Graph theory in disguise?
$begingroup$
There are $2n-1$ two-element subsets of set ${1,2,...,n}$. Prove that one can choose $n$ out of these such that their union contains no more than $frac{2}{3}n+1$ elements.
I was trying this one too and with no success. Then I read the official solution and it is a kind of not natural to me. Can someone try to walk another way?
combinatorics graph-theory optimization extremal-combinatorics probabilistic-method
asked Jul 15 '17 at 16:10
greedoidgreedoid
44k1155109
$endgroup$
add a comment |
$begingroup$
There are $2n-1$ two-element subsets of set ${1,2,...,n}$. Prove that one can choose $n$ out of these such that their union contains no more than $frac{2}{3}n+1$ elements.
I was trying this one too and with no success. Then I read the official solution and it is a kind of not natural to me. Can someone try to walk another way?
combinatorics graph-theory optimization extremal-combinatorics probabilistic-method
asked Jul 15 '17 at 16:10
greedoidgreedoid
44k1155109
$endgroup$
add a comment |
$begingroup$
There are $2n-1$ two-element subsets of set ${1,2,...,n}$. Prove that one can choose $n$ out of these such that their union contains no more than $frac{2}{3}n+1$ elements.
I was trying this one too and with no success. Then I read the official solution and it is a kind of not natural to me. Can someone try to walk another way?
combinatorics graph-theory optimization extremal-combinatorics probabilistic-method
asked Jul 15 '17 at 16:10
greedoidgreedoid
44k1155109
$endgroup$
There are $2n-1$ two-element subsets of set ${1,2,...,n}$. Prove that one can choose $n$ out of these such that their union contains no more than $frac{2}{3}n+1$ elements.
I was trying this one too and with no success. Then I read the official solution and it is a kind of not natural to me. Can someone try to walk another way?
combinatorics graph-theory optimization extremal-combinatorics probabilistic-method
combinatorics graph-theory optimization extremal-combinatorics probabilistic-method
asked Jul 15 '17 at 16:10
greedoidgreedoid
44k1155109
asked Jul 15 '17 at 16:10
greedoidgreedoid
44k1155109
edited Dec 11 '18 at 15:30
greedoid
asked Jul 15 '17 at 16:10
greedoidgreedoid
44k1155109
asked Jul 15 '17 at 16:10
greedoidgreedoid
44k1155109
asked Jul 15 '17 at 16:10
greedoidgreedoid
44k1155109
44k1155109
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
We are given a graph $G = (V, E)$ where $V = {v_1,dots,v_n}$ and $|E| = 2n - 1$. What we are trying to do is to remove as few edges as possible to leave as many isolated vertices as possible. To do this, we try to find a vertex of least degree and remove its edges.
The degree formula says that
$$ sum_{i = 1}^n deg(v_i) = 2|E| = 4n - 2. $$
It follows that there is a vertex in $G$ of degree $le 3$. If every vertex had degree $ge 4$ then $sum_{i = 1}^n deg(v_i) ge 4n$, which is impossible.
Without loss of generality, let this vertex be $v_n$. Then form the graph $G_1 = Gsetminus v = (V_1, E_1)$. Now $|E_1| ge 2n - 4$. Again by the degree formula we have
$$ sum_{i = 1}^{n - 1} deg(v_i) = 2|E_1| ge 4(n - 1) - 4 tag{1}$$
We want to conclude that there is a vertex of degree $le 3$. To conclude this, we would like to have an equality in $(1)$. To achieve this we allow ourselves to remove exactly three edges at each step (which might be more than the degree of the vertex we removed). Doing this, we have $2|E_1| = 4(n - 1) - 4$ and we get a vertex of degree $le 3$.
Inductively, suppose we remove the vertices $v_{n - k + 1}, dots, v_n$ (after possibly relabeling) and $3k$ edges to get the graph $G_k = (V_k, E_k)$ where $|E_k| = 2n - 1 - 3k$. Then we have
$$ sum_{i = 1}^{n - k} deg(v_i) = 2|E_k| = 4(n - k) - 2 - 2k $$
so there is a vertex of degree $le 3$.
answered Jul 15 '17 at 17:19
Trevor GunnTrevor Gunn
14.7k32047
$endgroup$
$begingroup$
How can you deduce this: $$ sum_{i = 1}^{n - 1} deg(v_i) = 2|E_1| ge 4(n - 1) - 4 $$ so again there is a vertex of degree $le 3$.
$endgroup$
– greedoid
Jul 15 '17 at 17:51
$begingroup$
And, what then? Where is a contradiction? With what?
$endgroup$
– greedoid
Jul 15 '17 at 17:56
$begingroup$
@guest It should make sense now.
$endgroup$
– Trevor Gunn
Jul 15 '17 at 18:03
1
$begingroup$
You didn't finished explicitly but I can see where are you going (everybody does). So a process will terminate when $|E_k|le 0$ and this is when $nle {3k+1over 2}$. So when $k = {2n+1over 3}$ we get the desired union. Great solution. Thanks! Do you have any idea how to approach with the probabilistic method?
$endgroup$
– greedoid
Jul 15 '17 at 18:12
$begingroup$
@guest It's the same solution as given in the booklet but with graph terminology. Unfortunately I don't know a lot about probabilistic methods in combinatorics.
$endgroup$
– Trevor Gunn
Jul 15 '17 at 18:19
|
show 1 more comment
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1 Answer
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$begingroup$
We are given a graph $G = (V, E)$ where $V = {v_1,dots,v_n}$ and $|E| = 2n - 1$. What we are trying to do is to remove as few edges as possible to leave as many isolated vertices as possible. To do this, we try to find a vertex of least degree and remove its edges.
The degree formula says that
$$ sum_{i = 1}^n deg(v_i) = 2|E| = 4n - 2. $$
It follows that there is a vertex in $G$ of degree $le 3$. If every vertex had degree $ge 4$ then $sum_{i = 1}^n deg(v_i) ge 4n$, which is impossible.
Without loss of generality, let this vertex be $v_n$. Then form the graph $G_1 = Gsetminus v = (V_1, E_1)$. Now $|E_1| ge 2n - 4$. Again by the degree formula we have
$$ sum_{i = 1}^{n - 1} deg(v_i) = 2|E_1| ge 4(n - 1) - 4 tag{1}$$
We want to conclude that there is a vertex of degree $le 3$. To conclude this, we would like to have an equality in $(1)$. To achieve this we allow ourselves to remove exactly three edges at each step (which might be more than the degree of the vertex we removed). Doing this, we have $2|E_1| = 4(n - 1) - 4$ and we get a vertex of degree $le 3$.
Inductively, suppose we remove the vertices $v_{n - k + 1}, dots, v_n$ (after possibly relabeling) and $3k$ edges to get the graph $G_k = (V_k, E_k)$ where $|E_k| = 2n - 1 - 3k$. Then we have
$$ sum_{i = 1}^{n - k} deg(v_i) = 2|E_k| = 4(n - k) - 2 - 2k $$
so there is a vertex of degree $le 3$.
answered Jul 15 '17 at 17:19
Trevor GunnTrevor Gunn
14.7k32047
$endgroup$
$begingroup$
How can you deduce this: $$ sum_{i = 1}^{n - 1} deg(v_i) = 2|E_1| ge 4(n - 1) - 4 $$ so again there is a vertex of degree $le 3$.
$endgroup$
– greedoid
Jul 15 '17 at 17:51
$begingroup$
And, what then? Where is a contradiction? With what?
$endgroup$
– greedoid
Jul 15 '17 at 17:56
$begingroup$
@guest It should make sense now.
$endgroup$
– Trevor Gunn
Jul 15 '17 at 18:03
1
$begingroup$
You didn't finished explicitly but I can see where are you going (everybody does). So a process will terminate when $|E_k|le 0$ and this is when $nle {3k+1over 2}$. So when $k = {2n+1over 3}$ we get the desired union. Great solution. Thanks! Do you have any idea how to approach with the probabilistic method?
$endgroup$
– greedoid
Jul 15 '17 at 18:12
$begingroup$
@guest It's the same solution as given in the booklet but with graph terminology. Unfortunately I don't know a lot about probabilistic methods in combinatorics.
$endgroup$
– Trevor Gunn
Jul 15 '17 at 18:19
|
show 1 more comment
$begingroup$
We are given a graph $G = (V, E)$ where $V = {v_1,dots,v_n}$ and $|E| = 2n - 1$. What we are trying to do is to remove as few edges as possible to leave as many isolated vertices as possible. To do this, we try to find a vertex of least degree and remove its edges.
The degree formula says that
$$ sum_{i = 1}^n deg(v_i) = 2|E| = 4n - 2. $$
It follows that there is a vertex in $G$ of degree $le 3$. If every vertex had degree $ge 4$ then $sum_{i = 1}^n deg(v_i) ge 4n$, which is impossible.
Without loss of generality, let this vertex be $v_n$. Then form the graph $G_1 = Gsetminus v = (V_1, E_1)$. Now $|E_1| ge 2n - 4$. Again by the degree formula we have
$$ sum_{i = 1}^{n - 1} deg(v_i) = 2|E_1| ge 4(n - 1) - 4 tag{1}$$
We want to conclude that there is a vertex of degree $le 3$. To conclude this, we would like to have an equality in $(1)$. To achieve this we allow ourselves to remove exactly three edges at each step (which might be more than the degree of the vertex we removed). Doing this, we have $2|E_1| = 4(n - 1) - 4$ and we get a vertex of degree $le 3$.
Inductively, suppose we remove the vertices $v_{n - k + 1}, dots, v_n$ (after possibly relabeling) and $3k$ edges to get the graph $G_k = (V_k, E_k)$ where $|E_k| = 2n - 1 - 3k$. Then we have
$$ sum_{i = 1}^{n - k} deg(v_i) = 2|E_k| = 4(n - k) - 2 - 2k $$
so there is a vertex of degree $le 3$.
answered Jul 15 '17 at 17:19
Trevor GunnTrevor Gunn
14.7k32047
$endgroup$
$begingroup$
How can you deduce this: $$ sum_{i = 1}^{n - 1} deg(v_i) = 2|E_1| ge 4(n - 1) - 4 $$ so again there is a vertex of degree $le 3$.
$endgroup$
– greedoid
Jul 15 '17 at 17:51
$begingroup$
And, what then? Where is a contradiction? With what?
$endgroup$
– greedoid
Jul 15 '17 at 17:56
$begingroup$
@guest It should make sense now.
$endgroup$
– Trevor Gunn
Jul 15 '17 at 18:03
1
$begingroup$
You didn't finished explicitly but I can see where are you going (everybody does). So a process will terminate when $|E_k|le 0$ and this is when $nle {3k+1over 2}$. So when $k = {2n+1over 3}$ we get the desired union. Great solution. Thanks! Do you have any idea how to approach with the probabilistic method?
$endgroup$
– greedoid
Jul 15 '17 at 18:12
$begingroup$
@guest It's the same solution as given in the booklet but with graph terminology. Unfortunately I don't know a lot about probabilistic methods in combinatorics.
$endgroup$
– Trevor Gunn
Jul 15 '17 at 18:19
|
show 1 more comment
$begingroup$
We are given a graph $G = (V, E)$ where $V = {v_1,dots,v_n}$ and $|E| = 2n - 1$. What we are trying to do is to remove as few edges as possible to leave as many isolated vertices as possible. To do this, we try to find a vertex of least degree and remove its edges.
The degree formula says that
$$ sum_{i = 1}^n deg(v_i) = 2|E| = 4n - 2. $$
It follows that there is a vertex in $G$ of degree $le 3$. If every vertex had degree $ge 4$ then $sum_{i = 1}^n deg(v_i) ge 4n$, which is impossible.
Without loss of generality, let this vertex be $v_n$. Then form the graph $G_1 = Gsetminus v = (V_1, E_1)$. Now $|E_1| ge 2n - 4$. Again by the degree formula we have
$$ sum_{i = 1}^{n - 1} deg(v_i) = 2|E_1| ge 4(n - 1) - 4 tag{1}$$
We want to conclude that there is a vertex of degree $le 3$. To conclude this, we would like to have an equality in $(1)$. To achieve this we allow ourselves to remove exactly three edges at each step (which might be more than the degree of the vertex we removed). Doing this, we have $2|E_1| = 4(n - 1) - 4$ and we get a vertex of degree $le 3$.
Inductively, suppose we remove the vertices $v_{n - k + 1}, dots, v_n$ (after possibly relabeling) and $3k$ edges to get the graph $G_k = (V_k, E_k)$ where $|E_k| = 2n - 1 - 3k$. Then we have
$$ sum_{i = 1}^{n - k} deg(v_i) = 2|E_k| = 4(n - k) - 2 - 2k $$
so there is a vertex of degree $le 3$.
answered Jul 15 '17 at 17:19
Trevor GunnTrevor Gunn
14.7k32047
$endgroup$
We are given a graph $G = (V, E)$ where $V = {v_1,dots,v_n}$ and $|E| = 2n - 1$. What we are trying to do is to remove as few edges as possible to leave as many isolated vertices as possible. To do this, we try to find a vertex of least degree and remove its edges.
The degree formula says that
$$ sum_{i = 1}^n deg(v_i) = 2|E| = 4n - 2. $$
It follows that there is a vertex in $G$ of degree $le 3$. If every vertex had degree $ge 4$ then $sum_{i = 1}^n deg(v_i) ge 4n$, which is impossible.
Without loss of generality, let this vertex be $v_n$. Then form the graph $G_1 = Gsetminus v = (V_1, E_1)$. Now $|E_1| ge 2n - 4$. Again by the degree formula we have
$$ sum_{i = 1}^{n - 1} deg(v_i) = 2|E_1| ge 4(n - 1) - 4 tag{1}$$
We want to conclude that there is a vertex of degree $le 3$. To conclude this, we would like to have an equality in $(1)$. To achieve this we allow ourselves to remove exactly three edges at each step (which might be more than the degree of the vertex we removed). Doing this, we have $2|E_1| = 4(n - 1) - 4$ and we get a vertex of degree $le 3$.
Inductively, suppose we remove the vertices $v_{n - k + 1}, dots, v_n$ (after possibly relabeling) and $3k$ edges to get the graph $G_k = (V_k, E_k)$ where $|E_k| = 2n - 1 - 3k$. Then we have
$$ sum_{i = 1}^{n - k} deg(v_i) = 2|E_k| = 4(n - k) - 2 - 2k $$
so there is a vertex of degree $le 3$.
answered Jul 15 '17 at 17:19
Trevor GunnTrevor Gunn
14.7k32047
edited Jul 15 '17 at 18:03
answered Jul 15 '17 at 17:19
Trevor GunnTrevor Gunn
14.7k32047
answered Jul 15 '17 at 17:19
Trevor GunnTrevor Gunn
14.7k32047
answered Jul 15 '17 at 17:19
Trevor GunnTrevor Gunn
14.7k32047
14.7k32047
$begingroup$
How can you deduce this: $$ sum_{i = 1}^{n - 1} deg(v_i) = 2|E_1| ge 4(n - 1) - 4 $$ so again there is a vertex of degree $le 3$.
$endgroup$
– greedoid
Jul 15 '17 at 17:51
$begingroup$
And, what then? Where is a contradiction? With what?
$endgroup$
– greedoid
Jul 15 '17 at 17:56
$begingroup$
@guest It should make sense now.
$endgroup$
– Trevor Gunn
Jul 15 '17 at 18:03
1
$begingroup$
You didn't finished explicitly but I can see where are you going (everybody does). So a process will terminate when $|E_k|le 0$ and this is when $nle {3k+1over 2}$. So when $k = {2n+1over 3}$ we get the desired union. Great solution. Thanks! Do you have any idea how to approach with the probabilistic method?
$endgroup$
– greedoid
Jul 15 '17 at 18:12
$begingroup$
@guest It's the same solution as given in the booklet but with graph terminology. Unfortunately I don't know a lot about probabilistic methods in combinatorics.
$endgroup$
– Trevor Gunn
Jul 15 '17 at 18:19
|
show 1 more comment
$begingroup$
How can you deduce this: $$ sum_{i = 1}^{n - 1} deg(v_i) = 2|E_1| ge 4(n - 1) - 4 $$ so again there is a vertex of degree $le 3$.
$endgroup$
– greedoid
Jul 15 '17 at 17:51
$begingroup$
And, what then? Where is a contradiction? With what?
$endgroup$
– greedoid
Jul 15 '17 at 17:56
$begingroup$
@guest It should make sense now.
$endgroup$
– Trevor Gunn
Jul 15 '17 at 18:03
1
$begingroup$
You didn't finished explicitly but I can see where are you going (everybody does). So a process will terminate when $|E_k|le 0$ and this is when $nle {3k+1over 2}$. So when $k = {2n+1over 3}$ we get the desired union. Great solution. Thanks! Do you have any idea how to approach with the probabilistic method?
$endgroup$
– greedoid
Jul 15 '17 at 18:12
$begingroup$
@guest It's the same solution as given in the booklet but with graph terminology. Unfortunately I don't know a lot about probabilistic methods in combinatorics.
$endgroup$
– Trevor Gunn
Jul 15 '17 at 18:19
$begingroup$
How can you deduce this: $$ sum_{i = 1}^{n - 1} deg(v_i) = 2|E_1| ge 4(n - 1) - 4 $$ so again there is a vertex of degree $le 3$.
$endgroup$
– greedoid
Jul 15 '17 at 17:51
$begingroup$
How can you deduce this: $$ sum_{i = 1}^{n - 1} deg(v_i) = 2|E_1| ge 4(n - 1) - 4 $$ so again there is a vertex of degree $le 3$.
$endgroup$
– greedoid
Jul 15 '17 at 17:51
$begingroup$
And, what then? Where is a contradiction? With what?
$endgroup$
– greedoid
Jul 15 '17 at 17:56
$begingroup$
And, what then? Where is a contradiction? With what?
$endgroup$
– greedoid
Jul 15 '17 at 17:56
$begingroup$
@guest It should make sense now.
$endgroup$
– Trevor Gunn
Jul 15 '17 at 18:03
$begingroup$
@guest It should make sense now.
$endgroup$
– Trevor Gunn
Jul 15 '17 at 18:03
1
1
$begingroup$
You didn't finished explicitly but I can see where are you going (everybody does). So a process will terminate when $|E_k|le 0$ and this is when $nle {3k+1over 2}$. So when $k = {2n+1over 3}$ we get the desired union. Great solution. Thanks! Do you have any idea how to approach with the probabilistic method?
$endgroup$
– greedoid
Jul 15 '17 at 18:12
$begingroup$
You didn't finished explicitly but I can see where are you going (everybody does). So a process will terminate when $|E_k|le 0$ and this is when $nle {3k+1over 2}$. So when $k = {2n+1over 3}$ we get the desired union. Great solution. Thanks! Do you have any idea how to approach with the probabilistic method?
$endgroup$
– greedoid
Jul 15 '17 at 18:12
$begingroup$
@guest It's the same solution as given in the booklet but with graph terminology. Unfortunately I don't know a lot about probabilistic methods in combinatorics.
$endgroup$
– Trevor Gunn
Jul 15 '17 at 18:19
$begingroup$
@guest It's the same solution as given in the booklet but with graph terminology. Unfortunately I don't know a lot about probabilistic methods in combinatorics.
$endgroup$
– Trevor Gunn
Jul 15 '17 at 18:19
|
show 1 more comment
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