The problem asks to generate a code word using the Hamming code method for the data "0101." It also asks to detect and correct any errors in the received code word "0101010."

Step 1: Generating the Hamming Code

Hamming code is a linear error-correcting code that adds redundancy bits to detect and correct single-bit errors in data transmission. Here's how it's done:

For a 4-bit data word "0101", we will need to add redundancy (parity) bits at positions that are powers of 2 (i.e., positions 1, 2, 4, 8, etc.). For this 4-bit data, we will add 3 parity bits at positions 1, 2, and 4.

Step 2: Arranging Data and Parity Bits

The positions of the parity bits are marked as "P", and the data bits are placed in the remaining positions:

We now place the data bits (0101) in the positions labeled D:

Step 3: Calculating Parity Bits

• P1 (covers bits 1, 3, 5, 7): $P1 = D1 \oplus D2 \oplus D4 = 0 \oplus 1 \oplus 1 = 0$
• P2 (covers bits 2, 3, 6, 7): $P2 = D1 \oplus D3 \oplus D4 = 0 \oplus 0 \oplus 1 = 1$
• P3 (covers bits 4, 5, 6, 7): $P3 = D2 \oplus D3 \oplus D4 = 1 \oplus 0 \oplus 1 = 0$

Now, placing the parity bits into their respective positions, the final Hamming code (encoded word) is:

The generated Hamming code is 0100101.

Step 4: Detecting the Error in the Received Code Word

The received code word is given as 0101010. We will check each parity bit to determine if there is an error.

Recalculate the Parity Bits for the Received Code Word:

• P1 (bits 1, 3, 5, 7): $P1 = D1 \oplus D2 \oplus D4 = 0 \oplus 1 \oplus 0 = 1$
• P2 (bits 2, 3, 6, 7): $P2 = D1 \oplus D3 \oplus D4 = 1 \oplus 1 \oplus 0 = 0$
• P3 (bits 4, 5, 6, 7): $P3 = D2 \oplus D3 \oplus D4 = 1 \oplus 1 \oplus 0 = 0$

The error syndrome (binary number formed by the incorrect parity checks) is 101, which equals 5 in decimal. This means there is an error in position 5.

Step 5: Correcting the Error

The bit in position 5 is currently 1, so we flip it to 0. The corrected code word becomes 0100010.

Thus, the original transmitted code word was 0100101, and the corrected version of the received word is 0100010.

Let me know if you'd like further clarification or have any questions.

Here are five related questions to explore further:

1. How does the Hamming distance help in detecting multiple-bit errors?
2. Can Hamming code correct more than one error in a received message?
3. What is the difference between even and odd parity in Hamming codes?
4. How would the Hamming code method change if the data word had more than 4 bits?
5. How does the position of parity bits in Hamming code vary with different data lengths?

Tip: Always double-check parity calculations when working with error-correcting codes to avoid missing subtle mistakes.