如果你想要在图中找到所有基本电路，你可以使用JAMES C. TIERNAN的EC算法，该算法在1970年的一篇论文中发现。

非常原始的EC算法，因为我设法在php中实现它(希望没有错误如下所示)。如果有循环，它也可以找到。这个实现中的电路(试图克隆原始电路)是非零元素。0在这里代表不存在(我们知道它是空的)。

除此之外，下面的实现使算法更具独立性，这意味着节点可以从任何地方开始，甚至从负数开始，例如-4，-3，-2，..等。

在这两种情况下，都要求节点是顺序的。

你可能需要研究原始论文，James C. Tiernan基本电路算法

<?php
echo  "<pre><br><br>";

$G = array(
        1=>array(1,2,3),
        2=>array(1,2,3),
        3=>array(1,2,3)
);


define('N',key(array_slice($G, -1, 1, true)));
$P = array(1=>0,2=>0,3=>0,4=>0,5=>0);
$H = array(1=>$P, 2=>$P, 3=>$P, 4=>$P, 5=>$P );
$k = 1;
$P[$k] = key($G);
$Circ = array();


#[Path Extension]
EC2_Path_Extension:
foreach($G[$P[$k]] as $j => $child ){
    if( $child>$P[1] and in_array($child, $P)===false and in_array($child, $H[$P[$k]])===false ){
    $k++;
    $P[$k] = $child;
    goto EC2_Path_Extension;
}   }

#[EC3 Circuit Confirmation]
if( in_array($P[1], $G[$P[$k]])===true ){//if PATH[1] is not child of PATH[current] then don't have a cycle
    $Circ[] = $P;
}

#[EC4 Vertex Closure]
if($k===1){
    goto EC5_Advance_Initial_Vertex;
}
//afou den ksana theoreitai einai asfales na svisoume
for( $m=1; $m<=N; $m++){//H[P[k], m] <- O, m = 1, 2, . . . , N
    if( $H[$P[$k-1]][$m]===0 ){
        $H[$P[$k-1]][$m]=$P[$k];
        break(1);
    }
}
for( $m=1; $m<=N; $m++ ){//H[P[k], m] <- O, m = 1, 2, . . . , N
    $H[$P[$k]][$m]=0;
}
$P[$k]=0;
$k--;
goto EC2_Path_Extension;

#[EC5 Advance Initial Vertex]
EC5_Advance_Initial_Vertex:
if($P[1] === N){
    goto EC6_Terminate;
}
$P[1]++;
$k=1;
$H=array(
        1=>array(1=>0,2=>0,3=>0,4=>0,5=>0),
        2=>array(1=>0,2=>0,3=>0,4=>0,5=>0),
        3=>array(1=>0,2=>0,3=>0,4=>0,5=>0),
        4=>array(1=>0,2=>0,3=>0,4=>0,5=>0),
        5=>array(1=>0,2=>0,3=>0,4=>0,5=>0)
);
goto EC2_Path_Extension;

#[EC5 Advance Initial Vertex]
EC6_Terminate:
print_r($Circ);
?>

然后这是另一个实现，更独立于图形，没有goto和数组值，而是使用数组键，路径，图形和电路存储为数组键(如果你喜欢使用数组值，只需更改所需的行)。示例图从-4开始，以显示其独立性。

<?php

$G = array(
        -4=>array(-4=>true,-3=>true,-2=>true),
        -3=>array(-4=>true,-3=>true,-2=>true),
        -2=>array(-4=>true,-3=>true,-2=>true)
);


$C = array();


EC($G,$C);
echo "<pre>";
print_r($C);
function EC($G, &$C){

    $CNST_not_closed =  false;                          // this flag indicates no closure
    $CNST_closed        = true;                         // this flag indicates closure
    // define the state where there is no closures for some node
    $tmp_first_node  =  key($G);                        // first node = first key
    $tmp_last_node  =   $tmp_first_node-1+count($G);    // last node  = last  key
    $CNST_closure_reset = array();
    for($k=$tmp_first_node; $k<=$tmp_last_node; $k++){
        $CNST_closure_reset[$k] = $CNST_not_closed;
    }
    // define the state where there is no closure for all nodes
    for($k=$tmp_first_node; $k<=$tmp_last_node; $k++){
        $H[$k] = $CNST_closure_reset;   // Key in the closure arrays represent nodes
    }
    unset($tmp_first_node);
    unset($tmp_last_node);


    # Start algorithm
    foreach($G as $init_node => $children){#[Jump to initial node set]
        #[Initial Node Set]
        $P = array();                   // declare at starup, remove the old $init_node from path on loop
        $P[$init_node]=true;            // the first key in P is always the new initial node
        $k=$init_node;                  // update the current node
                                        // On loop H[old_init_node] is not cleared cause is never checked again
        do{#Path 1,3,7,4 jump here to extend father 7
            do{#Path from 1,3,8,5 became 2,4,8,5,6 jump here to extend child 6
                $new_expansion = false;
                foreach( $G[$k] as $child => $foo ){#Consider each child of 7 or 6
                    if( $child>$init_node and isset($P[$child])===false and $H[$k][$child]===$CNST_not_closed ){
                        $P[$child]=true;    // add this child to the path
                        $k = $child;        // update the current node
                        $new_expansion=true;// set the flag for expanding the child of k
                        break(1);           // we are done, one child at a time
            }   }   }while(($new_expansion===true));// Do while a new child has been added to the path

            # If the first node is child of the last we have a circuit
            if( isset($G[$k][$init_node])===true ){
                $C[] = $P;  // Leaving this out of closure will catch loops to
            }

            # Closure
            if($k>$init_node){                  //if k>init_node then alwaya count(P)>1, so proceed to closure
                $new_expansion=true;            // $new_expansion is never true, set true to expand father of k
                unset($P[$k]);                  // remove k from path
                end($P); $k_father = key($P);   // get father of k
                $H[$k_father][$k]=$CNST_closed; // mark k as closed
                $H[$k] = $CNST_closure_reset;   // reset k closure
                $k = $k_father;                 // update k
        }   } while($new_expansion===true);//if we don't wnter the if block m has the old k$k_father_old = $k;
        // Advance Initial Vertex Context
    }//foreach initial


}//function

?>

我已经分析并记录了EC，但不幸的是，文档是希腊语。

我在搜索中找到了这个页面，由于循环与强连接组件不相同，我继续搜索，最后，我找到了一个高效的算法，它列出了有向图的所有(基本)循环。这篇论文来自唐纳德·b·约翰逊(Donald B. Johnson)，可以在以下链接中找到:

http://www.cs.tufts.edu/comp/150GA/homeworks/hw1/Johnson%2075.PDF

java实现可以在下面找到:

http://normalisiert.de/code/java/elementaryCycles.zip

约翰逊算法的Mathematica演示可以在这里找到，实现可以从右边下载(“下载作者代码”)。

注:实际上，这个问题有很多算法。本文列举了其中一些:

http://dx.doi.org/10.1137/0205007

根据文章，Johnson的算法是最快的。

我发现解决这个问题的最简单的选择是使用名为networkx的python库。

它实现了这个问题的最佳答案中提到的约翰逊算法，但它的执行非常简单。

简而言之，你需要以下几点:

import networkx as nx
import matplotlib.pyplot as plt

# Create Directed Graph
G=nx.DiGraph()

# Add a list of nodes:
G.add_nodes_from(["a","b","c","d","e"])

# Add a list of edges:
G.add_edges_from([("a","b"),("b","c"), ("c","a"), ("b","d"), ("d","e"), ("e","a")])

#Return a list of cycles described as a list o nodes
list(nx.simple_cycles(G))

答案:[['a'， 'b'， 'd'， 'e']， ['a'， 'b'， 'c']]

深度优先搜索和回溯应该在这里工作。 保存一个布尔值数组，以跟踪您以前是否访问过某个节点。如果您没有新节点可访问(不涉及已经访问过的节点)，那么只需返回并尝试不同的分支。

如果你有一个邻接表来表示图，DFS很容易实现。例如adj[A] = {B,C}表示B和C是A的子结点。

例如，下面的伪代码。“start”是开始的节点。

dfs(adj,node,visited):  
  if (visited[node]):  
    if (node == start):  
      "found a path"  
    return;  
  visited[node]=YES;  
  for child in adj[node]:  
    dfs(adj,child,visited)
  visited[node]=NO;

用开始节点调用上面的函数:

visited = {}
dfs(adj,start,visited)

http://www.me.utexas.edu/~bard/IP/Handouts/cycles.pdf

在DAG中查找所有循环涉及两个步骤(算法)。

第一步是使用Tarjan的算法找到强连接组件的集合。

从任意顶点开始。 这个顶点的DFS。每个节点x保留两个数字，dfs_index[x]和dfs_lowval[x]。 Dfs_index [x]存储访问节点的时间，而dfs_lowval[x] = min(dfs_low[k]) where K是x的所有子结点在dfs生成树中不是x的父结点。 具有相同dfs_lowval[x]的所有节点都在同一个强连接组件中。

第二步是在连接的组件中找到循环(路径)。我的建议是使用改进版的Hierholzer算法。

这个想法是:

Choose any starting vertex v, and follow a trail of edges from that vertex until you return to v. It is not possible to get stuck at any vertex other than v, because the even degree of all vertices ensures that, when the trail enters another vertex w there must be an unused edge leaving w. The tour formed in this way is a closed tour, but may not cover all the vertices and edges of the initial graph. As long as there exists a vertex v that belongs to the current tour but that has adjacent edges not part of the tour, start another trail from v, following unused edges until you return to v, and join the tour formed in this way to the previous tour.

下面是带有测试用例的Java实现的链接:

http://stones333.blogspot.com/2013/12/find-cycles-in-directed-graph-dag.html