0 votes 0 votes Shashi Shekhar 1 asked Dec 10, 2018 Shashi Shekhar 1 590 views answer comment Share Follow See all 8 Comments See all 8 8 Comments reply Satbir commented Dec 10, 2018 reply Follow Share since arrows are not present so option 1 is wrong since we could not place 1st graph on top of 2nd graph so it is not isomorphic 0 votes 0 votes Hemanth_13 commented Dec 10, 2018 reply Follow Share See the cycle length it's 4 for 1st graph and 3for 2nd graph so it is not isomorphic 0 votes 0 votes Shashi Shekhar 1 commented Dec 10, 2018 reply Follow Share are they homomorphic ? @Hemanth_13 @Satbir 0 votes 0 votes Hemanth_13 commented Dec 10, 2018 reply Follow Share No, they are not isomorphic 0 votes 0 votes Shashi Shekhar 1 commented Dec 10, 2018 reply Follow Share i am asking are they homomorphic? i already know they are not isomorphic. @Hemanth_13 0 votes 0 votes Satbir commented Dec 11, 2018 reply Follow Share they are not homomorphic bcoz we could not place one graph on top of other. Homomorphic means that a graph is a part of another graph. For eg: a traingle is homomorphic to a rectangle with daigonals mathematicaly, if G=(V(G),E(G)), G'=(V(G'),E(G')) then a mapping G->G' such that f(x)f(y) Є G' iff xy Є => G' is homomorphic to G. 0 votes 0 votes pream sagar commented Dec 13, 2018 reply Follow Share Degree of each vertices is same in both graph so it may be isomorphic. cycle in both graph is same is this necessary condition of isomorphic? 0 votes 0 votes Satbir commented Dec 13, 2018 reply Follow Share @pream sagar In isomorphism the relative positions of vertices and edges should also be same. Eg:- If a,b,c are vertices in graph 1 and m,n,o are vertices in graph 2 such that we have mapped a->m,b->n and c->o then if a and b are connected (i.e. have edge between them) in graph 1 then m and n should also have an edge between them. The above question is not satisfying this property. Cycle has nothing to do with it. We just have to transform one graph into another by redrawing the edges but without changing the starting and ending vertex of each edge such that one graph could be placed on top of other. 0 votes 0 votes Please log in or register to add a comment.