def test():
# Create an example graph
vertices = {'A', 'B', 'C', 'D', 'E'}
edges = {('A', 'B', 4), ('A', 'C', 2), ('B', 'C', 1), ('B', 'D', 3), ('C', 'E', 6), ('D', 'E', 1)}
graph = WeightedGraph(vertices, edges)
# What's the cheapest path from A to everywhere?
parents = graph.dijkstra('A')
for v in graph.vertices:
path = Graph.path(v, parents)
print(v, path, graph.path_weight(path))
def passThrough(verticies, edges, startVertex, endVertex, passByPoint):
graph = WeightedGraph(verticies, edges)
result = list()
parents1 = graph.dijkstra(startVertex)
path1 = Graph.path(passByPoint, parents1)
parents2 = graph.dijkstra(passByPoint)
path2 = Graph.path(endVertex, parents2)
if path1 is None or path2 is None:
return "The 3 verticies are not in the same compound, so it is not possible."
for anItem in path1:
result.append(anItem)
for i in range(1, len(path2)):
result.append(path2[i])
return " -> ".join(result)
def notPassThrough(verticies, edges, startVertex, endVertex, notPassByPoint):
edgesCopy = set()
result = list()
for aTuple in edges:
if notPassByPoint not in aTuple:
edgesCopy.add(aTuple)
graph = WeightedGraph(verticies, edgesCopy)
parents = graph.dijkstra(endVertex)
path = Graph.path(startVertex, parents)
for anItem in reversed(path):
result.append(anItem)
return " -> ".join(result)
for aWord in words:
vertices.add(aWord)
for i in range(len(aWord)):
newWord = aWord[0:i] + "*" + aWord[i + 1:]
if newWord not in specialDict:
specialDict[newWord] = [aWord]
else:
for aConnection in specialDict[newWord]:
edges.add((aWord, aConnection))
specialDict[newWord].append(aWord)
# Construct the graph
graph = Graph(vertices, edges)
# Find word ladders
while True:
source = input("Source word: ")
if source not in vertices:
print("Unknown word:", source, "\n")
continue
target = input("Target word: ")
if target not in vertices:
print("Unknown word:", target, "\n")
continue
parents = graph.bfs(source)
ladder = Graph.path(target, parents)
print("Shortest ladder:", ladder, "\n")
