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Comparing elements of sets
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Let $a_1, a_2, a_3, a_4$ be real numbers.
Consider the following sets
$$
mathcalU_1equiv -a_1, a_1, -a_2, a_2, 0, infty, -infty
$$
$$
mathcalU_2equiv -a_3, a_3, -a_4, a_4, 0, infty, -infty
$$
$$
mathcalU_3equiv (a_3-a_1), (a_4-a_2), infty, (-a_3+a_1), (-a_4+a_2), -infty, 0
$$
$$
mathcalUequiv mathcalU_1times mathcalU_2times mathcalU_3
$$
where $times$ denotes the Cartesian product. $mathcalU_1, mathcalU_2, mathcalU_3$ have cardinality $7$.
Let $uequiv (u_1, u_2, u_3)$ denote a generic element (triplet) of $mathcalU$ with $u_1in mathcalU_1$, $u_2in mathcalU_2$, $u_3in mathcalU_3$.
Take any pair of triplets $u, u'in mathcalU$. In what follows $uleq u'$ means that $u_1leq u'_1$,$u_2leq u'_2$,$u_3leq u'_3$.
Take any triplet $uin mathcalU$. In what follows, $(i_u_1, j_u_2, h_u_3)$ denote the position of $u_1, u_2, u_3$ respectively in $mathcalU_1, mathcalU_2, mathcalU_3$ For example, suppose $uequiv (a_1, a_4, 0)$; then, $(i_u_1, j_u_2, h_u_3)=(2,4,7)$.
Fix $aequiv (a_1, a_2, a_3, a_4)in mathbbR^4$. Suppose we have an algorithm that creates the following two matrices
1) $D(a)$ listing $(i_u_1, j_u_2, h_u_3),(i_u'_1, j_u'_2, h_u'_3)$ $forall u, u'in mathcalU$ such that $uleq u'$
1) $E(a)$ listing $(i_u_1, j_u_2, h_u_3),(i_u'_1, j_u'_2, h_u'_3)$ $forall u, u'in mathcalU$ such that $ugeq u'$.
[Remark: when $u=u'$ then $(i_u_1, j_u_2, h_u_3),(i_u'_1, j_u'_2, h_u'_3)$ appear in $D(a)$ and $E(a)$]
Question: Take any $tildeain mathbbR^4$ with $aneq tildea$. I want to find some sufficient (and possibly also necessary) conditions such that $D(a)=D(tildea)$ and $E(a)=E(tildea)$.
My thoughts After many attempts I realised the following:
Take $aequiv (a_1, a_2, a_3, a_4)in mathbbR^4$ and suppose $a$ is such that
$$
(star)
begincases
-infty< a_1= -a_2<0< a_2= -a_1<infty\
-infty< -a_4< a_3<0< -a_3< a_4<infty\
-infty< a_2-a_4< a_1-a_3<0< a_3-a_1< a_4-a_2<infty\
endcases
$$
Then, every $tildeaequiv (tildea_1, tildea_2, tildea_3, tildea_4)in mathbbR^4$ such that
$$
begincases
-infty< tildea_1= -tildea_2<0< tildea_2= -tildea_1<infty\
-infty< -tildea_4< tildea_3<0< -tildea_3< tildea_4<infty\
-infty< tildea_2-tildea_4< tildea_1-tildea_3<0< tildea_3-tildea_1< tildea_4-tildea_2<infty\
endcases
$$
will have $D(a)=D(tildea)$ and $E(a)=E(tildea)$. This can be generalised to any ordering of $a$ in $(star)$: if $tildea$ respects the same ordering as $a$, then $D(a)=D(tildea)$ and $E(a)=E(tildea)$.
Is this correct? Can you think of other sufficient conditions? Can I get some sufficient and necessary conditions?
general-topology inequality algorithms vectors real-numbers
asked Aug 2 at 12:28
TEX
1419
This question has an open bounty worth +100
reputation from TEX ending ending at 2018-08-11 19:00:44Z">in 4 days.
The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns.
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Let $a_1, a_2, a_3, a_4$ be real numbers.
Consider the following sets
$$
mathcalU_1equiv -a_1, a_1, -a_2, a_2, 0, infty, -infty
$$
$$
mathcalU_2equiv -a_3, a_3, -a_4, a_4, 0, infty, -infty
$$
$$
mathcalU_3equiv (a_3-a_1), (a_4-a_2), infty, (-a_3+a_1), (-a_4+a_2), -infty, 0
$$
$$
mathcalUequiv mathcalU_1times mathcalU_2times mathcalU_3
$$
where $times$ denotes the Cartesian product. $mathcalU_1, mathcalU_2, mathcalU_3$ have cardinality $7$.
Let $uequiv (u_1, u_2, u_3)$ denote a generic element (triplet) of $mathcalU$ with $u_1in mathcalU_1$, $u_2in mathcalU_2$, $u_3in mathcalU_3$.
Take any pair of triplets $u, u'in mathcalU$. In what follows $uleq u'$ means that $u_1leq u'_1$,$u_2leq u'_2$,$u_3leq u'_3$.
Take any triplet $uin mathcalU$. In what follows, $(i_u_1, j_u_2, h_u_3)$ denote the position of $u_1, u_2, u_3$ respectively in $mathcalU_1, mathcalU_2, mathcalU_3$ For example, suppose $uequiv (a_1, a_4, 0)$; then, $(i_u_1, j_u_2, h_u_3)=(2,4,7)$.
Fix $aequiv (a_1, a_2, a_3, a_4)in mathbbR^4$. Suppose we have an algorithm that creates the following two matrices
1) $D(a)$ listing $(i_u_1, j_u_2, h_u_3),(i_u'_1, j_u'_2, h_u'_3)$ $forall u, u'in mathcalU$ such that $uleq u'$
1) $E(a)$ listing $(i_u_1, j_u_2, h_u_3),(i_u'_1, j_u'_2, h_u'_3)$ $forall u, u'in mathcalU$ such that $ugeq u'$.
[Remark: when $u=u'$ then $(i_u_1, j_u_2, h_u_3),(i_u'_1, j_u'_2, h_u'_3)$ appear in $D(a)$ and $E(a)$]
Question: Take any $tildeain mathbbR^4$ with $aneq tildea$. I want to find some sufficient (and possibly also necessary) conditions such that $D(a)=D(tildea)$ and $E(a)=E(tildea)$.
My thoughts After many attempts I realised the following:
Take $aequiv (a_1, a_2, a_3, a_4)in mathbbR^4$ and suppose $a$ is such that
$$
(star)
begincases
-infty< a_1= -a_2<0< a_2= -a_1<infty\
-infty< -a_4< a_3<0< -a_3< a_4<infty\
-infty< a_2-a_4< a_1-a_3<0< a_3-a_1< a_4-a_2<infty\
endcases
$$
Then, every $tildeaequiv (tildea_1, tildea_2, tildea_3, tildea_4)in mathbbR^4$ such that
$$
begincases
-infty< tildea_1= -tildea_2<0< tildea_2= -tildea_1<infty\
-infty< -tildea_4< tildea_3<0< -tildea_3< tildea_4<infty\
-infty< tildea_2-tildea_4< tildea_1-tildea_3<0< tildea_3-tildea_1< tildea_4-tildea_2<infty\
endcases
$$
will have $D(a)=D(tildea)$ and $E(a)=E(tildea)$. This can be generalised to any ordering of $a$ in $(star)$: if $tildea$ respects the same ordering as $a$, then $D(a)=D(tildea)$ and $E(a)=E(tildea)$.
Is this correct? Can you think of other sufficient conditions? Can I get some sufficient and necessary conditions?
general-topology inequality algorithms vectors real-numbers
asked Aug 2 at 12:28
TEX
1419
This question has an open bounty worth +100
reputation from TEX ending ending at 2018-08-11 19:00:44Z">in 4 days.
The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns.
If $V_1,V_2,V_3$ have cardinality 7, then $V$ doesn't.
– Gerry Myerson
Aug 2 at 12:41
Yes, sorry, my mistake
– TEX
Aug 2 at 12:45
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $a_1, a_2, a_3, a_4$ be real numbers.
Consider the following sets
$$
mathcalU_1equiv -a_1, a_1, -a_2, a_2, 0, infty, -infty
$$
$$
mathcalU_2equiv -a_3, a_3, -a_4, a_4, 0, infty, -infty
$$
$$
mathcalU_3equiv (a_3-a_1), (a_4-a_2), infty, (-a_3+a_1), (-a_4+a_2), -infty, 0
$$
$$
mathcalUequiv mathcalU_1times mathcalU_2times mathcalU_3
$$
where $times$ denotes the Cartesian product. $mathcalU_1, mathcalU_2, mathcalU_3$ have cardinality $7$.
Let $uequiv (u_1, u_2, u_3)$ denote a generic element (triplet) of $mathcalU$ with $u_1in mathcalU_1$, $u_2in mathcalU_2$, $u_3in mathcalU_3$.
Take any pair of triplets $u, u'in mathcalU$. In what follows $uleq u'$ means that $u_1leq u'_1$,$u_2leq u'_2$,$u_3leq u'_3$.
Take any triplet $uin mathcalU$. In what follows, $(i_u_1, j_u_2, h_u_3)$ denote the position of $u_1, u_2, u_3$ respectively in $mathcalU_1, mathcalU_2, mathcalU_3$ For example, suppose $uequiv (a_1, a_4, 0)$; then, $(i_u_1, j_u_2, h_u_3)=(2,4,7)$.
Fix $aequiv (a_1, a_2, a_3, a_4)in mathbbR^4$. Suppose we have an algorithm that creates the following two matrices
1) $D(a)$ listing $(i_u_1, j_u_2, h_u_3),(i_u'_1, j_u'_2, h_u'_3)$ $forall u, u'in mathcalU$ such that $uleq u'$
1) $E(a)$ listing $(i_u_1, j_u_2, h_u_3),(i_u'_1, j_u'_2, h_u'_3)$ $forall u, u'in mathcalU$ such that $ugeq u'$.
[Remark: when $u=u'$ then $(i_u_1, j_u_2, h_u_3),(i_u'_1, j_u'_2, h_u'_3)$ appear in $D(a)$ and $E(a)$]
Question: Take any $tildeain mathbbR^4$ with $aneq tildea$. I want to find some sufficient (and possibly also necessary) conditions such that $D(a)=D(tildea)$ and $E(a)=E(tildea)$.
My thoughts After many attempts I realised the following:
Take $aequiv (a_1, a_2, a_3, a_4)in mathbbR^4$ and suppose $a$ is such that
$$
(star)
begincases
-infty< a_1= -a_2<0< a_2= -a_1<infty\
-infty< -a_4< a_3<0< -a_3< a_4<infty\
-infty< a_2-a_4< a_1-a_3<0< a_3-a_1< a_4-a_2<infty\
endcases
$$
Then, every $tildeaequiv (tildea_1, tildea_2, tildea_3, tildea_4)in mathbbR^4$ such that
$$
begincases
-infty< tildea_1= -tildea_2<0< tildea_2= -tildea_1<infty\
-infty< -tildea_4< tildea_3<0< -tildea_3< tildea_4<infty\
-infty< tildea_2-tildea_4< tildea_1-tildea_3<0< tildea_3-tildea_1< tildea_4-tildea_2<infty\
endcases
$$
will have $D(a)=D(tildea)$ and $E(a)=E(tildea)$. This can be generalised to any ordering of $a$ in $(star)$: if $tildea$ respects the same ordering as $a$, then $D(a)=D(tildea)$ and $E(a)=E(tildea)$.
Is this correct? Can you think of other sufficient conditions? Can I get some sufficient and necessary conditions?
general-topology inequality algorithms vectors real-numbers
asked Aug 2 at 12:28
TEX
1419
Let $a_1, a_2, a_3, a_4$ be real numbers.
Consider the following sets
$$
mathcalU_1equiv -a_1, a_1, -a_2, a_2, 0, infty, -infty
$$
$$
mathcalU_2equiv -a_3, a_3, -a_4, a_4, 0, infty, -infty
$$
$$
mathcalU_3equiv (a_3-a_1), (a_4-a_2), infty, (-a_3+a_1), (-a_4+a_2), -infty, 0
$$
$$
mathcalUequiv mathcalU_1times mathcalU_2times mathcalU_3
$$
where $times$ denotes the Cartesian product. $mathcalU_1, mathcalU_2, mathcalU_3$ have cardinality $7$.
Let $uequiv (u_1, u_2, u_3)$ denote a generic element (triplet) of $mathcalU$ with $u_1in mathcalU_1$, $u_2in mathcalU_2$, $u_3in mathcalU_3$.
Take any pair of triplets $u, u'in mathcalU$. In what follows $uleq u'$ means that $u_1leq u'_1$,$u_2leq u'_2$,$u_3leq u'_3$.
Take any triplet $uin mathcalU$. In what follows, $(i_u_1, j_u_2, h_u_3)$ denote the position of $u_1, u_2, u_3$ respectively in $mathcalU_1, mathcalU_2, mathcalU_3$ For example, suppose $uequiv (a_1, a_4, 0)$; then, $(i_u_1, j_u_2, h_u_3)=(2,4,7)$.
Fix $aequiv (a_1, a_2, a_3, a_4)in mathbbR^4$. Suppose we have an algorithm that creates the following two matrices
1) $D(a)$ listing $(i_u_1, j_u_2, h_u_3),(i_u'_1, j_u'_2, h_u'_3)$ $forall u, u'in mathcalU$ such that $uleq u'$
1) $E(a)$ listing $(i_u_1, j_u_2, h_u_3),(i_u'_1, j_u'_2, h_u'_3)$ $forall u, u'in mathcalU$ such that $ugeq u'$.
[Remark: when $u=u'$ then $(i_u_1, j_u_2, h_u_3),(i_u'_1, j_u'_2, h_u'_3)$ appear in $D(a)$ and $E(a)$]
Question: Take any $tildeain mathbbR^4$ with $aneq tildea$. I want to find some sufficient (and possibly also necessary) conditions such that $D(a)=D(tildea)$ and $E(a)=E(tildea)$.
My thoughts After many attempts I realised the following:
Take $aequiv (a_1, a_2, a_3, a_4)in mathbbR^4$ and suppose $a$ is such that
$$
(star)
begincases
-infty< a_1= -a_2<0< a_2= -a_1<infty\
-infty< -a_4< a_3<0< -a_3< a_4<infty\
-infty< a_2-a_4< a_1-a_3<0< a_3-a_1< a_4-a_2<infty\
endcases
$$
Then, every $tildeaequiv (tildea_1, tildea_2, tildea_3, tildea_4)in mathbbR^4$ such that
$$
begincases
-infty< tildea_1= -tildea_2<0< tildea_2= -tildea_1<infty\
-infty< -tildea_4< tildea_3<0< -tildea_3< tildea_4<infty\
-infty< tildea_2-tildea_4< tildea_1-tildea_3<0< tildea_3-tildea_1< tildea_4-tildea_2<infty\
endcases
$$
will have $D(a)=D(tildea)$ and $E(a)=E(tildea)$. This can be generalised to any ordering of $a$ in $(star)$: if $tildea$ respects the same ordering as $a$, then $D(a)=D(tildea)$ and $E(a)=E(tildea)$.
Is this correct? Can you think of other sufficient conditions? Can I get some sufficient and necessary conditions?
general-topology inequality algorithms vectors real-numbers
asked Aug 2 at 12:28
TEX
1419
edited Aug 2 at 12:45
asked Aug 2 at 12:28
TEX
1419
asked Aug 2 at 12:28
TEX
1419
asked Aug 2 at 12:28
TEX
1419
1419
This question has an open bounty worth +100
reputation from TEX ending ending at 2018-08-11 19:00:44Z">in 4 days.
The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns.
This question has an open bounty worth +100
reputation from TEX ending ending at 2018-08-11 19:00:44Z">in 4 days.
The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns.
If $V_1,V_2,V_3$ have cardinality 7, then $V$ doesn't.
– Gerry Myerson
Aug 2 at 12:41
Yes, sorry, my mistake
– TEX
Aug 2 at 12:45
add a comment |Â
If $V_1,V_2,V_3$ have cardinality 7, then $V$ doesn't.
– Gerry Myerson
Aug 2 at 12:41
Yes, sorry, my mistake
– TEX
Aug 2 at 12:45
If $V_1,V_2,V_3$ have cardinality 7, then $V$ doesn't.
– Gerry Myerson
Aug 2 at 12:41
If $V_1,V_2,V_3$ have cardinality 7, then $V$ doesn't.
– Gerry Myerson
Aug 2 at 12:41
Yes, sorry, my mistake
– TEX
Aug 2 at 12:45
Yes, sorry, my mistake
– TEX
Aug 2 at 12:45
add a comment |Â
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If $V_1,V_2,V_3$ have cardinality 7, then $V$ doesn't.
– Gerry Myerson
Aug 2 at 12:41
Yes, sorry, my mistake
– TEX
Aug 2 at 12:45