Consider a $n \times m$ chessboard…
$$ \int{f(x)dx} $$
Since $94 = 4 + 5x$ for some $x \in \mathbb{N}$, my claim is that the first player wins when it chooses $4$ in the beginning. Then, whenever the second player choose $a, x \in\set{1, 2, 3, 4}$, the first player just add it to 5. So choose $5-a$. By doing this, the first player always reaches $10x + 9$ or $10x + 4$ for $x \in \mathbb{N}$. Since $94$ is included, the first player must win.
$\mathfrak{ABCDEFG}$
$\mathscr{ABCDEFG}$
Hi $z = x + y$.
$$a^2 + b^2 = c^2$$
$$\begin{vmatrix}a & b\\ c & d \end{vmatrix}=ad-bc$$
from: https://yihui.org/en/2018/07/latex-math-markdown/
complex math (array)
$$ \begin{array} {lcl} L(p,w_i) &=& \dfrac{1}{N}\Sigma_{i=1}^N(\underbrace{f_r(x_2 \rightarrow x_1 \rightarrow x_0)G(x_1 \longleftrightarrow x_2)f_r(x_3 \rightarrow x_2 \rightarrow x_1)}_{sample\, radiance\, evaluation\, in\, stage2} \\\\\\ &=& \prod_{i=3}^{k-1}(\underbrace{\dfrac{f_r(x_{i+1} \rightarrow x_i \rightarrow x_{i-1})G(x_i \longleftrightarrow x_{i-1})}{p_a(x_{i-1})}}_{stored\,in\,vertex\, during\,light\, path\, tracing\, in\, stage1})\dfrac{G(x_k \longleftrightarrow x_{k-1})L_e(x_k \rightarrow x_{k-1})}{p_a(x_{k-1})p_a(x_k)}) \end{array} $$
Testing alignments
$$ \begin{align*} \frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{u^{s-1}}{e^{u}-1}\mathrm{d}u \end{align*} $$
Hx tests
$$ \mathcal{F} $$
$$ \lambda_i \xlongequal[]{\text{eq.}(1)}\int_{\set{i}} $$
$$ \begin{align*} \text{(1) } 0\oplus 0 & = 0\\ \text{(2) } 1\oplus 0 & = 1\\ \text{(3) } 0\oplus 1 & = 1\\ \text{(4) } 1\oplus 1 & = 0 \end{align*} $$
We observe that:
$$ \begin{itemize} \item for $y=x\oplus z$, $y$ agreees with $x$ if (1) or (3) happen; disagrees if (2) or (4) happen. \item $y=0$ iff $x$ and $z$ agree; $y=1$ iff $x$ and $z$ disagree. \end{itemize} $$
Theorem
Proof
Definition
Notation
Exercise
Proposition
Corollary
Observation
If you need these source codes:
Consider a $n \times m$ chessboard...
$$
\int{f(x)dx}
$$
Since $94 = 4 + 5x$ for some $x \in \mathbb{N}$, my claim is that the first player wins when it chooses $4$ in the beginning. Then, whenever the second player choose $a, x \in\set{1, 2, 3, 4}$, the first player just add it to 5. So choose $5-a$. By doing this, the first player always reaches $10x + 9$ or $10x + 4$ for $x \in \mathbb{N}$. Since $94$ is included, the first player must win.
$\mathfrak{ABCDEFG}$
Hi `$z = x + y$`.
`$$a^2 + b^2 = c^2$$`
`$$\begin{vmatrix}a & b\\
c & d
\end{vmatrix}=ad-bc$$`
from: https://yihui.org/en/2018/07/latex-math-markdown/
### complex math (array)
`$$
\begin{array} {lcl}
L(p,w_i) &=& \dfrac{1}{N}\Sigma_{i=1}^N(\underbrace{f_r(x_2
\rightarrow x_1
\rightarrow x_0)G(x_1
\longleftrightarrow x_2)f_r(x_3
\rightarrow x_2
\rightarrow x_1)}_{sample\, radiance\, evaluation\, in\, stage2}
\\\\\\ &=&
\prod_{i=3}^{k-1}(\underbrace{\dfrac{f_r(x_{i+1}
\rightarrow x_i
\rightarrow x_{i-1})G(x_i
\longleftrightarrow x_{i-1})}{p_a(x_{i-1})}}_{stored\,in\,vertex\, during\,light\, path\, tracing\, in\, stage1})\dfrac{G(x_k
\longleftrightarrow x_{k-1})L_e(x_k
\rightarrow x_{k-1})}{p_a(x_{k-1})p_a(x_k)})
\end{array}
$$`
### Testing alignments
`$$
\begin{align*}
\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{u^{s-1}}{e^{u}-1}\mathrm{d}u
\end{align*}
$$`
### Hx tests
1.
$$
\mathcal{F}
$$
2.
$$
\lambda_i \xlongequal[]{\text{eq.}(1)}\int_{\set{i}}
$$
3.
`$$
\begin{align*}
\text{(1) } 0\oplus 0 & = 0\\
\text{(2) } 1\oplus 0 & = 1\\
\text{(3) } 0\oplus 1 & = 1\\
\text{(4) } 1\oplus 1 & = 0
\end{align*}
$$`
We observe that:
4.
$$
\begin{itemize}
\item for $y=x\oplus z$, $y$ agreees with $x$ if (1) or (3) happen; disagrees
if (2) or (4) happen.
\item $y=0$ iff $x$ and $z$ agree; $y=1$ iff $x$ and $z$ disagree.
\end{itemize}
$$