使用 bfs 着色算法检查图是否在 Python 中是二分图

Using bfs coloring algorithm to check if graph is bipartite in Python

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

这是我正在使用的图表:

{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}}

我把它画出来了,这个图可以想象成一棵树,以 1 为根,然后分支到节点 2 和 4,其中 4 是叶子,但 2 继续前进。我正在使用颜色计数器为邻居着色相同的颜色(0 或 1)。 2 和 4 被赋予相同的颜色,然后算法正确地赋予 3 和 5 与其父 2 相反的颜色,但是当返回一个级别直到检查 4 时,颜色计数器增加,所以当它达到 8 时, 8 得到了错误的颜色。

我不知道如何最好地解决这个问题。

你应该根据你当前的顶点颜色选择颜色,比如colors[neighbor] = (colors[open_list[0]] + 1) % 2

此外,len(set(zeros_array) & set(ones_array)) == 0 将始终是 true,因此您不会检查二分法是否运行良好。您可以在 if neighbor not in colors: 的其他分支中检查它:只需断言您的邻居与当前顶点具有不同的颜色。