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Prove that in a tree with a maximum degree for each vertex is $k$, there are at least $k$ leaves.

So I said:

$2|E| = \sum_{v \in V} {\deg(v)} \leq k $ which is, if we say that we have AT MOST $k-1$ leaves (I used the contradiction method to prove) \begin{align} \sum_{v \in V} {\deg(v)} &= \sum_{v \; \text{is a leaf}} {1} + \sum_{v \; \text{is not a leaf}} {\deg(v)} \\ &\leq (k-1) + k(n-k+1) \\ &= 2k - k^2 + kn -1. \end{align}

But that obviously tells us nothing. All extra information I know is that in a tree the sum of all degrees is $2|E| = 2(n-1) = 2n - 2$ so somehow I should get to an inequality/equality regarding $n$ only.

Any help is appreciated

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6 Answers 6

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Let me give you a hint.

Start off by proving that every tree with at least two vertices must have at least 2 leaves.

Now, if the maximum degree is $k$, then there is a vertex $v$ of degree $k$. Consider the graph obtained by deleting $v$ from your tree. What does it look like? Prove that it is a collection of $k$ non-empty trees.

Consider each of these components. If the component has only one vertex, then this vertex was a leaf in the original graph: its only neighbor was $v$.

If the component has two or more vertices, then by the first claim it must have two leaves. Of these, only one could have been connected to $v$, since otherwise they form a cycle; hence at least one of them must have been a leaf in the original tree.

So, we get at least one leaf from each of the $k$ components of the new graph.

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  • $\begingroup$ The new graph doesn't need to have $2k$ vertices of degree 1. Consider a star graph where the central vertex is connected to $k$ vertices. Deleting it leaves $k$ vertices of degree 0. $\endgroup$
    – Calvin Lin
    Jun 23, 2013 at 17:31
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    $\begingroup$ Fair again. Time to drink more coffee... $\endgroup$ Jun 23, 2013 at 17:32
  • $\begingroup$ There, that should do it. $\endgroup$ Jun 23, 2013 at 17:34
  • $\begingroup$ Wow, an amazing way. Thank you so much. $\endgroup$
    – TheNotMe
    Jun 23, 2013 at 17:51
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Let the maximum degree be $k$. Start at a vertex of degree $k$ and travel away from it. There are $k$ possible choices of starting edge, each of which leads to at least one leaf. The leaves must be distinct, for otherwise there would be a cycle.

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  • $\begingroup$ Thank you, but it is a bit missing, you need to prove that why if the leaves we get to are not distinct then there is a cycle. $\endgroup$
    – TheNotMe
    Jun 23, 2013 at 18:12
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    $\begingroup$ There is a unique path between any two vertices in a tree. $\endgroup$
    – Théophile
    Jun 23, 2013 at 18:49
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Suppose , that there are $l$ leaves. You know , from theory , that it has exactly $|E|= n-1 edges$. Every vertex in a tree , that's not a leaf , should have a degree of max value k and at least 2. So we want to create a lower bound for l. So there should be at least one vertex of degree k , l vertexs of degree 1 ( the leaves) and (n-l-1) of degree $>=$ 2. Then $2|E| =2(n-1)= \sum deg(v) >= l + (n-l-1)2 +k \rightarrow 2n-2l-2+k+l=2n-l+k-2<= 2n -2 \rightarrow -l +k <=0 \rightarrow l>=k$

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More generally, if some vertex has degree $k$, then there are at least $k$ leaves, as the following proof shows.

The $k=1$ case is a tautology, so assume $k \ge 2$.

Let $n_d$ be the number of nodes of degree $d$, and let $n$ be the total number of nodes. By counting nodes and edges, we have \begin{align} \sum_d n_d &= n \tag1 \\ \sum_d d n_d &= 2(n-1) \tag2 \end{align} Combining $(1)$ and $(2)$ yields $$\sum_d d n_d = 2\left(\sum_d n_d-1\right).$$ Hence the number $n_1$ of leaves satisfies $$n_1 = 2 + \sum_{d\ge 2} (d-2) n_d \ge 2 + (k-2) n_k \ge 2 + (k-2) = k .$$

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Let $T = (V, E)$ be a tree with a vertex of degree $k$. Let $v_0$ be the vertex of degree $k$. Now let $v_i$ be a vertex such that $\{ v_0, v_i \} \in E(T)$ and $i \in \{ 1$, ... , $k \}$. Let $P_i$ be the path of maximum length in $T$ such that $v_0$ is the first vertex in the path and $v_i$ is the second vertex in the path. Note that in the path $P_i$, a subgraph of $T$, $deg_T(v_0) = 1$. Using the leaf lemma (every non-edgeless tree has at least $2$ nodes) and the fact that $P_i$ has no branches because it's a path, there exists exactly $2$ leaves in $P_i$. Furthermore, there exists $1$ leaf in the path such that the leaf is not $v_0$. In other words, the minimum number of leaves, $n_i$, in $P_i$—not including $v_0$—is $1$. Because $i \in \{1, ..., k\}$, we can take the sum $\displaystyle\sum_{i=1}^{k} n_i$ = $\displaystyle\sum_{i=1}^{k} 1= k$, we see that $T$ contains at least $k$ leaves (if you're wondering "Can't there be more leaves?", you would be correct, however, we only need to show the tree has at least $k$ leaves). One thing to note is that we can guarantee that all the counted leaves are distinct because if they weren't, we would either obtain a cycle in $T$ or a path longer than $P_i$, both of which are contradictions since cycles cannot exist in a tree and $P_i$ has been defined as a path of maximum length containing $v_0$ and $v_i$.

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we can directly prove a better result if a tree has a vertex of degree m then tree has at least m leaves solution is simple as tree on n vertices has n-1 edges sum of degrees of vertices in tree is 2(n-1) now if r is number of leaves in tree then we have n-r-1 vertices in tree whose degree is at least 2
hence we have m+r+2(n-r-1) less than or equal to 2(n-1)
that would give r is at least m i.e there are at least m leaves

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    $\begingroup$ I looked carefully to see how you might be proving "a better result". It seems to me you are claiming "if a tree has a vertex of degree $m$ then tree has at least $m$ leaves". Compare the Question: "Prove that in a tree with maximum degree $k$, there are at least $k$ leafs[sic]". Since you are posting a reply to a three year old Question, there is no reason to hurry your Answer. If you see a way to prove a stronger claim, that would be interesting, but I don't see how your claim is any stronger. $\endgroup$
    – hardmath
    Mar 16, 2017 at 0:46

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