Thanks for getting back to me. Smokie12 will have two skipped plaintext with an original message of 323. I will make smokie13 with an unknown message length if you want, with one skipped plaintext.

We can call them period 19 bigrams if you want, but I am taking my definition from practicalcryptography.com: http://practicalcryptography.com/crypta … id-cipher/. It is definitely a source of confusion, but I am glad that we both know what we are talking about.

I have a question for you. I count 25 bigrams repeats for the 340 in normal orientation and 29 when mirrored. 37 period 19 bigrams normal orientation and 39 period 15 bigrams. Should we consider the stronger response from the mirrored version?

That is a very good question. I am going to have to look into that further and really think it over. The relative position of B to A in a period 15 bigram is the mirror image of the relative position of B to A in a period 19 bigram. That is really weird. No doubt others have discussed this elsewhere. I would like to look at that a bit closer and mull it over. Maybe in the near future try to think of a possible method of encryption that may explain both.

I will have smokie12 done tonight.

The relative position of B to A in a period 15 bigram is the mirror image of the relative position of B to A in a period 19 bigram. That is really weird. No doubt others have discussed this elsewhere.

I don’t think I’ve heard of this relationship between repeating bigrams of different periods. Is there a description / illustration somewhere?

Maybe I don’t know what a mirror image period 15 bigram is. Not sure. Isn’t that the same as a normal period 19 bigram?

Here is smokie12. Same transposition map. Very similar key and randomization of symbol selection. Different plaintext. Two plaintext skipped during transposition.

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

About bigram repeats,

In a 17 by 20 grid, a cipher with a peak at period 19 will also (very usually) peak at period 15 in it’s mirrored version. It’s the same only that it is ready from right to left starting at a different offset. The highest peak is usually the correct orientation, and with the 340 the mirrored version of period 1 is also higher. That’s why I brought it up, if we want to do some work on this observation, then what orientations of the 340 should we consider.

smokie I will work on solving your latest cipher. Thanks for constructing it.

Update: some of the message came through (plaintext p2) when just considering 1 skip in the routine. Although the scores are very low, the top 10-20 results had the same structure. Nothing like that with the 340 and its scores were even lower. But I’ve explored only 1 orientation and assumed a length of 323 (17 by 19). I’ll also consider 2 skips and will get back with the results tomorrow.

177

Score: 20770 Ioc: 668 M: 195 C: 323 S: 63

mplyremarkahingso
hasnoldreadbleint
nkistomelymicethi
onthewaytamuchout
rabbitsaytosalthe
ohdearofdearsspdn
dbetoodatedhishal
thoughtssomeeneme
rwaresitocculante
amerthastheorredt
omamedonepleughts
isbutitthetreatth
ldeepmeequitmessa
lidbutwhenthenasu
stactuallysoprabb
schousonitewakidi
aitlocketindirstc
easitinethenlooke
rthelpdisnotmurri

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O.k., Jarlve. I will have to take a look at the actual period 15 bigram repeats highlighted on my spreadsheet, and compare them to the actual period 19 bigram repeats highlighted on my spreadsheet. I need to see them so that I can give a constructive answer to your question. It may take a little time. In the meantime, I am making misalignment lines, scrolling the portions of the rows to the left of the line, and comparing solve scores. Trying to see if I can find a way to detect a misalignment line. I wonder if there is another way. Let me know if you want something other than one symbol skipped with unknown message length for smokie13, which I will also start on soon.

had a thought for z in a simplistic approach to confuse the decipherment could he have started with a blank page and then filled in column one with junk. not sure if that’s been tested yet. so simply run it through without the first column in place. If its been thought of before my apologies I will keep thinking of ways he has simply confused the process.

Succesfully recovered the plaintext of smokie12 when considering 2 skips.

100,177

Score: 23194 Ioc: 648 M: 198 C: 323 S: 64

therewassothingso
veryremarkableint
hannordidalicethi
nkingoverymuchout
ofthewaytomparthe
rabbitsaytoinself
ohdearofdearishal
lbetoolatewhensme
thoughtinoverafte
rwardsitoccurredt
omerthangheoughtn
omavewonderedatth
isbutatthetimeina
llseemedquitesanu
ralbutwhentherabb
itactuallynookawa
nchounofitswaistc
oatlocketandlooke
danitandthenmurri

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No such returns for the 340, very low scores and top solves did not share a similar structure.

24,187

Score: 20792 Ioc: 623 M: 201 C: 323 S: 65

brcdicreasaddrest
henyotesofmytheig
hingourewindofshe
ctinasebudullmeun
themystordingproc
liventthertoscalr
dsareofgmehaclean
dsoomlotssahasham
ehewaiianyoudsheg
reesheantstobeind
ayitsheisriscount
husesancleiohourd
ieinthsycouventfi
nedpieciallandtoi
ngbutrierisalecia
llyidalolenorscho
icroflunitselucid
ftshasbltonewhats
willandowbutfinac

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O.k., Jarlve. That was some very good information. Tell me what you want next, if you want to stay with the plan of making smokie13 with unknown message length and only one plaintext skipped, or anything else that you see a need for based on your experience decoding smokie 9, 11 and 12.

With the scheme you originally suggested you can do the transposition part in 4 different ways:

1: start in the upper-left corner and move downwards,
2: start in the upper-right corner and move downwards,
3: start in the bottom-left corner and move upwards,
4: start in the bottom-right corner and move upwards.

For smokie13 I’d like you to make a cipher that has:

– An unknown transposition (1, 2, 3, 4).
– An unknown message length (from 313 to 333).
– During transposition, one letter/symbol added and one letter/symbol skipped, both at an unknown position and not on the same transposition line.

Then I’ll make a routine that considers everything mentioned and if the message recovers I will run the same on the 340 and get back with the results. If that sounds about right to you then go ahead and make smokie13. Thanks.

EDIT: smokie12 info coming here soon:

Here is the smokie12 tracing map.

That sounds good. Will begin soon.

Here is smokie13 pursuant to your instructions. The message is between 313 and 333. One plaintext is skipped, and one added.

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

Thanks for making the cipher to my specifications smokie!

The routine is coded and is being tested on the smokie11. If successful I’ll move on to the smokie13 and 340 with more depth. It will be the largest test I’ve ever run with nearly 4.000.000 variatons per cipher. Since it will take weeks to finish I’ll get back to the forum with any interesting developments (see update).

Update: smokie11 solved successfuly many times over (100+ scoring over 22k) with 1 iteration per variation, because many of these variations only differ slightly. So only about a tenth of the processing time I originally considered is needed, give or take about 24 hours.

Jarlve, I think that what you are doing is a worthwhile effort.

I just ran a short test of 1 million randomizations and count 37 appears just 1 time. That’s one in a million! This can no longer be ignored.

Let me know if you need anything.

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?