问题描述
我正在尝试对当前依赖于两个 for 循环的最大加权独立集问题矢量化一个目标函数。
目前，它采用networkx图，节点权重存储为'node_weight'。这是我目前的代码，它给出了我试图最小化的客观值。
def mwis_objective(x,G):
'''
Takes in networkx graph G and a bit string x from the Qasm output and calculates the < psi | C | psi >
Need to take note of the order of the bit string.
'''
array_of_x = np.array(list(x),dtype=int) # this takes the bit string 1001 to a numpy array for faster access
objective = 0
print(np.array(G.edges()))
for i,j in G.edges(): # independent set
if array_of_x[i] == 1 and array_of_x[j] ==1: # interconnecting nodes are independent set
objective += 2
# getting the maximum weight of nodes
node_weights = G.nodes(data='node_weight') # this is how the node weight is stored in my graph attributes
just_weights = np.array([weight[1] for weight in node_weights]) #gets just the node weight and converts it to a np array
scale = np.amax(just_weights) # gets the maximum weight so the node weights are
scaled_weights = just_weights/scale # used as J_i,j must be greater than weight of node; all node weights are scaled to below 0 and J_ij is put as 2
for i in np.array(G.nodes()):
if array_of_x[i] == 1:
objective -= scaled_weights[i]
return objective
print(mwis_objective('111',weighted_path_graph(3,[5,5,5])))
位串作为字符串输入，对于长度为 3 的路径图，'101' 表示节点 0 和 2 在集合中，但节点 1 不在。
第一个 for 循环检查集合中的两个节点是否连接，如果是，则对目标函数施加 2 的惩罚。第二个 for 检查节点是否在集合中，如果是，则从目标中减去其缩放权重。
我能想到的向量化第二个和的最佳方法是使用 np.where 函数。如果有一种方法可以将其配置为给出一个 1 的数组，其中节点是该集合的成员，例如 in_set = [ 1 0 1 ] 这可以是 scaled_weights 数组的时间，然后求和为一个。不过，我不确定如何执行此操作，也无法想出对第一个 for 循环进行类似加速的方法。
def weighted_path_graph(number_of_nodes,graph_weights):
"""
Creates a weighted path graph of default length three with different weights
Graph_weights is a list with the desired node weights
"""
path = nx.path_graph(number_of_nodes)
#graph_weights=[1,3,1]
for i in range(0,number_of_nodes):
path.nodes[i]["node_weight"] = graph_weights[i]
return path
解决方法
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