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ASLEC Cipher – The Entanglement.

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  ASLEC Cipher – The Entanglement. austin obyrne <ao402468@gmail.com> - 2021-10-25 08:38 -0700
    Re: ASLEC Cipher – The Entanglement. Richard Heathfield <rjh@cpax.org.uk> - 2021-10-25 17:10 +0100
    Re: ASLEC Cipher – The Entanglement. austin obyrne <ao402468@gmail.com> - 2021-10-25 11:06 -0700
    Re: ASLEC Cipher – The Entanglement. "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2021-10-25 12:51 -0700
      Re: ASLEC Cipher ? The Entanglement. Rich <rich@example.invalid> - 2021-10-25 20:09 +0000
        Re: ASLEC Cipher ? The Entanglement. MM <mrvmurray@gmail.com> - 2021-10-26 00:51 -0700
          Re: ASLEC Cipher ? The Entanglement. Richard Heathfield <rjh@cpax.org.uk> - 2021-10-26 09:22 +0100
        Re: ASLEC Cipher ? The Entanglement. austin obyrne <ao402468@gmail.com> - 2021-10-27 08:10 -0700
          Re: ASLEC Cipher ? The Entanglement. Richard Heathfield <rjh@cpax.org.uk> - 2021-10-27 17:25 +0100
          Re: ASLEC Cipher ? The Entanglement. MM <mrvmurray@gmail.com> - 2021-10-28 01:08 -0700
            Re: ASLEC Cipher ? The Entanglement. Richard Heathfield <rjh@cpax.org.uk> - 2021-10-28 09:19 +0100
              Re: ASLEC Cipher ? The Entanglement. MM <mrvmurray@gmail.com> - 2021-10-28 02:51 -0700
                Re: ASLEC Cipher ? The Entanglement. Rich <rich@example.invalid> - 2021-10-28 13:29 +0000
                Re: ASLEC Cipher ? The Entanglement. Richard Heathfield <rjh@cpax.org.uk> - 2021-10-28 15:10 +0100
                  Re: ASLEC Cipher ? The Entanglement. MM <mrvmurray@gmail.com> - 2021-10-28 09:22 -0700
        Re: ASLEC Cipher ? The Entanglement. austin obyrne <ao402468@gmail.com> - 2021-10-28 14:23 -0700
          Re: ASLEC Cipher ? The Entanglement. MM <mrvmurray@gmail.com> - 2021-10-28 14:47 -0700
          Re: ASLEC Cipher ? The Entanglement. Richard Heathfield <rjh@cpax.org.uk> - 2021-10-29 00:56 +0100
            Re: ASLEC Cipher ? The Entanglement. daleJ <daletejas@gmail.com> - 2021-10-29 17:22 -0700
      Re: ASLEC Cipher – The Entanglement. Max <maxturv26@gmx.net> - 2021-10-25 22:24 +0200
        Re: ASLEC Cipher – The Entanglement. "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2021-10-26 21:52 -0700
          Re: ASLEC Cipher – The Entanglement. "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2021-10-26 21:53 -0700

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#50565 — ASLEC Cipher – The Entanglement.

The encryption/ decryption model to go with this text may be seen at https://www.aslec.uk.

In the model let’say (postulated  here for discussion purposes only)
that Eve has overcome the insurmountable problem 
of finding the correct displacement of the plaintext in hand as
the position vector of a point in R^3 relative to the universal 
zero at (0, 0, 0).

She still has lots more miracle hurdles to overcome and it is this 
that I want to discuss here.

So far, allowing against all odds that she has discovered the 
vector ‘Pn’ that appears in the encryption/decryption model
she realises that in addition to defining ‘n’ in the model this Pn 
is also the direction vector of a straight line along which an 
infinite set  of hidden planes all intersect i.e. every plane in
 the set contains Pn.

Her problem now is to find which of that infinite set of planes is 
the right one that will yield ‘n’ when when it's normal vector 'N' is 
processed in the algorithm of the encryption/decryption model.

her task is impossible.

Again allowing once more that she has against all odds 
zoomed in on the right vector 'N' of the infinite set  
her final problem is in decoding this number 'n' (derived and belonging to N)
according to the substitution alphabet that Alice and Bob are using.

Again her task is impossible.

In general, her task amounts to guessing the data that resides instantaneously in the minds of the legitimate entities Alice and Bob.

Comment:

Clearly, no computer will ever be able to transfer data from a human memory to a computer memory.

Comment: 

This cipher is protected by at least one set of infinite keys (i.e. the normals of the hidden intersecting planes).

Comment :

This cipher is protected by a vast domain of vectors that represent the Pn’s in the cipher text. The unique piece of information i.e. a particular vector that would flag an end to the search  cannot be deduced by brute force since any search program written to do that is indeterminate because of having no terminating information that will flag an end to the search.

This cipher is without any shadow of doubt unbreakable by the unique  definition in all the handbooks.

Comment:

I am only here for the cryptography. Ada is a langiuage that I was introduced to at a seat-of-learning.some thirty years ago.

Austin O’Byrne

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#50566

On 25/10/2021 16:38, austin obyrne wrote:
> The encryption/ decryption model to go with this text may be seen at https://www.aslec.uk.
> 
> In the model let’say (postulated  here for discussion purposes only)
> that Eve has overcome the insurmountable problem
> of finding the correct displacement of the plaintext

No, let's not. Eve almost certainly won't bother with that. She'll find 
some other way in, like Colin did, and MM did, and I did, and goodness 
knows who else did.

<snip>

> She still has lots more miracle hurdles to overcome and it is this
> that I want to discuss here.

No, all she needs to do is look at the problem through cryptanalyst eyes 
instead of through Austin's blinkers.

<snip>

> her task is impossible.

Demonstrably false. Has been demonstrated at least three times.

<snip>

> In general, her task amounts to guessing the data that resides instantaneously in the minds of the legitimate entities Alice and Bob.

Not so. No such guesses were required last time, or the time before, or 
the time before that.

> This cipher is protected by at least one set of infinite keys (i.e. the normals of the hidden intersecting planes).

So was Shuttlepads.

Didn't help.

> This cipher is without any shadow of doubt unbreakable by the unique  definition in all the handbooks.

Except that we will keep breaking it.

> Comment:
> 
> I am only here for the cryptography.

Then why after all this time do you know so little about it?

-- 
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

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#50567

On Monday, 25 October 2021 at 16:38:02 UTC+1, austin obyrne wrote:
> The encryption/ decryption model to go with this text may be seen at https://www.aslec.uk. 
> 
> In the model let’say (postulated here for discussion purposes only) 
> that Eve has overcome the insurmountable problem 
> of finding the correct displacement of the plaintext in hand as 
> the position vector of a point in R^3 relative to the universal 
> zero at (0, 0, 0). 
> 
> She still has lots more miracle hurdles to overcome and it is this 
> that I want to discuss here. 
> 
> So far, allowing against all odds that she has discovered the 
> vector ‘Pn’ that appears in the encryption/decryption model 
> she realises that in addition to defining ‘n’ in the model this Pn 
> is also the direction vector of a straight line along which an 
> infinite set of hidden planes all intersect i.e. every plane in 
> the set contains Pn. 
> 
> Her problem now is to find which of that infinite set of planes is 
> the right one that will yield ‘n’ when when it's normal vector 'N' is 
> processed in the algorithm of the encryption/decryption model. 
> 
> her task is impossible. 
> 
> Again allowing once more that she has against all odds 
> zoomed in on the right vector 'N' of the infinite set 
> her final problem is in decoding this number 'n' (derived and belonging to N) 
> according to the substitution alphabet that Alice and Bob are using. 
> 
> Again her task is impossible. 
> 
> In general, her task amounts to guessing the data that resides instantaneously in the minds of the legitimate entities Alice and Bob. 
> 
> Comment: 
> 
> Clearly, no computer will ever be able to transfer data from a human memory to a computer memory. 
> 
> Comment: 
> 
> This cipher is protected by at least one set of infinite keys (i.e. the normals of the hidden intersecting planes). 
> 
> Comment : 
> 
> This cipher is protected by a vast domain of vectors that represent the Pn’s in the cipher text. The unique piece of information i.e. a particular vector that would flag an end to the search cannot be deduced by brute force since any search program written to do that is indeterminate because of having no terminating information that will flag an end to the search. 
> 
> This cipher is without any shadow of doubt unbreakable by the unique definition in all the handbooks. 
> 
> Comment: 
> 
> I am only here for the cryptography. Ada is a langiuage that I was introduced to at a seat-of-learning.some thirty years ago. 
> 
> Austin O’Byrne

Hi all,
Take this post and apply it to the encryption/decryption model at https://www.aslec.uk 
to get a full understanding of how my cipher called ASLEC works.

ASLEC stands for Alternating Skew Line Encryption Cipher.
Its the very first vector cipher ever invented and is driven by my invention of Vector Factoring,

It will enrich your cryptographic standing in any company.

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#50568

On 10/25/2021 8:38 AM, austin obyrne wrote:
> The encryption/ decryption model to go with this text may be seen at https://www.aslec.uk.
> 
> In the model let’say (postulated  here for discussion purposes only)
> that Eve has overcome the insurmountable problem
> of finding the correct displacement of the plaintext in hand as
> the position vector of a point in R^3 relative to the universal
> zero at (0, 0, 0).
> 
> She still has lots more miracle hurdles to overcome and it is this
> that I want to discuss here.
[...]

What about being able to break a ciphertext without even worrying about 
trying to get at the secret key? Sure, getting at the secret key would 
be nice, but who cares if one can bust a ciphertext, right?

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#50569 — Re: ASLEC Cipher ? The Entanglement.

Chris M. Thomasson <chris.m.thomasson.1@gmail.com> wrote:
> On 10/25/2021 8:38 AM, austin obyrne wrote:
>> The encryption/ decryption model to go with this text may be seen at 
>> https://www.aslec.uk.
>> 
>> In the model let?say (postulated  here for discussion purposes only) 
>> that Eve has overcome the insurmountable problem of finding the 
>> correct displacement of the plaintext in hand as the position vector 
>> of a point in R^3 relative to the universal zero at (0, 0, 0).
>> 
>> She still has lots more miracle hurdles to overcome and it is this 
>> that I want to discuss here.
> [...]
> 
> What about being able to break a ciphertext without even worrying 
> about trying to get at the secret key?  Sure, getting at the secret 
> key would be nice, but who cares if one can bust a ciphertext, right?

Given that Austin's "displacement" is likely fixed per character, his 
cipher likely devolves into a "Caesar cipher" (substitute one letter 
for another, always the same substitution every time the same letter 
appears).

If so, then frequency analysis will allow one to pick it apart without 
ever knowing the key.

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#50572 — Re: ASLEC Cipher ? The Entanglement.

On Monday, 25 October 2021 at 21:09:42 UTC+1, Rich wrote:
> Given that Austin's "displacement" is likely fixed per character, his 
> cipher likely devolves into a "Caesar cipher" (substitute one letter 
> for another, always the same substitution every time the same letter 
> appears). 

Its not fixed per character, but it comes from a limited set of baked-in
constants that are mixed by a scramble algorithm that is trivial to undo.
This is key reuse.

The remnant is slightly harder work to undo, but for reasons I've
already explained the algorithm itself lets you know what works and
what doesn't. There are more baked-in constants here, also mixed
with the same bad algorithm, so if ever brand-new ones are deployed,
it will take some sigint to build up the new set from scratch. As key
reuse is again at play, this will be tedious but inevitable.

The algorithm may actually be secure if key reuse was removed, but
then it would just be a ridiculously expensive OTP. The amount of
key material that goes into encrypting each character is staggering.
Still, as it is all eventually reused by current design, it all becomes
known to Eve in time, and all she needs to do is brute force the
scrambling parameters, which can be done layer-by-layer, with
the algorithm helpfully letting you know when you have the right
answer.

The displacement parameters are found dimension-by-
dimension using a least-squares search.

The vector multiplication can be attacked by factoring each
dimension and keeping the common factors in each. This removes
a sizeable amount of the encrypting complexity. In the meanwhile,
a brute force search of the scrambling parameters will yield leads
in a not-entirely-ridiculous length of time, giving confirmatory
trial decryptions.

The notion that these vectors cannot be searched as the vector
space is infinite is laughable. The scrambling parameters are extremely
limited in scope, and integer-only, and are thus a simple brute-force
search can be very heavily shortened to only the parameter space
that is possible. This turns out to be very doable in practice. This may
be some days, but its not the 2^99.5 tries that is the current worst-case
that AES128 is known to take. If you could do a trial of AES every
nanosecond, all 2^99.5 trials would take you 2.8*10^10 millennia
(please check my arithmetic).

If real numbers were used, then the part of the algorithm that is
used to construct the cross product using (currently) integer
parameters fails due to floating-point imprecision on practical
computers. If computers had infinite precision, the algorithm
could possibly work, at the risk of infinitely long ciphergrams
and keys. Code that uses floating-point numbers has never
been offered.

M
-- 

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#50573 — Re: ASLEC Cipher ? The Entanglement.

On 26/10/2021 08:51, MM wrote:
> If you could do a trial of AES every
> nanosecond, all 2^99.5 trials would take you 2.8*10^10 millennia
> (please check my arithmetic).

Looks good. I checked the order of magnitude for you by working out the 
(easier) 2^99 and 2^100 calculations:

Lower bound: 2.008*10e10 millennia,
Upper bound: 4.017*10e10 millennia.

-- 
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

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#50577 — Re: ASLEC Cipher ? The Entanglement.

On Monday, 25 October 2021 at 21:09:42 UTC+1, Rich wrote:
> Chris M. Thomasson <chris.m.t...@gmail.com> wrote: 
> > On 10/25/2021 8:38 AM, austin obyrne wrote: 
> >> The encryption/ decryption model to go with this text may be seen at 
> >> https://www.aslec.uk. 
> >>
> >> In the model let?say (postulated here for discussion purposes only)
> >> that Eve has overcome the insurmountable problem of finding the 
> >> correct displacement of the plaintext in hand as the position vector 
> >> of a point in R^3 relative to the universal zero at (0, 0, 0). 
> >> 
> >> She still has lots more miracle hurdles to overcome and it is this 
> >> that I want to discuss here. 
> > [...] 
> > 
> > What about being able to break a ciphertext without even worrying 
> > about trying to get at the secret key? Sure, getting at the secret 
> > key would be nice, but who cares if one can bust a ciphertext, right?
> Given that Austin's "displacement" is likely fixed per character, his 
> cipher likely devolves into a "Caesar cipher" (substitute one letter 
> for another, always the same substitution every time the same letter 
> appears). 
> 
> If so, then frequency analysis will allow one to pick it apart without 
> ever knowing the key.

> Given that Austin's "displacement" is likely fixed per character, his
> cipher likely devolves into a "Caesar cipher" (substitute one letter
> for another, always the same substitution every time the same letter
> appears).

>If so, then frequency analysis will allow one to pick it apart without
>>ever knowing the key.

I do not believe that anybody in this group iis  naive enough to think that !

Maybe you would like a lengthy test sample of his work so that you could 
demonstarte your cryptanalyic skills???? 

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#50578 — Re: ASLEC Cipher ? The Entanglement.

On 27/10/2021 16:10, austin obyrne wrote:
> On Monday, 25 October 2021 at 21:09:42 UTC+1, Rich wrote:
>> Chris M. Thomasson <chris.m.t...@gmail.com> wrote:
>>> On 10/25/2021 8:38 AM, austin obyrne wrote:
>>>> The encryption/ decryption model to go with this text may be seen at
>>>> https://www.aslec.uk.
>>>>
>>>> In the model let?say (postulated here for discussion purposes only)
>>>> that Eve has overcome the insurmountable problem of finding the
>>>> correct displacement of the plaintext in hand as the position vector
>>>> of a point in R^3 relative to the universal zero at (0, 0, 0).
>>>>
>>>> She still has lots more miracle hurdles to overcome and it is this
>>>> that I want to discuss here.
>>> [...]
>>>
>>> What about being able to break a ciphertext without even worrying
>>> about trying to get at the secret key? Sure, getting at the secret
>>> key would be nice, but who cares if one can bust a ciphertext, right?
>> Given that Austin's "displacement" is likely fixed per character, his
>> cipher likely devolves into a "Caesar cipher" (substitute one letter
>> for another, always the same substitution every time the same letter
>> appears).
>>
>> If so, then frequency analysis will allow one to pick it apart without
>> ever knowing the key.
> 
>> Given that Austin's "displacement" is likely fixed per character, his
>> cipher likely devolves into a "Caesar cipher" (substitute one letter
>> for another, always the same substitution every time the same letter
>> appears).
> 
>> If so, then frequency analysis will allow one to pick it apart without
>>> ever knowing the key.
> 
> I do not believe that anybody in this group iis  naive enough to think that !

We have already demonstrated how to do this, right here in sci.crypt. If 
you think it cannot be done, it is you who are naïf.

> Maybe you would like a lengthy test sample of his work so that you could
> demonstarte your cryptanalyic skills????

The weakness of your cryptography lies in key re-use, inevitable because 
the key is so vast and so difficult to change (requiring as it does the 
editing and recompiling of source code), and your cipher is 
insufficiently robust to withstand the cryptanalysis of multiple 
messages encrypted with the same key).

And I doubt whether anyone here would trust you to produce multiple 
ciphertext messages all encrypted with the same key.

-- 
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

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#50595 — Re: ASLEC Cipher ? The Entanglement.

On Wednesday, 27 October 2021 at 16:10:36 UTC+1, austin obyrne wrote:
> Maybe you would like a lengthy test sample of his work so that you could 
> demonstarte your cryptanalyic skills????

That's already been done.

Ciphergrams produced by you are of little value as you do not provide
them with sufficient integrity.

You also fail to accept outcomes that don't suit you.

M
-- 

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#50596 — Re: ASLEC Cipher ? The Entanglement.

On 28/10/2021 09:08, MM wrote:
> On Wednesday, 27 October 2021 at 16:10:36 UTC+1, austin obyrne wrote:
>> Maybe you would like a lengthy test sample of his work so that you could
>> demonstarte your cryptanalyic skills????
> 
> That's already been done.

Thrice, methinks; by you, by Colin, by me. Possibly also wizz? Wizz will 
know.

> 
> Ciphergrams produced by you are of little value as you do not provide
> them with sufficient integrity.

That's why I produce my own. Not easy to do without knowing the 
plaintext but possible with a little farting around.

> You also fail to accept outcomes that don't suit you.

Which is basically all of them.

But let us give him credit for a temporary (and later rescinded) 
acceptance that I'd cracked Shuttlepads wide open.

-- 
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

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#50597 — Re: ASLEC Cipher ? The Entanglement.

On Thursday, 28 October 2021 at 09:20:02 UTC+1, Richard Heathfield wrote:

> But let us give him credit for a temporary (and later rescinded) 
> acceptance that I'd cracked Shuttlepads wide open.

He did finally throw Shuttlepads in the trash not long after I
posted a full decrypt.

He has no clue how I did it. The last I remember was him
thinking that hexdump was some kind of cracking tool.

I shortened that 70000-line monster to somewhere in
the very low 100’s of lines. I also fixed it to read the key
material from a file and to optionally handle any file
instead of his ASCII-only compatible default. He declared
it to be “The public face of Shuttlepads”, but made no
effort whatsoever to learn from it.

With tips from that, and minimal further help, he could
have improved ASLEC from something worthy only of
derision into at least “worth a serious look”.

Not even vaguely interested.

M
— 

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#50601 — Re: ASLEC Cipher ? The Entanglement.

MM <mrvmurray@gmail.com> wrote:
> On Thursday, 28 October 2021 at 09:20:02 UTC+1, Richard Heathfield wrote:
> 
>> But let us give him credit for a temporary (and later rescinded) 
>> acceptance that I'd cracked Shuttlepads wide open.
> 
> He did finally throw Shuttlepads in the trash not long after I
> posted a full decrypt.
> 
> He has no clue how I did it. The last I remember was him
> thinking that hexdump was some kind of cracking tool.

"Any sufficiently advanced technology is indistinguishable from magic" 
(Arthur C. Clarke).

And given Austin's extreme lack of computer knowledge, he would very 
much see hexdump as some form of "magic".

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#50603 — Re: ASLEC Cipher ? The Entanglement.

On 28/10/2021 10:51, MM wrote:
> The last I remember was him
> thinking that hexdump was some kind of cracking tool.

Heaven knows what he'd make of cat, dd, and grep!

-- 
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

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#50609 — Re: ASLEC Cipher ? The Entanglement.

On Thursday, 28 October 2021 at 15:10:30 UTC+1, Richard Heathfield wrote:
> Heaven knows what he'd make of cat, dd, and grep!

Or an Ada compiler.

M
-- 

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#50612 — Re: ASLEC Cipher ? The Entanglement.

On Monday, 25 October 2021 at 21:09:42 UTC+1, Rich wrote:
> Chris M. Thomasson <chris.m.t...@gmail.com> wrote: 
> > On 10/25/2021 8:38 AM, austin obyrne wrote: 
> >> The encryption/ decryption model to go with this text may be seen at 
> >> https://www.aslec.uk. 
> >>
> >> In the model let?say (postulated here for discussion purposes only)
> >> that Eve has overcome the insurmountable problem of finding the 
> >> correct displacement of the plaintext in hand as the position vector 
> >> of a point in R^3 relative to the universal zero at (0, 0, 0). 
> >> 
> >> She still has lots more miracle hurdles to overcome and it is this 
> >> that I want to discuss here. 
> > [...] 
> > 
> > What about being able to break a ciphertext without even worrying 
> > about trying to get at the secret key? Sure, getting at the secret 
> > key would be nice, but who cares if one can bust a ciphertext, right?
> Given that Austin's "displacement" is likely fixed per character, his 
> cipher likely devolves into a "Caesar cipher" (substitute one letter 
> for another, always the same substitution every time the same letter 
> appears). 
> 
> If so, then frequency analysis will allow one to pick it apart without 
> ever knowing the key.

> If so, then frequency analysis will allow one to pick it apart without
> ever knowing the key.

You'll get your chance very, very soon - let there be no excuses !

AOB

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#50613 — Re: ASLEC Cipher ? The Entanglement.

On Thursday, 28 October 2021 at 22:23:30 UTC+1, austin obyrne wrote:
> You'll get your chance very, very soon - let there be no excuses ! 

We know it's not a monoalphabetic cipher. Not even Shuttlepads was
monoalphabetic, and it was almost trivial to break. You still don't know
how that happened.

What is more important is that you don't know how your vector cipher
was busted.
See here: https://groups.google.com/g/sci.crypt/c/GmJrmCW-dTI/m/6mboBBcvBQAJ

Printing numbers from some unknown source proves nothing. If you
want your algorithm to be treated seriously, then take the time to
write it up properly. This means someone reading the description
has sufficient information to write a program to implement it.
This program should be able to reproduce test results sufficiently
well to be interchangeable with your reference implementation.

This is how it works in real life. It takes hard work. Life is tough.

Go ahead - post your numbers. Maybe they will get busted, maybe
they won't. It won't change the fact that your cipher is ALREADY
busted. Game over.

M
-- 

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#50614 — Re: ASLEC Cipher ? The Entanglement.

On 28/10/2021 22:23, austin obyrne wrote:
> You'll get your chance very, very soon - let there be no excuses !

We had our chance. We took it. We busted VC wide open. You have no 
excuse for refusing to face this fact. Yet you continue so to do.

-- 
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

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#50615 — Re: ASLEC Cipher ? The Entanglement.

On Thursday, October 28, 2021 at 6:56:20 PM UTC-5, Richard Heathfield wrote:
> On 28/10/2021 22:23, austin obyrne wrote: 
> > You'll get your chance very, very soon - let there be no excuses !
> We had our chance. We took it. We busted VC wide open. You have no 
> excuse for refusing to face this fact. Yet you continue so to do.
> -- 
> Richard Heathfield 
> Email: rjh at cpax dot org dot uk 
> "Usenet is a strange place" - dmr 29 July 1999 
> Sig line 4 vacant - apply within

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#50570

On 25.10.21 21:51, Chris M. Thomasson wrote:
> On 10/25/2021 8:38 AM, austin obyrne wrote:
>> The encryption/ decryption model to go with this text may be seen at 
>> https://www.aslec.uk.
>>
>> In the model let’say (postulated  here for discussion purposes only)
>> that Eve has overcome the insurmountable problem
>> of finding the correct displacement of the plaintext in hand as
>> the position vector of a point in R^3 relative to the universal
>> zero at (0, 0, 0).
>>
>> She still has lots more miracle hurdles to overcome and it is this
>> that I want to discuss here.
> [...]
> 
> What about being able to break a ciphertext without even worrying about 
> trying to get at the secret key? Sure, getting at the secret key would 
> be nice, but who cares if one can bust a ciphertext, right?
> 

Absolutely. It would be interesting to see, how good the algorithm is in 
obscuring the relations between plaintext and ciphertext. I would like 
to give it a try, but, sadly, there isn't any complete, comprehensible 
description of the cipher and no usable implementation to generate 
controlled plaintext-key-ciphertext groups for statistical analysis.

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