Reputation
Top tag
Next privilege 250 Rep.
View close votes
Badges
4
Newest
 Critic
Impact
~2k people reached

  • 0 posts edited
  • 1 helpful flag
  • 4 votes cast
Jan
31
comment Is H'(M)=5*M + 9*H(M) mod 2^n a secure hash?
is this question on topic her? I dont think so: Do we accept questions asking for cryptanalysis of your cipher (hash function, ...) design?
Jan
30
comment How to calculate plaintext from modpower cipher
So I think answers should concentrate on the fact that e is small an not on well known requirements of the RSA.
Jan
30
comment How to calculate plaintext from modpower cipher
This is related to the RSA cryptosystem en.wikipedia.org/wiki/RSA_(cryptosystem)#Key_generation . You can decrypt a message if you can factor the modulus, which is easy to factor if it has only small prime factors. If k is the number of prime factors then the smallest prime factor cannot have more than #bits(n)/k bits. So k=2 is optimal. The advantage to use a small exponent e is that encryption is very fast. But does a small e have drawbacks? It seems so: check footnote 8 of the wiki article or paragraph 4 of crypto.stanford.edu/~dabo/abstracts/RSAattack-survey.html .
Jan
14
awarded  Critic
Jan
14
awarded  Citizen Patrol
Jan
10
awarded  Teacher
Jan
7
comment Hash function that allows to decide if A > B if you only have hash(A) and hash(B)?
I think that prove does not really make sense but it is to long to describe this in a comment so I post an answer crypto.stackexchange.com/a/31765/1915
Jan
7
answered Hash function that allows to decide if A > B if you only have hash(A) and hash(B)?
Jan
7
comment Hash function that allows to decide if A > B if you only have hash(A) and hash(B)?
No, that does not mean that $H(a)<H(b)$ or $H(a)>H(b)$. It means that you have a function $f$ and $f(H(a),H(b))$ returns 'true' if $a<b$ and 'false' if not. Of course this function reduces a total order relation on $\{H(0),...,H(2^{n-1})\}$ but it does definitely not mean that $0≤H(0)<H(1)<…<H(2n−1)<2^n$
Jan
7
comment Hash function that allows to decide if A > B if you only have hash(A) and hash(B)?
That is wrong. The OP does not require that H(A)<H(B) if A<B.
Jul
18
awarded  Supporter