LaTeX写伪代码

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{$i<middle$ \textbf{and} $j<right$}                    \If{$Array[i]<Array[j]$}                        \State $B[k++]\gets Array[i++]$                    \Else                        \State $B[k++] \gets Array[j++]$                        \State $result \gets result + (middle - i)$                    \EndIf                \EndWhile                \While{$i<middle$}                    \State $B[k++] \gets Array[i++]$                \EndWhile                \While{$j<right$}                    \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{$i=1$; $i<n$; $i++$ }      \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}

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}