This notebook was prepared by Donne Martin. Source and license info is on GitHub.

Solution Notebook

Problem: Find the shortest path between two nodes in a graph.

Constraints

  • Is the graph directed?
    • Yes
  • Is the graph weighted?
    • No
  • Can we assume we already have Graph and Node classes?
    • Yes
  • Are the inputs two Nodes?
    • Yes
  • Is the output a list of Node keys that make up the shortest path?
    • Yes
  • If there is no path, should we return None?
    • Yes
  • Can we assume this is a connected graph?
    • Yes
  • Can we assume the inputs are valid?
    • Yes
  • Can we assume this fits memory?
    • Yes

Test Cases

Input:

  • add_edge(source, destination, weight)
graph.add_edge(0, 1)
graph.add_edge(0, 4)
graph.add_edge(0, 5)
graph.add_edge(1, 3)
graph.add_edge(1, 4)
graph.add_edge(2, 1)
graph.add_edge(3, 2)
graph.add_edge(3, 4)

Result:

  • search_path(start=0, end=2) -> [0, 1, 3, 2]
  • search_path(start=0, end=0) -> [0]
  • search_path(start=4, end=5) -> None

Algorithm

To determine the shorted path in an unweighted graph, we can use breadth-first search keeping track of the previous nodes ids for each node. Previous nodes ids can be a dictionary of key: current node id and value: previous node id.

  • If the start node is the end node, return True
  • Add the start node to the queue and mark it as visited
    • Update the previous node ids, the previous node id of the start node is None
  • While the queue is not empty
    • Dequeue a node and visit it
    • If the node is the end node, return the previous nodes
    • Set the previous node to the current node
    • Iterate through each adjacent node
      • If the node has not been visited, add it to the queue and mark it as visited
        • Update the previous node ids
  • Return None

Walk the previous node ids backwards to get the path.

Complexity:

  • Time: O(V + E), where V = number of vertices and E = number of edges
  • Space: O(V + E)

Code

In [1]:
%run ../graph/graph.py
In [2]:
from collections import deque


class GraphShortestPath(Graph):

    def shortest_path(self, source_key, dest_key):
        if source_key is None or dest_key is None:
            return None
        if source_key is dest_key:
            return [source_key]
        prev_node_keys = self._shortest_path(source_key, dest_key)
        if prev_node_keys is None:
            return None
        else:
            path_ids = [dest_key]
            prev_node_key = prev_node_keys[dest_key]
            while prev_node_key is not None:
                path_ids.append(prev_node_key)
                prev_node_key = prev_node_keys[prev_node_key]
            return path_ids[::-1]

    def _shortest_path(self, source_key, dest_key):
        queue = deque()
        queue.append(self.nodes[source_key])
        prev_node_keys = {source_key: None}
        self.nodes[source_key].visit_state = State.visited
        while queue:
            node = queue.popleft()
            if node.key is dest_key:
                return prev_node_keys
            prev_node = node
            for adj_node in node.adj_nodes.values():
                if adj_node.visit_state == State.unvisited:
                    queue.append(adj_node)
                    prev_node_keys[adj_node.key] = prev_node.key
                    adj_node.visit_state = State.visited
        return None

Unit Test

In [3]:
%%writefile test_shortest_path.py
import unittest


class TestShortestPath(unittest.TestCase):

    def test_shortest_path(self):
        nodes = []
        graph = GraphShortestPath()
        for id in range(0, 6):
            nodes.append(graph.add_node(id))
        graph.add_edge(0, 1)
        graph.add_edge(0, 4)
        graph.add_edge(0, 5)
        graph.add_edge(1, 3)
        graph.add_edge(1, 4)
        graph.add_edge(2, 1)
        graph.add_edge(3, 2)
        graph.add_edge(3, 4)

        self.assertEqual(graph.shortest_path(nodes[0].key, nodes[2].key), [0, 1, 3, 2])
        self.assertEqual(graph.shortest_path(nodes[0].key, nodes[0].key), [0])
        self.assertEqual(graph.shortest_path(nodes[4].key, nodes[5].key), None)

        print('Success: test_shortest_path')


def main():
    test = TestShortestPath()
    test.test_shortest_path()


if __name__ == '__main__':
    main()
Overwriting test_shortest_path.py
In [4]:
%run -i test_shortest_path.py
Success: test_shortest_path