二部图的所有可能最大匹配

Question

我用networkx来寻找二部图的maximum cardinality matching。在

对于特定的图，匹配的边不是唯一的。在

有没有办法让我找到所有的最大匹配？在

对于以下示例，以下所有边都可以是最大匹配：

{1: 2, 2: 1}或{1: 3, 3: 1}或{}

import networkx as nx
import matplotlib.pyplot as plt

G = nx.MultiDiGraph()
edges = [(1,3), (1,4), (1,2)]

nx.is_bipartite(G)
True

nx.draw(G, with_labels=True)
plt.show()

不幸的是

^{pr2}$

只有回报

{1: 2, 2: 1}

有没有办法我也能得到其他的组合？在

Answer 1

我阅读了Uno的工作，并试图提出一个实现。下面是我的非常长的代码和一个工作示例。在这个特定的例子中，有4个“可行”的顶点（根据Uno的术语），因此每个顶点与一个已经覆盖的顶点进行切换，您就有了2^4 = 16不同的可能最大匹配。在

我不得不承认，我对图论非常陌生，我并没有完全遵循Uno的流程，有一些细微的差异，而且大多数时候我没有尝试做任何优化。我确实很难理解这篇论文，因为我认为解释不太完美，而且数字可能有错误。所以请小心使用，如果你能帮助优化它，那将是太好了！在

import networkx as nx
from networkx import bipartite

def plotGraph(graph):
    import matplotlib.pyplot as plt
    fig=plt.figure()
    ax=fig.add_subplot(111)

    pos=[(ii[1],ii[0]) for ii in graph.nodes()]
    pos_dict=dict(zip(graph.nodes(),pos))
    nx.draw(graph,pos=pos_dict,ax=ax,with_labels=True)
    plt.show(block=False)
    return


def formDirected(g,match):
    '''Form directed graph D from G and matching M.

    <g>: undirected bipartite graph. Nodes are separated by their
         'bipartite' attribute.
    <match>: list of edges forming a matching of <g>. 

    Return <d>: directed graph, with edges in <match> pointing from set-0
                (bipartite attribute ==0) to set-1 (bipartite attrbiute==1),
                and the other edges in <g> but not in <matching> pointing
                from set-1 to set-0.
    '''

    d=nx.DiGraph()

    for ee in g.edges():
        if ee in match or (ee[1],ee[0]) in match:
            if g.node[ee[0]]['bipartite']==0:
                d.add_edge(ee[0],ee[1])
            else:
                d.add_edge(ee[1],ee[0])
        else:
            if g.node[ee[0]]['bipartite']==0:
                d.add_edge(ee[1],ee[0])
            else:
                d.add_edge(ee[0],ee[1])

    return d


def enumMaximumMatching(g):
    '''Find all maximum matchings in an undirected bipartite graph.

    <g>: undirected bipartite graph. Nodes are separated by their
         'bipartite' attribute.

    Return <all_matches>: list, each is a list of edges forming a maximum
                          matching of <g>. 
    '''

    all_matches=[]

    #        Find one matching M        
    match=bipartite.hopcroft_karp_matching(g)

    #       -Re-orient match arcs       -
    match2=[]
    for kk,vv in match.items():
        if g.node[kk]['bipartite']==0:
            match2.append((kk,vv))
    match=match2
    all_matches.append(match)

    #        -Enter recursion        -
    all_matches=enumMaximumMatchingIter(g,match,all_matches,None)

    return all_matches


def enumMaximumMatchingIter(g,match,all_matches,add_e=None):
    '''Recurively search maximum matchings.

    <g>: undirected bipartite graph. Nodes are separated by their
         'bipartite' attribute.
    <match>: list of edges forming one maximum matching of <g>.
    <all_matches>: list, each is a list of edges forming a maximum
                   matching of <g>. Newly found matchings will be appended
                   into this list.
    <add_e>: tuple, the edge used to form subproblems. If not None,
             will be added to each newly found matchings.

    Return <all_matches>: updated list of all maximum matchings.
    '''

    #       -Form directed graph D       -
    d=formDirected(g,match)

    #        -Find cycles in D        -
    cycles=list(nx.simple_cycles(d))

    if len(cycles)==0:

        #    -If no cycle, find a feasible path    -
        all_uncovered=set(g.node).difference(set([ii[0] for ii in match]))
        all_uncovered=all_uncovered.difference(set([ii[1] for ii in match]))
        all_uncovered=list(all_uncovered)

        #       If no path, terminiate       
        if len(all_uncovered)==0:
            return all_matches

        #     Find a length 2 feasible path     
        idx=0
        uncovered=all_uncovered[idx]
        while True:

            if uncovered not in nx.isolates(g):
                paths=nx.single_source_shortest_path(d,uncovered,cutoff=2)
                len2paths=[vv for kk,vv in paths.items() if len(vv)==3]

                if len(len2paths)>0:
                    reversed=False
                    break

                #        Try reversed path        
                paths_rev=nx.single_source_shortest_path(d.reverse(),uncovered,cutoff=2)
                len2paths=[vv for kk,vv in paths_rev.items() if len(vv)==3]

                if len(len2paths)>0:
                    reversed=True
                    break

            idx+=1
            if idx>len(all_uncovered)-1:
                return all_matches

            uncovered=all_uncovered[idx]

        #      -Create a new matching M'      -
        len2path=len2paths[0]
        if reversed:
            len2path=len2path[::-1]
        len2path=zip(len2path[:-1],len2path[1:])

        new_match=[]
        for ee in d.edges():
            if ee in len2path:
                if g.node[ee[1]]['bipartite']==0:
                    new_match.append((ee[1],ee[0]))
            else:
                if g.node[ee[0]]['bipartite']==0:
                    new_match.append(ee)

        if add_e is not None:
            for ii in add_e:
                new_match.append(ii)

        all_matches.append(new_match)

        #          -Select e          -
        e=set(len2path).difference(set(match))
        e=list(e)[0]

        #        -Form subproblems        -
        g_plus=g.copy()
        g_minus=g.copy()
        g_plus.remove_node(e[0])
        g_plus.remove_node(e[1])

        g_minus.remove_edge(e[0],e[1])

        add_e_new=[e,]
        if add_e is not None:
            add_e_new.extend(add_e)

        all_matches=enumMaximumMatchingIter(g_minus,match,all_matches,add_e)
        all_matches=enumMaximumMatchingIter(g_plus,new_match,all_matches,add_e_new)

    else:
        #        Find a cycle in D        
        cycle=cycles[0]
        cycle.append(cycle[0])
        cycle=zip(cycle[:-1],cycle[1:])

        #      -Create a new matching M'      -
        new_match=[]
        for ee in d.edges():
            if ee in cycle:
                if g.node[ee[1]]['bipartite']==0:
                    new_match.append((ee[1],ee[0]))
            else:
                if g.node[ee[0]]['bipartite']==0:
                    new_match.append(ee)

        if add_e is not None:
            for ii in add_e:
                new_match.append(ii)

        all_matches.append(new_match)

        #        -Choose an edge E        -
        e=set(match).intersection(set(cycle))
        e=list(e)[0]

        #        -Form subproblems        -
        g_plus=g.copy()
        g_minus=g.copy()
        g_plus.remove_node(e[0])
        g_plus.remove_node(e[1])
        g_minus.remove_edge(e[0],e[1])

        add_e_new=[e,]
        if add_e is not None:
            add_e_new.extend(add_e)

        all_matches=enumMaximumMatchingIter(g_plus,match,all_matches,add_e_new)
        all_matches=enumMaximumMatchingIter(g_minus,new_match,all_matches,add_e)

    return all_matches

if __name__=='__main__':
    g=nx.Graph()
    edges=[
            [(1,0), (0,0)],
            [(1,1), (0,0)],
            [(1,2), (0,2)],
            [(1,3), (0,2)],
            [(1,4), (0,3)],
            [(1,4), (0,5)],
            [(1,5), (0,2)],
            [(1,5), (0,4)],
            [(1,6), (0,1)],
            [(1,6), (0,4)],
            [(1,6), (0,6)]
            ]

    for ii in edges:
        g.add_node(ii[0],bipartite=0)
        g.add_node(ii[1],bipartite=1)

    g.add_edges_from(edges)
    plotGraph(g)

    all_matches=enumMaximumMatching(g)

    for mm in all_matches:
        g_match=nx.Graph()
        for ii in mm:
            g_match.add_edge(ii[0],ii[1])
        plotGraph(g_match)

Answer 2

Takeaki-Uno的论文“二部图中所有完美、最大和最大匹配的枚举算法”有一个算法。http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.107.8179&rep=rep1&type=pdf

定理2说 “二部图中的最大匹配可以用O（mn^1/2）来计算+ nNm）时间和O（m）空间，其中Nm是最大匹配数（G）。”