0 votes 0 votes The following graph has no Euler circuit because It has $7$ vertices. It is even-valent (all vertices have even valence). It is not connected. It does not have a Euler circuit. Graph Theory nielit2017july-scientistb-cs discrete-mathematics graph-theory euler-graph + – admin asked Mar 30, 2020 • edited Nov 6, 2020 by soujanyareddy13 admin 2.3k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
3 votes 3 votes The above given Graph is not connected that's why Euler circuit doesn't exist. Abhishek Kumar Singh answered Oct 3, 2017 Abhishek Kumar Singh comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes An undirected graph has an Eulerian cycle if and only if every vertex has even degree, and all of its vertices with nonzero degree belong to a single connected component. topper98 answered Mar 19, 2020 topper98 comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes A. Makes no sense B. This is one of the condition for circuit being euler's circuit. C. Even disconnected graph can be euler circuit but the the vertices which aren't part of main component shouldn't have any edge between them. C is most appropriate option. Image Source: http://mathonline.wikidot.com/isolated-vertices-leaves-and-pendant-edges smsubham answered Apr 3, 2020 smsubham comment Share Follow See 1 comment See all 1 1 comment reply anurags228 commented Sep 16, 2020 reply Follow Share How can a disconnected graph be an Euler circuit? An Euler circuit in a graph G is a simple circuit containing every edge of G. If the graph is disconnected it's impossible to contain all the edges. Alright, I understood it. Sorry 😅 0 votes 0 votes Please log in or register to add a comment.