Is there any work on partition a partial order set into minimum number total order subsets?












3












$begingroup$


The problem is what's the minimum number of total order subsets can a partial order set partition into?



For example, (1,2) and (3,4) are comparable i.e. (1,2) < (3,4), and (1,2) and (2,1) are incomparable. For the partial order set { (1,2), (3,4), (2,1)} we can partition into two subsets {(1,2), (3,4)} and {(2,1)}, which both are total order sets.



I searched on the web and found little related documents. The question is that whether any works or solutions on this problem exist?






discrete-mathematics order-theory set-partition







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Consider Dilworth's theorem.
    $endgroup$
    – William Elliot
    Jan 4 at 9:10








  • 1




    $begingroup$
    @WilliamElliot Why not make that an answer?
    $endgroup$
    – Arthur
    Jan 4 at 9:33










  • $begingroup$
    @Arthur. Too short.
    $endgroup$
    – William Elliot
    Jan 4 at 11:31










  • $begingroup$
    @WilliamElliot Not at all.
    $endgroup$
    – Arthur
    Jan 4 at 11:36
















3












$begingroup$


The problem is what's the minimum number of total order subsets can a partial order set partition into?



For example, (1,2) and (3,4) are comparable i.e. (1,2) < (3,4), and (1,2) and (2,1) are incomparable. For the partial order set { (1,2), (3,4), (2,1)} we can partition into two subsets {(1,2), (3,4)} and {(2,1)}, which both are total order sets.



I searched on the web and found little related documents. The question is that whether any works or solutions on this problem exist?






discrete-mathematics order-theory set-partition







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Consider Dilworth's theorem.
    $endgroup$
    – William Elliot
    Jan 4 at 9:10








  • 1




    $begingroup$
    @WilliamElliot Why not make that an answer?
    $endgroup$
    – Arthur
    Jan 4 at 9:33










  • $begingroup$
    @Arthur. Too short.
    $endgroup$
    – William Elliot
    Jan 4 at 11:31










  • $begingroup$
    @WilliamElliot Not at all.
    $endgroup$
    – Arthur
    Jan 4 at 11:36














3












3








3


0



$begingroup$


The problem is what's the minimum number of total order subsets can a partial order set partition into?



For example, (1,2) and (3,4) are comparable i.e. (1,2) < (3,4), and (1,2) and (2,1) are incomparable. For the partial order set { (1,2), (3,4), (2,1)} we can partition into two subsets {(1,2), (3,4)} and {(2,1)}, which both are total order sets.



I searched on the web and found little related documents. The question is that whether any works or solutions on this problem exist?






discrete-mathematics order-theory set-partition







share|cite|improve this question











$endgroup$




The problem is what's the minimum number of total order subsets can a partial order set partition into?



For example, (1,2) and (3,4) are comparable i.e. (1,2) < (3,4), and (1,2) and (2,1) are incomparable. For the partial order set { (1,2), (3,4), (2,1)} we can partition into two subsets {(1,2), (3,4)} and {(2,1)}, which both are total order sets.



I searched on the web and found little related documents. The question is that whether any works or solutions on this problem exist?






discrete-mathematics order-theory set-partition




discrete-mathematics order-theory set-partition






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 8:58









Asaf Karagila♦

305k33436766




305k33436766










asked Jan 4 at 8:23









wang zhihaowang zhihao

1166




1166








  • 4




    $begingroup$
    Consider Dilworth's theorem.
    $endgroup$
    – William Elliot
    Jan 4 at 9:10








  • 1




    $begingroup$
    @WilliamElliot Why not make that an answer?
    $endgroup$
    – Arthur
    Jan 4 at 9:33










  • $begingroup$
    @Arthur. Too short.
    $endgroup$
    – William Elliot
    Jan 4 at 11:31










  • $begingroup$
    @WilliamElliot Not at all.
    $endgroup$
    – Arthur
    Jan 4 at 11:36














  • 4




    $begingroup$
    Consider Dilworth's theorem.
    $endgroup$
    – William Elliot
    Jan 4 at 9:10








  • 1




    $begingroup$
    @WilliamElliot Why not make that an answer?
    $endgroup$
    – Arthur
    Jan 4 at 9:33










  • $begingroup$
    @Arthur. Too short.
    $endgroup$
    – William Elliot
    Jan 4 at 11:31










  • $begingroup$
    @WilliamElliot Not at all.
    $endgroup$
    – Arthur
    Jan 4 at 11:36








4




4




$begingroup$
Consider Dilworth's theorem.
$endgroup$
– William Elliot
Jan 4 at 9:10






$begingroup$
Consider Dilworth's theorem.
$endgroup$
– William Elliot
Jan 4 at 9:10






1




1




$begingroup$
@WilliamElliot Why not make that an answer?
$endgroup$
– Arthur
Jan 4 at 9:33




$begingroup$
@WilliamElliot Why not make that an answer?
$endgroup$
– Arthur
Jan 4 at 9:33












$begingroup$
@Arthur. Too short.
$endgroup$
– William Elliot
Jan 4 at 11:31




$begingroup$
@Arthur. Too short.
$endgroup$
– William Elliot
Jan 4 at 11:31












$begingroup$
@WilliamElliot Not at all.
$endgroup$
– Arthur
Jan 4 at 11:36




$begingroup$
@WilliamElliot Not at all.
$endgroup$
– Arthur
Jan 4 at 11:36










1 Answer
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$begingroup$

Given a finite partially ordered set, Dilworth's theorem says that the minimal number of totally ordered subsets needed to cover the entire set is equal to the size of the largest antichain (also known as the width of the set). An antichain is a subset where no element is comparable to any of the other elements in the antichain.






share|cite|improve this answer











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    $begingroup$

    Given a finite partially ordered set, Dilworth's theorem says that the minimal number of totally ordered subsets needed to cover the entire set is equal to the size of the largest antichain (also known as the width of the set). An antichain is a subset where no element is comparable to any of the other elements in the antichain.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Given a finite partially ordered set, Dilworth's theorem says that the minimal number of totally ordered subsets needed to cover the entire set is equal to the size of the largest antichain (also known as the width of the set). An antichain is a subset where no element is comparable to any of the other elements in the antichain.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Given a finite partially ordered set, Dilworth's theorem says that the minimal number of totally ordered subsets needed to cover the entire set is equal to the size of the largest antichain (also known as the width of the set). An antichain is a subset where no element is comparable to any of the other elements in the antichain.






        share|cite|improve this answer











        $endgroup$



        Given a finite partially ordered set, Dilworth's theorem says that the minimal number of totally ordered subsets needed to cover the entire set is equal to the size of the largest antichain (also known as the width of the set). An antichain is a subset where no element is comparable to any of the other elements in the antichain.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 4 at 12:32

























        answered Jan 4 at 11:36









        ArthurArthur

        116k7116199




        116k7116199






























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