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Fast Software Encryption: 15th International Workshop, FSE ...

We have that ADi+3 = 0, AHi+3 = 0, AE0(Ai+3) = 0, ASi(Ei+3) = 0. We require AAi+4 = 0, AEi+4 = 0. So we deduce: AMaf+3(0,0,l) = 0, (7) AWi+3 = -AChi+3{0,-l ...

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