I have a tree (v is number of vertex and d is the maximum  degree)whose maximum degree is $\geq 2$. I want to show that, if I take out an specific edge from this tree, I'll get $2$ trees whose number of the vertex in each tree is no more than $$\frac{v(d-1)}d$$
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I was asked the following question:  Let $T=\left(V,E\right)$ be a tree, and let $d$ be the maximum degree of a vertex in that tree. Assume that $d \geq 2$. Prove there an edge in the tree such that, when we remove it from $T$, we will get two trees each of which has at most $\left\lceil \dfrac{d-1}{d}\left|V\right|\right\rceil$ vertices.  I tried going by induction on $d$ the maximum degree while I set the number of vertices and it didn't worked. also tried to do induction on the number of vertices and it didn't worked too.  Thanks for the help, Yoav
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On my book it says that the maximum number of edges E, in a simply connected undirected unweighted graph $G(V,E)$ is $\dfrac{|V|(|V|-1)}{2}$.  How can I show that it is true? Thanks!
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I'm reading the book Graphs and Their Uses which contains the following theorem and proof:     THEOREM 2.3. A connected graph with 2k odd vertices contains a family   of k distinct trails which, together, traverse all edges of the graph   exactly once.       PROOF. Let the odd vertices in the graph be denoted by   in some order.       $a_1,a_2,\dots,a_k$ and $b_1,b_2,\dots,b_k$      When we add the $k$ edges $a_lb_l, a_2b_2 ,\dots, a_kb_k$ to the graph, all   vertices become even and there is an Eulerian trail T. When these   edges are dropped out again, T falls into k separate trails covering   the edges in the original graph.   However this doesn't seem to make sense since in the graph whose vertices have degrees 3, 1, 1, 1 there is no way to add 2 edges in such a way that the degree of all odd vertices becomes even.  What am I missing here?
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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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I'm having some trouble proving the following theorem:     Theorem:   A connected graph on $n$ vertices is a tree iff it has $n-1$ edges   I sort of proved it one way, and I'm stuck the other way around.  Necessity ($\Rightarrow$):  A tree with $n$ vertices has $n-1$ edges.  Basis step ($n=1$):  A tree with one vertex has no edges  Assume a tree with $k\leq n$ vertices has $n-1$ edges  For $k+1$,  Let the added vertex be $u$, if $deg(u)=0$ then we don't have a tree, so $deg(u)&gt;0$. If $deg(u)&gt;1$ then we have a cycle (how do I argue about this?) so again we don't have a tree , so $deg(u)=1$.   Total number of edges = $(n-1) +1 = n \therefore$   $\therefore  P(1) \land \forall k,  P(k)\Rightarrow P(k+1)$  Sufficiency ($\Leftarrow$)  A connected graph with $n-1$ edges is a tree  How can I prove this part? Also, is my induction ok?
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Let $T = (V, E)$ be a rooted tree with an even number of vertices. Prove by induction that there exists a set of edges $S \subseteq E$ such that every $v \in V$ is incident with an odd number of edges in $S$.  I was wondering if my argument is sufficient to prove the statement above:  Base case: $|V|=2$, there is only 1 edge and both vertices are incident with 1 edge $-&gt;$ $S \subseteq E$.  Inductive step:  Let $|V|&gt;2$, implying that $T$ has at least 2 external vertices. Let us choose two arbitrary external vertices $w,u \in V$. Such vertices are only incident to 1 edge each, $e_w$ and $e_u$. We can say that $e_w,e_u \in S$.   By induction, there exists a set $S$ in $T'=T-\{w,u\}=(V',E')$ such that every $v \in V'$ is incident with an odd number of edges in $S$ because removing 2 vertices from a tree with an even number of vertices maintains the property $|V|=2n$.
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Reading online, it generally seems accepted that TREE(n) where n >= 3 is a finite number, but large enough to be incomputable and only has extremely loose lower bounds today. TREE(n) is the function defined by Harvey Friedman, based on Joseph Kruskal's tree theorem. A simplified definition:  "TREE(n) = the maximum length of a set of rooted trees Tm with n possible vertex labels, where Tm has m or fewer vertices, and subsequent trees do not homeomorphically embed a preceding tree."  We can trivially show that TREE(1) = 1 and TREE(2) = 3.  My question is, how can it be known that for values of n greater than 2, the series of possible trees is finite? If we can't compute the solution, how do we know that it ends? If there is an existing write up online describing proof that these numbers are finite, please link me to it, I'm finding limited reading material on the TREE function. Thanks.
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I’m stuck on this question in graph theory. The question is:  How many labeled trees are there over $V={0,1,2,...n}$ with which vertices 1,2,3 are leaves, and distance between any two of these leaves is 3 or more.  I tried using Cayley theorem but I don’t know how to apply it in this specific question.
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The number of (non-equal) forests on the vertex set V = {1, 2, ...,n} that contains exactly 2 connected components is given by $\sum_{k=1}^{n-1} {n-1 \choose k-1} k^{k-2} (n-k)^{n-k-2}$.  I am unsure how to approach this question. Could someone guide me through the proof? If G is a spanning tree and it is disconnected then I know it is a spanning forest. Also, I know by Cayley's thm that there are $n^{n-2}$ nonequal tree on a vertex set V = {1, 2, ...,n} which is where i'm assuming the $k^{k-2}$ comes from but I do not understand where the rest of it is coming from.
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I'm trying to pick up a little graph theory out of Bondy and Murty's Graph Theory as suggested .  Problem 1.1.12 has given me a little hitch.     Let $G$ be a simple graph of order $n$ and size $m$. (So there are $n$ vertices and $m$ edges). If $m&gt;\binom{n-1}{2}$, then $G$ is connected.   I'm following a hint in an appendix which says if $G$ is not connected, we can partition the vertices into parts $(X,Y)$ such that no edge joins a vertex in $X$ to a vertex in $Y$. What is the largest number of edges in $G$ if $|X|=r$ and $|Y|=n-r$?  I suppose the graphs on $X$ and $Y$ are then complete graphs, for a total of $\binom{r}{2}+\binom{n-r}{2}$ edges. Simplifying, this is $\frac{2r^2+n^2-2rn-n}{2}$, so I'm trying to show $$ \frac{2r^2+n^2-2rn-n}{2}\leq\frac{(n-1)(n-2)}{2} $$ to get the contrapositive.   The above is equivalent, if my algebra is correct, to showing $r\geq\frac{n-1}{n-r}$ for any $0\lt r\lt n$. This seems reasonable, but I can't quite show it. How would I do so?  Thanks.  I'd also be happy to see a solution that doesn't necessarily make use of the hint.
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Can anybody give me any hints about how to prove that for any graph $G$ the number of edges in it is at least $\chi(G)(\chi(G)-1)/2$?  $\chi(G)$ is the minimal number of colors we need to use to color the graph (it's chromatic index in other words).  I tried induction but I'm stuck at an induction step.  I do the induction on the number of edges.  When $|E(G)|=0$ then obviously we can color the graph using only one color, so the theorem holds.  Now induction step from $n \Rightarrow n+1$ edges.  Let's erase any edge $e=\{v,u\}$ from $G$. Then $\chi(G-e)$ is either the same, or $\chi(G-e)&lt;\chi(G)$, because we relaxed the graph an $v$ and $u$ no longer need to have different colors. From the first case the thesis is clear, because $G-e$ has more then $\chi(G)(\chi(G)-1)/2$ edges so with the one we erased it's still more. But from the second case we only know that $G \geq \chi'(G)(\chi'(G)-1)/2$ and $\chi'(G)(\chi'(G)-1)/2 \leq \chi(G)(\chi(G)-1)/2$ so that gives me nothing...
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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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Can you prove via induction that there exists a node in a directed graph of n nodes that can be reached in at most two edges from every other node in the graph. Every node in the graph is required to have an edge with every other node that either points out or in (For every pair of nodes X and Y, either X points to Y or Y points to X).
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If  $H$ has maximum vertex degree at most $3$ (so $\Delta(H)\leq 3$) and $H$ is a minor of $G$, then $H$ is a topological minor of $G$.   The converse follows by definition. However, most sources state this proposition without a proof. Any help is greatly appreciated.
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I am trying to prove the following     $d_n$ is a degree-sequence of the multigraph  $G$ (without loops) if and only if $\sum_{i=1}^{n}{d_i}$ is even and $d_1 \le\sum_{i=2}^{n}{d_i}$   I have no problems showing the first part.  By the "handshake"-lemma $\sum_{i=1}^{n}{d_i}=2e$ where $e$ is the number of edges in $G$  $$d_1 \le e $$  $$d_1\le \frac{\sum_{i=1}^{n}{d_i}}{2}$$  Hence $d_1 \le \sum_{i=2}^{n}{d_i}$  However, I don't know how to show the other part. Would appreciate some tips or hints
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I'm studying graph theory and got stuck in one of the problems. It's about graph  connectivity with specific degree sequence condition. Please help me!   My Try: Vertex of degree $d_n$ and $d_n$ other vertices which is connected to it by $d_n$ edges form a connected graph of $d_n+1$ vertices. Then a vertice $d_{n-d_n-_1}$ should be also in a component that contains vertice of $d_n$ degree.(Since there are at most $n-d_n-1$ vertices.) I don't know how to progress from here...     Q: Suppose $G$ is simple with degree sequence $d_1\le d_2\le \cdots d_n$, and for $k\le n-d_n-1, d_k\ge k$. Show $G$ is connected.
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Say $X$ and $Y$ are two different spanning trees of a simple and connected and undirected graph $G$. Prove that if $e_1$ is an edge in $X$ that isn't in $Y$, then there exists an edge $e_2$ in $Y$ that isn't in $X$... so the graph $(X − e_1) + e_2$ (obtained from $X$ on replacing $e_1$ by $e_2$) is also a spanning tree of the original graph, $G$.