我想在加权有向图上找到最小生成树（MST）。我一直在尝试使用我在Python中实现的Chu-Liu/Edmond's algorithm（下面的代码）。可以找到简单，清晰的算法描述here。我有两个问题。
Edmond的算法是否能保证收敛于解决方案？
我担心删除一个循环会增加另一个循环。如果发生这种情况，算法将继续尝试永久删除循环。
我似乎找到了一个这样的例子。输入图如下所示（在代码中）。该算法永远不会完成，因为它在周期[1,2]和[1,3]以及[5,4]和[5,6]之间切换。添加到图中的边以解决周期[5,4]创建周期[5,6]，反之亦然，对于[1,2]和[1,3]也类似。
我应该注意到，我不确定我的实施是否正确。
为解决此问题，我介绍了一个临时补丁。当删除边以移除循环时，我会从我们正在搜索MST的基础图G中永久删除该边。因此，不能再添加该边缘，这应该可以防止算法卡住。有了这个改变，我保证找到一个MST吗？
我怀疑人们可以找到一个病态案例，这一步骤会导致结果不是MST，但我无法想到一个。它似乎适用于我尝试的所有简单测试用例。
代码：
import sys
# --------------------------------------------------------------------------------- #
def _reverse(graph):
r = {}
for src in graph:
for (dst,c) in graph[src].items():
if dst in r:
r[dst][src] = c
else:
r[dst] = { src : c }
return r
# Finds all cycles in graph using Tarjan's algorithm
def strongly_connected_components(graph):
"""
Tarjan's Algorithm (named for its discoverer, Robert Tarjan) is a graph theory algorithm
for finding the strongly connected components of a graph.
Based on: http://en.wikipedia.org/wiki/Tarjan%27s_strongly_connected_components_algorithm
"""
index_counter = [0]
stack = []
lowlinks = {}
index = {}
result = []
def strongconnect(node):
# set the depth index for this node to the smallest unused index
index[node] = index_counter[0]
lowlinks[node] = index_counter[0]
index_counter[0] += 1
stack.append(node)
# Consider successors of `node`
try:
successors = graph[node]
except:
successors = []
for successor in successors:
if successor not in lowlinks:
# Successor has not yet been visited; recurse on it
strongconnect(successor)
lowlinks[node] = min(lowlinks[node],lowlinks[successor])
elif successor in stack:
# the successor is in the stack and hence in the current strongly connected component (SCC)
lowlinks[node] = min(lowlinks[node],index[successor])
# If `node` is a root node, pop the stack and generate an SCC
if lowlinks[node] == index[node]:
connected_component = []
while True:
successor = stack.pop()
connected_component.append(successor)
if successor == node: break
component = tuple(connected_component)
# storing the result
result.append(component)
for node in graph:
if node not in lowlinks:
strongconnect(node)
return result
def _mergeCycles(cycle,G,RG,g,rg):
allInEdges = [] # all edges entering cycle from outside cycle
minInternal = None
minInternalWeight = sys.maxint
# Find minimal internal edge weight
for n in cycle:
for e in RG[n]:
if e in cycle:
if minInternal is None or RG[n][e] < minInternalWeight:
minInternal = (n,e)
minInternalWeight = RG[n][e]
continue
else:
allInEdges.append((n,e)) # edge enters cycle
# Find the incoming edge with minimum modified cost
# modified cost c(i,k) = c(i,j) - (c(x_j, j) - min{j}(c(x_j, j)))
minExternal = None
minModifiedWeight = 0
for j,i in allInEdges: # j is vertex in cycle, i is candidate vertex outside cycle
xj, weight_xj_j = rg[j].popitem() # xj is vertex in cycle that currently goes to j
rg[j][xj] = weight_xj_j # put item back in dictionary
w = RG[j][i] - (weight_xj_j - minInternalWeight) # c(i,k) = c(i,j) - (c(x_j, j) - min{j}(c(x_j, j)))
if minExternal is None or w <= minModifiedWeight:
minExternal = (j,i)
minModifiedWeight = w
w = RG[minExternal[0]][minExternal[1]] # weight of edge entering cycle
xj,_ = rg[minExternal[0]].popitem() # xj is vertex in cycle that currently goes to j
rem = (minExternal[0], xj) # edge to remove
rg[minExternal[0]].clear() # popitem() should delete the one edge into j, but we ensure that
# Remove offending edge from RG
# RG[minExternal[0]].pop(xj, None) #highly experimental. throw away the offending edge, so we never get it again
if rem[1] in g:
if rem[0] in g[rem[1]]:
del g[rem[1]][rem[0]]
if minExternal[1] in g:
g[minExternal[1]][minExternal[0]] = w
else:
g[minExternal[1]] = { minExternal[0] : w }
rg = _reverse(g)
# --------------------------------------------------------------------------------- #
def mst(root,G):
""" The Chu-Liu/Edmond's algorithm
arguments:
root - the root of the MST
G - the graph in which the MST lies
returns: a graph representation of the MST
Graph representation is the same as the one found at:
http://code.activestate.com/recipes/119466/
Explanation is copied verbatim here:
The input graph G is assumed to have the following
representation: A vertex can be any object that can
be used as an index into a dictionary. G is a
dictionary, indexed by vertices. For any vertex v,
G[v] is itself a dictionary, indexed by the neighbors
of v. For any edge v->w, G[v][w] is the length of
the edge.
"""
RG = _reverse(G)
g = {}
for n in RG:
if len(RG[n]) == 0:
continue
minimum = sys.maxint
s,d = None,None
for e in RG[n]:
if RG[n][e] < minimum:
minimum = RG[n][e]
s,d = n,e
if d in g:
g[d][s] = RG[s][d]
else:
g[d] = { s : RG[s][d] }
cycles = [list(c) for c in strongly_connected_components(g)]
cycles_exist = True
while cycles_exist:
cycles_exist = False
cycles = [list(c) for c in strongly_connected_components(g)]
rg = _reverse(g)
for cycle in cycles:
if root in cycle:
continue
if len(cycle) == 1:
continue
_mergeCycles(cycle, G, RG, g, rg)
cycles_exist = True
return g
# --------------------------------------------------------------------------------- #
if __name__ == "__main__":
# an example of an input that works
root = 0
g = {0: {1: 23, 2: 22, 3: 22}, 1: {2: 1, 3: 1}, 3: {1: 1, 2: 0}}
# an example of an input that causes infinite cycle
root = 0
g = {0: {1: 17, 2: 16, 3: 19, 4: 16, 5: 16, 6: 18}, 1: {2: 3, 3: 3, 4: 11, 5: 10, 6: 12}, 2: {1: 3, 3: 4, 4: 8, 5: 8, 6: 11}, 3: {1: 3, 2: 4, 4: 12, 5: 11, 6: 14}, 4: {1: 11, 2: 8, 3: 12, 5: 6, 6: 10}, 5: {1: 10, 2: 8, 3: 11, 4: 6, 6: 4}, 6: {1: 12, 2: 11, 3: 14, 4: 10, 5: 4}}
h = mst(int(root),g)
print h
for s in h:
for t in h[s]:
print "%d-%d" % (s,t)
答案 0 :(得分：7)
不要做临时补丁。我承认实现收缩/收缩逻辑并不直观，并且在某些情况下递归是不可取的，因此这是一个适当的Python实现，可以使生产质量。我们不是在每个递归级别执行非约束步骤，而是将其推迟到最后并使用深度优先搜索，从而避免递归。 （这种修改的正确性最终来自互补松弛，是线性规划理论的一部分。）
下面的命名约定是_rep表示超级节点（即一个或多个签约节点的块）。
#!/usr/bin/env python3
from collections import defaultdict, namedtuple
Arc = namedtuple('Arc', ('tail', 'weight', 'head'))
def min_spanning_arborescence(arcs, sink):
good_arcs = []
quotient_map = {arc.tail: arc.tail for arc in arcs}
quotient_map[sink] = sink
while True:
min_arc_by_tail_rep = {}
successor_rep = {}
for arc in arcs:
if arc.tail == sink:
continue
tail_rep = quotient_map[arc.tail]
head_rep = quotient_map[arc.head]
if tail_rep == head_rep:
continue
if tail_rep not in min_arc_by_tail_rep or min_arc_by_tail_rep[tail_rep].weight > arc.weight:
min_arc_by_tail_rep[tail_rep] = arc
successor_rep[tail_rep] = head_rep
cycle_reps = find_cycle(successor_rep, sink)
if cycle_reps is None:
good_arcs.extend(min_arc_by_tail_rep.values())
return spanning_arborescence(good_arcs, sink)
good_arcs.extend(min_arc_by_tail_rep[cycle_rep] for cycle_rep in cycle_reps)
cycle_rep_set = set(cycle_reps)
cycle_rep = cycle_rep_set.pop()
quotient_map = {node: cycle_rep if node_rep in cycle_rep_set else node_rep for node, node_rep in quotient_map.items()}
def find_cycle(successor, sink):
visited = {sink}
for node in successor:
cycle = []
while node not in visited:
visited.add(node)
cycle.append(node)
node = successor[node]
if node in cycle:
return cycle[cycle.index(node):]
return None
def spanning_arborescence(arcs, sink):
arcs_by_head = defaultdict(list)
for arc in arcs:
if arc.tail == sink:
continue
arcs_by_head[arc.head].append(arc)
solution_arc_by_tail = {}
stack = arcs_by_head[sink]
while stack:
arc = stack.pop()
if arc.tail in solution_arc_by_tail:
continue
solution_arc_by_tail[arc.tail] = arc
stack.extend(arcs_by_head[arc.tail])
return solution_arc_by_tail
print(min_spanning_arborescence([Arc(1, 17, 0), Arc(2, 16, 0), Arc(3, 19, 0), Arc(4, 16, 0), Arc(5, 16, 0), Arc(6, 18, 0), Arc(2, 3, 1), Arc(3, 3, 1), Arc(4, 11, 1), Arc(5, 10, 1), Arc(6, 12, 1), Arc(1, 3, 2), Arc(3, 4, 2), Arc(4, 8, 2), Arc(5, 8, 2), Arc(6, 11, 2), Arc(1, 3, 3), Arc(2, 4, 3), Arc(4, 12, 3), Arc(5, 11, 3), Arc(6, 14, 3), Arc(1, 11, 4), Arc(2, 8, 4), Arc(3, 12, 4), Arc(5, 6, 4), Arc(6, 10, 4), Arc(1, 10, 5), Arc(2, 8, 5), Arc(3, 11, 5), Arc(4, 6, 5), Arc(6, 4, 5), Arc(1, 12, 6), Arc(2, 11, 6), Arc(3, 14, 6), Arc(4, 10, 6), Arc(5, 4, 6)], 0))
答案 1 :(得分：0)