(i) Find the code words generated by the encoding function with respect to the parity check matrix:
The given parity check matrix 
𝐻
H is:

𝐻
=
[
1
1
0
1
0
0
1
1
0
1
1
0
0
1
0
0
1
0
1
1
0
0
1
0
1
]
H= 
​
  
1
0
1
0
0
​
  
1
1
0
1
0
​
  
0
1
0
0
1
​
  
1
0
1
1
0
​
  
0
1
0
1
1
​
  
​
 


To find the code words generated by the encoding function with respect to the given parity check matrix \( H \), we need to perform the following steps:

### Step 1: Understanding the Parity Check Matrix
The given parity check matrix \( H \) is a \( 3 \times 5 \) matrix, represented as:

\[
H = 
\begin{bmatrix}
1 & 1 & 0 & 1 & 0 \\
0 & 0 & 1 & 1 & 0 \\
1 & 0 & 0 & 1 & 1
\end{bmatrix}
\]

This matrix defines the constraints for the valid code words. Any valid code word \( c \) must satisfy the equation:

\[
H \cdot c^T = 0
\]

where \( c^T \) is the transpose of the code word \( c \), and the result of the matrix-vector multiplication must give the zero vector (because valid code words lie in the null space of \( H \)).

### Step 2: Determine the Generator Matrix \( G \)
To generate code words, we need the generator matrix \( G \). Since \( H \) is a \( 3 \times 5 \) matrix, the code words will be of length 5, and the number of information bits (message bits) will be \( 5 - 3 = 2 \).

To find \( G \), we must express the 5-bit code words as linear combinations of the information bits, using the following relationship:

\[
G \cdot m^T = c^T
\]

where \( m \) is the message vector and \( c \) is the corresponding code word.

To construct \( G \), we start by reducing \( H \) to systematic form (if needed) and deducing the relationship between the information and parity bits. But first, let's solve for code words directly.

### Step 3: Finding Code Words
To find the code words, we can brute-force this by checking all possible 5-bit vectors \( c = (c_1, c_2, c_3, c_4, c_5) \) and testing if they satisfy \( H \cdot c^T = 0 \).

We will test this calculation using the matrix \( H \):

\[
H = 
\begin{bmatrix}
1 & 1 & 0 & 1 & 0 \\
0 & 0 & 1 & 1 & 0 \\
1 & 0 & 0 & 1 & 1
\end{bmatrix}
\]

Let's now find the set of code words satisfying the condition \( H \cdot c^T = 0 \).

I'll calculate the valid codewords based on this condition.The valid codewords generated by the given parity check matrix \( H \) are:

\[
\begin{aligned}
    c_1 &= (0, 0, 0, 0, 0) \\
    c_2 &= (0, 1, 1, 1, 1) \\
    c_3 &= (1, 0, 1, 1, 0) \\
    c_4 &= (1, 1, 0, 0, 1)
\end{aligned}
\]

These four codewords satisfy the condition \( H \cdot c^T = 0 \), meaning they are valid according to the given parity check matrix.

Would you like further details or clarification on any of the steps?

Here are five related questions to expand on this:
1. How can we determine the generator matrix \( G \) from the parity check matrix \( H \)?
2. What is the relationship between the parity check matrix and error detection?
3. How can we encode arbitrary messages using these codewords?
4. What is the significance of the null space in relation to codewords?
5. Can we calculate the Hamming distance between these codewords?

**Tip:** The codewords form a linear block code, and the number of valid codewords is determined by the null space of the parity check matrix \( H \).

Find the code words generated by the encoding function with respect to the parity check matrix H: H = [1 1 0 1 0; 0 0 1 1 0; 1 0 0 1 1]

Learn how to find the codewords generated by a given parity check matrix H = [1 1 0 1 0; 0 0 1 1 0; 1 0 0 1 1]. This tutorial covers key concepts in linear algebra and coding theory, including null spaces and linear block codes. Suitable for undergraduate students studying coding theory.

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