Mixing
Show that: $chi(G) + chi(overline{G}) leq |V| + 1$
$begingroup$
Show that: $chi(G) + chi(overline{G}) leq |V| + 1$
I have problem with starting with this task. I have already done similar ones like for example $chi(G) * chi(overline{G}) geq |V|$, but I can't really find the proper way of thinking to solve the task mentioned at the beginning.
Any tips?
discrete-mathematics graph-theory coloring
edited Jan 11 at 14:02
EdOverflow
25319
asked Jan 11 at 14:00
James SmithJames Smith
34817
$endgroup$
add a comment |
$begingroup$
Show that: $chi(G) + chi(overline{G}) leq |V| + 1$
I have problem with starting with this task. I have already done similar ones like for example $chi(G) * chi(overline{G}) geq |V|$, but I can't really find the proper way of thinking to solve the task mentioned at the beginning.
Any tips?
discrete-mathematics graph-theory coloring
edited Jan 11 at 14:02
EdOverflow
25319
asked Jan 11 at 14:00
James SmithJames Smith
34817
$endgroup$
$begingroup$
Have you tried induction on the number of vertices?
$endgroup$
– bof
Jan 11 at 14:03
$begingroup$
Not yet. Don't really know how to use it in this case as I haven't seen usage of induction in tasks of this type, but I will try to check if I will be to figure out something with it. Thanks.
$endgroup$
– James Smith
Jan 11 at 14:12
add a comment |
$begingroup$
Show that: $chi(G) + chi(overline{G}) leq |V| + 1$
I have problem with starting with this task. I have already done similar ones like for example $chi(G) * chi(overline{G}) geq |V|$, but I can't really find the proper way of thinking to solve the task mentioned at the beginning.
Any tips?
discrete-mathematics graph-theory coloring
edited Jan 11 at 14:02
EdOverflow
25319
asked Jan 11 at 14:00
James SmithJames Smith
34817
$endgroup$
Show that: $chi(G) + chi(overline{G}) leq |V| + 1$
I have problem with starting with this task. I have already done similar ones like for example $chi(G) * chi(overline{G}) geq |V|$, but I can't really find the proper way of thinking to solve the task mentioned at the beginning.
Any tips?
discrete-mathematics graph-theory coloring
discrete-mathematics graph-theory coloring
edited Jan 11 at 14:02
EdOverflow
25319
asked Jan 11 at 14:00
James SmithJames Smith
34817
edited Jan 11 at 14:02
EdOverflow
25319
asked Jan 11 at 14:00
James SmithJames Smith
34817
edited Jan 11 at 14:02
EdOverflow
25319
edited Jan 11 at 14:02
EdOverflow
25319
edited Jan 11 at 14:02
EdOverflow
25319
25319
asked Jan 11 at 14:00
James SmithJames Smith
34817
asked Jan 11 at 14:00
James SmithJames Smith
34817
asked Jan 11 at 14:00
James SmithJames Smith
34817
34817
$begingroup$
Have you tried induction on the number of vertices?
$endgroup$
– bof
Jan 11 at 14:03
$begingroup$
Not yet. Don't really know how to use it in this case as I haven't seen usage of induction in tasks of this type, but I will try to check if I will be to figure out something with it. Thanks.
$endgroup$
– James Smith
Jan 11 at 14:12
add a comment |
$begingroup$
Have you tried induction on the number of vertices?
$endgroup$
– bof
Jan 11 at 14:03
$begingroup$
Not yet. Don't really know how to use it in this case as I haven't seen usage of induction in tasks of this type, but I will try to check if I will be to figure out something with it. Thanks.
$endgroup$
– James Smith
Jan 11 at 14:12
$begingroup$
Have you tried induction on the number of vertices?
$endgroup$
– bof
Jan 11 at 14:03
$begingroup$
Have you tried induction on the number of vertices?
$endgroup$
– bof
Jan 11 at 14:03
$begingroup$
Not yet. Don't really know how to use it in this case as I haven't seen usage of induction in tasks of this type, but I will try to check if I will be to figure out something with it. Thanks.
$endgroup$
– James Smith
Jan 11 at 14:12
$begingroup$
Not yet. Don't really know how to use it in this case as I haven't seen usage of induction in tasks of this type, but I will try to check if I will be to figure out something with it. Thanks.
$endgroup$
– James Smith
Jan 11 at 14:12
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint. This can be proved by induction on $n(G)$, the number of vertices.
Choose any vertex $v$ of $G$. By the inductive hypothesis we have:
$$chi(G-v)+chi(overline{G-v})le n(G-v)+1=n(G)tag1$$
and we also have
$$chi(G)lechi(G-v)+1tag2$$
and
$$chi(overline G)lechi(overline{G-v})+1tag3$$
whence
$$chi(G)+chi(overline G)le n(G)+2.tag4$$
All we have to do now is show that equality can't hold in $(4)$. Well, for equality to hold in $(4)$, we must have equality in $(1)$, $(2)$, and $(3)$. But then . . .
Further hint:
Equality in $(2)$ implies $deg_G(v)gechi(G-v)$
answered Jan 11 at 14:28
bofbof
51.5k558120
$endgroup$
$begingroup$
Emm... A lot of time has passed since I asked that question, but was trying to focus on other classes. So if we would like to have equality in (2) then $deg(v) = Delta(G)$ or am I wrong? In that case it would not be possible to have equality in (3), right?
$endgroup$
– James Smith
Jan 22 at 19:18
1
$begingroup$
As I said in my Further hint, equality in $(2)$ implies $deg_G(v)gechi(G-v)$. If all the inequalities are equalities, then we have $$n(G)-1=deg_G(v)+deg_{overline G}(v)gechi(G-v)+chi(overline{G-v})=n(G),$$ i.e., $n(G)-1ge n(G)$, a contradiction.
$endgroup$
– bof
Jan 23 at 0:40
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Hint. This can be proved by induction on $n(G)$, the number of vertices.
Choose any vertex $v$ of $G$. By the inductive hypothesis we have:
$$chi(G-v)+chi(overline{G-v})le n(G-v)+1=n(G)tag1$$
and we also have
$$chi(G)lechi(G-v)+1tag2$$
and
$$chi(overline G)lechi(overline{G-v})+1tag3$$
whence
$$chi(G)+chi(overline G)le n(G)+2.tag4$$
All we have to do now is show that equality can't hold in $(4)$. Well, for equality to hold in $(4)$, we must have equality in $(1)$, $(2)$, and $(3)$. But then . . .
Further hint:
Equality in $(2)$ implies $deg_G(v)gechi(G-v)$
answered Jan 11 at 14:28
bofbof
51.5k558120
$endgroup$
$begingroup$
Emm... A lot of time has passed since I asked that question, but was trying to focus on other classes. So if we would like to have equality in (2) then $deg(v) = Delta(G)$ or am I wrong? In that case it would not be possible to have equality in (3), right?
$endgroup$
– James Smith
Jan 22 at 19:18
1
$begingroup$
As I said in my Further hint, equality in $(2)$ implies $deg_G(v)gechi(G-v)$. If all the inequalities are equalities, then we have $$n(G)-1=deg_G(v)+deg_{overline G}(v)gechi(G-v)+chi(overline{G-v})=n(G),$$ i.e., $n(G)-1ge n(G)$, a contradiction.
$endgroup$
– bof
Jan 23 at 0:40
add a comment |
$begingroup$
Hint. This can be proved by induction on $n(G)$, the number of vertices.
Choose any vertex $v$ of $G$. By the inductive hypothesis we have:
$$chi(G-v)+chi(overline{G-v})le n(G-v)+1=n(G)tag1$$
and we also have
$$chi(G)lechi(G-v)+1tag2$$
and
$$chi(overline G)lechi(overline{G-v})+1tag3$$
whence
$$chi(G)+chi(overline G)le n(G)+2.tag4$$
All we have to do now is show that equality can't hold in $(4)$. Well, for equality to hold in $(4)$, we must have equality in $(1)$, $(2)$, and $(3)$. But then . . .
Further hint:
Equality in $(2)$ implies $deg_G(v)gechi(G-v)$
answered Jan 11 at 14:28
bofbof
51.5k558120
$endgroup$
$begingroup$
Emm... A lot of time has passed since I asked that question, but was trying to focus on other classes. So if we would like to have equality in (2) then $deg(v) = Delta(G)$ or am I wrong? In that case it would not be possible to have equality in (3), right?
$endgroup$
– James Smith
Jan 22 at 19:18
1
$begingroup$
As I said in my Further hint, equality in $(2)$ implies $deg_G(v)gechi(G-v)$. If all the inequalities are equalities, then we have $$n(G)-1=deg_G(v)+deg_{overline G}(v)gechi(G-v)+chi(overline{G-v})=n(G),$$ i.e., $n(G)-1ge n(G)$, a contradiction.
$endgroup$
– bof
Jan 23 at 0:40
add a comment |
$begingroup$
Hint. This can be proved by induction on $n(G)$, the number of vertices.
Choose any vertex $v$ of $G$. By the inductive hypothesis we have:
$$chi(G-v)+chi(overline{G-v})le n(G-v)+1=n(G)tag1$$
and we also have
$$chi(G)lechi(G-v)+1tag2$$
and
$$chi(overline G)lechi(overline{G-v})+1tag3$$
whence
$$chi(G)+chi(overline G)le n(G)+2.tag4$$
All we have to do now is show that equality can't hold in $(4)$. Well, for equality to hold in $(4)$, we must have equality in $(1)$, $(2)$, and $(3)$. But then . . .
Further hint:
Equality in $(2)$ implies $deg_G(v)gechi(G-v)$
answered Jan 11 at 14:28
bofbof
51.5k558120
$endgroup$
Hint. This can be proved by induction on $n(G)$, the number of vertices.
Choose any vertex $v$ of $G$. By the inductive hypothesis we have:
$$chi(G-v)+chi(overline{G-v})le n(G-v)+1=n(G)tag1$$
and we also have
$$chi(G)lechi(G-v)+1tag2$$
and
$$chi(overline G)lechi(overline{G-v})+1tag3$$
whence
$$chi(G)+chi(overline G)le n(G)+2.tag4$$
All we have to do now is show that equality can't hold in $(4)$. Well, for equality to hold in $(4)$, we must have equality in $(1)$, $(2)$, and $(3)$. But then . . .
Further hint:
Equality in $(2)$ implies $deg_G(v)gechi(G-v)$
answered Jan 11 at 14:28
bofbof
51.5k558120
edited Jan 12 at 6:54
answered Jan 11 at 14:28
bofbof
51.5k558120
answered Jan 11 at 14:28
bofbof
51.5k558120
answered Jan 11 at 14:28
bofbof
51.5k558120
51.5k558120
$begingroup$
Emm... A lot of time has passed since I asked that question, but was trying to focus on other classes. So if we would like to have equality in (2) then $deg(v) = Delta(G)$ or am I wrong? In that case it would not be possible to have equality in (3), right?
$endgroup$
– James Smith
Jan 22 at 19:18
1
$begingroup$
As I said in my Further hint, equality in $(2)$ implies $deg_G(v)gechi(G-v)$. If all the inequalities are equalities, then we have $$n(G)-1=deg_G(v)+deg_{overline G}(v)gechi(G-v)+chi(overline{G-v})=n(G),$$ i.e., $n(G)-1ge n(G)$, a contradiction.
$endgroup$
– bof
Jan 23 at 0:40
add a comment |
$begingroup$
Emm... A lot of time has passed since I asked that question, but was trying to focus on other classes. So if we would like to have equality in (2) then $deg(v) = Delta(G)$ or am I wrong? In that case it would not be possible to have equality in (3), right?
$endgroup$
– James Smith
Jan 22 at 19:18
1
$begingroup$
As I said in my Further hint, equality in $(2)$ implies $deg_G(v)gechi(G-v)$. If all the inequalities are equalities, then we have $$n(G)-1=deg_G(v)+deg_{overline G}(v)gechi(G-v)+chi(overline{G-v})=n(G),$$ i.e., $n(G)-1ge n(G)$, a contradiction.
$endgroup$
– bof
Jan 23 at 0:40
$begingroup$
Emm... A lot of time has passed since I asked that question, but was trying to focus on other classes. So if we would like to have equality in (2) then $deg(v) = Delta(G)$ or am I wrong? In that case it would not be possible to have equality in (3), right?
$endgroup$
– James Smith
Jan 22 at 19:18
$begingroup$
Emm... A lot of time has passed since I asked that question, but was trying to focus on other classes. So if we would like to have equality in (2) then $deg(v) = Delta(G)$ or am I wrong? In that case it would not be possible to have equality in (3), right?
$endgroup$
– James Smith
Jan 22 at 19:18
1
1
$begingroup$
As I said in my Further hint, equality in $(2)$ implies $deg_G(v)gechi(G-v)$. If all the inequalities are equalities, then we have $$n(G)-1=deg_G(v)+deg_{overline G}(v)gechi(G-v)+chi(overline{G-v})=n(G),$$ i.e., $n(G)-1ge n(G)$, a contradiction.
$endgroup$
– bof
Jan 23 at 0:40
$begingroup$
As I said in my Further hint, equality in $(2)$ implies $deg_G(v)gechi(G-v)$. If all the inequalities are equalities, then we have $$n(G)-1=deg_G(v)+deg_{overline G}(v)gechi(G-v)+chi(overline{G-v})=n(G),$$ i.e., $n(G)-1ge n(G)$, a contradiction.
$endgroup$
– bof
Jan 23 at 0:40
add a comment |
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$begingroup$
Have you tried induction on the number of vertices?
$endgroup$
– bof
Jan 11 at 14:03
$begingroup$
Not yet. Don't really know how to use it in this case as I haven't seen usage of induction in tasks of this type, but I will try to check if I will be to figure out something with it. Thanks.
$endgroup$
– James Smith
Jan 11 at 14:12