Is there a Mathematician Cryptographr in the House.
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From: adacrypt on 19 Jul 2010 02:24 On Jul 19, 4:04 am, David Eather <eat...(a)tpg.com.au> wrote: > On 19/07/2010 5:44 AM, adacrypt wrote: > > > > > > > On Jul 18, 6:19 pm, Mok-Kong Shen<mok-kong.s...(a)t-online.de> wrote: > >> adacrypt wrote: > >>> Huge typo omission here, > > >>> I should have stated that N is in the range (X +127) and 2(X+32). > > >>> Then X works out to 63 (=> N =190) and the number of N's (as keys) > >>> works out to 14000 - 63. > > >>> The strength of this cipher is then in the decryption equation being > >>> one equation in three unknowns - two of the unknowns are the random > >>> keys (Key and N) in the equation - being random makes them totally > >>> indeterminable to an adversary. > > >> Your formulation, also in the first post, is not clear for me. Anyway, > >> if you want to exploit indeterminancy to enhance security, then simply > >> xoring two pseudo-random strams R1 and R2 (assumed independent, both, > >> say, of 32 bit units) will do the job: > > >> C = R1 ^ R2 ^ P > > >> where P and C are the plaintext and ciphertext units. This is of course > >> equivalent to: > > >> R = R1 ^ R2 C = R ^ P > > >> So the xoring is properly to be considered to be internal to the > >> single PRNG that generates R. One could however profitably do something > >> more in the combination for achieving higer security, see my thread > >> "A simple scheme of combining PRNGs" of 01.06.2010. > > >> M. K. Shen > > > Hi, There is much more to it than that - I am not sure really what you > > mean because there is a suggestion of PRNGS in your methods which are > > taboo to me - I don't use PRNGS ever. > > Yes you do. You just don't understand that you do. You have some shared > secret data. When you send a message than that shared data the process > to create more "key pad" to protect the message *is* a PRNG. FULL STOP. > PERIOD.- Hide quoted text - > > - Show quoted text - HBi David, >Yes you do. You just don't understand that you do. You have some shared >secret data. When you send a message than that shared data the process >to create more "key pad" to protect the message *is* a PRNG. FULL STOP. >PERIOD. I think there's a few words missing here, should it read "Yes you do. You just don't understand that you do. You have some shared secret data. When you send a message GREATER than that THE shared data VOLUME the process to create more "key pad" to protect the message *is* a PRNG. FULL STOP.? On the premise that you mean this I proceed: As it stands I can make provision for any message length up to 2^31 - 63 by providing a key length of Moduli (N's) of that same length and a keypad of KEYS (drawn from ASCII printable subset) in round multiples ( 'm' say) of the basic 95 elements that will overlap the real messagelength by some small amount that requires padding of the message length (probably) i.e. 2147483647 - 63 = 2147483584 or over half of 1 million pages ? surely it is acceptable that anything greater than this is done in a fresh start as a fresh block of text that will use new permutations of the random key sets. The random set of keys made up from modules of the 95 elements of ASCII must be made a round multiple ( 'm' ) of the basic 95 elements so as to ensure that it is stiil random (every element has still has equal probability despite being replicated 'm' times - it is then still as truly random as if there was only one module). There is nothing pseudo about this as PRNGS implies - RNGS yes without the pseudo bit. I am totally focused on the need for randomness in both sets of keys. Although this algorithm is designed round a dynamic Vigenere square I see not reason once that is understood of simply treating this execise of key generation as a configuration algorithm that produces random keys ad hoc without explanation - no need to labour the origins? - regards - adacrypt
From: adacrypt on 19 Jul 2010 03:01 On Jul 19, 7:09 am, Mok-Kong Shen <mok-kong.s...(a)t-online.de> wrote: > adacrypt wrote: > > Mok-Kong Shen wrote: > >> Your formulation, also in the first post, is not clear for me. Anyway, > > [snip] > > > Hi, There is much more to it than that - I am not sure really what you > > mean because there is a suggestion of PRNGS in your methods which are > > taboo to me - I don't use PRNGS ever. > > > The set of N's that can be paired with every possible pairing of Key > > and Plaintext as moduli in the algorithm [(X +Key) + (X +Plaintext)] > > (Mod N) = a residue (Mod N)>=0 > > are not found by any mathematical formula - they were found by > > validating in a specially designed test program -[snip] > > Your X and Key are two secret values that are randomly or pseudo- > randomly determined, right? So, I was using the practically more often > situation of employing PRNGs to get random values to illustrate my > point. So, independent of how you obtain these values, what is exactly > you point then? Is it 'somewhat' (because I am yet unclear of your > formulation) like that there are for the analyst two unknowns, namely > X and Key, but only one equation and that therefore there is > indeterminancy, which renders his work hard/infeasible? If that's the > case, then please re-read my previous post and compare your point with > what I described and tell if there is fundamental difference regarding > the issue of difficulty for the analyst. If not, please explain what > is 'special' of your point more clearly and use 'concrete' numbers to > illustrate one case of encryption and decryption according to your > scheme. > > M. K. Shen Hi, >Is it 'somewhat' (because I am yet unclear of your >formulation) like that there are for the analyst two unknowns, namely >X and Key, but only one equation and that therefore there is >indeterminancy, which renders his work hard/infeasible? If that's the The crypto strength is in the decrytpion algorithm being a single equation in four unknowns' Plaintext ( as messagetext now) = Cipher text + 2N (Key) 2X gives the value in ASCII of the current plaintext. KEY (as the key from ASCII subset), N as the modulus, and X are all unknowns to an adversary. Why don't you download my working models and go through them. I will try and get round to a longhand worked example later but you should be able to project waht's onthe table so afr easily enough. I tink what may be troubling you is the connection with the Vigener square - That is very difficult to describe in posts - indeed I wrestled with it myself for weeks before I could justify it to myself. The square is normally docked at (0,0) in the fourth quadrant of the XY plane - I decide to undock it and make it dynamically mobile in the same quadrant - this means I cause the top left hand corner i.e. the origin of the square to move around (giving rise to X as an increment of both the x and y coordinates) but x and y are ineffect the key and plaintext in the context of the mathematical equation of the square (see Bruce Schneir also on P. 15 I think, on the equation of the square as applied to the historic OTP - I remind you again that this cipher is not an OTP on any account). This ploy means the square is made to move along a line Y = - (X +x). It could be made to move along other similar lines also giving rise to (X +x ) and (Y +y) - it becomes difficult to assimilate at this point but once it all goes under the bonnet as sorcecode the driver doesn't need to think about it anymore - cheers - adacrypt
From: Mok-Kong Shen on 19 Jul 2010 03:09 adacrypt wrote: > Mok-Kong Shen wrote: >> Is it 'somewhat' (because I am yet unclear of your >> formulation) like that there are for the analyst two unknowns, namely >> X and Key, but only one equation and that therefore there is >> indeterminancy, which renders his work hard/infeasible? If that's the > > The crypto strength is in the decrytpion algorithm being a single > equation in four unknowns' > Plaintext ( as messagetext now) = Cipher text + 2N � (Key) � 2X > gives the value in ASCII of the current plaintext. > > KEY (as the key from ASCII subset), N as the modulus, and X are all > unknowns to an adversary. > > Why don't you download my working models and go through them. > > I will try and get round to a longhand worked example later but you > should be able to project waht's onthe table so afr easily enough. > > I tink what may be troubling you is the connection with the Vigener > square - That is very difficult to describe in posts - indeed I > wrestled with it myself for weeks before I could justify it to myself. > > The square is normally docked at (0,0) in the fourth quadrant of the > XY plane - I decide to undock it and make it dynamically mobile in the > same quadrant - this means I cause the top left hand corner i.e. the > origin of the square to move around (giving rise to X as an increment > of both the x and y coordinates) but x and y are ineffect the key and > plaintext in the context of the mathematical equation of the square > (see Bruce Schneir also on P. 15 I think, on the equation of the > square as applied to the historic OTP - I remind you again that this > cipher is not an OTP on any account). > > This ploy means the square is made to move along a line Y = - (X +x). > It could be made to move along other similar lines also giving rise to > (X +x ) and (Y +y) - it becomes difficult to assimilate at this point > but once it all goes under the bonnet as sorcecode the driver doesn't > need to think about it anymore - cheers - adacrypt You described you scheme using terms in a way that IMHO make your stuff very difficult to comprehend. Why couldn't you once use a 'concrete' example with actual numbers (like 5, 333 etc.) to explain how your scheme really works? (Please indicate thereby which are the secret values that are unknown to the analyst and that he has to figure out to break the scheme.) M. K. Shen
From: adacrypt on 19 Jul 2010 03:47 On Jul 19, 8:09 am, Mok-Kong Shen <mok-kong.s...(a)t-online.de> wrote: > adacrypt wrote: > > Mok-Kong Shen wrote: > >> Is it 'somewhat' (because I am yet unclear of your > >> formulation) like that there are for the analyst two unknowns, namely > >> X and Key, but only one equation and that therefore there is > >> indeterminancy, which renders his work hard/infeasible? If that's the > > > The crypto strength is in the decrytpion algorithm being a single > > equation in four unknowns' > > Plaintext ( as messagetext now) = Cipher text + 2N (Key) 2X > > gives the value in ASCII of the current plaintext. > > > KEY (as the key from ASCII subset), N as the modulus, and X are all > > unknowns to an adversary. > > > Why don't you download my working models and go through them. > > > I will try and get round to a longhand worked example later but you > > should be able to project waht's onthe table so afr easily enough. > > > I tink what may be troubling you is the connection with the Vigener > > square - That is very difficult to describe in posts - indeed I > > wrestled with it myself for weeks before I could justify it to myself. > > > The square is normally docked at (0,0) in the fourth quadrant of the > > XY plane - I decide to undock it and make it dynamically mobile in the > > same quadrant - this means I cause the top left hand corner i.e. the > > origin of the square to move around (giving rise to X as an increment > > of both the x and y coordinates) but x and y are ineffect the key and > > plaintext in the context of the mathematical equation of the square > > (see Bruce Schneir also on P. 15 I think, on the equation of the > > square as applied to the historic OTP - I remind you again that this > > cipher is not an OTP on any account). > > > This ploy means the square is made to move along a line Y = - (X +x). > > It could be made to move along other similar lines also giving rise to > > (X +x ) and (Y +y) - it becomes difficult to assimilate at this point > > but once it all goes under the bonnet as sorcecode the driver doesn't > > need to think about it anymore - cheers - adacrypt > > You described you scheme using terms in a way that IMHO make your stuff > very difficult to comprehend. Why couldn't you once use a 'concrete' > example with actual numbers (like 5, 333 etc.) to explain how your > scheme really works? (Please indicate thereby which are the secret > values that are unknown to the analyst and that he has to figure out to > break the scheme.) > > M. K. Shen- Hide quoted text - > > - Show quoted text - Hi, will do in time - very busy right now - adacrypt.
From: adacrypt on 19 Jul 2010 04:04
On Jul 19, 4:04 am, David Eather <eat...(a)tpg.com.au> wrote: > On 19/07/2010 5:44 AM, adacrypt wrote: > > > > > > > On Jul 18, 6:19 pm, Mok-Kong Shen<mok-kong.s...(a)t-online.de> wrote: > >> adacrypt wrote: > >>> Huge typo omission here, > > >>> I should have stated that N is in the range (X +127) and 2(X+32). > > >>> Then X works out to 63 (=> N =190) and the number of N's (as keys) > >>> works out to 14000 - 63. > > >>> The strength of this cipher is then in the decryption equation being > >>> one equation in three unknowns - two of the unknowns are the random > >>> keys (Key and N) in the equation - being random makes them totally > >>> indeterminable to an adversary. > > >> Your formulation, also in the first post, is not clear for me. Anyway, > >> if you want to exploit indeterminancy to enhance security, then simply > >> xoring two pseudo-random strams R1 and R2 (assumed independent, both, > >> say, of 32 bit units) will do the job: > > >> C = R1 ^ R2 ^ P > > >> where P and C are the plaintext and ciphertext units. This is of course > >> equivalent to: > > >> R = R1 ^ R2 C = R ^ P > > >> So the xoring is properly to be considered to be internal to the > >> single PRNG that generates R. One could however profitably do something > >> more in the combination for achieving higer security, see my thread > >> "A simple scheme of combining PRNGs" of 01.06.2010. > > >> M. K. Shen > > > Hi, There is much more to it than that - I am not sure really what you > > mean because there is a suggestion of PRNGS in your methods which are > > taboo to me - I don't use PRNGS ever. > > Yes you do. You just don't understand that you do. You have some shared > secret data. When you send a message than that shared data the process > to create more "key pad" to protect the message *is* a PRNG. FULL STOP. > PERIOD.- Hide quoted text - > > - Show quoted text - Hi again, >Yes you do. You just don't understand that you do. You have some shared >secret data. When you send a message than that shared data the process >to create more "key pad" to protect the message *is* a PRNG. FULL STOP. >PERIOD I think I should explain that in the development model of cipher that is on the table I have designed with a certain 'scope' of message lengths up to about 14250 characters - that's about four good well filled pages of text. Any messagelength up to this size is catered for as standard. The creator of a crypto system must create the set of modules as keys (N's) to cover their expected requirements ahead of time - it is quite easy to change this anytime. Then in the case of shorter messages of say 2000 characters they will use only a part of the full key set - note well; this subset of the full random keyset of N's is itself also random by the same definition that the parent no-repeating set has i.e. equal probability of each element. In this case, the kepad of keys a la ASCII printables must be 'sized' to cover the current message each time so as to be random i.e it must be comprised of a number of modules in excess or equal to the message length - that is done automatically by the software. There is no such thing as generating keys as you go along - they are preformed in arrays to cater for anticipated requirements but are very easy to change as required to suit special cases that always is expected arise - adacrypt |