update

Dockerfile

@@ -3,6 +3,8 @@
 
 FROM node:18-alpine AS base
 
+ENV NODE_OPTIONS="--max-old-space-size=8192"
+
 # 1. Install dependencies only when needed
 FROM base AS deps
 # Check https://github.com/nodejs/docker-node/tree/b4117f9333da4138b03a546ec926ef50a31506c3#nodealpine to understand why libc6-compat might be needed.

Jenkinsfile

@@ -2,6 +2,7 @@ pipeline {
     environment {
         registry = "trance0/notenextra"
         version = "1.0"
+        NODE_OPTIONS = "--max-old-space-size=8192"
     }
     
     agent any

docker-compose.yaml

@@ -3,7 +3,7 @@ services:
     build:
       context: ./
       dockerfile: ./Dockerfile
-    image: trance0/notenextra:v1.1.6
+    image: trance0/notenextra:v1.1.7
     restart: on-failure:5
     ports:
       - 13000:3000

pages/CSE559A/CSE559A_L14.md

@@ -0,0 +1,77 @@
+# CSE559A Lecture 14
+
+## Neural Network Training
+
+## Object Detection
+
+AP (Average Precision)
+
+### Benchmarks
+
+#### PASCAL VOC Challenge
+
+20 Challenge classes.
+
+CNN increases the accuracy of object detection.
+
+#### COCO dataset
+
+Common objects in context.
+
+Semantic segmentation. Every pixel is classified to tags.
+
+Instance segmentation. Every pixel is classified and grouped into instances.
+
+### Object detection: outline
+
+Proposal generation
+
+Object recognition
+
+#### R-CNN
+
+Proposal generation
+
+Use CNN to extract features from proposals.
+
+with SVM to classify proposals.
+
+Use selective search to generate proposals.
+
+Use AlexNet finetuned on PASCAL VOC to extract features.
+
+Pros:
+
+- Much more accurate than previous approaches
+- Andy deep architecture can immediately be "plugged in"
+
+Cons:
+
+- Not a single end-to-end trainable system
+  - Fine-tune network with softmax classifier (log loss)
+  - Train post-hoc linear SVMs (hinge loss)
+  - Train post-hoc bounding box regressors (least squares)
+- Training is slow 2000CNN passes for each image
+- Inference (detection) was slow
+
+#### Fast R-CNN
+
+Proposal generation
+
+Use CNN to extract features from proposals.
+
+##### ROI pooling and ROI alignment
+
+ROI pooling:
+
+- Pooling is applied to the feature map.
+- Pooling is applied to the proposal.
+
+ROI alignment:
+
+- Align the proposal to the feature map.
+- Align the proposal to the feature map.
+
+Use bounding box regression to refine the proposal.
+
+

pages/CSE559A/_meta.js

@@ -16,4 +16,5 @@ export default {
     CSE559A_L11: "Computer Vision (Lecture 11)",
     CSE559A_L12: "Computer Vision (Lecture 12)",
     CSE559A_L13: "Computer Vision (Lecture 13)",
+    CSE559A_L14: "Computer Vision (Lecture 14)",
 }

pages/Math4121/Math4121_L22.md

@@ -1 +1,123 @@
-# Lecture 22
+# Math 4121 Lecture 22
+
+## Continue on Arzela-Osgood Theorem
+
+
+
+Proof:
+
+Part 2: Control the integral on $\mathcal{U}$
+
+If $[x_i,x_{i+1}]\cap G_k\neq \emptyset$, then $\inf_{x\in [x_i,x_{i+1}]} |f_n(x)| < \frac{\alpha}{2}$ for all $n\geq K$. Denote such set as $P_1$.
+
+Otherwise, we denote such set as $P_2$.
+
+So $\ell(\mathcal{U})=\ell(P_1)+\ell(P_2)\geq c_e(G_K)+\ell(P_2)$.
+
+This implies $\ell(P_2)\leq \frac{\alpha}{4B}$ since $c_e(G_K)\leq c_e(\mathcal{U})+\frac{\alpha}{2B}$.
+
+Thus, for $n\geq K$,
+
+$$
+L(P,f_n)\leq \ell(P_1)\frac{\alpha}{2}+\ell(P_2)B
+$$
+
+So
+
+$$
+\int_\mathcal{U} |f_n(x)| dx \leq c_e(\mathcal{U})\frac{\alpha}{2}+\frac{\alpha}{2}
+$$
+
+All in all,
+
+$$
+\begin{aligned}
+\left\vert \int_\mathcal{U} f_n(x) dx\right\vert &\leq \frac{\alpha}{2}+\frac{\alpha}{2}\\
+&= \int_0^1 |f_n(x)| dx\\
+&\leq \int_\mathcal{U} |f_n(x)| dx + \int_\mathcal{C} |f_n(x)|dx\\
+&\leq c_e(\mathcal{U})\frac{\alpha}{2}+\frac{\alpha}{2}+c_e(\mathcal{C})\frac{\alpha}{2}\\
+&= \alpha
+\end{aligned}
+$$
+
+$\forall N\geq K$.
+
+QED
+
+### Baire Category Theorem
+
+Nowhere dense sets can be large, but they canot cover an open (or closed) interval.
+
+#### Theorem 4.7 (Baire Category Theorem)
+
+An open interval cannot be covered by a countable union of nowhere dense sets.
+
+Proof:
+
+Suppose $(0,1)\subset \bigcup_{n=1}^\infty S_n$ where each $S_n$ is nowhere dense. In particular, $\exists I_1$ closed interval such that $I_1\subset (0,1)$ and $I_1\cap S_1=\emptyset$.
+
+Now for each $k\geq 2$, $S_k$ is not dense in $I_{k-1}$ so $\exists I_k\subsetneq I_{k-1}$ such that $I_k\cap S_k=\emptyset$ for all $j\leq k$.
+
+By nested interval property, $\exists x\in \bigcap_{n=1}^\infty I_n$.
+
+Then $x\in (0,1)$ and $x\notin \bigcup_{n=1}^\infty S_n$.
+
+Contradiction with the assumption that $(0,1)\subset \bigcup_{n=1}^\infty S_n$.
+
+QED
+
+#### Definition First Category
+
+A countable union of nowhere dense sets is called a set of **first category**.
+
+#### Corollary 4.8
+
+Complement of a set of first category in $\mathbb{R}$ is dense in $\mathbb{R}$.
+
+Proof:
+
+We need to show that for every interval $I$, $\exists x\in I\cap S^c$. ($\exists x\in I$ and $x\notin S$)
+
+This is equivalent to the Baire Category Theorem.
+
+QED
+
+Recall a function is pointwise discontinuous if $\mathcal{C}=\{c\in [a,b]: f\text{ is continuous at } c\}$ is dense in $[a,b]$.
+
+$\mathcal{D}=[a,b]\setminus \mathcal{C}$ is called the set of points of discontinuity of $f$.
+
+#### Corollary 4.9
+
+$f$ is pointwise discontinuous if and only if $\mathcal{D}$ is of first category.
+
+Proof:
+
+Part 1: If $\mathcal{D}$ is of first category, then $f$ is pointwise discontinuous.
+
+Immediate from Corollary 4.8.
+
+Part 2: If $f$ is pointwise discontinuous, then $\mathcal{D}$ is of first category.
+
+Let $P_k=\{x\in [a,b]: w(f;x)\geq \frac{1}{k}\}$, $\mathcal{D}=\bigcup_{k=1}^\infty P_k$.
+
+Need to show that each $P_k$ is nowhere dense. (under the assumption that $\mathcal{C)$ is dense).
+
+Let $I\subseteq [a,b]$ so $\exists c\in \mathcal{C}\cap I$. So by definition of $w(f;c)$, $\exists J\subseteq I$ and $c\in J$ such that $w(f;J)\leq \frac{1}{k}$ so for all $x\in J$, $w(f;x)\leq \frac{1}{k}$. so $J\subseteq P_k=\emptyset$.
+
+Thus, $P_k$ is nowhere dense.
+
+QED
+
+#### Corollary 4.10
+
+Let $\{f_n\}$ be a sequence of pointwise discontinuous functions. The set of points at which all $f_n$ are simultaneously continuous is dense (it's also uncountable).
+
+Proof:
+
+$$
+\bigcap_{n=1}^\infty \mathcal{C}_n=\left(\bigcup_{n=1}^\infty \mathcal{D}_n\right)^c
+$$
+
+The complement of a set of first category is dense.
+
+QED