Open Problems:50

Suggested by	Piotr Indyk
Source	Bertinoro 2011
Short link	https://sublinear.info/50

For any $n=2^{h}-1$, we can identify the coordinates of a vector $v \in \mathbb R^n$ with the nodes of a full binary tree $B_h$ of height $h$ and root $1$. We define a $k$-sparse tree model ${\mathcal T}_k$ to be a set of all $T \subset [n]$ of size $k$ which form a connected subtree in $B_h$ rooted at $1$.

We want to design an $m \times n$ matrix $A$ such that for any $x \in \mathbb R^n$, one can recover from $Ax$ a vector $x^* \in \mathbb R^n$ such that:

\[ \left\|x^*-x\right\|_1 \leq \min_{x'\in\mathbb R^n,\ \operatorname{supp}(x') \subset T \mbox{ for some } T \in {\mathcal T}_k } C \left\|x'-x\right\|_1, \] where $\operatorname{supp}(x')$ is the set of non-zero coefficients of $x'$, and $C>0$ is a constant.

Question: Is it possible to achieve $m=O(k)$ for some constant $C>0$?

Background: It is known that a weaker bound of $m=O(k \log(n/k))$ is possible to achieve even if ${\mathcal T}_k$ is replaced by the set of all $k$-subsets of $[n]$ [CandesRT-06a]. However, since $|{\mathcal T}_k | =\exp(O(k))$, one can expect a better bound for ${\mathcal T}_k$. By using model-based compressive sensing framework of [BaraniukCDH-10] (cf. [IndykP-11]), one can achieve the desired bound of $m=O(k)$ but with superconstant $C$.