info
Google Groups no longer supports new Usenet posts or subscriptions. Historical content remains viewable.
Dismiss
Triangulation Cryptography.
31 views
Skip to first unread message
austin obyrne
Sep 5, 2021, 6:10:13 AM9/5/21
to
This is an alternative exposition of my so-called
“ Vector Cryptography”. I am introducing it here
because it does not use the expression ‘change-of-origin ‘
description of one of the key elements of secrecy
anymore and should be an easier model to understand.
It uses terminology instead that emanates entirely from
everyday classroom mathematics..
In this cryptography the model is a triangle whose sides
are defined by three vectors. Let’s call these V1. V2.
And V3. They represent the sides of the triangle.
V1 (Veeone) is the entities' computed ciphertext as a vector,
V2(Veetwo) is an arbitrary vector (a key element) that is added to V1.
V3 (veethree) is the sum of V1 and V. It closes the triangle.
How it Works.
V3 is the public (apparent) ciphertext. V2 must be subtracted
from V3 to get V1 (the real ciphertext) in order for systematic
decyption to take place.
Eve.
let’s say Eve has successfully intercepted a batch of public
ciphertext. She must then guess V2 and subtract it from V3
so as to get V1, Her task is impossible but allowing that she
is successful in doing she must next guess another three
vectors that are tertiary key elements embedded in V1 before she can decrypt and realise the messagetext. Again, her task is impossible.
This cipher is the ultimate in strong cryptography.
I commend it to the group.
Austin O’Byrne
“ Vector Cryptography”. I am introducing it here
because it does not use the expression ‘change-of-origin ‘
description of one of the key elements of secrecy
anymore and should be an easier model to understand.
It uses terminology instead that emanates entirely from
everyday classroom mathematics..
In this cryptography the model is a triangle whose sides
are defined by three vectors. Let’s call these V1. V2.
And V3. They represent the sides of the triangle.
V1 (Veeone) is the entities' computed ciphertext as a vector,
V2(Veetwo) is an arbitrary vector (a key element) that is added to V1.
V3 (veethree) is the sum of V1 and V. It closes the triangle.
How it Works.
V3 is the public (apparent) ciphertext. V2 must be subtracted
from V3 to get V1 (the real ciphertext) in order for systematic
decyption to take place.
Eve.
let’s say Eve has successfully intercepted a batch of public
ciphertext. She must then guess V2 and subtract it from V3
so as to get V1, Her task is impossible but allowing that she
is successful in doing she must next guess another three
vectors that are tertiary key elements embedded in V1 before she can decrypt and realise the messagetext. Again, her task is impossible.
This cipher is the ultimate in strong cryptography.
I commend it to the group.
Austin O’Byrne
Richard Heathfield
Sep 5, 2021, 7:04:51 AM9/5/21
to
On 05/09/2021 11:10, austin obyrne wrote:
> let’s say Eve has successfully intercepted a batch of public
> ciphertext. She must then guess V2
So you claim. Eve may be smarter than that. History has shown us that
> let’s say Eve has successfully intercepted a batch of public
> ciphertext. She must then guess V2
Eves can be clever little buggers, and the cleverest of them have
invented techniques the Alices and Bobs of this world never dreamed of.
Maybe your Eve thinks in ways you haven't anticipated.
Do you remember when I stripped off your text rotation from Shuttlepads?
It never occurred to you when designing that rotation that anyone would
find a practical way to strip it off. You were thinking then, too, along
the lines that Eve must behave in a certain way. But she just doesn't.
--
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
Leo
Sep 5, 2021, 9:07:51 AM9/5/21
to
Hi Austin,
This is indeed an easier to understand model compared to your previous
posts. There are a few things you could clarify to turn it into a
short description that people could implement and play around with.
> In this cryptography the model is a triangle whose sides are defined by
> three vectors. Let’s call these V1. V2.
> And V3. They represent the sides of the triangle.
How many dimensions are these vectors? Does it matter for the security
of your cipher? How big would the numbers have to be for acceptable
security? 64-bits, 128-bits, more?
> V1 (Veeone) is the entities' computed ciphertext as a vector, V2(Veetwo)
> is an arbitrary vector (a key element) that is added to V1.
> V3 (veethree) is the sum of V1 and V. It closes the triangle.
These operations sound simple to implement. I'd be interested to
visualize what the vectors look like after they are added to the key
element.
>
> How it Works.
>
> V3 is the public (apparent) ciphertext. V2 must be subtracted from V3
> to get V1 (the real ciphertext) in order for systematic decyption to
> take place.
This is a little confusing to me. When you say "public ciphertext",
you mean what we send out over the wire when communicating right?
After we subtract the key element and get to V1, why is V1 still a
ciphertext? Does V1 already need to be encrypted? In that case, what
benefit does this triangular cryptography with a key element add?
--
Leo
This is indeed an easier to understand model compared to your previous
posts. There are a few things you could clarify to turn it into a
short description that people could implement and play around with.
> In this cryptography the model is a triangle whose sides are defined by
> three vectors. Let’s call these V1. V2.
> And V3. They represent the sides of the triangle.
of your cipher? How big would the numbers have to be for acceptable
security? 64-bits, 128-bits, more?
> V1 (Veeone) is the entities' computed ciphertext as a vector, V2(Veetwo)
> is an arbitrary vector (a key element) that is added to V1.
> V3 (veethree) is the sum of V1 and V. It closes the triangle.
visualize what the vectors look like after they are added to the key
element.
>
> How it Works.
>
> V3 is the public (apparent) ciphertext. V2 must be subtracted from V3
> to get V1 (the real ciphertext) in order for systematic decyption to
> take place.
you mean what we send out over the wire when communicating right?
After we subtract the key element and get to V1, why is V1 still a
ciphertext? Does V1 already need to be encrypted? In that case, what
benefit does this triangular cryptography with a key element add?
--
Leo
Max
Sep 5, 2021, 9:31:45 AM9/5/21
to
On 05.09.21 12:10, austin obyrne wrote:
> This is an alternative exposition of my so-called
> “ Vector Cryptography”. I am introducing it here
> because it does not use the expression ‘change-of-origin ‘
> description of one of the key elements of secrecy
> anymore and should be an easier model to understand.
> It uses terminology instead that emanates entirely from
> everyday classroom mathematics..
>
> In this cryptography the model is a triangle whose sides
> are defined by three vectors. Let’s call these V1. V2.
> And V3. They represent the sides of the triangle.
>
> V1 (Veeone) is the entities' computed ciphertext as a vector,
> V2(Veetwo) is an arbitrary vector (a key element) that is added to V1.
> V3 (veethree) is the sum of V1 and V. It closes the triangle.
>
> How it Works.
>
> V3 is the public (apparent) ciphertext. V2 must be subtracted
> from V3 to get V1 (the real ciphertext) in order for systematic
> decyption to take place.
You're basically describing a one-time-pad. So, if you use a new random
> This is an alternative exposition of my so-called
> “ Vector Cryptography”. I am introducing it here
> because it does not use the expression ‘change-of-origin ‘
> description of one of the key elements of secrecy
> anymore and should be an easier model to understand.
> It uses terminology instead that emanates entirely from
> everyday classroom mathematics..
>
> In this cryptography the model is a triangle whose sides
> are defined by three vectors. Let’s call these V1. V2.
> And V3. They represent the sides of the triangle.
>
> V1 (Veeone) is the entities' computed ciphertext as a vector,
> V2(Veetwo) is an arbitrary vector (a key element) that is added to V1.
> V3 (veethree) is the sum of V1 and V. It closes the triangle.
>
> How it Works.
>
> V3 is the public (apparent) ciphertext. V2 must be subtracted
> from V3 to get V1 (the real ciphertext) in order for systematic
> decyption to take place.
V2 for each V1, this is unbreakable, but as soon as you re-use a V2 for
a second V1, it's easily broken (just like the OTP is if a key is re-used).
So, unless your suggesting to use this scheme as an OTP, in which case
there is no need for encrypting the plaintext to V1 in the first place,
this scheme just adds a meaningless extra layer on the underlying
encryption scheme that generates V1 from the plaintext. How is this done
anyway? You skipped that part in your post.
Leo
Sep 5, 2021, 9:40:37 AM9/5/21
to
> You're basically describing a one-time-pad. So, if you use a new random
> V2 for each V1, this is unbreakable, but as soon as you re-use a V2 for
> a second V1, it's easily broken (just like the OTP is if a key is
> re-used).
>
> So, unless your suggesting to use this scheme as an OTP, in which case
> there is no need for encrypting the plaintext to V1 in the first place,
> this scheme just adds a meaningless extra layer on the underlying
> encryption scheme that generates V1 from the plaintext. How is this done
> anyway? You skipped that part in your post.
This seems to be the case. The only sensible way I thought it could be
> V2 for each V1, this is unbreakable, but as soon as you re-use a V2 for
> a second V1, it's easily broken (just like the OTP is if a key is
> re-used).
>
> So, unless your suggesting to use this scheme as an OTP, in which case
> there is no need for encrypting the plaintext to V1 in the first place,
> this scheme just adds a meaningless extra layer on the underlying
> encryption scheme that generates V1 from the plaintext. How is this done
> anyway? You skipped that part in your post.
implemented is with modular addition and subtraction, which is
basically a one time pad.
You could possibly use this with a key derivation function to make new
keys for each block based on a real key and counter, but then you
could just take that KDF and xor it with your block to use it as an
OTP.
This cipher feels like xor with extra steps.
For reference, here's some output I get with 8D vectors and 32-bit
values.
Plaintext b'Hello world, and all its people.'
V1 (plaintext): (1214606444, 1864398703, 1919706156, 543256164,
543255660, 543782003, 544236911, 1886151982)
V2 (key): (745555363, 1887577460, 174688883, 1987969223,
3504861512, 1186523942, 3680487385, 167527441)
V3 (ciphertext): (1960161807, 3751976163, 2094395039, 2531225387,
4048117172, 1730305945, 4224724296, 2053679423)
V3 - V2 (plaintext): (1214606444, 1864398703, 1919706156, 543256164,
543255660, 543782003, 544236911, 1886151982)
V3 - V2: b'Hello world, and all its people.'
--
Leo
Chris M. Thomasson
Sep 5, 2021, 3:28:25 PM9/5/21
to
some time tomorrow to create a program that generates graphics that can
help visualize the process. Think of visualizing the steps that comprise
the act of encrypting a plaintext into ciphertext, and decrypting a
ciphertext into plaintext.
http://fractallife247.com/test/hmac_cipher/ver_0_0_0_1?ct_hmac_cipher=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
Chris M. Thomasson
Sep 5, 2021, 3:30:57 PM9/5/21
to
On 9/5/2021 3:10 AM, austin obyrne wrote:
help you visually plot your work on the screen. And, even animate it!
http://fractallife247.com/test/vfield/
Look at the code.
Rich
Sep 6, 2021, 11:51:02 AM9/6/21
to
>> let?s say Eve has successfully intercepted a batch of public
forget that.
> It never occurred to you when designing that rotation that anyone would
> find a practical way to strip it off.
If memory serves, he also espoused that his "shuttle" (what he termed
the rotation he was doing) was such that Eve would be unable to deduce
the original data. You, of course, proved that opinion wrong in short
order.
> You were thinking then, too, along the lines that Eve must behave in
> a certain way. But she just doesn't.
He is constrained by his terrible lack of knowledge of anything crypto,
such that he can't even imagine that Eve might do something he's never
imagined. I.e., if he can't imagine it, he believes "it" to be
impossible to do.
His description sounds like his new 'triangle crypto' idea may be
nothing more than this:
Vk (vector key)
Vp (vector plaintext)
Vc (vector ciphertext)
Vc(i) = Vp(i) + Vk
And if Vk is as above, fixed (and, this is AOB, Vk is likely a
constant), this is just an addition of the same Vk to every Vp. Which
looks way too much like a Caesar cipher to be secure. The only
difference is "vectors" instead of characters.
>> ciphertext. She must then guess V2
>
> So you claim. Eve may be smarter than that. History has shown us that
> Eves can be clever little buggers, and the cleverest of them have
> invented techniques the Alices and Bobs of this world never dreamed of.
> Maybe your Eve thinks in ways you haven't anticipated.
>
> Do you remember when I stripped off your text rotation from Shuttlepads?
He very well may have forgotten. Or else he would very much like to
>
> So you claim. Eve may be smarter than that. History has shown us that
> Eves can be clever little buggers, and the cleverest of them have
> invented techniques the Alices and Bobs of this world never dreamed of.
> Maybe your Eve thinks in ways you haven't anticipated.
>
> Do you remember when I stripped off your text rotation from Shuttlepads?
forget that.
> It never occurred to you when designing that rotation that anyone would
> find a practical way to strip it off.
the rotation he was doing) was such that Eve would be unable to deduce
the original data. You, of course, proved that opinion wrong in short
order.
> You were thinking then, too, along the lines that Eve must behave in
> a certain way. But she just doesn't.
such that he can't even imagine that Eve might do something he's never
imagined. I.e., if he can't imagine it, he believes "it" to be
impossible to do.
His description sounds like his new 'triangle crypto' idea may be
nothing more than this:
Vk (vector key)
Vp (vector plaintext)
Vc (vector ciphertext)
Vc(i) = Vp(i) + Vk
And if Vk is as above, fixed (and, this is AOB, Vk is likely a
constant), this is just an addition of the same Vk to every Vp. Which
looks way too much like a Caesar cipher to be secure. The only
difference is "vectors" instead of characters.
Richard Heathfield
Sep 6, 2021, 1:02:29 PM9/6/21
to
On 06/09/2021 16:50, Rich wrote:
> His description sounds like his new 'triangle crypto' idea may be
> nothing more than this:
>
> Vk (vector key)
> Vp (vector plaintext)
> Vc (vector ciphertext)
>
> Vc(i) = Vp(i) + Vk
>
> And if Vk is as above, fixed (and, this is AOB, Vk is likely a
> constant), this is just an addition of the same Vk to every Vp. Which
> looks way too much like a Caesar cipher to be secure. The only
> difference is "vectors" instead of characters.
I've lost count of how many times sci.crypt has broken this.
> His description sounds like his new 'triangle crypto' idea may be
> nothing more than this:
>
> Vk (vector key)
> Vp (vector plaintext)
> Vc (vector ciphertext)
>
> Vc(i) = Vp(i) + Vk
>
> And if Vk is as above, fixed (and, this is AOB, Vk is likely a
> constant), this is just an addition of the same Vk to every Vp. Which
> looks way too much like a Caesar cipher to be secure. The only
> difference is "vectors" instead of characters.
MM
Sep 6, 2021, 3:12:14 PM9/6/21
to
On Monday, 6 September 2021 at 18:02:29 UTC+1, Richard Heathfield wrote:
> I've lost count of how many times sci.crypt has broken this.
All you need is once:
> I've lost count of how many times sci.crypt has broken this.
https://groups.google.com/g/sci.crypt/c/GmJrmCW-dTI/m/6mboBBcvBQAJ
M
--
0 new messages