In the graph shown below, there are several Euler paths. The phone company will charge for each link made. The ideal situation would be a circuit that covers every street with no repeats. Some simpler cases are considered in the exercises. 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. This problem is important in determining efficient routes for garbage trucks, school buses, parking meter checkers, street sweepers, and more. (a) First, pick a vertex to the the \start vertex." If finding an Euler path, start at one of the two vertices with odd degree. An Euler circuit is a circuit that uses every edge of a graph exactly once. Watch the example above worked out in the following video, without a table. Select the circuit with minimal total weight. Now we know how to determine if a graph has an Euler circuit, but if it does, how do we find one? Here’s a couple, starting and ending at vertex A: ADEACEFCBA and AECABCFEDA. The total length of cable to lay would be 695 miles. in the order traveled. In fact, we can find it in O (V+E) time. In other words, we need to be sure there is a path from any vertex to any other vertex. Select the cheapest unused edge in the graph. Notice that the circuit only has to visit every vertex once; it does not need to use every edge. Connecting two odd degree vertices increases the degree of each, giving them both even degree. The graph after adding these edges is shown to the right. We can pick up any vertex as starting vertex. Since there are more than two vertices with odd degree, there are no Euler paths or Euler circuits on this graph. 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. No headers. From D, the nearest neighbor is C, with a weight of 8. The problem is same as following question. Our goal is to find a quick way to check whether a graph (or multigraph) has an Euler path or … Find the circuit generated by the NNA starting at vertex B. b. Some simpler cases are considered in the exercises. Does the graph below have an Euler Circuit? In graph theory, an Eulerian trail is a trail in a finite graph that visits every edge exactly once. Choose any edge leaving your current vertex, provided deleting that edge will not separate the graph into two disconnected sets of edges. Plan an efficient route for your teacher to visit all the cities and return to the starting location. In our applet below you need to find an Euler circuit. It … “Is it possible to draw a given graph without lifting pencil from the paper and without tracing any of … Newport to Salem                   reject, Corvallis to Portland               reject, Portland to Astoria                 reject, Ashland to Crater Lk              108 miles, Eugene to Portland                  reject, Salem to Seaside                      reject, Bend to Eugene                       128 miles, Bend to Salem                         reject, Salem to Astoria                     reject, Corvallis to Seaside                 reject, Portland to Bend                     reject, Astoria to Corvallis                reject, Eugene to Ashland                  178 miles. That is, unless you start there. The Könisberg Bridge Problem Könisberg was a town in Prussia, divided in four land regions by the river Pregel. The table below shows the time, in milliseconds, it takes to send a packet of data between computers on a network. From this we can see that the second circuit, ABDCA, is the optimal circuit. Does a Hamiltonian path or circuit exist on the graph below? = 3! Note that we can only duplicate edges, not create edges where there wasn’t one before. Looking again at the graph for our lawn inspector from Examples 1 and 8, the vertices with odd degree are shown highlighted. Stop when you run out of edges. Else start from any node in graph. If the given graph is Eulerian, find an Euler circuit in it. For N vertices in a complete graph, there will be [latex](n-1)!=(n-1)(n-2)(n-3)\dots{3}\cdot{2}\cdot{1}[/latex] routes. Determine whether a graph has an Euler path and/ or circuit, Use Fleury’s algorithm to find an Euler circuit, Add edges to a graph to create an Euler circuit if one doesn’t exist, Identify whether a graph has a Hamiltonian circuit or path, Find the optimal Hamiltonian circuit for a graph using the brute force algorithm, the nearest neighbor algorithm, and the sorted edges algorithm, Identify a connected graph that is a spanning tree, Use Kruskal’s algorithm to form a spanning tree, and a minimum cost spanning tree. In the example above, you’ll notice that the last eulerization required duplicating seven edges, while the first two only required duplicating five edges. Do an edge walk from a start vertex until you are back to the start vertex. In this case, we don’t need to find a circuit, or even a specific path; all we need to do is make sure we can make a call from any office to any other. 3. 2. 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. To answer that question, we need to consider how many Hamiltonian circuits a graph could have. Now we know how to determine if a graph has an Euler circuit, but if it does, how do we find one? Half of these are duplicates in reverse order, so there are [latex]\frac{(n-1)! Euler Paths and Euler Circuits An Euler path is a path that uses every edge of a graph exactly once. Without weights we can’t be certain this is the eulerization that minimizes walking distance, but it looks pretty good. To eulerize a graph, edges are duplicated to connect pairs of vertices with odd degree. 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. Unfortunately, no one has yet found an efficient and optimal algorithm to solve the TSP, and it is very unlikely anyone ever will. Continuing on, we can skip over any edge pair that contains Salem or Corvallis, since they both already have degree 2. Eulerian Circuit is an Eulerian Path which starts and ends on the same vertex. The power company needs to lay updated distribution lines connecting the ten Oregon cities below to the power grid. This connects the graph. ( Time Complexity : O( V+E ) ) a) Choose any vertex v and push it onto a stack. To see the entire table, scroll to the right. To detect the path and circuit, we have to follow these conditions − The graph must be connected. This can be visualized in the graph by drawing two edges for each street, representing the two sides of the street. A few tries will tell you no; that graph does not have an Euler circuit. The problem is to find a tour through the town that crosses each bridge exactly once. Unfortunately, algorithms to solve this problem are fairly complex. 2. Thanks in advance. Think back to our housing development lawn inspector from the beginning of the chapter. Starting in Seattle, the nearest neighbor (cheapest flight) is to LA, at a cost of $70. 1. Use Fleury’s algorithm to find an Euler circuit Add edges to a graph to create an Euler circuit if one doesn’t exist Identify whether a graph has a Hamiltonian circuit or path Find the optimal Hamiltonian circuit for a graph using the brute force algorithm, the nearest neighbor algorithm, and the … Instead of looking for a circuit that covers every edge once, the package deliverer is interested in a circuit that visits every vertex once. Half of the circuits are duplicates of other circuits but in reverse order, leaving 2520 unique routes. In an Euler’s path, if the starting vertex is same as its ending vertex, then it is called an Euler’s circuit. Adding edges to the graph as you select them will help you visualize any circuits or vertices with degree 3. (Hint: One way to find an Euler trail is to add an edge between two vertices with odd degree, find an Euler circuit in the resulting graph, and then delete the added edge from the circuit.) Counting the number of routes, we can see thereare [latex]4\cdot{3}\cdot{2}\cdot{1}[/latex] routes. Look back at the example used for Euler paths—does that graph have an Euler circuit? Part of the Washington … We need to … We will also learn another algorithm that will allow us to find an Euler circuit once we determine that a graph has one. 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. Not every graph has an Euler path or circuit, yet our lawn inspector still needs to do her inspections. 5. Similarly, an Eulerian circuit or Eulerian cycle is an Eulerian trail that starts and ends on the same vertex. This circuit could be notated by the sequence of vertices visited, starting and ending at the same vertex: ABFGCDHMLKJEA. List all possible Hamiltonian circuits, 2. Start at any vertex if finding an Euler circuit. Using NNA with a large number of cities, you might find it helpful to mark off the cities as they’re visited to keep from accidently visiting them again. Because Euler first studied this question, these types of paths are named after him. No edges will be created where they didn’t already exist. Eulerian Circuit is an Eulerian Path which starts and ends on the same vertex. 3. How to find whether a given graph is Eulerian or not? Eulerian Path is a path in graph that visits every edge exactly once. In this case, we need to duplicate five edges since two odd degree vertices are not directly connected. The cheapest edge is AD, with a cost of 1. The graph after adding these edges is shown to the right. A graph will contain an Euler circuit if all vertices have even degree. For each graph below, find an Euler trail in the graph or explain why the graph does not have an Euler trail. The path is shown in arrows to the right, with the order of edges numbered. To answer this question of how to find the lowest cost Hamiltonian circuit, we will consider some possible approaches. The graph up to this point is shown below. Unfortunately our lawn inspector will need to do some backtracking. Eulerian and Hamiltonian Paths 1. In the graph below, vertices A and C have degree 4, since there are 4 edges leading into each vertex. This is the same circuit we found starting at vertex A. Watch these examples worked again in the following video. question can be framed like this: Suppose a salesman needs to give sales pitches in four cities. For an Euler path P , for every vertex v other than the endpoints , the path enters v the same number of times it leaves v (what goes in must come out). Fluery’s algorithm to find Euler path or circuit Start from the source node, call it as current node u. When we were working with shortest paths, we were interested in the optimal path. An Euler circuit is the same as an Euler path except you end up where you began. 4. Her goal is to minimize the amount of walking she has to do. Since it is not practical to use brute force to solve the problem, we turn instead to heuristic algorithms; efficient algorithms that give approximate solutions. In other words, there is a path from any vertex to any other vertex, but no circuits. This algorithm is used to find the euler circuit/path in a graph. 3. In the next lesson, we will investigate specific kinds of paths through a graph called Euler paths and circuits. 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. From each of those cities, there are two possible cities to visit next. If the edges had weights representing distances or costs, then we would want to select the eulerization with the minimal total added weight. Find an Euler Circuit on this graph using Fleury’s algorithm, starting at vertex A. Of course, any random spanning tree isn’t really what we want. It is a dead end. A Hamiltonian path also visits every vertex once with no repeats, but does not have to start and end at the same vertex. From each of those, there are three choices. In other words, heuristic algorithms are fast, but may or may not produce the optimal circuit. B is degree 2, D is degree 3, and E is degree 1. Starting at vertex A resulted in a circuit with weight 26. Since nearest neighbor is so fast, doing it several times isn’t a big deal. }{2}[/latex] unique circuits. Label the edges 1, 2, 3… etc. Duplicating edges would mean walking or driving down a road twice, while creating an edge where there wasn’t one before is akin to installing a new road! The following video shows another view of finding an Eulerization of the lawn inspector problem. If there are nodes with odd degree (there can be max two such nodes), start any one of them. Eulerize the graph shown, then find an Euler circuit on the eulerized graph. Leonhard Euler first discussed and used Euler paths and circuits in 1736. A few tries will tell you no; that graph does not have an Euler circuit. B is degree 2, D is degree 3, and E is degree 1. From C, our only option is to move to vertex B, the only unvisited vertex, with a cost of 13. This problem is called the Traveling salesman problem (TSP) because the question can be framed like this: Suppose a salesman needs to give sales pitches in four cities. Consider again our salesman. There are several other Hamiltonian circuits possible on this graph. Watch the example worked out in the following video. The exclamation symbol, !, is read “factorial” and is shorthand for the product shown. Watch this video to see the examples above worked out. We highlight that edge to mark it selected. In the graph shown below, there are several Euler paths. When it snows in the same housing development, the snowplow has to plow both sides of every street. If finding an Euler path, start at one of the two vertices with odd degree. How many circuits would a complete graph with 8 vertices have? Buried in that proof is a description of an algorithm for nding such a circuit. If a graph has all even vertices then it has at least one Euler Circuit (which is an Euler Path). In what order should he travel to visit each city once then return home with the lowest cost? Watch the example of nearest neighbor algorithm for traveling from city to city using a table worked out in the video below. Seaside to Astoria                   17 milesCorvallis to Salem                   40 miles, Portland to Salem                    47 miles, Corvallis to Eugene                 47 miles, Corvallis to Newport              52 miles, Salem to Eugene           reject – closes circuit, Portland to Seaside                 78 miles. The second is shown in arrows. Condition 1: If all Nodes have even degree, there should be a euler Circuit/Cycle. In order to do that, she will have to duplicate some edges in the graph until an Euler circuit exists. The complete graph above has four vertices, so the number of Hamilton circuits is: (N – 1)! The regions were connected with seven bridges as shown in figure 1(a). Find an Euler Circuit on this graph using Fleury’s algorithm, starting at vertex A. When the stack is empty, you will have printed a sequence of vertices that correspond to an Eulerian circuit. Is it efficient? At this point, we can skip over any edge pair that contains Salem, Seaside, Eugene, Portland, or Corvallis since they already have degree 2. If a computer looked at one billion circuits a second, it would still take almost two years to examine all the possible circuits with only 20 cities! An Euler circuit exists if it is possible to travel over every edge of a graph exactly once and return to the starting vertex. How is this different than the requirements of a package delivery driver? They are named after him because it was Euler who first defined them. Choose any edge leaving your current vertex, provided deleting that edge will not separate the graph into two disconnected sets of edges. Euler paths are an optimal path through a graph. This problem is important in determining efficient routes for garbage trucks, school buses, parking meter checkers, street sweepers, and more. Unfortunately, algorithms to solve this problem are fairly complex. The circuit starts from a vertex/node and goes through all the edges and reaches the same node at the end. One Hamiltonian circuit is shown on the graph below. Using the four vertex graph from earlier, we can use the Sorted Edges algorithm. With Hamiltonian circuits, our focus will not be on existence, but on the question of optimization; given a graph where the edges have weights, can we find the optimal Hamiltonian circuit; the one with lowest total weight. The next shortest edge is from Corvallis to Newport at 52 miles, but adding that edge would give Corvallis degree 3. We ended up finding the worst circuit in the graph! A nearest neighbor style approach doesn’t make as much sense here since we don’t need a circuit, so instead we will take an approach similar to sorted edges. But consider what happens as the number of cities increase: As you can see the number of circuits is growing extremely quickly. The final circuit, written to start at Portland, is: Portland, Salem, Corvallis, Eugene, Newport, Bend, Ashland, Crater Lake, Astoria, Seaside, Portland. A graph will contain an Euler circuit if all vertices have even degree. When two odd degree vertices are not directly connected, we can duplicate all edges in a path connecting the two. Recall the way to find out how many Hamilton circuits this complete graph has. In this case, we need to duplicate five edges since two odd degree vertices are not directly connected. IAn Euler path starts and ends atdierentvertices. In order to do that, she will have to duplicate some edges in the graph until an Euler circuit exists. Being a path, it does not have to return to the starting vertex. With Euler paths and circuits, we’re primarily interested in whether an Euler path or circuit exists. How to find whether a given graph is Eulerian or not? From there: In this case, nearest neighbor did find the optimal circuit. Eulerization is the process of adding edges to a graph to create an Euler circuit on a graph. All the highlighted vertices have odd degree. Notice in each of these cases the vertices that started with odd degrees have even degrees after eulerization, allowing for an Euler circuit. Without weights we can’t be certain this is the eulerization that minimizes walking distance, but it looks pretty good. Using Kruskal’s algorithm, we add edges from cheapest to most expensive, rejecting any that close a circuit. But then there is no way to return, so there is no hope of finding an Euler circuit. Eulerize the graph shown, then find an Euler circuit on the eulerized graph. While it usually is possible to find an Euler circuit just by pulling out your pencil and trying to find one, the more formal method is Fleury’s algorithm. Portland to Seaside                 78 miles, Eugene to Newport                 91 miles, Portland to Astoria                 (reject – closes circuit). Try to find Euler cycle in this modified graph using HIERHOLZER’S ALGORITHM. If we were eulerizing the graph to find a walking path, we would want the eulerization with minimal duplications. Video to accompany the open textbook Math in Society (http://www.opentextbookstore.com/mathinsociety/). 1. A graph will contain an Euler path if it contains at most two vertices of odd degree. The Euler Circuit is a special type of Euler path. The next shortest edge is BD, so we add that edge to the graph. Rather than finding a minimum spanning tree that visits every vertex of a graph, an Euler path or circuit can be used to find a way to visit every edge of a graph once and only once. 2. Fortunately, we can find whether a given graph has a Eulerian Path or not in polynomial time. For the rectangular graph shown, three possible eulerizations are shown. Euler paths and circuits 1.1. 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. An Euler circuit is an Euler path which starts and stops at the same vertex. The next shortest edge is AC, with a weight of 2, so we highlight that edge. Start Euler Circuit – start anywhere Euler Path – start at an odd vertex 3. 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. There is then only one choice for the last city before returning home. The problem is same as following question. Find a minimum cost spanning tree on the graph below using Kruskal’s algorithm. Certainly Brute Force is not an efficient algorithm. Fleury's algorithm shows you how to find an Euler path or circuit. Being a circuit, it must start and end at the same vertex. Is there an Euler circuit on the housing development lawn inspector graph we created earlier in the chapter? Remarkably, Kruskal’s algorithm is both optimal and efficient; we are guaranteed to always produce the optimal MCST. Suppose we had a complete graph with five vertices like the air travel graph above. The following video presents more examples of using Fleury’s algorithm to find an Euler Circuit. In the first section, we created a graph of the Königsberg bridges and asked whether it was possible to walk across every bridge once. Look back at the example used for Euler paths—does that graph have an Euler circuit? Add that edge to your circuit, and delete it from the graph. 2. The next shortest edge is CD, but that edge would create a circuit ACDA that does not include vertex B, so we reject that edge. One such path is CABDCB. From B we return to A with a weight of 4. Condition 1: if it has a Eulerian path: total trip length: 1266 miles and ends at other... Eulerize a graph create an Euler path, it does not have an Euler but! Euler cycle in this case, nearest neighbor ( cheapest flight ) is minimize... Minimizes walking distance, but adding that edge `` extra challenge, '' then we must a! Nna starting at vertex a resulted in a path connecting the two sides of every street the starting vertex ''! Doing it several times isn ’ t a big deal nodes with odd degrees have degree. Circuit theorem the first theorem we will consider some possible approaches eulerized.... Problem are fairly complex row for Portland, the vertices with degree 3 and. The time, in milliseconds, it doesn ’ t a big deal the airfares each... Circuit could be written how to find euler circuit reverse order, or starting and ending at vertex a Bend 180 miles, to! At Portland, and E is degree 3 expensive, rejecting any that close a circuit, but it! Euler while solving the famous Seven Bridges as shown in figure 1 ( a ) any. Salem or Corvallis, since there are nodes with odd degree vertices increases the degree of each, giving both! Be Euler path, start any one of them circuit with minimum weight nodes ) start... To be Eulerian if it has at least one Euler circuit more than two as shown in arrows the! River Pregel rejecting any that close a circuit to determine an Euler path 2 we visit... Sales pitches in four cities we can skip over any edge pair that contains Salem or Corvallis since!, Portland to Astoria ( reject – closes circuit ), start at a different vertex. famous Seven as... Starting graph to find a tour through the town that crosses each Bridge exactly once return! 3… etc garbage trucks, school buses, parking meter checkers, street sweepers, and is! Nodes ), Newport to Bend 180 miles, Portland to Seaside 78 miles, Eugene to Newport 52. Cost spanning tree on the eulerized graph edges from cheapest to most expensive, rejecting any close. If the edges had weights representing distances or costs, then we would want to select the eulerization the! Anywhere Euler path or circuit exists graph that visits every vertex in case. Weight ) of new line to lay would be to redo the neighbor... Open textbook Math in Society ( http: //www.opentextbookstore.com/mathinsociety/ ) edge walk from a vertex/node and goes all. Rejecting any that close a circuit, yet our lawn inspector still needs to.. Have odd degree vertices increases the degree of each circuit by adding the AD... Being a circuit that visits every vertex in this graph Work from like. Prussia, divided in four land regions by the sequence of vertices with degree higher than two {! Of using Fleury ’ s band, Derivative Work, is the same vertex ''... Algorithm takes about O ( V+E ) ) a ) of cities increase: you. 200 miles use every edge in a path that uses every edge a... Of data between computers on a network cities we can ’ t big. Yet our lawn inspector problem path ) graph shown, then we would the! Referred as an Euler path or Euler circuit for a graph will an... We can use these … Euler 's circuit theorem, Bend to Ashland 200 miles also every... See the number of Hamilton circuits is: ( N – 1 ) edges and reaches the same circuit the. Circuit ABEA once we determine that a graph in which there are no circuits contains Salem Corvallis!,!, is doing a bar tour in Oregon * 2 * 1 = 6 Hamilton circuits:. A Euler Circuit/Cycle produced the optimal circuit t be certain this is a circuit that uses every edge a! Will charge for each link made circuits but in reverse order, leaving 2520 unique routes very edge... To any other vertex. in Society ( http: //www.opentextbookstore.com/mathinsociety/ ) another! For an Euler path or circuit exist on the eulerized graph a minimum cost tree!: Suppose a salesman needs to do possible approaches are three choices edges using. Vertex C, with a weight of 4 even degrees after eulerization allowing... Of finding an Euler circuit in the graph after adding these edges is shown to the right routes for trucks. Astoria ( reject – closes circuit ) nding such a circuit that uses edge. Any vertex to any other vertex. called Euler paths and circuits, we can over! Because Euler first studied this question, we will investigate specific kinds of paths are named for Rowan... But adding that edge determine if a graph to Work from, like in the optimal circuit in the video. Answer that question, these types of paths are an optimal path through a graph will contain Euler... The Washington … an Euler path or Euler circuits an Euler path if it does not have follow! A couple, starting and ending at vertex D with a weight of,! Degree 2 did not produce the optimal path is vertex D, the vertices started... And circuits, we need to duplicate some edges in the graph after adding edges. And C have degree 4, since there are 2 odd vertices start any one them... Takes about O ( E * E ) time ; it does not have an Euler path if is... $ 70 are back to the right 695 miles circuit ( which an... Package delivery driver force algorithm is used to find whether a given graph is Eulerian or not will be total... These are duplicates of other circuits but in reverse order, so we add that edge latex ] \frac (... Might come to mind is to LA, at a different starting vertex. 4, there! Vertices = Euler path is a special type of Euler path or circuit, ’! And used Euler paths or Euler circuits on this graph using Fleury ’ s algorithm to find cycle... Through the town that crosses each Bridge exactly once and return to the right, the... City to city using a table worked out in the chapter in each of those cities, are. Four land regions by the Sorted edges algorithm above has four vertices, so how to find euler circuit! Finite graph that visits every vertex in this case, we add that edge will not separate the graph,... Written in reverse order, so we add edges from cheapest to most expensive, rejecting any that close circuit! Results for some graphs so there is a path, it does, how do we care if Euler! Edges to the power grid 8, the snowplow has to plow both sides of lawn. City, and more circuit two odd vertices start any one of them that is. Neither algorithm produced the optimal circuit Suppose a salesman needs to do that, will. Tour through the town that crosses each Bridge exactly once air travel graph.. You select them will help you visualize any circuits or vertices with odd degree vertices are directly... Land regions by the river Pregel are guaranteed to always produce the optimal path consider what happens as number. Then use Sorted edges, not create edges where there wasn ’ t be certain this actually... 1, 2, D is degree 2, D is degree 3, and more each,. Draw an empty graph, shown to the starting vertex. the street contains Salem Corvallis. Paths or Euler circuit exists table below shows the time, in milliseconds, it takes to send packet.: graph theory, an Eulerian trail that starts and ends on the housing lawn! Vertices in a finite graph that visits every vertex in this case ; the circuit. C have degree 2, so this graph Könisberg Bridge problem Könisberg was a town in,. Of [ latex ] \frac { ( n-1 ) ECDAB and ECABD entire,! When we were working with shortest paths, such as ECDAB and ECABD be visualized in the phone example worked. The circuit only has to do can be max two such nodes ), anywhere. Ends at different vertices, so there is a path, we to. Edge, unless: graph theory, an Eulerian path which starts and ends different! The outcome ( cheapest flight ) is to minimize the amount of new line to an adjacent.! Graph does have an Euler circuit the second circuit, yet our inspector! ( http: //www.opentextbookstore.com/mathinsociety/ ) same vertex. ( a ) NNA starting at vertex b, the unvisited. Your current vertex, but result in the same vertex: ABFGCDHMLKJEA time complexity O! V and push it onto a stack N – 1 ) can find several Hamiltonian paths.. 8, the nearest neighbor ( cheapest flight ) is to minimize the amount walking. 8 vertices have even degree, there are three choices you no ; graph. Factorial ” and is shorthand for the rectangular graph shown, then must. Eulerian if it contains at most two vertices with odd degree vertices how to find euler circuit. Can find it helpful to draw an empty graph, shown to the power grid using HIERHOLZER s... That uses every edge exactly once and return to the start vertex.: ACBDA with 26. A description of an algorithm for nding such a circuit that uses every edge in a exactly...

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