Question

目标：尝试将用python编写的算法的某些行转换为伪代码。

给定算法的目标：查找带有周期的有向图中的所有周期。

我的立场：我很好理解算法背后的理论，我自己也编写了不同的版本，但是我不能自己编写一个小，高效和正确的算法。

到目前为止我做了什么：我无法描述花了多少星期，编写了Tarjan，各种版本的DFS，Flloyd等在php中，但不幸的是它们只是部分解决方案而且必须扩展它们。

另外：我已经在线运行了这个算法并且它有效，我需要它用于我堆叠的学校项目，不能继续进行。

这是算法：

def paths_rec(path,edges):
    if len(path) > 0 and path[0][0] == path[-1][1]:
        print "cycle", path
        return #cut processing when find a cycle

    if len(edges) == 0:
        return

    if len(path) == 0:
        #path is empty so all edges are candidates for next step
        next_edges = edges
    else:
        #only edges starting where the last one finishes are candidates
        next_edges = filter(lambda x: path[-1][1] == x[0], edges)

    for edge in next_edges:
        edges_recursive = list(edges)
        edges_recursive.remove(edge)
        #recursive call to keep on permuting possible path combinations
        paths_rec(list(path) + [edge], edges_recursive)


def all_paths(edges):
    paths_rec(list(),edges)

if __name__ == "__main__":
    #edges are represented as (node,node) 
    # so (1,2) represents 1->2 the edge from node 1 to node 2. 
    edges = [(1,2),(2,3),(3,4),(4,2),(2,1)]
    all_paths(edges)

这是我设法用伪代码写的，我用#?标记了我不理解的行。一旦我用伪代码将它们编码，我就可以在php中对它进行编码，我非常熟悉它。

procedure paths_rec (path, edges)

if size(path) > 0 and path[0][0] equals path[-1][1] 
    print "cycle"
    for each element in path
        print element
    end of for
    return
end of if


if size(edges) equals 0
    return
end of if

if size(path) equals 0
    next_edges equals edges
else
    next edges equals filter(lambda x: path[-1][1] == x[0], edges) #?
end of else  

for each edge in next_edges
edges_recursive = list(edges) #?
edges_recursive.remove(edge)#?
#recursive call to keep on permuting possible path combinations
paths_rec(list(path) + [edge], edges_recursive)#?

Answer 1

第next_edges = filter(lambda x: path[-1][1] == x[0], edges)行创建一个新列表，其中包含edges中的第一个点与当前路径的最后一个元素的第二个点相同的边，并等同于

next_edges = []
for x in edges:
    if path[len(path) - 1][1] == x[0]
        next_edges.append[x]

这些行创建了边缘列表edges的新副本，因此当从此副本中删除edge时，它不会更改列表edges

edges_recursive = list(edges)
edges_recursive.remove(edge)

paths_rec(list(path) + [edge], edges_recursive)只是recursive call，其中添加了edge的路径以及从中删除了edge的新边缘列表。

以伪代码转换一些python行

1 个答案: