Spanning Trees

A Spanning Tree is a tree created from a graph that contains all of the nodes of the graph, but no cycles. In other words, you can create a minimum spanning tree from a graph, by removing one edge from any path that forms a cycle.

In the end, this will leave you with another graph, that is also a tree. If this graph contains N nodes, it will contain N-1 edges.

It contains at least N-1 edges, because this many are required to attach each node to the graph. Think of it this way. If the graph contains 1 node, there is nothing to attach it to -- as a result, no edges are needed. If we add nodes to the graph one at a time, we will need to use an edge to attach the new node to the graph. If we add a node, without adding an edge, we've really created a new graph with only one node -- we haven't expanded the old one, because we haven't established a relationship between the new node and any node in the old graph.

It contains at most N-1 edges, because more than one edge per node will create cycles, which would violate the definition of a tree. Imagine that you have a graph, any graph, with the minimum number, N-1, edges. Draw out your imaginary graph.

Now, try to add another edge to your graph, without creating a cycle. You can't do it. The reason is that each node, is already connected to another node of the graph. Adding an additional edge will connected it to a second point on the same graph. As a result, the graph will have a cycle -- it is no longer a tree.

Creating Spanning trees: Traversing Graphs

Today we are going to talk about two different ways of creating minimum spanning trees: the Depth-First Search and the Breadth-First Search. These algorithms are going to be very similar to the algorithms we studied for traversing trees.

The big difference is that we are going to mark the edges as we visit them, and avoid revisiting marked nodes. This is necessary, because graphs may have cycles -- and we don't want to go around-and-around ground we have already covered. So, if we leave a trail of breadcrumbs, and avoid recovering the ground we've already walked.

We can construct a spanning tree by keeping track of the path we follow to reach each node. In other words, we can start out with an empty graph, in addition to the one that we want to traverse. Then, instead of, or in addition to, printing each node we reach during the traversal, we can also add it to the new graph, by creating the same edge that we followed in the traversal to reach it.

So, if we get to Node-X from Node-Y, we then add an edge to Node-X from Node-Y in our new graph. Then, in the end, the new graph has the same nodes as the old graph -- but only one edge connecting each pair.

Depth First Search

Depth First Search (DFS) is a generalization of the preorder traversal. Starting at some arbitrarily chosen vertex v, we mark v so that we know we've visited it, process v, and then recursively traverse all unmarked vertices adjacent to v (v will be a different vertex with every new method call).

When we visit a vertex in which all of its neighbors have been visited, we return to its calling vertex, and visit one of its unvisited neighbors, repeating the recursion in the same manner. We continue until we have visited all of the starting vertex's neighbors, which means that we're done. The recursion (stack) guides us through the graph.

Performing a DFS on the following graph:

1. Any vertex can be the starting vertex. We choose to visit 1 first. (push 1)
2. From 1, we can go on to 0, 2, or 3.
3. We visit 0. (push 0) [0]
4. From 0, we can go on to 4 or 5 (we've already been to 1). We visit 4. (push 4) [4 0]
5. From 4, we can go on to 3 or 5 (we've already been to 0). We visit 3. (push 3) [3 4 0]
6. From 3, we can go on to 2 or 4 (we've already been to 1). We visit 2. (push 2) [2 3 4 0]
   
   
7. Now we've gone as far as we can go (from 2, we've already visited both 1 and 3). We can start returning.
   
   
8. We return to 3. (pop 2) [3 4 0]
9. We've already been to 1, 2, and 4, so we return.
10. We return to 4. (pop 3) [4 0]
    
    
11. We've already been to 0 and 3, but not to 5. We visit 5. (push 5) [5 4 0] (pop 5, pop 4, pop 0) []
    
    
12. We're done.

Pseudocode for DFS

public void depthFirstSearch(Vertex v)
{
	v.visited = true;
        // print the node or add it to the new spanning tree here
	for(each vertex w adjacent to v)
		if(!w.visited)
			depthFirstSearch(w);		
}

Breadth First Search

Breadth First Search (BFS) searches the graph one level (one edge away from the starting vertex) at a time. In this respect, it is very similar to the level order traversal that we discussed for trees.

Starting at some arbitrarily chosen vertex v, we mark v so that we know we've visited it, process v, and then visit and process all of v's neighbors.

Now that we've visited and processed all of v's neighbors, we need to visit and process all of v's neighbors neighbors. So we go to the first neighbor we visited and visit all of its neighbors, then the second neighbor we visited, and so on. We continue this process until we've visited all vertices in the graph. We don't use recursion in a BFS because we don't want to traverse recursively. We want to traverse one level at a time.

So imagine that you visit a vertex v, and then you visit all of v's neighbors w. Now you need to visit each w's neighbors. How are you going to remember all of your w's so that you can go back and visit their neighbors? You're already marked and processed all of the w's. How are you going to find each w's neighbors if you don't remember where the w's are? After all, you're not using recursion, so there's no stack to keep track of them.

To perform a BFS, we use a queue. Every time we visit vertex w's neighbors, we dequeue w and enqueue w's neighbors. In this way, we can keep track of which neighbors belong to which vertex. This is the same technique that we saw for the level-order traversal of a tree. The only new trick is that we need to makr the verticies, so we don't visit them more than once -- and this isn't even new, since this technique was used for the blobs problem during our discussion of recursion.

Performing a BFS on the same graph:

1. We choose to start by visiting 1. (enqueue 1) [1]
   
   
2. We visit 0, 2, and 3 because they are all one step away. (dequeue 1, enqueue 0, enqueue 2, enqueue 3) [0 2 3]
   
   
3. Because we visited 0 first, we go back to 0 and visit its neighbors, 4 and 5. (dequeue 0, enqueue 4, enqueue 5) [2 3 4 5] 2 and 3 have no unvisited neighbors. (dequeue 2, dequeue 3, dequeue 4, dequeue 5) [ ]
   
   
4. We're done.

Pseudocode

public void breadthFirstSearch(vertex v)
{
	Queue q = new Queue();
	
	v.visited = true;
	q.enQueue(v);
	
	while( !q.isEmpty() )
	{
  		Vertex w = (Vertex)q.deQueue();
                // Pritn the node or add it to the spanning tree here.
  		
		for(each vertex x adjacent to w)
		{			
			if( !x.visited )
			{
				x.visited = true;
				q.enQueue(x);
			}
		}
	}
}

Finding a Minimum Spanning Tree of an Undirected, Connected Graph

A minimum spanning tree of an undirected graph is a tree formed from that graph's edges that connects all the vertices of that graph at the lowest total cost. You can make a spanning tree of a graph only if the graph is connected. There may be more than one spanning tree of a particular graph.

The number of edges in a minimum spanning tree of a graph will be the number of vertices it has - 1. A minimum spanning tree is a tree because it's acyclic. It's spanning because it reaches every vertex in the graph, and it's minimum for the obvious reason. If we need to wire a house with a minimum of cable, then a we need to find a minimum spanning tree of a graph of the electrical layout of the house.

Minimum Spanning Trees: Why Do We Care?

Many real situations can be modeled with graphs. And, many real situations can be solved by finding the minimum spanning tree of graphs.

My favorite example involves an electrician and a house. Imagine that a collection of electrical outlets have been installed in the walls of a house, and that the electicity enters the house at a single electrical box. We can model this situation as a graph, where each of the outlets and the electrical box is a node, and the walls are the edges. The length of each wall, or segment thereof, is the distance along the wall between two of the electrical connections.

As a result, the electrician may have many different routes he can use to wire the outlets -- they may be reachable by different paths along the walls. So, the electrician wants to find the path that requires the least amount of wire. This saves money, becuase less wire is needed. And, it saves time, because the runs along the wal are shorter and, as a consequence, take less time to install.

To solve this problem, the electrician can model the outlets and wall segments conecting them as a graph rooted at the electrical box. Then, the electrician can find the minimum spanning tree of the graph. The edges in this tree give the paths that the electrician should use to run the wires -- they will reach each node, while requiring the least amount of wire.

Greedy Algorithms

Today we are going to study one algorithm for finding the minimum spanning tree of a graph. It is known as Kruskal's Algorithm. Kruskal's algorithm is an example of a Greedy Algorithm. Greedy Algorithms operate by breaking a decision making process down into small steps, and making the best decision at each step. For many problems, this approach will lead to the best possible overall solution -- for others, it will not.

For example, we "make change", by using a greedy algorithm. We hand back $10 bills, until handing back another $10 would be giving back too much money. Then we hand out $5 bills, then $1 bills, then quarters, then dimes, then nickels, then pennies. In the end, we are guaranteed that we have returned the correct amount of change -- with the fewest possible bills or coins.

But, this algorithm would not work, for example, if we had 12-cent coins. Normall, if we have to return 21-cents of change, we return dime-dime-penny. But, with a 12-cent coin, let's call it the "dozen", we'd return dozen-nickel-penny-penny-penny-penny. From this example, we can see that greedy algorithms are appropriate for some problems -- but not all problems.

But, Kruskal's Algorithm, is a greedy algorithm -- and does actually work. Let's take a look at a different example, the sidewalk example, and see how.

Kruskal's Algorithm

The basic algorithm is very straight-forward. We simple attempt to add each edge from the original graph to the minimum spanning tree, beginning with the lowest-weight edge and finishing with the greatest-weight edge. We add the edge if it doesn't cause a cycle and passs it up, if it does. We continue to add edges, until we've added N-1 edges, where N is the number of verticies. Remember, spanning trees have exactly N-1 edges -- never more, never less.

In order to make it easy to select the candidates in the right order, from the lowest weight to the highest weight, we store the edges in a priority queue (heap). Then, selecting an edge is simply the deleteMin() operation.

Another way of viewing the algorithm is to views the inital configuration as a forrest of trees, with each vertex in its own, independent tree. If we take this view, then adding an edge merges two trees into one. When Kruskal's terminates, there is only one tree - the minimum spanning tree.

Let's take a look at the algorithm in operation, using the smae graph as we used last class:

The following table shows the verticies sorted by weight and whetehr or not each vertex was accepted. Remember, we evaluate each vertex, one-at-a-time, from the top of this list down. In a real implemention, we would have added them to a heap, and be using deleteMin() to get to the top one for each iteration.

Edge	Weight	Action
(1,4)	1	Accepted
(6,7)	1	Accepted
(1,2)	2	Accepted
(3,4)	2	Accepted
(2,4)	3	Rejected
(1,3)	4	Rejected
(4,7)	4	Accepted
(3,6)	5	Rejected
(5,7)	6	Accepted

Using Sets to Detect Cycles

So, how can we figure out if adding a particular edge to a tree will create a cycle? This is very important to Kruskal's Algorithm, because we only add an edge, if doing so won't create a cycle.

Imagine that each vertex in a graph is its own set (remember the Set class you created for the lab?).

If you connect two vertices in the same set together, you'll create a cycle. As long as you connect vertices from two different sets, you won't create a cycle. A and B are in different sets. I'll connect them.

Now AB is a set. Can I connect C to AB? C and AB are in different sets, I'll connect them. Now I have a set called ABC.

Can I connect C to B? C and B are in the same set, so connecting them will create a cycle.

Union-Find

Earlier this semester, we created a Set class using LinkedLists. This class supported a very broad array of Set operations. But, for this particular problem, we only need to do two things: look in a set to see if an item is there (find) and unite two sets (union). And, since the Set lab, we've learned about trees, which added a powerful tool to our kit. Let's see how we can use trees to represent sets in a way that leads to efficient Union and Find operations.

Again, imagine a graph, with each vertex as its own set. Now imagine that each set of vertices is a tree. So before we connect any edges, each vertex is its own tree, and the graph is a forest of trees.

We'll use an array to represent the trees. Create an array, with each index of the array representing the corresponding vertex of the graph. Place a sentinel value, -1, into each array element. We will use this sentinel value to denote the root of the tree. Before we connect any edges, each vertex is its own tree, so its the root. -1 represents the root of the tree set.

  0    1    2    3    4    5    6    7
  [-1] [-1] [-1] [-1] [-1] [-1] [-1] [-1]
  

Now I decide to connect 7 to 1. They are in different sets, so I connect them.

  0    1    2    3    4    5    6    7
  [-1] [-1] [-1] [-1] [-1] [-1] [-1] [ 1]
  

The parent of 7 is now 1, and 7 is no longer the root of the tree. If we want to find 7's parent, we simply look at its value, which is 1. To find 1's parent, we look at its value, which is -1, indicating that 1 has no parent and is the root of the tree.

Now I decide to connect 2 to 1. They are in different sets, so I connect them.

  0    1    2    3    4    5    6    7
  [-1] [-1] [ 1] [-1] [-1] [-1] [-1] [ 1]
  

Now I decide to connect 0 to 7. They are in different sets, so I connect them.

  0    1    2    3    4    5    6    7
  [ 7] [-1] [ 1] [-1] [-1] [-1] [-1] [ 1]
  

I want to connect 4 to 6. They are in different sets, so I connect them.

  0    1    2    3    4    5    6    7
  [ 7] [-1] [ 1] [-1] [ 6] [-1] [-1] [ 1]
  

I want to connect 5 to 6. They are in different sets, so I connect them.

  0    1    2    3    4    5    6    7
  [ 7] [-1] [ 1] [-1] [ 6] [ 6] [-1] [ 1]
  

Note that connect 4 to 5 would now create a cycle.

I want to connect 1 to 4. They are in different sets, so I connect them.

  0    1    2    3    4    5    6    7
  [ 7] [ 4] [ 1] [-1] [ 6] [ 6] [-1] [ 1]
  

How do you find out what set a particular vertex of the graph belongs to? You simply follow its ancestors up until you find an array element with a value of -1. At this point, all vertices except 3 are in the same set -- 6. 6 is the only element with a value of -1. Vertex 6 is the root of the tree consisting of all of the vertices in the graph.

The operation that connects to vertices by changing the array element value of one to be the other is called union. The operation that finds the root of a particular vertex's tree is called find.