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. The encoding function $$R$$ maps 3-bit vectors from $$B^3$$ to 5-bit code words in $$B^5$$. 3. **Key Concept:** Code words $$\mathbf{c}$$ satisfy $$H \mathbf{c} = \mathbf{0}$$ over $$B$$ (binary field). Since $$H$$ is $$5 \times 3$$, the code words are in the image of $$R$$, which is a subspace of $$B^5$$. 4. **Step to find code words:** Since $$R$$ maps $$B^3$$ to $$B^5$$, and $$H$$ is $$5 \times 3$$, the code words are generated by the null space of $$H^T$$ (transpose of $$H$$), i.e., $$\mathbf{c} = R(\mathbf{m})$$ for $$\mathbf{m} \in B^3$$. 5. **Calculate $$H^T$$:** $$H^T = \begin{pmatrix}1 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 1 \end{pmatrix}$$ 6. **Find generator matrix $$G$$ such that $$G H^T = 0$$:** Since $$H$$ is parity check matrix, $$G$$ is a $$3 \times 5$$ matrix whose rows form a basis for the null space of $$H^T$$. 7. **Express $$G$$ in systematic form:** We want $$G = [I_3 | P]$$ where $$I_3$$ is $$3 \times 3$$ identity and $$P$$ is $$3 \times 2$$ matrix. 8. **From $$H$$, identify $$P$$:** Rewrite $$H$$ as $$H = \begin{pmatrix}P^T & I_2\end{pmatrix}$$ where $$I_2$$ is $$2 \times 2$$ identity. Given $$H$$ is $$5 \times 3$$, this is not standard form, so we find $$G$$ by solving $$H G^T = 0$$. 9. **Enumerate all $$\mathbf{m} \in B^3$$:** $$B^3 = \{(0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,0,1), (1,1,0), (1,1,1)\}$$ 10. **Calculate code words $$\mathbf{c} = R(\mathbf{m})$$:** Since $$R$$ is encoding function, and $$H$$ is parity check matrix, code words satisfy $$H \mathbf{c} = 0$$. 11. **Find all $$\mathbf{c}$$ such that $$H \mathbf{c} = 0$$:** Let $$\mathbf{c} = (c_1, c_2, c_3, c_4, c_5)^T$$. From $$H \mathbf{c} = 0$$: $$\begin{cases} 1 c_1 + 1 c_2 + 0 c_3 = 0 \\ 0 c_1 + 1 c_2 + 1 c_3 = 0 \\ 1 c_1 + 0 c_2 + 0 c_3 = 0 \\ 0 c_1 + 1 c_2 + 0 c_3 = 0 \\ 0 c_1 + 0 c_2 + 1 c_3 = 0 \end{cases}$$ Simplify: - From row 3: $$c_1 = 0$$ - From row 4: $$c_2 = 0$$ - From row 5: $$c_3 = 0$$ Then from row 1: $$c_1 + c_2 = 0$$ is true. From row 2: $$c_2 + c_3 = 0$$ is true. So $$c_1 = c_2 = c_3 = 0$$. 12. **Conclusion:** The only code word is the zero vector $$\mathbf{c} = (0,0,0,0,0)^T$$. **Final answer:** The code words generated by $$H$$ under $$R$$ are only the zero vector $$\boxed{(0,0,0,0,0)}$$. This means the code defined by $$H$$ and $$R$$ is trivial with only one code word.