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Vector Cryptography - Parallelogram Resolution of the Ciphertext Vector.
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Austin Obyrne
Apr 25, 2013, 4:54:04 AM4/25/13
to
There is no hard sell about this cryptography. The intractability comes from a very well-known proof test in mathematics and science i.e. a vector must satisfy the parallelogram theory of addition before it is ascribed a vector. A corollary of this is that any vector may be taken as the diagonal of an arbitrary parallelogram in the basic parallelogram theory of vector addition and countless other new parallelograms constructed around it .
Then, there are *infinitely* many other parallelograms that may be constructed around this same diagonal using different pairs of vectors to do it.
In my cryptography Alice and Bob use one particular pair and the diagonal of their parallelogram is the public ciphertext vector. They alone know the pair in question and there is no implicit means (by cryptanalysis) that can find that pair. The particular pair of sides resides figuratively in the inner recessses of their minds and short of putting them on the rack to extract it there is no means of ascertaining which pair it is.
Clearly, this cryptography is unbreakable.
No amount of denigrating of me will change this truism.
That’s what I like most about mathematics – there’s no bull – there can’t be ever - why people try it on I will never understand.
It’s time to bite the bullet – go for it - it’s going to happen anyway no matter what anybody says - this is unbreakable cryptography.
- adacrypt
Then, there are *infinitely* many other parallelograms that may be constructed around this same diagonal using different pairs of vectors to do it.
In my cryptography Alice and Bob use one particular pair and the diagonal of their parallelogram is the public ciphertext vector. They alone know the pair in question and there is no implicit means (by cryptanalysis) that can find that pair. The particular pair of sides resides figuratively in the inner recessses of their minds and short of putting them on the rack to extract it there is no means of ascertaining which pair it is.
Clearly, this cryptography is unbreakable.
No amount of denigrating of me will change this truism.
That’s what I like most about mathematics – there’s no bull – there can’t be ever - why people try it on I will never understand.
It’s time to bite the bullet – go for it - it’s going to happen anyway no matter what anybody says - this is unbreakable cryptography.
- adacrypt
Austin Obyrne
Apr 25, 2013, 12:19:07 PM4/25/13
to
On Thursday, April 25, 2013 9:54:04 AM UTC+1, Austin Obyrne wrote:
> There is no hard sell about this cryptography. The intractability comes from a very well-known proof test in mathematics and science i.e. a vector must satisfy the parallelogram theory of addition before it is ascribed a vector. A corollary of this is that any vector may be taken as the diagonal of an arbitrary parallelogram in the basic parallelogram theory of vector addition and countless other new parallelograms constructed around it . Then, there are *infinitely* many other parallelograms that may be constructed around this same diagonal using different pairs of vectors to do it. In my cryptography Alice and Bob use one particular pair and the diagonal of their parallelogram is the public ciphertext vector. They alone know the pair in question and there is no implicit means (by cryptanalysis) that can find that pair. The particular pair of sides resides figuratively in the inner recessses of their minds and short of putting them on the rack to extract it there is no means of ascertaining which pair it is. Clearly, this cryptography is unbreakable. No amount of denigrating of me will change this truism. That’s what I like most about mathematics – there’s no bull – there can’t be ever - why people try it on I will never understand. It’s time to bite the bullet – go for it - it’s going to happen anyway no matter what anybody says - this is unbreakable cryptography. - adacrypt
How it unfolds.
> There is no hard sell about this cryptography. The intractability comes from a very well-known proof test in mathematics and science i.e. a vector must satisfy the parallelogram theory of addition before it is ascribed a vector. A corollary of this is that any vector may be taken as the diagonal of an arbitrary parallelogram in the basic parallelogram theory of vector addition and countless other new parallelograms constructed around it . Then, there are *infinitely* many other parallelograms that may be constructed around this same diagonal using different pairs of vectors to do it. In my cryptography Alice and Bob use one particular pair and the diagonal of their parallelogram is the public ciphertext vector. They alone know the pair in question and there is no implicit means (by cryptanalysis) that can find that pair. The particular pair of sides resides figuratively in the inner recessses of their minds and short of putting them on the rack to extract it there is no means of ascertaining which pair it is. Clearly, this cryptography is unbreakable. No amount of denigrating of me will change this truism. That’s what I like most about mathematics – there’s no bull – there can’t be ever - why people try it on I will never understand. It’s time to bite the bullet – go for it - it’s going to happen anyway no matter what anybody says - this is unbreakable cryptography. - adacrypt
Alice and Bob know two sides of the parallelogram. One of these is a position vector that (vectorially) represents Alice’s chosen number. That position vector is one key in a ‘cascading’ group of three (or possibly more) interacting keys.
The next key is the defining 'normal' vector of the plane that contains this position vector. Again this vector (N) is one of a set of normal vectors that are privy to Alice and Bob alone, it is key number 2.
Finally, key number 3 is an integer that is Alice’s numerical representation of the plaintext in her primary encryption alphabet.
These three keys are interdependent and are used in turn as the operands of the several steps in the decryption process.
While it is being said here that Alice and Bob alone are privy to the keysets and these keys reside figuratively in the inner recesses of their minds, in practice these keysets are stored in arrays of the computer program and are called in sequence at encryption/decryption time by both Alice and Bob as part of their mutual databases.
- adacrypt
Austin Obyrne
Apr 26, 2013, 8:10:56 AM4/26/13
to
On Thursday, April 25, 2013 9:54:04 AM UTC+1, Austin Obyrne wrote:
> There is no hard sell about this cryptography. The intractability comes from a very well-known proof test in mathematics and science i.e. a vector must satisfy the parallelogram theory of addition before it is ascribed a vector. A corollary of this is that any vector may be taken as the diagonal of an arbitrary parallelogram in the basic parallelogram theory of vector addition and countless other new parallelograms constructed around it . Then, there are *infinitely* many other parallelograms that may be constructed around this same diagonal using different pairs of vectors to do it. In my cryptography Alice and Bob use one particular pair and the diagonal of their parallelogram is the public ciphertext vector. They alone know the pair in question and there is no implicit means (by cryptanalysis) that can find that pair. The particular pair of sides resides figuratively in the inner recessses of their minds and short of putting them on the rack to extract it there is no means of ascertaining which pair it is. Clearly, this cryptography is unbreakable. No amount of denigrating of me will change this truism. That’s what I like most about mathematics – there’s no bull – there can’t be ever - why people try it on I will never understand. It’s time to bite the bullet – go for it - it’s going to happen anyway no matter what anybody says - this is unbreakable cryptography. - adacrypt
Vector Cryptography - Topics Arising in Applied Vector Methods.
1) Vector addition.
2) Vector subtraction.
3) The vector or cross product in vector multiplication.
4) The vector equation of a straight line.
5) Orthogonal Skew Lines.
6) The equation of a plane.
7) Factoring of a vector.
‘Vector’ in every case here means a physical vector being used to represent a physical quantity such as moment, acceleration etc.
Factoring of a vector (6 above) is my private contribution to mathematics. Should any reader wish to contest this new invention let him/her thoroughly disprove it publicly or desist from saying so.
Should any reader say this new piece of vector methodology by me is true but is ‘old hat’ then they should produce proof of it’s being in existence before now in mathematics.
Vector factoring was copyright recorded by me in London and Washingon USA some 20 years ago.
Complete newbies to vectors as a topic in mathematics will need a general foundation course and a customised foundation course in applied vector methods that is geared towards this cryptography.
- adacrypt.
Daniel Kruyt
May 4, 2013, 6:24:48 PM5/4/13
to
How does one factorize a vector? The only thing that comes up on Google is your website and some Brazillean thing, so I highly doubt this was publicly announced 20 years ago.
Austin Obyrne
May 5, 2013, 2:10:41 AM5/5/13
to
On Saturday, May 4, 2013 11:24:48 PM UTC+1, Daniel Kruyt wrote:
> How does one factorize a vector? The only thing that comes up on Google is your website and some Brazillean thing, so I highly doubt this was publicly announced 20 years ago.
You'll find it here among other related PDF's.
> How does one factorize a vector? The only thing that comes up on Google is your website and some Brazillean thing, so I highly doubt this was publicly announced 20 years ago.
http://www.adacryptpages.com/downloads/Vector-Cryptography-Combined-Package.pdf
This is a difficult topic to understand first time round - it was several years in the making and I did not have cryptography in mind - it is well covered in this PDF.
The application to cryptography is almost innocuous compared to the as-yet-unused applications in other sciences in my opinion but only time well tell that.
Best Wishes,
Austin O'Byrne.
Antti Louko
May 5, 2013, 3:42:23 AM5/5/13
to
readable by any other PDF reader than Acroread. His incompetence is
even greater in cryptography and mathematics.
Daniel Kruyt
May 5, 2013, 4:50:34 AM5/5/13
to
I'm well aware of his history and endless posting, but can one be blamed for being curious?
Your PDF is of little interest to me as it contains too much on-the-side information for me to find the placing of vector factorization. So, please describe it in full here, if you can.
Your PDF is of little interest to me as it contains too much on-the-side information for me to find the placing of vector factorization. So, please describe it in full here, if you can.
Daniel Kruyt
May 5, 2013, 5:24:15 AM5/5/13
to
Okay, having read quite a few of your posts, PDFs, etc, I have come to the
conclusion that you are a little delusional.
Each normal DOES, in fact, have an infinite number of parallelograms. But not in
deterministic cryptography being executed on current technology.
We are limited to a certain bit-depth of the numbers we work with. Therefore,
saying there are an infinite number of them is highly incorrect. Take, for
instance, the angle 89.999999999999999999999999999999999999999999999999999999999
degrees. This cannot be represented using vectors that have 32 bits assigned to
each of their elements ( or whatever you call them ). You have reached the
upper bound of what you can represent. And so, there is now a finite number
available. It is relatively basic mathematics.
That said, however, it seems a pretty novel idea. The only thing is that you
have to perform integer factorization, and at the bit-lengths that you will
need to give high-security to the algorithm, this will quickly become computationally infeasible. Your idea is kind of moot at this stage. :/
Kind regards,
Daniel R. Kruyt
conclusion that you are a little delusional.
Each normal DOES, in fact, have an infinite number of parallelograms. But not in
deterministic cryptography being executed on current technology.
We are limited to a certain bit-depth of the numbers we work with. Therefore,
saying there are an infinite number of them is highly incorrect. Take, for
instance, the angle 89.999999999999999999999999999999999999999999999999999999999
degrees. This cannot be represented using vectors that have 32 bits assigned to
each of their elements ( or whatever you call them ). You have reached the
upper bound of what you can represent. And so, there is now a finite number
available. It is relatively basic mathematics.
That said, however, it seems a pretty novel idea. The only thing is that you
have to perform integer factorization, and at the bit-lengths that you will
need to give high-security to the algorithm, this will quickly become computationally infeasible. Your idea is kind of moot at this stage. :/
Kind regards,
Daniel R. Kruyt
rossum
May 5, 2013, 6:36:11 AM5/5/13
to
http://www.psoriasissociety.ttsg.org/pdfs/Dunning-Kruger%20Effect.pdf
rossum
Austin Obyrne
May 5, 2013, 10:07:21 AM5/5/13
to
On Sunday, May 5, 2013 10:24:15 AM UTC+1, Daniel Kruyt wrote:
> Okay, having read quite a few of your posts, PDFs, etc, I have come to the conclusion that you are a little delusional. Each normal DOES, in fact, have an infinite number of parallelograms. But not in deterministic cryptography being executed on current technology. We are limited to a certain bit-depth of the numbers we work with. Therefore, saying there are an infinite number of them is highly incorrect. Take, for instance, the angle 89.999999999999999999999999999999999999999999999999999999999 degrees. This cannot be represented using vectors that have 32 bits assigned to each of their elements ( or whatever you call them ). You have reached the upper bound of what you can represent. And so, there is now a finite number available. It is relatively basic mathematics. That said, however, it seems a pretty novel idea. The only thing is that you have to perform integer factorization, and at the bit-lengths that you will need to give high-security to the algorithm, this will quickly become computationally infeasible. Your idea is kind of moot at this stage. :/ Kind regards, Daniel R. Kruyt
You haven't a clue.
> Okay, having read quite a few of your posts, PDFs, etc, I have come to the conclusion that you are a little delusional. Each normal DOES, in fact, have an infinite number of parallelograms. But not in deterministic cryptography being executed on current technology. We are limited to a certain bit-depth of the numbers we work with. Therefore, saying there are an infinite number of them is highly incorrect. Take, for instance, the angle 89.999999999999999999999999999999999999999999999999999999999 degrees. This cannot be represented using vectors that have 32 bits assigned to each of their elements ( or whatever you call them ). You have reached the upper bound of what you can represent. And so, there is now a finite number available. It is relatively basic mathematics. That said, however, it seems a pretty novel idea. The only thing is that you have to perform integer factorization, and at the bit-lengths that you will need to give high-security to the algorithm, this will quickly become computationally infeasible. Your idea is kind of moot at this stage. :/ Kind regards, Daniel R. Kruyt
You are digressing into irrelevant semantics and a misguided analysis (which is analytic instead of being the required *graphic analysis)- this crypto is intensely geometric and your demonstration of rounding error analysis doesn't come on the radar ever.
This not something that can be rationalised as an exercise in pure mathematics by some anlytic corrolary. You either know the mathemtics of the geometry of planes or you don't. If you did you would not be making such daft accusatins like my being delusional etc.
The mathematical proof that is supplied everywhere along the line by me is unassailable
Engineers and Applied mathematicians will have no problem with this algorithm.
There is nothing for me to say - this is all old hat basic vector arithmetic being used to validate a very transparent model that any person can understand.
The spatial model is my invention and this is the core intellectual platform of this cryptography.
Basically, I don't think you know enough about vector mathematics and you are of the belief that you can work it out instead by some replacement corollary in linear analysis.
It won't work.
Your'e not up to this crypto I'm afraid.
I wouldn't be saying this if I wasn't supremely sure of my ground.
- adacrypt
David Eather
May 5, 2013, 9:06:22 PM5/5/13
to
On Sun, 05 May 2013 19:24:15 +1000, Daniel Kruyt <daniel...@gmail.com>
wrote:
Why would you even try to factor this kind of cipher? The technique is
irrelevant to the cipher in question. Just imagine IBM claiming it's
design of DES is secure because it can't be factored. It makes no sense
except as a signpost to the ignorance of AO. It has been pointed out to
the OP that IF he could use infinite sized numbers than the cipher might
be secure, but it would also be unusable as there is no possible way to
encode or transmit such numbers (the average length of a number chosen
from the range 0 - infinity is infinite).
AO knows nothing about crypto or computers. There is no point telling him
about limits and resolution of 32-bit numbers - especially since he uses
16 bit numbers! Have a look at his crypto code. You have to edit his
source code to change the key because he doesn't know how to make it a
variable and get it from the user - even though he can load a string from
a user. Is such rigid and limited thinking even possible - you betcha!
wrote:
irrelevant to the cipher in question. Just imagine IBM claiming it's
design of DES is secure because it can't be factored. It makes no sense
except as a signpost to the ignorance of AO. It has been pointed out to
the OP that IF he could use infinite sized numbers than the cipher might
be secure, but it would also be unusable as there is no possible way to
encode or transmit such numbers (the average length of a number chosen
from the range 0 - infinity is infinite).
AO knows nothing about crypto or computers. There is no point telling him
about limits and resolution of 32-bit numbers - especially since he uses
16 bit numbers! Have a look at his crypto code. You have to edit his
source code to change the key because he doesn't know how to make it a
variable and get it from the user - even though he can load a string from
a user. Is such rigid and limited thinking even possible - you betcha!
Daniel Kruyt
May 20, 2013, 9:36:49 AM5/20/13
to
Austin, rounding error is not to be over looked. Saying that there is an infinite number of vectors that can be represented is incorrect when using finite data sets, no?
I plan on doing a short cryptanalysis of your cipher, I'll post it on this thread in a little while if you'll agree to it.
A "break" shall be defined as something requiring less time/memory than a brute-force, are you okay with that?
I plan on doing a short cryptanalysis of your cipher, I'll post it on this thread in a little while if you'll agree to it.
A "break" shall be defined as something requiring less time/memory than a brute-force, are you okay with that?
Paulo Marques
May 20, 2013, 10:00:15 AM5/20/13
to
Daniel Kruyt wrote:
> Austin, rounding error is not to be over looked. Saying that there is
> an infinite number of vectors that can be represented is incorrect when
> using finite data sets, no?
He's never going to understand that...
> Austin, rounding error is not to be over looked. Saying that there is
> an infinite number of vectors that can be represented is incorrect when
> using finite data sets, no?
> I plan on doing a short cryptanalysis of your cipher, I'll post it on
> this thread in a little while if you'll agree to it.
>
> A "break" shall be defined as something requiring less time/memory
> than a brute-force, are you okay with that?
the net for threads with me and Austin).
A while back I completely broke his cypher and it was not a theoretical
break. I actually recovered plaintext from an encrypted message.
There is no way Austin will ever realize is cypher is nonsense. The only
thing we can do is stop feeding the troll... :(
--
Paulo Marques - www.grupopie.com
"Nostalgia isn't what it used to be."
Austin Obyrne
May 20, 2013, 10:18:58 AM5/20/13
to
On Monday, May 20, 2013 2:36:49 PM UTC+1, Daniel Kruyt wrote:
> Austin, rounding error is not to be over looked. Saying that there is an infinite number of vectors that can be represented is incorrect when using finite data sets, no? I plan on doing a short cryptanalysis of your cipher, I'll post it on this thread in a little while if you'll agree to it. A "break" shall be defined as something requiring less time/memory than a brute-force, are you okay with that?
< Austin, rounding error is not to be over looked.
1) Rounding error cannot arise using integers only in the cross product
> Austin, rounding error is not to be over looked. Saying that there is an infinite number of vectors that can be represented is incorrect when using finite data sets, no? I plan on doing a short cryptanalysis of your cipher, I'll post it on this thread in a little while if you'll agree to it. A "break" shall be defined as something requiring less time/memory than a brute-force, are you okay with that?
< Austin, rounding error is not to be over looked.
2) < is incorrect when using using finite data sets,
Using 32–bit arithmetic in an Ada-programmed cipher.
There are three columns of coefficients that can each be populated with an entire set i.e. the integers 1 – 2147483647 (this being the max positive integer number that can be stored without overflow in the computer)
The number of permutations of ways in which a single column may be populated is therefore [2147483647(factorial)]
Combinations of each of the three columns with the other two increases the number of possible vectors that may be written when each column is populated in all its different ways.
The final vector space is {[ 2147483647 (factorial) ]}^3.
OK - I have held my hands up already and said this is not infinite per se (by the expected definition).
With the final vector space at {[ 2147483647 (factorial) ]}^3 however this number is incalculable and is out of this world in magnitude and albeit not by definition ‘infinite’ (as one would expect in pure academic number theory circles), it is to all intents and purposes ‘practically’ infinite to any cryptanalyst.
3) I plan on doing a short cryptanalysis of your cipher, I'll post it on this thread in a little while if you'll agree to it.
A "break" shall be defined as something requiring less time/memory than a brute-force, are you okay with that?
4) A "break" shall be defined as something requiring less time/memory than a brute-force, are you okay with that?
Anything that looks like an accurate decryption by whatever name you wish to call it would be disastrous to me but go ahead – I am open to everything – I have to be to earn credibility.
I am ready to send more ciphertext samples to anyone who wants them - PM was made to look stupid - he did not break anything.
- Austin
Robert Wessel
May 21, 2013, 4:22:37 AM5/21/13
to
the key was a single five (decimal) digit number. At other times it's
been two four digit numbers, two integers with a product less than
14250, and a few other things.
Austin Obyrne
May 21, 2013, 6:45:54 AM5/21/13
to
On Tuesday, May 21, 2013 9:22:37 AM UTC+1, robert...@yahoo.com wrote:
> On Mon, 20 May 2013 06:36:49 -0700 (PDT), Daniel Kruyt <daniel...@gmail.com> wrote: >Austin, rounding error is not to be over looked. Saying that there is an infinite number of vectors that can be represented is incorrect when using finite data sets, no? > >I plan on doing a short cryptanalysis of your cipher, I'll post it on this thread in a little while if you'll agree to it. > >A "break" shall be defined as something requiring less time/memory than a brute-force, are you okay with that? Why bother improving on brute force? Last I heard the secret part of the key was a single five (decimal) digit number. At other times it's been two four digit numbers, two integers with a product less than 14250, and a few other things.
Hi,
> On Mon, 20 May 2013 06:36:49 -0700 (PDT), Daniel Kruyt <daniel...@gmail.com> wrote: >Austin, rounding error is not to be over looked. Saying that there is an infinite number of vectors that can be represented is incorrect when using finite data sets, no? > >I plan on doing a short cryptanalysis of your cipher, I'll post it on this thread in a little while if you'll agree to it. > >A "break" shall be defined as something requiring less time/memory than a brute-force, are you okay with that? Why bother improving on brute force? Last I heard the secret part of the key was a single five (decimal) digit number. At other times it's been two four digit numbers, two integers with a product less than 14250, and a few other things.
It’s impossible for me to comment on isolated fragments of the whole like this one from you without muddying the waters even more.
A lot of readers are still thinking and arguing with a scalar mentality.
I can assure you that this crypto has hugely geometric connotations that are implemented by a collection of random keys that must all be known and used in what is essentially vector arithmetic.
Unless a reader is totally au fait with vector methods and plane geometry (this is what provides the essential substitution methodoly) there is no hope of me informing him by posts here in sci crypt and even then the academic people (outside of this group) that I have presented this stuff to are having difficulty getting their heads around it.
It needs a lot of one-to-one.
I am not being evasive when I say that it is impossible for me to convey a proper understanding by way of posts here – hence these supplements - but even that must stop now because a lot of people are simply not into vectors sufficiently to absorb what comes next and are still arguing along scalar lines which they understand but wrongly think will suffice.
Just out of interest - Are *you* totally au fait with vector methods applied to plane geometry?? – would like to know.
- adacrypt
Austin Obyrne
May 22, 2013, 3:41:20 AM5/22/13
to
On Tuesday, May 21, 2013 11:45:54 AM UTC+1, Austin Obyrne wrote:
> On Tuesday, May 21, 2013 9:22:37 AM UTC+1, robert...@yahoo.com wrote: > On Mon, 20 May 2013 06:36:49 -0700 (PDT), Daniel Kruyt <daniel...@gmail.com> wrote: >Austin, rounding error is not to be over looked. Saying that there is an infinite number of vectors that can be represented is incorrect when using finite data sets, no? > >I plan on doing a short cryptanalysis of your cipher, I'll post it on this thread in a little while if you'll agree to it. > >A "break" shall be defined as something requiring less time/memory than a brute-force, are you okay with that? Why bother improving on brute force? Last I heard the secret part of the key was a single five (decimal) digit number. At other times it's been two four digit numbers, two integers with a product less than 14250, and a few other things. Hi, It’s impossible for me to comment on isolated fragments of the whole like this one from you without muddying the waters even more. A lot of readers are still thinking and arguing with a scalar mentality. I can assure you that this crypto has hugely geometric connotations that are implemented by a collection of random keys that must all be known and used in what is essentially vector arithmetic. Unless a reader is totally au fait with vector methods and plane geometry (this is what provides the essential substitution methodoly) there is no hope of me informing him by posts here in sci crypt and even then the academic people (outside of this group) that I have presented this stuff to are having difficulty getting their heads around it. It needs a lot of one-to-one. I am not being evasive when I say that it is impossible for me to convey a proper understanding by way of posts here – hence these supplements - but even that must stop now because a lot of people are simply not into vectors sufficiently to absorb what comes next and are still arguing along scalar lines which they understand but wrongly think will suffice. Just out of interest - Are *you* totally au fait with vector methods applied to plane geometry?? – would like to know. - adacrypt
Supplement - 4
> On Tuesday, May 21, 2013 9:22:37 AM UTC+1, robert...@yahoo.com wrote: > On Mon, 20 May 2013 06:36:49 -0700 (PDT), Daniel Kruyt <daniel...@gmail.com> wrote: >Austin, rounding error is not to be over looked. Saying that there is an infinite number of vectors that can be represented is incorrect when using finite data sets, no? > >I plan on doing a short cryptanalysis of your cipher, I'll post it on this thread in a little while if you'll agree to it. > >A "break" shall be defined as something requiring less time/memory than a brute-force, are you okay with that? Why bother improving on brute force? Last I heard the secret part of the key was a single five (decimal) digit number. At other times it's been two four digit numbers, two integers with a product less than 14250, and a few other things. Hi, It’s impossible for me to comment on isolated fragments of the whole like this one from you without muddying the waters even more. A lot of readers are still thinking and arguing with a scalar mentality. I can assure you that this crypto has hugely geometric connotations that are implemented by a collection of random keys that must all be known and used in what is essentially vector arithmetic. Unless a reader is totally au fait with vector methods and plane geometry (this is what provides the essential substitution methodoly) there is no hope of me informing him by posts here in sci crypt and even then the academic people (outside of this group) that I have presented this stuff to are having difficulty getting their heads around it. It needs a lot of one-to-one. I am not being evasive when I say that it is impossible for me to convey a proper understanding by way of posts here – hence these supplements - but even that must stop now because a lot of people are simply not into vectors sufficiently to absorb what comes next and are still arguing along scalar lines which they understand but wrongly think will suffice. Just out of interest - Are *you* totally au fait with vector methods applied to plane geometry?? – would like to know. - adacrypt
This supplement is included here (as well as elsewhere) because it says a lot.
Number-theoretic cryptography took off in earnest when computers came into vogue in the 1970’s.
People thoughtlessly (why shouldn’t they? one may well ask) used the traditional data systems as the selection domain for raw data in cryptography because after all, there was no need to question its suitability. The fact that it had worked for them in billions of other cases since the year dot was probably so entrenched that it just didn’t bear questioning.
However, with hindsight, it can be seen today that this was a mistake for several reasons.
The raw data used for encryption transformations in cryptography should ideally come with a lot of *manageable disorder as an intrinsic property (call this innate ‘starting’ entropy if you like) such that a design cryptographer has only to invent minimal extra 'make-up' entropy to add to it by means of an encryption algorithm that will produce an unbreakable cipher.
That is not the case with the popular data sources that have been used however as the selection domains by cryptographers, namely, our number system , our natural alphabets and even the ASCII code that was created specially with computers in mind.
This data is so perfectly ordered that it has *zero entropy (as disorder) which means the cryptographer has a mountain to climb in providing the necessary extra entropy that will barely make it into ‘strong’ cryptography with little chance of making it totally unbreakable cryptography.
Nobody seems to have noticed this and cryptographers have instinctively filled the void with studiously contrived compensating complexity. Their entire modus operandi has been spent on thinking up (complexity-intensive) difficult-to-invert algorithms. These algorithms are horrendously difficult to invent and so far there has not been even one successful cipher that is totally immune to inversion by brute force (even AES isn’t completely safe). This is disregarding the impractical OTP.
Going down this wrong road by cryptographers in the 70's can be easily understood given that our number system has served mathematics perfectly for thousands of years even to the extent of a moon landing. Why should anybody stop to think that it might not be suitable for some purpose like cryptography and require changing – but that is what is required now – cryptographers must stop using these highly ordered data systems, to wit, the integer set, natural alphabets and information codes like ASCII are all taboo to cryptography in their unscrambled state.
Not mentioned so far and hugely important also is the fact that cryptanalysts have been given the same access to these same data systems that were being mutually used by both cryptographers and cryptanalysats all along, which is tantamount to giving the latter a head start in their nefarious trade of cryptanalysis. This is key information available to adversaries that is there simply for the picking up, an absolute and total folly by cryptographers is the only way to describe it but what was the alternative one may well ask.
Summarising.
The use of integers and alphanumeric data taken directly (i.e. without reversible substitution into any other form) from the highly ordered traditional sources as the selection domain by cryptographers for encryption transformations must stop forthwith because clearly, it is unsuitable to cryptography. Being used as the selection domain for the raw data that will be used in encryption transformations is tantamount to directly handing ‘giveaway’ information on a plate to adversaries.
It is amazing that this has gone on so long unnoticed.
The huge list of papers, past, present and future still being read in the establishment to day makes one wonder how much longer this fruitless search , i.e. for the essential compensating complexity that is required to mask the natural transparency of numbers and alphanumeric data, will go on when the only and proper solution is to stop using numbers from ordered number lines altogether.
Vector cryptography puts all of this right and guarantees unbreakable cryptography. No amount of computer power can break it.
There is no triumphalism or egoism on my part in what I say here – this is a mathematical and scientific fact – a part of the Universe.
I think most readers will see the sense of what I am saying here.
- adacrypt
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