Latex：插入伪代码

*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}  

1. \begin{algorithm}[h]  
2.   \caption{An example for format For \& While Loop in Algorithm}  
3.   \begin{algorithmic}[1]  
4.     \For{each $i\in [1,9]$}  
5.       \State initialize a tree $T_{i}$ with only a leaf (the root);  
6.       \State $T=T\cup T_{i};$  
7.     \EndFor  
8.     \ForAll {$c$ such that $c\in RecentMBatch(E_{n-1})$}  
9.       \label{code:TrainBase:getc}  
10.       \State $T=T\cup PosSample(c)$;  
11.       \label{code:TrainBase:pos}  
12.     \EndFor;  
13.     \For{$i=1$; $i<n$; $i++$ }  
14.       \State $//$ Your source here;  
15.     \EndFor  
16.     \For{$i=1$ to $n$}  
17.       \State $//$ Your source here;  
18.     \EndFor  
19.     \State $//$ Reusing recent base classifiers.  
20.     \label{code:recentStart}  
21.     \While {$(|E_n| \leq L_1 )and( D \neq \phi)$}  
22.       \State Selecting the most recent classifier $c_i$ from $D$;  
23.       \State $D=D-c_i$;  
24.       \State $E_n=E_n+c_i$;  
25.     \EndWhile  
26.     \label{code:recentEnd}  
27.   \end{algorithmic}  
28. \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{$g_{i_jr}^{(j)}\cdot g_{i_jr}^{(j)'}<0$}  
        {  
            \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}