Advanced Encryption Standard - AES: 4th International by Hans Dobbertin, Vincent Rijmen, Visit Amazon's Aleksandra

By Hans Dobbertin, Vincent Rijmen, Visit Amazon's Aleksandra Sowa Page, search results, Learn about Author Central, Aleksandra Sowa,

This ebook constítutes the completely refereed postproceedings of the 4th overseas convention at the complex Encryption regular, AES 2004, held in Bonn, Germany in may perhaps 2004. the ten revised complete papers provided including an introductory survey and four invited papers via major researchers have been rigorously chosen in the course of rounds of reviewing and development. The papers are geared up in topical sections on cryptanalytic assaults and similar subject matters, algebraic assaults and similar effects, implementations, and different themes. All in all, the papers represent a newest evaluation of the cutting-edge of information encryption utilizing the complicated Encryption regular AES, the de facto international usual for info encryption.

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Additional resources for Advanced Encryption Standard - AES: 4th International Conference, AES 2004, Bonn, Germany, May 10-12, 2004, Revised Selected and Invited Papers

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If we denote this byte the k th (0 ≤ k ≤ 3), j is then defined by j = (k + 1 mod 4) + 12 (7) By computing C ⊕ D, we determine k and thus obtain j. , 15}, Ci = SubByte(MShif tRow−1 (i) ) ⊕ Ki (8) More precisely: - if i = 0: 9 9 Ci = SubByte(MShif tRow−1 (i) ) ⊕ SubByte(K(i+1 mod 4)+12 ) ⊕ Ki9 ⊕ 0x36 (9) - if i ∈ {1, 2, 3}: 9 9 Ci = SubByte(MShif tRow−1 (i) ) ⊕ SubByte(K(i+1 mod 4)+12 ) ⊕ Ki9 (10) We also have for the faulty ciphertext: 9 10 Dj = SubByte(MShif tRow−1 (j) ) ⊕ Kj ⊕ ej (11) 34 C. Giraud and - if k = 0: 9 9 9 Dk = SubByte(MShif tRow−1 (k) ) ⊕ SubByte(Kj ⊕ ej ) ⊕ Kk ⊕ 0x36 (12) - if k ∈ {1, 2, 3}: 9 9 9 Dk = SubByte(MShif tRow−1 (k) ) ⊕ SubByte(Kj ⊕ ej ) ⊕ Kk (13) It is easy to see, from (8) and (11), that the value of the fault ej is equal to Cj ⊕ Dj .

Proof. Contained in the proof of Theorem 1 in [23]. Remark 5. Clearly if wt(γa ) + wt(γb ) = Bl , then Wl [γa , γb ] ≤ (2n − 1). Further, the values χ(w,i) and υ (w,j) depend only on γa and γb , not on the specific values of a and b. Lemma 4. Given a, b ∈ {0, 1}N \ 0 that satisfy wt(γa ) + wt(γb ) > Bl , let W = Wl [γa , γb ], f = wt(γa ), = wt(γb ), and let χ(w,i) , υ (w,j) be defined as above. Consider the vectors Vw in (10). Select any (f + −Bl ) vector positions, and fix a value in {0, 1}n \ 0 for each position.

Kuhn. Low cost attacks on tamper resistant devices. In B. Christianson, B. Crispo, T. Mark, A. Lomas, and M. Roe, editors, 5th Security Protocols Workshop, volume 1361 of Lecture Notes in Computer Science, pages 125–136. Springer-Verlag, 1997. 3. I. Biehl, B. Meyer, and V. M¨ uller. Differential Fault Analysis on Elliptic Curve Cryptosystems. In M. Bellare, editor, Advances in Cryptology – CRYPTO 2000, volume 1880 of Lecture Notes in Computer Science, pages 131–146. SpringerVerlag, 2000. 4. E. Biham and A.

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