How to find the minimum number of rooks present in the given situation on the chess board?












2












$begingroup$


I have the following question with me:




"Rooks are placed on the $n * n$ chess board satisfying the following condition:



If the square $(i,j)$ is free then there are at least n rooks on the $i^{th}$ row and $j^{th}$ column together. "



Show that there are at least $n^2/2$ rooks present on the board




I consider the row/column which has the minimum number of rooks, say $k$. If $kgeq n/2$ it is easy to show that



But I am unable to prove the statement for $k < n/2$. Can somebody please explan how to proceed?






combinatorics graph-theory optimization







share|cite|improve this question











$endgroup$












  • $begingroup$
    sorry that was major mistake in typing the question. I have edited it now
    $endgroup$
    – saisanjeev
    Jan 3 at 6:31
















2












$begingroup$


I have the following question with me:




"Rooks are placed on the $n * n$ chess board satisfying the following condition:



If the square $(i,j)$ is free then there are at least n rooks on the $i^{th}$ row and $j^{th}$ column together. "



Show that there are at least $n^2/2$ rooks present on the board




I consider the row/column which has the minimum number of rooks, say $k$. If $kgeq n/2$ it is easy to show that



But I am unable to prove the statement for $k < n/2$. Can somebody please explan how to proceed?






combinatorics graph-theory optimization







share|cite|improve this question











$endgroup$












  • $begingroup$
    sorry that was major mistake in typing the question. I have edited it now
    $endgroup$
    – saisanjeev
    Jan 3 at 6:31














2












2








2


0



$begingroup$


I have the following question with me:




"Rooks are placed on the $n * n$ chess board satisfying the following condition:



If the square $(i,j)$ is free then there are at least n rooks on the $i^{th}$ row and $j^{th}$ column together. "



Show that there are at least $n^2/2$ rooks present on the board




I consider the row/column which has the minimum number of rooks, say $k$. If $kgeq n/2$ it is easy to show that



But I am unable to prove the statement for $k < n/2$. Can somebody please explan how to proceed?






combinatorics graph-theory optimization







share|cite|improve this question











$endgroup$




I have the following question with me:




"Rooks are placed on the $n * n$ chess board satisfying the following condition:



If the square $(i,j)$ is free then there are at least n rooks on the $i^{th}$ row and $j^{th}$ column together. "



Show that there are at least $n^2/2$ rooks present on the board




I consider the row/column which has the minimum number of rooks, say $k$. If $kgeq n/2$ it is easy to show that



But I am unable to prove the statement for $k < n/2$. Can somebody please explan how to proceed?






combinatorics graph-theory optimization




combinatorics graph-theory optimization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 6:30







saisanjeev

















asked Jan 2 at 11:51









saisanjeevsaisanjeev

997212




997212












  • $begingroup$
    sorry that was major mistake in typing the question. I have edited it now
    $endgroup$
    – saisanjeev
    Jan 3 at 6:31


















  • $begingroup$
    sorry that was major mistake in typing the question. I have edited it now
    $endgroup$
    – saisanjeev
    Jan 3 at 6:31
















$begingroup$
sorry that was major mistake in typing the question. I have edited it now
$endgroup$
– saisanjeev
Jan 3 at 6:31




$begingroup$
sorry that was major mistake in typing the question. I have edited it now
$endgroup$
– saisanjeev
Jan 3 at 6:31










2 Answers
2






active

oldest

votes


















1












$begingroup$

Let $c_j$ be the number of rooks placed on $j$-th column and $r_i$ be the number of rooks placed on $i$-th row. The condition can be phrased as
$$
(i,j) text{ is free} Rightarrow r_i + c_j geq n.
$$ Now, let $F $ be the family of free pairs $(i,j)$. Then, we have
$$
sum_{i=1}^n r_i = sum_{j=1}^n c_j =n^2 -|F|,
$$ and
$$
n|F| le sum_{(i,j)in F} (r_i + c_j).tag{*}
$$On the other hand, the RHS of $(*)$ can be computed as
$$
sum_{(i,j)in F} (r_i + c_j) = sum_{i=1}^n r_i(n-r_i) + sum_{j=1}^n c_j(n-c_j) = 2n(n^2-|F|) -sum_{i=1}^n r_i^2-sum_{j=1}^n c_j^2.
$$ since for each $i$-th row, the number of $j$ for which $(i,j)$ is free is $n-r_i$ and for each $j$-th column, it is $n-c_j$. By Cauchy-Schwarz, we have
$$
sum_{i=1}^n r_i^2ge frac{(sum_{i=1}^nr_i)^2}{n} = frac{(n^2-|F|)^2}{n}
$$ and similarly for $sum_{j=1}^n c_j^2$. Therefore, we get the bound
$$
n|F| leq 2n(n^2-|F|)-2frac{(n^2-|F|)^2}{n}.
$$ Multiply $n$ on both sides and we have
$$
2|F|^2 le n^2 |F|.
$$ This proves $|F|le frac{n^2}{2}$ as desired.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Let $c_i$ be the number of free cells in $i$-th column and



    let $r_j$ be the number of free cells in $i$-th row.



    We make a graph between free cells so that two are connected if both are in the same line (row or column). So we see that each cell has a degree at most $n-2$. Let $f$ be a number of all cells. Then by handshake lemma we have $$2varepsilon leq fcdot (n-2)$$
    where $varepsilon$ is total number of edges.



    Notice that $$ f = sum _{j= 1}^n r_j = sum _{i= 1}^n c_i$$



    Let us count $varepsilon $ on each row and column. Since we have $r_j$ free cells in $j-$ th row, we have ${r_jchoose 2}$ edges in $j$-th row. The same story goes for columns. So $$ varepsilon = sum _{j= 1}^n {r_jchoose 2}+sum _{i= 1}^n{c_ichoose 2}$$
    By Cauchy inequality we have: $$sum _{j= 1}^n {r_jchoose 2} geq {{1over n}f^2-fover 2}$$
    and $$sum _{i= 1}^n {c_ichoose 2} geq {{1over n}f^2-fover 2}$$



    So $$(n-2)cdot fgeq 4{{1over n}f^2-fover 2} implies fleq {n^2over 2}$$
    and we are done.






    share|cite|improve this answer











    $endgroup$













      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%2f3059391%2fhow-to-find-the-minimum-number-of-rooks-present-in-the-given-situation-on-the-ch%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












      $begingroup$

      Let $c_j$ be the number of rooks placed on $j$-th column and $r_i$ be the number of rooks placed on $i$-th row. The condition can be phrased as
      $$
      (i,j) text{ is free} Rightarrow r_i + c_j geq n.
      $$       Now, let $F $ be the family of free pairs $(i,j)$. Then, we have
      $$
      sum_{i=1}^n r_i = sum_{j=1}^n c_j =n^2 -|F|,
      $$       and
      $$
      n|F| le sum_{(i,j)in F} (r_i + c_j).tag{*}
      $$      On the other hand, the RHS of $(*)$ can be computed as
      $$
      sum_{(i,j)in F} (r_i + c_j) = sum_{i=1}^n r_i(n-r_i) + sum_{j=1}^n c_j(n-c_j) = 2n(n^2-|F|) -sum_{i=1}^n r_i^2-sum_{j=1}^n c_j^2.
      $$       since for each $i$-th row, the number of $j$ for which $(i,j)$ is free is $n-r_i$ and for each $j$-th column, it is $n-c_j$. By Cauchy-Schwarz, we have
      $$
      sum_{i=1}^n r_i^2ge frac{(sum_{i=1}^nr_i)^2}{n} = frac{(n^2-|F|)^2}{n}
      $$       and similarly for $sum_{j=1}^n c_j^2$. Therefore, we get the bound
      $$
      n|F| leq 2n(n^2-|F|)-2frac{(n^2-|F|)^2}{n}.
      $$       Multiply $n$ on both sides and we have
      $$
      2|F|^2 le n^2 |F|.
      $$       This proves $|F|le frac{n^2}{2}$ as desired.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        Let $c_j$ be the number of rooks placed on $j$-th column and $r_i$ be the number of rooks placed on $i$-th row. The condition can be phrased as
        $$
        (i,j) text{ is free} Rightarrow r_i + c_j geq n.
        $$         Now, let $F $ be the family of free pairs $(i,j)$. Then, we have
        $$
        sum_{i=1}^n r_i = sum_{j=1}^n c_j =n^2 -|F|,
        $$         and
        $$
        n|F| le sum_{(i,j)in F} (r_i + c_j).tag{*}
        $$        On the other hand, the RHS of $(*)$ can be computed as
        $$
        sum_{(i,j)in F} (r_i + c_j) = sum_{i=1}^n r_i(n-r_i) + sum_{j=1}^n c_j(n-c_j) = 2n(n^2-|F|) -sum_{i=1}^n r_i^2-sum_{j=1}^n c_j^2.
        $$         since for each $i$-th row, the number of $j$ for which $(i,j)$ is free is $n-r_i$ and for each $j$-th column, it is $n-c_j$. By Cauchy-Schwarz, we have
        $$
        sum_{i=1}^n r_i^2ge frac{(sum_{i=1}^nr_i)^2}{n} = frac{(n^2-|F|)^2}{n}
        $$         and similarly for $sum_{j=1}^n c_j^2$. Therefore, we get the bound
        $$
        n|F| leq 2n(n^2-|F|)-2frac{(n^2-|F|)^2}{n}.
        $$         Multiply $n$ on both sides and we have
        $$
        2|F|^2 le n^2 |F|.
        $$         This proves $|F|le frac{n^2}{2}$ as desired.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          Let $c_j$ be the number of rooks placed on $j$-th column and $r_i$ be the number of rooks placed on $i$-th row. The condition can be phrased as
          $$
          (i,j) text{ is free} Rightarrow r_i + c_j geq n.
          $$           Now, let $F $ be the family of free pairs $(i,j)$. Then, we have
          $$
          sum_{i=1}^n r_i = sum_{j=1}^n c_j =n^2 -|F|,
          $$           and
          $$
          n|F| le sum_{(i,j)in F} (r_i + c_j).tag{*}
          $$          On the other hand, the RHS of $(*)$ can be computed as
          $$
          sum_{(i,j)in F} (r_i + c_j) = sum_{i=1}^n r_i(n-r_i) + sum_{j=1}^n c_j(n-c_j) = 2n(n^2-|F|) -sum_{i=1}^n r_i^2-sum_{j=1}^n c_j^2.
          $$           since for each $i$-th row, the number of $j$ for which $(i,j)$ is free is $n-r_i$ and for each $j$-th column, it is $n-c_j$. By Cauchy-Schwarz, we have
          $$
          sum_{i=1}^n r_i^2ge frac{(sum_{i=1}^nr_i)^2}{n} = frac{(n^2-|F|)^2}{n}
          $$           and similarly for $sum_{j=1}^n c_j^2$. Therefore, we get the bound
          $$
          n|F| leq 2n(n^2-|F|)-2frac{(n^2-|F|)^2}{n}.
          $$           Multiply $n$ on both sides and we have
          $$
          2|F|^2 le n^2 |F|.
          $$           This proves $|F|le frac{n^2}{2}$ as desired.






          share|cite|improve this answer











          $endgroup$



          Let $c_j$ be the number of rooks placed on $j$-th column and $r_i$ be the number of rooks placed on $i$-th row. The condition can be phrased as
          $$
          (i,j) text{ is free} Rightarrow r_i + c_j geq n.
          $$           Now, let $F $ be the family of free pairs $(i,j)$. Then, we have
          $$
          sum_{i=1}^n r_i = sum_{j=1}^n c_j =n^2 -|F|,
          $$           and
          $$
          n|F| le sum_{(i,j)in F} (r_i + c_j).tag{*}
          $$          On the other hand, the RHS of $(*)$ can be computed as
          $$
          sum_{(i,j)in F} (r_i + c_j) = sum_{i=1}^n r_i(n-r_i) + sum_{j=1}^n c_j(n-c_j) = 2n(n^2-|F|) -sum_{i=1}^n r_i^2-sum_{j=1}^n c_j^2.
          $$           since for each $i$-th row, the number of $j$ for which $(i,j)$ is free is $n-r_i$ and for each $j$-th column, it is $n-c_j$. By Cauchy-Schwarz, we have
          $$
          sum_{i=1}^n r_i^2ge frac{(sum_{i=1}^nr_i)^2}{n} = frac{(n^2-|F|)^2}{n}
          $$           and similarly for $sum_{j=1}^n c_j^2$. Therefore, we get the bound
          $$
          n|F| leq 2n(n^2-|F|)-2frac{(n^2-|F|)^2}{n}.
          $$           Multiply $n$ on both sides and we have
          $$
          2|F|^2 le n^2 |F|.
          $$           This proves $|F|le frac{n^2}{2}$ as desired.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 2 at 12:34

























          answered Jan 2 at 12:28









          SongSong

          16.7k1941




          16.7k1941























              1












              $begingroup$

              Let $c_i$ be the number of free cells in $i$-th column and



              let $r_j$ be the number of free cells in $i$-th row.



              We make a graph between free cells so that two are connected if both are in the same line (row or column). So we see that each cell has a degree at most $n-2$. Let $f$ be a number of all cells. Then by handshake lemma we have $$2varepsilon leq fcdot (n-2)$$
              where $varepsilon$ is total number of edges.



              Notice that $$ f = sum _{j= 1}^n r_j = sum _{i= 1}^n c_i$$



              Let us count $varepsilon $ on each row and column. Since we have $r_j$ free cells in $j-$ th row, we have ${r_jchoose 2}$ edges in $j$-th row. The same story goes for columns. So $$ varepsilon = sum _{j= 1}^n {r_jchoose 2}+sum _{i= 1}^n{c_ichoose 2}$$
              By Cauchy inequality we have: $$sum _{j= 1}^n {r_jchoose 2} geq {{1over n}f^2-fover 2}$$
              and $$sum _{i= 1}^n {c_ichoose 2} geq {{1over n}f^2-fover 2}$$



              So $$(n-2)cdot fgeq 4{{1over n}f^2-fover 2} implies fleq {n^2over 2}$$
              and we are done.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                Let $c_i$ be the number of free cells in $i$-th column and



                let $r_j$ be the number of free cells in $i$-th row.



                We make a graph between free cells so that two are connected if both are in the same line (row or column). So we see that each cell has a degree at most $n-2$. Let $f$ be a number of all cells. Then by handshake lemma we have $$2varepsilon leq fcdot (n-2)$$
                where $varepsilon$ is total number of edges.



                Notice that $$ f = sum _{j= 1}^n r_j = sum _{i= 1}^n c_i$$



                Let us count $varepsilon $ on each row and column. Since we have $r_j$ free cells in $j-$ th row, we have ${r_jchoose 2}$ edges in $j$-th row. The same story goes for columns. So $$ varepsilon = sum _{j= 1}^n {r_jchoose 2}+sum _{i= 1}^n{c_ichoose 2}$$
                By Cauchy inequality we have: $$sum _{j= 1}^n {r_jchoose 2} geq {{1over n}f^2-fover 2}$$
                and $$sum _{i= 1}^n {c_ichoose 2} geq {{1over n}f^2-fover 2}$$



                So $$(n-2)cdot fgeq 4{{1over n}f^2-fover 2} implies fleq {n^2over 2}$$
                and we are done.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Let $c_i$ be the number of free cells in $i$-th column and



                  let $r_j$ be the number of free cells in $i$-th row.



                  We make a graph between free cells so that two are connected if both are in the same line (row or column). So we see that each cell has a degree at most $n-2$. Let $f$ be a number of all cells. Then by handshake lemma we have $$2varepsilon leq fcdot (n-2)$$
                  where $varepsilon$ is total number of edges.



                  Notice that $$ f = sum _{j= 1}^n r_j = sum _{i= 1}^n c_i$$



                  Let us count $varepsilon $ on each row and column. Since we have $r_j$ free cells in $j-$ th row, we have ${r_jchoose 2}$ edges in $j$-th row. The same story goes for columns. So $$ varepsilon = sum _{j= 1}^n {r_jchoose 2}+sum _{i= 1}^n{c_ichoose 2}$$
                  By Cauchy inequality we have: $$sum _{j= 1}^n {r_jchoose 2} geq {{1over n}f^2-fover 2}$$
                  and $$sum _{i= 1}^n {c_ichoose 2} geq {{1over n}f^2-fover 2}$$



                  So $$(n-2)cdot fgeq 4{{1over n}f^2-fover 2} implies fleq {n^2over 2}$$
                  and we are done.






                  share|cite|improve this answer











                  $endgroup$



                  Let $c_i$ be the number of free cells in $i$-th column and



                  let $r_j$ be the number of free cells in $i$-th row.



                  We make a graph between free cells so that two are connected if both are in the same line (row or column). So we see that each cell has a degree at most $n-2$. Let $f$ be a number of all cells. Then by handshake lemma we have $$2varepsilon leq fcdot (n-2)$$
                  where $varepsilon$ is total number of edges.



                  Notice that $$ f = sum _{j= 1}^n r_j = sum _{i= 1}^n c_i$$



                  Let us count $varepsilon $ on each row and column. Since we have $r_j$ free cells in $j-$ th row, we have ${r_jchoose 2}$ edges in $j$-th row. The same story goes for columns. So $$ varepsilon = sum _{j= 1}^n {r_jchoose 2}+sum _{i= 1}^n{c_ichoose 2}$$
                  By Cauchy inequality we have: $$sum _{j= 1}^n {r_jchoose 2} geq {{1over n}f^2-fover 2}$$
                  and $$sum _{i= 1}^n {c_ichoose 2} geq {{1over n}f^2-fover 2}$$



                  So $$(n-2)cdot fgeq 4{{1over n}f^2-fover 2} implies fleq {n^2over 2}$$
                  and we are done.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 2 at 12:42

























                  answered Jan 2 at 12:34









                  greedoidgreedoid

                  45.9k1160116




                  45.9k1160116






























                      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%2f3059391%2fhow-to-find-the-minimum-number-of-rooks-present-in-the-given-situation-on-the-ch%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

                      Cabo Verde

                      Image
                        Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
                      Read more

                      Gyllenstierna

                      Image
                      .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 .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
                      Read more

                      Karlovacs län

                      Image
                      .mw-parser-output table.ambox{margin:0 10%;border-collapse:collapse;background:#fbfbfb;border:1px solid #aaa;border-left:10px solid #608ec2}.mw-parser-output table.ambox th.ambox-text,.mw-parser-output table.ambox td.ambox-text{padding:.25em .5em;width:100%}.mw-parser-output table.ambox td.ambox-image{padding:2px 0 2px .5em;text-align:center;vertical-align:middle}.mw-parser-output table.ambox td.ambox-imageright{padding:2px 4px 2px 0;text-align:center;vertical-align:middle}.mw-parser-output table.ambox-notice{border-left:10px solid #608ec2}.mw-parser-output table.ambox-delete,.mw-parser-output table.ambox-serious{border-left:10px solid #b22222}.mw-parser-output table.ambox-content{border-left:10px solid #f28500}.mw-parser-output table.ambox-style{border-left:10px solid #f4c430}.mw-parser-output table.ambox-merge{border-left:10px solid #9932cc}.mw-parser-output table.ambox-protection{border-left:10px solid #bba}.mw-parser-output .ambox+.ambox,.mw-parser-output .topbox+.ambox,.mw-pars
                      Read more