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Concerning $(k_1, k_2, k_1)$ vs. $(k_1, k_1, k_2)$ $(k_1, k_1, k_2)$ can be split into $(k_1, k_1)$ with $2^{56}$ possibilities and $(k_2)$ with $2^{56}$ possibilities, so the meet-in-the-middle attack has cost $2^{56}$. $(k_1, k_2, k_1)$ can be split into $(k_1, k_2)$ with $2^{112}$ possibilities and $(k_2)$ with $2^{56}$ possibilities or into $(k_1)$ ...



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