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4 votes
4 votes

Which of the following Graph has Euler Path but is not an Euler Graph?
A. K1,1 
B.K2,10 
C.K2,11
D.K10,11.

in Graph Theory
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4 Comments

K2,11 is complete bipartite graph ryt .?

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Yes, it is and two nodes are having degree 11.And that is odd , so there can be Euler Path in That K2,11 also. That's what I am thinking. :)

Sir, is there any wrong in that logic?

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A connected graph is Euler graph iff it has at most two odd degree vertices.
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2 Answers

1 vote
1 vote
option A is right

euler path is possible but euler cycle is not possible so it is not a euler graph

2 Comments

What About Option C?

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edited by
Option C is also satisfying the criteria so isn't it Eular path ?
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0 votes
0 votes

The degree-sequence of a Complete Bipartite Graph $K_{p,q}$ is  $(p,p,...,p,q,...,q)$  where p is repeated q times and q is repeated p times. Therefore, $K_{2,11}$ would have a degree sequence (2,2,2,2,2,2,2,2,2,2,2,11,11). 

Since $K_{2,11}$ is a connected graph with exactly two of its vertices having odd degrees while remaining having even degrees, it will have a Euler path.

Hence options A and C would be right.

Also read,

https://math.stackexchange.com/questions/221512/hamilton-euler-circuit-path

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