When you surf the web, send an email, or log in to a laboratory computer from another location on campus a lot of work is going on behind the scenes to get the information on your computer transferred to another computer. The in-depth study of how information flows from one computer to another over the Internet is the primary topic for a class in computer networking. However, we will talk about how the Internet works just enough to understand another very important graph algorithm.
Figure 1 shows you a high-level overview of how communication on the Internet works. When you use your browser to request a web page from a server, the request must travel over your local area network and out onto the Internet through a router. The request travels over the Internet and eventually arrives at a router for the local area network where the server is located. The web page you requested then travels back through the same routers to get to your browser. Inside the cloud labelled “Internet” in Figure 1 are additional routers. The job of all of these routers is to work together to get your information from place to place. You can see there are many routers for yourself if your computer supports the traceroute command. The text below shows the output of the traceroute command which illustrates that there are 13 routers between the web server at Luther College and the mail server at the University of Minnesota.
1 188.8.131.52 2 hilda.luther.edu (184.108.40.206) 3 ICN-Luther-Ether.icn.state.ia.us (220.127.116.11) 4 ICN-ISP-1.icn.state.ia.us (18.104.22.168) 5 p3-0.hsa1.chi1.bbnplanet.net (22.214.171.124) 6 ae-1-54.bbr2.Chicago1.Level3.net (126.96.36.199) 7 so-3-0-0.mpls2.Minneapolis1.Level3.net (188.8.131.52) 8 ge-3-0.hsa2.Minneapolis1.Level3.net (184.108.40.206) 9 p1-0.minnesota.bbnplanet.net (220.127.116.11) 10 TelecomB-BR-01-V4002.ggnet.umn.edu (18.104.22.168) 11 TelecomB-BN-01-Vlan-3000.ggnet.umn.edu (22.214.171.124) 12 TelecomB-CN-01-Vlan-710.ggnet.umn.edu (126.96.36.199) 13 baldrick.cs.umn.edu (188.8.131.52)(N!) 88.631 ms (N!) Routers from One Host to the Next over the Internet
Each router on the Internet is connected to one or more other routers. So if you run the traceroute command at different times of the day, you are likely to see that your information flows through different routers at different times. This is because there is a cost associated with each connection between a pair of routers that depends on the volume of traffic, the time of day, and many other factors. By this time it will not surprise you to learn that we can represent the network of routers as a graph with weighted edges.
Figure 2 shows a small example of a weighted graph that represents the interconnection of routers in the Internet. The problem that we want to solve is to find the path with the smallest total weight along which to route any given message. This problem should sound familiar because it is similar to the problem we solved using a breadth first search, except that here we are concerned with the total weight of the path rather than the number of hops in the path. It should be noted that if all the weights are equal, the problem is the same.
The algorithm we are going to use to determine the shortest path is called “Dijkstra’s algorithm.” Dijkstra’s algorithm is an iterative algorithm that provides us with the shortest path from one particular starting node to all other nodes in the graph. Again this is similar to the results of a breadth first search.
To keep track of the total cost from the start node to each destination we will make use of the dist instance variable in the Vertex class. The dist instance variable will contain the current total weight of the smallest weight path from the start to the vertex in question. The algorithm iterates once for every vertex in the graph; however, the order that we iterate over the vertices is controlled by a priority queue. The value that is used to determine the order of the objects in the priority queue is dist. When a vertex is first created dist is set to a very large number. Theoretically you would set dist to infinity, but in practice we just set it to a number that is larger than any real distance we would have in the problem we are trying to solve.
The code for Dijkstra’s algorithm is shown in Listing 1. When the algorithm finishes the distances are set correctly as are the predecessor links for each vertex in the graph.
from pythonds.graphs import PriorityQueue, Graph, Vertex def dijkstra(aGraph,start): pq = PriorityQueue() start.setDistance(0) pq.buildHeap([(v.getDistance(),v) for v in aGraph]) while not pq.isEmpty(): currentVert = pq.delMin() for nextVert in currentVert.getConnections(): newDist = currentVert.getDistance() \ + currentVert.getWeight(nextVert) if newDist < nextVert.getDistance(): nextVert.setDistance( newDist ) nextVert.setPred(currentVert) pq.decreaseKey(nextVert,newDist)
Dijkstra’s algorithm uses a priority queue. You may recall that a priority queue is based on the heap that we implemented in the Tree Chapter. There are a couple of differences between that simple implementation and the implementation we use for Dijkstra’s algorithm. First, the PriorityQueue class stores tuples of key, value pairs. This is important for Dijkstra’s algorithm as the key in the priority queue must match the key of the vertex in the graph. Secondly the value is used for deciding the priority, and thus the position of the key in the priority queue. In this implementation we use the distance to the vertex as the priority because as we will see when we are exploring the next vertex, we always want to explore the vertex that has the smallest distance. The second difference is the addition of the decreaseKey method. As you can see, this method is used when the distance to a vertex that is already in the queue is reduced, and thus moves that vertex toward the front of the queue.
Let’s walk through an application of Dijkstra’s algorithm one vertex at a time using the following sequence of figures as our guide. We begin with the vertex \(u\). The three vertices adjacent to \(u\) are \(v,w,\) and \(x\). Since the initial distances to \(v,w,\) and \(x\) are all initialized to sys.maxint, the new costs to get to them through the start node are all their direct costs. So we update the costs to each of these three nodes. We also set the predecessor for each node to \(u\) and we add each node to the priority queue. We use the distance as the key for the priority queue. The state of the algorithm is shown in Figure 3.
In the next iteration of the while loop we examine the vertices that are adjacent to \(x\). The vertex \(x\) is next because it has the lowest overall cost and therefore bubbled its way to the beginning of the priority queue. At \(x\) we look at its neighbors \(u,v,w\) and \(y\). For each neighboring vertex we check to see if the distance to that vertex through \(x\) is smaller than the previously known distance. Obviously this is the case for \(y\) since its distance was sys.maxint. It is not the case for \(u\) or \(v\) since their distances are 0 and 2 respectively. However, we now learn that the distance to \(w\) is smaller if we go through \(x\) than from \(u\) directly to \(w\). Since that is the case we update \(w\) with a new distance and change the predecessor for \(w\) from \(u\) to \(x\). See Figure 4 for the state of all the vertices.
The next step is to look at the vertices neighboring \(v\) (see Figure 5). This step results in no changes to the graph, so we move on to node \(y\). At node \(y\) (see Figure 6) we discover that it is cheaper to get to both \(w\) and \(z\), so we adjust the distances and predecessor links accordingly. Finally we check nodes \(w\) and \(z\) (see see Figure 6 and see Figure 8). However, no additional changes are found and so the priority queue is empty and Dijkstra’s algorithm exits.
It is important to note that Dijkstra’s algorithm works only when the weights are all positive. You should convince yourself that if you introduced a negative weight on one of the edges to the graph that the algorithm would never exit.
We will note that to route messages through the Internet, other algorithms are used for finding the shortest path. One of the problems with using Dijkstra’s algorithm on the Internet is that you must have a complete representation of the graph in order for the algorithm to run. The implication of this is that every router has a complete map of all the routers in the Internet. In practice this is not the case and other variations of the algorithm allow each router to discover the graph as they go. One such algorithm that you may want to read about is called the “distance vector” routing algorithm.
Finally, let us look at the running time of Dijkstra’s algorithm. We first note that building the priority queue takes \(O(V)\) time since we initially add every vertex in the graph to the priority queue. Once the queue is constructed the while loop is executed once for every vertex since vertices are all added at the beginning and only removed after that. Within that loop each call to delMin, takes \(O(\log V)\) time. Taken together that part of the loop and the calls to delMin take \(O(V \log(V))\). The for loop is executed once for each edge in the graph, and within the for loop the call to decreaseKey takes time \(O(E \log(V))\). So the combined running time is :math:` O((V+E) log(V))`.
For our last graph algorithm let’s consider a problem that online game designers and Internet radio providers face. The problem is that they want to efficiently transfer a piece of information to anyone and everyone who may be listening. This is important in gaming so that all the players know the very latest position of every other player. This is important for Internet radio so that all the listeners that are tuned in are getting all the data they need to reconstruct the song they are listening to. Figure 9 illustrates the broadcast problem.
There are some brute force solutions to this problem, so let’s look at them first to help understand the broadcast problem better. This will also help you appreciate the solution that we will propose when we are done. To begin, the broadcast host has some information that the listeners all need to receive. The simplest solution is for the broadcasting host to keep a list of all of the listeners and send individual messages to each. In Figure 9 we show a small network with a broadcaster and some listeners. Using this first approach, four copies of every message would be sent. Assuming that the least cost path is used, let’s see how many times each router would handle the same message.
All messages from the broadcaster go through router A, so A sees all four copies of every message. Router C sees only one copy of each message for its listener. However, routers B and D would see three copies of every message since routers B and D are on the cheapest path for listeners 1, 2, and 3. When you consider that the broadcast host must send hundreds of messages each second for a radio broadcast, that is a lot of extra traffic.
A brute force solution is for the broadcast host to send a single copy of the broadcast message and let the routers sort things out. In this case, the easiest solution is a strategy called uncontrolled flooding. The flooding strategy works as follows. Each message starts with a time to live (ttl) value set to some number greater than or equal to the number of edges between the broadcast host and its most distant listener. Each router gets a copy of the message and passes the message on to all of its neighboring routers. When the message is passed on the ttl is decreased. Each router continues to send copies of the message to all its neighbors until the ttl value reaches 0. It is easy to convince yourself that uncontrolled flooding generates many more unnecessary messages than our first strategy.
The solution to this problem lies in the construction of a minimum weight spanning tree. Formally we define the minimum spanning tree \(T\) for a graph \(G = (V,E)\) as follows. \(T\) is an acyclic subset of \(E\) that connects all the vertices in \(V\). The sum of the weights of the edges in T is minimized.
Figure 10 shows a simplified version of the broadcast graph and highlights the edges that form a minimum spanning tree for the graph. Now to solve our broadcast problem, the broadcast host simply sends a single copy of the broadcast message into the network. Each router forwards the message to any neighbor that is part of the spanning tree, excluding the neighbor that just sent it the message. In this example A forwards the message to B. B forwards the message to D and C. D forwards the message to E, which forwards it to F, which forwards it to G. No router sees more than one copy of any message, and all the listeners that are interested see a copy of the message.
The algorithm we will use to solve this problem is called Prim’s algorithm. Prim’s algorithm belongs to a family of algorithms called the “greedy algorithms” because at each step we will choose the cheapest next step. In this case the cheapest next step is to follow the edge with the lowest weight. Our last step is to develop Prim’s algorithm.
The basic idea in constructing a spanning tree is as follows:
While :math:`T` is not yet a spanning tree Find an edge that is safe to add to the tree Add the new edge to :math:`T`
The trick is in the step that directs us to “find an edge that is safe.” We define a safe edge as any edge that connects a vertex that is in the spanning tree to a vertex that is not in the spanning tree. This ensures that the tree will always remain a tree and therefore have no cycles.
The Python code to implement Prim’s algorithm is shown in Listing 2. Prim’s algorithm is similar to Dijkstra’s algorithm in that they both use a priority queue to select the next vertex to add to the growing graph.
from pythonds.graphs import PriorityQueue, Graph, Vertex def prim(G,start): pq = PriorityQueue() for v in G: v.setDistance(sys.maxsize) v.setPred(None) start.setDistance(0) pq.buildHeap([(v.getDistance(),v) for v in G]) while not pq.isEmpty(): currentVert = pq.delMin() for nextVert in currentVert.getConnections(): newCost = currentVert.getWeight(nextVert) \ + currentVert.getDistance() if v in pq and newCost<nextVert.getDistance(): nextVert.setPred(currentVert) nextVert.setDistance(newCost) pq.decreaseKey(nextVert,newCost)
The following sequence of figures (Figure 11 thru Figure 17) shows the algorithm in operation on our sample tree. We begin with the starting vertex as A. The distances to all the other vertices are initialized to infinity. Looking at the neighbors of A we can update distances to two of the additional vertices B and C because the distances to B and C through A are less than infinite. This moves B and C to the front of the priority queue. Update the predecessor links for B and C by setting them to point to A. It is important to note that we have not formally added B or C to the spanning tree yet. A node is not considered to be part of the spanning tree until it is removed from the priority queue.
Since B has the smallest distance we look at B next. Examining B’s neighbors we see that D and E can be updated. Both D and E get new distance values and their predecessor links are updated. Moving on to the next node in the priority queue we find C. The only node C is adjacent to that is still in the priority queue is F, thus we can update the distance to F and adjust F’s position in the priority queue.
Now we examine the vertices adjacent to node D. We find that we can update E and reduce the distance to E from 6 to 4. When we do this we change the predecessor link on E to point back to D, thus preparing it to be grafted into the spanning tree but in a different location. The rest of the algorithm proceeds as you would expect, adding each new node to the tree.