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An Eulerian trail, [note 1] or Euler walk, in an undirected graph is a walk that uses each edge exactly once. If such a walk exists, the graph is called traversable or semi-eulerian. [3] 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
Apply dynamic programming to this path decomposition to find a longest path in time (!), where is the number of vertices in the graph. Since the output path has length at least as large as d {\displaystyle d} , the running time is also bounded by O ( ℓ ! 2 ℓ n ) {\displaystyle O(\ell !2^{\ell }n)} , where ℓ {\displaystyle \ell } is the ...
After corresponding edges are added (red), the length of the Eulerian circuit is found. In graph theory and combinatorial optimization , Guan's route problem , the Chinese postman problem , postman tour or route inspection problem is to find a shortest closed path or circuit that visits every edge of an (connected) undirected graph at least once.
Every three-digit sequence occurs exactly once if one visits every vertex exactly once (a Hamiltonian path). The de Bruijn sequences can be constructed by taking a Hamiltonian path of an n-dimensional de Bruijn graph over k symbols (or equivalently, an Eulerian cycle of an (n − 1)-dimensional de Bruijn graph). [5]
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In fact in order to answer a level ancestor query, the algorithm needs to jump from a path to another until it reaches the root and there could be Θ(√ n) of such paths on a leaf-to-root path. This leads us to an algorithm that can pre-process the tree in O( n ) time and answers queries in O( √ n ).
This analogy can be made rigorous: the n-dimensional m-symbol De Bruijn graph is a model of the Bernoulli map The Bernoulli map (also called the 2x mod 1 map for m = 2) is an ergodic dynamical system, which can be understood to be a single shift of a m-adic number. [5]
For the Hamiltonian Path problem, c would consist of a string of vertices where the first vertex is the start of the proposed path and the last is the end. [22] The algorithm will determine if c is a valid Hamiltonian Path in G and if so, accept. To decide this, the algorithm first verifies that all of the vertices in G appear exactly once in c.