Is the maximal irredundant set a dominating set?
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$W$ is irredundant in G=(V,E) if $ forall v in W, N[v] - N[W -
{v}] not= emptyset $
But this means maximal irredundant set is a dominating set!
Proof: suppose not then take $W$ a maximal irredundant set: $N[W] not= V$ then $exists $u$ in (V - W) s.t. N[u] - N[W ] not= emptyset $. Contradiction.
With that the irredundant number (which is the cardinality of a minimum maximal irredundant set) denoted by $ir(G)$ equals the domination number $γ(G)$. Where's the mistake?
The definition is from the Handbook of Graph Theory by Jonathan Gross and Jay Yellen. Similar definitions are found all over the internet.
combinatorics graph-theory
edited Nov 23 at 2:00
Misha Lavrov
43.3k555103
asked Nov 22 at 20:10
one1
174
add a comment |
up vote
1
down vote
favorite
$W$ is irredundant in G=(V,E) if $ forall v in W, N[v] - N[W -
{v}] not= emptyset $
But this means maximal irredundant set is a dominating set!
Proof: suppose not then take $W$ a maximal irredundant set: $N[W] not= V$ then $exists $u$ in (V - W) s.t. N[u] - N[W ] not= emptyset $. Contradiction.
With that the irredundant number (which is the cardinality of a minimum maximal irredundant set) denoted by $ir(G)$ equals the domination number $γ(G)$. Where's the mistake?
The definition is from the Handbook of Graph Theory by Jonathan Gross and Jay Yellen. Similar definitions are found all over the internet.
combinatorics graph-theory
edited Nov 23 at 2:00
Misha Lavrov
43.3k555103
asked Nov 22 at 20:10
one1
174
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$W$ is irredundant in G=(V,E) if $ forall v in W, N[v] - N[W -
{v}] not= emptyset $
But this means maximal irredundant set is a dominating set!
Proof: suppose not then take $W$ a maximal irredundant set: $N[W] not= V$ then $exists $u$ in (V - W) s.t. N[u] - N[W ] not= emptyset $. Contradiction.
With that the irredundant number (which is the cardinality of a minimum maximal irredundant set) denoted by $ir(G)$ equals the domination number $γ(G)$. Where's the mistake?
The definition is from the Handbook of Graph Theory by Jonathan Gross and Jay Yellen. Similar definitions are found all over the internet.
combinatorics graph-theory
edited Nov 23 at 2:00
Misha Lavrov
43.3k555103
asked Nov 22 at 20:10
one1
174
$W$ is irredundant in G=(V,E) if $ forall v in W, N[v] - N[W -
{v}] not= emptyset $
But this means maximal irredundant set is a dominating set!
Proof: suppose not then take $W$ a maximal irredundant set: $N[W] not= V$ then $exists $u$ in (V - W) s.t. N[u] - N[W ] not= emptyset $. Contradiction.
With that the irredundant number (which is the cardinality of a minimum maximal irredundant set) denoted by $ir(G)$ equals the domination number $γ(G)$. Where's the mistake?
The definition is from the Handbook of Graph Theory by Jonathan Gross and Jay Yellen. Similar definitions are found all over the internet.
combinatorics graph-theory
combinatorics graph-theory
edited Nov 23 at 2:00
Misha Lavrov
43.3k555103
asked Nov 22 at 20:10
one1
174
edited Nov 23 at 2:00
Misha Lavrov
43.3k555103
asked Nov 22 at 20:10
one1
174
edited Nov 23 at 2:00
Misha Lavrov
43.3k555103
edited Nov 23 at 2:00
Misha Lavrov
43.3k555103
edited Nov 23 at 2:00
Misha Lavrov
43.3k555103
43.3k555103
asked Nov 22 at 20:10
one1
174
asked Nov 22 at 20:10
one1
174
asked Nov 22 at 20:10
one1
174
174
add a comment |
add a comment |
1 Answer
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There exist maximal irredundant sets which are not dominating sets.
For example, let $G$ be a $5$-cycle with vertices $v_1, v_2, v_3, v_4, v_5$ adjacent (cyclically) in that order. Then ${v_1, v_2}$ is a maximal irredundant set, even though it is not dominating, because:
${v_1, v_2, v_3}$ is not irredundant: $v_2$ is a redundant vertex.
${v_1, v_2, v_5}$ is not irredundant: $v_1$ is a redundant vertex.
${v_1, v_2, v_4}$ is not irredundant - this is the surprising case! Here, $v_1$ is redundant, because ${v_2, v_4}$ is a dominating set, and in particular $N[{v_2,v_4}]$ contains $N[v_1]$. Also, $v_2$ is redundant for similar reasons.- Then, of course, no superset of any of these three sets can be irredundant.
In general, when $W$ is an irredundant set and $u notin N[W]$, then $u$ is definitely not going to be redundant in $W cup {u}$, sure - but that doesn't mean that $W cup {u}$ is an irredundant set! Some vertices of $W$ may become redundant with the addition of $u$.
Here is an example of a graph $G$ in which $ir(G) ne gamma(G)$:
In this graph, the two vertices of degree $3$ form a maximal irredundant set, so $ir(G) =2$, but there is no dominating set of size $2$, and $gamma(G) = 3$.
answered Nov 23 at 1:50
Misha Lavrov
43.3k555103
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
up vote
1
down vote
accepted
There exist maximal irredundant sets which are not dominating sets.
For example, let $G$ be a $5$-cycle with vertices $v_1, v_2, v_3, v_4, v_5$ adjacent (cyclically) in that order. Then ${v_1, v_2}$ is a maximal irredundant set, even though it is not dominating, because:
${v_1, v_2, v_3}$ is not irredundant: $v_2$ is a redundant vertex.
${v_1, v_2, v_5}$ is not irredundant: $v_1$ is a redundant vertex.
${v_1, v_2, v_4}$ is not irredundant - this is the surprising case! Here, $v_1$ is redundant, because ${v_2, v_4}$ is a dominating set, and in particular $N[{v_2,v_4}]$ contains $N[v_1]$. Also, $v_2$ is redundant for similar reasons.- Then, of course, no superset of any of these three sets can be irredundant.
In general, when $W$ is an irredundant set and $u notin N[W]$, then $u$ is definitely not going to be redundant in $W cup {u}$, sure - but that doesn't mean that $W cup {u}$ is an irredundant set! Some vertices of $W$ may become redundant with the addition of $u$.
Here is an example of a graph $G$ in which $ir(G) ne gamma(G)$:
In this graph, the two vertices of degree $3$ form a maximal irredundant set, so $ir(G) =2$, but there is no dominating set of size $2$, and $gamma(G) = 3$.
answered Nov 23 at 1:50
Misha Lavrov
43.3k555103
add a comment |
up vote
1
down vote
accepted
There exist maximal irredundant sets which are not dominating sets.
For example, let $G$ be a $5$-cycle with vertices $v_1, v_2, v_3, v_4, v_5$ adjacent (cyclically) in that order. Then ${v_1, v_2}$ is a maximal irredundant set, even though it is not dominating, because:
${v_1, v_2, v_3}$ is not irredundant: $v_2$ is a redundant vertex.
${v_1, v_2, v_5}$ is not irredundant: $v_1$ is a redundant vertex.
${v_1, v_2, v_4}$ is not irredundant - this is the surprising case! Here, $v_1$ is redundant, because ${v_2, v_4}$ is a dominating set, and in particular $N[{v_2,v_4}]$ contains $N[v_1]$. Also, $v_2$ is redundant for similar reasons.- Then, of course, no superset of any of these three sets can be irredundant.
In general, when $W$ is an irredundant set and $u notin N[W]$, then $u$ is definitely not going to be redundant in $W cup {u}$, sure - but that doesn't mean that $W cup {u}$ is an irredundant set! Some vertices of $W$ may become redundant with the addition of $u$.
Here is an example of a graph $G$ in which $ir(G) ne gamma(G)$:
In this graph, the two vertices of degree $3$ form a maximal irredundant set, so $ir(G) =2$, but there is no dominating set of size $2$, and $gamma(G) = 3$.
answered Nov 23 at 1:50
Misha Lavrov
43.3k555103
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
There exist maximal irredundant sets which are not dominating sets.
For example, let $G$ be a $5$-cycle with vertices $v_1, v_2, v_3, v_4, v_5$ adjacent (cyclically) in that order. Then ${v_1, v_2}$ is a maximal irredundant set, even though it is not dominating, because:
${v_1, v_2, v_3}$ is not irredundant: $v_2$ is a redundant vertex.
${v_1, v_2, v_5}$ is not irredundant: $v_1$ is a redundant vertex.
${v_1, v_2, v_4}$ is not irredundant - this is the surprising case! Here, $v_1$ is redundant, because ${v_2, v_4}$ is a dominating set, and in particular $N[{v_2,v_4}]$ contains $N[v_1]$. Also, $v_2$ is redundant for similar reasons.- Then, of course, no superset of any of these three sets can be irredundant.
In general, when $W$ is an irredundant set and $u notin N[W]$, then $u$ is definitely not going to be redundant in $W cup {u}$, sure - but that doesn't mean that $W cup {u}$ is an irredundant set! Some vertices of $W$ may become redundant with the addition of $u$.
Here is an example of a graph $G$ in which $ir(G) ne gamma(G)$:
In this graph, the two vertices of degree $3$ form a maximal irredundant set, so $ir(G) =2$, but there is no dominating set of size $2$, and $gamma(G) = 3$.
answered Nov 23 at 1:50
Misha Lavrov
43.3k555103
There exist maximal irredundant sets which are not dominating sets.
For example, let $G$ be a $5$-cycle with vertices $v_1, v_2, v_3, v_4, v_5$ adjacent (cyclically) in that order. Then ${v_1, v_2}$ is a maximal irredundant set, even though it is not dominating, because:
${v_1, v_2, v_3}$ is not irredundant: $v_2$ is a redundant vertex.
${v_1, v_2, v_5}$ is not irredundant: $v_1$ is a redundant vertex.
${v_1, v_2, v_4}$ is not irredundant - this is the surprising case! Here, $v_1$ is redundant, because ${v_2, v_4}$ is a dominating set, and in particular $N[{v_2,v_4}]$ contains $N[v_1]$. Also, $v_2$ is redundant for similar reasons.- Then, of course, no superset of any of these three sets can be irredundant.
In general, when $W$ is an irredundant set and $u notin N[W]$, then $u$ is definitely not going to be redundant in $W cup {u}$, sure - but that doesn't mean that $W cup {u}$ is an irredundant set! Some vertices of $W$ may become redundant with the addition of $u$.
Here is an example of a graph $G$ in which $ir(G) ne gamma(G)$:
In this graph, the two vertices of degree $3$ form a maximal irredundant set, so $ir(G) =2$, but there is no dominating set of size $2$, and $gamma(G) = 3$.
answered Nov 23 at 1:50
Misha Lavrov
43.3k555103
edited Nov 23 at 2:12
answered Nov 23 at 1:50
Misha Lavrov
43.3k555103
answered Nov 23 at 1:50
Misha Lavrov
43.3k555103
answered Nov 23 at 1:50
Misha Lavrov
43.3k555103
43.3k555103
add a comment |
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