*本文属于转载。
*转载链接：https://blog.csdn.net/lwb102063/article/details/53046265
目录
clrscode
algorithm
algorithm2e
algorithmic
clrscode
前期准备
\usepackage{clrscode3e} \begin{codebox}
\Procname{$\proc{Insertion-Sort(A)}$}
\li \For $j \gets 2$ \To $\id{length}[A]$ \label{li:for}
\li \Do $\id{key} \gets A[j]$ \label{li:for-begin}
\li \Comment Insert $A[j]$ into the sorted sequence $A[1 \twodots j-1]$.
\li $i \gets j-1$
\li \While $i>0$ and $A[i]>\id{key}$ \label{li:while}
\li \Do $A[i+1] \gets A[i]$ \label{li:while-begin}
\li $i \gets i-1$ \label{li:while-end}
\End
\li $A[i+1] \gets \id{key}$ \label{li:for-end}
\End
\end{codebox}algorithm
前期准备
\usepackage{algorithm}
\usepackage{algpseudocode}
\usepackage{amsmath}
\renewcommand{\algorithmicrequire}{\textbf{Input:}} % Use Input in the format of Algorithm
\renewcommand{\algorithmicensure}{\textbf{Output:}} % Use Output in the format of Algorithm \begin{algorithm}[htb]
\caption{ Framework of ensemble learning for our system.}
\label{alg:Framwork}
\begin{algorithmic}[1]
\Require
The set of positive samples for current batch, $P_n$;
The set of unlabelled samples for current batch, $U_n$;
Ensemble of classifiers on former batches, $E_{n-1}$;
\Ensure
Ensemble of classifiers on the current batch, $E_n$;
\State Extracting the set of reliable negative and/or positive samples $T_n$ from $U_n$ with help of $P_n$;
\label{code:fram:extract}
\State Training ensemble of classifiers $E$ on $T_n \cup P_n$, with help of data in former batches;
\label{code:fram:trainbase}
\State $E_n=E_{n-1}cup E$;
\label{code:fram:add}
\State Classifying samples in $U_n-T_n$ by $E_n$;
\label{code:fram:classify}
\State Deleting some weak classifiers in $E_n$ so as to keep the capacity of $E_n$;
\label{code:fram:select} \\
\Return $E_n$;
\end{algorithmic}
\end{algorithm} - \begin{algorithm}[h]
- \caption{An example for format For \& While Loop in Algorithm}
- \begin{algorithmic}[1]
- \For{each $i\in [1,9]$}
- \State initialize a tree $T_{i}$ with only a leaf (the root);
- \State $T=T\cup T_{i};$
- \EndFor
- \ForAll {$c$ such that $c\in RecentMBatch(E_{n-1})$}
- \label{code:TrainBase:getc}
- \State $T=T\cup PosSample(c)$;
- \label{code:TrainBase:pos}
- \EndFor;
- \For{$i=1$; $i<n$; $i++$ }
- \State $//$ Your source here;
- \EndFor
- \For{$i=1$ to $n$}
- \State $//$ Your source here;
- \EndFor
- \State $//$ Reusing recent base classifiers.
- \label{code:recentStart}
- \While {$(|E_n| \leq L_1 )and( D \neq \phi)$}
- \State Selecting the most recent classifier $c_i$ from $D$;
- \State $D=D-c_i$;
- \State $E_n=E_n+c_i$;
- \EndWhile
- \label{code:recentEnd}
- \end{algorithmic}
- \end{algorithm}
\begin{algorithm}[h]
\caption{Conjugate Gradient Algorithm with Dynamic Step-Size Control}
\label{alg::conjugateGradient}
\begin{algorithmic}[1]
\Require
$f(x)$: objective funtion;
$x_0$: initial solution;
$s$: step size;
\Ensure
optimal $x^{*}$
\State initial $g_0=0$ and $d_0=0$;
\Repeat
\State compute gradient directions $g_k=\bigtriangledown f(x_k)$;
\State compute Polak-Ribiere parameter $\beta_k=\frac{g_k^{T}(g_k-g_{k-1})}{\parallel g_{k-1} \parallel^{2}}$;
\State compute the conjugate directions $d_k=-g_k+\beta_k d_{k-1}$;
\State compute the step size $\alpha_k=s/\parallel d_k \parallel_{2}$;
\Until{($f(x_k)>f(x_{k-1})$)}
\end{algorithmic}
\end{algorithm} \makeatletter
\def\BState{\State\hskip-\ALG@thistlm}
\makeatother
\begin{algorithm}
\caption{My algorithm}\label{euclid}
\begin{algorithmic}[1]
\Procedure{MyProcedure}{}
\State $\textit{stringlen} \gets \text{length of }\textit{string}$
\State $i \gets \textit{patlen}$
\BState \emph{top}:
\If {$i > \textit{stringlen}$} \Return false
\EndIf
\State $j \gets \textit{patlen}$
\BState \emph{loop}:
\If {$\textit{string}(i) = \textit{path}(j)$}
\State $j \gets j-1$.
\State $i \gets i-1$.
\State \textbf{goto} \emph{loop}.
\State \textbf{close};
\EndIf
\State $i \gets i+\max(\textit{delta}_1(\textit{string}(i)),\textit{delta}_2(j))$.
\State \textbf{goto} \emph{top}.
\EndProcedure
\end{algorithmic}
\end{algorithm}
algorithm2e
algorithm2e包可能会与其它包产生冲突，一个常见的错误提示是“Too many }'...”。为了解决这个问题，要在引入algorithm2e包之前加入下面的命令：
\makeatletter
\newif\if@restonecol
\makeatother
\let\algorithm\relax
\let\endalgorithm\relax
前期准备：
\makeatletter
\newif\if@restonecol
\makeatother
\let\algorithm\relax
\let\endalgorithm\relax
\usepackage[linesnumbered,ruled,vlined]{algorithm2e}%[ruled,vlined]{
\usepackage{algpseudocode}
\usepackage{amsmath}
\renewcommand{\algorithmicrequire}{\textbf{Input:}} % Use Input in the format of Algorithm
\renewcommand{\algorithmicensure}{\textbf{Output:}} % Use Output in the format of Algorithm
\begin{algorithm}
\caption{identify Row Context}
\KwIn{$r_i$, $Backgrd(T_i)$=${T_1,T_2,\ldots ,T_n}$ and similarity threshold $\theta_r$}
\KwOut{$con(r_i)$}
$con(r_i)= \Phi$\;
\For{$j=1;j \le n;j \ne i$}
{
float $maxSim=0$\;
$r^{maxSim}=null$\;
\While{not end of $T_j$}
{
compute Jaro($r_i,r_m$)($r_m\in T_j$)\;
\If{$(Jaro(r_i,r_m) \ge \theta_r)\wedge (Jaro(r_i,r_m)\ge r^{maxSim})$}
{
replace $r^{maxSim}$ with $r_m$\;
}
}
$con(r_i)=con(r_i)\cup {r^{maxSim}}$\;
}
return $con(r_i)$\;
\end{algorithm} \begin{algorithm}
\caption{Service checkpoint image storage node and routing path selection}
\LinesNumbered
\KwIn{host server $PM_s$ that $SerImg_k$ is fetched from, $subnet_s$ that $PM_s$ belongs to, $pod_s$ that $PM_s$ belongs to}
\KwOut{Service image storage server $storageserver$,and the image transfer path $path$}
$storageserver$ = Storage node selection($PM_s$, $SerImg_k$,$subnet_s$,$pod_s$)\;
\If{ $storageserver$ $\neq$ null}
{
select a path from $storageserver$ to $PM_s$ and assign the path to $path$\;
}
\textbf{final} \;
\textbf{return} $storageserver$ and $path$;
\end{algorithm} \begin{algorithm}
\caption{Storage node selection}
\LinesNumbered
\KwIn{host server $PM_s$ that the checkpoint image $Img$ is fetched from, $subnet_s$ that $PM_s$ belongs to, $pod_s$ that $PM_s$ belongs to}
\KwOut{Image storage server $storageserver$}
\For{ each host server $PM_i$ in the same subnet with $PM_s$ }
{
\If{ $PM_i$ is not a service providing node or checkpoint image storage node of $S_k$ }
{
add $PM_i$ to $candidateList$ \;
}
}
sort $candidateList$ by reliability desc\;
init $storageserver$ ;
\For{ each $PM_k$ in $candidateList$}
{
\If{ $SP(PM_k)$ $\geq$ $E(SP)$ of $pod_i$ and $BM_k$ $\le$ size of $Img$ }
{
assign $PM_k$ to $storageserver$\;
goto final\;
}
}
clear $candidateList$\;
add all other subnets in $pod_s$ to $netList$\;
\For{ each subnet $subnet_j$ in $netList$}
{
clear $candidateList$\;
\For {each $PM_i$ in $subnet_j$ }
{
\If{ $PM_i$ is not a service providing node or checkpoint image storage node of $S_k$ }
{
add $PM_i$ to $candidateList$\;
}
}
sort all host in $candidateList$ by reliability desc\;
\For{ each $PM_k$ in $candidateList$}
{
\If{$SP(PM_k)$ $\geq$ $E(SP)$ of $pod_i$ and $BM_k$ $\le$ size of $Img$}
{
assign $PM_k$ to $storageserver$ \;
goto final\;
}
}
}
\textbf{final} \;
\textbf{return} $storageserver$;
\end{algorithm} \begin{algorithm}
\caption{Delta checkpoint image storage node and routing path selection}
\LinesNumbered
\KwIn{host server $PM_s$ that generates the delta checkpoint image $DImg_{kt}$, $subnet_s$ that $PM_s$ belongs to, $pod_s$ that $PM_s$ belongs to}
\KwOut{Delta image storage server $storageserver$,and the image transfer path $Path$}
$storageserver$ = Storage node selection($PM_s$, $DImg_{kt}$,$subnet_s$,$pod_s$)\;
\If{ $storageserver$ $\equiv$ null}
{
the delta checkpoint image is stored in the central storage server\;
goto final\;
}
construct weighted topological graph $graph_s$ of $pod_s$\;
calculate the shortest path from $storageserver$ to $PM_s$ in $graph_s$ by using the Dijkstra algorithm\;
\textbf{final} \;
\textbf{return} $storageserver$ and $path$;
\end{algorithm} \documentclass[8pt,twocolumn]{ctexart}
\usepackage{amssymb}
\usepackage{bm}
\usepackage{textcomp} %命令\textacutedbl的包,二阶导符号
% Page length commands go here in the preamble
\setlength{\oddsidemargin}{-0.25in} % Left margin of 1 in + 0 in = 1 in
\setlength{\textwidth}{9in} % 纸张宽度Right margin of 8.5 in - 1 in - 6.5 in = 1 in
\setlength{\topmargin}{-.75in} % Top margin of 2 in -0.75 in = 1 in
\setlength{\textheight}{9.2in} % Lower margin of 11 in - 9 in - 1 in = 1 in
\setlength{\parindent}{0in}
\makeatletter
\newif\if@restonecol
\makeatother
\let\algorithm\relax
\let\endalgorithm\relax
\usepackage[linesnumbered,ruled,vlined]{algorithm2e}%[ruled,vlined]{
\usepackage{algpseudocode}
\renewcommand{\algorithmicrequire}{\textbf{Input:}}
\renewcommand{\algorithmicensure}{\textbf{Output:}}
\begin{document}
\begin{algorithm}
\caption{component matrices computing}
\LinesNumbered
\KwIn{$\mathcal{X}\in\mathbb{R}^{l_1\times l_2\times\cdots\times l_N},\varepsilon,\lambda,\delta,R$}
\KwOut{$A^{(j)}s$ for $j=1$ to $N$}
\textbf{Initialize} all $A^{(j)}s$ //which can be seen as the $0^{th}$ round iterations\;
{$l$\hspace*{-1pt}\textacutedbl}$=L$ //if we need to judge whether $(11)$ is true then {$l$\hspace*{-1pt}\textacutedbl} denotes $L|_{t-1}$\;
\For{ each $A_{i_jr}^{{j}}(1\le j\le N,1\le i_j\le I_j,1\le r\le R)$ }
{//$1^{st}$ round iterations\;
$g_{i_jr}^{(j)'}=g_{i_jr}^{(j)}$\;
$A_{i_jr}^{(j)'}=A_{i_jr}^{(j)}$//if the rollback shown as $(12)$ is needed,$A_{i_jr}^{(j)'}$ denotes $A_{i_jr}^{(j)}|_{t-1}$\;
$A_{i_jr}^{(j)}=A_{i_jr}^{(j)}-\mathrm{{\bf sign}}\left(g_{i_jr}^{(j)}\right)\cdot\delta_{i_jr}^{(j)}$\;
}
\Repeat(//other rounds of iterations for computing component matrices){$\bm{L\le \varepsilon}$ or maximum iterations exhausted}
{
$l'=L$ //if we need to judge whether $(11)$ is true then $l'$ denotes $L|_t$\;
\For{ each $A_{i_jr}^{{j}}(1\le j\le N,1\le i_j\le I_j,1\le r\le R)$}
{
\If{$g_{i_jr}^{(j)}\cdot g_{i_jr}^{(j)'}>0$}
{
$A_{i_jr}^{(j)'}=A_{i_jr}^{(j)} $\;
$g_{i_jr}^{(j)'}=g_{i_jr}^{(j)} $\;
$\delta_{i_jr}^{(j)}=\bm{\min}\left(\delta_{i_jr}^{(j)}\cdot\eta^{+},Max\_Step\_Size\right)$\;
$A_{i_jr}^{(j)}=A_{i_jr}^{(j)}-\mathrm{{\bf sign}}\left(g_{i_jr}^{(j)}\right)\cdot\delta_{i_jr}^{(j)}$\;
}
\ElseIf{$g_{i_jr}^{(j)}\cdot g_{i_jr}^{(j)'}<$}
{
\If{$l'>l$\hspace*{-1pt}\textacutedbl}
{
$g_{i_jr}^{(j)'}=g_{i_jr}^{(j)}$\;
$A_{i_jr}^{(j)}=A_{i_jr}^{(j)'}$// if $(11)$ is true then rollback as $(12)$\;
$\delta_{i_jr}^{(j)}=\bm{\max}\left(\delta_{i_jr}^{(j)}\times\eta^{-},Min\_Step\_Size\right)$\;
}
\Else
{
$A_{i_jr}^{(j)'}=A_{i_jr}^{(j)} $\;
$g_{i_jr}^{(j)'}=g_{i_jr}^{(j)} $\;
$\delta_{i_jr}^{(j)}=\bm{\max}\left(\delta_{i_jr}^{(j)}\cdot\eta^{-},Min\_Step\_Size\right)$\;
$A_{i_jr}^{(j)}=A_{i_jr}^{(j)}-\mathrm{{\bf sign}}\left(g_{i_jr}^{(j)}\right)\cdot\delta_{i_jr}^{(j)}$\;
}
}
\Else
{
$A_{i_jr}^{(j)'}=A_{i_jr}^{(j)} $\;
$g_{i_jr}^{(j)'}=g_{i_jr}^{(j)} $\;
$A_{i_jr}^{(j)}=A_{i_jr}^{(j)}-\mathrm{{\bf sign}}\left(g_{i_jr}^{(j)}\right)\cdot\delta_{i_jr}^{(j)}$\;
}
}
$l$\hspace*{-1pt}\textacutedbl$=l'$\;
}
\end{algorithm}
\end{document} \usepackage[ruled,linesnumbered]{algorithm2e}
\usepackage{amsmath}
\begin{algorithm}
\caption{Learning algorithm of R2P}
\label{alg:r2p}
\KwIn{ratings $R$, joint demographic representations $Y$,learning rate $\eta$,maximum iterative number $maxIter$, negative sampling number $k$\;}
\KwOut{interaction matrix $\bm{W}$, movie vectors $V$\;}
Initialize $\bm{W},V$ randomly\;
$t = 0$\;
For convenience, define $\vec{\varphi}_n = \sum_{m\in S_n}r_{m,n}\vec{v}_m$\; %\varphi_n\bm{W}\vec{y}_n
\While{not converged \rm{or} $t>maxIter$}
{
t = t+1\;
\For{$n=1;n \le N;n++$}
{
$\bm{W} = \bm{W}+\eta\big(1-\sigma\left(\vec{\varphi}_n^T\bm{W}\vec{y}_n\right)\big)\vec{\varphi}_n\vec{y}_n^T$\;\label{algline:W}
\For{$m\in S_n$}
{
$\vec{v}_m=\vec{v}_m+ \eta\left(1-\sigma\left(\vec{\varphi}_n^T\bm{W}\vec{y}_n\right)\right)r_{m,n}\bm{W}\vec{y}_n$\;\label{algline:V}
}
\For{$i=1;i\le k;i++$}
{
sample negative sample $\vec{y}_i$ from $P_n$\;
$\bm{W} = \bm{W}-\eta\big(1-\sigma\left(-\vec{\varphi}_n^T\bm{W}\vec{y}_n\right)\big)\vec{\varphi}_n\vec{y}_i^T$\;
\For{$m\in S_n$}
{
$\vec{v}_m=\vec{v}_m- \eta\left(1-\sigma\left(-\vec{\varphi}_n^T\bm{W}\vec{y}_n\right)\right)r_{m,n}\bm{W}\vec{y}_i$\;
}
}
}
$\bm{W} = \bm{W}-2\lambda\eta\bm{W}$\;
$V=V-2\lambda\eta V$
}
return $\bm{W},V$\;
%\end{algorithmic}
\end{algorithm} algorithmic
1 \documentclass[11pt]{ctexart}
2 \usepackage[top=2cm, bottom=2cm, left=2cm, right=2cm]{geometry}
3 \usepackage{algorithm}
4 \usepackage{algorithmicx}
5 \usepackage{algpseudocode}
6 \usepackage{amsmath}
7
8 \floatname{algorithm}{算法}
9 \renewcommand{\algorithmicrequire}{\textbf{输入:}}
10 \renewcommand{\algorithmicensure}{\textbf{输出:}}
11
12 \begin{document}
13 \begin{algorithm}
14 \caption{用归并排序求逆序数}
15 \begin{algorithmic}[1] %每行显示行号
16 \Require $Array$数组，$n$数组大小
17 \Ensure 逆序数
18 \Function {MergerSort}{$Array, left, right$}
19 \State $result \gets 0$
20 \If {$left < right$}
21 \State $middle \gets (left + right) / 2$
22 \State $result \gets result +$ \Call{MergerSort}{$Array, left, middle$}
23 \State $result \gets result +$ \Call{MergerSort}{$Array, middle, right$}
24 \State $result \gets result +$ \Call{Merger}{$Array,left,middle,right$}
25 \EndIf
26 \State \Return{$result$}
27 \EndFunction
28 \State
29 \Function{Merger}{$Array, left, middle, right$}
30 \State $i\gets left$
31 \State $j\gets middle$
32 \State $k\gets 0$
33 \State $result \gets 0$
34 \While{$i<middle$ \textbf{and} $j<right$}
35 \If{$Array[i]<Array[j]$}
36 \State $B[k++]\gets Array[i++]$
37 \Else
38 \State $B[k++] \gets Array[j++]$
39 \State $result \gets result + (middle - i)$
40 \EndIf
41 \EndWhile
42 \While{$i<middle$}
43 \State $B[k++] \gets Array[i++]$
44 \EndWhile
45 \While{$j<right$}
46 \State $B[k++] \gets Array[j++]$
47 \EndWhile
48 \For{$i = 0 \to k-1$}
49 \State $Array[left + i] \gets B[i]$
50 \EndFor
51 \State \Return{$result$}
52 \EndFunction
53 \end{algorithmic}
54 \end{algorithm}
55 \end{document}