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Route Transposition and Phenomenon

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(@mr-lowe)
Posts: 1197
Noble Member
 

Thanks for the visuals Jarlve. i will have to get some time aside to learn how to do that.

 
Posted : August 2, 2020 3:26 am
smokie treats
(@smokie-treats)
Posts: 1626
Noble Member
Topic starter
 

If you haven’t yet, then pick up programming, it is actually a very relaxing and rewarding thing to start with.

I did. When the bookstores opened back up, I got a couple of books about Javascript. I only put a few hours a week into the new part of my hobby, but so far I have a program that does corpus training and makes and solves simple substitution ciphers ( most of the time ) in Google Chrome. I have a lot of work to do, but it is fun and rewarding. So far I just get my data from the console.log, so it is a very rudimentary program. Rather than go to homophonic, I aspire to fine tune and perfect my simple substitution solver, then work on transposition. We will see how it goes, and I might ask some questions later.

 
Posted : August 5, 2020 8:17 pm
(@fishermansfriend)
Posts: 132
Estimable Member
 

Jarlve, I really like this visual. I have wondered (like everyone), why so many +’s? At one point I wondered, well, what if Z wrote out text into each line, but only did so with complete words, then making the leftover spaces at the end of each row null, so let’s say + symbols.
That would obviously give too many +’s though. What I like about this is that 24 could be one for each row, plus a few at the end where the message actually runs out.

Sometimes I wonder if there’s just a way to run this grid backwards, like a rubik’s cube and get things to line up… The + symbols seem like a tantalizing tool to look for patterns.
Is there a way to do this? Like run a program with a parameter – x # of x type of rearrangements, to produce x pattern of + symbols…

I feel like a period transposition will never be enough. I think it’s some kind of regular(ish) transposition that has been chopped up and rearranged after. Hopefully just two transformations!

For example, let’s say he did period 19 ( a board favorite), but then cut the cipher into 4 unequal pieces – once vertically, once horizontally…

Anyway that’s sort of along the lines of what I’m currently looking into – although not with a period 19.

On a side note – What if some of the symbols stand for numbers and what’s encoded just happens to use a lot of some particular number? He was after starting to talk about bombs and maps etc…
What if this is actually the first of those communications?

 
Posted : August 6, 2020 12:54 am
Jarlve
(@jarlve)
Posts: 2547
Famed Member
 

If you haven’t yet, then pick up programming, it is actually a very relaxing and rewarding thing to start with.

I did. When the bookstores opened back up, I got a couple of books about Javascript. I only put a few hours a week into the new part of my hobby, but so far I have a program that does corpus training and makes and solves simple substitution ciphers ( most of the time ) in Google Chrome. I have a lot of work to do, but it is fun and rewarding. So far I just get my data from the console.log, so it is a very rudimentary program. Rather than go to homophonic, I aspire to fine tune and perfect my simple substitution solver, then work on transposition. We will see how it goes, and I might ask some questions later.

Nice to hear from you smokie!

Seems you are doing great with the programming. That’s not bad at all. Have fun with it and take it easy. Of course feel free to ask anything.

AZdecrypt

 
Posted : August 6, 2020 6:12 pm
Jarlve
(@jarlve)
Posts: 2547
Famed Member
 

Jarlve, I really like this visual. I have wondered (like everyone), why so many +’s? At one point I wondered, well, what if Z wrote out text into each line, but only did so with complete words, then making the leftover spaces at the end of each row null, so let’s say + symbols. That would obviously give too many +’s though. What I like about this is that 24 could be one for each row, plus a few at the end where the message actually runs out.

I still think it is most likely a 1:1 substitute.

Sometimes I wonder if there’s just a way to run this grid backwards, like a rubik’s cube and get things to line up… The + symbols seem like a tantalizing tool to look for patterns.
Is there a way to do this? Like run a program with a parameter – x # of x type of rearrangements, to produce x pattern of + symbols…

Largo did it but didn’t find anything. Though such analysis would be really hard to well since there are so many patterns one could look for.

I feel like a period transposition will never be enough. I think it’s some kind of regular(ish) transposition that has been chopped up and rearranged after. Hopefully just two transformations!

We’ve tried so much in this direction. Just nothing. I am growing skeptic, maybe there is non-trivial polyalphabetism, but so hard to explore.

On a side note – What if some of the symbols stand for numbers and what’s encoded just happens to use a lot of some particular number? He was after starting to talk about bombs and maps etc…
What if this is actually the first of those communications?

Try it. We welcome help and original ideas.

AZdecrypt

 
Posted : August 6, 2020 6:33 pm
doranchak
(@doranchak)
Posts: 2614
Member Admin
 

I’m attending the ACA convention today (virtually), and just saw an interesting talk about an encrypted diary.
It uses a very simple plaintext shifting system.
Example:

Plaintext: FLEE AT DAWN

The system is basically to go up or down 2 letters, completely arbitrarily.
So, F + 2 = H
L + 2 = N
E + 2 = G
E – 2 = C

A – 2 = Y
T + 2 = V

D + 2 = F
A – 2 = Y
W – 2 = U
N + 2 = P

Ciphertext: HNGC YV FYUP

To decipher, you perform the process in reverse and work out the words that fit the best.

HNGC + 2 = JPIE
HNGC – 2 = FLEA

So you have to pick the best shift at each position based on which ones make words that make sense in context.
Since it’s only limited to 2 shift values, it doesn’t suffer too heavily from an "any answer can fit" problem.

What if Zodiac did that to his plaintext before substituting the letters for symbols?
Could a solver like AZDecrypt still work out the correct solution?

The diary the system was used in was written in the early 20th century, and I’m not aware of any other examples of the use of this method.

http://zodiackillerciphers.com

 
Posted : September 12, 2020 10:38 pm
(@4on4off)
Posts: 64
Trusted Member
 

I’m attending the ACA convention today (virtually)

Look forward to viewing if someone is kind enough to record it and share.

 
Posted : September 13, 2020 3:12 am
Jarlve
(@jarlve)
Posts: 2547
Famed Member
 

What if Zodiac did that to his plaintext before substituting the letters for symbols?
Could a solver like AZDecrypt still work out the correct solution?

Do you have a 340 length plain text like this?

AZdecrypt

 
Posted : September 13, 2020 10:38 am
doranchak
(@doranchak)
Posts: 2614
Member Admin
 

What if Zodiac did that to his plaintext before substituting the letters for symbols?
Could a solver like AZDecrypt still work out the correct solution?

Do you have a 340 length plain text like this?

Here are 10 ciphers. Each is a different plaintext subjected to a shift of +/- N, where + or – is selected randomly. N is constant for each text and is in the range [1,13].
Then the shifted plaintext is encoded homophonically using 63 cipher symbols.

S6R*z|R%1U;UOzq4C
z#_TbcLMUGBLLZ^<D
)V<6U2Bc^qGtV9-G3
T6|@-CR;UWR7#zb^z
R-GJ_J>4&z+2F5ND+
+zZ.1BL&)5|qd2Fy;
B542pty;.Z.7YE(%-
2K|B-7+q>A&38O5z@
P_LXt2)A/q.G2J4f1
CR-5tP;5&5-HzU&BL
HlA;j_22AG2Jb6Uz&
2S6zWq:4&qbRLZJJW
82)bUU&C23kFzc2Ob
42YE*G@bGJDMWJz-f
VfAU:BC1jM45Fq&%8
14.Gd#Rb5L6E(612)
AR3.25@S11zVpU77#
ztq|5^W+6(PzL*9*-
6OX&f@P_W/+|y;TXk
Fkd.9+H.q:WR712&W

BF6JCNA@t)GK)_N*#
(>/|X(G8JGVX.AqRA
(@OFc-8FE8W.DJlWG
_)BH-(^&.y8Ljzf2E
Gc8J.EBG9*%c5_7Xj
7(c&(c5_#CEjz4^&D
<H_X(5L8@R9lHU5Wf
A@((@&_fzJ+bE8Xtp
JC|95XB5bd(Md(._J
yd(GfTB(|)_py*cjL
1DTA63C-JU*(_CK;y
(%flH9R*FDlzpbldC
clc(1JB@9(H36CGDA
cAy8@J6KCS-#4by@z
f)5%DdK<(BY-12AG@
Z*C%B^:7__3A.k3C_
kXBjfXC_7Gcl-BVf4
c896TC99(7dTPDWc-
*^JCtVqMP<@j#(c2N
)BF1T*L|(354pZHz3

H@.At5qOU%V7#RLDL
+kU./7OA+%G^|WZ_E
DBV%qU/J.@J8k6q)F
PF#b:L^b^U>8HG:OL
8/DLNK+/U6fj39pHK
/q^129t:Ep#6J6l%Z
NSUL:4S-DG<++|:T7
>(*q:LkN+d@qzOM.B
>8BHUj@|9%NTcBb@9
)&>dlM2DN%9M-8bSW
qM(Z).*+O+..LBq.-
Z#NVf%YRd6|6AbL#D
kNz.*Z-^MM8>+%AVX
Mq7f_+qMHc6_-V^lM
8WX6:;EFL+)>qATf<
@M.9X|DAMMC>1-bt@
@OSCbZWT^CSJLt8Wk
BJ@q*LV@:4A;KD(Sq
b36jK6|O%y^Hq<@T.
dO>OOB3>&:37HtqWY

7tV7G:k@j|V^Af)cY
37TpA)A5/d/)c6d>N
jqKY&VAV.GAG6GK5k
^HOcWTUC8)R;*<)Tl
13<_V*V_N@7)WGk3)
bTZO/Pb._O)E:cyRV
4l2#C5f4AVP1/DGJE
ORUSlMYZUXd7;k&#b
XE@zTjCA-ck9X;4D*
%;k(7t+#tOy:7t5+D
cHdk3;*pWCfM:5jFp
E.qbk9lU*OqbkV2Hf
XzVk)V>T@DkWcNY9B
>dULz)YZNPCE)tc7)
DfHBS6DXObBXqGEOt
X8NV/U:Y)ZDOlBWtj
t9_W<q)Uk)CO)tYHV
qc2.&PE9RdRLX1NBX
ZG6)zp8TU7):k5q:*
6VqcHqq)8PS14Vk9R

PpT>Z_k.qfYd>&#1k
(WE>A/SpqPBkf7XV4
kY@U4(XV>S%>AGT8D
:1*l^1+8DcpkHD3%y
YLqX%(k1.Wk:bMLdX
Hf%>^Zl@H)#@btU>^
SkGZ1&H.z8NtNfC5f
8XzZp4z>^F*ZMtH*#
qcX%bJtElBc%%bYOd
DcP:H>-k6PL*FSf*G
b#A:13lBcUUb2d.AA
N:;9X.M)p4E.1At@*
pE>^c*W6fcyf5737K
H>Sp&z8*(N:1V>q<H
D6:6fAXS:l6W(kjb|
^%U;Xl-)1*T%X4qlA
:kU>*G&#CqE(^pV7:
>8M:AlStR|UqcPl.K
ZYzN:@;fG;lPV|>y1
;5DK6)Z;A:lz%4AA:

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MKUzd#-F3&.C>yKT:
zlV5j9DM9:;7yj2bZ
/XHGAXNlfqFG3X-3T
kS9<ZN2ATYZy7Mf%)
H8pUj|RD^Z5zB(dM.
%MXtNq%M5_j&P:>)>
+SO&c@:q&y3:(>./5
Y(/Bb9Z/-;+6SBzU8
P2LpB%X:/NkAPEjMO
3X4jO&d6^pP8/Hd#2
R9_#fP*Dtyk2PK/Z_
E1F#2R7t:@.2;Ky:Z
VD5j8MdJCDVAJ7tVY
@AdKJG+d#pPtjBU;T
*F+MJD3VMJ9R:P7AD
*XXASVNdBVkG3&+Dq
RN9%_b&VO_T7P4W:S
OFlN^+5;C/UkcDTzd
qEON1721k1Cq%)Tkt

H.5A)CA1GG6BW-dTN
9LDVc.DzTf%|pLCN1
j%c41<LG<4*XO5bE.
_(Y)(@SBdW1.LL7c.
E._XS6WC.ct&jEPG.
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W-.1.&BYBdD<AkNS.
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7PPPZEd*MN1Ej*b1+
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&D5J)8y;WT;&T>Y_j
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-19y^Gz.NjYXRW3(5
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(KV9j(/#/-4^5&WtO
9fV97;&RF/1V>y6)Y
7l.U6N7KK/Jp%VS<V
y:WS*WddFK/1^UT3q
*(ydqXp/NE%3W).P*
GXW8G.+HN><TB)NVK

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-zk@F*5>U:+dW2)k)
c)+KHTVbyH/Sq<86N
1E))1lK&Gj)dY:2/|
&1<7<YAHK@q%dWD38
#^HBK3<BtLd)ZR9Y7
Jl&7zM^FJYbEOL<-U
/OLT81ENBl4M*jP_T
:kCOb|&F-:W1ktj>*
7<d3UB4LO:K:(cYUV
-z/%C<tF<&#B3Ll<:
L3FU-zK<H1#<-:Ojb
d1Pb<T>Tj6WpLd-W2
><>%SAH%ZJ>2fU/OZ
3>R<&6:zD7l><&FRR
kU4:WNA;YLtk&.:9p
:9fW2jP6tJzMSHAk;
:Lt-)Ap*F8F:<jlkS
bRU#A:9fdSAJb.>b-
@C55A&%cT+k3%KtOf

y2_HfpyY|LH8|_LKY
>_MUL>L7+|&S:Hlcl
P9_V#:q;R3S4%2UK@
VE*f2(b1WR*Uf;X<O
Oy/DKj2##H/d;y99_
5jB|U7qTUP(O%JA&V
NK1Wq(MyVtYZ&H/GO
S8<M8_/8^_bk/#NT4
L^DJB>Hc5Y#F->f4W
9JLlMWGB9WU@F(<%D
26A-/O>D:F8RStGOJ
t78*B2f_D(9Pj;:OB
@|4UB5Olt;@FCPLO4
*.7Y(7E|LO(zM#R5y
NOW8+J8L&HSbG|GM4
-yYFO8Ol+P:t;+Kl*
RUSlEMzCKA);51:>+
#F(d><tzP9;l>:/lF
AzCbK|+bJ:*UU&kO&
l&YW8AO:T4H-ZB>Bt

If you need a shifted plaintext, too, here is the one corresponding to the last cipher:

QLILCWQEBPLUBIPBE
RIBVPRPWYBZXQLKWK
HJILOQKLPMXUSLVBY
LGHCLDIODPHVCLOSF
FQROBVLOOLRNLQJJI
XVPBVWKHVHDFSHXZL
ABODKDBQLQEHZLRAF
XUSBUIRUXIIEROAHU
PXOHPRLWXEOBWRCUD
JHPKBDAPJDVYBDSSO
LTXWRFROQBUPXQAFH
QWUHPLCIODJHVLQFP
YBUVPXFKQLYBGHPFU
HGWEDWGBPFDKBOPXQ
AFDUYHUPZLXIABABU
WQEBFUFKYHQQLYBKH
PVXKGBKGBXSLXOQRY
OBDNRSQKHJLKRQRKB
XKGIBBYIHQHVVZEFZ
KZEDUXFQHULWHPRPQ

http://zodiackillerciphers.com

 
Posted : September 13, 2020 1:47 pm
Jarlve
(@jarlve)
Posts: 2547
Famed Member
 

Thanks. The plain text solves easily doing the following in AZdecrypt:

1) Set solver to "Substitution + polyphones [user]".
2) Go to Functions -> Symbols. In the Symbols window select "Set plaintext letters for all symbols", fill out 2 in the A1 field underneath it and click on "Apply manipulation" (need to redo this every time the cipher changes!).
3) Go to Options -> Solver. Change Multiplicity weight to 1 (not required but it is generally better with this solver since it will then penalize the use of extra letters when not needed).
4) Click on Solve.

In general, while allowing polyalphabetism, the solution usually has ambiguous sections.

Same procedure should also work for the homophonics but 7 or 8-grams will be needed with a much longer run time. The good thing about solving it this way is that you don’t have to care about the values of N. The bad thing is that it is does not take in account the constraints of the fixed N values, inflating its difficulty. Thus, a proper approach would be the change to solver to have fixed N values, with an additional hill-climber on top to figure out the values of N.

My computer is currently running with 8-grams on the last cipher but it hasn’t solved it yet, judging from the scores I feel confident that the solution will pop out, it is just a matter of run time.

AZdecrypt

 
Posted : September 13, 2020 3:59 pm
(@4on4off)
Posts: 64
Trusted Member
 

In general, while allowing polyalphabetism, the solution usually has ambiguous sections.

Most definitely. When using it on the 340 many words pop out but are obviously not words the Zodiac would use and do not form a coherent message.

They may if rearranged but definitely not Z speak, albeit my attempts have been short runs as I have been playing with that feature recently with limited time available.

 
Posted : September 13, 2020 4:28 pm
Jarlve
(@jarlve)
Posts: 2547
Famed Member
 

@4on4off,

Thanks for using my program. The ambiguity I was referring to relates to the actual (real) solution of the cipher. Entire word combinations (sections) can get replaced by other word combination here and there. And yes, using that solver will cause more words to pop out because it has a much larger amount of freedom. I’m not sure what settings you are running but you may at least want to move up to 6-grams.

My computer is currently running with 8-grams on the last cipher but it hasn’t solved it yet, judging from the scores I feel confident that the solution will pop out, it is just a matter of run time.

Solution hasn’t popped out yet.

AZdecrypt

 
Posted : September 14, 2020 10:11 am
doranchak
(@doranchak)
Posts: 2614
Member Admin
 

Thanks. The plain text solves easily doing the following in AZdecrypt:

1) Set solver to "Substitution + polyphones [user]".
2) Go to Functions -> Symbols. In the Symbols window select "Set plaintext letters for all symbols", fill out 2 in the A1 field underneath it and click on "Apply manipulation" (need to redo this every time the cipher changes!).
3) Go to Options -> Solver. Change Multiplicity weight to 1 (not required but it is generally better with this solver since it will then penalize the use of extra letters when not needed).
4) Click on Solve.

In general, while allowing polyalphabetism, the solution usually has ambiguous sections.

Incredible! Thanks for the instructions. I was able to use them to successfully solve a few other plaintexts constructed the same way.
Seems like the substitution key has gotten much, much larger due to the limited polyalphabetism, and yet your solver can still break it.

By the way, I started a new page to collect AZdecrypt "tips and tricks" here: http://zodiackillerciphers.com/wiki/ind … and_tricks
I want to add more stuff to it, like examples of ways to solve other cipher types, and other instructions and info you’ve given in other forum posts. Can you recommend some other of your forum posts for me to link to from there?

Same procedure should also work for the homophonics but 7 or 8-grams will be needed with a much longer run time. The good thing about solving it this way is that you don’t have to care about the values of N. The bad thing is that it is does not take in account the constraints of the fixed N values, inflating its difficulty. Thus, a proper approach would be the change to solver to have fixed N values, with an additional hill-climber on top to figure out the values of N.
My computer is currently running with 8-grams on the last cipher but it hasn’t solved it yet, judging from the scores I feel confident that the solution will pop out, it is just a matter of run time.

Solution hasn’t popped out yet.

I wonder if you will need to adjust the solver to take the fixed N values into account.
Do you want me to post the plaintext, and the plaintext with modified shift applied for that last cipher, just to rule out any mistakes in encipherment?

http://zodiackillerciphers.com

 
Posted : September 14, 2020 5:32 pm
(@4on4off)
Posts: 64
Trusted Member
 

@4on4off,

Thanks for using my program. The ambiguity I was referring to relates to the actual (real) solution of the cipher. Entire word combinations (sections) can get replaced by other word combination here and there. And yes, using that solver will cause more words to pop out because it has a much larger amount of freedom. I’m not sure what settings you are running but you may at least want to move up to 6-grams.

My computer is currently running with 8-grams on the last cipher but it hasn’t solved it yet, judging from the scores I feel confident that the solution will pop out, it is just a matter of run time.

Solution hasn’t popped out yet.

My apologies, I was messing around with Substitution + sparse polyalphabetism with 8-grams_english_beijinghouse_v6.txt loaded without adjusting any settings.

As you have stated, the larger amount of freedom generates more words but clearly not Z speak:

I DERNET WAS A RELATION
THE DISTANCES WE
REPORT TO MAKE RED
PICKING UP FOR UN INTENTION
BACK IN THAT HIS LIFE THE ISSUES
OF THE DAY JUST FORWARD
A GREAT CONDITIONALLY
DIDNT HAVE TO EXPLAINS
AND BUILDING THE PROJECT
CHANNEL HANNITY REQUESTLING
THE HEAD AND THAT HERE THEY
ARE GETTING PRICE THE PROVIDED
FROM THE ARGUMENT IN THE NUMBERS
TER AND THERE ARE THERE A REAL KILLER
JOBS INDUCE IN VES SERVATIONS FAMILY ANTT

 
Posted : September 14, 2020 6:50 pm
greenglow2
(@greenglow2)
Posts: 7
Active Member
 

we all know what got "transposed"

this is good stuff

 
Posted : September 15, 2020 7:03 am
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