Implications of multiple ways to order eight numbers












1














Consider two sets $A,B$ composed of four real numbers each.



These eight real numbers are in $[0,1]$.



Consider other four real numbers $c,d,e,f$ each in $[0,1]$, all different between each other.



Assume there exists a way of ordering the four numbers in each set $A,B$ such that
$$
begin{cases}
w^A_1+w^B_1=c\
w^A_2+w^B_2=d\
w^A_3+w^B_3=e\
w^A_4+w^B_4=f\
end{cases}
$$

where




  • $w^A_h$ denotes the $h$th element of set $A$ once we have ordered its 4 elements


  • $w^B_h$ denotes the $h$th element of set $B$ once we have ordered its 4 elements



Claim: if such an ordering is not unique, then it should be that two numbers in $A$ are equal and/or that two numbers in $B$ are equal.



Is this claim correct? If yes, how can I prove it? If not, can you provide a counterexample?



(similar question here but with 2 elements per set)





Maybe the claim is wrong? Let ${a_1,a_2,a_3,a_4}$ be the elements of $A$ and ${b_1,b_2,b_3,b_4}$ be the elements of $B$. We could have:



order I
$$
begin{cases}
a_2+b_3=c\
a_4+b_4=d\
a_1+b_1=e\
a_3+b_2=f
end{cases}
$$



and



order II
$$
begin{cases}
a_1+b_2=c\
a_2+b_1=d\
a_3+b_4=e\
a_4+b_3=f
end{cases}
$$



which implies
$$
begin{cases}
a_2+b_3=a_1+b_2\
a_4+b_4=a_2+b_1\
a_1+b_1=a_3+b_4\
a_3+b_2=a_4+b_3
end{cases}
$$

Does this imply that two numbers in $A$ are equal and/or that two numbers in $B$ are equal?










share|cite|improve this question





























    1














    Consider two sets $A,B$ composed of four real numbers each.



    These eight real numbers are in $[0,1]$.



    Consider other four real numbers $c,d,e,f$ each in $[0,1]$, all different between each other.



    Assume there exists a way of ordering the four numbers in each set $A,B$ such that
    $$
    begin{cases}
    w^A_1+w^B_1=c\
    w^A_2+w^B_2=d\
    w^A_3+w^B_3=e\
    w^A_4+w^B_4=f\
    end{cases}
    $$

    where




    • $w^A_h$ denotes the $h$th element of set $A$ once we have ordered its 4 elements


    • $w^B_h$ denotes the $h$th element of set $B$ once we have ordered its 4 elements



    Claim: if such an ordering is not unique, then it should be that two numbers in $A$ are equal and/or that two numbers in $B$ are equal.



    Is this claim correct? If yes, how can I prove it? If not, can you provide a counterexample?



    (similar question here but with 2 elements per set)





    Maybe the claim is wrong? Let ${a_1,a_2,a_3,a_4}$ be the elements of $A$ and ${b_1,b_2,b_3,b_4}$ be the elements of $B$. We could have:



    order I
    $$
    begin{cases}
    a_2+b_3=c\
    a_4+b_4=d\
    a_1+b_1=e\
    a_3+b_2=f
    end{cases}
    $$



    and



    order II
    $$
    begin{cases}
    a_1+b_2=c\
    a_2+b_1=d\
    a_3+b_4=e\
    a_4+b_3=f
    end{cases}
    $$



    which implies
    $$
    begin{cases}
    a_2+b_3=a_1+b_2\
    a_4+b_4=a_2+b_1\
    a_1+b_1=a_3+b_4\
    a_3+b_2=a_4+b_3
    end{cases}
    $$

    Does this imply that two numbers in $A$ are equal and/or that two numbers in $B$ are equal?










    share|cite|improve this question



























      1












      1








      1


      1





      Consider two sets $A,B$ composed of four real numbers each.



      These eight real numbers are in $[0,1]$.



      Consider other four real numbers $c,d,e,f$ each in $[0,1]$, all different between each other.



      Assume there exists a way of ordering the four numbers in each set $A,B$ such that
      $$
      begin{cases}
      w^A_1+w^B_1=c\
      w^A_2+w^B_2=d\
      w^A_3+w^B_3=e\
      w^A_4+w^B_4=f\
      end{cases}
      $$

      where




      • $w^A_h$ denotes the $h$th element of set $A$ once we have ordered its 4 elements


      • $w^B_h$ denotes the $h$th element of set $B$ once we have ordered its 4 elements



      Claim: if such an ordering is not unique, then it should be that two numbers in $A$ are equal and/or that two numbers in $B$ are equal.



      Is this claim correct? If yes, how can I prove it? If not, can you provide a counterexample?



      (similar question here but with 2 elements per set)





      Maybe the claim is wrong? Let ${a_1,a_2,a_3,a_4}$ be the elements of $A$ and ${b_1,b_2,b_3,b_4}$ be the elements of $B$. We could have:



      order I
      $$
      begin{cases}
      a_2+b_3=c\
      a_4+b_4=d\
      a_1+b_1=e\
      a_3+b_2=f
      end{cases}
      $$



      and



      order II
      $$
      begin{cases}
      a_1+b_2=c\
      a_2+b_1=d\
      a_3+b_4=e\
      a_4+b_3=f
      end{cases}
      $$



      which implies
      $$
      begin{cases}
      a_2+b_3=a_1+b_2\
      a_4+b_4=a_2+b_1\
      a_1+b_1=a_3+b_4\
      a_3+b_2=a_4+b_3
      end{cases}
      $$

      Does this imply that two numbers in $A$ are equal and/or that two numbers in $B$ are equal?










      share|cite|improve this question















      Consider two sets $A,B$ composed of four real numbers each.



      These eight real numbers are in $[0,1]$.



      Consider other four real numbers $c,d,e,f$ each in $[0,1]$, all different between each other.



      Assume there exists a way of ordering the four numbers in each set $A,B$ such that
      $$
      begin{cases}
      w^A_1+w^B_1=c\
      w^A_2+w^B_2=d\
      w^A_3+w^B_3=e\
      w^A_4+w^B_4=f\
      end{cases}
      $$

      where




      • $w^A_h$ denotes the $h$th element of set $A$ once we have ordered its 4 elements


      • $w^B_h$ denotes the $h$th element of set $B$ once we have ordered its 4 elements



      Claim: if such an ordering is not unique, then it should be that two numbers in $A$ are equal and/or that two numbers in $B$ are equal.



      Is this claim correct? If yes, how can I prove it? If not, can you provide a counterexample?



      (similar question here but with 2 elements per set)





      Maybe the claim is wrong? Let ${a_1,a_2,a_3,a_4}$ be the elements of $A$ and ${b_1,b_2,b_3,b_4}$ be the elements of $B$. We could have:



      order I
      $$
      begin{cases}
      a_2+b_3=c\
      a_4+b_4=d\
      a_1+b_1=e\
      a_3+b_2=f
      end{cases}
      $$



      and



      order II
      $$
      begin{cases}
      a_1+b_2=c\
      a_2+b_1=d\
      a_3+b_4=e\
      a_4+b_3=f
      end{cases}
      $$



      which implies
      $$
      begin{cases}
      a_2+b_3=a_1+b_2\
      a_4+b_4=a_2+b_1\
      a_1+b_1=a_3+b_4\
      a_3+b_2=a_4+b_3
      end{cases}
      $$

      Does this imply that two numbers in $A$ are equal and/or that two numbers in $B$ are equal?







      linear-algebra combinatorics permutations combinations linear-programming






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      edited Dec 23 '18 at 19:29

























      asked Dec 22 '18 at 20:47









      STF

      801420




      801420






















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          The claim is wrong. Pick the following partial solution of the last system: $(a_1,a_2,a_3,a_4)=(0,a,a’,a+a’)$ and $(b_1,b_2,b_3,b_4)=(a’,a+b,b,0)$. Then $(c,d,e,f)=(a+b,a+a’,a’,a+a’+b)$. For instance, we can put $a=0.1$, $a’=0.15$ and $b=0.2$. Then $(a_1,a_2,a_3,a_4)=(0,0.1,0.15,0.25)$, $(b_1,b_2,b_3,b_4)=(0.15,0.3,0.2,0)$, and $(c,d,e,f)=(0.3,0.25,0.15,0.45)$.






          share|cite|improve this answer





















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            1 Answer
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            The claim is wrong. Pick the following partial solution of the last system: $(a_1,a_2,a_3,a_4)=(0,a,a’,a+a’)$ and $(b_1,b_2,b_3,b_4)=(a’,a+b,b,0)$. Then $(c,d,e,f)=(a+b,a+a’,a’,a+a’+b)$. For instance, we can put $a=0.1$, $a’=0.15$ and $b=0.2$. Then $(a_1,a_2,a_3,a_4)=(0,0.1,0.15,0.25)$, $(b_1,b_2,b_3,b_4)=(0.15,0.3,0.2,0)$, and $(c,d,e,f)=(0.3,0.25,0.15,0.45)$.






            share|cite|improve this answer


























              1














              The claim is wrong. Pick the following partial solution of the last system: $(a_1,a_2,a_3,a_4)=(0,a,a’,a+a’)$ and $(b_1,b_2,b_3,b_4)=(a’,a+b,b,0)$. Then $(c,d,e,f)=(a+b,a+a’,a’,a+a’+b)$. For instance, we can put $a=0.1$, $a’=0.15$ and $b=0.2$. Then $(a_1,a_2,a_3,a_4)=(0,0.1,0.15,0.25)$, $(b_1,b_2,b_3,b_4)=(0.15,0.3,0.2,0)$, and $(c,d,e,f)=(0.3,0.25,0.15,0.45)$.






              share|cite|improve this answer
























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                The claim is wrong. Pick the following partial solution of the last system: $(a_1,a_2,a_3,a_4)=(0,a,a’,a+a’)$ and $(b_1,b_2,b_3,b_4)=(a’,a+b,b,0)$. Then $(c,d,e,f)=(a+b,a+a’,a’,a+a’+b)$. For instance, we can put $a=0.1$, $a’=0.15$ and $b=0.2$. Then $(a_1,a_2,a_3,a_4)=(0,0.1,0.15,0.25)$, $(b_1,b_2,b_3,b_4)=(0.15,0.3,0.2,0)$, and $(c,d,e,f)=(0.3,0.25,0.15,0.45)$.






                share|cite|improve this answer












                The claim is wrong. Pick the following partial solution of the last system: $(a_1,a_2,a_3,a_4)=(0,a,a’,a+a’)$ and $(b_1,b_2,b_3,b_4)=(a’,a+b,b,0)$. Then $(c,d,e,f)=(a+b,a+a’,a’,a+a’+b)$. For instance, we can put $a=0.1$, $a’=0.15$ and $b=0.2$. Then $(a_1,a_2,a_3,a_4)=(0,0.1,0.15,0.25)$, $(b_1,b_2,b_3,b_4)=(0.15,0.3,0.2,0)$, and $(c,d,e,f)=(0.3,0.25,0.15,0.45)$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 26 '18 at 21:48









                Alex Ravsky

                39.2k32080




                39.2k32080






























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