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19.5. Shortest-Paths Problems

19.5.1. Shortest-Paths Problems

On a road map, a road connecting two towns is typically labeled with its distance. We can model a road network as a directed graph whose edges are labeled with real numbers. These numbers represent the distance (or other cost metric, such as travel time) between two vertices. These labels may be called weights, costs, or distances, depending on the application. Given such a graph, a typical problem is to find the total length of the shortest path between two specified vertices. This is not a trivial problem, because the shortest path may not be along the edge (if any) connecting two vertices, but rather may be along a path involving one or more intermediate vertices.

Todo

type:Slideshow

Incorporate the following paragraph into a slideshow with the figure below it.

For example, in Figure 19.5.1, the cost of the path from \(A\) to \(B\) to \(D\) is 15. The cost of the edge directly from \(A\) to \(D\) is 20. The cost of the path from \(A\) to \(C\) to \(B\) to \(D\) is 10. Thus, the shortest path from \(A\) to \(D\) is 10 (rather than along the edge connecting \(A\) to \(D\)). We use the notation \(\mathbf{d}(A, D) = 10\) to indicate that the shortest distance from \(A\) to \(D\) is 10. In Figure 19.5.1, there is no path from \(E\) to \(B\), so we set \(\mathbf{d}(E, B) = \infty\). We define \(\mathbf{w}(A, D) = 20\) to be the weight of edge \((A, D)\), that is, the weight of the direct connection from \(A\) to \(D\). Because there is no edge from \(E\) to \(B\), \(\mathbf{w}(E, B) = \infty\). Note that \(\mathbf{w}(D, A) = \infty\) because the graph of Figure 19.5.1 is directed. We assume that all weights are positive.

Figure 19.5.1: Example graph for shortest-path definitions.

19.5.1.1. Single-Source Shortest Paths

We will now present an algorithm to solve the single-source shortest paths problem. Given Vertex \(S\) in Graph \(\mathbf{G}\), find a shortest path from \(S\) to every other vertex in \(\mathbf{G}\). We might want only the shortest path between two vertices, \(S\) and \(T\). However in the worst case, finding the shortest path from \(S\) to \(T\) requires us to find the shortest paths from \(S\) to every other vertex as well. So there is no better algorithm (in the worst case) for finding the shortest path to a single vertex than to find shortest paths to all vertices. The algorithm described here will only compute the distance to every such vertex, rather than recording the actual path. Recording the path requires only simple modifications to the algorithm.

Computer networks provide an application for the single-source shortest-paths problem. The goal is to find the cheapest way for one computer to broadcast a message to all other computers on the network. The network can be modeled by a graph with edge weights indicating time or cost to send a message to a neighboring computer.

For unweighted graphs (or whenever all edges have the same cost), the single-source shortest paths can be found using a simple breadth-first search. When weights are added, BFS will not give the correct answer.

Todo

type:Slideshow

Provide a slideshow to demonstrate the following example.

One approach to solving this problem when the edges have differing weights might be to process the vertices in a fixed order. Label the vertices \(v_0\) to \(v_{n-1}\), with \(S = v_0\). When processing Vertex \(v_1\), we take the edge connecting \(v_0\) and \(v_1\). When processing \(v_2\), we consider the shortest distance from \(v_0\) to \(v_2\) and compare that to the shortest distance from \(v_0\) to \(v_1\) to \(v_2\). When processing Vertex \(v_i\), we consider the shortest path for Vertices \(v_0\) through \(v_{i-1}\) that have already been processed. Unfortunately, the true shortest path to \(v_i\) might go through Vertex \(v_j\) for \(j > i\). Such a path will not be considered by this algorithm. However, the problem would not occur if we process the vertices in order of distance from \(S\). Assume that we have processed in order of distance from \(S\) to the first \(i-1\) vertices that are closest to \(S\); call this set of vertices \(\mathbf{S}\). We are now about to process the \(i\) th closest vertex; call it \(X\).

A shortest path from \(S\) to \(X\) must have its next-to-last vertex in \(S\). Thus,

\[\mathbf{d}(S, X) = \min_{U \in \mathbf{S}}(\mathbf{d}(S, U) + \mathbf{w}(U, X)).\]

In other words, the shortest path from \(S\) to \(X\) is the minimum over all paths that go from \(S\) to \(U\), then have an edge from \(U\) to \(X\), where \(U\) is some vertex in \(\mathbf{S}\).

This solution is usually referred to as Dijkstra's algorithm. It works by maintaining a distance estimate \(\mathbf{D}(X)\) for all vertices \(X\) in \(\mathbf{V}\). The elements of \(\mathbf{D}\) are initialized to the value INFINITE. Vertices are processed in order of distance from \(S\). Whenever a vertex \(v\) is processed, \(\mathbf{D}(X)\) is updated for every neighbor \(X\) of \(V\). Here is an implementation for Dijkstra's algorithm. At the end, array D will contain the shortest distance values.

// Compute shortest path distances from s, store them in D
static void Dijkstra(Graph G, int s, int[] D) {
  for (int i=0; i<G.nodeCount(); i++)    // Initialize
    D[i] = INFINITY;
  D[s] = 0;
  for (int i=0; i<G.nodeCount(); i++) {  // Process the vertices
    int v = minVertex(G, D);     // Find next-closest vertex
    G.setValue(v, VISITED);
    if (D[v] == INFINITY) return; // Unreachable
    int[] nList = G.neighbors(v);
    for (int j=0; j<nList.length; j++) {
      int w = nList[j];
      if (D[w] > (D[v] + G.weight(v, w)))
        D[w] = D[v] + G.weight(v, w);
    }
  }
}
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Todo

type:AV

Provide an AV that runs on a random graph. An initial version is in AV/Development/TopSort/dijkstraAV.*.

There are two reasonable solutions to the key issue of finding the unvisited vertex with minimum distance value during each pass through the main for loop. The first method is simply to scan through the list of \(|\mathbf{V}|\) vertices searching for the minimum value, as follows:

// Find the unvisited vertex with the smalled distance
static int minVertex(Graph G, int[] D) {
  int v = 0;  // Initialize v to any unvisited vertex;
  for (int i=0; i<G.nodeCount(); i++)
    if (G.getValue(i) != VISITED) { v = i; break; }
  for (int i=0; i<G.nodeCount(); i++)  // Now find smallest value
    if ((G.getValue(i) != VISITED) && (D[i] < D[v]))
      v = i;
  return v;
}

Todo

type:Code

Why does the code look for an unvisited value first? Is there an easier way?

Because this scan is done \(|\mathbf{V}|\) times, and because each edge requires a constant-time update to D, the total cost for this approach is \(\Theta(|\mathbf{V}|^2 + |\mathbf{E}|) = \Theta(|\mathbf{V}|^2)\), because \(|\mathbf{E}|\) is in \(O(|\mathbf{V}|^2)\).

Todo

type:AV

AV here to demonstrate the minVertex implementation.

An alternative approach is to store unprocessed vertices in a min-heap ordered by their distance from the processed vertices. The next-closest vertex can be found in the heap in \(\Theta(\log |\mathbf{V}|)\) time. Every time we modify \(\mathbf{D}(X)\), we could reorder \(X\) in the heap by deleting and reinserting it. This is an example of a priority queue with priority update. To implement true priority updating, we would need to store with each vertex its position within the heap so that we can remove its old distances whenever it is updated by processing new edges. A simpler approach is to add the new (always smaller) distance value for a given vertex as a new record in the heap. The smallest value for a given vertex currently in the heap will be found first, and greater distance values found later will be ignored because the vertex will already be marked as VISITED. The only disadvantage to repeatedly inserting distance values in this way is that it will raise the number of elements in the heap from \(\Theta(|\mathbf{V}|)\) to \(\Theta(|\mathbf{E}|)\) in the worst case. But in practice this only adds a slight increase to the depth of the heap. The time complexity is \(\Theta((|\mathbf{V}| + |\mathbf{E}|) \log |\mathbf{E}|)\), because for each edge that we process we must reorder the heap. We use the KVPair class to store key-value pairs in the heap, with the edge weight as the key and the target vertex as the value. here is the implementation for Dijkstra's algorithm using a heap.

// Dijkstra's shortest-paths: priority queue version
static void DijkstraPQ(Graph G, int s, int[] D) {
  int v;                                 // The current vertex
  KVPair[] E = new KVPair[G.edgeCount()];        // Heap for edges
  E[0] = new KVPair(0, s);               // Initial vertex
  MinHeap H = new MinHeap(E, 1, G.edgeCount());
  for (int i=0; i<G.nodeCount(); i++)            // Initialize distance
    D[i] = INFINITY;
  D[s] = 0;
  for (int i=0; i<G.nodeCount(); i++) {          // For each vertex
    do { KVPair temp = (KVPair)(H.removemin());
         if (temp == null) return;       // Unreachable nodes exist
         v = (Integer)temp.value(); } // Get position
      while (G.getValue(v) == VISITED);
    G.setValue(v, VISITED);
    if (D[v] == INFINITY) return;        // Unreachable
    int[] nList = G.neighbors(v);
    for (int j=0; j<nList.length; j++) {
      int w = nList[j];
      if (D[w] > (D[v] + G.weight(v, w))) { // Update D
        D[w] = D[v] + G.weight(v, w);
        H.insert(new KVPair(D[w], w));
      }
    }
  }
}

Todo

type:Slideshow

This slideshow illustrates Dijkstra's algorithm using the heap. The start vertex is A. All vertices except A have an initial value of \(\infty\). After processing Vertex A, its neighbors have their D estimates updated to be the direct distance from A. After processing C (the closest vertex to A), Vertices B and E are updated to reflect the shortest path through C. The remaining vertices are processed in order B, D, and E. Changes in the D array should be shown along with this.

Using MinVertex to scan the vertex list for the minimum value is more efficient when the graph is dense, that is, when \(|\mathbf{E}|\) approaches \(|\mathbf{V}|^2\). Using a heap is more efficient when the graph is sparse because its cost is \(\Theta((|\mathbf{V}| + |\mathbf{E}|) \log |\mathbf{E}|)\). However, when the graph is dense, this cost can become as great as \(\Theta(|\mathbf{V}|^2 \log |\mathbf{E}|) = \Theta(|V|^2 \log |V|)\).

Todo

type:Slideshow

Slideshow to demonstrate the relative costs of the two algorithms.

Now you can practice using Dijkstra's algorithm.

Todo

type:Exercise

Summary battery of questions for Dijkstra's algorithm

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