Find a sub-graph where each node in the sub-graph has minimum degree d.












0












$begingroup$


Given an undirected graph $G$ and integer $d$, find a sub-graph $H$ of $G$ such that every node in $H$ has minimum degree $d$.



I wonder if there is any condition to decide whether $H$ exists or not. If such $H$ exists, is there any polynomial algorithm to find such $H$.






graph-theory







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Given an undirected graph $G$ and integer $d$, find a sub-graph $H$ of $G$ such that every node in $H$ has minimum degree $d$.



    I wonder if there is any condition to decide whether $H$ exists or not. If such $H$ exists, is there any polynomial algorithm to find such $H$.






    graph-theory







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Given an undirected graph $G$ and integer $d$, find a sub-graph $H$ of $G$ such that every node in $H$ has minimum degree $d$.



      I wonder if there is any condition to decide whether $H$ exists or not. If such $H$ exists, is there any polynomial algorithm to find such $H$.






      graph-theory







      share|cite|improve this question









      $endgroup$




      Given an undirected graph $G$ and integer $d$, find a sub-graph $H$ of $G$ such that every node in $H$ has minimum degree $d$.



      I wonder if there is any condition to decide whether $H$ exists or not. If such $H$ exists, is there any polynomial algorithm to find such $H$.






      graph-theory




      graph-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Feb 2 at 23:14









      ParadoxParadox

      909




      909






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          Remove all vertices from $G$ of degree less than $d$ and call the resulting graph $G_1$; if there is such a subgraph $H$ of $G$, then $H$ contains no vertex in $G setminus G_1$. Then remove from $G_1$ all vertices of degree(in the smaller graph $G_1$) of degree less than $d$ and call the resulting graph $G_2$. If there is a graph $H$ as described in your OP it can contain none of the vertices in $G_1 setminus G_2$. Continue on in a similar fashion: For general $ell$,
          remove from $G_{ell}$ all vertices of degree(in $G_{ell}$) of degree less than $d$ and call the resulting graph $G_{ell+1}$. If there is a graph $H$ as described in your OP it can contain none of the vertices in $G_{ell} setminus G_{ell+1}$. Stop this process when either the resulting graph is empty or when every vertex in the resulting graph $H'$ has degree at least $d$.



          If there is indeed a subgraph of $G$ where every vertex has degree $d$, then the resulting graph $H'$ will be nonempty (as in each step above you didn't remove any vertices from that graph) and (as every vertex in $H'$ has degree at least $d$) is the graph you are seeking.






          share|cite|improve this answer











          $endgroup$














            Your Answer








            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3097923%2ffind-a-sub-graph-where-each-node-in-the-sub-graph-has-minimum-degree-d%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0












            $begingroup$

            Remove all vertices from $G$ of degree less than $d$ and call the resulting graph $G_1$; if there is such a subgraph $H$ of $G$, then $H$ contains no vertex in $G setminus G_1$. Then remove from $G_1$ all vertices of degree(in the smaller graph $G_1$) of degree less than $d$ and call the resulting graph $G_2$. If there is a graph $H$ as described in your OP it can contain none of the vertices in $G_1 setminus G_2$. Continue on in a similar fashion: For general $ell$,
            remove from $G_{ell}$ all vertices of degree(in $G_{ell}$) of degree less than $d$ and call the resulting graph $G_{ell+1}$. If there is a graph $H$ as described in your OP it can contain none of the vertices in $G_{ell} setminus G_{ell+1}$. Stop this process when either the resulting graph is empty or when every vertex in the resulting graph $H'$ has degree at least $d$.



            If there is indeed a subgraph of $G$ where every vertex has degree $d$, then the resulting graph $H'$ will be nonempty (as in each step above you didn't remove any vertices from that graph) and (as every vertex in $H'$ has degree at least $d$) is the graph you are seeking.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              Remove all vertices from $G$ of degree less than $d$ and call the resulting graph $G_1$; if there is such a subgraph $H$ of $G$, then $H$ contains no vertex in $G setminus G_1$. Then remove from $G_1$ all vertices of degree(in the smaller graph $G_1$) of degree less than $d$ and call the resulting graph $G_2$. If there is a graph $H$ as described in your OP it can contain none of the vertices in $G_1 setminus G_2$. Continue on in a similar fashion: For general $ell$,
              remove from $G_{ell}$ all vertices of degree(in $G_{ell}$) of degree less than $d$ and call the resulting graph $G_{ell+1}$. If there is a graph $H$ as described in your OP it can contain none of the vertices in $G_{ell} setminus G_{ell+1}$. Stop this process when either the resulting graph is empty or when every vertex in the resulting graph $H'$ has degree at least $d$.



              If there is indeed a subgraph of $G$ where every vertex has degree $d$, then the resulting graph $H'$ will be nonempty (as in each step above you didn't remove any vertices from that graph) and (as every vertex in $H'$ has degree at least $d$) is the graph you are seeking.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                Remove all vertices from $G$ of degree less than $d$ and call the resulting graph $G_1$; if there is such a subgraph $H$ of $G$, then $H$ contains no vertex in $G setminus G_1$. Then remove from $G_1$ all vertices of degree(in the smaller graph $G_1$) of degree less than $d$ and call the resulting graph $G_2$. If there is a graph $H$ as described in your OP it can contain none of the vertices in $G_1 setminus G_2$. Continue on in a similar fashion: For general $ell$,
                remove from $G_{ell}$ all vertices of degree(in $G_{ell}$) of degree less than $d$ and call the resulting graph $G_{ell+1}$. If there is a graph $H$ as described in your OP it can contain none of the vertices in $G_{ell} setminus G_{ell+1}$. Stop this process when either the resulting graph is empty or when every vertex in the resulting graph $H'$ has degree at least $d$.



                If there is indeed a subgraph of $G$ where every vertex has degree $d$, then the resulting graph $H'$ will be nonempty (as in each step above you didn't remove any vertices from that graph) and (as every vertex in $H'$ has degree at least $d$) is the graph you are seeking.






                share|cite|improve this answer











                $endgroup$



                Remove all vertices from $G$ of degree less than $d$ and call the resulting graph $G_1$; if there is such a subgraph $H$ of $G$, then $H$ contains no vertex in $G setminus G_1$. Then remove from $G_1$ all vertices of degree(in the smaller graph $G_1$) of degree less than $d$ and call the resulting graph $G_2$. If there is a graph $H$ as described in your OP it can contain none of the vertices in $G_1 setminus G_2$. Continue on in a similar fashion: For general $ell$,
                remove from $G_{ell}$ all vertices of degree(in $G_{ell}$) of degree less than $d$ and call the resulting graph $G_{ell+1}$. If there is a graph $H$ as described in your OP it can contain none of the vertices in $G_{ell} setminus G_{ell+1}$. Stop this process when either the resulting graph is empty or when every vertex in the resulting graph $H'$ has degree at least $d$.



                If there is indeed a subgraph of $G$ where every vertex has degree $d$, then the resulting graph $H'$ will be nonempty (as in each step above you didn't remove any vertices from that graph) and (as every vertex in $H'$ has degree at least $d$) is the graph you are seeking.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Feb 2 at 23:40

























                answered Feb 2 at 23:29









                MikeMike

                4,641512




                4,641512






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3097923%2ffind-a-sub-graph-where-each-node-in-the-sub-graph-has-minimum-degree-d%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

                    Image
                    28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
                    Read more

                    Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

                    Image
                    up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
                    Read more

                    A Topological Invariant for $pi_3(U(n))$

                    Image
                    3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
                    Read more