Jarlve: I think that you are saying that a misalignment line caused by a skipped or added symbol shifts the bigram periods that are cut by the line, and that you are finding evidence of misalignment lines in your statistics.
Doranchak: Thanks for the info. Maybe the prime phobia phenomenon and odd even phenomenon are caused by the same encoding process, considering that all primes but position 2 are odd. Last night I wrote one short paragraph to try to express two separate ideas. The other patterns that you were showing, the last few posts yesterday, those I think may be coincidence. I am just saying that if you shuffle all those symbols or make a practice message, you are likely to find a lot of interesting patterns that are repeated once or twice.
I can’t verify this untransposition. It starts the same as 340n_b19n but is misaligned from there.
Thanks for checking. I found the difference. Here’s where we start the same way:
At that last line, I continue the pattern by wrapping around to the R on line 1, column 3, whereas you continue your sequence from line 1, column 2.
My approach (of using modulo to wrap around to the beginning again) only works when the period doesn’t share factors with 340. So I should probably use your approach.
Yesterday, the thought entered my head that perhaps these prime devoid symbols signify overlapping of 2 periods or similar (example 14, 15). That the scheme he used at one point started to overlap on already filled squares and that instead of moving to another square he *overlapped* and created some symbols especially for this. I guess that when 2 periods overlap, these positions will very likely not be prime.
That’s an interesting hypothesis. It would be interesting if test ciphers produced under this hypothesis can be shown to have the same prime-phobic quality.
Why would it appear again the same way? Is it just the normal geometrical relationship between the period 15 scheme and the period 19 scheme? Or did I mess up my untransposition?
I just realized that flipping the 340 turns the period 19 sequences into period 15 sequences. So that explains it. Duhh. 
Jarlve: I think that you are saying that a misalignment line caused by a skipped or added symbol shifts the bigram periods that are cut by the line, and that you are finding evidence of misalignment lines in your statistics.
Yes. I’m thinking that comparing the conflicting periods (15 and 29 in the 340) may help to identify what went wrong and where it went wrong. The smokie12 is also a clear example of that so maybe you could start there and then move on to the 340 (check the images in my previous post). I would really like you to do that with your spreadsheet. From the perspective of your hypothesis (smokie9 etc) but also from the interpretation that I postulated (period 15 thing).
My approach (of using modulo to wrap around to the beginning again) only works when the period doesn’t share factors with 340. So I should probably use your approach.
Yes. I found that out a while ago. But I will take another look at it soon and also at the interpretation of wrapping around the cipher in 2 dimensions. Tomorrow I’ll have and share the results from the first interpretation (the 16 untranspositions) and then I’ll work on scoring other interpretations.
I just realized that flipping the 340 turns the period 19 sequences into period 15 sequences. So that explains it. Duhh.
You are excused, it’s counterintuitive.
Jarlve, I was working on this and didn’t get a chance to read your last post.
Just two rough ideas here.
A period 2 bigram becomes a period 38 bigram or mirrored period 30 bigram after transposition. But you are also finding a high count of mirrored period 29 bigrams.
OR
A period 1 bigram becomes a period 19 bigram or mirrored 15 bigram after transposition. And some type of vertical misalignment is making them into mirrored period 29 bigrams. The example below is by analogy only. It is just a rough idea.
What I need to do at this point is update my spreadsheet. If I want to find bigrams of a certain period, I have to change the formulas, which is stupid. I need to be able to pop a period value into a cell and have the bigrams and bigram repeats automatically listed, as well as highlighting of cells automatic. Some of it already is, but all of it. Then I can do more discovery and analysis. I have been wanting to do it anyway, so this is good.
It would be interesting to know more about these mirrored period 29 bigram repeats. Where are they exactly and what symbols do they have. I think that this is very significant evidence of transposition. You’re on this thing like a bloodhound right now!
It would be interesting to know more about these mirrored period 29 bigram repeats. Where are they exactly and what symbols do they have. I think that this is very significant evidence of transposition. You’re on this thing like a bloodhound right now!
Okay, please download and examine the following images (I have already shown these):
The x-axis is the period, the y-axis is the frequency of the total bigrams repeats for that period. The different colors are the orientations of the 340 (my routine considers all that), red=normal, green=mirrored, etc.
In image A, on top it says 408 bigram 19 scheme. This is just the 408 cipher transposed into a clean period 19 scheme, you can see that I put several peak numbers on there. 1 is the main peak and is consistent with the scheme, 2 is the mirrored peak of the scheme, 3 is a "reflection" of the main peak. Because there is such a high peak at 19 there is also a peak at 38 because it simply is 19 times 2. The peak at 38 is a watered down version of the main peak at 19, a reflection so to say, because 19 is a factor.
The middle graph of image A says 408 bigram 15 scheme. This is the 408 transposed into a clean period 15 scheme (I’ve illustrated these transpositions before). This is the scheme that so far is most consistent with the 340 (bigram wise). Peak 1 at 15, peak 2 at 19, peak 3 (reflection of the main peak) at 30.
Then at the bottom of image A there’s the 340. Which can be seen to be most consistent with the bigram 15 scheme though the reflection (peak 3) of the main peak is at 29 instead of 30.
In image B, at the bottom you can see the smokie12. I don’t remember it exactly but the smokie12 is also a misaligned transposition. Interestingly enough this has also caused the relfectionary peaks of 1 and 2 (that are 3 and 4) to shift away from their expected position. Because peak 3 should be at 38 (19 times 2) but is at 37 instead and peak 4 is at 31 instead of 30. The same shift can be observed in the 340 and I believe this is evidence of the misalignment of its scheme since these shifts don’t show up in the clean tests. Now I’m wondering if comparing these peaks (15 versus 29) will get us anywhere.
See what I’m getting at. And why this is so exciting. Because the bigram period 15, 19 thing is a very, very significant observation. But we haven’t solved the 340 with that hypothesis yet. So I’ve came to wonder why. And now it seems likely that there is misalignment because of the peak shift (which can also clearly be seen in the smoke12). So I’d like to you to compare bigram period 15 repeats versus bigram period 29 repeats and find out why the repeats are peaking at 29 and not at 30 (for the 340). I think your spreadsheet approach and knack for detail are ideal for this task. Take your time, find out as much as you can. And if you can’t at least we’ll know that comparing these peaks will get us nowhere. I don’t know how to compare them also, you’ll have to come up with something, if possible.
Could it be that you need to put the two ciphers together to check for patterns mirroring each other?
O.k., Jarlve, my spreadsheet is updated and I will begin my investigations.
My count of bigram repeats for the period 15 scheme for the 340 comes out to 41 instead of 40. What count do you guys get?
Here are the individual bigram counts I’m using to produce the total: z6 (2) 2p (2) #2 (3) G+ (2) cF (2) PY (2) BO (2) TB (2) ^D (2) MF (2) *5 (2) MV (2) N: (2) +B (3) ;+ (2) +4 (3) 9^ (2) (+ (3) OF (2) )+ (2) p+ (5) Xz (2) .L (2) |p (2) YA (2) |T (2) +M (2) K< (2) <S (3) k. (2) +l (3) +k (2)
Total = sum(count(bigram(k))-1) for all k
From a few pages earlier:
Update: also added the new measurement.
Bigram repeat counts, new measurement (5-grams):
340n_b19n: 37, 630
340n_b19m: 34, 592
340n_b19f: 34, 592
340n_b19r: 37, 630
340m_b15n: 41, 666
340m_b15m: 41, 664
340m_b15f: 41, 664
340m_b15r: 41, 666
340f_b15n: 41, 660
340f_b15m: 38, 666
340f_b15f: 38, 666
340f_b15r: 41, 660
340r_b19n: 37, 640
340r_b19m: 35, 608
340r_b19f: 35, 608
340r_b19r: 37, 640
Thanks. It’s getting harder to find info in this enormous thread!
Have you ever seen this show? They have this way of storing things. They just stack stuff on piles and roughly remember where everything is and what they have in stock. It’s the same thing for me with this thread.
OK so for the mirrored 340, we have a peak of 41 repeats at period 15, and a second peak of 34 repeats at period 29. I also see that there is another peak of 33 repeats at period 110. Do you see the same count for period 110?
110 does not seem to be near a multiple of 15 or 29, so is it just a phantom? Could the period 29 peak be a phantom and not a true reflection of the period 15 peak + misalignment? Or perhaps the period 110 peak is also some evidence of this misalignment?
Here is the 340 rewritten with the period 110 scheme:
dUzZ22+zG;+ST%tkL q4p1KVN|+cHk5.DP# bMV94>^L2l@5p+f>z ^RYNENFH_G)+lfO+K #-*j5<fZ.JUY2VWG> D(E%KCO49#64(A:By *+c1pBXNFBJ2yHR2z ZG#+qLbM)+f(M.W<^ GdpZ-OUy(+BKMFBc< z:73yp.BOc2+|KJLR ^)++|+VBOUWpll3zF V-+*O<8KClM6^T17b Lpp+pBTSYc/DRk|p4 E)&t(+5/P/TFRH5+S |UOdFPjBct&Wz+z+. CMMW94)_T+(5+D*c2 |C>J-FR*klBCcOd<Y *2_-RBO8tl;7F(<^c WRBd8Fkp5F|||LN+z +;.#KVy|GSAX98c6O
My count of bigram repeats for the period 15 scheme for the 340 comes out to 41 instead of 40. What count do you guys get?
Here are the individual bigram counts I’m using to produce the total: z6 (2) 2p (2) #2 (3) G+ (2) cF (2) PY (2) BO (2) TB (2) ^D (2) MF (2) *5 (2) MV (2) N: (2) +B (3) ;+ (2) +4 (3) 9^ (2) (+ (3) OF (2) )+ (2) p+ (5) Xz (2) .L (2) |p (2) YA (2) |T (2) +M (2) K< (2) <S (3) k. (2) +l (3) +k (2)
Total = sum(count(bigram(k))-1) for all k
I count 71 bigram repeats where AB, AB and AB count as three. But I have 32 pairs of symbols that repeat. So 71 – 32 = 39. My count may be off by one or two.
I had to make my own column chart for starters. 340 is blue and smokie12 is red. I will drill into smokie12 to see if what I did caused the spike with the red arrow.
I see that smokie12 looks more uniform and the 340 looks like it has more variance. More coming soon.
EDIT: This may take a little time so that I can put an analysis together. I am a bit tired lately, so bear with me.
