20_Shortest_path.pptx

# No implementation and analysis the initialization

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Implementation and analysis The initialization requires (| V |) memory and run time We iterate | V | – 1 times, each time finding next closest vertex to the source Iterating through the table requires is (| V |) time Each time we find a vertex, we must check all of its neighbors With an adjacency matrix, the run time is (| V |(| V | + | V |)) = (| V | 2 ) With an adjacency list, the run time is (| V | 2 + | E |) = (| V | 2 ) as | E | = O(| V | 2 ) Can we do better? How about using a priority queue to find the closest vertex? Assume we are using a binary heap
Implementation and analysis The initialization still requires (| V |) memory and run time The priority queue will also requires (| V |) memory We must use an adjacency list, not an adjacency matrix We iterate | V | times, each time finding the closest vertex to the source Place the distances into a priority queue The size of the priority queue is (| V |) Thus, the work required for this is O(| V | ln(| V |)) Is this all the work that is necessary? Recall that each edge visited may result in a distance being updated Thus, the work required for this is O(| E | ln(| V |)) Thus, the total run time is O(| V | ln(| V |) + | E | ln(| V |)) = O(| E | ln(| V |))

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Implementation and analysis Here is an example of a worst-case scenario: Immediately, all of the vertices are placed into the queue Each time a vertex is visited, all the remaining vertices are checked, and in succession, each is pushed to the top of the binary heap
Implementation and analysis We could use a different heap structure: A Fibonacci heap is a node-based heap Pop is still O(ln(| V |)) , but inserting and moving a key is (1) Thus, because we are only calling pop | V | – 1 times, the overall run-time reduces to O(| E | + | V | ln(| V |))

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Implementation and analysis Thus, we have two run times when using A binary heap: O(| E | ln(| V |)) A Fibonacci heap: O(| E | + | V | ln(| V |)) Questions : Which is faster if | E | = (| V |) ? How about if | E | = (| V | 2 ) ?
Summary We have seen an algorithm for finding single-source shortest paths Start with the initial vertex Continue finding the next vertex that is closest Dijkstra’s algorithm always finds the next closest vertex It solves the problem in O(| E | + | V | ln(| V |)) time

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Triangle Inequality If the distances satisfy the triangle inequality, That is, the distance between a and b is less than the distance from a to c plus the distance from c to b , we can use the A* search which is faster than Dijkstra’s algorithm All Euclidean distances satisfy the triangle inequality
Negative Weights If some of the edges have negative weight, so long as there are no cycles with negative weight, the Bellman-Ford algorithm will find the minimum distance It is slower than Dijkstra’s algorithm

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Outline Definition and applications Dijkstra’s algorithm Floyd-Warshall algorithm
Background Dijkstra’s algorithm finds the shortest path between two nodes Run time: O(| E | ln(| V |)) If we wanted to find the shortest path between all pairs of nodes, we could apply Dijkstra’s algorithm to each vertex: Run time: O(| V | | E | ln(| V |))

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