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An Eulerian cycle, [note 1] also called an Eulerian circuit or Euler tour, in an undirected graph is a cycle that uses each edge exactly once. If such a cycle exists, the graph is called Eulerian or unicursal. [4] The term "Eulerian graph" is also sometimes used in a weaker sense to denote a graph where every vertex has even degree.
An alternative form of the problem asks for a path that traverses all bridges and also has the same starting and ending point. Such a walk is called an Eulerian circuit or an Euler tour. Such a circuit exists if, and only if, the graph is connected and all nodes have even degree. All Eulerian circuits are also Eulerian paths, but not all ...
The knight's tour problem is the mathematical problem of finding a knight's tour. Creating a program to find a knight's tour is a common problem given to computer science students. [ 3 ] Variations of the knight's tour problem involve chessboards of different sizes than the usual 8 × 8 , as well as irregular (non-rectangular) boards.
The Euler tour technique (ETT), named after Leonhard Euler, is a method in graph theory for representing trees. The tree is viewed as a directed graph that contains two directed edges for each edge in the tree. The tree can then be represented as a Eulerian circuit of the directed graph, known as the Euler tour representation (ETR) of the tree
A few variants of the Chinese Postman Problem have been studied and shown to be NP-complete. [10] The windy postman problem is a variant of the route inspection problem in which the input is an undirected graph, but where each edge may have a different cost for traversing it in one direction than for traversing it in the other direction.
In mathematics, the inverse problem for Lagrangian mechanics is the problem of determining whether a given system of ordinary differential equations can arise as the Euler–Lagrange equations for some Lagrangian function. There has been a great deal of activity in the study of this problem since the early 20th century.
Finally, it will come back into the room through the fourth wall and end. If the solution line starts somewhere else, the observer will see the solution line come into and leave his room exactly twice, passing through all four walls in some order. There is no problem with any of this. Consider, however, the observers in the remaining three rooms.
Euler stated the fundamental results for this problem in terms of the number of odd vertices in the graph, which the handshaking lemma restricts to be an even number. If this number is zero, an Euler tour exists, and if it is two, an Euler path exists. Otherwise, the problem cannot be solved.