为什么 Floyd Warshall 算法的实现依赖于节点顺序？

Question

这是我对 Floyd Warshall 算法的实现：

def algorithm(self, graph):
    nodes = graph.keys()
    shuffle(nodes, lambda : 0.5)

    int_nodes = range(len(nodes))
    arcs = set((a,b) for a in nodes for b in graph[a])

    distances = {} 
    for i in int_nodes:
        distances[(i, i, 0)] = 0

    for i in int_nodes:
        for j in int_nodes:
            distances[(i, j, 0)] = 1 if (nodes[i], nodes[j]) in arcs else float("inf")

    for k in range(1, len(nodes)):
        for i in int_nodes:
            for j in int_nodes:
                distances[(i, j, k)] = min(distances[(i, j, k-1)], distances[(i, k, k-1)] + distances[(k, j, k-1)])


    return {(nodes[i], nodes[j]): distances[(i, j, len(nodes)-1)] for i in int_nodes for j in int_nodes}

如果我改变随机播放的种子，结果有时会改变。

为什么会这样？

编辑。

这是一个最小的工作示例：

from random import shuffle

def algorithm(graph, n):
nodes = graph.keys()
shuffle(nodes, lambda : n)

int_nodes = range(len(nodes))
arcs = set((a,b) for a in nodes for b in graph[a])

distances = {} 
for i in int_nodes:
    distances[(i, i, 0)] = 0

for i in int_nodes:
    for j in int_nodes:
        distances[(i, j, 0)] = 1 if (nodes[i], nodes[j]) in arcs else float("inf")

for k in range(1, len(nodes)):
    for i in int_nodes:
        for j in int_nodes:
            distances[(i, j, k)] = min(distances[(i, j, k-1)], distances[(i, k, k-1)] + distances[(k, j, k-1)])


return {(nodes[i], nodes[j]): distances[(i, j, len(nodes)-1)] for i in int_nodes for j in int_nodes}

if __name__ == "__main__":
graph = {'Z': ['B', 'H', 'G', 'O', 'I'], 'F': ['C', 'G', 'D', 'O'], 'L': ['M', 'C', 'D', 'E', 'H'], 'C': ['F', 'G', 'B', 'L', 'M', 'I'], 'B': ['C', 'Z', 'I', 'O', 'H', 'G'], 'D': ['F', 'L', 'G', 'M', 'E'], 'E': ['L', 'D', 'G', 'M'], 'H': ['B', 'L', 'Z', 'I', 'O'], 'G': ['C', 'F', 'D', 'E', 'Z', 'B'], 'O': ['B', 'H', 'F', 'I', 'Z'], 'M': ['L', 'D', 'E', 'C'], 'I': ['B', 'H', 'O', 'Z', 'C']}

for i in [0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9]:
    dis1 = algorithm(graph,i )
    print sum(dis1.values())

这是输出：

总长度应该是一样的，但随着种子的变化而变化。

Answer 1

你的最后一组循环没有考虑 k=0，它应该，否则你在搜索中忽略了形式 a->0->b 的路径。一般来说，使用 k 来索引节点和迭代的想法有点奇怪（并且使调试更难）。

您可以通过

轻松修复它

from random import shuffle

def algorithm(graph, n):
  nodes = graph.keys()
  shuffle(nodes, lambda : n)

  int_nodes = range(len(nodes))
  arcs = set((a,b) for a in nodes for b in graph[a])

  distances = {} 
  for i in int_nodes:
      distances[(i, i, -1)] = 0

  for i in int_nodes:
      for j in int_nodes:
          distances[(i, j, -1)] = 1 if (nodes[i], nodes[j]) in arcs else float("inf")

  for k in int_nodes:
      for i in int_nodes:
          for j in int_nodes:
              distances[(i, j, k)] = min(distances[(i, j, k-1)], distances[(i, k, k-1)] + distances[(k, j, k-1)])


  return {(nodes[i], nodes[j]): distances[(i, j, len(nodes)-1)] for i in int_nodes for j in int_nodes}

if __name__ == "__main__":
  graph = {'Z': ['B', 'H', 'G', 'O', 'I'], 'F': ['C', 'G', 'D', 'O'], 'L': ['M', 'C', 'D', 'E', 'H'], 'C': ['F', 'G', 'B', 'L', 'M', 'I'], 'B': ['C', 'Z', 'I', 'O', 'H', 'G'], 'D': ['F', 'L', 'G', 'M', 'E'], 'E': ['L', 'D', 'G', 'M'], 'H': ['B', 'L', 'Z', 'I', 'O'], 'G': ['C', 'F', 'D', 'E', 'Z', 'B'], 'O': ['B', 'H', 'F', 'I', 'Z'], 'M': ['L', 'D', 'E', 'C'], 'I': ['B', 'H', 'O', 'Z', 'C']}

  for i in [0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9]:
      dis1 = algorithm(graph,i )
      print sum(dis1.values())

这给出了

为什么 Floyd Warshall 算法的实现依赖于节点顺序？

Why is this implementation of Floyd Warshall algorithm dependent on node order?

python

algorithm

floyd-warshall