10.1 写出图10.20所示非赋权无向图的关联矩阵和邻接矩阵
绘制图 import networkx as nx
import pylab as plt
import numpy as np
A=np.zeros((6,6))
List=[(1,2),(1,4),(2,3),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)]
for i in List:
A[i[0]-1,i[1]-1]=1
G=nx.Graph(A)
pos=nx.spring_layout(G)
nx.draw(G,pos,with_labels=True,font_size=12)
plt.show()
10.2 计算图10.21所示赋权无向图中从
v
1
到
v
5
v_1到v_5
v1到v5的最短路径和最短距离
绘制图 import networkx as nx
import pylab as plt
import numpy as np
LIST=[(1,2,7),(1,3,3),(1,4,12),(2,3,3),(2,6,2),(3,4,8),(4,5,1),(5,6,3)]
G=nx.Graph()
G.add_nodes_from(range(1,7))
G.add_weighted_edges_from(LIST)
weight=nx.get_edge_attributes(G,'weight')#获取权重信息
pos=nx.shell_layout(G)
nx.draw(G,pos,font_size=12,font_weight='bold',with_labels=True)
nx.draw_networkx_edge_labels(G,pos,edge_labels=weight)
plt.show()
求解任意两点的最短路 distance=dict(nx.shortest_path_length(G,weight='weight'))
a=nx.to_numpy_matrix(G)
m,n=a.shape
for i in range(1,m+1):
for j in range(1,n+1):
if i!=j:
print('{0}到{1}的最短距离为:{2}\n'.format(i, j, distance[i][j]))
10.3 求图10.21 所示赋权无向图的最小生成树 import networkx as nx
import pylab as plt
import numpy as np
LIST=[(1,2,7),(1,3,3),(1,4,12),(2,3,3),(2,6,2),(3,4,8),(4,5,1),(5,6,3)]
G=nx.Graph()
G.add_nodes_from(range(1,7))
G.add_weighted_edges_from(LIST)
T=nx.minimum_spanning_tree(G)#默认为破圈法
w=nx.get_edge_attributes(T,'weight')
print("最小生成树的长度为:",sum(w.values()))
nx.draw(T,pos=nx.shell_layout(T),with_labels=True,node_color='blue')
nx.draw_networkx_edge_labels(T,pos=nx.shell_layout(T),edge_labels=w)
plt.show()
#最小生成树的长度为: 12
10.4 已知有6个村子,互相间道路的距离如图10.22所示,拟合建一所小学,已知A处有小学生100人,B处80人,C处60 人,D处40人,E处70人,F处90 人。问小学应再建在那个村庄,使学生上学最方便(走的总路程最短)
求每个点到其余各点的最短距离矩阵 (每个村庄)人数
×
\times
× 距离=总路程 import networkx as nx
import pylab as plt
import numpy as np
#绘制图
nodes='abcdef'.upper()
List=[(1,2,2),(1,3,7),(2,3,4),(2,4,6),(2,5,8),(3,4,1),(3,5,3),(4,5,1),(4,6,6),(5,6,3)]
G=nx.Graph()
G.add_nodes_from(range(1,7))
G.add_weighted_edges_from(List)
w=nx.get_edge_attributes(G,'weight')
pos=nx.shell_layout(G)
nx.draw(G,pos,node_color='red',labels=dict(zip(range(1,7),list(nodes))))
nx.draw_networkx_edge_labels(G,pos,font_size=10,edge_labels=w)
plt.show()
#构造最短距离矩阵
distance=nx.shortest_path_length(G,weight='weight')
matrix=np.zeros((6,6))
for i in distance:
a=i[0]-1
for j in range(1,7):
matrix[a][j-1]=i[1][j]
person=np.array([100,80,60,40,70,90]).reshape(6,1)
print((matrix@person).flatten())
#结果:
#[2350. 1870. 1550. 1590. 1710. 2490.] 由此可见应在C处建学校
10.5 已知95个目标点的数据(excel),第一列是这95个点的编号,第2,3列是这95个点
x
,
y
x,y
x,y 坐标,第4列这些点重要性分类,表明“1”的是第一类重要目标点,表面“2”的是第二列重要点,未标明类别的是一般目标点,第5,6,7列标明了这些点的链接关系。如第三行数据:
C
C
C -1160 587.5 D F 表示顶点
C
C
C 的坐标为
(
−
1160
,
587.5
)
(-1160,587.5 )
(−1160,587.5) ,它是一般目标点,
C
点和
D
点,
F
点相连
C点和D点,F点相连
C点和D点,F点相连 完成以下问题:
画出上面的无向图,第一类目标点用
⋆
\star
⋆ 表示,第二类目标点用*表示,一般目标点用‘.’这里要求画出无向图的度量图,即各个顶点的位置坐标必须准确(非拓扑图) 求顶点L到顶点M3的最短距离和最短路径,并画出最短路径 import networkx as nx
import pylab as plt
import math
import numpy as np
import pandas as pd
data = pd.read_excel('Pex10_5.xlsx',keep_default_na=False)
# 获取坐标构造pos
x_site = data['x坐标'].tolist()
y_site = data['y坐标'].tolist()
d = len(x_site)
position = zip(range(0, d), zip(x_site, y_site))
pos = dict(position) # 构造字典
nodes = data['顶点'].tolist() # 构造点的标签
matrix = pd.DataFrame(np.zeros((d, d)), index=nodes, columns=nodes) # 构造无向图的邻接矩阵
data1 = data.set_index('顶点')
pos_alph = dict(zip(nodes, zip(x_site, y_site))) # 插寻点的坐标
f = lambda X, Y: np.round(np.sqrt((X[0] - X[1]) ** 2 + (Y[0] - Y[1]) ** 2))
for i in nodes:
if len(data1['相邻的顶点1'][i]) != 0:
temp = data1['相邻的顶点1'][i]
x1, y1 = pos_alph[i]
x2, y2 = pos_alph[temp]
matrix[i][temp] = f([x1, x2], [y1, y2]) # 修改为两点之间的距离
if len(data1['相邻的顶点2'][i]) != 0:
temp = data1['相邻的顶点2'][i]
x1, y1 = pos_alph[i]
x2, y2 = pos_alph[temp]
matrix[i][temp] = f([x1, x2], [y1, y2]) # 修改为两点之间的距离
x3, y3 = pos_alph['P3']
x4, y4 = pos_alph['G3']
x5, y5 = pos_alph['O3']
matrix['P3']['G3'] = f([x3, x4], [y3, y4])
matrix['O3']['G3'] = f([x3, x5], [y3, y5])
mat = matrix.T.values # 转化为上三角矩阵
G=nx.Graph(mat)
w=nx.get_edge_attributes(G,'weight')
#求最短距离
source=nodes.index('L');end=nodes.index('M3')
dis=nx.shortest_path_length(G,source,end,weight='weight')
print('L到M3的最短距离为:',dis)
path=nx.shortest_path(G,source,end)
print('L到M3的路径为:',path)
path_edges=list(zip(path,path[1:]))
#绘制最短路径
nx.draw(G,pos,node_size=100,node_color='blue',width=0.7,labels=dict(zip(range(d),nodes)),alpha=0.7,font_size=10)
nx.draw_networkx_edge_labels(G,pos,edge_labels=w,font_size=4)
nx.draw_networkx_edges(G,pos,edgelist=path_edges,edge_color='red',width=2)
# 对应的点的标记修改
data2 = data.set_index('顶点类别')
point1=data2['顶点'][1].tolist()
point2=data2['顶点'][2].tolist()
ind1=[nodes.index(i) for i in point1 if i in nodes]
ind2=[nodes.index(i) for i in point2 if i in nodes]
nx.draw_networkx_nodes(G,pos,nodelist=ind1,node_color='red',node_shape='*')
nx.draw_networkx_nodes(G,pos,nodelist=ind2,node_color='red',node_shape='^')
plt.show()
#结果:
#L到M3的最短距离为: 2613.0
#L到M3的路径为: [10, 9, 22, 14, 26, 27, 49, 80, 79, 89]
当边的权值为两点之间的距离时,求上面无向图的最小生成树,并画出最小生成树 import networkx as nx
import pylab as plt
import math
import numpy as np
import pandas as pd
data = pd.read_excel('Pex10_5.xlsx',keep_default_na=False)
# 获取坐标构造pos
x_site = data['x坐标'].tolist()
y_site = data['y坐标'].tolist()
d = len(x_site)
position = zip(range(0, d), zip(x_site, y_site))
pos = dict(position) # 构造字典
nodes = data['顶点'].tolist() # 构造点的标签
matrix = pd.DataFrame(np.zeros((d, d)), index=nodes, columns=nodes) # 构造无向图的邻接矩阵
data1 = data.set_index('顶点')
pos_alph = dict(zip(nodes, zip(x_site, y_site))) # 插寻点的坐标
f = lambda X, Y: np.round(np.sqrt((X[0] - X[1]) ** 2 + (Y[0] - Y[1]) ** 2))
print(pos_alph)
print(data1)
for i in nodes:
if len(data1['相邻的顶点1'][i]) != 0:
temp = data1['相邻的顶点1'][i]
x1, y1 = pos_alph[i]
x2, y2 = pos_alph[temp]
matrix[i][temp] = f([x1, x2], [y1, y2]) # 修改为两点之间的距离
if len(data1['相邻的顶点2'][i]) != 0:
temp = data1['相邻的顶点2'][i]
x1, y1 = pos_alph[i]
x2, y2 = pos_alph[temp]
matrix[i][temp] = f([x1, x2], [y1, y2]) # 修改为两点之间的距离
x3, y3 = pos_alph['P3']
x4, y4 = pos_alph['G3']
x5, y5 = pos_alph['O3']
matrix['P3']['G3'] = f([x3, x4], [y3, y4])
matrix['O3']['G3'] = f([x3, x5], [y3, y5])
mat = matrix.T.values # 转化为上三角矩阵
G=nx.Graph(mat)
T=nx.minimum_spanning_tree(G)
w=nx.get_edge_attributes(T,'weight')
nx.draw(T,pos=pos,labels=dict(zip(range(d), nodes)),node_size=100,font_size=9,node_color='green',width=0.7)
nx.draw_networkx_edge_labels(T,pos,edge_labels=w,font_size='4')
print(w)
plt.show()
甲,乙两个煤矿分别生产煤500万吨,供应A,B,C三个电厂发电需要,各电厂用量分别为300,300,400(万吨)。已知煤矿之间,煤矿与电厂之间以及各电厂之间相互距离(单位:km)如表10.5,表10.6,表10.7所示。煤可以直接运达,也可以经转运抵达,试确定从煤矿到各电厂间煤的最优调运方案。
甲乙甲0120乙1000
ABC甲15012080乙6016040
ABCA070100B500120C1001500参考:2012年数学建模集训小题目 - 豆丁网 (docin.com)
考虑为线性规划问题+图论问题:
求解出在最短路约束下的分配情况 求解最短路:顶点集为
V
=
{
v
1
,
v
2
,
v
3
,
v
4
,
v
5
}
V=\{v_1,v_2,v_3,v_4,v_5\}
V={v1,v2,v3,v4,v5} (甲,乙,A,B,C)边集为相应的路径(权重) import networkx as nx
import pylab as plt
import numpy as np
import pandas as pd
def floyd(graph):
m = len(graph)
dis = graph
path = np.zeros((m, m)) # 路由矩阵初始化
for k in range(m):
for i in range(m):
for j in range(m):
if dis[i][k] + dis[k][j] < dis[i][j]:
dis[i][j] = dis[i][k] + dis[k][j]
path[i][j] = k#更新先驱点
return dis, path
plt.rcParams['font.sans-serif']=['SimHei']
inf=np.inf
if __name__ == "__main__":
W=np.array([[0,120,150,120,80],[100,0,60,160,40],[inf,inf,0,70,100],[inf,inf,50,0,120],[inf,inf,100,150,0]])
nodes=list('甲乙ABC')
edges=[(i,j,W[i-1,j-1])for i in range(1,6)for j in range (1,6) ]
edge=edges.copy()
for i in edges:
if i[2]==inf or i[2]==0:
edge.remove(i)
G=nx.DiGraph()
for k in range(len(edge)):
G.add_edge(edge[k][0]-1,edge[k][1]-1,weight=edge[k][2])
w=nx.get_edge_attributes(G,'weight')
nx.draw(G,pos=nx.shell_layout(G),labels=dict(zip(range(5),nodes)),node_color='red')
nx.draw_networkx_edge_labels(G,nx.shell_layout(G),edge_labels=w)
plt.show()
dis,path=floyd(W)#dis 为最短距离矩阵
mat=dis[0:2,2:]#切片处理
table=pd.DataFrame(mat,columns=list('ABC'),index=['甲','乙'])
print(table.to_markdown())#转化为markdown格式
最短距离:
ABC甲15012080乙6013040 线性规划:
x
i
j
x_{ij}
xij:第
i
i
i个煤矿到第
j
j
j个电厂的调运量
c
i
j
c_{ij}
cij:第
i
i
i个煤矿到第
j
j
j个电厂的最短距离
b
j
b_{j}
bj:第
j
j
j个电厂的需求量
a
i
a_{i}
ai:第
i
i
i个煤矿的产量目标函数:总吨公里数(吨数
×
\times
×公里数)约束条件:
产量约束需求量约束 线性规划模型如下:
m
i
n
z
=
∑
i
=
1
2
∑
j
=
3
3
c
i
j
x
i
j
s
.
t
.
{
∑
j
=
1
3
x
i
j
=
a
i
,
i
=
1
,
2
∑
i
=
1
2
x
i
j
=
b
j
,
j
=
1
,
2
,
3
x
i
j
≥
0
,
i
=
1
,
2
;
j
=
1
,
2
,
3
min~~z=\sum_{i=1}^2\sum_{j=3}^3c_{ij}x_{ij}\\s.t. \left\{ \begin{aligned} &\sum_{j=1}^3x_{ij}=a_i,i=1,2\\ &\sum_{i=1}^2x_{ij}=b_j,j=1,2,3\\ &x_{ij}\ge0,i=1,2;j=1,2,3 \end{aligned} \right.
min z=i=1∑2j=3∑3cijxijs.t.⎩
⎨
⎧j=1∑3xij=ai,i=1,2i=1∑2xij=bj,j=1,2,3xij≥0,i=1,2;j=1,2,3 LNGO: sets:
fac/1..2/:a;
plant/1..3/:b;
coo(fac,plant):x,c;
endsets
data:
a=500,500;
b=300,300,400;
c=150,120,80,60,130,40;
enddata
min=@sum(coo(i,j):c(i,j)*x(i,j));
@for(fac(i):@sum(plant(j):x(i,j))=a(i));
@for(plant(j):@sum(fac(i):x(i,j))=b(j));
Objective value: 78000.00
X( 1, 1) 0.00000
X( 1, 2) 300.0000
X( 1, 3) 200.0000
X( 2, 1) 300.0000
X( 2, 2) 0.000000
X( 2, 3) 200.0000
所以应该从甲地运往B地300万吨,C地200万吨;从乙地运往A地300 万吨,C地200万吨 10.7 图10.23 给出了6支球队的比赛结果,即1队战胜2,4,5,6对,而输给了3队;5队战胜3,6队,而输给1,2,4队;等等
利用竞赛图的适当方法,给出6支球队的一个排名顺序(竞赛图必存在哈密顿圈(不一定回到起始点))参考:《数学模型》(第五版)循环比赛问题
如果可以找到唯一的一条完全路径,则由它经过的顶点的顺序即为排名顺序如果存在多条路径:
双向连通图:接连矩阵的最大特征值对应的特征向量
→
\rightarrow
→ 排名顺序非双向连通图:无法确定所有队的排名 如图所示已经存在以下两条完全路径:排除第一种情况3-1-2-4-5-61-2-5-3-4-6 利用PageRank 算法,再次给出6支球队的排名顺序 import networkx as nx
import pylab as plt
import numpy as np
from scipy.sparse.linalg import eigs
# 绘制有向图
def floyd(graph):
m = len(graph)
dis = graph
path = np.zeros((m, m)) # 路由矩阵初始化
for k in range(m):
for i in range(m):
for j in range(m):
if dis[i][k] + dis[k][j] < dis[i][j]:
dis[i][j] = dis[i][k] + dis[k][j]
path[i][j] = k # 更新先驱点
return dis, path
if __name__ == '__main__':
L = [(1, 2), (1, 4), (1, 5), (1, 6), (2, 4), (2, 6), (2, 5), (3, 1), (3, 2), (3, 4), (4, 5), (4, 6), (5, 3), (5, 6),
(6, 3)]
G = nx.DiGraph()
G.add_nodes_from(range(1, 7))
G.add_edges_from(L)
matrix = np.array(nx.to_numpy_matrix(G)) # 接连矩阵
pos = nx.shell_layout(G)
nx.draw(G, pos, node_size=250, font_weight='bold', node_color='red', with_labels=True, arrowsize=17, width=0.5,
alpha=0.6)
A1 = matrix.copy()
A1[A1 == 0] = np.inf
dis, path = floyd(A1);
print(dis) # 可以看出竞赛图是双向连通图
# 竞赛图算法
w, v = np.linalg.eig(matrix)
print(v[:, 0].argsort()) # 1->3->2->5->4->6 排名顺序
# PageRank 算法
A2 = matrix / np.tile(matrix.sum(axis=1, keepdims=True), (1, matrix.shape[1])) # tile 扩展行和
A2 = 0.15 / matrix.shape[0] + 0.85 * A2 # 构造状态转移矩阵
W, V = eigs(A2.T, 1) # 特征值为1时的特征向量
V = V.real;
V = V.flatten() # 展开成一维数组
V = V / V.sum() # 向量归一化
# 绘制柱状图
plt.figure(2)
plt.bar(range(1, matrix.shape[0] + 1), V, width=0.7, color='b')
plt.show()
#排名为:1->2-> 5-> 4-> 6-> 3
计算如图10.24所示网络的度分布,网络直径,平均路径长度,各节点的聚类系数和整个网络的聚类系数
import networkx as nx
import pylab as plt
L = [(1, 2), (1, 5), (2, 4), (2, 3), (2, 5),
(3, 5), (3, 4), (5, 6)]
G = nx.Graph() # 构造无向图
G.add_nodes_from(range(1, 7)) # 添加顶点集
G.add_edges_from(L)
D = nx.diameter(G) # 求网络直径
LH = nx.average_shortest_path_length(G) # 求平均路径长度
Ci = nx.clustering(G) # 求各顶点的聚类系数
C = nx.average_clustering(G) # 求整个网络的聚类系数
print("网络直径为:", D, "\n平均路径长度为:", LH)
print("各顶点的聚类系数为:")
for index, value in enumerate(Ci.values()):
print("(顶点v{:d}: {:.4f});".format(index + 1, value), end=' ')
print("\n整个网络的聚类系数为:{:.4f}".format(C))
nx.draw(G, pos=nx.shell_layout(G), with_labels=True)
plt.show()
# 网络直径为: 3
# 平均路径长度为: 1.5333333333333334
# 各顶点的聚类系数为:
# (顶点v1: 1.0000); (顶点v2: 0.5000); (顶点v3: 0.6667); (顶点v4: 1.0000); (顶点v5: 0.3333); (顶点v6: 0.0000);
# 整个网络的聚类系数为:0.5833
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