Is there a Mathematician Cryptographr in the House.
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From: adacrypt on 18 Jul 2010 15:44 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. 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 - the program is called "Make_Moduli_Program_Mark_0" and is to be found in the download that is available from my website (see link earlier) - the program can be run just to satisfy a user's curiosity but once it is understood the range of N's can be written down directly in future - a general message length is decided first of all then the lower bound of N becomes (X +127) and the upperbound is 2(X +32) where X is max keylength +63. Why 63 - this is seen by experimenting - the general idea is to understand how to write this down by direct inspection of the scheme parameters - the theory behind it is related to the Vigenere square being made mobile in the fourth quadrant of the XY plane. Test this as much as like. The set of N's as moduli are simply consecutive positive integers in the given range above. Are these transparent you may well ask - naw its a random set that is scrambled and sliced. Secondary alphabets (as an alternative to the ASCII printable subset) are quite feasible also - once you think the whole thing through it is esay to see how the crypto strength is assured. This is quite difficult and certainly not something you can think through in a few seconds - the set of validated moduli could be said to be a sequence of integers that have a rule in modular arithmetic residue classes - cheers - adacrypt
From: David Eather on 18 Jul 2010 23:04 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.
From: unruh on 19 Jul 2010 01:20 On 2010-07-19, David Eather <eather(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. Not if he makes no attempt to make sure that the successive bytes are uncorrelated with earlier ones. Eg if he just keeps reusing the same key material.
From: David Eather on 19 Jul 2010 01:41 On 19/07/2010 3:20 PM, unruh wrote: > On 2010-07-19, David Eather<eather(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. > > Not if he makes no attempt to make sure that the successive bytes are > uncorrelated with earlier ones. Eg if he just keeps reusing the same key > material. > Oh come on Bill! He's not *that* stupid. Is he?? ;-)
From: Mok-Kong Shen on 19 Jul 2010 02:09
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 |