From 56e93bb0072abe3da038068265bc84f9d0c9e994 Mon Sep 17 00:00:00 2001
From: Trance-0 <60459821+Trance-0@users.noreply.github.com>
Date: Sun, 23 Nov 2025 12:06:09 -0600
Subject: [PATCH] Update CSE5313_L24.md

---
 content/CSE5313/CSE5313_L24.md | 43 +++++++++++++++++++++++++++-------
 1 file changed, 34 insertions(+), 9 deletions(-)

diff --git a/content/CSE5313/CSE5313_L24.md b/content/CSE5313/CSE5313_L24.md
index be8239a..9eddab0 100644
--- a/content/CSE5313/CSE5313_L24.md
+++ b/content/CSE5313/CSE5313_L24.md
@@ -37,7 +37,6 @@ Need $P=SM$ worker nodes, and index each one by $s\in [0,S-1], n\in [0,M-1]$.
 
 Worker node $(s,n)$ performs matrix multiplication $\tilde{A}_s^\top\cdot B_n$.
 
-
 $$
 \begin{bmatrix}
 A_0^\top\\
@@ -46,7 +45,7 @@ A_0^\top+A_1^\top
 \end{bmatrix}
 \begin{bmatrix}
 B_0 & B_1
-\enn{bmatrix}
+\end{bmatrix}
 $$
 
 Need $S-M$ responses from each column.
@@ -71,7 +70,7 @@ A_0^\top+A_1^\top
 \end{bmatrix}
 \begin{bmatrix}
 B_0 & B_1 & B_0+B_1
-\enn{bmatrix}
+\end{bmatrix}
 $$
 
 Decodability depends on the pattern.
@@ -95,6 +94,10 @@ Corollary:
 >
 > 1. $K_{1D-MDS}=P-S+M=\Theta(P)$ (linearly)
 > 2. $K_{2D-MDS}=P-(S-M+1)^2+1$.  
+>     - Consider $S\times S$ bipartite graph with $(S-M+1)\times (S-M+1)$ complete subgraph.
+>     - There exists subgraph with all degrees larger than $S-M\implies$ not decodable.
+>     - On the other hand: Fewer than $(S-M+1)^2$ edges cannot form a subgraph with all degrees $>S-M$.
+>     - $K$ scales sub-linearly with $P$.
 > 3. $K_{product}<P-M^2=S^2-M^2=\Theta(\sqrt{P})$
 >
 > Our goal is to get rid of $P$.
@@ -298,22 +301,44 @@ Suppose $P=(r+1)\cdot K$.
 
 ### Linear codes
 
-However, $f$ is a polynomial of degree $d$, not a linear transformation unless $d=1$.
+Recall previous linear computations (matrix-vector):
+
+- $[\tilde{A}_1,\tilde{A}_2,\tilde{A}_3]=[A_1,A_2,A_1+A_2]$ is the corresponding codeword of $[A_1,A_2]$.
+- Every worker node $i$ computes $f(\tilde{A}_i)=\tilde{A}_i x$.
+- $[\tilde{A}_1x, \tilde{A}_2x, \tilde{A}_3x]=[A_1x,A_2x,A_1x+A_2x]$ is the corresponding codeword of $[A_1x,A_2x]$.
+  - This enables to decode $[A_1x,A_2x]$ from $[\tilde{A}_1x,\tilde{A}_2x,\tilde{A}_3 x]$.
+
+However, $f$ is a **polynomial of degree $d$**, not a linear transformation unless $d=1$.
 
 - $f(cX)\neq cf(X)$, where $c$ is a constant.
 - $f(X_1+X_2)\neq f(X_1)+f(X_2)$.
 
+> [!CAUTION]
+> 
+> $[f(\tilde{X}_1),f(\tilde{X}_2),\ldots,f(\tilde{X}_K)]$ is not the codeword corresponding to $[f(X_1),f(X_2),\ldots,f(X_K)]$ in any linear code.
+
 Our goal is to create an encoder/decode such that:
 
 - Linear encoding: is the codeword of $[X_1,X_2,\ldots,X_K]$ for some linear code.
+  - i.e., $[\tilde{X}_1,\tilde{X}_2,\ldots,\tilde{X}_K]=[X_1,X_2,\ldots,X_K]G$ for some generator matrix $G$.
+  - Every $\tilde{X}_i$ is some linear combination of $X_1,\ldots,X_K$.
 - The $f(X_i)$ are decodable from some subset of $f(\tilde{X}_i)$'s.
+  - Some of coded results are missing, erroneous.
 - $X_i$'s are kept private.
 
 ### Lagrange Coded Computing
 
 Let $\ell(z)$ be a polynomial whose evaluations at $\omega_1,\ldots,\omega_{K}$ are $X_1,\ldots,X_K$.
 
-Then every $f(X_i)=f(\ell(\omega_i))$ is an evaluation of polynomial $f\cicc \ell(z)$ at $\omega_i$.
+- That is, $\ell(\omega_i)=X_i$ for every $\omega_i\in \mathbb{F}, i\in [K]$.
+
+Some example constructions:
+
+Given $X_1,\ldots,X_K$ with corresponding $\omega_1,\ldots,\omega_K$
+
+- $\ell(z)=\sum_{i=1}^K X_iL_i(z)$, where $L_i(z)=\prod_{j\in[K],j\neq i} \frac{z-\omega_j}{\omega_i-\omega_j}=\begin{cases} 0 & \text{if } j\neq i \\ 1 & \text{if } j=i \end{cases}$.
+
+Then every $f(X_i)=f(\ell(\omega_i))$ is an evaluation of polynomial $f\circ \ell(z)$ at $\omega_i$.
 
 If the master obtains the composition $h=f\circ \ell$, it can obtain every $f(X_i)=h(\omega_i)$.
 
@@ -334,19 +359,19 @@ Need polynomial $\ell(z)$ such that:
 
 Having obtained such $\ell$ we let $\tilde{X}_i=\ell(\alpha_i)$ for every $i\in [P]$.
 
-$\span{\tilde{X}_1,\tilde{X}_2,\ldots,\tilde{X}_P}=\span{\ell_1(x),\ell_2(x),\ldots,\ell_P(x)}$.
+$span{\tilde{X}_1,\tilde{X}_2,\ldots,\tilde{X}_P}=span{\ell_1(x),\ell_2(x),\ldots,\ell_P(x)}$.
 
 Want $X_k=\ell(\omega_k)$ for every $k\in [K]$.
 
 Tool: Lagrange interpolation.
 
 - $\ell_k(z)=\prod_{i\neq k} \frac{z-\omega_j}{\omega_k-\omega_j}$.
-- $\ell(z)=1$ and $\ell_k(\omega_k)=0$ for every $j\neq k$.
-- $\deg \ell(z)=K-1$.
+- $\ell_k(\omega_k)=1$ and $\ell_k(\omega_k)=0$ for every $j\neq k$.
+- $\deg \ell_k(z)=K-1$.
 
 Let $\ell(z)=\sum_{k=1}^K X_k\ell_k(z)$.
 
-- $\deg \ell=K-1$.
+- $\deg \ell\leq K-1$.
 - $\ell(\omega_k)=X_k$ for every $k\in [K]$.
 
 Let $\tilde{X}_i=\ell(\alpha_i)=\sum_{k=1}^K X_k\ell_k(\alpha_i)$.