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







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  • $begingroup$
    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

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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











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      2 Answers
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      2 Answers
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      active

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      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






























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                      .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 ...
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