Happy Eid Al-Adha

26
Nov/09
0
Wish you all the blessing Eid Al-Adha

Wish you all the blessing Eid Al-Adha

Filed under: Uncategorized

Hair Complexities

18
Nov/09
0

“Anyone has a mirror ???”

Filed under: Uncategorized

My-Wave :)

16
Nov/09
0

Google-Wave

Filed under: Uncategorized

Dialah, Purwaning Dumadi

7
Nov/09
0

Filed under: Refleksi

Little-Dragon-2

15
Oct/09
4

It is interesting to see the property of Little-Dragon-2 (LD2)[1] as the improved version of Little-Dragon[2] developed based on MI (Matsumoto-Imai) scheme as the pioneer in MPKC. One of the known problem in construction Multivariate Public Key Cryptosystem (MPKC) is the possibility to have several equivalent private-keys. The equivalent private-keys in MPKC occur when two unequal private-keys produce the same public-keys (Since the construction of the public-key in MPKC is a composition of  L_1 \circ P \circ L_2, where  P is a multivariate polynomial of the degree > 1 (generally is 2) over a prime field \mathbb{F}_q (q = p^n, where p must be a prime. In case of binary field, p = 2 ), L_1 and  L_2 are two invertible affine transformation of the form  L(x) = A(x) + C, where A is the (n \times n) invertible matrix over \mathbb{F}_q and C is a defined-constant on the same field).

The problem of equivalent keys in MPKC can be solved easily by generate permutation polynomial P over \mathbb{F}_q. Such characteristic is produced when the polynomial p(x) in \mathbb{F}_q[x] is a polynomial function from \mathbb{F}_q onto \mathbb{F}_q. Thus, there are four condition that make a polynomial p(x) is a permutation polynomial over \mathbb{F}_q, that are:

  1. p(x) is onto
  2. p(x) is one-to-one
  3. p(x)=a has a solution in \mathbb{F}_q, for each a \in \mathbb{F}_q
  4. p(x)=a has a unique solution in \mathbb{F}_q, for each a \in \mathbb{F}_q

Then, based on the fact that :

  1. Every linear polynomial that is a polynomial of the form ax + b with a \neq 0 over \mathbb{F}_q is a permutation polynomial of \mathbb{F}_q
  2. The monomial x^n is a permutation polynomial of \mathbb{F}_q if and only if n and q-1 are coprime (\gcd (n, q-1) = 1)

It is proven that polynomial over \mathbb{F}_q of the form x^{\left(2^{2^rk }+ 2^r\right)} + x^{2^{2^rk}} + x^{2^r}, where r, k \in \mathbb{Z}^+, is a permutation polynomial if and only if 2^{2^rk} + 2^r and  2^n-1 are coprime, which \gcd (2^{2^rk} + 2^r, 2^n-1) = 1. The extension of this idea is by having polynomial function of the form g(x) = \left( x^{2^{2^r}k} + x^{2^r} + \alpha \right)^l + x, where T(\alpha) is a trace function from \mathbb{F}_{2^{m}} to \mathbb{F}_2, in which T(\alpha) = \alpha + \alpha^2 + \alpha^{2^2} + ... + \alpha^{2^{m-1}} = 1 and l \cdot \left( 2^{2^r k} + 2^r\right) = 1 \mod 2^n - 1. However, the question is how to choose the value of r and k so that the polynomial generated will satisfy the conditions above ?

Choosing r and k

For the case of public-key cryptosystem, the permutation polynomial only exist where l is of the form  2^t \pm 1. Specifically, for r = 0 , \mathbb{F}_{2^{2m - 1}}, k=m and l = 2^m - 1 , g(x) is obviously a permutation polynomial, since 2^{2^r k} + 2^r = 2^m + 1, therefore  (2^m - 1) \cdot (2^m + 1) = 2^{2m} - 1 = 1 \mod 2^n - 1. So the public key is the form of g(x) = \left( x^{2^m} + x + \alpha\right)^{2^m -1} + x  where \alpha is the hidden monomial and must be kept as the private keys, together with L_1 and L_2 as the two invertible linear affine transformation.

However, there is no much improvement on the size of public key and the complexity of the encryption process \mathcal{O}(n^3). Moreover, its resistance against many type of algorithm than can solve multivariate polynomial problem (Buchberger, XL, XSL, Linearization, Relinearization, F_4 and F_5, Zhuang-Zhi, etc) as well as the key recovery attack against hidden monomial-based MPKC have not been examined that far.

So  as the closing question :  anyone interested ? :-)

\mathcal{ROEHM}

[1]. R.P. Singh, A. Saikia, B.K.Sarma., Little Dragon Two: An efficient Multivariate Public Key Cryptosystem. IACR ePrint Archive.

[2]. J.Patarin. Asymmetric cryptography with a hidden monomial. Advances in Cryptology-Crypto ‘96, Springer-Verlag, pp. 4560.

[3] C. Wolf., B. Preenel., Equivalent Keys in Multivariate Quadratic Public Key Systems. IACR ePrint Archive.

Filed under: Kriptografi

My Public Key

29
Sep/09
0

Dear folks,

I’ve posted my public key in this blog. Just in case you need to send confidential information or files, I strongly encourage you to encrypt it before send it to my email.

Regards,

 \mathcal{ROEHM}

Filed under: Kabar

Selamat Hari Raya Idul Fitri

19
Sep/09
0

Taqabbalallah minna wa minkum

 \mathcal{ROEHM}

Filed under: Uncategorized

English is Not That Difficult

29
Aug/09
0

engrish funny anyone free
see more Engrish

 \mathcal{ROEHM}

Filed under: Uncategorized

When word speak more than a voice

10
Aug/09
0
Choose, either pay or pray ...

Choose, either pay or pray ...

 \mathcal{ROEHM}

Filed under: Refleksi

Key Recovery Attacks of Practical Complexity on AES Variants With Up To 10 Rounds

4
Aug/09
0

Here we come everybody, an almost practical related-key attack against reduced round (10 rounds) of AES-256.

AES is the best known and most widely used block cipher. Its three versions (AES-128, AES-192, and AES-256) di ffer in their key sizes (128 bits, 192 bits and 256 bits) and in their number of rounds (10, 12, and 14, respectively). In the case of AES-128, there is no known attack which is faster than the 2^{128} complexity of exhaustive search. However, AES-192 and AES-256 were recently shown to be breakable by attacks which require 2^{176} and 2^{119} time, respectively. While these complexities are much faster than exhaustive search, they are completely non-practical, and do not seem to pose any real threat to the  security of AES-based systems. In this paper we describe several attacks which can break with practical complexity variants of AES-256 whose number of rounds are comparable to that of AES-128. One of our attacks uses only two related keys and 2^{39} time to recover the complete 256-bit key of a 9-round version of AES-256 (the best previous attack on this variant required 4 related keys and 2^{120} time). Another attack can break a 10 round version of AES-256 in 2^{45} time, but it uses a stronger type of related subkey attack (the best previous attack on this variant required 64 related keys and 2^{172} time). While neither AES-128 nor AES-256 can be directly broken by these attacks, the fact that their hybrid (which combines the smaller number of rounds from AES-128 along with the larger key size from AES-256) can be broken with such a low complexity raises serious concern about the remaining safety margin o ffered by the AES family of cryptosystems.

 \mathcal{ROEHM}

Filed under: Kriptografi