Given numbers $0 \leq d\_i \leq 1$ for $i=1,\ldots,m$, it is easy to see that you can always find signs $\varepsilon\_i \in \{-1,1\}$ such that the partial sums $\sum\_{i=1}^k \varepsilon\_i d\_i/2$, for $k=0,\ldots,m$(where $\sum\_{i=1}^0 = 0$) lie in an interval of size at most 1. Is this best possible? i.e. is there a choice of $m$ and $d\_i$ such that for any choice of signs the strip must be of size at least 1-$\delta$?


Such an example for $\delta = 1/4$ is to take $m = 3$, $d\_1 = 1 = d\_3$ and $d\_2 = 1/2$. What about for smaller $\delta$?



Balancing real numbers in one dimension
Of course, this is best possible. Choose $d\_{2k-1}=1$, $d\_{2k}=1-\delta$ for $k=1,2,\dots$. If two consecutive signs are the same, we already have two partial sums on a distance $(2-\delta)/2$. If signs alternate, partial sums tend to infinity (actually, we need $m=O(1/\delta)$ for getting too large partial sum).


