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Displacement Cryptography.
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adacrypt
Feb 16, 2012, 11:56:39 AM2/16/12
to
I don’t mind any criticism that may come from repeating this
description of new cryptography because it is so important to have its
potential fully understood.
Variously called “Displacement Cryptography” because that is what it
is compared with traditional scalar number-line cryptography and
“Vector Cryptography” because it uses vector methodology, it is also
being called “Skew Line encryptions” because the graphic model is a
pair of orthogonal skew lines.
A salient feature of this cryptography is that the internal algorithm
can be manipulated so that the external ciphertext string that ensues
is fashioned into a truly random additional key. This is unheard of
in any previous cryptography.
The upshot of using this new cryptography is,
1) Ciphertext Only Attack:
The elements of ciphertext are position vectors of 7 or 8 digits in
decimal positive integers as coefficients of (i,j,k). Each element of
ciphertext represents one plaintext. Each elemental ciphertext is the
resultant vector of two component vectors to which Alice and Bob alone
are privy.
In general, any position vector like these can be resolved into an
infinite number of components and in this case of Alice and Bob
encrypting, it can be resolved into an infinite number of pairs of
component vectors, only they alone know which particular pair is the
right one that will decrypt properly. Given unlimited computer power
and the whole of eternity to do so, no cryptanalyst can ever decrypt
this ciphertext by inverting the encryption algorithm. This form of
attack is completely foiled therefore for evermore.
2) Implicit Numerical Attack.
The ciphertext is a string of hugely disparate displacements being
represented by position vectors and as such they are definitively
discontinuous data as mathematical operands in any calculations that
assumes correlations between the separate elements. This form of
attack is simply a non-starter.
3) Implicit Statistical Attack.
This is the new jewel in the crown of the cryptographer in this
cryptology. The ciphertext string is a truly random keyset that
defeats any attempt at mapping the ciphertext directly to the
plaintext it represents. Such an attack is reminiscent of the Kasiski/
Babbage attack of yore – it seeks to circumvent the two previous
attack forms (1 and 2 above) by going straight for the jugular. When
the ciphertext string is a proven random key there is no hope for this
attack. The internal encryption algorithm churns out external truly
random ciphertext strings every time that are hugely invulnerable to
any attack.
Digressing.
There are many possibilities in vector methodology apart from this
instance of a displacement cipher here by me for cryptographers who
are interested in creating random keysets for their particular cipher
design what ever it might be.
- adacrypt
description of new cryptography because it is so important to have its
potential fully understood.
Variously called “Displacement Cryptography” because that is what it
is compared with traditional scalar number-line cryptography and
“Vector Cryptography” because it uses vector methodology, it is also
being called “Skew Line encryptions” because the graphic model is a
pair of orthogonal skew lines.
A salient feature of this cryptography is that the internal algorithm
can be manipulated so that the external ciphertext string that ensues
is fashioned into a truly random additional key. This is unheard of
in any previous cryptography.
The upshot of using this new cryptography is,
1) Ciphertext Only Attack:
The elements of ciphertext are position vectors of 7 or 8 digits in
decimal positive integers as coefficients of (i,j,k). Each element of
ciphertext represents one plaintext. Each elemental ciphertext is the
resultant vector of two component vectors to which Alice and Bob alone
are privy.
In general, any position vector like these can be resolved into an
infinite number of components and in this case of Alice and Bob
encrypting, it can be resolved into an infinite number of pairs of
component vectors, only they alone know which particular pair is the
right one that will decrypt properly. Given unlimited computer power
and the whole of eternity to do so, no cryptanalyst can ever decrypt
this ciphertext by inverting the encryption algorithm. This form of
attack is completely foiled therefore for evermore.
2) Implicit Numerical Attack.
The ciphertext is a string of hugely disparate displacements being
represented by position vectors and as such they are definitively
discontinuous data as mathematical operands in any calculations that
assumes correlations between the separate elements. This form of
attack is simply a non-starter.
3) Implicit Statistical Attack.
This is the new jewel in the crown of the cryptographer in this
cryptology. The ciphertext string is a truly random keyset that
defeats any attempt at mapping the ciphertext directly to the
plaintext it represents. Such an attack is reminiscent of the Kasiski/
Babbage attack of yore – it seeks to circumvent the two previous
attack forms (1 and 2 above) by going straight for the jugular. When
the ciphertext string is a proven random key there is no hope for this
attack. The internal encryption algorithm churns out external truly
random ciphertext strings every time that are hugely invulnerable to
any attack.
Digressing.
There are many possibilities in vector methodology apart from this
instance of a displacement cipher here by me for cryptographers who
are interested in creating random keysets for their particular cipher
design what ever it might be.
- adacrypt
Karl-Uwe Frank
Feb 16, 2012, 12:04:48 PM2/16/12
to
I would like to send you a reminder, just in case you are not aware of
my question
under the thread topic **The Scalable Key Cipher – A Modernised OTP for
the Brave.**
http://groups.google.com/group/sci.crypt/msg/ed48da95a3f25e92
Cheers,
Karl-Uwe
my question
under the thread topic **The Scalable Key Cipher – A Modernised OTP for
the Brave.**
http://groups.google.com/group/sci.crypt/msg/ed48da95a3f25e92
Cheers,
Karl-Uwe
adacrypt
Feb 16, 2012, 1:25:57 PM2/16/12
to
then unthinkingly post it to the link - I do't know if i swill open
anywhere and I haven't the time or effort to re-do it.
Hang fire to another time and I will tryand do it again - a PDF is on
offer meanwhile that shows it all,
- adacrypt
adacrypt
Feb 16, 2012, 2:43:14 PM2/16/12
to
On Feb 16, 5:04 pm, Karl-Uwe Frank <karl.fr...@freecx.co.uk> wrote:
sends this code to Bob in number-theoretic ciphertext as follows.
POSTA DI FALCONE
58017 57956 57944 58013
57992 57967 57957 58056
57905 58020 57959 57871
58537 58531 58051 58248
This 100% random ciphertext.
- adacrypt
Mark Murray
Feb 16, 2012, 3:11:03 PM2/16/12
to
M
--
Mark "No Nickname" Murray
Notable nebbish, extreme generalist.
adacrypt
Feb 16, 2012, 3:48:14 PM2/16/12
to
6779575 8414058 5710770 8343172 5829092 9516599
6481753 6514610 6210888 9638522 5900245 5712656
4703257 2416107 8415651 6134280 6733446 9218460
2917424 5820305 5717520 7145397 2426573 4717796
8132780 5642881 8716960 9338208 7547381 6219394
6400589 9823603 5714888 3444102 5722021 9116621
6944013 5635063 5715425 1659091 9531550 8415322
6252880 3536138 7515535 8377235 5721386 6514824
Thos ciphertext is 100% random also.
- adacrypt
Gene and Debbie Styer
Feb 16, 2012, 3:40:37 PM2/16/12
to
On Feb 16, 2:43 pm, adacrypt <austin.oby...@hotmail.com> wrote:
>
> 58017 57956 57944 58013
> 57992 57967 57957 58056
> 57905 58020 57959 57871
> 58537 58531 58051 58248
>
> This 100% random ciphertext.
>
> - adacrypt
I don't know what definition of "random" you are using, but when I
>
> 58017 57956 57944 58013
> 57992 57967 57957 58056
> 57905 58020 57959 57871
> 58537 58531 58051 58248
>
> This 100% random ciphertext.
>
> - adacrypt
look at the distribution of digits I come up with:
First digit: All 5's - Not random
Second digit: Half 7, half 8 - Ok if you expect only 7's and 8's, but
not otherwise
Third digit: Seven 9's, five 0's, and other stuff - Not random
Fourth digit: Five 5's, and other stuff - Not random
Last digit: Four 7's, three 1's, and other stuff - I was expecting
this digit to be random by sheer chance, but even this fails.
---
Switching to a topic on a different thread - you talk about the
possibility of future quantum computers being used to attack AES, but
don't seem to realize that the same approach can be used on _ANY_
algorithm where the key is shorter than the message - and your
algorithm is not an exception.
Bruce Stephens
Feb 16, 2012, 4:24:25 PM2/16/12
to
Gene and Debbie Styer <Eugene...@eku.edu> writes:
[...]
> I don't know what definition of "random" you are using,
He seems to mean just non-repeating (probably based on a typo in Simon
Singh's "The Code Book").
[...]
> Switching to a topic on a different thread - you talk about the
> possibility of future quantum computers being used to attack AES, but
> don't seem to realize that the same approach can be used on _ANY_
> algorithm where the key is shorter than the message - and your
> algorithm is not an exception.
IIUC (that is, if wikipedia's page is correct) the generic attack halves
the effective key size. So even if (and this seems speculative) we ever
get quantum computers powerful enough, AES-256 will still have 128-bit
security.
Presuming adacrypt's ciphers use their keyspace to the full (and nobody
thinks that they do) that would still leave them with stupidly large
effective strengths.
[...]
> I don't know what definition of "random" you are using,
Singh's "The Code Book").
[...]
> Switching to a topic on a different thread - you talk about the
> possibility of future quantum computers being used to attack AES, but
> don't seem to realize that the same approach can be used on _ANY_
> algorithm where the key is shorter than the message - and your
> algorithm is not an exception.
the effective key size. So even if (and this seems speculative) we ever
get quantum computers powerful enough, AES-256 will still have 128-bit
security.
Presuming adacrypt's ciphers use their keyspace to the full (and nobody
thinks that they do) that would still leave them with stupidly large
effective strengths.
Karl-Uwe Frank
Feb 16, 2012, 4:57:24 PM2/16/12
to
T E M P L A R S
57727 57845 57711 57725
58882 57138 57606 57557
encrypted as number-theoretic ciphertext.
Can you please confirm that my calculation is correct?
Cheers,
Karl-Uwe
Robert Wessel
Feb 16, 2012, 5:15:30 PM2/16/12
to
to be *far* from random.
Mark Murray
Feb 16, 2012, 5:38:09 PM2/16/12
to
On 16/02/2012 22:15, Robert Wessel wrote:
>> 6779575 8414058 5710770 8343172 5829092 9516599
>> 6481753 6514610 6210888 9638522 5900245 5712656
>> 4703257 2416107 8415651 6134280 6733446 9218460
>> 2917424 5820305 5717520 7145397 2426573 4717796
>> 8132780 5642881 8716960 9338208 7547381 6219394
>> 6400589 9823603 5714888 3444102 5722021 9116621
>> 6944013 5635063 5715425 1659091 9531550 8415322
>> 6252880 3536138 7515535 8377235 5721386 6514824
>>
>> Thos ciphertext is 100% random also.
>
>
> What's your defintion of random? By simple inspection, these appear
> to be *far* from random.
His definition is "no two numbers are the same".
>> 6779575 8414058 5710770 8343172 5829092 9516599
>> 6481753 6514610 6210888 9638522 5900245 5712656
>> 4703257 2416107 8415651 6134280 6733446 9218460
>> 2917424 5820305 5717520 7145397 2426573 4717796
>> 8132780 5642881 8716960 9338208 7547381 6219394
>> 6400589 9823603 5714888 3444102 5722021 9116621
>> 6944013 5635063 5715425 1659091 9531550 8415322
>> 6252880 3536138 7515535 8377235 5721386 6514824
>>
>> Thos ciphertext is 100% random also.
>
>
> What's your defintion of random? By simple inspection, these appear
> to be *far* from random.
A years supply of fawning admiration from me (or a pint at the local,
your choice!) if you manage to shift this kookery.
adacrypt
Feb 16, 2012, 6:18:48 PM2/16/12
to
>
> - Show quoted text -
58011 57999 57960 58061
57974 57967 57949 58095
57869 58052 57959 57907
58522 58574 58046 58281
This checks out accurately by decryption.
- adacrypt
adacrypt
Feb 16, 2012, 6:33:35 PM2/16/12
to
>
> - Show quoted text -
I encrypt everything - spaces and all - hence sixteen items of
> - Show quoted text -
ciphertext to your eight - I also treat capitals as different to lower
case - i.e they are separate characters in the ASCII subset of
writables that I use as my domain of both keys and plaintext - maybe
that explains the difference. - I don't use keywords as keys.
The scarmbled set of 95 ASCII cgaraters is my instantaneous 'key' -
there are 95 factorial ways of scrambling the domain => no shortage of
fresh keys.
- adacrypt
Karl-Uwe Frank
Feb 16, 2012, 6:39:05 PM2/16/12
to
TEMPLARS
57727 57845 57711 57725
58882 57138 57606 57557
Cheers,
Karl-Uwe
adacrypt
Feb 16, 2012, 7:01:33 PM2/16/12
to
denary representation in ASCII as the key if it decrypts ok - its
extra work using a key word every time but each to his own - what is
important is that the algorithm works for the whole of ASCII and
indeed the whole of Unicode.
The security is provided by the OTP principles and a random internal
key but in todays world of computer driven number theoretic
cryptography the randomness of the ciphertext string has to be taken
into account also (in my belief at least) to prevent a stastical
attack on the ciphertext - this never arose inthe historic OTP because
obviously it was alpahabetic only.
Random means equal probability implied by no repeats in any set of
data - it translates to the uncertainty it causes in the mind of the
adversary who knows all else usually except the key and the state of
order of the key - random means he can only guess that the message
text is some string of plaintexts in a very large possibility space of
such plaintexts.
- adcrypt
adacrypt
Feb 16, 2012, 7:36:44 PM2/16/12
to
mine because we have no mutual arrangements in vogue and are out of
sync.
In a real world case Alice and Bob have a mutually synchronised
database that they use and they are fully agreed on the modus operandi
that they will use at all times so that there are no misunderstanding
such as now about encryption all spaces for instance, etc.
This requires a one-off initial secure delivery of the copy database
from Alice to Bob which might be seen as a nuisance but once its done
then that's it for all time - as long as their mutual database is kept
secure from theft they can have secure communications for evermore -
just like Mr Kerchoff said.
- adacrypt
Karl-Uwe Frank
Feb 16, 2012, 9:25:30 PM2/16/12
to
I ask, because after I synced myself with your encryption I wonder why
you submitted both messages containing errors?
Below please find them correctly with your errors marked
POSTA DI FALCONE
58017 57956 57946 (**) 58013 (** error 57944 **)
57986 (**) 57967 57957 58056 (** error 57992 **)
57905 58022 (**) 57959 57871 (** error 58020 **)
58537 58531 58051 58248
T E M P L A R S
57974 57967 57945 (**) 58095 (** error 57949 **)
57869 58052 57959 57907
58522 58574 58046 58281
In a real world such mistakes may be very problematic!
58522 58574 58046 58281
Cheers,
Karl-Uwe
Karl-Uwe Frank
Feb 16, 2012, 9:32:06 PM2/16/12
to
Uhps, don't know what happend with the spaces between each number
Here again the correct messages
POSTA DI FALCONE
58017 57956 57946 58013 (** error 57944 **)
57986 57967 57957 58056 (** error 57992 **)
57905 58022 57959 57871 (** error 58020 **)
57974 57967 57945 58095 (** error 57949 **)
POSTA DI FALCONE
58017 57956 57944 58013
57992 57967 57957 58056
57905 58020 57959 57871
58537 58531 58051 58248
Karl-Uwe
Here again the correct messages
POSTA DI FALCONE
58017 57956 57946 58013 (** error 57944 **)
57986 57967 57957 58056 (** error 57992 **)
57905 58022 57959 57871 (** error 58020 **)
58537 58531 58051 58248
T E M P L A R S
58013 57999 57960 58061 (** error 58011 **)
T E M P L A R S
57974 57967 57945 58095 (** error 57949 **)
57869 58052 57959 57907
58522 58574 58046 58281
and below your's containing the errors
58522 58574 58046 58281
POSTA DI FALCONE
58017 57956 57944 58013
57992 57967 57957 58056
57905 58020 57959 57871
58537 58531 58051 58248
T E M P L A R S
58011 57999 57960 58061
57974 57967 57949 58095
57869 58052 57959 57907
58522 58574 58046 58281
Cheers,
57974 57967 57949 58095
57869 58052 57959 57907
58522 58574 58046 58281
Karl-Uwe
adacrypt
Feb 17, 2012, 12:54:52 AM2/17/12
to
POSTA DI FALCONE
115597 115524 115486 115594
115520 115423 115501 115597
115442 115632 115471 115371
116036 116036 115601 115805
Why should the ciphertext from two different sources be identical -
that would be very transparent surely
- adacrypt
adacrypt
Feb 17, 2012, 1:22:39 AM2/17/12
to
>
> - Show quoted text -
Better use this on second thoughts - I didn't check the previous set
> - Show quoted text -
of ciphertext for decrypting for
POSTA DI FALCONE
115587 115540 115468 115543
115528 115361 115491 115560
115471 115608 115501 115409
116051 116033 115601 115788
Any further sets I send you for "POSTA DI FALCONE"
will not coincide with this or anything you generate because that's
the nature of the beast.
It will decrypt fine at Bob' s end which is the all important thing.
- adacrypt
adacrypt
Feb 17, 2012, 1:35:15 AM2/17/12
to
cryptography. I am not prepared to keep up a running defenece of what
you perceive to be *errors.
Good luck .
- adacrypt
Mark Murray
Feb 17, 2012, 4:17:53 AM2/17/12
to
On 17/02/2012 06:35, adacrypt wrote:
> In fairness,I think I have done enough now in tabling this new
> cryptography. I am not prepared to keep up a running defenece of what
> you perceive to be *errors.
For something you are so desperate to have accepted, you sure are going
> In fairness,I think I have done enough now in tabling this new
> cryptography. I am not prepared to keep up a running defenece of what
> you perceive to be *errors.
the extra mile in sorting out the niggles.
adacrypt
Feb 17, 2012, 4:55:40 AM2/17/12
to
behind this cryptography. I don't take kindly to spontaneous ill-
founded criticism unless I think it has been well thought through
first - I dislike people just pushing my buttons experimentally.
This ciphertext always travels unsigned,
Example,
n = -17
The plaintext message is "A short utility file. The rain in Spain is
mainly in the plains." (no inverted commas in message).
This is the ciphertext.
231600 231433 231319 231511 231494 231209
231339 231643 231256 231625 231374 231103
233097 233115 231697 232233 232143 232224
232327 231889 232973 232828 231927 232290
232203 232790 231843 232361 232204 232808
231747 232854 232139 232928 231750 232270
232792 231951 232396 232126 231703 232815
232363 232569 231786 232148 232429 231908
232311 231800 232866 232739 232634 231876
232457 232119 232565 232592 231681 232573
232171
This ciphertext is 100% random.
- adacrypt
adacrypt
Feb 17, 2012, 7:28:14 AM2/17/12
to
there is no point in sending a repetitious minus sign along with
integers - my program takes the minus sign off and puts it back on
when needed - a cryptanalyst can't tell that of course so it
represents a mild extra entanglement.
- adacrypt
Karl-Uwe Frank
Feb 17, 2012, 7:42:34 AM2/17/12
to
ciphertext *again*???
Didn't you realise that you send me T E M P L A R S back encrypted in
a correct way in comparison to my try???
>> According to your cryptographic scheme this is
>> T E M P L A R S
>>
>>
>> 57727 57845 57711 57725
>> 58882 57138 57606 57557
>>
>> encrypted as number-theoretic ciphertext.
>>
>> Can you please confirm that my calculation is correct?
>> 58882 57138 57606 57557
>>
>> encrypted as number-theoretic ciphertext.
>>
>> Can you please confirm that my calculation is correct?
> 58011 57999 57960 58061
> 57974 57967 57949 58095
> 57869 58052 57959 57907
> 58522 58574 58046 58281
> 57974 57967 57949 58095
> 57869 58052 57959 57907
> 58522 58574 58046 58281
> This checks out accurately by decryption.
>
>
> I encrypt everything - spaces and all - hence sixteen items of
> ciphertext to your eight - I also treat capitals as different to lower
> case - i.e they are separate characters in the ASCII subset of
> writables that I use as my domain of both keys and plaintext - maybe
> that explains the difference. - I don't use keywords as keys.
>
Therefore let me answer
> ciphertext to your eight - I also treat capitals as different to lower
> case - i.e they are separate characters in the ASCII subset of
> writables that I use as my domain of both keys and plaintext - maybe
> that explains the difference. - I don't use keywords as keys.
>
58013 57963 57956 58014
58017 57928 57942 58095
57878 58010 57959 57867
Cheers,
Karl-Uwe
adacrypt
Feb 17, 2012, 8:29:46 AM2/17/12
to
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
My Cipher Invention seems to be working for you - I am pleased about
> - Show quoted text -
that - it is in the public domain as "Dilating Key Encryptions ".
http://www.adacrypt.com/introduction.html
I have invited users there to modify it as they wish - I suggest you
do this to your heart's contentent.
I cannot comment on what you are finding by your own methods at
present - I repeat I am pleased that you are succeeding.
To me its "job done" in the creation of this cipher and I am open to
anybody else taking my invention a step further by way of on -going
development.
- adacrypyt
Karl-Uwe Frank
Feb 17, 2012, 11:15:39 AM2/17/12
to
On 17.02.12 13:29, adacrypt wrote:
> My Cipher Invention seems to be working for you - I am pleased about
> that - it is in the public domain as "Dilating Key Encryptions ".
>
> http://www.adacrypt.com/introduction.html
>
> I have invited users there to modify it as they wish - I suggest you
> do this to your heart's contentent.
>
> I cannot comment on what you are finding by your own methods at
> present - I repeat I am pleased that you are succeeding.
>
> To me its "job done" in the creation of this cipher and I am open to
> anybody else taking my invention a step further by way of on -going
> development.
>
> - adacrypyt
>
You don't really believe that any-one would ever use that "cipher", do you?
> My Cipher Invention seems to be working for you - I am pleased about
> that - it is in the public domain as "Dilating Key Encryptions ".
>
> http://www.adacrypt.com/introduction.html
>
> I have invited users there to modify it as they wish - I suggest you
> do this to your heart's contentent.
>
> I cannot comment on what you are finding by your own methods at
> present - I repeat I am pleased that you are succeeding.
>
> To me its "job done" in the creation of this cipher and I am open to
> anybody else taking my invention a step further by way of on -going
> development.
>
> - adacrypyt
>
Even when I, without any knowledge of cryptanalysis, can figure out the
correlation between the characters of those two messages "POSTA DI
FALCONE" and "T E M P L A R S" in an hour, without knowing the key or
anything else - not even downloaded you source-code because I'm not
working on Windows - just using a pocket calculator, what do you suppose
will happen if the girls and boys from these agencies with unlimited
resources and professional cryptanalyst's get hand on some messages
encrypted with your scheme?
Sorry, but I will not spend any further time on any matter of your
ideas, because you did not even are aware on how extreme weak your
"cipher" seems to be - even after a lot of help was offered to you from
crypto pro's around here. In fact, you are quite stubborn in your
believes of the strength of your schemes.
Like you, I was so convinced of the strength of my cipher algorithm when
I started with cryptography in June 2011. But, as time goes by, I was
reading, listening and learning about cryptogtaphy - and still do -
which obviously let me realise that my cipher algorithm's have some
flaws that any professional cryptanalyst can break them easily. Perhaps
you should register at "The Cryto Forum" and let the boys there take a
look at your cipher and they will break it in a matter of minutes.
POSTA DI FALCONE
================
P O S T
58017 57956 57944 58013
A D I
57992 57967 57957 58056
F A L
57905 58020 57959 57871
C O N E
58537 58531 58051 58248
T E M P L A R S
================
T E M P L A R S
T E
58011 57999 57960 58061
M P
57974 57967 57949 58095
L A
57869 58052 57959 57907
R S
58522 58574 58046 58281
ABCDEFGHIJKLMNOPQRSTUVWXYZ
| |
1 26
** does not apply for (space)
(58061) (57999) (57993) (58061) = (Base Value) -28 -Char**
P O S T
58017 57956 57946 58013
(58017) (57967) (57991) (58095) = (Base Value) -30 -Char
A D I
57986 57967 57957 58056
(57905) (58052) (57620) (57907) = (Base Value) -24 -Char
F A L
57905 58022 57959 57871
(58568) (58574) (58093) (58281) = (Base Value) -28 -Char
C O N E
58537 58531 58051 58248
===
(58061) (57999) (57993) (58061) = (Base Value) -28 -Char
T E
58013 57999 57960 58061
(58017) (57967) (57991) (58095) = (Base Value) -30 -Char
M P
57974 57967 57945 58095
(57905) (58052) (57620) (57907) = (Base Value) -24 -Char
L A
57869 58052 57959 57907
(58568) (58574) (58093) (58281) = (Base Value) -28 -Char
R S
58522 58574 58046 58281
I'm terribly sorry, but
58013 57963 57956 58014
58017 57928 57942 58095
57878 58010 57959 57867
my cipher algorithm are "unbreakable" ... and better grateful accept any
help and greater knowledge of other cryptographers.
Cheers,
Karl-Uwe
adacrypt
Feb 17, 2012, 12:15:36 PM2/17/12
to
> Karl-Uwe- Hide quoted text -
>
> - Show quoted text -
You haven't a clue!.
>
> - Show quoted text -
- adacrypt
Mark Murray
Feb 17, 2012, 12:34:44 PM2/17/12
to
On 17/02/2012 17:15, adacrypt wrote:
> You haven't a clue!.
Amazing. Utterly amazing. Just when I thought that stupidity was
> You haven't a clue!.
reaching an asymptote, a new order of magnitude term emerges.
Karl-Uwe Frank
Feb 17, 2012, 12:40:48 PM2/17/12
to
> You haven't a clue!.
>
> - adacrypt
... or go to "The Crypto Forum" where they will tear to shreds your
"secure cipher" scheme ... but perhaps you're afraid of exactly that :-)
Bye,
Karl-Uwe
Karl-Uwe Frank
Feb 17, 2012, 2:24:59 PM2/17/12
to
On 17.02.12 17:15, adacrypt wrote:
>
> You haven't a clue!.
>
> - adacrypt
Just for you a new cipher invented in an hour with the same strength as
>
> You haven't a clue!.
>
> - adacrypt
your scheme.
Addition-with-Drop-and-Carry-over-Cipher
========================================
(Secret Key = Difference between first and last character i.e. L-E =
12-5 = 7)
Encrypt + /* if Value > 26 then Value -26 */
Decrypt - /* if Value < 0 then Value +26 */
Plaintext = TEMPLARS
Key Characters = L E
Secret Key = L-E = 12-5 = 7
ABCDEFGHIJKLMNOPQRSTUVWXYZ
| |
1 26
How it works
1) Always calculate the first char twice with the difference value of
the secret key (i.e. 7)
2) Always only calculate each second step with the difference value (i.e. 7)
3) If encrypt and the added value is > 26 subtract 26,
if decrypt and the subtract value is < 0 add 26
4) The result of each first calculation is the carry over value
5) Build the cipher message in taking the final result of each column
Now encipher: ** (in brakets the carry-over-value of the previous
calculation)
T E M P L A R S
+ 7 (1) (6)(19) (9)(21)(22)(14)**
------------------------------
(1)*(6)(19) (9)(21)(22)(14) 7
+ 7 0 7 0 7 0 7 0
8 6 26 9 2 22 21 7
------------------------------
H F Z I B V U G
* (this is 1 because 20+7 = 27 -26 = 1)
The cipher message now read
HFZIBVUG
(L) HFZIBVUG (E)
Easy isn't it?
Only need to share two ASCII characters as secret key for each word!
Now decipher: ** (in brakets the carry-over-value of the previous
calculation)
(L) HFZIBVUG (E) (L-E = 12-5 = 7)
H F Z I B V U G
8 6 26 9 2 22 21 7
- 7 0 7 0 7 0 7 0
------------------------------
(1) (6) 19 9 21 22 14 7
- 7 1 (6)(19) (9)(21)(22)(14)**
------------------------------
20* 5 13 16 12 1 18 19
T E M P L A R S
* (this is 20 because 1-7 = -6 + 26 = 20)
And now imagine having a huge keyset with double-letter combinations,
shared between the communication partners.
And than imagine using for every word that has to be enciphered a
different double-letter combination!
This is an unbreakable cipher, because it's infeasible to know each
double-letter combination in advance!!!
And for sure the ciphertext is 100% random.
Cheers,
Karl-Uwe
Mark Murray
Feb 17, 2012, 2:51:17 PM2/17/12
to
On 17/02/2012 19:24, Karl-Uwe Frank wrote:
> And now imagine having a huge keyset with double-letter combinations,
> shared between the communication partners.
Adacrypt is going to say this is the same as his shared database.
> And now imagine having a huge keyset with double-letter combinations,
> shared between the communication partners.
> And than imagine using for every word that has to be enciphered a
> different double-letter combination!
>
> This is an unbreakable cipher, because it's infeasible to know each
> double-letter combination in advance!!!
the same as this and is thus also unbreakable.
> And for sure the ciphertext is 100% random.
duplicate cipertext characters and thus can't be random.
</advocate client="devil">
Karl-Uwe Frank
Feb 17, 2012, 3:47:53 PM2/17/12
to
On 17.02.12 19:51, Mark Murray wrote:
> On 17/02/2012 19:24, Karl-Uwe Frank wrote:
>> And now imagine having a huge keyset with double-letter combinations,
>> shared between the communication partners.
>
> Adacrypt is going to say this is the same as his shared database.
>
Yupp, that's true ... and as easy to share as his database, especially
> On 17/02/2012 19:24, Karl-Uwe Frank wrote:
>> And now imagine having a huge keyset with double-letter combinations,
>> shared between the communication partners.
>
> Adacrypt is going to say this is the same as his shared database.
>
with several hundreds of thousand or even millions of communication
partners.
>> And than imagine using for every word that has to be enciphered a
>> different double-letter combination!
>>
>> This is an unbreakable cipher, because it's infeasible to know each
>> double-letter combination in advance!!!
>
> Adacrypt is going to say that his pseudo-random sequence generator is
> the same as this and is thus also unbreakable.
>
And so easy to handle as well. But unfortunately it doesn't take me
decades of scientific mathematical studies and only may need a pocket
calculator and a printed paper with an individual matrix of
double-letter combinations as a secret key sheet for encryption :-)
For sure I could write a program that do the math, with a huge
obfuscated source in order to bury the calculation.
And for sure, the cryptanalyst's will never be able to break the cipher
because it's sooooo difficult to try all possible double-letter
combinations - that's in fact I suppose what he believe about the
integer range his cipher use for character substitution. Really a sad story.
>> And for sure the ciphertext is 100% random.
>
> Adacrypt is going to disagree because you system doesn't preclude
> duplicate cipertext characters and thus can't be random.
>
"scientific" :-)
It reminds me on another fantastic discussion about random numbers at
sci.crypt.random-numbers. Take a look at http://kingkonglomerate.com/
and then search for a thread "BulletProof Random Number Generator". This
guy has some strong similarity's with adacrypt regarding stubbornness
and learn resistance.
> </advocate client="devil">
>
> M
Sometimes it looks like he is bound to this mantra
"The first hurdle is that, by definition, there is no repetition in a random key, ..."
Maybe it's time to break the spell and we should ask Simon Singh to
clarify the typo in his book and release adacrypt's mind ;-)
Cheers,
Karl-Uwe
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