A graph $G$ with exactly one vertex of degree one contains a cycle.  Is there any counter example?   I could not find any graph for counterexample.  Please help.
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Could anyone give a counter example for that theorem :     A graph G has exactly one vertex of degree $1$, then it contains a cycle.   I am so confused. I wonder that may I give a counter example which considers trees.
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I'm trying to prove : Given a simple graph $G$ with $n$ vertices, where $n$ is even, prove that if every vertex has degree $\dfrac n2 + 1$, then $G$ must contain a (simple) $3$-cycle. A (simple) $3$-cycle is a set of $3$ (distinct) vertices, $a, b, c$ such that $ab$ is an edge, $bc$ is an edge and $ac$ is an edge.  I know there are similar ones already on here, but I can't comprehend any of the answers. This is what I have so far:  Base Case: Since the graph has to have at least $3$ vertices to have a simple $3$-cycle, the lowest possible even number of vertices is $n = 4$.   $\dfrac42 + 1 = 3$ degree (edges attached to node). This means each vertex connects to each of the other $3$ vertices since $G$ is a simple graph. Thus, any combination of $3$ vertices will be a simple $3$-cycle.  Inductive Hypothesis: Assume a simple graph with $k$ vertices, where $k$ is even, contains a simple $3$-cycle if each vertex has degree $\dfrac k2 + 1$.  Inductive Step: Since only concerned about when the graph has an even number of vertices, going to show a simple graph with $k + 2$ vertices, where $k$ is even, contains a simple $3$-cycle if every vertex has degree $\dfrac{k+2}{2} + 1$   This is as far as I get and don't know how to finish the Inductive Step (Also please feel free to let me know if I messed anything up, up to this point)
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The problem: let $(x_1, x_2, ..., x_r)$ be an r-cycle in $S_n$. Show that for every $h \in S_n$, $h \circ (x_1, x_2, \ldots, x_r) \circ h^{-1} = (h(x_1), h(x_2), \ldots, h(x_r))$
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The problem states that:     A simple (or strict) graph with $n$ vertices and $m&gt;\dfrac{3(n-1)}{2}$ edges, has two vertex   joined by three independent paths.   Any hints, ideas or useful results, to attack this problem?
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How do I prove that there exists (or does not exist) bipartite graph with this degree sequence: $(3,3,3,3,3,5,6,6,6,6,6,6,6,6)$ ?   The sum of the degree sequence is even and simplfied it looks like this: $(1,1,1,1,1,1)$ => it's a degree sequence of a graph.  The simplified degree sequence might suggest that it could be bipartite graph, but I have no idea how to prove it.
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How do i show this proof?  If $G$ is a connected graph then its center is the vertex $v$ such that the maximum of distances from $v$ to the other vertices of $G$ is minimal possible. Prove that a tree has either one center or two adjacent centers. Give an example of a tree of each type with $7$ vertices.
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Seems to be an easy question, but I can't find the right direction.  Let $G$ connected graph on at least 4 vertices, such that every edge in it, participates in a perfect matching. Prove that $G$ is 2-connected.  Any help will be appreciated.  Thanks in advance.
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i think if the graph G has an odd cycle, it's not two-colorable, otherwise it can be two colorable.   i read in one notes that the following is True: we couldent two-colorable any graph G that has cycle.    anyone could clarify me ?
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Given a simple graph $G$ [no loops or parallel edges] on six vertices,  let $G^c$ denote its complement. It is known that either $G$ or $G^c$ must contain a triangle  $T$ in it. [An example of a Ramsey number I think.]   My question came up when I tried to find such a graph on six vertices  which had only one $T$ in it or its complement.  I believe if I did things right there are none of these. This I tried to check by drawing a copy of each graph $G$  on six vertices which has a $T$ in it,  and then looking at the complements to see if any had  no $T$'s in them. I found none.  One way to pose my question for a general number $n$ of vertices  is then as follows: Consider the set of all graphs $G$ on $n$ vertices,  and for each one construct $G^c.$  Count the number of $T$ in $G$ and add that to the number of $T$ in $G^c.$  Finally find the minimal such sum for the given $n$  and call that say $M(n).$ So my above six vertex conjecture  is that $M(6)=2.$   By considering a cycle graph on five vertices we find $M(5)=0.$   So going further one could ask if anything is known  about this minimum function $M(n)$ for larger $n$ [such as bounds, etc.] And does it have a name?  I would also like to see a simpler way to handle the six vertex case, one not involving drawing all the graphs.
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Suppose that we are given two graphs on a relatively small number of vertices. Here's an example:    Are these two graphs isomorphic? (That is: is there a bijection $f$ from $\{A,B,\dots,I\}$, the vertex set of the first graph, to $\{1,2,\dots,9\}$, the vertex set of the second graph, such that $vw$ is an edge of the first graph if and only if $f(v)f(w)$ is an edge of the second graph?)   If not, how can we tell? If so, how can we tell—and how can we find an isomorphism?     This question comes up a lot, usually asking about specific examples of two graphs which may or may not be isomorphic. Sometimes, the fiendish teacher of graph theory will even give three or more graphs and ask "Which of these are isomorphic?"  I'm asking this question so that we can provide  to questions of this type which would be worth linking to when specific instances of this question come up. So all improvements and/or suggestions are welcome!
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Can somebody give me an example of a graph on eight vertices that has a pair of pseudosimilar vertices? Or, is it not possible? If it is not possible, please do not give me a proof, as I would like to figure out how to prove it myself.     Definition: vertices $u$ and $v$ in a graph $X$ are similar if there is an automorphism of $X$ that maps $u$ to $v$. If $u$ and $v$ are similar, then the vertex-deleted subgraphs $X-u$ and $X-v$ are isomorphic. If $X-u$ and $X-v$ are isomorphic but $u$ and $v$ are not similar, we say that they are pseudosimilar.
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I am given this problem: Suppose you have a complete graph with $n$ edges. How many vertices does your graph have?  I have this math figured out so far: We know that a complete graph has $m$ vertices, with $m-1$ edges connected to each. This makes the sum of the total number of degrees $m(m-1)$. Then, since this sum is twice the number of edges, the number of edges is $\frac{m(m-1)}{2}$.  But I don't think that is the answer. I think I am missing a piece of the math, but I can't figure out what else I need to do. Can anyone help me with what I'm missing or doing wrong?
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all I am asked to prove that given a loopless multigraph $G$ with vertices all having even degree, it is possible to orient the graph in such way that the in-degree of every vertex is equal to the out degree of every vertex.  Now for the case when the number of vertices finite, this is easy using Eulerian trails.  However, I am curious if the same statement holds for a graph having countably infinite vertices all with even degree and finitely many edges?  Any help in see why or why the infinite graph case is true would be awesome.
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Theorem: Given a graph G has a Euler Circuit, then every vertex of G   has a even degree      Proof: We must show that for an arbitrary vertex v of G, v has a   positive even degree.   What does it mean by every even degree? When I think of an even degree I think of polynomial functions.   What I am trying to prove?
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If a graph $G$ has at least three vertices, and has a Hamilton Path starting at every vertex, must it contain a Hamilton Circuit?    I have been struggling with this problem.  It seems that because every vertex both starts and does not start a Hamilton Path, every vertex is contained in a circuit.  But I cannot extend this idea to show that there is a Hamilton Circuit.
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On the  it is mentioned that it is not a Cayley graph.    How it this proved? Honestly I don't even know how to start this. The only criteria I can think of is that all vertices must have the same degree. Also the degree being odd should imply that one generator of the group has order 2. But then how do I proceed?  Edit: As already mentioned in the comments, the link did not answer my question.
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I have a graph $G(V,E)$ where each vertex has degree at most $k$. How can I prove that I need at most $2k + 1$ disjoint sets of vertices in which there is no connection between two vertices from the same set to create the set of vertices $V$.