[Graph theory] Euler circle and Hamiltonian ring correlation

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Euler traces and loops
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Contains traces of all edges only once.

General diagram the necessary conditions for the existence of Central Europe: each vertex has an even number of degrees.
General connectivity diagram the necessary and sufficient conditions for the existence of a central European pull ring: an even number of each vertex.
The number of open traces in a general connected graph: Set the number of odd vertices m>0, it can be divided into not less than M/2 open track.
Special case: There is an open-Euler condition with only 2 odd vertices. And the trace takes these 2 points as the starting and ending points.

China postman problem: General connectivity graphs contain the shortest path at least once for all edges.
Solution: The general connectivity graph of K-bar, there is a closed path of 2K length, twice times the number of repetitions per side.

To find Euler's circle algorithm:
W Vertex set F edge set
I) Make I=1
II) Make w={x0,x1}
III) Make F={A1}
IV) when xi<>x0, perform
1) Find a side ai+1={xi,xi+1} that is not in F.
2) put the xi+1 in W (perhaps Xi+1 already in W)
3) put the ai+1 into F
4) Make i=i+1

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Hamiltonian chains and loops
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Hamiltonian chain: A chain that contains all vertices only once.
Hamiltonian ring: Closed Hamiltonian chain.
In kn there are n! and n! of different hahaha rings.
The problem of the existence calculation method of the hardy chain and the ring is not solved in the present graph theory.

There is no decision on the ring: There is a bridge in the connected graph.
There are sufficient conditions for the HA circle: a simple diagram satisfying the ore nature and order n>=3.
Ore properties: N-order simple diagram, for all non-contiguous different vertex pairs, there is D (x) +d (y) >=n.
That is: 1 all vertices have a large degree or 2 few vertices are small, while other fixed points are of great degree
Simplified ore: the degree of all points >=N/2
There are sufficient conditions in the Ring 2: (according to the simplified Ore nature) Order n>=3 simple diagram, the degree of all points >=n/2.
The HA chain has sufficient conditions: N-Order simple graphs, if the different vertex pairs of all non-contiguous, there is D (x) +d (y) >=n-1.

To find the Hamiltonian loop algorithm:
(if the given figure satisfies the ore nature, the result is a ring of ha)
I) Start the construction chain at any vertex, until it can no longer be extended
II) Check whether the end is adjacent
1) No Turn III, is the next step
2) If the chain vertex number is =n, complete. Otherwise, move down.
3) Find a vertex that is not on the chain adjacent to the vertex on the chain. Links the original chain to the end, disconnects from the vertices at the previous chain, and connects the new vertices to the corresponding vertices. Turn II.
III) Find a pair of adjacency vertices from the chain, with the previous one adjacent to the tail vertex and the latter adjacent to the first vertex. Break the chain from the next vertex and the second half

The part flips back to the chain. The new chain is contiguous. Turn II

Focus: Construct the chain to the longest, then add a vertex to each part, and loop it in to continue adding