Show that: $chi(G) + chi(overline{G}) leq |V| + 1$












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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












1








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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 14:02









EdOverflow

25519




25519










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




















  • $begingroup$
    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












1 Answer
1






active

oldest

votes


















3












$begingroup$

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 $deg_G(v)gechi(G-v)$







share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

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 $deg_G(v)gechi(G-v)$







share|cite|improve this answer











$endgroup$













  • $begingroup$
    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


















3












$begingroup$

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 $deg_G(v)gechi(G-v)$







share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















3












3








3





$begingroup$

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 $deg_G(v)gechi(G-v)$







share|cite|improve this answer











$endgroup$



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 $deg_G(v)gechi(G-v)$








share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 12 at 6:54

























answered Jan 11 at 14:28









bofbof

52.6k559121




52.6k559121












  • $begingroup$
    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








  • 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




















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