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Networkx:所有生成樹和給定一個簡單無向網格網絡這樣它們相關聯的總重量

import networkx as nx 
from pylab import * 
import matplotlib.pyplot as plt 
%pylab inline 

ncols=3 
N=3 
G=nx.grid_2d_graph(N,N) 
labels = dict(((i,j), i + (N-1-j) * N) for i, j in G.nodes()) 
nx.relabel_nodes(G,labels,False) 
inds=labels.keys() 
vals=labels.values() 
inds=[(N-j-1,N-i-1) for i,j in inds] 
pos2=dict(zip(vals,inds)) 
nx.draw_networkx(G, pos=pos2, with_labels=True, node_size = 200, node_color='orange',font_size=10) 
plt.axis('off') 
plt.title('grid') 
plt.show() 

而鑑於每個邊緣具有對應於它的長度的重量:

#Weights 
from math import sqrt 

weights = dict() 
for source, target in G.edges(): 
    x1, y1 = pos2[source] 
    x2, y2 = pos2[target] 
    weights[(source, target)] = round((math.sqrt((x2-x1)**2 + (y2-y1)**2)),3) 

for e in G.edges(): 
    G[e[0]][e[1]] = weights[e] #Assigning weights to G.edges() 

怎麼可能計算網格中的所有生成樹,以及它們相關的總重量?

注意:這是一個所有權重= 1的微不足道的情況。

+1

我知道代碼.... ;-)是否網格保證是正常的? – Paul

+0

此外,大概你已經看到,networkx有一個'minimum_spanning_tree'函數? – Paul

+0

是的,你可以自己繪製它來檢查。但是你有希望的回答讓我覺得這隻適用於電網,對嗎?因爲我不得不爲其他網絡模型做這個:) – FaCoffee

回答

2

這樣做比預期的要長,但下面的代碼找到了一般情況下的所有生成樹。獲取關聯的總重量應該是微不足道的,因爲您可以訪問每棵樹的邊界列表。

不要在很大的樹上使用它 - 即使是玩具的例子也會產生192棵生成樹。

import numpy as np 
import matplotlib.pyplot as plt 
import networkx as nx 

def _expand(G, explored_nodes, explored_edges): 
    """ 
    Expand existing solution by a process akin to BFS. 

    Arguments: 
    ---------- 
    G: networkx.Graph() instance 
     full graph 

    explored_nodes: set of ints 
     nodes visited 

    explored_edges: set of 2-tuples 
     edges visited 

    Returns: 
    -------- 
    solutions: list, where each entry in turns contains two sets corresponding to explored_nodes and explored_edges 
     all possible expansions of explored_nodes and explored_edges 

    """ 
    frontier_nodes = list() 
    frontier_edges = list() 
    for v in explored_nodes: 
     for u in nx.neighbors(G,v): 
      if not (u in explored_nodes): 
       frontier_nodes.append(u) 
       frontier_edges.append([(u,v), (v,u)]) 

    return zip([explored_nodes | frozenset([v]) for v in frontier_nodes], [explored_edges | frozenset(e) for e in frontier_edges]) 

def find_all_spanning_trees(G, root=0): 
    """ 
    Find all spanning trees of a Graph. 

    Arguments: 
    ---------- 
    G: networkx.Graph() instance 
     full graph 

    Returns: 
    ST: list of networkx.Graph() instances 
     list of all spanning trees 

    """ 

    # initialise solution 
    explored_nodes = frozenset([root]) 
    explored_edges = frozenset([]) 
    solutions = [(explored_nodes, explored_edges)] 
    # we need to expand solutions number_of_nodes-1 times 
    for ii in range(G.number_of_nodes()-1): 
     # get all new solutions 
     solutions = [_expand(G, nodes, edges) for (nodes, edges) in solutions] 
     # flatten nested structure and get unique expansions 
     solutions = set([item for sublist in solutions for item in sublist]) 

    return [nx.from_edgelist(edges) for (nodes, edges) in solutions] 


if __name__ == "__main__": 

    N = 3 
    G = nx.grid_2d_graph(N,N) 
    labels = dict(((i,j), i + (N-1-j) * N) for i, j in G.nodes()) 
    nx.relabel_nodes(G,labels,False) 
    inds=labels.keys() 
    vals=labels.values() 
    inds=[(N-j-1,N-i-1) for i,j in inds] 
    pos2=dict(zip(vals,inds)) 

    fig, ax = plt.subplots(1,1) 
    nx.draw_networkx(G, pos=pos2, with_labels=True, node_size = 200, node_color='orange',font_size=10,ax=ax) 
    plt.axis('off') 
    plt.title('grid') 

    ST = find_all_spanning_trees(G) 
    print len(ST) 

    for g in ST: 
     fig, ax = plt.subplots(1,1) 
     nx.draw_networkx(g, pos=pos2, with_labels=True, node_size = 200, node_color='orange',font_size=10,ax=ax) 
     plt.axis('off') 
     plt.title('grid') 
     plt.show() 
+0

我認爲原始版本是按照預期工作,但這絕對有可能起作用。 – Paul

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