update build and types
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17
Jenkinsfile
vendored
17
Jenkinsfile
vendored
@@ -20,12 +20,25 @@ pipeline {
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steps {
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steps {
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script {
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script {
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echo "Building docker image ${registry}:${version}.${env.BUILD_ID}"
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echo "Building docker image ${registry}:${version}.${env.BUILD_ID}"
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def customImage = docker.build("${registry}:${version}.${env.BUILD_ID}")
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def customImage = docker.build("${registry}:v${version}.${env.BUILD_ID}")
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echo "Pushing docker image ${registry}:${version}.${env.BUILD_ID}"
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echo "Pushing docker image ${registry}:v${version}.${env.BUILD_ID}"
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customImage.push()
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customImage.push()
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}
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}
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}
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}
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}
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}
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stage('Deploy') {
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steps {
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echo "Deploying docker image ${registry}:v${version}.${env.BUILD_ID}"
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echo "Stopping existing container"
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sh 'docker stop notenextra'
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echo "Removing existing container"
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sh 'docker rm notenextra'
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echo "Running new docker container"
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sh 'docker run -d -p 13000:3000 --name notenextra ${registry}:v${version}.${env.BUILD_ID}'
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}
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}
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}
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}
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}
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}
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}
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@@ -166,11 +166,11 @@ y \\
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\end{pmatrix}
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\end{pmatrix}
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$$
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$$
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Constraint from a match $(x_i,x_i^′)$: $x_i^′≅Hx_i$
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Constraint from a match $(x_i,x_i^')$: $x_i^'≅Hx_i$
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How can we get rid of the scale ambiguity?
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How can we get rid of the scale ambiguity?
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Cross product trick: $x_i^′ × Hx_i=0$
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Cross product trick: $x_i^' × Hx_i=0$
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The cross product is defined as:
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The cross product is defined as:
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@@ -181,9 +181,9 @@ $$
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Let $h_1^T, h_2^T, h_3^T$ be the rows of $H$. Then
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Let $h_1^T, h_2^T, h_3^T$ be the rows of $H$. Then
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$$
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$$
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x_i^′ × Hx_i=\begin{pmatrix}
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x_i^' × Hx_i=\begin{pmatrix}
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x_i^′ \\
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x_i^' \\
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y_i^′ \\
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y_i^' \\
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1
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1
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\end{pmatrix} \times \begin{pmatrix}
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\end{pmatrix} \times \begin{pmatrix}
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h_1^T x_i \\
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h_1^T x_i \\
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@@ -192,18 +192,18 @@ x_i^′ × Hx_i=\begin{pmatrix}
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\end{pmatrix}
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\end{pmatrix}
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=
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=
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\begin{pmatrix}
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\begin{pmatrix}
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y_i^′ h_3^T x_i−h_2^T x_i \\
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y_i^' h_3^T x_i−h_2^T x_i \\
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h_1^T x_i−x_i^′ h_3^T x_i \\
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h_1^T x_i−x_i^' h_3^T x_i \\
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x_i^′ h_2^T x_i−y_i^′ h_1^T x_i
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x_i^' h_2^T x_i−y_i^' h_1^T x_i
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\end{pmatrix}
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\end{pmatrix}
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$$
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$$
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Constraint from a match $(x_i,x_i^′)$:
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Constraint from a match $(x_i,x_i^')$:
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$$
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$$
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x_i^′ × Hx_i=\begin{pmatrix}
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x_i^' × Hx_i=\begin{pmatrix}
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x_i^′ \\
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x_i^' \\
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y_i^′ \\
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y_i^' \\
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1
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1
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\end{pmatrix} \times \begin{pmatrix}
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\end{pmatrix} \times \begin{pmatrix}
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h_1^T x_i \\
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h_1^T x_i \\
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@@ -212,9 +212,9 @@ x_i^′ × Hx_i=\begin{pmatrix}
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\end{pmatrix}
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\end{pmatrix}
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=
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=
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\begin{pmatrix}
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\begin{pmatrix}
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y_i^′ h_3^T x_i−h_2^T x_i \\
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y_i^' h_3^T x_i−h_2^T x_i \\
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h_1^T x_i−x_i^′ h_3^T x_i \\
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h_1^T x_i−x_i^' h_3^T x_i \\
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x_i^′ h_2^T x_i−y_i^′ h_1^T x_i
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x_i^' h_2^T x_i−y_i^' h_1^T x_i
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\end{pmatrix}
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\end{pmatrix}
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$$
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$$
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@@ -222,9 +222,9 @@ Rearranging the terms:
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$$
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$$
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\begin{bmatrix}
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\begin{bmatrix}
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0^T &-x_i^T &y_i^′ x_i^T \\
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0^T &-x_i^T &y_i^' x_i^T \\
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x_i^T &0^T &-x_i^′ x_i^T \\
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x_i^T &0^T &-x_i^' x_i^T \\
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y_i^′ x_i^T &x_i^′ x_i^T &0^T
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y_i^' x_i^T &x_i^' x_i^T &0^T
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\end{bmatrix}
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\end{bmatrix}
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\begin{bmatrix}
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\begin{bmatrix}
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h_1 \\
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h_1 \\
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33
pages/Math4121/Math4121_L35.md
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33
pages/Math4121/Math4121_L35.md
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@@ -0,0 +1,33 @@
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# Math4121 Lecture 35
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## Continue on Lebesgue Integration
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### Lebesgue Integration
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#### Definition of Lebesgue Integral
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For simple functions $\phi = \sum_{i=1}^{n} a_i \chi_{S_i}$, given a measure $E$, the Lebesgue integral is defined as:
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$$
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\int_{\mathbb{R}^n} \phi \, dm = \sum_{i=1}^{n} a_i m(S_i\cap E)
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$$
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Given a non-negative measurable function $f$ and a measurable set $E$.
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Define $\int_E f \, dm = \sup \left\{ \int_E \phi \, dm : \phi \text{ is a simple function and } \phi \leq f \right\}$
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(**We do allows $\int_E f \, dm = \infty$**)
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For general measurable function $f$, we can define $f^-(x)=\max\{0,-f(x)\}$, $f^+(x)=\max\{0,f(x)\}$. (The positive part of the function and the negative part of the function, both non-negative)
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Then $f=f^+-f^-$.
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We say $f$ is integrable if $\int_E f^+ \, dm < \infty$ and $\int_E f^- \, dm < \infty$. (both finite) If at least one is finite, define
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$$
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\int_E f \, dm = \int_E f^+ \, dm - \int_E f^- \, dm
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$$
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We allow for $A-\infty = -\infty$ and $A+\infty = \infty$ for any $A\in \mathbb{R}$. But not $\infty-\infty$.
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