Minimum number of colors enough to color the vertices of any graph whose vertex lies on at most $k$-odd...












1












$begingroup$


I've got no time to solve this problem during the exam as it's the last one.




What is the minimum number of colors that would suffice to color a graph so that adjacent nodes get different colors if each node lies on at most $k$-odd cycles?




But I still have no idea about it (even unsure about the answer). Hints will also be appreciated.






discrete-mathematics graph-theory coloring







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I've got no time to solve this problem during the exam as it's the last one.




    What is the minimum number of colors that would suffice to color a graph so that adjacent nodes get different colors if each node lies on at most $k$-odd cycles?




    But I still have no idea about it (even unsure about the answer). Hints will also be appreciated.






    discrete-mathematics graph-theory coloring







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      3



      $begingroup$


      I've got no time to solve this problem during the exam as it's the last one.




      What is the minimum number of colors that would suffice to color a graph so that adjacent nodes get different colors if each node lies on at most $k$-odd cycles?




      But I still have no idea about it (even unsure about the answer). Hints will also be appreciated.






      discrete-mathematics graph-theory coloring







      share|cite|improve this question











      $endgroup$




      I've got no time to solve this problem during the exam as it's the last one.




      What is the minimum number of colors that would suffice to color a graph so that adjacent nodes get different colors if each node lies on at most $k$-odd cycles?




      But I still have no idea about it (even unsure about the answer). Hints will also be appreciated.






      discrete-mathematics graph-theory coloring




      discrete-mathematics graph-theory coloring






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 6 at 16:59









      amWhy

      1




      1










      asked Jan 6 at 7:13









      Oolong milk teaOolong milk tea

      787




      787






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          Not sure if this is correct:



          Let $G$ be the graph. Assume that there is a vertex $vin V(G)$ that lie on $k$-odd cycles, otherwise each node could lie on $k+n$ cycles for any $nin mathbb{N}$. And this just says the chromatic number could be anything really.



          So choose the node $v in V(G)$ which lie on $k$ odd cycle, removes it. then all $k$ odd cycles are gone. Any node on those cycles cannot be on the $k$ removed cycles in the remaining graph. So if you see another node $v_2$ that is in $k$ odd cycles, then it is not adjacent to $v$. So removes $v_2$ to remove the $k$-odd cycles that $v_2$ is in. Continue this way until stuck. Notice all the removed nodes can be color using the same color so let's just color them with color $k$. The resulting graph has the property that all node are in at most $k-1$ cycles. So as before, pick a vertex $v^2$ that is adjacent to $k-1$ cycles, if there is any. Removes those one by one as before and notice they are all pairwise no adjacent and hence can be color using the same color. Color the removed vertices using color $k-1$. OK then you do this at most $k$ steps to get a no odd cycle graph, which can be colored using two colors as it would be bipartite. These 2 colors together with the $k$ color used, we get a $k+2$ coloring of the graph?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Please explain the part "then it is not adjacent to $v$". How do you know it's not adjacent to $v$?
            $endgroup$
            – bof
            Jan 6 at 23:50










          • $begingroup$
            Ah right sorry...I'm actually not sure about it either...
            $endgroup$
            – nafhgood
            Jan 7 at 0:42











          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%2f3063569%2fminimum-number-of-colors-enough-to-color-the-vertices-of-any-graph-whose-vertex%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












          $begingroup$

          Not sure if this is correct:



          Let $G$ be the graph. Assume that there is a vertex $vin V(G)$ that lie on $k$-odd cycles, otherwise each node could lie on $k+n$ cycles for any $nin mathbb{N}$. And this just says the chromatic number could be anything really.



          So choose the node $v in V(G)$ which lie on $k$ odd cycle, removes it. then all $k$ odd cycles are gone. Any node on those cycles cannot be on the $k$ removed cycles in the remaining graph. So if you see another node $v_2$ that is in $k$ odd cycles, then it is not adjacent to $v$. So removes $v_2$ to remove the $k$-odd cycles that $v_2$ is in. Continue this way until stuck. Notice all the removed nodes can be color using the same color so let's just color them with color $k$. The resulting graph has the property that all node are in at most $k-1$ cycles. So as before, pick a vertex $v^2$ that is adjacent to $k-1$ cycles, if there is any. Removes those one by one as before and notice they are all pairwise no adjacent and hence can be color using the same color. Color the removed vertices using color $k-1$. OK then you do this at most $k$ steps to get a no odd cycle graph, which can be colored using two colors as it would be bipartite. These 2 colors together with the $k$ color used, we get a $k+2$ coloring of the graph?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Please explain the part "then it is not adjacent to $v$". How do you know it's not adjacent to $v$?
            $endgroup$
            – bof
            Jan 6 at 23:50










          • $begingroup$
            Ah right sorry...I'm actually not sure about it either...
            $endgroup$
            – nafhgood
            Jan 7 at 0:42
















          0












          $begingroup$

          Not sure if this is correct:



          Let $G$ be the graph. Assume that there is a vertex $vin V(G)$ that lie on $k$-odd cycles, otherwise each node could lie on $k+n$ cycles for any $nin mathbb{N}$. And this just says the chromatic number could be anything really.



          So choose the node $v in V(G)$ which lie on $k$ odd cycle, removes it. then all $k$ odd cycles are gone. Any node on those cycles cannot be on the $k$ removed cycles in the remaining graph. So if you see another node $v_2$ that is in $k$ odd cycles, then it is not adjacent to $v$. So removes $v_2$ to remove the $k$-odd cycles that $v_2$ is in. Continue this way until stuck. Notice all the removed nodes can be color using the same color so let's just color them with color $k$. The resulting graph has the property that all node are in at most $k-1$ cycles. So as before, pick a vertex $v^2$ that is adjacent to $k-1$ cycles, if there is any. Removes those one by one as before and notice they are all pairwise no adjacent and hence can be color using the same color. Color the removed vertices using color $k-1$. OK then you do this at most $k$ steps to get a no odd cycle graph, which can be colored using two colors as it would be bipartite. These 2 colors together with the $k$ color used, we get a $k+2$ coloring of the graph?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Please explain the part "then it is not adjacent to $v$". How do you know it's not adjacent to $v$?
            $endgroup$
            – bof
            Jan 6 at 23:50










          • $begingroup$
            Ah right sorry...I'm actually not sure about it either...
            $endgroup$
            – nafhgood
            Jan 7 at 0:42














          0












          0








          0





          $begingroup$

          Not sure if this is correct:



          Let $G$ be the graph. Assume that there is a vertex $vin V(G)$ that lie on $k$-odd cycles, otherwise each node could lie on $k+n$ cycles for any $nin mathbb{N}$. And this just says the chromatic number could be anything really.



          So choose the node $v in V(G)$ which lie on $k$ odd cycle, removes it. then all $k$ odd cycles are gone. Any node on those cycles cannot be on the $k$ removed cycles in the remaining graph. So if you see another node $v_2$ that is in $k$ odd cycles, then it is not adjacent to $v$. So removes $v_2$ to remove the $k$-odd cycles that $v_2$ is in. Continue this way until stuck. Notice all the removed nodes can be color using the same color so let's just color them with color $k$. The resulting graph has the property that all node are in at most $k-1$ cycles. So as before, pick a vertex $v^2$ that is adjacent to $k-1$ cycles, if there is any. Removes those one by one as before and notice they are all pairwise no adjacent and hence can be color using the same color. Color the removed vertices using color $k-1$. OK then you do this at most $k$ steps to get a no odd cycle graph, which can be colored using two colors as it would be bipartite. These 2 colors together with the $k$ color used, we get a $k+2$ coloring of the graph?






          share|cite|improve this answer











          $endgroup$



          Not sure if this is correct:



          Let $G$ be the graph. Assume that there is a vertex $vin V(G)$ that lie on $k$-odd cycles, otherwise each node could lie on $k+n$ cycles for any $nin mathbb{N}$. And this just says the chromatic number could be anything really.



          So choose the node $v in V(G)$ which lie on $k$ odd cycle, removes it. then all $k$ odd cycles are gone. Any node on those cycles cannot be on the $k$ removed cycles in the remaining graph. So if you see another node $v_2$ that is in $k$ odd cycles, then it is not adjacent to $v$. So removes $v_2$ to remove the $k$-odd cycles that $v_2$ is in. Continue this way until stuck. Notice all the removed nodes can be color using the same color so let's just color them with color $k$. The resulting graph has the property that all node are in at most $k-1$ cycles. So as before, pick a vertex $v^2$ that is adjacent to $k-1$ cycles, if there is any. Removes those one by one as before and notice they are all pairwise no adjacent and hence can be color using the same color. Color the removed vertices using color $k-1$. OK then you do this at most $k$ steps to get a no odd cycle graph, which can be colored using two colors as it would be bipartite. These 2 colors together with the $k$ color used, we get a $k+2$ coloring of the graph?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 6 at 23:44

























          answered Jan 6 at 23:37









          nafhgoodnafhgood

          1,803422




          1,803422












          • $begingroup$
            Please explain the part "then it is not adjacent to $v$". How do you know it's not adjacent to $v$?
            $endgroup$
            – bof
            Jan 6 at 23:50










          • $begingroup$
            Ah right sorry...I'm actually not sure about it either...
            $endgroup$
            – nafhgood
            Jan 7 at 0:42


















          • $begingroup$
            Please explain the part "then it is not adjacent to $v$". How do you know it's not adjacent to $v$?
            $endgroup$
            – bof
            Jan 6 at 23:50










          • $begingroup$
            Ah right sorry...I'm actually not sure about it either...
            $endgroup$
            – nafhgood
            Jan 7 at 0:42
















          $begingroup$
          Please explain the part "then it is not adjacent to $v$". How do you know it's not adjacent to $v$?
          $endgroup$
          – bof
          Jan 6 at 23:50




          $begingroup$
          Please explain the part "then it is not adjacent to $v$". How do you know it's not adjacent to $v$?
          $endgroup$
          – bof
          Jan 6 at 23:50












          $begingroup$
          Ah right sorry...I'm actually not sure about it either...
          $endgroup$
          – nafhgood
          Jan 7 at 0:42




          $begingroup$
          Ah right sorry...I'm actually not sure about it either...
          $endgroup$
          – nafhgood
          Jan 7 at 0:42


















          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%2f3063569%2fminimum-number-of-colors-enough-to-color-the-vertices-of-any-graph-whose-vertex%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

          1300-talet

          Image
          Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
          Read more

          1300-talet

          Image
          Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
          Read more

          Display a custom attribute below product name in the front-end Magento 1.9.3.8

          Image
          0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
          Read more