Let $a_1, a_2, a_3, a_4$ be real numbers.
Consider the following sets
$$ \mathcal{U}_1\equiv \{-a_1, a_1, -a_2, a_2, 0, \infty, -\infty\} $$ $$ \mathcal{U}_2\equiv \{-a_3, a_3, -a_4, a_4, 0, \infty, -\infty\} $$ $$ \mathcal{U}_3\equiv \{(a_3-a_1), (a_4-a_2), \infty, (-a_3+a_1), (-a_4+a_2), -\infty, 0\} $$ $$ \mathcal{U}\equiv \mathcal{U}_1\times \mathcal{U}_2\times \mathcal{U}_3 $$ where $\times$ denotes the Cartesian product. $\mathcal{U}_1, \mathcal{U}_2, \mathcal{U}_3$ have cardinality $7$.
Let $u\equiv (u_1, u_2, u_3)$ denote a generic element (triplet) of $\mathcal{U}$ with $u_1\in \mathcal{U}_1$, $u_2\in \mathcal{U}_2$, $u_3\in \mathcal{U}_3$.
Take any pair of triplets $u, u'\in \mathcal{U}$. In what follows $u\leq u'$ means that $u_{1}\leq u'_1$,$u_{2}\leq u'_2$,$u_{3}\leq u'_3$.
Take any triplet $u\in \mathcal{U}$. In what follows, $(i_{u_1}, j_{u_2}, h_{u_3})$ denote the position of $u_1, u_2, u_3$ respectively in $\mathcal{U}_1, \mathcal{U}_2, \mathcal{U}_3$ For example, suppose $u\equiv (a_1, a_4, 0)$; then, $(i_{u_1}, j_{u_2}, h_{u_3})=(2,4,7)$.
Fix $a\equiv (a_1, a_2, a_3, a_4)\in \mathbb{R}^4$. Consider the following two matrices
1) $D(a)$ listing $(i_{u_1}, j_{u_2}, h_{u_3}),(i_{u'_1}, j_{u'_2}, h_{u'_3})$ $\forall u, u'\in \mathcal{U}$ such that $u\leq u'$
1) $E(a)$ listing $(i_{u_1}, j_{u_2}, h_{u_3}),(i_{u'_1}, j_{u'_2}, h_{u'_3})$ $\forall u, u'\in \mathcal{U}$ such that $u\geq u'$.
[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 $\tilde{a}\in \mathbb{R}^4$ with $a\neq \tilde{a}$. I want to find some sufficient (and possibly also necessary) conditions such that $D(a)=D(\tilde{a})$ and $E(a)=E(\tilde{a})$.
My thoughts After many attempts I realised the following: Take $a\equiv (a_1, a_2, a_3, a_4)\in \mathbb{R}^4$ and suppose $a$ is such that $$ (\star) \begin{cases} -\infty< a_1= -a_2<0< a_2= -a_1<\infty\\ -\infty< -a_4< a_3<0< -a_3< a_4<\infty\\ -\infty< a_2-a_4< a_1-a_3<0< a_3-a_1< a_4-a_2<\infty\\ \end{cases} $$ Then, every $\tilde{a}\equiv (\tilde{a}_1, \tilde{a}_2, \tilde{a}_3, \tilde{a}_4)\in \mathbb{R}^4$ such that $$ \begin{cases} -\infty< \tilde{a}_1= -\tilde{a}_2<0< \tilde{a}_2= -\tilde{a}_1<\infty\\ -\infty< -\tilde{a}_4< \tilde{a}_3<0< -\tilde{a}_3< \tilde{a}_4<\infty\\ -\infty< \tilde{a}_2-\tilde{a}_4< \tilde{a}_1-\tilde{a}_3<0< \tilde{a}_3-\tilde{a}_1< \tilde{a}_4-\tilde{a}_2<\infty\\ \end{cases} $$ will have $D(a)=D(\tilde{a})$ and $E(a)=E(\tilde{a})$. This can be generalised to any ordering of $a$ in $(\star)$: if $\tilde{a}$ respects the same ordering as $a$, then $D(a)=D(\tilde{a})$ and $E(a)=E(\tilde{a})$.
Is this correct? Can you think of other sufficient conditions? Can I get some sufficient and necessary conditions?