Exists a bound for threewidth in $Delta(G)leq 3$ graphs?
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Is it true that $Delta(G)leq3$ implies $TW(G)leq6$? (where $TW$ is the "treewidth"). I know that for $Delta(G)leq4$ there is no bound for treewidth because the $ntimes n$ grid has $TW=n$. For $Deltaleq2$ it is easy to see that $TWleq 2$, and the case $Deltaleq1$ is trivial. I think this is the right bound only because I cannot find bigger counterexamples; but I dont have any idea to prove this.
graph-theory
asked Dec 20 '18 at 2:22
Veridian DynamicsVeridian Dynamics
1,909312
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add a comment |
$begingroup$
Is it true that $Delta(G)leq3$ implies $TW(G)leq6$? (where $TW$ is the "treewidth"). I know that for $Delta(G)leq4$ there is no bound for treewidth because the $ntimes n$ grid has $TW=n$. For $Deltaleq2$ it is easy to see that $TWleq 2$, and the case $Deltaleq1$ is trivial. I think this is the right bound only because I cannot find bigger counterexamples; but I dont have any idea to prove this.
graph-theory
asked Dec 20 '18 at 2:22
Veridian DynamicsVeridian Dynamics
1,909312
$endgroup$
add a comment |
$begingroup$
Is it true that $Delta(G)leq3$ implies $TW(G)leq6$? (where $TW$ is the "treewidth"). I know that for $Delta(G)leq4$ there is no bound for treewidth because the $ntimes n$ grid has $TW=n$. For $Deltaleq2$ it is easy to see that $TWleq 2$, and the case $Deltaleq1$ is trivial. I think this is the right bound only because I cannot find bigger counterexamples; but I dont have any idea to prove this.
graph-theory
asked Dec 20 '18 at 2:22
Veridian DynamicsVeridian Dynamics
1,909312
$endgroup$
Is it true that $Delta(G)leq3$ implies $TW(G)leq6$? (where $TW$ is the "treewidth"). I know that for $Delta(G)leq4$ there is no bound for treewidth because the $ntimes n$ grid has $TW=n$. For $Deltaleq2$ it is easy to see that $TWleq 2$, and the case $Deltaleq1$ is trivial. I think this is the right bound only because I cannot find bigger counterexamples; but I dont have any idea to prove this.
graph-theory
graph-theory
asked Dec 20 '18 at 2:22
Veridian DynamicsVeridian Dynamics
1,909312
asked Dec 20 '18 at 2:22
Veridian DynamicsVeridian Dynamics
1,909312
asked Dec 20 '18 at 2:22
Veridian DynamicsVeridian Dynamics
1,909312
asked Dec 20 '18 at 2:22
Veridian DynamicsVeridian Dynamics
1,909312
asked Dec 20 '18 at 2:22
Veridian DynamicsVeridian Dynamics
1,909312
1,909312
add a comment |
add a comment |
2 Answers
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As part of their work on graph minors leading to the Robertson–Seymour theorem and the graph structure theorem, Neil Robertson and Paul Seymour proved that a family $F$ of finite graphs has unbounded treewidth if and only if the minors of graphs in $F$ include arbitrarily large square grid graphs, or equivalently subgraphs of the hexagonal tiling formed by intersecting it with arbitrarily large disks. (Wikipedia)
answered Dec 20 '18 at 7:54
Alex RavskyAlex Ravsky
40.4k32282
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add a comment |
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A simple proof goes this way:
Take $K_n$ the complete graph in $n$ vertices. Take $vin G$, replace it with a cycle of length $d(v)=n-1$ and connect the $d(v)$ vertices in a 1-to-1 fashion to a vertex of the cycle. The result is that every vertex in the cycle has degree $3$. Repeat this until no vertex of degree $>3$ left. The graph $G$ obtained is $3$-regular and (clearly) $K_nleq G$. Then, $text{TW}(G)geqtext{TW}(K_n)=n-1$. So, the treewidth is unbounded even in the class of $3$-regular graphs.
answered Dec 22 '18 at 17:14
Veridian DynamicsVeridian Dynamics
1,909312
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
As part of their work on graph minors leading to the Robertson–Seymour theorem and the graph structure theorem, Neil Robertson and Paul Seymour proved that a family $F$ of finite graphs has unbounded treewidth if and only if the minors of graphs in $F$ include arbitrarily large square grid graphs, or equivalently subgraphs of the hexagonal tiling formed by intersecting it with arbitrarily large disks. (Wikipedia)
answered Dec 20 '18 at 7:54
Alex RavskyAlex Ravsky
40.4k32282
$endgroup$
add a comment |
$begingroup$
As part of their work on graph minors leading to the Robertson–Seymour theorem and the graph structure theorem, Neil Robertson and Paul Seymour proved that a family $F$ of finite graphs has unbounded treewidth if and only if the minors of graphs in $F$ include arbitrarily large square grid graphs, or equivalently subgraphs of the hexagonal tiling formed by intersecting it with arbitrarily large disks. (Wikipedia)
answered Dec 20 '18 at 7:54
Alex RavskyAlex Ravsky
40.4k32282
$endgroup$
add a comment |
$begingroup$
As part of their work on graph minors leading to the Robertson–Seymour theorem and the graph structure theorem, Neil Robertson and Paul Seymour proved that a family $F$ of finite graphs has unbounded treewidth if and only if the minors of graphs in $F$ include arbitrarily large square grid graphs, or equivalently subgraphs of the hexagonal tiling formed by intersecting it with arbitrarily large disks. (Wikipedia)
answered Dec 20 '18 at 7:54
Alex RavskyAlex Ravsky
40.4k32282
$endgroup$
As part of their work on graph minors leading to the Robertson–Seymour theorem and the graph structure theorem, Neil Robertson and Paul Seymour proved that a family $F$ of finite graphs has unbounded treewidth if and only if the minors of graphs in $F$ include arbitrarily large square grid graphs, or equivalently subgraphs of the hexagonal tiling formed by intersecting it with arbitrarily large disks. (Wikipedia)
answered Dec 20 '18 at 7:54
Alex RavskyAlex Ravsky
40.4k32282
answered Dec 20 '18 at 7:54
Alex RavskyAlex Ravsky
40.4k32282
answered Dec 20 '18 at 7:54
Alex RavskyAlex Ravsky
40.4k32282
answered Dec 20 '18 at 7:54
Alex RavskyAlex Ravsky
40.4k32282
40.4k32282
add a comment |
add a comment |
$begingroup$
A simple proof goes this way:
Take $K_n$ the complete graph in $n$ vertices. Take $vin G$, replace it with a cycle of length $d(v)=n-1$ and connect the $d(v)$ vertices in a 1-to-1 fashion to a vertex of the cycle. The result is that every vertex in the cycle has degree $3$. Repeat this until no vertex of degree $>3$ left. The graph $G$ obtained is $3$-regular and (clearly) $K_nleq G$. Then, $text{TW}(G)geqtext{TW}(K_n)=n-1$. So, the treewidth is unbounded even in the class of $3$-regular graphs.
answered Dec 22 '18 at 17:14
Veridian DynamicsVeridian Dynamics
1,909312
$endgroup$
add a comment |
$begingroup$
A simple proof goes this way:
Take $K_n$ the complete graph in $n$ vertices. Take $vin G$, replace it with a cycle of length $d(v)=n-1$ and connect the $d(v)$ vertices in a 1-to-1 fashion to a vertex of the cycle. The result is that every vertex in the cycle has degree $3$. Repeat this until no vertex of degree $>3$ left. The graph $G$ obtained is $3$-regular and (clearly) $K_nleq G$. Then, $text{TW}(G)geqtext{TW}(K_n)=n-1$. So, the treewidth is unbounded even in the class of $3$-regular graphs.
answered Dec 22 '18 at 17:14
Veridian DynamicsVeridian Dynamics
1,909312
$endgroup$
add a comment |
$begingroup$
A simple proof goes this way:
Take $K_n$ the complete graph in $n$ vertices. Take $vin G$, replace it with a cycle of length $d(v)=n-1$ and connect the $d(v)$ vertices in a 1-to-1 fashion to a vertex of the cycle. The result is that every vertex in the cycle has degree $3$. Repeat this until no vertex of degree $>3$ left. The graph $G$ obtained is $3$-regular and (clearly) $K_nleq G$. Then, $text{TW}(G)geqtext{TW}(K_n)=n-1$. So, the treewidth is unbounded even in the class of $3$-regular graphs.
answered Dec 22 '18 at 17:14
Veridian DynamicsVeridian Dynamics
1,909312
$endgroup$
A simple proof goes this way:
Take $K_n$ the complete graph in $n$ vertices. Take $vin G$, replace it with a cycle of length $d(v)=n-1$ and connect the $d(v)$ vertices in a 1-to-1 fashion to a vertex of the cycle. The result is that every vertex in the cycle has degree $3$. Repeat this until no vertex of degree $>3$ left. The graph $G$ obtained is $3$-regular and (clearly) $K_nleq G$. Then, $text{TW}(G)geqtext{TW}(K_n)=n-1$. So, the treewidth is unbounded even in the class of $3$-regular graphs.
answered Dec 22 '18 at 17:14
Veridian DynamicsVeridian Dynamics
1,909312
answered Dec 22 '18 at 17:14
Veridian DynamicsVeridian Dynamics
1,909312
answered Dec 22 '18 at 17:14
Veridian DynamicsVeridian Dynamics
1,909312
answered Dec 22 '18 at 17:14
Veridian DynamicsVeridian Dynamics
1,909312
1,909312
add a comment |
add a comment |
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