Mixing
Find the bounds of the number of leaves in a tree
$begingroup$
Problem: Let $n ≥ 3$ be an integer, $T$ be tree on $n$ vertices, with no vertex of degree $2$, and m be the number of leaves in $T$. It is true that we always have one of the following:
- $m geq ( n + 2 ) / 2$
- $( n + 4 ) / 3 leq m leq ( n + 2 ) / 2$
- $1 leq m leq ( n + 4 ) / 3$
- $m = ( n + 4 ) / 3$
My question: I know that for every graph $2 | E | = sum _ { v in V } d ( v )$. More specifically we have for a tree $2(n-1) = sum _ { v in V } d ( v )$. Now since there are no vertices of degree $2$ and I know we have $m$ vertices of degree 1:
$$mcdot 1 + 0cdot 2+ n_3cdot 3+n_4cdot 4+ldots=2(n-1)$$
with $n_k$ the number of vertices of degree $k$. Clearly we have an upper bound quiet generous:
$$mleq 2(n-1)$$
I can't seem to find another way to bound the value of m.
discrete-mathematics graph-theory trees
asked Jan 20 at 8:06
NotAbelianGroupNotAbelianGroup
18211
$endgroup$
add a comment |
$begingroup$
Problem: Let $n ≥ 3$ be an integer, $T$ be tree on $n$ vertices, with no vertex of degree $2$, and m be the number of leaves in $T$. It is true that we always have one of the following:
- $m geq ( n + 2 ) / 2$
- $( n + 4 ) / 3 leq m leq ( n + 2 ) / 2$
- $1 leq m leq ( n + 4 ) / 3$
- $m = ( n + 4 ) / 3$
My question: I know that for every graph $2 | E | = sum _ { v in V } d ( v )$. More specifically we have for a tree $2(n-1) = sum _ { v in V } d ( v )$. Now since there are no vertices of degree $2$ and I know we have $m$ vertices of degree 1:
$$mcdot 1 + 0cdot 2+ n_3cdot 3+n_4cdot 4+ldots=2(n-1)$$
with $n_k$ the number of vertices of degree $k$. Clearly we have an upper bound quiet generous:
$$mleq 2(n-1)$$
I can't seem to find another way to bound the value of m.
discrete-mathematics graph-theory trees
asked Jan 20 at 8:06
NotAbelianGroupNotAbelianGroup
18211
$endgroup$
add a comment |
$begingroup$
Problem: Let $n ≥ 3$ be an integer, $T$ be tree on $n$ vertices, with no vertex of degree $2$, and m be the number of leaves in $T$. It is true that we always have one of the following:
- $m geq ( n + 2 ) / 2$
- $( n + 4 ) / 3 leq m leq ( n + 2 ) / 2$
- $1 leq m leq ( n + 4 ) / 3$
- $m = ( n + 4 ) / 3$
My question: I know that for every graph $2 | E | = sum _ { v in V } d ( v )$. More specifically we have for a tree $2(n-1) = sum _ { v in V } d ( v )$. Now since there are no vertices of degree $2$ and I know we have $m$ vertices of degree 1:
$$mcdot 1 + 0cdot 2+ n_3cdot 3+n_4cdot 4+ldots=2(n-1)$$
with $n_k$ the number of vertices of degree $k$. Clearly we have an upper bound quiet generous:
$$mleq 2(n-1)$$
I can't seem to find another way to bound the value of m.
discrete-mathematics graph-theory trees
asked Jan 20 at 8:06
NotAbelianGroupNotAbelianGroup
18211
$endgroup$
Problem: Let $n ≥ 3$ be an integer, $T$ be tree on $n$ vertices, with no vertex of degree $2$, and m be the number of leaves in $T$. It is true that we always have one of the following:
- $m geq ( n + 2 ) / 2$
- $( n + 4 ) / 3 leq m leq ( n + 2 ) / 2$
- $1 leq m leq ( n + 4 ) / 3$
- $m = ( n + 4 ) / 3$
My question: I know that for every graph $2 | E | = sum _ { v in V } d ( v )$. More specifically we have for a tree $2(n-1) = sum _ { v in V } d ( v )$. Now since there are no vertices of degree $2$ and I know we have $m$ vertices of degree 1:
$$mcdot 1 + 0cdot 2+ n_3cdot 3+n_4cdot 4+ldots=2(n-1)$$
with $n_k$ the number of vertices of degree $k$. Clearly we have an upper bound quiet generous:
$$mleq 2(n-1)$$
I can't seem to find another way to bound the value of m.
discrete-mathematics graph-theory trees
discrete-mathematics graph-theory trees
asked Jan 20 at 8:06
NotAbelianGroupNotAbelianGroup
18211
asked Jan 20 at 8:06
NotAbelianGroupNotAbelianGroup
18211
edited Jan 20 at 8:25
NotAbelianGroup
asked Jan 20 at 8:06
NotAbelianGroupNotAbelianGroup
18211
asked Jan 20 at 8:06
NotAbelianGroupNotAbelianGroup
18211
asked Jan 20 at 8:06
NotAbelianGroupNotAbelianGroup
18211
18211
add a comment |
add a comment |
1 Answer
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$begingroup$
$m$ could not be greater than $n$ as the number of nodes is $n$. The upper bound for $m$ is $n-1$ (as an example a star). Hence, $m leq n-1$.
answered Jan 20 at 8:15
OmGOmG
2,502822
$endgroup$
$begingroup$
Oh sorry you are right, I don't know why I wrote this nonsense. However I don't see how $n_2 =0$ impacts the calculation.
$endgroup$
– NotAbelianGroup
Jan 20 at 8:25
$begingroup$
@NotAbelianGroup it's ok. $n_2 = 0$ has no impact on the upper bound of $m$. Except, you can't have a tree with $n = 3$ and $n_2 = 0$!
$endgroup$
– OmG
Jan 20 at 8:37
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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votes
$begingroup$
$m$ could not be greater than $n$ as the number of nodes is $n$. The upper bound for $m$ is $n-1$ (as an example a star). Hence, $m leq n-1$.
answered Jan 20 at 8:15
OmGOmG
2,502822
$endgroup$
$begingroup$
Oh sorry you are right, I don't know why I wrote this nonsense. However I don't see how $n_2 =0$ impacts the calculation.
$endgroup$
– NotAbelianGroup
Jan 20 at 8:25
$begingroup$
@NotAbelianGroup it's ok. $n_2 = 0$ has no impact on the upper bound of $m$. Except, you can't have a tree with $n = 3$ and $n_2 = 0$!
$endgroup$
– OmG
Jan 20 at 8:37
add a comment |
$begingroup$
$m$ could not be greater than $n$ as the number of nodes is $n$. The upper bound for $m$ is $n-1$ (as an example a star). Hence, $m leq n-1$.
answered Jan 20 at 8:15
OmGOmG
2,502822
$endgroup$
$begingroup$
Oh sorry you are right, I don't know why I wrote this nonsense. However I don't see how $n_2 =0$ impacts the calculation.
$endgroup$
– NotAbelianGroup
Jan 20 at 8:25
$begingroup$
@NotAbelianGroup it's ok. $n_2 = 0$ has no impact on the upper bound of $m$. Except, you can't have a tree with $n = 3$ and $n_2 = 0$!
$endgroup$
– OmG
Jan 20 at 8:37
add a comment |
$begingroup$
$m$ could not be greater than $n$ as the number of nodes is $n$. The upper bound for $m$ is $n-1$ (as an example a star). Hence, $m leq n-1$.
answered Jan 20 at 8:15
OmGOmG
2,502822
$endgroup$
$m$ could not be greater than $n$ as the number of nodes is $n$. The upper bound for $m$ is $n-1$ (as an example a star). Hence, $m leq n-1$.
answered Jan 20 at 8:15
OmGOmG
2,502822
answered Jan 20 at 8:15
OmGOmG
2,502822
answered Jan 20 at 8:15
OmGOmG
2,502822
answered Jan 20 at 8:15
OmGOmG
2,502822
2,502822
$begingroup$
Oh sorry you are right, I don't know why I wrote this nonsense. However I don't see how $n_2 =0$ impacts the calculation.
$endgroup$
– NotAbelianGroup
Jan 20 at 8:25
$begingroup$
@NotAbelianGroup it's ok. $n_2 = 0$ has no impact on the upper bound of $m$. Except, you can't have a tree with $n = 3$ and $n_2 = 0$!
$endgroup$
– OmG
Jan 20 at 8:37
add a comment |
$begingroup$
Oh sorry you are right, I don't know why I wrote this nonsense. However I don't see how $n_2 =0$ impacts the calculation.
$endgroup$
– NotAbelianGroup
Jan 20 at 8:25
$begingroup$
@NotAbelianGroup it's ok. $n_2 = 0$ has no impact on the upper bound of $m$. Except, you can't have a tree with $n = 3$ and $n_2 = 0$!
$endgroup$
– OmG
Jan 20 at 8:37
$begingroup$
Oh sorry you are right, I don't know why I wrote this nonsense. However I don't see how $n_2 =0$ impacts the calculation.
$endgroup$
– NotAbelianGroup
Jan 20 at 8:25
$begingroup$
Oh sorry you are right, I don't know why I wrote this nonsense. However I don't see how $n_2 =0$ impacts the calculation.
$endgroup$
– NotAbelianGroup
Jan 20 at 8:25
$begingroup$
@NotAbelianGroup it's ok. $n_2 = 0$ has no impact on the upper bound of $m$. Except, you can't have a tree with $n = 3$ and $n_2 = 0$!
$endgroup$
– OmG
Jan 20 at 8:37
$begingroup$
@NotAbelianGroup it's ok. $n_2 = 0$ has no impact on the upper bound of $m$. Except, you can't have a tree with $n = 3$ and $n_2 = 0$!
$endgroup$
– OmG
Jan 20 at 8:37
add a comment |
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