From afd5a3bf4c0f72e73277fadcd2aed7691186f719 Mon Sep 17 00:00:00 2001 From: Trance-0 <60459821+Trance-0@users.noreply.github.com> Date: Tue, 30 Sep 2025 14:22:24 -0500 Subject: [PATCH] updates --- content/CSE510/_meta.js | 1 + content/CSE5313/CSE5313_L10.md | 232 +++++++++++++++++++++++++++++++++ content/CSE5313/_meta.js | 1 + 3 files changed, 234 insertions(+) create mode 100644 content/CSE5313/CSE5313_L10.md diff --git a/content/CSE510/_meta.js b/content/CSE510/_meta.js index 3a5efda..aa4ef7e 100644 --- a/content/CSE510/_meta.js +++ b/content/CSE510/_meta.js @@ -13,4 +13,5 @@ export default { CSE510_L8: "CSE510 Deep Reinforcement Learning (Lecture 8)", CSE510_L9: "CSE510 Deep Reinforcement Learning (Lecture 9)", CSE510_L10: "CSE510 Deep Reinforcement Learning (Lecture 10)", + CSE510_L11: "CSE510 Deep Reinforcement Learning (Lecture 11)", } \ No newline at end of file diff --git a/content/CSE5313/CSE5313_L10.md b/content/CSE5313/CSE5313_L10.md new file mode 100644 index 0000000..db7ce6c --- /dev/null +++ b/content/CSE5313/CSE5313_L10.md @@ -0,0 +1,232 @@ +# CSE5313 Coding and information theory for data science (Recitation 10) + +## Question 2 + +Let $C$ be a Reed-Solomon code generated by + +$$ +G=\begin{bmatrix} +1 & 1 & \cdots & 1\\ +\alpha_1 & \alpha_2 & \cdots & \alpha_n\\ +\alpha_1^2 & \alpha_2^2 & \cdots & \alpha_n^2\\ +\vdots & \vdots & \cdots & \vdots\\ +\alpha_1^{k-1} & \alpha_2^{k-1} & \cdots & \alpha_n^{k-1} +\end{bmatrix} +$$ + +prove that there exists $v_1,v_2,\ldots,v_n\in \mathbb{F}_q\setminus \{0\}$ such that the parity check matrix is + +$$ +H=\begin{bmatrix} +1 & 1 & \cdots & 1\\ +\alpha_1 & \alpha_2 & \cdots & \alpha_n\\ +\alpha_1^2 & \alpha_2^2 & \cdots & \alpha_n^2\\ +\vdots & \vdots & \cdots & \vdots\\ +\alpha_1^{k-1} & \alpha_2^{k-1} & \cdots & \alpha_n^{k-1} +\end{bmatrix}\begin{bmatrix} +v_1 & 0 & \cdots & 0\\ +0 & v_2 & \cdots & 0\\ +\vdots & \vdots & \ddots & \vdots\\ +0 & 0 & \cdots & v_n +\end{bmatrix} +$$ + +### Some lemmas for linear codes + +First we introduce the following lemmas for linear codes + +Let $G$ and $H$ be the generator and parity-check matrices of (any) linear code + +#### Lemma 1 + +$$ +H G^T = 0 +$$ + +
+Proof + +By definition of generator matrix and parity-check matrix, $forall e_i\in H$, $e_iG^T=0$. + +So $H G^T = 0$. +
+ +#### Lemma 2 + +Any matrix $M\in \mathbb{F}_q^{(n-k)\times n}$ such that $\operatorname{rank}(M) = n - k$ and $M G^T = 0$ is a parity-check matrix for $C$ (i.e. $C = \ker M$). + +
+Proof + +It is sufficient to show that the two statements + +1. $\forall c\in C, c=uG, u\in \mathbb{F}^k$ + +$M c^T = M(uG)^T = M(G^T u^T) = 0$ since $M G^T = 0$. + +Thus $C \subseteq \ker M$. + +2. $\dim (\ker M) +\operatorname{rank}(M) = n$ + +We proceed by showing that $\dim (\ker M) =\dim (C)$. + +Suppose $C$ does not span $\ker M$. Let $u_1,...,u_k$ be a basis for $C$. Then there exists $v\in \ker M\setminus C$. + +By linear independence, if we have scalar $a_1,...,a_k$ and $b$ such that $a_1u_1+...+a_ku_k+bv=0$, then $a_1=...=a_k=b=0$. So $v=a_1u_1+...+a_ku_k$ for some $a_1,...,a_k$. + +By definition of linear code, we have $v\in C$, contradicting the assumption. + +
+ +### Solution + +We proceed by applying the lemma 2. + +1. $\operatorname{rank}(H) = n - k$ since $H$ is a Vandermonde matrix times a diagonal matrix with no zero entries, so $H$ is invertible. + +2. $H G^T = 0$. + +note that $\forall$ row $i$ of $H$, $0\leq i\leq n-k-1$, $\forall$ column $j$ of $G^T$, $0\leq j\leq k-1$ + +So + +$$ +\begin{aligned} +H G^T &= \begin{bmatrix} +1 & 1 & \cdots & 1\\ +\alpha_1 & \alpha_2 & \cdots & \alpha_n\\ +\alpha_1^2 & \alpha_2^2 & \cdots & \alpha_n^2\\ +\vdots & \vdots & \cdots & \vdots\\ +\alpha_1^{k-1} & \alpha_2^{k-1} & \cdots & \alpha_n^{k-1} +\end{bmatrix}\begin{bmatrix} +v_1 & 0 & \cdots & 0\\ +0 & v_2 & \cdots & 0\\ +\vdots & \vdots & \ddots & \vdots\\ +0 & 0 & \cdots & v_n +\end{bmatrix}\begin{bmatrix} +1 & \alpha_1 & \alpha_1^2 & \cdots & \alpha_1^{k-1}\\ +1 & \alpha_2 & \alpha_2^2 & \cdots & \alpha_2^{k-1}\\ +\vdots & \vdots & \vdots & \cdots & \vdots\\ +1 & \alpha_n & \alpha_n^2 & \cdots & \alpha_n^{k-1} +\end{bmatrix}\\ +&= +\begin{bmatrix} +1 & 1 & \cdots & 1\\ +\alpha_1 & \alpha_2 & \cdots & \alpha_n\\ +\alpha_1^2 & \alpha_2^2 & \cdots & \alpha_n^2\\ +\vdots & \vdots & \cdots & \vdots\\ +\alpha_1^{k-1} & \alpha_2^{k-1} & \cdots & \alpha_n^{k-1} +\end{bmatrix} +\begin{bmatrix} +v_1\\ +v_2\\ +\vdots\\ +v_n +\end{bmatrix}\\ +&=\sum_{l=1}^n\alpha_l^{r}v_l=0 +\end{aligned} +$$ + +## Question 3 + +Show that in an MDS code $[n,k,d]_{\mathbb{F}_q}$, $d=n-k+1$, every $k$ entries determine the remaining $n-k$ entries of $G$. + +That is, every $k\times k$ submatrix of $G$ is invertible. + +
+Proof + +Let $G$ be the generator matrix, and $G'$ be any $k\times k$ submatrix of $G$. + +$$ +G=\begin{bmatrix} +G'\in \mathbb{F}^{k\times k}|G''\in \mathbb{F}^{k\times (n-k)} +\end{bmatrix} +$$ + +We proceed by contradiction, suppose $G'$ is not invertible. + +Then there exists $m',m''\in \mathbb{F}_q^k$ such that $m'\neq m''$ but $m'G'=m''G'$. + +Note that $m'G=[m'G'|m'G'']$ and $m''G=[m''G'|m''G'']$. + +So there are only $n-k$ entries of $m'G$ and $m''G$ are different, so $d\leq n-k$. + +That violate with the assumption that $d=n-k+1$. +
+ +## Reed-Muller code + +### Definition of Reed-Muller code (binary case) + +$$ +RM(r,m)=\left\{(f(\alpha_1),\ldots,f(\alpha_2^m))|\alpha_i\in \mathbb{F}_2^m,\deg f\leq r\right\} +$$ + +Length of $RM(r,m)$ is $2^m$. + +
+Example of Reed-Muller code + +Let $r=2$, $m=3$. + +$\alpha_1=(0,0,0)$, $\alpha_2=(0,0,1)$, $\alpha_3=(0,1,0)$, $\alpha_4=(0,1,1)$, $\alpha_5=(1,0,0)$, $\alpha_6=(1,0,1)$, $\alpha_7=(1,1,0)$, $\alpha_8=(1,1,1)$. + +$p(x)\deg\leq 2=\{1,x_1,x_2,x_3,x_1x_2,x_1x_3,x_2x_3\}$. + +So $p(x)=1+x_1+x_2+x_3+x_1x_2+x_1x_3+x_2x_3$. + +The generator matrix is defined by + +$$ +G=\begin{bmatrix}1\\x_1\\x_2\\x_3\\x_1x_2\\x_1x_3\\x_2x_3\end{bmatrix}\begin{bmatrix} +1& 1&1&1&1&1&1&1\\ +0& 0&0&0&1&1&1&1\\ +0& 0&1&1&0&0&1&1\\ +0& 1&0&1&0&1&0&1\\ +0& 0&0&0&0&0&1&1\\ +0& 0&0&0&0&1&0&1\\ +0& 0&0&1&0&0&0&1\\ +\end{bmatrix} +$$ + +
+ +So $\dim RM(r,m)=\sum_{i=0}^{r}\binom{m}{i}$. + +## Question 4 + +$RM(m-1,m)$ is the parity check code. + +
+Proof + +By previous lemma, it is sufficient to show that $\dim (RM(m-1,m))=n-1$ and $RM(m-1,m)\subseteq \text{ parity code}$. + +For the first property, + +$$ +\dim (RM(m-1,m))=\sum_{i=0}^{m-1}\binom{m}{i}=\sum_{i=0}^{m}\binom{m}{i}-\binom{m}{m}=2^m-1=n-1 +$$ + +For the second property, + +recall that $c=(c_1,c_2,\ldots,c_n)\in$ parity if $\sum_{i=1}^n c_i=0$. + +So we need to show that $\sum_{i=1}^{2^m}f(\alpha_i)=0$ for every $f$ with $\deg f\leq m-1$. + +Note that $\forall f\in RM(m-1,m)$, we can write $f(x_1,x_2,\ldots,x_m)=\sum_{S\subseteq [m],|S|\leq m-1}f_Sx_S$ + +So $\sum_{i=1}^{2^m}f(\alpha_i)=\sum_{S\subseteq [m],|S|\leq m-1}f_S\sum_{i=1}^{2^m}x_S(\alpha_i)=0$. + +We denote $J(s)=\sum_{i=1}^{2^m}x_S(\alpha_i)$. + +Note that $J(S)=\begin{cases} +1 & \text{if number of 1 in $S$ is odd} +0 & \text{otherwise} +\end{cases}$ + +And number of 1 in $S$ is $2^{m-|S|}$ which is even. + +So $J(S)=0$ for all $S\subseteq [m],|S|\leq m-1$. +
\ No newline at end of file diff --git a/content/CSE5313/_meta.js b/content/CSE5313/_meta.js index 35560ed..3ca12dc 100644 --- a/content/CSE5313/_meta.js +++ b/content/CSE5313/_meta.js @@ -12,4 +12,5 @@ export default { CSE5313_L7: "CSE5313 Coding and information theory for data science (Lecture 7)", CSE5313_L8: "CSE5313 Coding and information theory for data science (Lecture 8)", CSE5313_L9: "CSE5313 Coding and information theory for data science (Lecture 9)", + CSE5313_L10: "CSE5313 Coding and information theory for data science (Recitation 10)", } \ No newline at end of file