update build and types

This commit is contained in:
Zheyuan Wu
2025-04-16 10:28:17 -05:00
parent 99ab8173af
commit aca0058a8e
3 changed files with 66 additions and 20 deletions

17
Jenkinsfile vendored
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@@ -20,12 +20,25 @@ pipeline {
steps { steps {
script { script {
echo "Building docker image ${registry}:${version}.${env.BUILD_ID}" echo "Building docker image ${registry}:${version}.${env.BUILD_ID}"
def customImage = docker.build("${registry}:${version}.${env.BUILD_ID}") def customImage = docker.build("${registry}:v${version}.${env.BUILD_ID}")
echo "Pushing docker image ${registry}:${version}.${env.BUILD_ID}" echo "Pushing docker image ${registry}:v${version}.${env.BUILD_ID}"
customImage.push() customImage.push()
} }
} }
} }
stage('Deploy') {
steps {
echo "Deploying docker image ${registry}:v${version}.${env.BUILD_ID}"
echo "Stopping existing container"
sh 'docker stop notenextra'
echo "Removing existing container"
sh 'docker rm notenextra'
echo "Running new docker container"
sh 'docker run -d -p 13000:3000 --name notenextra ${registry}:v${version}.${env.BUILD_ID}'
}
}
} }
} }
}

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@@ -166,11 +166,11 @@ y \\
\end{pmatrix} \end{pmatrix}
$$ $$
Constraint from a match $(x_i,x_i^)$: $x_i^≅Hx_i$ Constraint from a match $(x_i,x_i^')$: $x_i^'≅Hx_i$
How can we get rid of the scale ambiguity? How can we get rid of the scale ambiguity?
Cross product trick: $x_i^ × Hx_i=0$ Cross product trick: $x_i^' × Hx_i=0$
The cross product is defined as: The cross product is defined as:
@@ -181,9 +181,9 @@ $$
Let $h_1^T, h_2^T, h_3^T$ be the rows of $H$. Then Let $h_1^T, h_2^T, h_3^T$ be the rows of $H$. Then
$$ $$
x_i^ × Hx_i=\begin{pmatrix} x_i^' × Hx_i=\begin{pmatrix}
x_i^ \\ x_i^' \\
y_i^ \\ y_i^' \\
1 1
\end{pmatrix} \times \begin{pmatrix} \end{pmatrix} \times \begin{pmatrix}
h_1^T x_i \\ h_1^T x_i \\
@@ -192,18 +192,18 @@ x_i^ × Hx_i=\begin{pmatrix}
\end{pmatrix} \end{pmatrix}
= =
\begin{pmatrix} \begin{pmatrix}
y_i^ h_3^T x_ih_2^T x_i \\ y_i^' h_3^T x_ih_2^T x_i \\
h_1^T x_ix_i^ h_3^T x_i \\ h_1^T x_ix_i^' h_3^T x_i \\
x_i^ h_2^T x_iy_i^ h_1^T x_i x_i^' h_2^T x_iy_i^' h_1^T x_i
\end{pmatrix} \end{pmatrix}
$$ $$
Constraint from a match $(x_i,x_i^)$: Constraint from a match $(x_i,x_i^')$:
$$ $$
x_i^ × Hx_i=\begin{pmatrix} x_i^' × Hx_i=\begin{pmatrix}
x_i^ \\ x_i^' \\
y_i^ \\ y_i^' \\
1 1
\end{pmatrix} \times \begin{pmatrix} \end{pmatrix} \times \begin{pmatrix}
h_1^T x_i \\ h_1^T x_i \\
@@ -212,9 +212,9 @@ x_i^ × Hx_i=\begin{pmatrix}
\end{pmatrix} \end{pmatrix}
= =
\begin{pmatrix} \begin{pmatrix}
y_i^ h_3^T x_ih_2^T x_i \\ y_i^' h_3^T x_ih_2^T x_i \\
h_1^T x_ix_i^ h_3^T x_i \\ h_1^T x_ix_i^' h_3^T x_i \\
x_i^ h_2^T x_iy_i^ h_1^T x_i x_i^' h_2^T x_iy_i^' h_1^T x_i
\end{pmatrix} \end{pmatrix}
$$ $$
@@ -222,9 +222,9 @@ Rearranging the terms:
$$ $$
\begin{bmatrix} \begin{bmatrix}
0^T &-x_i^T &y_i^ x_i^T \\ 0^T &-x_i^T &y_i^' x_i^T \\
x_i^T &0^T &-x_i^ x_i^T \\ x_i^T &0^T &-x_i^' x_i^T \\
y_i^ x_i^T &x_i^ x_i^T &0^T y_i^' x_i^T &x_i^' x_i^T &0^T
\end{bmatrix} \end{bmatrix}
\begin{bmatrix} \begin{bmatrix}
h_1 \\ h_1 \\

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@@ -0,0 +1,33 @@
# Math4121 Lecture 35
## Continue on Lebesgue Integration
### Lebesgue Integration
#### Definition of Lebesgue Integral
For simple functions $\phi = \sum_{i=1}^{n} a_i \chi_{S_i}$, given a measure $E$, the Lebesgue integral is defined as:
$$
\int_{\mathbb{R}^n} \phi \, dm = \sum_{i=1}^{n} a_i m(S_i\cap E)
$$
Given a non-negative measurable function $f$ and a measurable set $E$.
Define $\int_E f \, dm = \sup \left\{ \int_E \phi \, dm : \phi \text{ is a simple function and } \phi \leq f \right\}$
(**We do allows $\int_E f \, dm = \infty$**)
For general measurable function $f$, we can define $f^-(x)=\max\{0,-f(x)\}$, $f^+(x)=\max\{0,f(x)\}$. (The positive part of the function and the negative part of the function, both non-negative)
Then $f=f^+-f^-$.
We say $f$ is integrable if $\int_E f^+ \, dm < \infty$ and $\int_E f^- \, dm < \infty$. (both finite) If at least one is finite, define
$$
\int_E f \, dm = \int_E f^+ \, dm - \int_E f^- \, dm
$$
We allow for $A-\infty = -\infty$ and $A+\infty = \infty$ for any $A\in \mathbb{R}$. But not $\infty-\infty$.