The chromatic number of a triangle-free graph.If G is a graph such that $Delta(G) = 3$ then $chi'(G) le 4$Upper bound on $chi(G)$ for a triangle-free graphEdge chromatic number question (Graph theory)Problem in Chromatic NumberFind an example of a regular triangle-free $4$-chromatic graphGraph with chromatic index 1 billion more than max degree?A triangle-free, 6-chromatic graph with 44 verticesAny planar graph of maximum degree $4$ has chromatic number at most $4$.Prove that the chromatic number of a graph is the same as the maximum of the chromatic numbers its blocks.Total chromatic number bound implied by list coloring conjecture

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The chromatic number of a triangle-free graph.


If G is a graph such that $Delta(G) = 3$ then $chi'(G) le 4$Upper bound on $chi(G)$ for a triangle-free graphEdge chromatic number question (Graph theory)Problem in Chromatic NumberFind an example of a regular triangle-free $4$-chromatic graphGraph with chromatic index 1 billion more than max degree?A triangle-free, 6-chromatic graph with 44 verticesAny planar graph of maximum degree $4$ has chromatic number at most $4$.Prove that the chromatic number of a graph is the same as the maximum of the chromatic numbers its blocks.Total chromatic number bound implied by list coloring conjecture













6












$begingroup$



Let $G$ be a triangle-free graph. Prove that $$chi(G)leq3leftlceildfracDelta(G)+14rightrceil$$




What's the relationship between the chromatic number and the maximum degree of a triangle-free graph $G$. I got a hint that I could apply the Brook's Theorem but I have no clue where to start.










share|cite|improve this question











$endgroup$











  • $begingroup$
    Note that (for $nge 3$) the inequality holds for every odd cycle that is not a triangle and that furthemore, a triangle-free graph is certainly not a complete graph.
    $endgroup$
    – Studentmath
    May 20 '14 at 11:29






  • 1




    $begingroup$
    @Studentmath: please elaborate. How does this help to show that e.g. a triangle-free graph with maximum degree 23 only requires 18 colors.
    $endgroup$
    – Leen Droogendijk
    May 20 '14 at 13:32











  • $begingroup$
    I still can't figure this out. Can anyone post an answer?
    $endgroup$
    – Josh Ng
    Mar 14 at 17:29















6












$begingroup$



Let $G$ be a triangle-free graph. Prove that $$chi(G)leq3leftlceildfracDelta(G)+14rightrceil$$




What's the relationship between the chromatic number and the maximum degree of a triangle-free graph $G$. I got a hint that I could apply the Brook's Theorem but I have no clue where to start.










share|cite|improve this question











$endgroup$











  • $begingroup$
    Note that (for $nge 3$) the inequality holds for every odd cycle that is not a triangle and that furthemore, a triangle-free graph is certainly not a complete graph.
    $endgroup$
    – Studentmath
    May 20 '14 at 11:29






  • 1




    $begingroup$
    @Studentmath: please elaborate. How does this help to show that e.g. a triangle-free graph with maximum degree 23 only requires 18 colors.
    $endgroup$
    – Leen Droogendijk
    May 20 '14 at 13:32











  • $begingroup$
    I still can't figure this out. Can anyone post an answer?
    $endgroup$
    – Josh Ng
    Mar 14 at 17:29













6












6








6


1



$begingroup$



Let $G$ be a triangle-free graph. Prove that $$chi(G)leq3leftlceildfracDelta(G)+14rightrceil$$




What's the relationship between the chromatic number and the maximum degree of a triangle-free graph $G$. I got a hint that I could apply the Brook's Theorem but I have no clue where to start.










share|cite|improve this question











$endgroup$





Let $G$ be a triangle-free graph. Prove that $$chi(G)leq3leftlceildfracDelta(G)+14rightrceil$$




What's the relationship between the chromatic number and the maximum degree of a triangle-free graph $G$. I got a hint that I could apply the Brook's Theorem but I have no clue where to start.







graph-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 15 at 15:47









MarianD

1,4741616




1,4741616










asked May 20 '14 at 10:06









user137988user137988

675




675











  • $begingroup$
    Note that (for $nge 3$) the inequality holds for every odd cycle that is not a triangle and that furthemore, a triangle-free graph is certainly not a complete graph.
    $endgroup$
    – Studentmath
    May 20 '14 at 11:29






  • 1




    $begingroup$
    @Studentmath: please elaborate. How does this help to show that e.g. a triangle-free graph with maximum degree 23 only requires 18 colors.
    $endgroup$
    – Leen Droogendijk
    May 20 '14 at 13:32











  • $begingroup$
    I still can't figure this out. Can anyone post an answer?
    $endgroup$
    – Josh Ng
    Mar 14 at 17:29
















  • $begingroup$
    Note that (for $nge 3$) the inequality holds for every odd cycle that is not a triangle and that furthemore, a triangle-free graph is certainly not a complete graph.
    $endgroup$
    – Studentmath
    May 20 '14 at 11:29






  • 1




    $begingroup$
    @Studentmath: please elaborate. How does this help to show that e.g. a triangle-free graph with maximum degree 23 only requires 18 colors.
    $endgroup$
    – Leen Droogendijk
    May 20 '14 at 13:32











  • $begingroup$
    I still can't figure this out. Can anyone post an answer?
    $endgroup$
    – Josh Ng
    Mar 14 at 17:29















$begingroup$
Note that (for $nge 3$) the inequality holds for every odd cycle that is not a triangle and that furthemore, a triangle-free graph is certainly not a complete graph.
$endgroup$
– Studentmath
May 20 '14 at 11:29




$begingroup$
Note that (for $nge 3$) the inequality holds for every odd cycle that is not a triangle and that furthemore, a triangle-free graph is certainly not a complete graph.
$endgroup$
– Studentmath
May 20 '14 at 11:29




1




1




$begingroup$
@Studentmath: please elaborate. How does this help to show that e.g. a triangle-free graph with maximum degree 23 only requires 18 colors.
$endgroup$
– Leen Droogendijk
May 20 '14 at 13:32





$begingroup$
@Studentmath: please elaborate. How does this help to show that e.g. a triangle-free graph with maximum degree 23 only requires 18 colors.
$endgroup$
– Leen Droogendijk
May 20 '14 at 13:32













$begingroup$
I still can't figure this out. Can anyone post an answer?
$endgroup$
– Josh Ng
Mar 14 at 17:29




$begingroup$
I still can't figure this out. Can anyone post an answer?
$endgroup$
– Josh Ng
Mar 14 at 17:29










2 Answers
2






active

oldest

votes


















3





+50







$begingroup$

First take a partition of $V$ into $k=lceil(Delta+1)/4rceil$ classes $V_1,dots,V_k$ that minimizes the number of monochromatic edges. Note $Delta<4k.$ By the pigeonhole principle, for each vertex $vin V_j,$ there is some class $V_i$ such that at most three of the at most $Delta$ neighbors of $v$ lie in $V_i.$ By minimality, putting $v$ in $V_i$ instead of $V_j$ cannot decrease the number of monochromatic edges, so $v$ has at most three neighbors in $V_j.$ In other words $Delta(G[V_j])leq 3$ for each $j.$



By Brooks' theorem and the triangle-free assumption, each $G[V_j]$ has a 3-coloring. Using disjoint color sets for each $V_j$ we get a $3k$-coloring of $G$ as required.



A history of this result is given in the introduction to "Chromatic number, girth and maximal degree" by B. Bollobás. Improving the bound is an open problem.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Do you mean $k = left lceil(Delta + 1)/4right rceil$?
    $endgroup$
    – ensbana
    Mar 17 at 17:11










  • $begingroup$
    @ensbana: I did mean that, thank you
    $endgroup$
    – Dap
    Mar 17 at 18:37


















1












$begingroup$

Hint: If there exists a partition $V(G)$ into $k$ subsets $V_1,ldots,V_k$ such that $chi(G_i)leq ell$ for every $i$ (where $G_i$ is the subgraph of $G$ induced by $V_i$), then $chi(G)leq ell k$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    I still can't figure this out. Could you elaborate? Thanks.
    $endgroup$
    – Josh Ng
    Mar 14 at 17:28










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3





+50







$begingroup$

First take a partition of $V$ into $k=lceil(Delta+1)/4rceil$ classes $V_1,dots,V_k$ that minimizes the number of monochromatic edges. Note $Delta<4k.$ By the pigeonhole principle, for each vertex $vin V_j,$ there is some class $V_i$ such that at most three of the at most $Delta$ neighbors of $v$ lie in $V_i.$ By minimality, putting $v$ in $V_i$ instead of $V_j$ cannot decrease the number of monochromatic edges, so $v$ has at most three neighbors in $V_j.$ In other words $Delta(G[V_j])leq 3$ for each $j.$



By Brooks' theorem and the triangle-free assumption, each $G[V_j]$ has a 3-coloring. Using disjoint color sets for each $V_j$ we get a $3k$-coloring of $G$ as required.



A history of this result is given in the introduction to "Chromatic number, girth and maximal degree" by B. Bollobás. Improving the bound is an open problem.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Do you mean $k = left lceil(Delta + 1)/4right rceil$?
    $endgroup$
    – ensbana
    Mar 17 at 17:11










  • $begingroup$
    @ensbana: I did mean that, thank you
    $endgroup$
    – Dap
    Mar 17 at 18:37















3





+50







$begingroup$

First take a partition of $V$ into $k=lceil(Delta+1)/4rceil$ classes $V_1,dots,V_k$ that minimizes the number of monochromatic edges. Note $Delta<4k.$ By the pigeonhole principle, for each vertex $vin V_j,$ there is some class $V_i$ such that at most three of the at most $Delta$ neighbors of $v$ lie in $V_i.$ By minimality, putting $v$ in $V_i$ instead of $V_j$ cannot decrease the number of monochromatic edges, so $v$ has at most three neighbors in $V_j.$ In other words $Delta(G[V_j])leq 3$ for each $j.$



By Brooks' theorem and the triangle-free assumption, each $G[V_j]$ has a 3-coloring. Using disjoint color sets for each $V_j$ we get a $3k$-coloring of $G$ as required.



A history of this result is given in the introduction to "Chromatic number, girth and maximal degree" by B. Bollobás. Improving the bound is an open problem.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Do you mean $k = left lceil(Delta + 1)/4right rceil$?
    $endgroup$
    – ensbana
    Mar 17 at 17:11










  • $begingroup$
    @ensbana: I did mean that, thank you
    $endgroup$
    – Dap
    Mar 17 at 18:37













3





+50







3





+50



3




+50



$begingroup$

First take a partition of $V$ into $k=lceil(Delta+1)/4rceil$ classes $V_1,dots,V_k$ that minimizes the number of monochromatic edges. Note $Delta<4k.$ By the pigeonhole principle, for each vertex $vin V_j,$ there is some class $V_i$ such that at most three of the at most $Delta$ neighbors of $v$ lie in $V_i.$ By minimality, putting $v$ in $V_i$ instead of $V_j$ cannot decrease the number of monochromatic edges, so $v$ has at most three neighbors in $V_j.$ In other words $Delta(G[V_j])leq 3$ for each $j.$



By Brooks' theorem and the triangle-free assumption, each $G[V_j]$ has a 3-coloring. Using disjoint color sets for each $V_j$ we get a $3k$-coloring of $G$ as required.



A history of this result is given in the introduction to "Chromatic number, girth and maximal degree" by B. Bollobás. Improving the bound is an open problem.






share|cite|improve this answer











$endgroup$



First take a partition of $V$ into $k=lceil(Delta+1)/4rceil$ classes $V_1,dots,V_k$ that minimizes the number of monochromatic edges. Note $Delta<4k.$ By the pigeonhole principle, for each vertex $vin V_j,$ there is some class $V_i$ such that at most three of the at most $Delta$ neighbors of $v$ lie in $V_i.$ By minimality, putting $v$ in $V_i$ instead of $V_j$ cannot decrease the number of monochromatic edges, so $v$ has at most three neighbors in $V_j.$ In other words $Delta(G[V_j])leq 3$ for each $j.$



By Brooks' theorem and the triangle-free assumption, each $G[V_j]$ has a 3-coloring. Using disjoint color sets for each $V_j$ we get a $3k$-coloring of $G$ as required.



A history of this result is given in the introduction to "Chromatic number, girth and maximal degree" by B. Bollobás. Improving the bound is an open problem.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 17 at 18:37

























answered Mar 15 at 15:25









DapDap

18.9k842




18.9k842











  • $begingroup$
    Do you mean $k = left lceil(Delta + 1)/4right rceil$?
    $endgroup$
    – ensbana
    Mar 17 at 17:11










  • $begingroup$
    @ensbana: I did mean that, thank you
    $endgroup$
    – Dap
    Mar 17 at 18:37
















  • $begingroup$
    Do you mean $k = left lceil(Delta + 1)/4right rceil$?
    $endgroup$
    – ensbana
    Mar 17 at 17:11










  • $begingroup$
    @ensbana: I did mean that, thank you
    $endgroup$
    – Dap
    Mar 17 at 18:37















$begingroup$
Do you mean $k = left lceil(Delta + 1)/4right rceil$?
$endgroup$
– ensbana
Mar 17 at 17:11




$begingroup$
Do you mean $k = left lceil(Delta + 1)/4right rceil$?
$endgroup$
– ensbana
Mar 17 at 17:11












$begingroup$
@ensbana: I did mean that, thank you
$endgroup$
– Dap
Mar 17 at 18:37




$begingroup$
@ensbana: I did mean that, thank you
$endgroup$
– Dap
Mar 17 at 18:37











1












$begingroup$

Hint: If there exists a partition $V(G)$ into $k$ subsets $V_1,ldots,V_k$ such that $chi(G_i)leq ell$ for every $i$ (where $G_i$ is the subgraph of $G$ induced by $V_i$), then $chi(G)leq ell k$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    I still can't figure this out. Could you elaborate? Thanks.
    $endgroup$
    – Josh Ng
    Mar 14 at 17:28















1












$begingroup$

Hint: If there exists a partition $V(G)$ into $k$ subsets $V_1,ldots,V_k$ such that $chi(G_i)leq ell$ for every $i$ (where $G_i$ is the subgraph of $G$ induced by $V_i$), then $chi(G)leq ell k$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    I still can't figure this out. Could you elaborate? Thanks.
    $endgroup$
    – Josh Ng
    Mar 14 at 17:28













1












1








1





$begingroup$

Hint: If there exists a partition $V(G)$ into $k$ subsets $V_1,ldots,V_k$ such that $chi(G_i)leq ell$ for every $i$ (where $G_i$ is the subgraph of $G$ induced by $V_i$), then $chi(G)leq ell k$.






share|cite|improve this answer









$endgroup$



Hint: If there exists a partition $V(G)$ into $k$ subsets $V_1,ldots,V_k$ such that $chi(G_i)leq ell$ for every $i$ (where $G_i$ is the subgraph of $G$ induced by $V_i$), then $chi(G)leq ell k$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered May 21 '14 at 7:42









CasteelsCasteels

10k42234




10k42234











  • $begingroup$
    I still can't figure this out. Could you elaborate? Thanks.
    $endgroup$
    – Josh Ng
    Mar 14 at 17:28
















  • $begingroup$
    I still can't figure this out. Could you elaborate? Thanks.
    $endgroup$
    – Josh Ng
    Mar 14 at 17:28















$begingroup$
I still can't figure this out. Could you elaborate? Thanks.
$endgroup$
– Josh Ng
Mar 14 at 17:28




$begingroup$
I still can't figure this out. Could you elaborate? Thanks.
$endgroup$
– Josh Ng
Mar 14 at 17:28

















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