Mixing
proof of theorem: if $G$ is a graph in which all its cycle lengths are divisible by 4, then $sq(G) le...
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Theorem: If $G$ is a graph in which all its cycle lengths are divisible by 4, then $sq(G) le Delta^2(G)$.
$sq(G)$ denote strong chromatics index and $Delta(G)$ denote maximum degree of graph $G$
Proof of this theorem is by induction on $|E(G)|$.
In base step $|E(G)|=1$, then $sq(G)=1$ and $Delta^2(G)=1$
In inductive step we assume that: for any graph $H$ with $m$ edges $sq(H)le Delta^2(H)$
inductive hypothesis: for any graph $G$ with $m+1$ edges $sq(G)le Delta^2(G)$.
How can I prove inductive step?
graph-theory
asked Jan 25 at 17:10
KateKate
104
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add a comment |
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Theorem: If $G$ is a graph in which all its cycle lengths are divisible by 4, then $sq(G) le Delta^2(G)$.
$sq(G)$ denote strong chromatics index and $Delta(G)$ denote maximum degree of graph $G$
Proof of this theorem is by induction on $|E(G)|$.
In base step $|E(G)|=1$, then $sq(G)=1$ and $Delta^2(G)=1$
In inductive step we assume that: for any graph $H$ with $m$ edges $sq(H)le Delta^2(H)$
inductive hypothesis: for any graph $G$ with $m+1$ edges $sq(G)le Delta^2(G)$.
How can I prove inductive step?
graph-theory
asked Jan 25 at 17:10
KateKate
104
$endgroup$
$begingroup$
What have you tried for the induction? Have you looked at minors of $G$, edge contraction, edge deletion? removing a (non-isolated) vertex?
$endgroup$
– Thomas Lesgourgues
Jan 26 at 11:15
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Please check this paper: GYARFAS, A., & TUZA, Z. (1990). The strong chromatic index of graphs. Ars Combinatoria, 29, 205-211.
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– W.R.P.S
Jan 27 at 8:50
$begingroup$
@ThomasLesgourgues if $x$ is a vertex with degree $1$ then we can remove this vertex with incident egde
$endgroup$
– Kate
Jan 27 at 16:00
add a comment |
$begingroup$
Theorem: If $G$ is a graph in which all its cycle lengths are divisible by 4, then $sq(G) le Delta^2(G)$.
$sq(G)$ denote strong chromatics index and $Delta(G)$ denote maximum degree of graph $G$
Proof of this theorem is by induction on $|E(G)|$.
In base step $|E(G)|=1$, then $sq(G)=1$ and $Delta^2(G)=1$
In inductive step we assume that: for any graph $H$ with $m$ edges $sq(H)le Delta^2(H)$
inductive hypothesis: for any graph $G$ with $m+1$ edges $sq(G)le Delta^2(G)$.
How can I prove inductive step?
graph-theory
asked Jan 25 at 17:10
KateKate
104
$endgroup$
Theorem: If $G$ is a graph in which all its cycle lengths are divisible by 4, then $sq(G) le Delta^2(G)$.
$sq(G)$ denote strong chromatics index and $Delta(G)$ denote maximum degree of graph $G$
Proof of this theorem is by induction on $|E(G)|$.
In base step $|E(G)|=1$, then $sq(G)=1$ and $Delta^2(G)=1$
In inductive step we assume that: for any graph $H$ with $m$ edges $sq(H)le Delta^2(H)$
inductive hypothesis: for any graph $G$ with $m+1$ edges $sq(G)le Delta^2(G)$.
How can I prove inductive step?
graph-theory
graph-theory
asked Jan 25 at 17:10
KateKate
104
asked Jan 25 at 17:10
KateKate
104
asked Jan 25 at 17:10
KateKate
104
asked Jan 25 at 17:10
KateKate
104
asked Jan 25 at 17:10
KateKate
104
104
$begingroup$
What have you tried for the induction? Have you looked at minors of $G$, edge contraction, edge deletion? removing a (non-isolated) vertex?
$endgroup$
– Thomas Lesgourgues
Jan 26 at 11:15
$begingroup$
Please check this paper: GYARFAS, A., & TUZA, Z. (1990). The strong chromatic index of graphs. Ars Combinatoria, 29, 205-211.
$endgroup$
– W.R.P.S
Jan 27 at 8:50
$begingroup$
@ThomasLesgourgues if $x$ is a vertex with degree $1$ then we can remove this vertex with incident egde
$endgroup$
– Kate
Jan 27 at 16:00
add a comment |
$begingroup$
What have you tried for the induction? Have you looked at minors of $G$, edge contraction, edge deletion? removing a (non-isolated) vertex?
$endgroup$
– Thomas Lesgourgues
Jan 26 at 11:15
$begingroup$
Please check this paper: GYARFAS, A., & TUZA, Z. (1990). The strong chromatic index of graphs. Ars Combinatoria, 29, 205-211.
$endgroup$
– W.R.P.S
Jan 27 at 8:50
$begingroup$
@ThomasLesgourgues if $x$ is a vertex with degree $1$ then we can remove this vertex with incident egde
$endgroup$
– Kate
Jan 27 at 16:00
$begingroup$
What have you tried for the induction? Have you looked at minors of $G$, edge contraction, edge deletion? removing a (non-isolated) vertex?
$endgroup$
– Thomas Lesgourgues
Jan 26 at 11:15
$begingroup$
What have you tried for the induction? Have you looked at minors of $G$, edge contraction, edge deletion? removing a (non-isolated) vertex?
$endgroup$
– Thomas Lesgourgues
Jan 26 at 11:15
$begingroup$
Please check this paper: GYARFAS, A., & TUZA, Z. (1990). The strong chromatic index of graphs. Ars Combinatoria, 29, 205-211.
$endgroup$
– W.R.P.S
Jan 27 at 8:50
$begingroup$
Please check this paper: GYARFAS, A., & TUZA, Z. (1990). The strong chromatic index of graphs. Ars Combinatoria, 29, 205-211.
$endgroup$
– W.R.P.S
Jan 27 at 8:50
$begingroup$
@ThomasLesgourgues if $x$ is a vertex with degree $1$ then we can remove this vertex with incident egde
$endgroup$
– Kate
Jan 27 at 16:00
$begingroup$
@ThomasLesgourgues if $x$ is a vertex with degree $1$ then we can remove this vertex with incident egde
$endgroup$
– Kate
Jan 27 at 16:00
add a comment |
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$begingroup$
What have you tried for the induction? Have you looked at minors of $G$, edge contraction, edge deletion? removing a (non-isolated) vertex?
$endgroup$
– Thomas Lesgourgues
Jan 26 at 11:15
$begingroup$
Please check this paper: GYARFAS, A., & TUZA, Z. (1990). The strong chromatic index of graphs. Ars Combinatoria, 29, 205-211.
$endgroup$
– W.R.P.S
Jan 27 at 8:50
$begingroup$
@ThomasLesgourgues if $x$ is a vertex with degree $1$ then we can remove this vertex with incident egde
$endgroup$
– Kate
Jan 27 at 16:00