Can we balance $2$-powers?

Question

If a sequence of reals $-1<x_1,\dots,x_k<1$ satisfies

\begin{equation*} x_{i+1}= \begin{cases} 2x_i, & \text{if } 2|x_i|<1 \\ 2x_i-2, & \text{if } 2x_i>1 \\ 2x_i+2, & \text{if } 2x_i<-1 \\ \end{cases} \end{equation*}
for $i=1,\dots,k$, where $x_{k+1}=x_1$, then is there a permutation $\pi$ of $\{1,\dots,k\}$ such that $0\le \sum_{i=1}^j x_{\pi(i)}<1$ for every $j$?

Motivation. The problem came up in a recent joint work with Gábor Damásdi, Nóra Frankl and János Pach.
This question is similar to Steinitz's theorem and to other vector balancing problems.
Indeed, it can be proved for any $x_i$'s satisfying the conditions of the conjecture, that $\sum_{i=1}^k x_i=0$.
Note that if the $x_i$'s are any sequence satisfying $\sum_{i=1}^k x_i=0$ and $|x_i|<1/2$ for every $i$, then one can easily find a permutation for which $0\le \sum_{i=1}^j x_{\pi(i)}<1$ for every $j$. But without this bound, we have to exploit that $x_{i+1}=2x_i$, as otherwise there would be counterexamples (e.g., $0.6,0.6,0.6,-0.9,-0.9$).
Could it be that the statement is true because we always have many $i$'s such that $|x_i|<1/2$, and these can be used somehow to take care of the other $x_i$'s?
Some weaker versions, where $1$ is replaced with $1.2$, have been proved by others (unpublished).

Answer 1

UPDATED. I've changed the verification strategy and computationally proved the statement for all $k\leq 30$, using the following randomized algorithm.

Randomized algorithm. For a given list of $k$ values of $x$'s sorted in increasing order, we set initially $s=0$, and at each of $k$ steps we identify the range of list elements that belong to the interval $[-s,1-s)$, select one of them, say $v$, randomly, remove it from the list, and add $v$ to $s$ (thus keeping $s$ in the interval $[0,1)$). The algorithm succeeds if we are able to make all $k$ steps, and fails otherwise (when at some step there are no suitable elements in the list).

From Fedor's observation it follows that for a fixed $k$, we can restrict our attention to values $x_1 = \frac{m}{2^k-1}$ for $m\equiv 2\pmod4$ in the interval $[2,2^k-2]$ and the sequences of $x_i$ emerging from them. For each such sequence (sorted in increasing order), we run the randomized algorithm multiple times until it succeeds.

It's worth to note that for a fixed $k$, the average number of algorithm runs (over all $m$) used in my computation was about $k/2$.

Answer 2

Fedor Petrov's comment shows than for each $k$ there are only finitely many cases for $x_1$ to check, so I wrote a program to do this. The positive answer is already obtained for all natural $k\le 10$. Almost most of the considered sequences $(x_i)$ can be balanced by the following algorithm. Fist pick the maximal positive $x_i$ and at each next step pick some allowed remaining $x_i$ with the maximal absolute value. So now I am thinking how to balance a few remaining exceptional sequences.