python - "Hamiltonian"路径使用 Python

Question

我正在尝试使用 Python 实现对遍历所有图形顶点的任意路径(不一定是循环)的递归搜索。这是我的代码:

def hamilton(G, size, pt, path=[]):
    if pt not in set(path):
        path.append(pt)
        if len(path)==size:
            return path
        for pt_next in G[pt]:
            res_path = [i for i in path]
            hamilton (G, size, pt_next, res_path)

这里，pt是起点，path是之前遍历的所有顶点的列表，不包括pt，默认为空。问题是，无论何时找到这样的路径，返回语句都会引用过程的某些内部调用，因此程序不会终止或返回路径。

例如，取 G = {1:[2,3,4], 2:[1,3,4], 3:[1,2,4], 4:[1,2, 3]}(即完整的 4 图)并运行 hamilton(G,4,1,[])。它返回 None，但如果您打印路径而不是将其作为值返回，您会发现它实际上找到了从 1 开始的所有六个路径。

如果我告诉程序打印路径和 return 语句，它最终会打印所有这些路径，因此运行时间比需要的长得多。

如何修复代码，以便在找到第一个合适的路径后终止执行？

Answer 1

基本错误是递归调用的结果需要返回，如果它没有导致死胡同的话。

此外，如果点pt 没有邻居，G[pt] 将引发IndexError。使用 dict.get 可以很容易地解决这个问题。

def hamilton(G, size, pt, path=[]):
    print('hamilton called with pt={}, path={}'.format(pt, path))
    if pt not in set(path):
        path.append(pt)
        if len(path)==size:
            return path
        for pt_next in G.get(pt, []):
            res_path = [i for i in path]
            candidate = hamilton(G, size, pt_next, res_path)
            if candidate is not None:  # skip loop or dead end
                return candidate
        print('path {} is a dead end'.format(path))
    else:
        print('pt {} already in path {}'.format(pt, path))
    # loop or dead end, None is implicitly returned

例子

>>> G = {1:[2,3,4], 2:[1,3,4], 3:[1,2,4], 4:[1,2,3]}
>>> hamilton(G, 4, 1)
hamilton called with pt=1, path=[]
hamilton called with pt=2, path=[1]
hamilton called with pt=1, path=[1, 2]
pt 1 already in path [1, 2]
hamilton called with pt=3, path=[1, 2]
hamilton called with pt=1, path=[1, 2, 3]
pt 1 already in path [1, 2, 3]
hamilton called with pt=2, path=[1, 2, 3]
pt 2 already in path [1, 2, 3]
hamilton called with pt=4, path=[1, 2, 3]
[1, 2, 3, 4]

>>> G = {1:[2], 2:[3,4], 4:[3]}
>>> hamilton(G, 4, 1)
hamilton called with pt=1, path=[]
hamilton called with pt=2, path=[1]
hamilton called with pt=3, path=[1, 2]
path [1, 2, 3] is a dead end
hamilton called with pt=4, path=[1, 2]
hamilton called with pt=3, path=[1, 2, 4]
[1, 2, 4, 3]

>>> G = {1:[2], 2:[3,4], 4:[3]}
>>> hamilton(G, 4, 2)
hamilton called with pt=2, path=[]
hamilton called with pt=3, path=[2]
path [2, 3] is a dead end
hamilton called with pt=4, path=[2]
hamilton called with pt=3, path=[2, 4]
path [2, 4, 3] is a dead end
path [2, 4] is a dead end
path [2] is a dead end
None

请注意，由于“可变默认参数”问题，多次使用该函数可能会导致错误结果。但这不是这个答案的范围。