Well done smokie, it is a high scoring solution that you came up with. I transcribed two 10 by 17 irregular rectangles (1 character missing at bottom-right) using columnar transposition (starting upper-right corner) into a 17 by 20 rectangle (stacked left-to-right, top-to-bottom) and added 2 random characters at the end and mirrored it.

I’m really liking this and would like to create 3 simple single rectangle route transposition ciphers with 63 symbol for you to solve if you are interested. If you are interested I would also like you to create 3 such ciphers for me. If we can solve these I’d like add some difficulties but still stick with single rectangles for a while.

Do you know the case "The Old Lady Killer"? Here is one excerpt that sticked for me:

"Then an odd coincidence distracted the investigation: at least three of ********* victims owned a print of an eighteenth-century painting by the French artist Jean-Baptiste Greuze, Boy in Red Waistcoat."

Yes I noticed that the misalignments were in the same place for both rectangles, but I thought that you skipped a plaintext in the same place for both. I locked in on period 17 at first a while ago, and tried everything up to 6 columns. It wouldn’t solve and I just assumed that you wouldn’t make the inscription rectangles bigger so I spent some time looking at mirrored period 8. Then I realized my assumption must be wrong, went back to period 17 and had it partially solved in just a few minutes. Here is a period 17 untransposition of the first 8 rows, and some of the words show through.

SIMUCHOU
SEHERFIC
KERALRER
NSACTUAA
FORSTIFL
IKEHERSF
ASBEENTH
UGHSIMUH
NDOSINEC
HESTRINI
TFEMALEG
IONESOLH
THEFACTI
TSHEHAST
BLEDINBD
SODESIFI
ELLTHANP

So a route transposition untransposed not exactly right will show bits and pieces of some of the correct words, and they won’t change from solve to solve but they will appear in different places. I think that is very important. I knew I had something when I saw bits of words "strongest females" and a misalignment with the letter "g". Two words like that used together gave context and I knew I was one the right track.

I don’t know "The Old Lady Killer" but did have fun with the three messages. I used them to update my spreadsheets. But I have been suffering from burnout at times and have other pressing concerns. I want to continue, and like your idea. How about you make me one message and I make you three for now. Give me a little time to locate some original plaintext. Thanks.

Okay smokie,

I included the excerpt from "The Old Lady Killer" case as it is a good example of how odd coincidences can sidetrack investigations and can only hope that period 15/19 is not such a coincidence.

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

Actually this is a lot more fun than trying to solve the 340 because there is someone who knows what the cipher and solution are and a rewarding experience when I get a solution. Here is one for you. I am not sure how difficult it is to solve.

Smokie39

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

Jarlve, I have been poking around in your message, making observations and taking notes. I am not yet at the point of making a plan to solve.

But I have been thinking. Pretty much all I do, after I think I know what variations a transposition has, is redraft into column count = suspected period. Then select a chunk of rows, rotate it 90 degrees, then mirror, flip, and flip and mirror the rotated chunk. Try to solve all four. If that doesn’t work, then I select another chunk and so on. If that doesn’t work, then try a different period or variation. With 1:1 substitution the method works pretty well, even with a misalignment, which is not too difficult to visually identify. I know if I have a solution within a second using your program; I really like being able to paste different sized messages into the paste window. It has improved my productivity in using the program.

I suggest that we also exchange versions of the same message with the exact same transposition matrix, but 1:1 substitution. It would be interesting to see the difference between what the program does with the correctly selected chunk 1:1 versus homophonic. I want to keep going with the message that I have for now, but letting you know that I can easily make a 1:1 version if you want.

Jarlve, I have been poking around in your message, making observations and taking notes. I am not yet at the point of making a plan to solve.

But I have been thinking. Pretty much all I do, after I think I know what variations a transposition has, is redraft into column count = suspected period. Then select a chunk of rows, rotate it 90 degrees, then mirror, flip, and flip and mirror the rotated chunk. Try to solve all four. If that doesn’t work, then I select another chunk and so on. If that doesn’t work, then try a different period or variation. With 1:1 substitution the method works pretty well, even with a misalignment, which is not too difficult to visually identify. I know if I have a solution within a second using your program; I really like being able to paste different sized messages into the paste window. It has improved my productivity in using the program.

I suggest that we also exchange versions of the same message with the exact same transposition matrix, but 1:1 substitution. It would be interesting to see the difference between what the program does with the correctly selected chunk 1:1 versus homophonic. I want to keep going with the message that I have for now, but letting you know that I can easily make a 1:1 version if you want.

Hey smokie,

If you have any other suggestions for AZdecrypt let me know. Perhaps something could be added which would further automate your workflow?

Ofcourse the reason why 1:1 substitution is easier lies in the lowered multiplicity. To offset homophonic substitution (higher multiplicity) one could move up to 6 or 7-grams, but statistically (bigrams and such) there is much less to go on. It is a good suggestion to exchange lower multiplicity versions also. But for now I’m good with smokie39 and don’t want any hints, it is an excellent cipher.

Jarlve, I have been poking around in your message, making observations and taking notes. I am not yet at the point of making a plan to solve.

But I have been thinking. Pretty much all I do, after I think I know what variations a transposition has, is redraft into column count = suspected period. Then select a chunk of rows, rotate it 90 degrees, then mirror, flip, and flip and mirror the rotated chunk. Try to solve all four. If that doesn’t work, then I select another chunk and so on. If that doesn’t work, then try a different period or variation. With 1:1 substitution the method works pretty well, even with a misalignment, which is not too difficult to visually identify. I know if I have a solution within a second using your program; I really like being able to paste different sized messages into the paste window. It has improved my productivity in using the program.

I suggest that we also exchange versions of the same message with the exact same transposition matrix, but 1:1 substitution. It would be interesting to see the difference between what the program does with the correctly selected chunk 1:1 versus homophonic. I want to keep going with the message that I have for now, but letting you know that I can easily make a 1:1 version if you want.

Hey smokie,

If you have any other suggestions for AZdecrypt let me know. Perhaps something could be added which would further automate your workflow?

Ofcourse the reason why 1:1 substitution is easier lies in the lowered multiplicity. To offset homophonic substitution (higher multiplicity) one could move up to 6 or 7-grams, but statistically (bigrams and such) there is much less to go on. It is a good suggestion to exchange lower multiplicity versions also. But for now I’m good with smokie39 and don’t want any hints, it is an excellent cipher.

I am happy with my message for now as well. I am very excited about the attack plan that I am thinking about. I am quite happy with the program. Thanks.

I was watching the discussion about merging symbols to increase the count of period bigram repeats. I am wondering if using this together with cycle scores could help to find true cycles. An experiment with test messages, where symbols are merged to find the percentage increase in bigram repeats. And then multiply the cycle score by the percentage increase. Use the new cycle scores to compare all cycles, and see if the method would help to isolate true cycles. Or something like that. Make test messages similar to 340 cycle stats. Rank them by score. Then multiply the score with percentage increase in bigram repeats and rank again. I added it to my to do list. But am hoping that the 340 will be solved this year. I would like to see it solved this year if possible.

I tried to answer the question if symbol number 7 in the smokie39 is a wildcard or a 1:1 substitute. The wildcard could disturb some cycles and on average show a greater improvement when randomized. Though in case a wildcard is collection of perfectly merged cycles (a 3-symbol cycle with 8 occurances per symbol for example) there would be no difference since no cycles are disturbed.

Thus, I developed a test which per symbol, assigns new random symbol numbers (as randomization) iterated and measured a 1000 times. The average score per symbol is then substracted by the base score. The measurement used is my 2-symbol cycles score with a modification to ignore cycles that have a symbol which only occur once.

The lists are sorted with the best improvements on top, basicly the symbols on top are most likely to be not involved in any cycles. In brackets are the symbol frequencies. My interpretation is that there is a good chance that symbol number 7 in the smokie39 is a wildcard which disturbed some cycles. Smokie, could you please confirm that this is the case or not? That I would like to know for now, but nothing else yet. I’m not very sure that this is the case but the numbers seem to suggest it.

The average improvement of the 408 is (-149.37), for the 340 it’s (-78.05) and for the smokie39 it is (-66.61). These numbers seem to fall directly in line with their 2-symbol cycle scores, as these rate the smokie39 as less cyclic then the 340, and the 408 as more cyclic than both. It can be observed that the "+" symbol in the 340 shows an improvement of (-54.51) while in the smokie39 symbol number 7 (which also occurs 24 times) shows a higher improvement (-23.29).

I’ve also done some extra tests which showed that when a symbol with 24 occurances is created by merging random symbols (thus highly disturbing cycles), positive improvements were noted, telling us that in the 340 that likely did not happen.

408.txt:

Symbols versus 2-symbol cycles weight=5:
--------------------------------------------------
Average improvement: -149.37
j: -5.29 (1)
7: -44.38 (3)
f: -46.91 (3)
_: -49.48 (8)
z: -74.59 (4)
c: -85.19 (6)
%: -96.80 (11)
J: -96.93 (6)
A: -98.62 (8)
!: -102.45 (5)
S: -105.34 (6)
=: -106.65 (7)
Q: -108.57 (5)
Y: -111.76 (10)
/: -116.43 (6)
(: -120.01 (4)
q: -120.05 (16)
X: -127.34 (9)
d: -128.75 (6)
#: -132.18 (10)
8: -137.64 (8)
F: -140.69 (6)
l: -141.22 (5)
R: -151.46 (12)
B: -152.52 (12)
r: -154.17 (7)
K: -158.92 (5)
: -162.32 (5)
e: -164.67 (10)
V: -166.04 (9)
L: -166.27 (8)
N: -168.14 (6)
G: -170.58 (7)
6: -176.24 (8)
I: -176.81 (11)
@: -180.51 (6)
E: -181.97 (9)
T: -185.09 (7)
p: -185.74 (6)
9: -186.75 (14)
k: -190.25 (9)
^: -192.58 (6)
+: -193.48 (8)
M: -195.27 (8)
H: -195.45 (8)
W: -200.45 (9)
): -201.90 (8)
5: -207.47 (8)
P: -207.52 (11)
D: -208.47 (6)
t: -211.33 (7)
Z: -220.44 (8)
U: -220.98 (10)
O: -234.99 (7)


340:

Symbols versus 2-symbol cycles weight=5:
--------------------------------------------------
Average improvement: -78.05
@: -4.26 (1)
W: -26.33 (6)
): -28.97 (5)
A: -36.61 (2)
j: -41.39 (2)
%: -42.17 (2)
b: -47.58 (3)
H: -48.73 (4)
4: -50.49 (6)
1: -52.62 (3)
:: -53.50 (2)
F: -53.91 (10)
+: -54.51 (24)
5: -54.98 (7)
B: -55.27 (12)
3: -56.88 (2)
T: -57.15 (5)
&: -57.54 (2)
6: -57.72 (3)
C: -58.26 (5)
X: -60.36 (2)
p: -64.15 (11)
k: -64.17 (5)
E: -64.20 (3)
q: -66.77 (2)
L: -66.97 (6)
P: -67.41 (3)
#: -71.01 (5)
;: -74.70 (3)
J: -75.66 (4)
y: -75.80 (5)
S: -77.59 (4)
8: -80.38 (4)
d: -81.16 (5)
.: -82.60 (6)
N: -83.18 (5)
-: -86.51 (5)
G: -87.62 (6)
_: -88.32 (3)
c: -89.08 (10)
U: -90.99 (5)
/: -91.46 (3)
<: -92.19 (6)
9: -92.46 (4)
2: -93.98 (9)
V: -97.93 (6)
O: -99.14 (10)
7: -99.80 (3)
(: -99.87 (7)
R: -100.65 (8)
|: -104.44 (10)
f: -105.88 (4)
z: -106.23 (9)
>: -107.83 (4)
Y: -109.26 (4)
t: -112.72 (4)
^: -116.46 (6)
Z: -118.54 (4)
D: -119.42 (4)
*: -123.28 (6)
l: -123.57 (7)
K: -130.57 (7)
M: -133.86 (7)


smokie39:

Symbols versus 2-symbol cycles weight=5:
--------------------------------------------------
Average improvement: -66.61
46: -5.93 (1)
22: -9.70 (2)
63: -19.16 (4)
7: -23.29 (24)
10: -33.81 (3)
31: -35.04 (2)
29: -36.95 (2)
36: -37.07 (5)
61: -37.28 (5)
50: -38.91 (3)
12: -39.38 (6)
1: -39.84 (5)
62: -40.14 (3)
53: -41.33 (5)
41: -41.40 (8)
54: -42.70 (4)
51: -43.11 (2)
47: -43.34 (2)
49: -45.27 (5)
45: -46.87 (2)
58: -47.55 (2)
60: -51.95 (4)
43: -53.87 (5)
52: -54.24 (3)
33: -56.04 (4)
17: -56.38 (4)
15: -56.38 (10)
11: -57.01 (5)
3: -57.20 (4)
32: -60.57 (7)
44: -62.49 (6)
57: -62.80 (4)
25: -63.27 (10)
4: -65.02 (6)
37: -65.83 (8)
59: -66.33 (3)
26: -66.67 (3)
2: -69.57 (8)
38: -70.35 (8)
48: -71.01 (3)
20: -71.69 (11)
42: -78.23 (6)
13: -79.73 (5)
55: -80.33 (5)
40: -81.22 (7)
9: -85.04 (4)
56: -85.79 (5)
6: -86.08 (7)
27: -86.60 (9)
14: -88.78 (10)
39: -94.76 (4)
34: -104.22 (5)
30: -105.23 (7)
18: -107.00 (4)
28: -108.03 (7)
35: -109.01 (6)
23: -109.38 (6)
5: -114.30 (5)
19: -115.67 (5)
16: -116.11 (6)
21: -118.55 (6)
8: -126.56 (5)
24: -129.30 (5)

I used the new test to look at rows and columns.

Row wise, the average improvements are, for the 408 (-210.93), for the 340 (-104.88) and for the smokie39 (-75.86). Row 14 in the 340 scores high with row 12 as follow up. In the 408, row 24 and 23 are on top. Row 20 may contain filler in the smokie39.

Column wise, the average improvements are, for the 408 (-406.88), for the 340 (-177.66) and for the smokie39 (-130.96). Column 10 in the smokie39 scores high and will need further investigation. In the 340 all columns yield negative scores with column 13 and 8 slightly on top.

This test reconfirms that in the 340, the rows and columns at which the pivots intersect (row 12 and 14 and column 8 and 13) are the ones that most likely could be considered as filler/nulls or disturbances of encoding per the measurement. Especially row 14 is highly suspicious.

408:

Rows versus 2-symbol cycles weight=5:
--------------------------------------------------
Average improvement: -210.93
24: 14.54
23: -33.10
20: -36.59
22: -57.48
1: -68.76
18: -90.55
19: -114.57
21: -130.43
14: -140.32
17: -162.14
8: -187.82
13: -212.22
15: -217.78
7: -219.94
9: -244.92
11: -264.02
2: -284.34
10: -328.40
16: -331.68
3: -345.52
6: -346.59
12: -364.32
4: -370.04
5: -525.30

Columns versus 2-symbol cycles weight=5:
--------------------------------------------------
Average improvement: -406.88
4: -204.19
9: -211.85
6: -312.76
17: -338.58
16: -364.67
5: -375.02
12: -395.47
15: -396.56
7: -414.40
8: -421.60
2: -439.44
11: -457.81
14: -489.02
13: -490.69
3: -511.94
1: -523.58
10: -569.42


340:

Rows versus 2-symbol cycles weight=5:
--------------------------------------------------
Average improvement: -104.88
14: 62.12
12: -10.64
10: -20.54
18: -20.97
13: -56.16
4: -57.63
17: -62.05
11: -62.37
6: -80.00
20: -103.30
8: -119.32
19: -130.33
9: -140.79
1: -143.13
16: -157.25
7: -162.19
2: -178.99
15: -205.53
3: -206.51
5: -242.10

Columns versus 2-symbol cycles weight=5:
--------------------------------------------------
Average improvement: -177.66
13: -92.80
8: -98.22
2: -98.60
17: -107.32
5: -142.18
3: -143.45
9: -146.50
12: -150.60
1: -164.41
11: -182.28
10: -191.21
4: -212.23
16: -217.78
15: -235.31
7: -260.38
14: -278.54
6: -298.33


smokie39:

Rows versus 2-symbol cycles weight=5:
--------------------------------------------------
Average improvement: -75.86
20: 29.00
13: 15.97
17: 10.79
9: 10.14
10: -1.70
12: -4.48
18: -5.83
1: -11.92
19: -14.93
4: -74.04
14: -82.50
2: -88.58
11: -108.64
15: -123.48
7: -128.63
16: -140.68
6: -171.28
8: -173.89
3: -204.71
5: -247.87

Columns versus 2-symbol cycles weight=5:
--------------------------------------------------
Average improvement: -130.96
10: 82.83
15: -0.77
11: -11.07
13: -78.47
17: -82.63
7: -106.01
3: -132.79
4: -137.07
12: -140.93
6: -151.48
9: -160.44
5: -171.79
1: -178.61
16: -184.46
14: -203.53
8: -269.47
2: -299.62

Symbol 7 maps to three plaintext. I should give you smokie39, the exact same message, but with symbol 7 mapping to only one plaintext. I will wait until you say it is o.k., but smokie39 is very difficult with symbol 7 as a polyphone.

EDIT: Smokie39 is difficult even with no polyphones.

Okay, thanks for the feedback smokie. That’s what I like about the smokie39, it is hard but I am in the unique position to ask questions about observations that I feel confident about.

Smokie,

Could you also create a route transposition for me in a numbered plaintext format? That is transposition without homophonic substitution, but with numbers as substitution. The idea is that you start simple and every time I solve it you create a new one that is slightly more difficult. So that eventually we have a series of transposition ciphers with varying difficulties.

A few notes on the smokie39. No questions at the moment.

– While going through more tests to identify nulls I spotted something familiar, a potential filler segment at position 218. In the 340 these segments are strongly connected to the pivots. After going through the list of top segments I found 2 pivots visually. One at (row 10, column 6) and another one at (row 13, column 12), and a smaller one at (row 12 column 4). There may be more. For now, I’m assuming that smokie put these in somehow and I need to get a pivot measurement going.

– I also found strange diagonal repeats around the second pivot involving symbol numbers 25 and 49.

– There’s a strong possibility that column 10 (or column 8 after reversing the cipher) is a null and possibly there is also nulls at the end of the cipher (from position 334 or so). While column 10 may be a null, removing it destroys the second pivot.

Smokie,

Could you also create a route transposition for me in a numbered plaintext format? That is transposition without homophonic substitution, but with numbers as substitution. The idea is that you start simple and every time I solve it you create a new one that is slightly more difficult. So that eventually we have a series of transposition ciphers with varying difficulties.

EDIT: Yes, I agree that we should work our way up from easy to difficult and we could work our way up from 1:1 to 63 symbols. I apologize for the delay; somehow I missed the post until now. I will have a message for you shortly.

EDIT: Here is smokie40, not the same as smokie39. I can make these very easily.

1 2 3 4 5 4 4 5 1 5 6 7 8 1 4 9 1
10 11 1 8 12 6 12 7 9 13 13 14 4 7 15 8 7
15 16 17 6 15 1 18 14 4 19 12 5 1 16 6 6 4
17 11 3 1 19 8 8 3 5 13 4 12 13 5 17 7 19
5 20 6 17 8 5 5 4 15 7 4 1 1 19 9 18 4
4 15 5 13 8 5 14 13 12 13 1 6 1 17 1 20 12
4 6 17 1 6 5 15 3 13 8 3 13 20 19 12 15 12
1 1 6 4 15 8 15 7 1 3 20 3 12 1 19 4 1
4 4 17 4 12 11 1 7 17 10 5 8 17 2 3 1 20
12 13 9 12 15 13 3 9 1 3 5 1 18 13 1 3 17
15 8 17 1 4 4 6 4 17 1 6 6 6 8 1 7 1
17 11 4 15 13 7 21 12 4 5 8 19 18 12 14 8 13
12 19 10 12 16 8 19 19 17 7 17 4 14 15 17 13 1
4 3 6 1 4 19 13 4 7 4 1 15 20 15 16 6 18
11 1 4 12 11 6 5 12 6 8 5 12 1 1 13 17 6
5 12 15 13 7 3 1 3 4 13 13 13 4 1 7 20 20
12 12 6 3 13 8 18 1 3 15 1 19 13 19 5 13 11
7 1 1 3 6 3 4 14 8 4 1 14 20 1 12 12 6
11 13 19 6 1 6 10 5 15 13 8 21 3 12 15 12 13
8 12 6 4 1 7 1 6 5 12 15 4 1 10 20 6 16

EDIT: smokie40 is 340 plaintext; incomplete inscription rectangle 16 x 22 with only four plaintext in the 22nd row LRTB. Transcription RLBT, but row 3 is mirrored and transcribed RL.