1. **Problem Statement:** Find the code words generated by the parity check matrix $$H=\begin{pmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$$ when the encoding function is $$R: B^3 \to B^5$$. 2. **Understanding the problem:** The parity check matrix $$H$$ is a $$5 \times 3$$ matrix, and the encoding function maps 3-bit vectors to 5-bit code words. Usually, the parity check matrix $$H$$ satisfies $$H \cdot c = 0$$ for any code word $$c$$ in the code. 3. **Step to find code words:** Since $$H$$ is $$5 \times 3$$, the code words $$c$$ are length 3 vectors in $$B^3$$. The encoding function $$R$$ maps these 3-bit vectors to 5-bit code words in $$B^5$$. 4. **Find the generator matrix $$G$$:** The generator matrix $$G$$ is a $$3 \times 5$$ matrix such that $$H \cdot G^T = 0$$. Since $$H$$ is $$5 \times 3$$, $$G$$ must be $$3 \times 5$$. 5. **Given $$H$$, find $$G$$:** The matrix $$H$$ can be written as $$H = \begin{pmatrix}P^T & I_5\end{pmatrix}$$ if it is in standard form, but here $$H$$ is $$5 \times 3$$, so the standard form is not straightforward. Instead, since $$R: B^3 \to B^5$$, the code words are generated by $$c = R(m)$$ for message $$m \in B^3$$. 6. **Calculate code words:** For each $$m = (m_1, m_2, m_3) \in B^3$$, the code word $$c = R(m)$$ is a 5-bit vector. Since $$H$$ is the parity check matrix, code words satisfy $$H \cdot c = 0$$. 7. **Find all code words:** The code words are the null space of $$H$$. Since $$H$$ is $$5 \times 3$$, the null space is the set of all $$c \in B^3$$ such that $$H \cdot c = 0$$. 8. **Calculate $$H \cdot c$$ for all $$c \in B^3$$:** The 8 possible vectors in $$B^3$$ are: $$000, 001, 010, 011, 100, 101, 110, 111$$. 9. **Compute $$H \cdot c$$ for each:** - For $$c=000$$: $$H \cdot 000 = 0$$ - For $$c=001$$: $$H \cdot 001 = \begin{pmatrix}0 \\ 1 \\ 0 \\ 0 \\ 1\end{pmatrix} \neq 0$$ - For $$c=010$$: $$H \cdot 010 = \begin{pmatrix}1 \\ 1 \\ 0 \\ 1 \\ 0\end{pmatrix} \neq 0$$ - For $$c=011$$: $$H \cdot 011 = H \cdot (010 + 001) = H \cdot 010 + H \cdot 001 = \begin{pmatrix}1 \\ 0 \\ 0 \\ 1 \\ 1\end{pmatrix} \neq 0$$ - For $$c=100$$: $$H \cdot 100 = \begin{pmatrix}1 \\ 0 \\ 1 \\ 0 \\ 0\end{pmatrix} \neq 0$$ - For $$c=101$$: $$H \cdot 101 = H \cdot (100 + 001) = H \cdot 100 + H \cdot 001 = \begin{pmatrix}1 \\ 1 \\ 1 \\ 0 \\ 1\end{pmatrix} \neq 0$$ - For $$c=110$$: $$H \cdot 110 = H \cdot (100 + 010) = H \cdot 100 + H \cdot 010 = \begin{pmatrix}0 \\ 1 \\ 1 \\ 1 \\ 0\end{pmatrix} \neq 0$$ - For $$c=111$$: $$H \cdot 111 = H \cdot (100 + 010 + 001) = H \cdot 100 + H \cdot 010 + H \cdot 001 = \begin{pmatrix}1 \\ 0 \\ 1 \\ 1 \\ 1\end{pmatrix} \neq 0$$ 10. **Conclusion:** Only $$c=000$$ satisfies $$H \cdot c = 0$$, so the only code word generated by $$H$$ is $$000$$. **Final answer:** The code words generated by the parity check matrix $$H$$ under the encoding function $$R: B^3 \to B^5$$ is only $$\{000\}$$.