Implications of multiple ways to order eight numbers
$begingroup$
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
asked Dec 22 '18 at 20:47
STFSTF
431422
$endgroup$
add a comment |
$begingroup$
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
asked Dec 22 '18 at 20:47
STFSTF
431422
$endgroup$
add a comment |
$begingroup$
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
asked Dec 22 '18 at 20:47
STFSTF
431422
$endgroup$
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
linear-algebra combinatorics permutations combinations linear-programming
asked Dec 22 '18 at 20:47
STFSTF
431422
asked Dec 22 '18 at 20:47
STFSTF
431422
edited Dec 23 '18 at 19:29
STF
asked Dec 22 '18 at 20:47
STFSTF
431422
asked Dec 22 '18 at 20:47
STFSTF
431422
asked Dec 22 '18 at 20:47
STFSTF
431422
431422
add a comment |
add a comment |
1 Answer
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$begingroup$
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 $(c,d,e,f)=(0.3,0.25,0.15,0.45)$.
answered Dec 26 '18 at 21:48
Alex RavskyAlex Ravsky
42.4k32383
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
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 $(c,d,e,f)=(0.3,0.25,0.15,0.45)$.
answered Dec 26 '18 at 21:48
Alex RavskyAlex Ravsky
42.4k32383
$endgroup$
add a comment |
$begingroup$
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 $(c,d,e,f)=(0.3,0.25,0.15,0.45)$.
answered Dec 26 '18 at 21:48
Alex RavskyAlex Ravsky
42.4k32383
$endgroup$
add a comment |
$begingroup$
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 $(c,d,e,f)=(0.3,0.25,0.15,0.45)$.
answered Dec 26 '18 at 21:48
Alex RavskyAlex Ravsky
42.4k32383
$endgroup$
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 $(c,d,e,f)=(0.3,0.25,0.15,0.45)$.
answered Dec 26 '18 at 21:48
Alex RavskyAlex Ravsky
42.4k32383
answered Dec 26 '18 at 21:48
Alex RavskyAlex Ravsky
42.4k32383
answered Dec 26 '18 at 21:48
Alex RavskyAlex Ravsky
42.4k32383
answered Dec 26 '18 at 21:48
Alex RavskyAlex Ravsky
42.4k32383
42.4k32383
add a comment |
add a comment |
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