numbers must be some sort of score. I don’t know how we could accurately score an n-gram with a wildcard symbol if the wildcard symbol could represent more than one letter. For example, THAN is 83 and THEN is 87. So how could accurately score THZN? But I do think that we found a way to examine one polyalphabetic symbol. If there are four symbols with a total count of 57, I don’t know. I’ll do some more experiments later.

EDIT: I found bigraphs TH which scores 124, and ZH which scores 51. A TH must go with unigraph E to make the word "the." If we use the wildcard approach, the scoring would definitely be incorrect. But for now finding the word with some score is better than no word or score at all.

@glurk, you paved the way!

For example, THAN is 83 and THEN is 87. So how could accurately score THZN? But I do think that we found a way to examine one polyalphabetic symbol. If there are four symbols with a total count of 57, I don’t know. I’ll do some more experiments later.

I pondered over it also.

If you remember I also tried removing all wildcards from the cipher, and this worked (barely) for the 408. For which I found a couple of decent transcriptions.

Expected plaintext:

ilikkillingpeople
auseitissomucfuni
ismorefunthaillin
gwlameinteforrete
cauemaisthemotdan
geousanimalofallt
okillsoethiggives
ehemtthrillingepe
eceiseenbetterang
ettingyourrocksfi
thagilthestpartfi
tisthatwhedeiwill
berebornnparaicen
dalltheihvekildwl
lbomemyslavesiwil
otgiveoumnamebeca
ueyouwilltr

Recovered:

ilittallingpeople
auceitionbrunhusa
idrourfunshaillin
gwlerdistahorsete
neusraintherostag
grtudeneralofallt
otilloodthaggiaan
ehersthrillingspe
rndicaenmetssuesg
ettingyoursontohi
shegalthdntparthi
sietherwhateiwill
messmounspareanen
tallthrihadtilywl
lmorerycleasoiwil
osgaasoursaremana
udyouwilltu

Recovered:

ilaffallindpeople
auseinatherumhusa
idrourfunshaillan
dwleroistthorvene
meusraihtherostan
drtuoeneralofallt
ofalltoonhaddinth
ehersthrillandspe
rmoistencetssuesd
entindyoursomftha
shedalthohnparthi
sietherwhtteiwill
cesscounspareamen
tallthrihnofilewl
lcoreryslenstiwil
osdansoursarectma
uoyouwilltu

Experiment 3 ZZ Wildcard n-graphs

Tried the perfect cycle message and the zz wildcard n-graphs. Perfect solve in 15 seconds.

Then I substituted Symbol 63 where almost all of the +’s are found in the 340. I had to leave out two of them so that I could work with all 63 symbols.

25 30 26 29 12 29 27 31 32 28 35 20 44 13 39 44 30
14 6 63 7 1 58 49 16 25 53 26 50 51 40 33 58 8
21 19 58 36 27 63 28 52 34 41 45 17 19 58 37 55 22
2 38 29 25 31 32 26 35 20 60 27 30 63 63 3 33 18
28 36 56 63 12 19 42 46 47 13 49 57 63 14 7 4 58
50 15 34 5 37 25 51 53 24 16 33 43 52 54 10 1 38
20 17 63 39 58 49 2 35 26 34 3 31 40 19 4 32 30
55 41 29 27 31 32 50 42 63 18 56 21 28 63 20 20 25
59 12 51 63 13 63 22 14 33 43 52 53 54 23 45 26 30
31 27 37 20 15 61 44 16 63 17 38 8 18 28 49 12 59
13 63 6 14 55 56 15 47 57 24 5 36 20 16 53 54 25
37 20 62 39 58 48 45 40 7 29 50 41 19 63 60 26 55
21 1 20 27 46 32 63 22 17 6 18 51 57 44 2 47 53
42 19 28 54 25 52 55 23 3 56 60 24 12 38 26 63 63
30 31 6 13 48 14 6 43 45 35 28 36 44 4 46 5 11
25 8 15 1 37 9 2 32 30 57 21 16 26 22 3 59 17
29 27 31 63 18 10 60 28 30 63 6 12 7 39 34 13 33
63 63 32 4 59 14 50 25 60 26 30 31 38 40 53 20 27
59 15 62 41 58 34 62 35 5 33 16 6 17 8 1 58 51
18 62 42 58 60 28 32 30 54 47 62 55 43 52 31 39 63

There are three occurrences of double zz and ZKDecrypto solved with them. However, two of those occurrences split between words, but one does not: "will" becomes "zzll." No locking and the program found a solution in about one minute.

ILIKEKILLINGPEOPL
EBZCAUSEITISSOMUC
HFUNIZISMOREFUNTH
ANKILLINGWILZZAME
INTZEFORRESTZECAU
SEMANISTHEMOSTDAN
GEZOUSANIMALOFALL
TOKILLSOZETHIZGGI
VESZEZHEMOSTTHRIL
LINGEXPEZENCEISEV
EZBETTERTHANGETTI
NGYOURROCKSOFZWIT
HAGIRLZHEBESTPART
OFITISTHATWHENIZZ
LLBEREBORNINPARAD
ICEANDALLTHEIHAVE
KILZEDWILZBECOMEM
ZZLAVESIWILLNOTGI
VEYOUMYNAMEBECAUS
EYOUWILLTRYTOSLOZ

Let me move some stuff around now; make a totally original double symbol experiment…

Experiment 4 ZZ Wildcard n-graphs

The same message with perfect cycles. I substituted ten pairs of 63. Six of them are in words, and four of them split between words.

25 30 26 29 12 29 27 31 32 28 35 20 44 13 39 44 30
14 6 15 63 63 58 49 16 25 53 26 50 51 40 33 58 8
21 19 58 36 27 54 28 52 34 41 45 17 19 58 37 55 22
2 38 29 25 63 63 26 35 20 60 27 30 63 63 3 33 18
28 36 56 23 12 19 42 46 47 13 49 57 6 14 7 4 58
50 15 34 5 37 25 51 53 24 16 33 43 52 54 10 1 38
20 17 63 63 58 49 2 35 26 34 3 31 40 19 4 32 30
55 41 29 27 31 32 50 42 33 18 56 21 28 36 20 20 25
59 12 51 34 13 57 22 14 33 43 52 63 63 23 45 26 30
31 27 37 20 15 61 44 16 46 17 38 8 18 28 49 12 59
13 35 6 14 55 56 15 47 57 24 5 36 20 16 53 54 25
37 20 62 39 58 48 45 40 63 63 50 41 19 19 60 26 55
21 63 63 27 46 32 56 22 17 6 18 51 57 44 2 47 53
42 19 28 54 25 52 55 23 3 56 60 24 12 38 26 60 27
30 31 6 13 48 14 6 43 45 35 28 63 63 4 46 5 11
25 8 15 1 37 9 2 32 30 57 21 16 26 22 3 59 17
29 27 31 32 18 10 60 28 30 31 6 12 63 63 34 13 33
62 49 63 63 59 14 50 25 60 26 30 31 38 40 53 20 27
59 15 62 41 58 34 62 35 5 33 16 6 17 8 1 58 51
18 62 42 58 60 28 32 30 54 47 62 55 43 52 31 39 60

At 2:40 I had this, where the program thought that the Z’s were S’s again:

ILIKEKILLINGPEOPL
EBESSUREITISSOMUC
HFUNITISMOREFUNTH
ANKISSINGWILSSAME
INTHEFORRERTBECAU
SEMANISTHEMOSTDAN
GESSURANIMALOFALL
TOKILLSOMETHINGGI
VESMETHEMOSSSHRIL
LINGEXPERENCEIREV
ENBETTERTHANGETTI
NGYOUZROSSSOFFWIT
HSSIRLTHEBESTPART
OFITISTHATWHENIWI
LLBEZEBORNISSARAD
ICEANDALLTHEIHAVE
KILLEDWILLBESSMEM
YRSSVESIWILLNOTGI
VEYOUMYNAMEBECAUS
EYOUWILLTRYTOSLOW

Zodiac thinks that killing people is "more fun that kissing wils same in the forrer". But most of it is readable. Score 40079.

No change at 20 minutes. I will do some housework and return to check in a bit. Maybe I will try to lock 63 to letter Z if the program doesn’t figure it out.

EDIT: At thirty minutes I locked 63 to Z and the score was exactly the same: 40079. Then I changed the R at the end of "because" to an S and no score change. Then I stopped it and restarted it and the program changed the ZZ’s back to SS’s. Maybe a scoring issue.

If we can fine tune this process a little, I don’t see why we couldn’t merge more than one APS into one symbol.

Smokie

WOW! I am just now (after reading and pondering this since approx. 2009) if Zodiac really said in his code that all he has killed will become SAINTS rather than SLAVES in Paradise?

"Slaves in Paradise" never made any sense to me. In the Catholic faith, it is believed that everyone who dies and goes to heaven becomes a "saint" and does not have to be formally canonized. Anyone interested could Google search about this. "Saints in Paradise" makes more sense imo. There is a term "the communion of saints" that Zodiac could have heard and been referring to in his twisted thinking. Or,
Could the code read "the best part is all that I have killed will be remembered? (Now those unsolved lines come to mind, beorie, etc.)

Sorry if this is "off the wall" as these cipher threads are totally over my head and I could never make any sense at all out of that decrypto thingy. Almost didn’t even read this these posts and have skipped many of them but just happened to today and read the last one quickly for context or as you would read notes that may have been taken using "speed writing" method (usng spd rtng mthd).

capricorn: You have to read more of the thread to know what is going on. We aren’t trying to figure out what Z said in the 408, we all think that he said slaves. But we are using the text of the 408 to conduct some experiments with a computer program that solves coded messages. smokie

Mea culpa! Sorry will try to comprehend these threads and just wondering what computer would do with "saints" rather than "slaves" or the rest I posted since different words are obviously coming up with the various inputs being used. Wondering if Harden’s solution was slightly off with a few words as it might make a big difference.

What if the text of the 408 is wrong? That would affect everything, no?

Also thinking of lyrics to song that was big hit in the day "Strangers in Paradise" when reading the last "solve."

No problem capricorn!

Jarlve, are you out there? I checked your ngram files, but I will need some with letters between the wildcards. Like this:

ABCDE
ZBCDE
AZCDE
ABZDE
ABCZE
ABCDZ
ZBCDE
ZZCDE
ZBZDE
ZBCZE
ZBCDZ
AZZDE
AZCZE
AZCDZ
ABZZE
ABZDZ
ABCZZ

If I am correct about the wildcards, then they are here:

Should we just start with two Z’s maximum per ngram or have ngrams with three Z’s?

I think that there will be a lot of duplicates with different scores. ABCD and EBCF will both look the same when they are both ZBCZ. But maybe we can consolidate the list and average the scores? Just an idea. Maybe that wouldn’t work because some are going to appear in higher frequency and we wouldn’t want to lower their score.

What do you think?