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by pclmulqdq
516 days ago
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I believe that the polynomial for GF2P8AFFINEQB is user-defined. One argument is an 8x8 matrix in GF(2) and the result is [A.x + b] in GF(2)^8 for each 8-bit section. Don't quote me on this, but I believe that matrix multiply in GF(2)^8 gets you a transform in GF(2^8). |
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