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
$begingroup$
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.
graph-theory
$endgroup$
add a comment |
$begingroup$
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.
graph-theory
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
graph-theory
$endgroup$
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
graph-theory
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $chi(G)leq ell k$.
$endgroup$
$begingroup$
I still can't figure this out. Could you elaborate? Thanks.
$endgroup$
– Josh Ng
Mar 14 at 17:28
add a comment |
Your Answer
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2 Answers
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oldest
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2 Answers
2
active
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active
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votes
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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 $chi(G)leq ell k$.
$endgroup$
$begingroup$
I still can't figure this out. Could you elaborate? Thanks.
$endgroup$
– Josh Ng
Mar 14 at 17:28
add a comment |
$begingroup$
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 $chi(G)leq ell k$.
$endgroup$
$begingroup$
I still can't figure this out. Could you elaborate? Thanks.
$endgroup$
– Josh Ng
Mar 14 at 17:28
add a comment |
$begingroup$
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 $chi(G)leq ell k$.
$endgroup$
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 $chi(G)leq ell k$.
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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