Here I ask a question on permutations of $n$ distinct real numbers.

QUESTION: Let $a_1,a_2,\ldots,a_n\ (n>1)$ be (pairwise) distinct real numbers. Is there a permutation $b_1,\ldots,b_n$ of $a_1,\ldots,a_n$ with $b_1 = a_1$ such that the $n−1$ numbers $$|b_1 −b_2|,\ |b_2 −b_3|,\ \ldots,\ |b_{n−1} −b_n|$$ are pairwise distinct ?

Actually, I formulated this question in 2013 and conjectured that the answer should be positive. If $a_1$ is the smallest (or largest) number among $a_1,\ldots,a_n$, then it is easy to construct a desired permutation $(b_1,\ldots,b_n)$. In fact, when $a_1<a_2<\ldots<a_n$ we may take $$(b_1,b_2,\ldots,b_n)=(a_1,a_n,a_2,a_{n-1},\ldots,a_{\lfloor n/2\rfloor+1})$$ to meet the purpose.

In 2015, Francesco Monopoli [Electron. J. Combin. 22(2015), no. 3, #P3.20] showed that my question for $a_1,\ldots,a_n$ has a positive answer if the set $A=\{a_1,a_2,\ldots,a_n\}$ forms an arithmetic progression.

As any path is a tree, in 2014 I made the following conjecture which unifies my question and Ringel and Kotzig's graceful tree conjecture.

**Conjecture**. Let $a_1,a_2,\ldots,a_n$ be $n > 1$ distinct real numbers, and let $T$ be any tree with vertices $v_1,\ldots,v_n$, where $v_1$ is a leaf (i.e., $\deg_T(v_1) = 1$). Then there is a bijection $f$ from the vertex set $V(T)$ of the tree $T$ to $A = \{a_1,\ldots,a_n\}$ with $f (v_1) = a_1$ such that those $|f (v_i)−f (v_j)|$ with $v_i$ and $v_j$ adjacent are pairwise distinct.

Any ideas towards the final solution of my question? Comments on the above general conjecture are also welcome!