Suppose we want an error-correcting code that will allow all single-bit errors to be

corrected for memory words of length 12.

a) How many parity bits are necessary?

b) Assuming we are using the Hamming algorithm presented in this chapter to design our error-correcting code, find the code word to represent the 12-bit information word:

100100011010

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.

\[
\begin{aligned}
&\text{Bit positions:} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 \\
&\text{Data/parity:} & P1 & P2 & d1 & P4 & d2 & d3 & d4 & P8 & d5 & d6 & d7 & d8 & d9 & d10 & d11 & P16 & d12 \\
&\text{Data bits:} &  &  & 1 &  & 0 & 0 & 1 &  & 0 & 0 & 1 & 1 & 0 & 1 &  & 0
\end{aligned}
\]

#### 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.

Suppose we want an error-correcting code that will allow all single-bit errors to be corrected for memory words of length 12. a) How many parity bits are necessary? b) Assuming we are using the Hamming algorithm presented in this chapter to design our error-correcting code, find the code word to represent the 12-bit information word: 100100011010

Discover how to design an error-correcting code using the Hamming algorithm for a 12-bit memory word. Learn how to calculate the number of necessary parity bits and find the 17-bit code word for the information word 100100011010. This problem covers Hamming codes, binary systems, and error correction, suitable for Grades 10-12 and university-level students.

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