CODEWITHSUNDEEP
python-3.xalgorithmgraph-algorithmshortest-path
Stoner
Stoner

Reputation: 906

(Algorithms) Finding the shortest path that passes through a required set of nodes (possibly with BFS) and returns to the origin in Python

I am trying to find a shortest path that passes through a set of nodes [4,7,9] (order does not need to be preserved) and then returns to the origin (node 1). I have the set of edges:

E = [(1, 10), (1, 11), (2, 3), (2, 10), (3, 2), (3, 12), (4, 5), (4, 12), (5, 4), (5, 14), (6, 7), (6, 11), (7, 6), (7, 13), (8, 9), (8, 13), (9, 8), (9, 15), (10, 1), (10, 11), (10, 2), (11, 1), (11, 10), (11, 6), (12, 13), (12, 3), (12, 4), (13, 12), (13, 7), (13, 8), (14, 15), (14, 5), (15, 14), (15, 9)]

and I tried adapting the answer at How can I use BFS to get a path containing some given nodes in order? but yielded the error:

Traceback (most recent call last):
  File "C:/Users/../rough-work.py", line 41, in <module>
    graph[edge[0]].link(graph[edge[-1]])
KeyError: 15

My adapted code is as follows:

class Node:
    def __init__(self, name):
        self.name = name
        self.neighbors = []

    def link(self, node): 
        # The edge is undirected: implement it as two directed edges
        self.neighbors.append(node)
        node.neighbors.append(self)

    def shortestPathTo(self, target):
        # A BFS implementation which retains the paths
        queue = [[self]]
        visited = set()
        while len(queue):
            path = queue.pop(0) # Get next path from queue (FIFO)
            node = path[-1] # Get last node in that path
            for neighbor in node.neighbors:
                if neighbor == target:
                    # Found the target node. Return the path to it
                    return path + [target]
                # Avoid visiting a node that was already visited
                if not neighbor in visited:
                    visited.add(neighbor)
                    queue.append(path + [neighbor])

###
n = 15                    
nodes = list(range(1,n))
E = [(1, 10), (1, 11), (2, 3), (2, 10), (3, 2), (3, 12), (4, 5), (4, 12), (5, 4), (5, 14), (6, 7), (6, 11), (7, 6), (7, 13), (8, 9), (8, 13), (9, 8), (9, 15), (10, 1), (10, 11), (10, 2), (11, 1), (11, 10), (11, 6), (12, 13), (12, 3), (12, 4), (13, 12), (13, 7), (13, 8), (14, 15), (14, 5), (15, 14), (15, 9)]

# Create the nodes of the graph (indexed by their names)
graph = {}
for letter in nodes:
    graph[letter] = Node(letter)

print(graph)

# Create the undirected edges
for edge in E:
    graph[edge[0]].link(graph[edge[-1]])

# Concatenate the shortest paths between each of the required node pairs 
start = 1
path = [graph[1]]
for end in [4,7,9,1]:
    path.extend( graph[start].shortestPathTo(graph[end])[1:] )
    start = end

# Print result: the names of the nodes on the path
print([node.name for node in path])

What could possibly be the problem with the code? I will like to extend the graph to a arbitrarily large number of nodes, greater than 26 - the number of alphabets (as I infer that the previous implementation was only for character-based nodes). Or, if there is a more straightforward way in doing this that will be great!

Thanks and some help will be deeply appreciated!

Upvotes: 0

Views: 467

Answers (1)

al-dev
al-dev

Reputation: 256

The KeyError: 15 and your line print(graph) should have given you the clue: the latter shows that your graph dictionary contains only 14 entries, whereas your edges in E clearly make reference to 15 separate indices. Change n = 15 to n = 16 and it works:

[1, 10, 2, 3, 12, 4, 12, 13, 7, 13, 8, 9, 8, 13, 7, 6, 11, 1]

Remember that:

>>> len(list(range(1,16)))
15

Upvotes: 1

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