• is it connected?
• is there a path from node j to node k?
• does it contain a cycle?
• what nodes are reachable from node j?
• can the nodes be ordered so that for every node j, j comes before all of its successors in the ordering?

The basic idea of a depth-first search is to start at some node n, and then to follow an edge out of n, then another edge out, etc, getting as far away from n as possible before visiting any more of n's neighbors. To prevent infinite loops in graphs with cycles, we must keep track of which nodes have been visited. Here is the basic algorithm for a depth-first seach from node n:

1. mark n "visited"
2. recursively do a depth-first search from each of n's unvisited successors

static void dfs (node n) {
   n.visited = true;
   Sequence S = n.successors;
   for (S.start(); S.isCurrent(); S.advance()) {
      node m = S.getCurrent();
      if (! m.visited) dfs(m);
   }
}

Note that in the example illustrated above, the order in which the nodes are visited is: 0, 2, 3, 1, 4. Another possible order (if node 4 were the first successor of node 0) is: 0, 4, 2, 3, 1.

To analyze the time required for depth-first search, note that one call is made to dfs for each node that is reachable from the start node. Each call looks at all successors of the current node, so the time is O(# reachable nodes + total # of outgoing edges from those nodes). In the worst case, this is all nodes and all edges, so the worst-case time is O(N + E).

1. is it connected?
2. is there a path from node j to node k?
3. does it contain a cycle?
4. what nodes are reachable from node j?
5. can the nodes be ordered so that for every node j, j comes before all of its successors in the ordering?

Path Detection

• When the graph represents airline routes, and we want to ask "Can I fly from Madison to London (maybe w/ some connections)?", or
• when the graph represent CS course prerequisites, and we want to ask "Is CS 367 a (transitive) prerequisite for CS 640?"

• step 1: mark all nodes "not visited"
• step 2: dfs(j)
• step 3: there is a path from j to k iff k is marked "visited"

Cycle Detection

1. does a graph contain a cycle?
2. is there a cyclic path starting from node j?

Consider the example given above to illustrate depth-first search. There is a cycle in that graph starting from node 0. Is there something that happens during the depth-first search that indicates the presence of that cycle?? Note that during dfs(1), 0 is a successor of 1, but is already visited. But that isn't quite enough to say that there's a cycle, because during dfs(3), node 4 is a successor of 3 that has already been visited, but there is no cycle starting from node 4.

What's the difference? The answer is that when node 0 is considered as a successor of node 1, the call dfs(0) is still "active" (i.e., its activation record is still on the stack); however, when node 4 is considered as a successor of node 3, the call dfs(4) has already finished. How can we tell the difference?? The answer is to keep track of when a node is "inProgress" (as well as whether it has been visited or not). We can do this by using a "mark" field with three possible values:

1. unvisited
2. inProgress
3. done

Here's the code for cycle detection:

static boolean hasCycle(node n) {
   n.mark = inProgress;
   Sequence S = n.successors;
   for (S.start(); S.isCurrent(); S.advance()) {
      node m = S.getCurrent();
      if (m.mark == inProgress) return true;
      if (m.mark != done) {
         if (hasCycle(m)) return true;
      }
   }
   n.mark = done;
   return false;
}

static boolean graphHasCycle(Graph G) {
    mark all nodes unvisited;
    for each node k in the graph {
       if (node k is marked unvisited) {
          if (hasCycle(k)) return true;
       }
    }
    return false;
}

Topological Numbering

Topological numbering addresses exactly this problem. The goal is to assign numbers to nodes so that for every edge j -> k, the number assigned to j is less than the number assigned to k. A topolgoical numbering of the prerequisites graph would tell you one legal order in which to take the CS courses. For example:

To find a topological numbering, we use a variation of depth-first search. The intuition is as follows:

• As long as there are no cycles in the graph, there must be at least one node with no outgoing edges.
• The last number (N) can be given to any such node (310, 577, or 640 in our example).
• Once all of a node's successors have numbers, the node itself can get the next smallest number.

static int topNum (node n, int num) {
    n.mark = inProgress;
    Sequence S = n.successors;
    for (S.start(); S.isCurrent(); S.advance()) {
       node m = S.getCurrent();
       if (m.mark == inProgress) {
           // no topological ordering for a cyclic graph!
           throw new CycleException();
       }
       if (k.mark != done) num = topNum(k, num);
    }
    // here when n has no more successors
    j.mark = done;
    j.number = num;
    return num-1;
}

Breadth-first search uses a queue rather than recursion (which actually uses a stack); the queue holds "nodes to be visited". If the graph is a tree, breadth-first search gives you a level-order traversal. Here's the pseudo code:

static void bfs (node n) {
  Queue Q = new Queue();
  node  current;

  n.visited = true;
  Q.enqueue(n);
  while (! Q.isEmpty())){
     current = Q.dequeue();
     Sequence S = current.successors;
     for (S.start(); S.isCurrent(); S.advance()) {
        node k = S.getCurrent();
        if (! k.visited){
            k.visited = true;
            Q.enqueue (k);
        } // end if k not visited
     } // end for every successor k
  } // end while Q not empty
}

The order in which nodes are "visited" as a result of bfs(0) is:

As with depth-first search, all nodes marked "visited" are reachable from the start node, but nodes are visited in a different order then they would be using depth-first search.

We can use a variation of bfs to find the shortest distance (the length of the shortest path) to each reachable node:

• keep an array of distances; fill in distance[n] = 0
• when a node k is about to be enqueued, set distance[k] = distance[current] + 1

This technique only works in unweighted graphs (i.e., in graphs in which all edges are assumed to have length 1). An interesting problem is how to find shortest paths in a weighted graph; i.e., given a "start" node n, to find, for each other node m, the path from n to m for which the sum of the weights on the edges is minimal (assuming that no edge has a negative weight). For example, in the following graph, nodes represent cities, edges represent highways, and the weights on the edges represent distances (the length of the highway between the two cities). Breadth-first search can only tell you which route from Madison to Green Bay goes through the fewest other cities; it cannot tell you which route is the shortest.

A clever algorithm that can be used to solve this problem (to find shortest paths in a weighted graph with non-negative edge weights) has been defined by Edsgar Dijkstra (and so is called "Dijkstra's algorithm"). The worst-case running time of the algorithm is O(E log N), assuming that edges are represented using adjacency lists, not an adjacency matrix, where E is the number of edges and N is the number of nodes. You can find a description of the algorithm in most data structures or algorithms textbooks; you are not responsible for understanding it for this class.

• A graph a set of nodes and a set of edges.
• There are two kinds of graphs: directed and undirected.
• Low-level operations include:
  • add a node
  • add an edge
  • remove a node
  • remove an edge
  • determine whether there is an edge j->k
  • find all successors / predecessors / neighbors of node j
• High-level operations include:
  • depth-first search, which can be done on the entire graph (e.g., to find cycles or to produce a topological ordering), or on part of a graph (e.g., to determine which nodes are reachable from a given node)
  • breadth-first search, which can also be used to determine reachability, and can be used to find shortest paths in unweighted graphs
  • Dijkstra's algorithm, which finds shortest paths in weighted graphs