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for i in reversed(range(16)): # At this point we know A_i (= state). We need to find the byte c[i] # and the previous accumulator A_i-1. # The relation is: # state = rol64(prev,5) ^ (c * M) # => rol64(prev,5) = state ^ (c * M) # Since rol64 is bijective we can invert it: # prev = ror64(state ^ (c * M),5) # We just need to find the unique c (0..255) that makes the next iteration # consistent. Because the algorithm is linear, *any* c yields a valid prev, # but only one will lead to a final accumulator of zero after 16 steps. # The easiest way: brute‑force the 256 possibilities for this byte, keep the # candidate that makes the accumulator after processing *all* remaining bytes # equal to zero. Since we go backwards, the first candidate we find is the # correct one. dosprn crack

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Let B_i = A_i // known (starting from TARGET_HASH) For i = 15 downto 0: // We need to find A_i-1 such that: // B_i = ROL64(A_i-1,5) XOR (c[i] * M) // => ROL64(A_i-1,5) = B_i XOR (c[i] * M) // => A_i-1 = ROR64( B_i XOR (c[i] * M), 5 )