Endomorphism of a graph and a maximal clique
Let $G= (V, E)$ be a simple graph satisfy the following:
- Independence number is $3$,
- Clique number is $frac{n^2}{4}$,
- Number of vertices is $n^2$, where $n$ is even.
Let $f in$ End$(G)$, where End$(G)$ is the collection of all homomorphism from $G$ to $G$.
I want to show that atmost three clique of maximum size can be mapped under $f$ to a single clique of maximum size.
Since independence number of a graph is $3$ so that atmost three vertices can be mapped into a single single vertex. Also, it is easy to verified that image of maximal clique is a maximal clique under a homomorphism. I am stuck here, how to prove that atmost three clique of maximum size can be mapped to a single clique of maximum size.
graph-theory algebraic-graph-theory
asked Dec 9 at 8:29
user120386
1,067820
add a comment |
Let $G= (V, E)$ be a simple graph satisfy the following:
- Independence number is $3$,
- Clique number is $frac{n^2}{4}$,
- Number of vertices is $n^2$, where $n$ is even.
Let $f in$ End$(G)$, where End$(G)$ is the collection of all homomorphism from $G$ to $G$.
I want to show that atmost three clique of maximum size can be mapped under $f$ to a single clique of maximum size.
Since independence number of a graph is $3$ so that atmost three vertices can be mapped into a single single vertex. Also, it is easy to verified that image of maximal clique is a maximal clique under a homomorphism. I am stuck here, how to prove that atmost three clique of maximum size can be mapped to a single clique of maximum size.
graph-theory algebraic-graph-theory
asked Dec 9 at 8:29
user120386
1,067820
add a comment |
Let $G= (V, E)$ be a simple graph satisfy the following:
- Independence number is $3$,
- Clique number is $frac{n^2}{4}$,
- Number of vertices is $n^2$, where $n$ is even.
Let $f in$ End$(G)$, where End$(G)$ is the collection of all homomorphism from $G$ to $G$.
I want to show that atmost three clique of maximum size can be mapped under $f$ to a single clique of maximum size.
Since independence number of a graph is $3$ so that atmost three vertices can be mapped into a single single vertex. Also, it is easy to verified that image of maximal clique is a maximal clique under a homomorphism. I am stuck here, how to prove that atmost three clique of maximum size can be mapped to a single clique of maximum size.
graph-theory algebraic-graph-theory
asked Dec 9 at 8:29
user120386
1,067820
Let $G= (V, E)$ be a simple graph satisfy the following:
- Independence number is $3$,
- Clique number is $frac{n^2}{4}$,
- Number of vertices is $n^2$, where $n$ is even.
Let $f in$ End$(G)$, where End$(G)$ is the collection of all homomorphism from $G$ to $G$.
I want to show that atmost three clique of maximum size can be mapped under $f$ to a single clique of maximum size.
Since independence number of a graph is $3$ so that atmost three vertices can be mapped into a single single vertex. Also, it is easy to verified that image of maximal clique is a maximal clique under a homomorphism. I am stuck here, how to prove that atmost three clique of maximum size can be mapped to a single clique of maximum size.
graph-theory algebraic-graph-theory
graph-theory algebraic-graph-theory
asked Dec 9 at 8:29
user120386
1,067820
asked Dec 9 at 8:29
user120386
1,067820
edited Dec 9 at 11:25
asked Dec 9 at 8:29
user120386
1,067820
asked Dec 9 at 8:29
user120386
1,067820
asked Dec 9 at 8:29
user120386
1,067820
1,067820
add a comment |
add a comment |
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