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







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    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







    share|cite|improve this question


























      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







      share|cite|improve this question
















      $W$ 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






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 23 at 2:00









      Misha Lavrov

      43.3k555103




      43.3k555103










      asked Nov 22 at 20:10









      one1

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      174






















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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 $ir(G) ne gamma(G)$:



          enter image description here



          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 $gamma(G) = 3$.






          share|cite|improve this answer























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            1 Answer
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            active

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            up vote
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            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 $ir(G) ne gamma(G)$:



            enter image description here



            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 $gamma(G) = 3$.






            share|cite|improve this answer



























              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 $ir(G) ne gamma(G)$:



              enter image description here



              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 $gamma(G) = 3$.






              share|cite|improve this answer

























                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 $ir(G) ne gamma(G)$:



                enter image description here



                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 $gamma(G) = 3$.






                share|cite|improve this answer














                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 $ir(G) ne gamma(G)$:



                enter image description here



                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 $gamma(G) = 3$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 23 at 2:12

























                answered Nov 23 at 1:50









                Misha Lavrov

                43.3k555103




                43.3k555103






























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