By Hans Dobbertin, Vincent Rijmen, Visit Amazon's Aleksandra Sowa Page, search results, Learn about Author Central, Aleksandra Sowa,
This publication const?tutes the completely refereed postproceedings of the 4th overseas convention at the complicated Encryption ordinary, AES 2004, held in Bonn, Germany in may well 2004.
The 10 revised complete papers provided including an introductory survey and four invited papers through prime researchers have been conscientiously chosen in the course of rounds of reviewing and development. The papers are prepared in topical sections on cryptanalytic assaults and comparable themes, algebraic assaults and comparable effects, implementations, and different subject matters. All in all, the papers represent a most recent review of the cutting-edge of information encryption utilizing the complex Encryption general AES, the de facto global average for facts encryption.
Read Online or Download Advanced Encryption Standard – AES: 4th International Conference, AES 2004, Bonn, Germany, May 10-12, 2004, Revised Selected and Invited Papers PDF
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Extra resources for Advanced Encryption Standard – AES: 4th International Conference, AES 2004, Bonn, Germany, May 10-12, 2004, Revised Selected and Invited Papers
Example text
Due to the ShiftRows and MixColumns transformations, 8 8 , M18 , M68 or on M11 (resp. on M88 , we know that if we induce a fault on M12 8 8 8 M13 , M2 or on M7 ), the result of these bytes after M C ◦SR ◦SB will be XORed 9 9 9 to K15 (resp. K89 to K11 ). So, we want a fault to occur on one of these with K12 8 8 bytes of M and to test if this happens, we look at the faulty ciphertext: if only the 4 bytes (D12 , D9 , D6 , D3 ) (resp. (D8 , D5 , D2 , D15 )) differ from (C12 , C9 , C6 , C3 ) (resp. (C8 , C5 , C2 , C15 )) of the correct ciphertext, this shows that 8 8 8 , M18 , M68 , M11 ) (resp.
In that case, two of the last 4 bytes of the faulty ciphertext will be different from those of the correct ciphertext. We must hence check if this condition is true: if it is not, we abandon this faulty ciphertext and we generate another faulty ciphertext with a fault on K 9 and we test it again. Key Scheduling Key Scheduling MC o SR o SB SR o SB Round 9 Round 10 Fig. 3. Fault on the 14th byte of the penultimate round key K 9 Now, we will see that it is possible to identify: – the position j of the byte on which the fault occurred – and the value ej of this fault.
W, ) , (10) for 1 ≤ w ≤ W . Each coordinate of Vw is an element of {0, 1}n \ 0 (recall that n is the s-box input/output size). Lemma 3. 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. Then for fixed i (1 ≤ i ≤ f ), the values χ(1,i) , . . , χ(W,i) are distinct, and for fixed j (1 ≤ j ≤ ), the values υ (1,j) , . . , υ (W,j) are distinct. In other words, for the set of vectors {Vw }W w=1 , all the values in any one position are distinct.