Now, at line 6 we set M2 = ( IH3 a D0 a B29) al 9 a G(B29 , C29 , D29) a A29 a K30 (8) With this new value of M2 we get ... In short, the attack is based on the fact that M2 appears at the very beginning and that M6 and M9 appear at the very endanbsp;...
|Title||:||Selected Areas in Cryptography|
|Author||:||Roberto Avanzi, Liam Keliher, Francesco Sica|
|Publisher||:||Springer - 2009-08-22|