To solve this problem, we'll need to break it into two parts: determining how many parity bits are required and then constructing the code word using the Hamming algorithm.

Part a: How many parity bits are necessary?

To determine the number of parity bits needed for error correction, we use the Hamming code formula:

$2^r \geq n + r + 1$

Where:

• $r$ is the number of parity bits,
• $n$ is the number of data bits (which is 12 in this case),
• $r + n$ is the total number of bits (data + parity bits),
• $1$ is added to account for the "syndrome" bit (to detect the error position).

We need to find the smallest $r$ that satisfies this inequality.

Substituting $n = 12$:

$2^r \geq 12 + r + 1$ $2^r \geq 13 + r$

Now, we test values of $r$:

• For $r = 3$, $2^3 = 8$, but $13 + 3 = 16$ (so this is not enough).
• For $r = 4$, $2^4 = 16$, but $13 + 4 = 17$ (still not enough).
• For $r = 5$, $2^5 = 32$, and $13 + 5 = 18$ (this satisfies the condition).

Thus, 5 parity bits are needed.

---

Part b: Find the code word using the Hamming algorithm.

Now, we'll construct the 17-bit code word (12 data bits + 5 parity bits).

The positions of the parity bits are powers of 2: 1, 2, 4, 8, and 16. The data bits will fill the remaining positions. The parity bits will be calculated later.

The 12-bit data word is:
$100100011010$

Step 1: Place data bits in non-parity positions

Start by positioning the data bits (represented by d1, d2, ...) in the code word.

&\text{Bit positions:} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 \\ &\text{Data/parity:} & P1 & P2 & d1 & P4 & d2 & d3 & d4 & P8 & d5 & d6 & d7 & d8 & d9 & d10 & d11 & P16 & d12 \\ &\text{Data bits:} & & & 1 & & 0 & 0 & 1 & & 0 & 0 & 1 & 1 & 0 & 1 & & 0 \end{aligned}$$ #### Step 2: Calculate the parity bits Each parity bit checks certain positions: - **P1** checks positions 1, 3, 5, 7, 9, 11, 13, 15, 17. - **P2** checks positions 2, 3, 6, 7, 10, 11, 14, 15. - **P4** checks positions 4, 5, 6, 7, 12, 13, 14, 15. - **P8** checks positions 8, 9, 10, 11, 12, 13, 14, 15. - **P16** checks positions 16, 17. Now, calculate each parity bit so that the total number of 1s in the positions it checks is even. 1. **P1** checks positions: 1, 3, 5, 7, 9, 11, 13, 15, 17. - Bits: 1, 0, 0, 1, 0, 1, 0 (sum of 1s is 3, so P1 = 1). 2. **P2** checks positions: 2, 3, 6, 7, 10, 11, 14, 15. - Bits: 0, 1, 0, 1, 0, 1, 0 (sum of 1s is 3, so P2 = 1). 3. **P4** checks positions: 4, 5, 6, 7, 12, 13, 14, 15. - Bits: 0, 0, 0, 1, 1, 1, 0 (sum of 1s is 3, so P4 = 1). 4. **P8** checks positions: 8, 9, 10, 11, 12, 13, 14, 15. - Bits: 0, 0, 0, 1, 1, 1, 0 (sum of 1s is 3, so P8 = 1). 5. **P16** checks positions: 16, 17. - Bits: 0, 0 (sum of 1s is 0, so P16 = 0). #### Final Code Word Now that we have the parity bits, the complete code word is: $$\boxed{11011001001101000}$$ --- ### Summary: - a) We need **5 parity bits**. - b) The code word for the 12-bit information word $$100100011010$$ is $$\boxed{11011001001101000}$$. Would you like more details or have any questions? Here are 5 related questions to deepen understanding: 1. What is the maximum number of errors a Hamming code can detect? 2. How would the code change if we used double-bit error correction? 3. Can a Hamming code correct burst errors? 4. What is the structure of the parity-check matrix in Hamming codes? 5. How does the syndrome decoding process work in Hamming codes? **Tip:** Always check the placement of parity bits in Hamming codes, as incorrect positions can lead to misidentification of errors.