 encryption. Plaintext 1001001 1101101 1101001 1110011 1110011 1111001 1101111 1110101 Alice's key 0001001 1010100 0101011 0111101 0111010 0001001 1001001 1100100 Step 1 1000000 0111001 1000010 1001110 1001001 1110000 0100110 0010001 Bob's key 1011001 1000111 0111010 1111101 0101110 0011001 1001000 0001011 Step 2 0011001 1111110 1111000 0110011 1100111 1101001 1101110 0011010 Alice's key 0001001 1010100 0101011 0111101 0111010 0001001 1001001 1100100 Step 3 0010000 0101010 1010011 0001110 1011101 1100000 0100111 1111110 Bob's key 1011001 1000111 0111010 1111101 0101110 0011001 1001000 0001011 Step 4 1001001 1101101 1101001 1110011 1110011 1111001 1101111 1110101 Plaintext Table 25.5 Intercepting the message in Table 25.4. Step 1 1000000 0111001 1000010 1001110 1001001 1110000 0100110 0010001 Step 2 0011001 1111110 1111000 0110011 1100111 1101001 1101110 0011010 1011001 1000111 0111010 1111101 0101110 0011001 1001000 0001011 Step 3 0010000 0101010 1010011 0001110 1011101 1100000 0100111 1111110 1001001 1101101 1101001 1110011 1110011 1111001 1101111 1110101 Plaintext I steps, Alice and Bob only used their own keys. The explanation of this successful retrieval is the following. Call M the plaintext message block and A, B the keys from Alice and Bob. The cipher in step 1 is, thus, M XOR A, or M + A. The cipher in step 2 is M + A + B. That in step 3 is M + B = M + B (using the above- described property of XOR). Finally, the result in step 4 is M + B + B = M, which is the plaintext. While the previous double-key encryption surely works without any key exchange, it is absolutely unsafe! Indeed, if Eve is ever able to intercept all three messages (ciphers in step 1, step 2, and step 3 in the above), the only thing she has to do to retrieve the plaintext is to XOR these together, as shown in Table 25.5. There is nothing surprising in this result if we consider from the previous definitions that what Eve does is the operation (M + + B) = MOMOM = M. Since Eve can SO effortlessly retrieve the plaintext, without even any cry