Construction of graph with degrees $d$ and $(d + 1)$












0














Let $n = a + b$ and $d$ be non-negative integers such that: $ad + b(d + 1)$ is even and $(d + 1) leq (n - 1)$. Does there exist a graph with $n$ vertices such that $a$ of them have degree $d$ and $b$ of them have degree $(d + 1)$? Is there an explicit construction?






graph-theory







share|cite|improve this question



























    0














    Let $n = a + b$ and $d$ be non-negative integers such that: $ad + b(d + 1)$ is even and $(d + 1) leq (n - 1)$. Does there exist a graph with $n$ vertices such that $a$ of them have degree $d$ and $b$ of them have degree $(d + 1)$? Is there an explicit construction?






    graph-theory







    share|cite|improve this question

























      0












      0








      0







      Let $n = a + b$ and $d$ be non-negative integers such that: $ad + b(d + 1)$ is even and $(d + 1) leq (n - 1)$. Does there exist a graph with $n$ vertices such that $a$ of them have degree $d$ and $b$ of them have degree $(d + 1)$? Is there an explicit construction?






      graph-theory







      share|cite|improve this question













      Let $n = a + b$ and $d$ be non-negative integers such that: $ad + b(d + 1)$ is even and $(d + 1) leq (n - 1)$. Does there exist a graph with $n$ vertices such that $a$ of them have degree $d$ and $b$ of them have degree $(d + 1)$? Is there an explicit construction?






      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 Nov 17 at 16:09









      frafour

      841212




      841212






















          2 Answers
          2






          active

          oldest

          votes


















          1














          This is not the answer but maybe that can give some idea:



          Fix an integer $kgeq3$. There exist a $(2k-1)$-vertex $(k-2)$-edge-connected simple graph $H_k=(V_k,E_k)$ with $V_k={y_1,y_2,...,y_{k+1},z_{1},z_{2},...,z_{k-2}}$, where all vertices $y_i$ have degree $k$ and all vertices $z_j$ have degree $(k-1)$.



          proof:
          Start with a k-vertex complete graph on the vertices ${y_1,...,y_k}$, plus a $(k-2)$-vertex complete graph on the vertices ${z_1,...,z_{k-2}}$. Next place an edge from the vertex $y_{k+1}$ to the vertices $y_{k-1}$ and $y_{k}$, then $(k-2)$ edges of the form $y_jz_j$ and $(k-2)$ edges of the form $y_{k+1}z_j$.






          share|cite|improve this answer





























            0














            We start with $k$-regular graphs on $n$ vertices constructed as follows. Put the $n$ vertices around a circle. If $k$ is even, connect each vertex to its $k$ closest neighbours. If $k$ is odd, then $n$ is even, so we can construct a $(k - 1)$-regular graph as before, and then add an edge between each vertex and its antipode.



            From this construction the one I was asking for follows. One has to proceed by cases, and I have checked all the details and it works, but I won't write everything here. The idea is that you either construct a $d$-regular graph, then check that you have a large enough matching in the complement, and add part of it to augment the degree of some vertices by 1; or you construct a $(d + 1)$-regular graph, then check you have a large enough matching, and eliminate part of it to decrease the degree of some vertices by 1. It's a bit tedious to check everything but it's quite straightforward.






            share|cite|improve this answer





















              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              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%2f3002518%2fconstruction-of-graph-with-degrees-d-and-d-1%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1














              This is not the answer but maybe that can give some idea:



              Fix an integer $kgeq3$. There exist a $(2k-1)$-vertex $(k-2)$-edge-connected simple graph $H_k=(V_k,E_k)$ with $V_k={y_1,y_2,...,y_{k+1},z_{1},z_{2},...,z_{k-2}}$, where all vertices $y_i$ have degree $k$ and all vertices $z_j$ have degree $(k-1)$.



              proof:
              Start with a k-vertex complete graph on the vertices ${y_1,...,y_k}$, plus a $(k-2)$-vertex complete graph on the vertices ${z_1,...,z_{k-2}}$. Next place an edge from the vertex $y_{k+1}$ to the vertices $y_{k-1}$ and $y_{k}$, then $(k-2)$ edges of the form $y_jz_j$ and $(k-2)$ edges of the form $y_{k+1}z_j$.






              share|cite|improve this answer


























                1














                This is not the answer but maybe that can give some idea:



                Fix an integer $kgeq3$. There exist a $(2k-1)$-vertex $(k-2)$-edge-connected simple graph $H_k=(V_k,E_k)$ with $V_k={y_1,y_2,...,y_{k+1},z_{1},z_{2},...,z_{k-2}}$, where all vertices $y_i$ have degree $k$ and all vertices $z_j$ have degree $(k-1)$.



                proof:
                Start with a k-vertex complete graph on the vertices ${y_1,...,y_k}$, plus a $(k-2)$-vertex complete graph on the vertices ${z_1,...,z_{k-2}}$. Next place an edge from the vertex $y_{k+1}$ to the vertices $y_{k-1}$ and $y_{k}$, then $(k-2)$ edges of the form $y_jz_j$ and $(k-2)$ edges of the form $y_{k+1}z_j$.






                share|cite|improve this answer
























                  1












                  1








                  1






                  This is not the answer but maybe that can give some idea:



                  Fix an integer $kgeq3$. There exist a $(2k-1)$-vertex $(k-2)$-edge-connected simple graph $H_k=(V_k,E_k)$ with $V_k={y_1,y_2,...,y_{k+1},z_{1},z_{2},...,z_{k-2}}$, where all vertices $y_i$ have degree $k$ and all vertices $z_j$ have degree $(k-1)$.



                  proof:
                  Start with a k-vertex complete graph on the vertices ${y_1,...,y_k}$, plus a $(k-2)$-vertex complete graph on the vertices ${z_1,...,z_{k-2}}$. Next place an edge from the vertex $y_{k+1}$ to the vertices $y_{k-1}$ and $y_{k}$, then $(k-2)$ edges of the form $y_jz_j$ and $(k-2)$ edges of the form $y_{k+1}z_j$.






                  share|cite|improve this answer












                  This is not the answer but maybe that can give some idea:



                  Fix an integer $kgeq3$. There exist a $(2k-1)$-vertex $(k-2)$-edge-connected simple graph $H_k=(V_k,E_k)$ with $V_k={y_1,y_2,...,y_{k+1},z_{1},z_{2},...,z_{k-2}}$, where all vertices $y_i$ have degree $k$ and all vertices $z_j$ have degree $(k-1)$.



                  proof:
                  Start with a k-vertex complete graph on the vertices ${y_1,...,y_k}$, plus a $(k-2)$-vertex complete graph on the vertices ${z_1,...,z_{k-2}}$. Next place an edge from the vertex $y_{k+1}$ to the vertices $y_{k-1}$ and $y_{k}$, then $(k-2)$ edges of the form $y_jz_j$ and $(k-2)$ edges of the form $y_{k+1}z_j$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 19 at 0:51









                  mathnoob

                  1,775322




                  1,775322























                      0














                      We start with $k$-regular graphs on $n$ vertices constructed as follows. Put the $n$ vertices around a circle. If $k$ is even, connect each vertex to its $k$ closest neighbours. If $k$ is odd, then $n$ is even, so we can construct a $(k - 1)$-regular graph as before, and then add an edge between each vertex and its antipode.



                      From this construction the one I was asking for follows. One has to proceed by cases, and I have checked all the details and it works, but I won't write everything here. The idea is that you either construct a $d$-regular graph, then check that you have a large enough matching in the complement, and add part of it to augment the degree of some vertices by 1; or you construct a $(d + 1)$-regular graph, then check you have a large enough matching, and eliminate part of it to decrease the degree of some vertices by 1. It's a bit tedious to check everything but it's quite straightforward.






                      share|cite|improve this answer


























                        0














                        We start with $k$-regular graphs on $n$ vertices constructed as follows. Put the $n$ vertices around a circle. If $k$ is even, connect each vertex to its $k$ closest neighbours. If $k$ is odd, then $n$ is even, so we can construct a $(k - 1)$-regular graph as before, and then add an edge between each vertex and its antipode.



                        From this construction the one I was asking for follows. One has to proceed by cases, and I have checked all the details and it works, but I won't write everything here. The idea is that you either construct a $d$-regular graph, then check that you have a large enough matching in the complement, and add part of it to augment the degree of some vertices by 1; or you construct a $(d + 1)$-regular graph, then check you have a large enough matching, and eliminate part of it to decrease the degree of some vertices by 1. It's a bit tedious to check everything but it's quite straightforward.






                        share|cite|improve this answer
























                          0












                          0








                          0






                          We start with $k$-regular graphs on $n$ vertices constructed as follows. Put the $n$ vertices around a circle. If $k$ is even, connect each vertex to its $k$ closest neighbours. If $k$ is odd, then $n$ is even, so we can construct a $(k - 1)$-regular graph as before, and then add an edge between each vertex and its antipode.



                          From this construction the one I was asking for follows. One has to proceed by cases, and I have checked all the details and it works, but I won't write everything here. The idea is that you either construct a $d$-regular graph, then check that you have a large enough matching in the complement, and add part of it to augment the degree of some vertices by 1; or you construct a $(d + 1)$-regular graph, then check you have a large enough matching, and eliminate part of it to decrease the degree of some vertices by 1. It's a bit tedious to check everything but it's quite straightforward.






                          share|cite|improve this answer












                          We start with $k$-regular graphs on $n$ vertices constructed as follows. Put the $n$ vertices around a circle. If $k$ is even, connect each vertex to its $k$ closest neighbours. If $k$ is odd, then $n$ is even, so we can construct a $(k - 1)$-regular graph as before, and then add an edge between each vertex and its antipode.



                          From this construction the one I was asking for follows. One has to proceed by cases, and I have checked all the details and it works, but I won't write everything here. The idea is that you either construct a $d$-regular graph, then check that you have a large enough matching in the complement, and add part of it to augment the degree of some vertices by 1; or you construct a $(d + 1)$-regular graph, then check you have a large enough matching, and eliminate part of it to decrease the degree of some vertices by 1. It's a bit tedious to check everything but it's quite straightforward.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 29 at 12:45









                          frafour

                          841212




                          841212






























                              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.





                              Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                              Please pay close attention to the following guidance:


                              • 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.


                              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%2f3002518%2fconstruction-of-graph-with-degrees-d-and-d-1%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

                              Berounka

                              Image
                              Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
                              Read more

                              Sphinx de Gizeh

                              Image
                              Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
                              Read more

                              Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

                              Image
                              3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
                              Read more