递归误差 - 分离度

Question

我试图找出电影数据库中任意两个演员之间的分离度。 当我达到我的基本情况时，我成功了，即 1 度分离（即演员与另一个演员在同一部电影中）但我使用递归来找到所有其他分离度，我得到：

runtime error: maximum recursion depth exceeded in cmp.

##gets file with movie information
f = open("filename.txt")
actedWith = {}
ActorList = []
movies = {}
actedIn = []
dos = 1

def getDegrees(target, base, dos):
    for actor in actedWith[base]:
        if target == actor:
            print base, "has ", dos, " degree(s) of separation from ", target
            return
    dos = dos+1
    for actor in actedWith[base]:
        getDegrees(target, actor, dos)


for l in f:
    ##strip of whitespace
    l = l.strip()
    ##split by where forward-slashes are
    l = l.split("/")
    ##add the first "word" on the line to the database of movie names
    movies = {l[0] : l[1:]}
    for e in l[1:]:
        if e in actedWith:
            actedWith[e] = actedWith[e]+movies[l[0]]
        else:
            actedWith[e] = movies[l[0]]

base = raw_input("Enter Actor Name (Last, First): ")
target = raw_input("Enter Second Actor Name (Last, First): ")
getDegrees(target, base, dos)

我使用的文本文件可以在http://www.mediafire.com/?qtryvkzmuv5jey3找到

为了测试基本情况，我使用：Bacon, Kevin 和 Pitt, Brad。

为了测试其他人，我使用 Bacon, Kevin 和 Gamble, Nathan。

Answer 1

两个建议（我没有查看文本文件，只是在这里介绍第一原则并快速阅读您的代码）：

1. 当您从 getDegrees 返回时，您仍在执行返回后的其余函数。您需要返回一个 True （或其他内容）以指示搜索结束并且应该回滚整个函数调用堆栈。第一个 return 将更改为“return True”，最后一行将更改为“if getDegrees(target, actor, dos): return True”。
2. 跟踪已搜索过哪些演员。如果两个演员互相演戏，或者关系中有一个循环，你就会来回循环。

此代码尝试修复返回和图形循环问题。但是，某处仍然存在逻辑错误； Kevin Bacon 和 James Belushi（分离度为 2）给出以下内容：

Siravo, Joseph 与 Belushi, James 有 179 度的分离度

编辑：通过添加“原始”参数修复。

但是递归问题是固定的。

##gets file with movie information
f = open("filename.txt")
actedWith = {}
ActorList = []
movies = {}
actedIn = []
dos = 1

def getDegrees(original, target, base, dos=0, seen=[]):
    dos = dos+1
    print "----> checking %s against %s" % (target, base)
    for actor in actedWith[base]:
        #print "\t" + actor
        if target == actor:
            print original, "has ", dos, " degree(s) of separation from ", target
            return True
    for actor in actedWith[base]:
        if actor in seen: continue
        seen = seen + [actor]
        if getDegrees(original, target, actor, dos, seen):
            return True
    return False


for l in f:
    ##strip of whitespace
    l = l.strip()
    ##split by where forward-slashes are
    l = l.split("/")
    ##add the first "word" on the line to the database of movie names
    movies = {l[0] : l[1:]}
    for e in l[1:]:
        if e in actedWith:
            actedWith[e] = actedWith[e]+movies[l[0]]
        else:
            actedWith[e] = movies[l[0]]

original = raw_input("Enter Actor Name (Last, First): ")
target = raw_input("Enter Second Actor Name (Last, First): ")
getDegrees(original, target, original)

例子：

Bacon, Kevin has  65  degree(s) of separation from  Kosaka, Masami

Answer 2

除非我没有看到 actedWith 的某些属性，否则您没有任何东西可以防止无限循环。例如，您的一个递归调用将是getDegrees("Gamble, Nathan", "Pitt, Brad", 2)，那么由于凯文培根与布拉德皮特合作，当您更深入时，您将调用getDegrees("Gamble, Nathan", "Bacon, Kevin", 3)。看到问题了吗？

Answer 3

这可能是无限递归。您正在搜索植根于目标的树；那棵树上的一些路径正在到达它们上游的点。您需要一种方法来识别这一点，并在它发生时停止往下看。

一种方法是在路径上保留祖先列表。比如：

def getDegrees(target, base, dos, ancestors):  # Also carry a list of "ancestors"
    for actor in actedWith[base]:
        if target == actor:
            print base, "has ", dos, " degree(s) of separation from ", target
            return
    dos = dos+1
    ancestors = ancestors + [base]  # Must be separate variable binding to avoid mutating the caller's copy
    for actor in actedWith[base]:
        if actor in ancestors: continue  # Check if on path, skip if so
        getDegrees(target, actor, dos, ancestors)

...

getDegrees(target, base, dos, [target])

请注意，“祖先”是指路径上的一个点，而不是演员可能与之相关的人。

这并不能避免演员自己拥有actedWith 的情况（希望输入文件永远不会包含它），但只需稍作改动即可。