Question

代码：

def bipartite(G):
    open_list = [1]
    colors = {}
    color_counter = 0
    # assign a color to the first node being visited
    colors[1] = 0

    while open_list:
        # up the counter here so that all neighbors get the same color
        color_counter += 1
        # use first elem for bfs
        current_neighbors = G[open_list[0]]
        current_color = color_counter % 2
        # prints used for debugging
        print open_list
        print "The current color is: %s" % (current_color,)
        for neighbor in current_neighbors:
            if neighbor not in colors:
                open_list.append(neighbor)
                colors[neighbor] = current_color
                # print used for debugging
                print "parent is: %s, child is: %s, %s's color is: %s" \
                % (open_list[0], neighbor, neighbor, colors[neighbor])
                # print used for debugging
            else: print "parent is: %s, child is: %s, already colored: %s" \
                % (open_list[0], neighbor, colors[neighbor])
        open_list.pop(0)
    # now, return array of values that has one of the two colors
    zeros_array = []
    ones_array = []
    for key in colors.keys():
        if colors[key] == 0:
            zeros_array.append(key)
        else:
            ones_array.append(key)

    if len(set(zeros_array) & set(ones_array)) == 0:
        return zeros_array
    else:
        return None

这是我使用的图表：

{1: {2: 1, 4: 1}, 2: {1: 1, 3: 1, 5: 1}, 3: {8: 1, 2: 1}, 4: {1: 1}, 5: {2: 1, 6: 1}, 6: {5: 1}, 8: {3: 1}}

我把它绘制出来，图形可以看作一棵树，其中1为根，并分支到节点2和4，其中4是叶子，但是2继续前进。我使用颜色计数器为相邻颜色（0或1）的邻居着色。 2和4给出相同的颜色，然后算法正确地给出3和5他们的父2的相反颜色，但是当返回一个级别直到检查4时，颜色计数器递增，所以当它到达8时， 8得到错误的颜色。

我坚持如何最好地解决这个问题。

Answer 1

您应该根据当前的顶点颜色选择颜色，例如colors[neighbor] = (colors[open_list[0]] + 1) % 2

此外，len(set(zeros_array) & set(ones_array)) == 0始终为true，因此您无需检查二分法是否顺利。你可以在if neighbor not in colors:的else分支中检查它：只是声明你的邻居有不同颜色的当前顶点。

使用bfs着色算法检查Python中的图是否为二分

1 个答案: