Number of Eulerian Graphs

Question

Find the number of connected Eulerian graphs with 6 unlabelled vertices.

Draw each of them.

Note: I'm looking for a fast procedure don't comment just the numerical answer.

Comments

can we consider #edges on our own? as its only implemented on edges. there is no role of vertex as it can be traversed any times.
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shall the graph can  be a circuit?
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can u plz be a bit brief?

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graph with 6 edges is eulerian and connected. Minimum number of edges for connected graph = $5$

Edges in complete graph = $15$

How many graphs containing edges between 5 and 15 are Eulerian?

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that i got it will be in that range but i think than it will be much of permutations and combination and will be more time consuming . but wait i must try one.

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answer is just 8.

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and vertices are unlabelled

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see first thing is that a graph is eulerian if it starts from a vertex and reaches to same vertex after travelling every edge exactly once.

so this means it must have atleast 6 edges.
and formula for #edges is  =(n(n-1)/2 = 15 atmost.

so u need to consider only 10 cases where u can start and end at same vertes.

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check formula for atmost edges in a simple graph again.

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so sorry bro :/ yes u are right

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@arvin, atleast 5 right?

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atleast 5 for connected and atleast 6 for eulerian

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then atmost is 14, how 15?

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@shaikh see here we are talking about #edges to be connected and eulerian(graph) so minimum is 6 it cannot be 5 because u cannot reach back to starting node with 5 vertices as it will be a tree. (condition for eulerian graph)

atmost 15 because a complete graph will have 15 edges.

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@mk utkarsh see the solution brother i have to really recall everything .

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yes you are correct..

Answer 1

for a graph to be an Eulerian

​​​​​​​1) it must start and end at same vertex with each edge covered exactly  once and 

2) the degree of each node must be of even degree.

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for a graph with #vertex= 6

possible degree values(to satisfy if its euler or not even degree sequences)=

{2,2,2,2,2,2} , {4,2,2,2,2,2}, {4,4,2,2,2,2} , { 4,4,4,2,2,2} , { 4,4,4,4,2,2} , {4,4,4,4,4,4}  =  #sequences =7 out of which only 3rd sequence {4,4,2,2,2,2} will have 2 different graphs.

#total =8

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you can use Havel hakimi algorithm to check the validity of the degree sequence... i found each valid.

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i tried each possible sequence but every graph i went for was ISOMORPHIC to the down mentioned 8 graphs. 

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Comments

@srestha mam even degree disconnected graphs cannot be eulerian..

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Note that a disconnected graph may also be Euler.

https://math.stackexchange.com/questions/1414667/is-it-possible-disconnected-graph-has-euler-circuit?rq=1

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my mistake, updated !