Interesting ideas Largo. That does sound rather hopeless to solve.

If I was committed to trying to write a hillclimber for this sort of problem, I’d probably start with a test cipher that only has one space removed. Then two spaces removed. Then three spaces removed. Etc. If those easy cases can be solved by a hillclimber then there’s some hope of making it work for a larger number of spaces. But I fear that the search landscape would be too bumpy, so that partially correct keys still yield terrible measurements that don’t give much useful direction to the search.

Another approach I would think about is to generate millions of test ciphers under your scheme, and then train a neural network to recognize where the spaces need to be re-inserted. It would be a long shot but it might be interesting.

Jarlve, didn’t you work on something similar for your solver? On a type of imperfect transposition, with the assumption that Zodiac might have skipped symbols during transposition, thus breaking any columnar solver that expects perfection?

That would be 100+ skips. I also don’t think it can be solved if that is the case.

Me and smokie worked on this for almost 6 months or so, we went up to a total of any combination of 10 nulls/skips for the period 15/19 hypothesis: viewtopic.php?f=81&t=4125

The solver that can do this in AZdecrypt is called Substitution + nulls and skips and its options can be found under Options, Solvers:

(Substitution + nulls and skips) Period: 1
(Substitution + nulls and skips) Nulls and skips: 5
(Substitution + nulls and skips) Manual nulls: 0
(Substitution + nulls and skips) Manual skips: 0

It gets real hard real fast and a periodic cipher with 10 nulls could take days to solve.

Another approach I would think about is to generate millions of test ciphers under your scheme, and then train a neural network to recognize where the spaces need to be re-inserted. It would be a long shot but it might be interesting.

I’m skeptic that a neural network would be able to unentangle/recognize it somehow if there are 100+ skips.

Jarlve, I have a question about the "Substitution + nulls and skips" solver:

When exploring skips, does the solver insert dummy cipher symbols at random locations in the cipher text? So it’s hillclimbing a solve for the cipher text as if those new symbols are present?

As opposed to nulls, where cipher symbols at random locations are selected to be ignored during the hillclimb?

Eventually I want to complete the descriptions of the solvers at: http://zodiackillerciphers.com/wiki/ind … ks#Solvers

Jarlve, I have a question about the "Substitution + nulls and skips" solver:

When exploring skips, does the solver insert dummy cipher symbols at random locations in the cipher text? So it’s hillclimbing a solve for the cipher text as if those new symbols are present?

As opposed to nulls, where cipher symbols at random locations are selected to be ignored during the hillclimb?

Eventually I want to complete the descriptions of the solvers at: http://zodiackillerciphers.com/wiki/ind … ks#Solvers

Thanks.

Yes, it inserts a new symbol for every skip and for it removes a single character for every null. If the period > 1 then it will do either of this before the period operation. There is allot more that goes into it under the hood though. Me and smokie spent a great deal of time optimizing it. But perhaps I need to give this solver a retune and menu for 1.20.

In case of Largo’s hypothesis I should make a change to the solver so that all the skip symbols are one symbol (space). And then perhaps try to solve it with n-grams that have spaces in it. AZdecrypt 1.19 has such n-grams under the /N-grams/Spaces/ directory.

It were 8 months!

XX. Beginning of 8 months and 40 pages of programming and testing for up to 10 nulls and skips in route transposition + homophonic substitution: viewtopic.php?f=81&t=3196&start=800

In case of Largo’s hypothesis I should make a change to the solver so that all the skip symbols are one symbol (space). And then perhaps try to solve it with n-grams that have spaces in it. AZdecrypt 1.19 has such n-grams under the /N-grams/Spaces/ directory.

Sounds good.
That reminds me – I was playing with a homophonic substitution cipher recently but the "twist" is that several of the symbols encode to word breaks (spaces). So the spaces are homophonically encoded, just like the high-frequency plaintext letters. I didn’t think azdecrypt can solve for word breaks directly so I wrote a brute force method that considers all possible selections of up to N symbols that might be spaces. Then it computes the average word length for each cipher with restored word breaks, and keeps the ones that are closest to the expected average (around 4.7 I think). Then I just fed those in a batch to azdecrypt, with the candidate "word break" symbols removed from the cipher text. Seemed to work, but if it directly solved for the word break symbols using the spaces n-grams, it would probably solve much faster.

I’m skeptic that a neural network would be able to unentangle/recognize it somehow if there are 100+ skips.

That is understandable. It would seem like magic if it worked!

It may be more beneficial to train the network to distinguish between two categories of ciphers: transpositions WITH skips, and transpositions WITHOUT skips.

If that network is accurate and it says Z340 is a transposition WITHOUT skips, then we don’t have to build the magical thing that can solve the ones WITH skips.

In case of Largo’s hypothesis I should make a change to the solver so that all the skip symbols are one symbol (space). And then perhaps try to solve it with n-grams that have spaces in it. AZdecrypt 1.19 has such n-grams under the /N-grams/Spaces/ directory.

Sounds good.
That reminds me – I was playing with a homophonic substitution cipher recently but the "twist" is that several of the symbols encode to word breaks (spaces). So the spaces are homophonically encoded, just like the high-frequency plaintext letters. I didn’t think azdecrypt can solve for word breaks directly so I wrote a brute force method that considers all possible selections of up to N symbols that might be spaces. Then it computes the average word length for each cipher with restored word breaks, and keeps the ones that are closest to the expected average (around 4.7 I think). Then I just fed those in a batch to azdecrypt, with the candidate "word break" symbols removed from the cipher text. Seemed to work, but if it directly solved for the word break symbols using the spaces n-grams, it would probably solve much faster.

That’s clever. I take it the cipher did not solve with the spaces?

Here’s a solve for a cipher with spaces without using space n-grams:

Score: 22223.85 IOC: 0.0722 Multiplicity: 0.1852 Seconds: 0.28
Repeats: SBECAUSES SWILLING LLINGS NSTHE THATS WILLD THANS
PC-cycles: 1856

IS LIKES WILLING A PEOPLES BECAUSES ITS AS 
ANOS MUCH S FUND ITS IND MORES FUN A THANS 
WILLINGS WILDS GAMED ANS THED FOR RESTS BECAUSES 
MAND INS THE AMONTS DANGEROUS DANIMALS OF 
A ALLS TO SWILL D NO ME THINGS GIVEN SMED 
THES MOST A THRALLINGS EXPERENCES IT DINS 
EVENS BETTERD THANS GETTINGS YOURS ROCK NS 
OFFS WITHATS GIRLS THE D BESTS PART A OF 
SITS AND THATS WHEN A IS DIE SID WILL D BES 
REBORN A IN A PARADICES AND SALL ST

H4($;84[H(($<E,V2
BNW&'?Z)JA.@*$"*7
.,%Q'5XRC*TUMI_>4
SFI5Q1&4TAL,":M'
/S99HLE'!$W6*EG5Z
I7#4>C@ITDP1Y.04?
8^+X.8*5J<I_%4>:2
,5BF0*6+ME&PQX.IJ
LS5W4DT,G9W'"B*/
H9(I%Q5Z"CH#E4E$K
@F*5ZI0:@45D.",>C
17W9_<E'Y-V&PZ<R@
*S0IH%4YK8M*?2>0Y
1I":GL4E8>0_LE'OQ
UP41D^;F*DTT4=S"C
,3'EHP(*>:&I?Y.04
]+1",BT'$>*7%I0CJ
"4!:8#,_'6S2*_I=_
W9I?84P2?Q1<,_#,V
P6_)Z'GL6*(9'>

Alternatively we could consider the spaces as nulls and use the Substitution + units solver with the standard settings, Unit: symbol, Mode: Remove and perhaps Multiplicity weight: 1:

Restart: 1 Hill climber: 4881/5000 @ 500000
Score: 20443.27 IOC: 0.0662 Multiplicity: 0.2140 Seconds: 10.82
Repeats: WILLING THEMOST BECAUSE ILLING METH ITIS (2) ANGE
PC-cycles: 2261

Unit: symbol
Mode: remove
Key(4,,,',*,I)

ILIKE WILLING PEOPLE BECAUSE IT IS SO MUCH 
FUN IT IS MORE FUN THAN WILLING WILD GAME 
IN THE FOR REST BECAUSE MAN IS THE MOST DANGEROUS 
ANIMAL OF ALL TO WILL SOMETHING GIVES ME 
THE MOST THRILLING EXPERENCE IT IS EVEN BETTER 
THAN GETTING YOUR ROCKS OFF WITHE GIRL THE 
BEST PART OF IT IS THAT WHEN I DIE I WILL 
BE REBORN IN PARADICE AND ALLT

I’m skeptic that a neural network would be able to unentangle/recognize it somehow if there are 100+ skips.

That is understandable. It would seem like magic if it worked!

It may be more beneficial to train the network to distinguish between two categories of ciphers: transpositions WITH skips, and transpositions WITHOUT skips.

If that network is accurate and it says Z340 is a transposition WITHOUT skips, then we don’t have to build the magical thing that can solve the ones WITH skips.

Say that the network is 95% accurate and unknowingly to us is actually just measuring bigrams. And the transpositions with skips (100+) are just recognized as shuffles?

I would try to differentiate between shuffles and transpositions with skips. But the random number generator for the shuffles and skip positions has to be cryptographic or better (true).

That’s clever. I take it the cipher did not solve with the spaces?

That’s correct.

Here’s a solve for a cipher with spaces without using space n-grams:

How many symbols did you use to represent space?

Alternatively we could consider the spaces as nulls and use the Substitution + units solver with the standard settings, Unit: symbol, Mode: Remove and perhaps Multiplicity weight: 1:

Ah! Nice idea. Can you remind me of the effect of multiplicity weight? Does it normally punish solutions that increase the multiplicity of the input cipher?

Say that the network is 95% accurate and unknowingly to us is actually just measuring bigrams. And the transpositions with skips (100+) are just recognized as shuffles?
I would try to differentiate between shuffles and transpositions with skips. But the random number generator for the shuffles and skip positions has to be cryptographic or better (true).

Good point. A more general classifier would be better. But so far it seems to be hard to achieve. Kopal’s research is encouraging. 90% accuracy on that test set seems really good, but I don’t remember how many cipher types are involved.

How many symbols did you use to represent space?

Automatically encoded, seems it were 5.

Ah! Nice idea. Can you remind me of the effect of multiplicity weight? Does it normally punish solutions that increase the multiplicity of the input cipher?

It follows this calculation: score/=1+((unique_symbols/cipher_length)*multiplicity_weight). The higher the multiplicity, the more the score will be reduced. It may prevent some of the solvers from inflating the score by changes that increase multiplicity. It will still do that but now there is a trade off.

Secondly it can also be used as a filter, in case of using the Substitution + units solver for removing the spaces as nulls, the key size (amount of null symbols) will increase with every hill-climber iteration and eventually the multiplicity will become very high and the solver may find non-solutions that exceed the score of the underlying plain text, in that case the multiplicity weight will punish these and the best solution (according to the multiplicity weight trade off) will still be on top, sort of.

Good point. A more general classifier would be better. But so far it seems to be hard to achieve. Kopal’s research is encouraging. 90% accuracy on that test set seems really good, but I don’t remember how many cipher types are involved.

I hope we will see more of this. I don’t think that AI is the one and only solution to achieving perfection in cryptography but it certainly will become a major player at some point.

Thank you for the explanations. I updated the Wiki. Maybe in 10 years, I’ll finish documenting all of AZdecrypt!

Always wondered if the + symbols might related to empty spaces – but the # of + symbols doesn’t neatly jive with what you would expect for the number of natural spaces in a text this long. Also doesn’t seem to fit with filler at the end – since that would be more than a full row of filler…

Another thought I had was what if Z only filled out each row with full words – no wrapping. Then filled those remaining spaces with + symbols. Then transposed. However I didn’t feel like this passed the smell test when I drew it up.

Anyway, curious if anyone has new thoughts on the +’s