Latex写算法的伪代码排版

algorithmicx例子

相应代码：

 \documentclass[11pt]{ctexart}
 \usepackage[top=2cm, bottom=2cm, left=2cm, right=2cm]{geometry}
 \usepackage{algorithm}
 \usepackage{algorithmicx}
 \usepackage{algpseudocode}
 \usepackage{amsmath}

 \floatname{algorithm}{算法}
 \renewcommand{\algorithmicrequire}{\textbf{输入:}}
 \renewcommand{\algorithmicensure}{\textbf{输出:}}

 \begin{document}
     \begin{algorithm}
         \caption{用归并排序求逆序数}
         \begin{algorithmic}[1] %每行显示行号
             \Require $Array$数组，$n$数组大小
             \Ensure 逆序数
             \Function {MergerSort}{$Array, left, right$}
                 \State $result \gets 0$
                 \If {$left < right$}
                     \State $middle \gets (left + right) / 2$
                     \State $result \gets result +$ \Call{MergerSort}{$Array, left, middle$}
                     \State $result \gets result +$ \Call{MergerSort}{$Array, middle, right$}
                     \State $result \gets result +$ \Call{Merger}{$Array,left,middle,right$}
                 \EndIf
                 \State \Return{$result$}
             \EndFunction
             \State
             \Function{Merger}{$Array, left, middle, right$}
                 \State $i\gets left$
                 \State $j\gets middle$
                 \State $k\gets 0$
                 \State $result \gets 0$
                 \While{$i
                     \If{$Array[i]
                         \State $B[k++]\gets Array[i++]$
                     \Else
                         \State $B[k++] \gets Array[j++]$
                         \State $result \gets result + (middle - i)$
                     \EndIf
                 \EndWhile
                 \While{$i
                     \State $B[k++] \gets Array[i++]$
                 \EndWhile
                 \While{$j
                     \State $B[k++] \gets Array[j++]$
                 \EndWhile
                 \For{$i = 0 \to k-1$}
                     \State $Array[left + i] \gets B[i]$
                 \EndFor
                 \State \Return{$result$}
             \EndFunction
         \end{algorithmic}
     \end{algorithm}
 \end{document}

algorithm例子

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

example 1

代码：

   \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}

example 2

代码：

 \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
       \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}

example 3

代码：

 \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}

example 4

代码：

 \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例子

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

所以前期准备：

 \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

example 1

代码：

 \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}

example 2

代码：

 \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}

example 3

代码：

 \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}

example 4

代码：

 \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}

example 5

 \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}

example 6

 \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}