In the graph below, vertices A and C have degree 4, since there are 4 edges leading into each vertex. Eulerian Circuit is an Eulerian Path which starts and ends on the same vertex. The cheapest edge is AD, with a cost of 1. How can they minimize the amount of new line to lay? Eulerian Circuit is an Eulerian Path which starts and ends on the same vertex. In graph theory, an Eulerian trail is a trail in a finite graph that visits every edge exactly once. In other words, there is a path from any vertex to any other vertex, but no circuits. For the third edge, we’d like to add AB, but that would give vertex A degree 3, which is not allowed in a Hamiltonian circuit. We then add the last edge to complete the circuit: ACBDA with weight 25. Think back to our housing development lawn inspector from the beginning of the chapter. One Hamiltonian circuit is shown on the graph below. A graph will contain an Euler path if it contains at most two vertices of odd degree. Use NNA starting at Portland, and then use Sorted Edges. Some simpler cases are considered in the exercises. Using Kruskal’s algorithm, we add edges from cheapest to most expensive, rejecting any that close a circuit. A few tries will tell you no; that graph does not have an Euler circuit. Notice that every vertex in this graph has even degree, so this graph does have an Euler circuit. 1. In what order should he travel to visit each city once then return home with the lowest cost? While the Sorted Edge algorithm overcomes some of the shortcomings of NNA, it is still only a heuristic algorithm, and does not guarantee the optimal circuit. Your teacher’s band, Derivative Work, is doing a bar tour in Oregon. Notice that every vertex in this graph has even degree, so this graph does have an Euler circuit. A circuit is a path that starts and ends at the same vertex. (a) First, pick a vertex to the the \start vertex." The graph after adding these edges is shown to the right. The following route can make the tour in 1069 miles: Portland, Astoria, Seaside, Newport, Corvallis, Eugene, Ashland, Crater Lake, Bend, Salem, Portland. Euler Paths and Euler Circuits An Euler path is a path that uses every edge of a graph exactly once. The path is shown in arrows to the right, with the order of edges numbered. Start Euler Circuit – start anywhere Euler Path – start at an odd vertex 3. To see the entire table, scroll to the right. The Könisberg Bridge Problem Könisberg was a town in Prussia, divided in four land regions by the river Pregel. Following that idea, our circuit will be: Total trip length:                     1266 miles. The graph after adding these edges is shown to the right. Graph Theory: Euler Paths and Euler Circuits . Here’s a couple, starting and ending at vertex A: ADEACEFCBA and AECABCFEDA. “Is it possible to draw a given graph without lifting pencil from the paper and without tracing any of … In the first section, we created a graph of the Königsberg bridges and asked whether it was possible to walk across every bridge once. A Hamiltonian circuit is a circuit that visits every vertex once with no repeats. Now we know how to determine if a graph has an Euler circuit, but if it does, how do we find one? question can be framed like this: Suppose a salesman needs to give sales pitches in four cities. Here’s a couple, starting and ending at vertex A: ADEACEFCBA and AECABCFEDA. We can pick up any vertex as starting vertex. Unfortunately, while it is very easy to implement, the NNA is a greedy algorithm, meaning it only looks at the immediate decision without considering the consequences in the future. 2. This is called a complete graph. From each of those cities, there are two possible cities to visit next. 2. Fortunately, we can find whether a given graph has a Eulerian Path or not in polynomial time. Its really very difficult to find the Eulerian path. If we start at vertex E we can find several Hamiltonian paths, such as ECDAB and ECABD. Eulerian Circuit is an Eulerian Path which starts and ends on the same vertex. In the example above, you’ll notice that the last eulerization required duplicating seven edges, while the first two only required duplicating five edges. Using the four vertex graph from earlier, we can use the Sorted Edges algorithm. The complete graph above has four vertices, so the number of Hamilton circuits is: (N – 1)! The first option that might come to mind is to just try all different possible circuits. Half of these are duplicates in reverse order, so there are [latex]\frac{(n-1)! A company requires reliable internet and phone connectivity between their five offices (named A, B, C, D, and E for simplicity) in New York, so they decide to lease dedicated lines from the phone company. The following video gives more examples of how to determine an Euler path, and an Euler Circuit for a graph. There is then only one choice for the last city before returning home. Luckily, Euler solved the question of whether or not an Euler path or circuit will exist. Notice that the same circuit could be written in reverse order, or starting and ending at a different vertex. Hamiltonian circuits are named for William Rowan Hamilton who studied them in the 1800’s. Does the graph below have an Euler Circuit? In order to do that, she will have to duplicate some edges in the graph until an Euler circuit exists. To eulerize a graph, edges are duplicated to connect pairs of vertices with odd degree. The path is shown in arrows to the right, with the order of edges numbered. The following video presents more examples of using Fleury’s algorithm to find an Euler Circuit. In this case, following the edge AD forced us to use the very expensive edge BC later. An Euler circuit is a circuit that uses every edge in a graph with no repeats. To select an edge click a vertex and drag the line to an adjacent vertex. }{2}[/latex] unique circuits. When two odd degree vertices are not directly connected, we can duplicate all edges in a path connecting the two. The problem can be stated mathematically like this: Given the graph in the image, is it possible to construct a path that visits each edge exactly once? The lawn inspector is interested in walking as little as possible. The ideal situation would be a circuit that covers every street with no repeats. The problem of finding the optimal eulerization is called the Chinese Postman Problem, a name given by an American in honor of the Chinese mathematician Mei-Ko Kwan who first studied the problem in 1962 while trying to find optimal delivery routes for postal carriers. We can see that once we travel to vertex E there is no way to leave without returning to C, so there is no possibility of a Hamiltonian circuit. An Euler path starts and ends at different vertices, whereas an Euler circuit starts and ends at the same vertex. Not every graph has an Euler path or circuit, yet our lawn inspector still needs to do her inspections. To apply the Brute force algorithm, we list all possible Hamiltonian circuits and calculate their weight: Note: These are the unique circuits on this graph. The following video gives more examples of how to determine an Euler path, and an Euler Circuit for a graph. Of course, any random spanning tree isn’t really what we want. Watch these examples worked again in the following video. Note that we can only duplicate edges, not create edges where there wasn’t one before. Each Euler Path will begin at one of the odd vertex and end at the other one. Better! After this conversion is performed, we must find a path in the graph that visits every edge exactly once. The book gives a proof that if a graph is connected, and if every vertex has even degree, then there is an Euler circuit in the graph. Notice that even though we found the circuit by starting at vertex C, we could still write the circuit starting at A: ADBCA or ACBDA. To answer this question of how to find the lowest cost Hamiltonian circuit, we will consider some possible approaches. The second is shown in arrows. Since there are more than two vertices with odd degree, there are no Euler paths or Euler circuits on this graph. “Is it possible to draw a given graph without lifting pencil from the paper and without tracing any of the edges more than once”. Choose any edge leaving your current vertex, provided deleting that edge will not separate the graph into two disconnected sets of edges. What happened? The problem is same as following question. Not every graph has an Euler path or circuit, yet our lawn inspector still needs to do her inspections. The problem is to find a tour through the town that crosses each bridge exactly once. Euler paths are an optimal path through a graph. The Euler Circuit is a special type of Euler path. Starting at vertex D, the nearest neighbor circuit is DACBA. From Seattle there are four cities we can visit first. This connects the graph. This graph contains two vertices with odd degree (D and E) and three vertices with even degree (A, B, and C), so Euler’s theorems tell us this graph has an Euler path, but not an Euler circuit. This can be visualized in the graph by drawing two edges for each street, representing the two sides of the street. Portland to Seaside                 78 miles, Eugene to Newport                 91 miles, Portland to Astoria                 (reject – closes circuit). How to find whether a given graph is Eulerian or not? If the edges had weights representing distances or costs, then we would want to select the eulerization with the minimal total added weight. Pick up a starting Vertex. In the first section, we created a graph of the Königsberg bridges and asked whether it was possible to walk across every bridge once. When it snows in the same housing development, the snowplow has to plow both sides of every street. The vertex a a has degree 1, and if you try to make an Euler circuit, you see that you will get stuck at the vertex. Notice that this is actually the same circuit we found starting at C, just written with a different starting vertex. The next shortest edge is AC, with a weight of 2, so we highlight that edge. From D, the nearest neighbor is C, with a weight of 8. Look back at the example used for Euler paths—does that graph have an Euler circuit? In this case, we form our spanning tree by finding a subgraph – a new graph formed using all the vertices but only some of the edges from the original graph. For each graph below, find an Euler trail in the graph or explain why the graph does not have an Euler trail. Some simpler cases are considered in the exercises. Unfortunately, algorithms to solve this problem are fairly complex. The following video shows another view of finding an Eulerization of the lawn inspector problem. The Road Inspector: Finding Euler Circuits Given a connected, undirected graph G = (V,E), find an Euler circuit in G Can check if one exists: • Check if all vertices have even degree Basic Euler Circuit Algorithm: 1. Stop when you run out of edges. The computers are labeled A-F for convenience. Add that edge to your circuit, and delete it from the graph. 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