I made a new measurement. It compares n-grams in the cipher and when a symbol matches it is counted, to give an example. N-gram size 5: "A1B1C" vs "A2B2C", 3 symbols match so the score for this comparison is 6, 3*(3-1). All n-grams are compared and the scores are summed.

I am trying to replicate this measurement but having trouble matching your numbers. In your example (3*(3-1)), the 3 is because 3 symbols match, but what does 3-1 signify?

Hopefully the numbers will add up:

If I did not skip the symbols at positions 80 and 100 there would have been:

34 period 37 bigram repeats; and
33 period 38 bigram repeats.

That is a difference of 1.

After I skipped the symbols at positions 80 and 100, there were:

47 period 37 bigram repeats; and
21 period 38 bigram repeats.

That is a difference of 26.

26 – 1 = 25 = total change in number of period 38 bigram repeats to period 37 bigram repeats.

4 count period 37 bigram repeats in the distorted areas, which would have been period 38 but for the distortion
+
30 count period 37 bigram repeats where where a true period 37 bigram in the distorted areas matched false period 37 bigrams outside of the distorted areas
=
34 count period 37 bigram repeats "created" by skipping the symbols at positions 80 and 110.
–
10 count period 38 bigram repeats that were destroyed by skipping the two symbols
=
24.

Somewhere I missed one pair of period 37 or period 38 bigrams with matching symbols. Or maybe it was the 49-23 period 37/38 bigram repeat with that mapped to VR because there were three total. Not sure.

Answer: Yes, skipping two symbols caused a big change in counts of period 38 to period 37 bigram repeats. But many of them were true period 37 bigrams in the distorted areas that matched up with false period 37 bigrams in the undistorted areas, which inflated the numbers. It would probably be very indefinite to try to use this type of analysis to detect period 37, 38 or 39 changes caused by skipping or adding a plaintext during transposition in the 340. However, it may be somewhat more feasible to make similar detections of period 18, 19 or 20 changes caused by skipping or adding a plaintext during transposition because the count of period 19 bigram repeats is much larger and the areas distorted are smaller. Multiple skipped or added symbols will make detection more difficult, and if the distorted areas overlap, then detection could be very difficult.

I will work on this subject some more in the near future.

EDIT: Here are the smokie12 period 1 bigram repeats with count of 69 after the message is untransposed and the missing symbols replaced. The original message as posted in its transposed form and missing the two symbols had 67 count period 19 bigram repeats:

EDIT 2: This exercise was my first try, and it is giving me a lot of ideas about how to detect what Zodiac did with his transposition scheme. I didn’t analyse period 18/19 bigrams. But let’s say for instance a 17×19 transposed message had an added plaintext causing a distortion and more period 20 and period 39 bigram repeats. I wonder if there is a way to locate a few rows in the message where a higher count of period 20 bigrams match period 19 bigrams in the other rows, AND a higher count of period 39 bigrams match period 38 bigrams in the other rows. And I wonder if the diagonals of period 19 bigrams would also be mainly interrupted at those rows. Maybe we can develop some other tools to help us decide whether there are skipped or added symbols, in case there are multiples. I would like to stay with transposition until we feel as confident as possible about the scheme.

.

.

@doranchak,

I made this rather quickly without really thinking about it, if you spot any room for improvement let me know. I fired it and up and it seemed to work fine.

dim as integer dis=wdt_operand(slot,1) 'n-gram size
for i=1 to p-(dis-1) 'p=cipher length
     for j=i+1 to p-(dis-1)
     c=0
     for k=0 to dis-1
          if a1(i+k)=a1(j+k) then c+=1
     next k
     b+=(c*(c-1))
     next j
next i
wdt_score(slot,id,od,s1,s2)=b 'total score

@smokie,

That looks very interesting, thanks allot for your analysis of my question. I would to add that the observation is minor for the smokie12 but stronger for the mirrored 340 (period 15 versus 29). So what basicly is happening is that some period 30, 38 bigrams are destroyed by the misalignment and some are created. Causing a shift.

EDIT: Oddly the number of period 19 bigram repeats stayed exactly the same and the number of period 18 bigram repeats didn’t change that much. There must have been an offset.

Yes, that is my primary observation. Period 15, 19 are unchanged but period 30, 38 shifts away.

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.

I think you are right by the way, that what I’ve been calling a reflection of 15 and 19 is actually a period 2 bigram in the untransposed cipher translated to 30, 38. That’s interesting because that could mean that the 340 had a strong period 2 bigram count just like the 408.

408 (340 characters) bigrams:
Period 1: 46; Period 2: 38 (still high); Period 3: 28; Period 4: 19.

Blue is smokie12 without skipping the symbols at positions 80 and 110. Red is smokie12 with skipping the symbols at positions 80 and 110. The x-axis is the period, and the y-axis is the repeat count. The distortions made subtle changes in a lot of areas around the lines that I drew. You can see that there were a lot of changes around the period 36 to period 41 range. The number of period 38 bigram repeats was somewhat reduced.

Yes, it’s very interesting to see the differences in the 36 to 41 range. You can see how strong bigram analysis is for homophonic substitution ciphers with not too many symbols.

I would like to stay with transposition until we feel as confident as possible about the scheme.

I didn’t think of period 30, 38 bigrams as untransposed period 2 bigrams. It adds to the evidence that the 340 is a transposition cipher in our current direction.

.
I think that the potential for a big shift in period 37, 38 and 39 bigram repeat counts is caused by all of the changes that happen when a symbol is skipped or added because of the width of the distorted areas. A lot of stuff changes when the distorted area is four rows wide. But with period 18, 19 and 20 bigram repeat counts, the distorted area is only two rows wide.

Two posts above shows smokie12 bigram repeats when the message is untransposed and repaired. The bigram repeats cluster together because those are words or sets of words that contain high English language frequency bigrams and/ or there aren’t enough symbols in the key mapping to the plaintext to diffuse their frequency. When the message is transposed, they show up in diagonal rows. If they pass through a two row wide distorted area caused by the skipping or adding of a plaintext during transposition, you should be able to see an offset.

EDIT: Here is one such clustering of period 1 bigram repeats in untransposed and repaired smokie12:

Here is the distortion map for period 18 and 19 bigram repeats that would be caused by skipping two symbols:

The clustering of period 1 bigram repeats becomes a diagonal row when the message is transposed. Below is smokie12 transposed and with the symbols skipped at positions 80 and 110:

The diagonal row is offset when it passes through the distorted area. I didn’t do a thorough examination of other clusterings because I have to get ready for work. There may be more.
.

@doranchak,

I made this rather quickly without really thinking about it, if you spot any room for improvement let me know. I fired it and up and it seemed to work fine.

dim as integer dis=wdt_operand(slot,1) 'n-gram size
for i=1 to p-(dis-1) 'p=cipher length
     for j=i+1 to p-(dis-1)
     c=0
     for k=0 to dis-1
          if a1(i+k)=a1(j+k) then c+=1
     next k
     b+=(c*(c-1))
     next j
next i
wdt_score(slot,id,od,s1,s2)=b 'total score

Thank you; that helps a lot. My numbers are closer to yours now but still off by a small amount for some reason (I’m ignoring this for now).

One thing I noticed is j starts at i+1, which means the compared ngrams overlap with each other. So the length 6 string AAAAAA would contribute to the score twice when counting for 5-grams.

Also, I was curious if the approach was over-counting bigrams so I modified it to avoid adding to your b variable unless c > 2. The result is that the peaks no longer occur at periods 19 and 15. Even so, period 19 still appears with the 7th best score for the unmodified 340, and period 15 appears with 8th best score for the mirrored 340.

http://zodiackillerciphers.com

 

Jarlve, see two posts above.

Here are the 340 period 19 bigram repeats. An examination of the diagonal rows may shed clues for where any skipped or added symbols are. Perhaps also the transposition scheme or shape of the original message. For example, if it was a wraparound transposition, you may see diagonal rows of period 19 bigram clusters continue from the bottom of the message to the top of the message.

.

I think that the potential for a big shift in period 37, 38 and 39 bigram repeat counts is caused by all of the changes that happen when a symbol is skipped or added because of the width of the distorted areas. A lot of stuff changes when the distorted area is four rows wide. But with period 18, 19 and 20 bigram repeat counts, the distorted area is only two rows wide.

Excellent explanation, I now understand the 15/29, 19/37 phenomenon.

One thing I noticed is j starts at i+1, which means the compared ngrams overlap with each other. So the length 6 string AAAAAA would contribute to the score twice when counting for 5-grams.

Also, I was curious if the approach was over-counting bigrams so I modified it to avoid adding to your b variable unless c > 2. The result is that the peaks no longer occur at periods 19 and 15. Even so, period 19 still appears with the 7th best score for the unmodified 340, and period 15 appears with 8th best score for the mirrored 340.

I think the j=i+1 is okay, as your trying to get as much information as possible. The c>2 modification is cool, though it seems like using higher "c", "c>1,c>2,c>3" makes the measurement unreliable. Because you are shifting away from measuring bigrams into trigrams and larger, which typically occur only a few times and thus are much more prone to outliers. So the whole measurement becomes more prone to outliers.

I’m currently running a test with randomizations of the 340 and counting bigrams at period 15 and 29.

Out of 26.000.000 randomizations only 16 have scored 41 @ 15 or more (highest 42). Out of these 16 none scored 34 @ 29 or more.

I did some testing on 10 plaintexts each 340 characters (p1 to p10).

Averages:

2-gram repeats: 185
3-grams: 73
4-grams: 33
5-grams: 18

This would suggest a ratio close to powers of 2.

But after encoding (63 symbols, cyclic):

2-grams: 25
3-grams: 2
4-grams: 0.2
5-grams: 0.2

The ratio changes and is (highly) subject to the quality of the encoding, in general a higher quality encoding (flatter distribution of symbols) will produce less repeating fragments. Though it’s interesting (or remarkebly?) to see that with high quality encoding 1 cipher produced 2 * 4-grams and 2 * 5-grams.

This would certainly suggest that my original n-gram approach is not biased enough towards longer repeating fragments. I’ve changed my code to reflect this, it’s probably still not exponential enough but that can be changed by increasing operand3.

operand1=5 'n-gram size
operand2=1 'if c>operand2 then...
operand3=2 'increasing this number will improve bias towards longer repeating fragments

for i=1 to cipher_length-(operand1-1)
     for j=i+1 to cipher_length-(operand1-1)
          for k=0 to operand1-1
               if a1(i+k).c=a1(j+k).c then c+=1
          next k
          if c>operand2 then b+=(operand3^(c-2))
          c=0
     next j
next i
score=b

.
Here is another such cluster of period 1 bigrams that become a distorted diagonal line after transposition and passing through the distortion area.

That was the only other one that I could find. But I am looking at the shape of the distortion area and how the missing plaintext should be at the inside of the distortion area elbow. I suggest that with the 340 any period 18 CD configuration with period 19 AB and BC configurations above or period 19 DE and EF configurations below could be evidence of a cluster of bigram repeats passing through a distortion area caused by a skipped symbol. Although there are a lot of period 19 bigram repeats in the 340 to look at, there may be only a few such configurations.

I want to make a distortion map for an added plaintext.
.

I’m glad to see when other areas of the Zodiac discussion slow down here, you guys carry on your cipher work and research. Keep up the good work!

Thanks, morf. We are going to get it.

.

I’m currently running a test with randomizations of the 340 and counting bigrams at period 15 and 29. Out of 26.000.000 randomizations only 16 have scored 41 @ 15 or more (highest 42). Out of these 16 none scored 34 @ 29 or more.

Jarlve, I believe that your mirrored period 29 is my normal period 39. Here are the period 39 bigram repeats with total count of 54. That does seem a bit high, and they appear to be concentrated at the bottom half of the message. I may overhaul my spreadsheets to show period x bigram repeats by half (currently only geared to period 19 bigram repeats in the first half versus the second half), but am a bit tired tonight. Can you easily find out how many period 39 bigram repeats there are in only the first half versus the second half? If not, I may be able to do it on Wednesday and probably will anyway.

I wonder if adding plaintext during transposition caused the high number of period 39 bigram repeats, or if something else is going on. If this is statistically significant or easily replicated with some type of transposition and encoding scheme.
.

Can you easily find out how many period 39 bigram repeats there are in only the first half versus the second half?

It may have something to do with the period itself. These are counts by row for p29, note that 1 bigram repeat maps to 4 squares.

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

rows 1-10: 38
rows 11-20: 98

I’m glad to see when other areas of the Zodiac discussion slow down here, you guys carry on your cipher work and research. Keep up the good work!

Thanks, join us! I think that solving the 340 will be a key to his other puzzles, and so on, his name. Case solved.

Here is another such cluster of period 1 bigrams that become a distorted diagonal line after transposition and passing through the distortion area.

Yes, you are right. As you say, perhaps these diagonal lines can then help identify the misalignment. Very interesting!

To give an update on my work, I’m still busy (and will be for quite a while) making somewhat of a list of possible schemes relating to the period 15, 19 thing. For every scheme I generate some basic strings and score them. It is my hope that perhaps hitting on the right scheme may produce a small bump in the scores and if not outright solve it, but hey, it’s the 340…

I also concluded yesterdays test. In 100.000.000 randomizations of the 340 only 55 had a bigram repeat count of 41 or more @ period 15. Out of these 55, none had a secondary peak of 34 or more @ period 29. As for the significance of the 15, 29 discrepancy, I think it’s quite big because I haven’t been able to reproduce it will a clean transposition scheme. In a clean transposition scheme it should only happen if the period by itself is an outlier.

.
K. I have been thinking about an idea. Wondering if possibly the first row was not transposed instead of the last row. As if he set up the first row to do columnar transposition where the first row was a key, and just never rearranged the columns. Instead he transposed the message causing period 19 bigram repeats.

I am thinking about an experiment to try to replicate the high number of period 39 bigram repeats.

I think that high count symbols have something to do with it. If you have A1B1C that are all part of the same word, and transpose, you get A19B19C and A38C. If you add a plaintext between B and C, then you get A19B20C and A39C where A, B and C are still all part of the same word. However, you could get a matching A39C where there isn’t an added plaintext between A and C and A and C were never part of the same word. If A or C or both are high count, then the probability of this happening is higher than if they are low count. The result is two fewer A38C and two more A39C. A handful of these would skew the statistics. And the 340 has several high count symbols that make up a large percentage of the message.

Here are the period 39 bigram repeats. My symbols 19 and 20, which are the + and the B, are heavily represented:

.

I just found a really, really interesting interpretation with 42 bigrams and 4 trigrams. That also scored (the bump I was looking for?) 250 points higher (18111) than anything in my current list with the Reddit n-grams. I will work on scoring it will all my n-gram sets.

Remember the card Zodiac sent to Paul Averly? viewtopic.php?f=63&t=55

Two times a reference to 14 is made. Take the 340, mirror it. Then cast it in a 14 * 25 grid and apply the following untransposition.

Which gives:

>OpW%l+(D^k+)W
V9/RLYPFMSK#.k
2l+p2z<|NUOzpB
H*1AYc-t7yJKL-
6zX*j^:NfTld4+
Gddlcp)GtU<K@V
C|8M+H25EVM(L.
k5*+BEdBT|++G9
zFFVU(Rtc4D^b2
#L>P3Z#l+(MYJ+
J5|2+pG1G+;.B+
Zf+pD&OK*Fz8+p
ORjK8(4+(:N6R&
bp2#qR_TFO4^92
BT7FO%^C/lp9fS
<FM<B+;Fc(W^C5
ycUKFc_Nk+)B.E
2#l+OBypd+p|f>
4+RR-ykW)R)MVb
NJ/zS<WcLqU.||
5z7CT|GZc5*Ztz
+c25VFOBWL.y-4
8_c13B+AYP6zX|
OOCBKBS<>;*HM+
-D|H

Do we have here a strange coincidence or are we one step closer to solving the 340?

Update: here are some of the basic variations on the scheme. Using either the mirrored or flipped 340 as a starting point. I’ll start running this through the solving modules.

results_sub_directory=340_test4

cipher_information=340m_14by25_ut13n
>pOl%W^D(+VW)+
kPYLR/9k.#KSMF
|<z2p+l21*HBpz
OUNLKJy7t-cYAT
fN:^j*Xz6-G)pc
lddG+4dl2H+M8|
CV@K<UtdEB+*5k
.L(MVE5R(UVFFz
9G++|TB#Z3P>L#
2b^D4ctGp+2|5J
+JYM(+lO&Dp+fZ
+B.;+G14(8KjRO
p+8zF*K_Rq#2pb
&R6N:(+^%OF7TB
29^4OFT;+B<MF<
Sf9pl/C_cFKUcy
5C^W(cFyBO+l#2
E.B)+kNy-RR+4>
f|p+dpz/JNbVM)
R)Wk5||.UqLcW<
S*5cZG|TC7zFV5
2c+ztZ4-y.LWBO
+B31c_8Xz6PYAB
COO|<SBKH*;>-+
M|DH

cipher_information=340m_14by25_ut13m
+)WV+(D^W%lOp>
FMSK#.k9/RLYPk
zpBH*12l+p2z<|
TAYc-t7yJKLNUO
cp)G-6zX*j^:Nf
|8M+H2ld4+Gddl
k5*+BEdtU<K@VC
zFFVU(R5EVM(L.
#L>P3Z#BT|++G9
J5|2+pGtc4D^b2
Zf+pD&Ol+(MYJ+
ORjK8(41G+;.B+
bp2#qR_K*Fz8+p
BT7FO%^+(:N6R&
<FM<B+;TFO4^92
ycUKFc_C/lp9fS
2#l+OByFc(W^C5
>4+RR-yNk+)B.E
)MVbNJ/zpd+p|f
<WcLqU.||5kW)R
5VFz7CT|GZc5*S
OBWL.y-4Ztz+c2
BAYP6zX8_c13B+
+->;*HKBS<|OOC
HD|M

cipher_information=340m_14by25_ut14n
>OpW%l+(D^k+)W
V9/RLYPFMSK#.k
2l+p2z<|NUOzpB
H*1AYc-t7yJKL-
6zX*j^:NfTld4+
Gddlcp)GtU<K@V
C|8M+H25EVM(L.
k5*+BEdBT|++G9
zFFVU(Rtc4D^b2
#L>P3Z#l+(MYJ+
J5|2+pG1G+;.B+
Zf+pD&OK*Fz8+p
ORjK8(4+(:N6R&
bp2#qR_TFO4^92
BT7FO%^C/lp9fS
<FM<B+;Fc(W^C5
ycUKFc_Nk+)B.E
2#l+OBypd+p|f>
4+RR-ykW)R)MVb
NJ/zS<WcLqU.||
5z7CT|GZc5*Ztz
+c25VFOBWL.y-4
8_c13B+AYP6zX|
OOCBKBS<>;*HM+
-D|H

cipher_information=340m_14by25_ut14m
W)+k^D(+l%WpO>
k.#KSMFPYLR/9V
BpzOUN|<z2p+l2
-LKJy7t-cYA1*H
+4dlTfN:^j*Xz6
V@K<UtG)pclddG
.L(MVE52H+M8|C
9G++|TBdEB+*5k
2b^D4ctR(UVFFz
+JYM(+l#Z3P>L#
+B.;+G1Gp+2|5J
p+8zF*KO&Dp+fZ
&R6N:(+4(8KjRO
29^4OFT_Rq#2pb
Sf9pl/C^%OF7TB
5C^W(cF;+B<MF<
E.B)+kN_cFKUcy
>f|p+dpyBO+l#2
bVM)R)Wky-RR+4
||.UqLcW<Sz/JN
ztZ*5cZG|TC7z5
4-y.LWBOFV52c+
|Xz6PYA+B31c_8
+MH*;><SBKBCOO
H|D-

cipher_information=340m_14by25_ut15n
H|D-+MH*;><SBK
BCOO|Xz6PYA+B3
1c_84-y.LWBOFV
52c+ztZ*5cZG|T
C7z5||.UqLcW<S
z/JNbVM)R)Wky-
RR+4>f|p+dpyBO
+l#2E.B)+kN_cF
KUcy5C^W(cF;+B
<MF<Sf9pl/C^%O
F7TB29^4OFT_Rq
#2pb&R6N:(+4(8
KjROp+8zF*KO&D
p+fZ+B.;+G1Gp+
2|5J+JYM(+l#Z3
P>L#2b^D4ctR(U
VFFz9G++|TBdEB
+*5k.L(MVE52H+
M8|CV@K<UtG)pc
lddG+4dlTfN:^j
*Xz6-LKJy7t-cY
A1*HBpzOUN|<z2
p+l2k.#KSMFPYL
R/9VW)+k^D(+l%
WpO>

cipher_information=340m_14by25_ut15m
KBS<>;*HM+-D|H
3B+AYP6zX|OOCB
VFOBWL.y-48_c1
T|GZc5*Ztz+c25
S<WcLqU.||5z7C
-ykW)R)MVbNJ/z
OBypd+p|f>4+RR
Fc_Nk+)B.E2#l+
B+;Fc(W^C5ycUK
O%^C/lp9fS<FM<
qR_TFO4^92BT7F
8(4+(:N6R&bp2#
D&OK*Fz8+pORjK
+pG1G+;.B+Zf+p
3Z#l+(MYJ+J5|2
U(Rtc4D^b2#L>P
BEdBT|++G9zFFV
+H25EVM(L.k5*+
cp)GtU<K@VC|8M
j^:NfTld4+Gddl
Yc-t7yJKL-6zX*
2z<|NUOzpBH*1A
LYPFMSK#.k2l+p
%l+(D^k+)WV9/R
>OpW

cipher_information=340m_14by25_ut16n
HD|M+->;*HKBS<
|OOCBAYP6zX8_c
13B+OBWL.y-4Zt
z+c25VFz7CT|GZ
c5*S<WcLqU.||5
kW)R)MVbNJ/zpd
+p|f>4+RR-yNk+
)B.E2#l+OByFc(
W^C5ycUKFc_C/l
p9fS<FM<B+;TFO
4^92BT7FO%^+(:
N6R&bp2#qR_K*F
z8+pORjK8(41G+
;.B+Zf+pD&Ol+(
MYJ+J5|2+pGtc4
D^b2#L>P3Z#BT|
++G9zFFVU(R5EV
M(L.k5*+BEdtU<
K@VC|8M+H2ld4+
Gddlcp)G-6zX*j
^:NfTAYc-t7yJK
LNUOzpBH*12l+p
2z<|FMSK#.k9/R
LYPk+)WV+(D^W%
lOp>

cipher_information=340m_14by25_ut16m
<SBKH*;>-+M|DH
c_8Xz6PYABCOO|
tZ4-y.LWBO+B31
ZG|TC7zFV52c+z
5||.UqLcW<S*5c
dpz/JNbVM)R)Wk
+kNy-RR+4>f|p+
(cFyBO+l#2E.B)
l/C_cFKUcy5C^W
OFT;+B<MF<Sf9p
:(+^%OF7TB29^4
F*K_Rq#2pb&R6N
+G14(8KjROp+8z
(+lO&Dp+fZ+B.;
4ctGp+2|5J+JYM
|TB#Z3P>L#2b^D
VE5R(UVFFz9G++
<UtdEB+*5k.L(M
+4dl2H+M8|CV@K
j*Xz6-G)pclddG
KJy7t-cYATfN:^
p+l21*HBpzOUNL
R/9k.#KSMF|<z2
%W^D(+VW)+kPYL
>pOl

cipher_information=340f_14by25_ut13n
|AY8_cOBWLZtz+
cz7CT|GS<WcLqU
kW)R)MVbpd+p|f
>4+Nk+)B.E2#lH
Fc(W^C5ycUD|C/
lp9fS<FMM+-TFO
4^92BT7>;*H+(:
N6R&bp2KBS<K*F
z8+pORjOOCB1G+
;.B+Zf+P6zXl+(
MYJ+J5|13B+tc4
D^b2#L>.y-4BT|
++G9zFF25VF5EV
M(L.k5*Zc5*tU<
K@VC|8M.||5ld4
+GddlcpNJ/z-6z
X*j^:NfRR-yAYc
-t7yJKL+OByNUO
zpBH*1KFc_2l+p
2z<|<B+;FMSK#.
kFO%^9/RLYP#qR
_k+)WVK8(4+(D^
pD&OW%l2+pGOpP
3Z#>VU(R+BEd+H
2)GT

cipher_information=340f_14by25_ut13m
+ztZLWBOc_8YA|
UqLcW<SG|TC7zc
f|p+dpbVM)R)Wk
Hl#2E.B)+kN+4>
/C|DUcy5C^W(cF
OFT-+MMF<Sf9pl
:(+H*;>7TB29^4
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cipher_information=340f_14by25_ut14n
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cipher_information=340f_14by25_ut14m
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cipher_information=340f_14by25_ut15n
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cipher_information=340f_14by25_ut15m
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cipher_information=340f_14by25_ut16n
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cipher_information=340f_14by25_ut16m
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