Reputation
517
Top tag
Next privilege 1,000 Rep.
See votes, expandable usercard
Badges
4 9
Newest
 Enthusiast
Impact
~14k people reached

  • 0 posts edited
  • 0 helpful flags
  • 16 votes cast
Mar
2
comment Ciphers providing Deniable encryption?
One method that is well known is: Take an OTP1 and obtain the cipertext C = M1 xor OTP1. Compute OPT2 = C xor M2. Now one has M1 = C xor OPT1 and M2 = C xor OTP2. OTP1 and OTP2 are the keys required.
Mar
2
comment Can shuffling plaintext produce a useable key?
@JPresperEckert: In your 1st comment above, you mentioned the case of adding key characters (similar in essence to addition mod 26, I suppose). The frequency distribution is of course changed, depending on the key stream. In general there will be some (more or less) improvements. (In the special case that the key stream is ideally random, the resulting stream will also be ideally random. But this fact is apparently not of practical relevance.)
Feb
29
comment When we turn Random shuffle to Pseudorandom Shuffle
A PRNG producing integers has commonly a range [0 .. N-1], dividing its output by N gives a real number in [0 .. 1).
Feb
29
awarded  Custodian
Feb
29
reviewed Approve Entropy calculation
Feb
29
asked Entropy calculation
Feb
28
comment Applicability of birthday attack to AES brute force
@YehudaLindell: Many thanks. I was completely wrong as well in my thinking during the past couple of days. It's a big shame of mine.
Feb
28
comment Applicability of birthday attack to AES brute force
@fgrieu: I have to sincerely ask your pardon for not having carefully read your writing in your main post above. That I have only now noticed that the birthday estimate is irrelevant to the chance of finding the key is a big shame of mine.
Feb
28
comment Applicability of birthday attack to AES brute force
@YehudaLindell: If I understand fgrieu's answer correctly, the chance is 0.5 if one tries a number of keys equal to that claimed in the book.
Feb
28
comment Applicability of birthday attack to AES brute force
@YehudaLindell: You are apparently confused by the volume of past discussions (some revised/deleted), I suppose. The birthday solution is the correct solution, see the discussions between fgrieu and me below.
Feb
28
comment Applicability of birthday attack to AES brute force
@fgrieu: [Addendum:] Since at the time of my earlier comment beginning with "Yes, many ..." I was yet confused by our discussions, i.e. not sure of whether the birthday solution is the long very well known correct solution, thus actually needing no discussions at all in this thread, I posted subsequently something to another group (in my sense of calling "attention to a larger circle of people" as indicated in that comment) and that post was (deservedly!) down-voted.
Feb
28
comment Applicability of birthday attack to AES brute force
[continued from above] is "redundant" and hence no good. Further, you critisized my description with the partyroom on the ground that with a number of keys in the room equal to the collision estimate there could be identical keys and hence my description is not valid. But that critique is invalid and my later rescue employing twice the collision estimate is "unnecessary". This is because the math underlying the birthday problem permits identical items from the outset. E.g. HAC p.53 illustrates the problem with an urn of balls "with replacement". So IMHO we had actually "Much Ado for Nothing".
Feb
28
comment Applicability of birthday attack to AES brute force
@fgrieu: I like to say that in the original version of your answer to the thread you mentioned sort of alternative viewpoints, one of which is the IMHO common saying that bruteforcing a block cipher needs n/2 trials in order to have 0.5 chance of success. (Note: "bruteforcing" is not further qualified with additional attributes.) But this common saying is blatantly a mistake due to the math known as the birthday problem. Further, the partyroom popularly used is a special case to render surprise to common people, so my use of it to argue for the general case [to be continued for space below]
Feb
27
comment Applicability of birthday attack to AES brute force
@fgrieu: Could you please say what's your opinion on Thomas Byrd's comment?
Feb
27
comment Applicability of birthday attack to AES brute force
@fgrieu: Yes, many thanks for that. I was subsequently thinking that very probably some other estimates should be similarly reviewed and perhaps someone in our community may eventually have the opportunity to call attention to that to a larger circle of people.
Feb
27
comment Applicability of birthday attack to AES brute force
@fgrieu: Would that imply that there might be other estimates relevant for crypto that are similarly affected?
Feb
27
comment Applicability of birthday attack to AES brute force
@fgrieu: Perhaps I yet misunderstood you. I meant that with key number of twice the collision estimate one highly probably ensures that all the keys in the room would contain distinct keys of a number equal to that estimate (which was the condition not satisfied in my first comment according to your original critique).
Feb
27
comment Applicability of birthday attack to AES brute force
@fgrieu: Ok. But, if the number of keys already in the room is twice the collision estimate, would my argument be ok?
Feb
27
comment How to perform AES MixColumns as matrix multiplication in GF(2) (boolean values)?
If I don't err, GF is unique only up to isomorphism. As layman in math I am afraid that the conversion you mentioned could be fairly specific for the case of AES and thus not "general". I suggest that you also post to the subcommunity "mathematics" with an appropriate formulation.
Feb
27
comment Are there any historic examples of breaking the one-time pad because of a careless key re-use? i.e simple attack on a multiple-time pad?
The Wiki page gives a highly significant real example. But how "simple" was that attack? It would be fine if there exists a paper detailing how one should advantageously proceed in practice to deal with a two-time pad IMHO.