The Computational Complexity of the Restricted Isometry Property, the Nullspace Property, and Related Concepts in Compressed Sensing

I Introduction

Acentral problem in compressed sensing (CS), see, e.g., , is the task of finding a sparsest solution to an underdetermined linear system, i.e.,

Many such conditions employ the famous restricted isometry property (RIP) (see and also ), which is satisfied with order $k$ and a constant $\delta_{k}$ by a given matrix $\bm{A}$ if

In the literature, it is often mentioned that evaluating the RIP, i.e., computing the constant $\underline{\delta}_{k}$ for some $\bm{A}$ and $k$, is presumably a computationally hard problem. Most papers seem to refer to NP-hardness, but this is often not explicitly stated. This motivated the development of several (polynomial-time) approximation algorithms for $\underline{\delta}_{k}$, e.g., the semidefinite relaxations in . However, while a widely accepted conjecture in the CS community, NP-hardness has, to the best of our knowledge, not been proven so far.

Recently, some first results in this direction have been obtained: In and , hardness and non-approximability results about the RIP were derived under certain (non-standard) complexity assumptions; see also . In work independent from the present paper, shows that it is NP-hard to verify (1) for given $\bm{A}$, $k$ and $\delta_{k}\in(0,1)$.

where $\lVert{\bm{x}}\rVert_{k,1}$ denotes the sum of the $k$ largest absolute values of entries in $\bm{x}$. The NSP guarantees exact recovery of $k$-sparse solutions to ( P 0 ) by solving ( P 1 ) whenever (2) holds with some constant $\alpha_{k}<1/2$.

Again, the computation of $\underline{\alpha}_{k}$ is suspected to be NP-hard, and several heuristics have been developed to compute good bounds on $\underline{\alpha}_{k}$, e.g., the semidefinite programming approaches in , or an LP-based relaxation in . However, as far as we know, no rigorous proof of (NP-)hardness has been given.

In this paper, we show that it is NP-hard to compute the RIC and NSC of a given matrix with given $k$; see Sections III and IV, respectively. More precisely, we show that unless P$=$NP, there is no polynomial time algorithm that computes $\underline{\delta}_{k}$ or $\underline{\alpha}_{k}$ for all given instances $(\bm{A},k)$. We also prove that certifying the RIP given $\bm{A}$, $k$ and some $\delta_{k}\in(0,1)$ is NP-hard.

Prior to this, in Section II, we prove NP-hardness of computing the spark of a matrix, i.e., the smallest number of linearly dependent columns. In fact, our main results concerning the complexity of determining the RIC or NSC follow from reduction of a decision problem concerning the existence of small linearly dependent column subsets. The term spark was first defined in , where strong results considering uniqueness of solutions to ( P 0 ) were proven. Ever since, its value has been claimed to be NP-hard to calculate, but, to the best of our knowledge, without a proof or reference for this fact. It seems to have escaped researchers’ notice that contains a proof that deciding whether the spark equals the number of rows is NP-hard, by reduction from the Subset Sum Problem (cf. [MP9] in ). Moreover, provides a different proof for this special case, by a reduction from the (homogeneous) Maximum Feasible Subsystem problem . Even earlier, in , the authors claim to have a proof, but give credit to the dissertation for establishing NP-hardness of spark computations. However, after closer inspection, the result in is in fact not about the spark, but the girth of so-called transversal matroids of bipartite graphs. Only recently, a variant of the latter proof has resurfaced in , where it is used to derive (non-deterministic) complexity results for constructing so-called full spark frames, i.e., matrices exhibiting the highest possible spark. Every transversal matroid can be represented by a matrix over any infinite field or finite field with sufficiently large cardinality , but there is no known deterministic way to construct such a matrix.

We adapt the proof idea from , a reduction from the $k$-Clique Problem, to vector matroids and thus establish that spark computation is NP-hard (without the restriction that the spark equals the row size). Our proof also makes use of results from , see Section II for more details.

Moreover, we gather several more complexity statements regarding problems related to the spark or RIP in Sections II and III, some of which are apparently new as well. In particular, we also show that solving the sparse principal component analysis problem (see, e.g., ) is strongly NP-hard, which is another widely accepted statement that appears to be lacking rigorous proof so far, and we extend this proof to show that the NP-hardness of RIC computation in fact holds in the strong sense; see Section III-B. Recall that strong NP-hardness implies that (unless P$=$NP) there cannot exist a fully polynomial-time approximation scheme (FPTAS), i.e., an algorithm that solves a minimization problem within a factor of $(1+\varepsilon)$ of the optimial value in polynomial time with respect to the input size and $1/\varepsilon$, see ; an FPTAS often exists for weakly NP-hard problems. Strong NP-hardness can also be understood as an indication that a problem’s intractability does not depend on ill-conditioning (due to the occurence of very large numbers) of the input data.

Throughout the article, for an $m\times n$ matrix $\bm{A}$ and a subset $S\subseteq\{1,\dots,n\}$, we denote by $\bm{A}_{S}$ the submatrix of $\bm{A}$ formed by the columns indexed by $S$. Sometimes we additionally restrict the rows to some index set $R$ and write $\bm{A}_{RS}$ for the resulting submatrix. Similarly, $\bm{x}_{S}$ denotes the part of a vector $\bm{x}$ containing the entries indexed by $S$. By $\bm{A}^{\top}$ and $\bm{x}^{\top}$, we denote the transpose of a matrix $\bm{A}$ or vector $\bm{x}$, respectively. For graph theoretic concepts and notation we refer to , for complexity theory to , and for matroid theory to .

II Complexity issues related to the spark

In this section, we deal with complexity issues related to linearly dependent columns of a given matrix $\bm{A}\in\mathds{Q}^{m\times n}$. Inclusion-wise minimal collections of linearly dependent columns are called circuits. More precisely, a circuit is a set $C\subseteq\{1,\dots,n\}$ of column indices such that $\bm{A}_{C}\bm{x}=0$ has a nonzero solution, but every proper subset of $C$ does not have this property, i.e., $\rank(\bm{A}_{C})=\lvert{C}\rvert-1=\rank(\bm{A}_{C\setminus\{j\}})$ for every $j\in C$. For notational simplicity, we will sometimes identify circuits $C$ with the associated solutions $\bm{x}\in\mathds{R}^{n}$ of $\bm{A}\bm{x}=0$ having support $C$. The spark of $\bm{A}$ is the size of its smallest circuit.

Clearly, the first two columns yield the minimum-size circuit (in fact, the only one), i.e., $\spark(\bm{A})=2$. In particular, note that generally, $\spark(\bm{A})\leq k$ does not guarantee that there also exists a vector with $k$ nonzeros in the nullspace of $\bm{A}$; e.g., take $k=3$ for the above $\bm{A}$. On the other hand, it is immediately clear that a nullspace vector with support size $k$ does not yield $\spark(\bm{A})=k$, but only $\spark(\bm{A})\leq k$. This distinction between circuits and nullspace vectors in general will be crucial in the proofs below.

The main result of this section is the following.

Given a matrix $\bm{A}\in\mathds{Q}^{m\times n}$ and a positive integer $k$, the problem to decide whether there exist a circuit of $\bm{A}$ of size at most $k$ is NP-complete.

For our proof, we employ several auxiliary results:

The vertex-edge incidence matrix of an undirected simple graph with $N$ vertices, $B$ bipartite components, and $Q$ isolated vertices has rank $N-B-Q$.

This result seems to be rediscovered every once in a while. The earliest proof we are aware of is due to van Nuffelen and works through various case distinctions considering linear dependencies of the rows and consequences of the existence of isolated or bipartite components.

Let $G=(V,E)$ be a simple undirected graph with vertex set $V$ and edge set $E$. Let $\bm{A}$ be its vertex-edge incidence matrix, and let $k>4$ be some integer. Suppose $G$ only has connected components with at least four vertices each, $\lvert{E}\rvert=\binom{k}{2}$, and $\rank(\bm{A})=k$. Then the graph $G$ has exactly $\lvert{V}\rvert=k$ vertices.

Let $G=(V,E)$ and $k>4$ be the graph and integer given in the statement of the lemma. Assume that $G$ has no component with less than four vertices, has $\lvert{E}\rvert=\binom{k}{2}$ edges, and that its incidence matrix $\bm{A}$ has $\rank(\bm{A})=k$.

Since consequently, $G$ has no isolated vertices, Lemma 1 tells us that the number of vertices is

where $B$ is the number of bipartite components in $G$. Assume that $B>0$, since otherwise the lemma is trivially true.

We claim that the number of edges in $G$ can be at most

To see this, recall that $G$ can have at most $\binom{N}{2}$ edges. Each connected component has at least four vertices. Since there are no edges between such a component and vertices outside, the total number of possible edges is reduced by at least $4(N-4)/2$ per component (the factor $1/2$ ensures that we do not count any edges twice). Since $G$ has at least $B$ connected components, the possible number of edges is hence decreased at least by the second term in (3). Moreover, since each bipartite component has at least four vertices, at least two of the potential edges cannot be present inside each such component, which yields the last term in (3). Note that the bound (3) is sharp if $G$ consists only of bipartite components with four vertices each.

if $B>0$. Thus, there are strictly less than $\binom{k}{2}$ edges, contradicting the requirement $\lvert{E}\rvert=\binom{k}{2}$. Hence, $B=0$. ∎

Let $\bm{H}=(h_{ij})\in\mathds{Z}^{m\times n}$ be a full-rank integer matrix with $m\leq n$ and let $\alpha\coloneqq\max\,\lvert{h_{ij}}\rvert$. Let $q\in\{m,\dots,n\}$ and define

For any column subset $S$ with $\lvert{S}\rvert=q$, if $\rank(\bm{H}_{S})=m$ and $\lvert{x}\rvert\geq\alpha^{m}q^{qn}+1$, then $\bm{H}(x)_{S}\in\mathds{Z}^{q\times q}$ has full rank $q$.

This result is a combination of Lemma 1 and (ideas from the proof of) Proposition 4 from . For clarity, we give the details here. Consider some $S\subseteq\{1,\dots,n\}$ with $\lvert{S}\rvert=q$ (w.l.o.g., $q>m\geq 1$; otherwise there is nothing to show). Assume that $\rank(\bm{H}_{S})=m$ and note that the last $q-m$ rows of $\bm{H}(x)_{S}$ form a submatrix of a generalized Vandermonde matrix with distinct nodes (see, e.g., ), which is easily seen to have full rank as well; see also . Consider the polynomial $p(x)\coloneqq{\rm det}(\bm{H}(x)_{S})$. From [17, Lemma 1], we know that there exists some $x$ for which the subspace spanned by the last $q-m$ rows of $\bm{H}(x)_{S}$ and the row space of $\bm{H}_{S}$ are transversal, i.e., they only intersect trivially. In particular, this shows that $p$ cannot be identical to the zero polynomial (both transversal parts have full rank). Let $d$ be the degree of $p(x)$ (which depends on the choice of $S$); thus, $p(x)=\beta_{0}+\beta_{1}x+\dots+\beta_{d}x^{d}$ with $\beta_{d}\neq 0$. Expanding the determinant ${\rm det}(\bm{H}(x)_{S})$ using Leibniz’s formula, one can derive that $\lvert{\beta_{i}}\rvert\leq\alpha^{m}q^{qn}$ for all $i$, by noting that the expansion consists of $q!<q^{q}$ summands which each are the product of precisely one entry per matrix row and column—in absolute value terms, we can extract a factor of $\alpha^{m}$ from this sum (from the $m$ rows corresponding to $\bm{H}_{S}$), and upper-bound all absolute values of coefficients of $x$ by the highest possible value (occurring when the last $q-m$ columns of $\bm{H}(x)$ are contained in $S$), which can be no larger than $(q-m)^{(n-1)(q-m)}<q^{qn-q}$. Moreover, it is easy to see that $\beta_{i}\in\mathds{Z}$ for all $i$. Applying Cauchy’s bound to the monic polynomial obtained from dividing $p(x)$ by $\beta_{d}$ yields

Thus, any $x$ with $\lvert{x}\rvert\geq\alpha^{m}q^{qn}+1$ is not a root of $p(x)$; equivalently, $\rank(\bm{H}(x)_{S})=q$ for such $x$. ∎

The problem is clearly in NP: Given a subset $C$ of column indices of $\bm{A}$, it can be verified in polynomial time that $\lvert{C}\rvert\leq k$ and that $\rank(\bm{A}_{C})=\lvert{C}\rvert-1=\rank(\bm{A}_{C\setminus\{j\}})$ for every $j\in C$ (by Gaussian elimination, see, e.g., ).

To show hardness, we reduce the NP-complete $k$-Clique Problem (cf. [GT19] in , or ): Given a simple undirected graph $G$, decide whether $G$ has a clique, i.e., a vertex-induced complete subgraph, of size $k$. We may assume without loss of generality that $k>4$.

For the given graph $G$ with $n$ vertices and $m$ edges construct a matrix $\bm{A}=(a_{ie})$ of size $(n+\binom{k}{2}-k-1)\times m$ as follows: Index the first $n$ rows of $\bm{A}$ by the vertices of $G$ and its columns by the edges of $G$ (we will also identify the vertices and edges with their indices). Let the first $n$ rows of $\bm{A}$ contain the vertex-edge incidence matrix of $G$ (i.e., set $a_{ie}=1$ if $i\in e$, and $0$ otherwise). For the non-vertex rows $n+i$, $i\in\{1,\dots,\binom{k}{2}-k-1\}$, set $a_{(n+i)e}=(U+i-1)^{e-1}$ with $U\coloneqq k^{2k^{2}m}+1$; note that this corresponds to the bottom part of $A$ consisting of a Vandermonde matrix (each row consists of increasing powers of the distinct numbers $U,\dots,U+\binom{k}{2}-k-2$). Clearly, this matrix $\bm{A}$ can be constructed in polynomial time, and its encoding length is polynomially related to that of the input (in particular, that of its largest entry is $\mathcal{O}(k^{2}m^{2}\log_{2}(k))$).

We first show that $G$ has a $k$-clique if and only if $\bm{A}$ has a circuit of size $\binom{k}{2}$. Suppose that $G$ has a $k$-clique, $k>4$, say on the vertices in the set $R$ (so that $\lvert{R}\rvert=k$), and with its $\binom{k}{2}$ edges in the set $C$. Since $\bm{A}_{C}$ has all-zero rows for each vertex outside of $R$, $\lvert{R}\rvert+\text{(number of non-vertex rows)}=\binom{k}{2}-1=\lvert{C}\rvert-1\geq\rank(\bm{A}_{C})$. Clearly, a clique is never bipartite (it always contains odd cycles, for $k\geq 3$). Hence, by Lemma 1, the rows of $\bm{A}_{C}$ indexed by $R$ are linearly independent. Now observe that removing any edge from a $k$-clique does not affect the rank of the associated incidence matrix, since the subgraph remains connected and non-bipartite with less vertices than edges (for $k\geq 4$). Thus, by Lemma 1, the rank of the nonzero vertex row part of $\bm{A}_{C}$ remains $k$ if any column from $C$ is removed. Therefore, since $a_{ie}\leq 1$ for all $i\leq n$ and all $e\leq m$, Lemma 3 applies to $\bm{A}_{C\setminus\{e\}}$ for every $e\in C$ (with $x=U$, $\bm{H}(x)=\bm{A}$, $S=C\setminus\{e\}$, and $q=\binom{k}{2}-1$) and yields $\rank(\bm{A}_{C\setminus\{e\}})=q=\lvert{C}\rvert-1$, whence also $\rank(\bm{A}_{C})=\lvert{C}\rvert-1$. Thus, $C$ is a circuit.

Conversely, suppose that $\bm{A}$ has a circuit $C$ of size $\lvert{C}\rvert=\binom{k}{2}$ with $k>4$. Then, by definition of a circuit, $\rank(\bm{A}_{C})=\lvert{C}\rvert-1$, so $\bm{A}_{C}$ has at least $\lvert{C}\rvert-1$ nonzero rows. Since these include the $\lvert{C}\rvert-k-1$ non-vertex rows, the set $R$ of nonzero vertex rows of $\bm{A}_{C}$ has size $\lvert{R}\rvert\geq\big(\lvert{C}\rvert-1\big)-\big(\lvert{C}\rvert-k-1\big)=k$. Let $\bm{A}_{RC}$ and $\bm{A}_{NC}$ denote the vertex and non-vertex row submatrices of $\bm{A}_{C}$, respectively. Since $\rank(\bm{A}_{C})\leq\rank(\bm{A}_{NC})+\rank(\bm{A}_{RC})=\lvert{C}\rvert-k-1+\rank(\bm{A}_{RC})$ and $\lvert{R}\rvert\geq k$, clearly $\rank(\bm{A}_{RC})\geq k$. Suppose that $\rank(\bm{A}_{RC})>k$; then there would exist a subset $R^{\prime}\subseteq R$ with $\lvert{R^{\prime}}\rvert=k+1$ and $\rank(\bm{A}_{R^{\prime}C})=k+1$. But by Lemma 3, the square matrix $(\bm{A}_{R^{\prime}C}^{\top},\bm{A}_{NC}^{\top})^{\top}$ would then have full rank $\lvert{C}\rvert$, whence $\rank(\bm{A}_{C})=\lvert{C}\rvert>\lvert{C}\rvert-1$, contradicting the fact that $C$ is a circuit. Thus, the upper part $\bm{A}_{RC}$ of $\bm{A}_{C}$ must in fact have rank exactly $k$.

Observe that the subgraph $(R,C)$ of $G$ with vertex set $R$ and edge set $C$ cannot contain components with less than $4$ vertices: such a subgraph $(R^{\prime},C^{\prime})$ could have at most as many edges as vertices, so that the associated incidence matrix $\bm{A}^{\prime}_{C^{\prime}}$ has full column rank. Removing a column corresponding to an edge $e\in C^{\prime}$ would reduce the rank, i.e., $\rank(\bm{A}^{\prime}_{C^{\prime}\setminus\{e\}})<\rank(\bm{A}^{\prime}_{C^{\prime}})$. Moreover, note that $\bm{A}_{RC}$ has block diagonal form where the blocks are the incidence matrices of the separate graph components within $(R,C)$, so that the rank is the sum of the ranks of the blocks (one of which is $\bm{A}_{R^{\prime}C^{\prime}}$). In particular, the non-vertex row part of $\bm{A}_{C}$ maintains full (row) rank when any column is removed from $C$, so that deleting $e\in C^{\prime}$ would yield $\rank(\bm{A}_{C\setminus\{e\}})=\rank(\bm{A}_{C})-1$, contradicting the fact that $C$ is a circuit. Thus, Lemma 2 applies to the graph $(R,C)$ and yields that $\lvert{R}\rvert=k$. This implies that the vertices in $R$ form a $k$-clique, because $R$ can induce at most $\binom{k}{2}$ edges and the $\binom{k}{2}$ edges in $C$ are among them.

We now show that each circuit of $\bm{A}$ has size at least $\binom{k}{2}$. This proves the claim, since by the arguments above, it shows that there exists a circuit of size at most (in fact, exactly) $\binom{k}{2}$ if and only if $G$ has a $k$-clique, i.e., for the given construction a solution to the spark problem yields a solution to the clique problem as well.

Suppose that $\bm{A}$ has a circuit $C$ with $c\coloneqq\lvert{C}\rvert<\binom{k}{2}$. Let $d\coloneqq\binom{k}{2}-c>0$. Clearly, not all vertex rows restricted to any column subset can be zero, and any submatrix of the non-vertex part of $\bm{A}$ with fewer than $\binom{k}{2}-k$ columns is of full (column) rank. Therefore, $c>\binom{k}{2}-k$ necessarily. Since $C$ is a circuit, $\bm{A}_{C}$ has $c-1$ nonzero rows (similar to the arguments above, it can be seen that the lower bound $c-1$ holds with equality). Because the $\binom{k}{2}-k-1$ non-vertex rows are among these, and by Lemmas 2 and 3, $\bm{A}_{C}$ has $(c-1)-\big(\binom{k}{2}-k-1\big)=k-d>0$ nonzero vertex rows. Denote the set of such rows by $R$, and let $r$ be the number of edges in the subgraph of $G$ induced by the vertices in $R$. Since $\lvert{R}\rvert=k-d$ vertices can induce at most $\binom{k-d}{2}$ edges, $r\leq\binom{k-d}{2}$. But surely all the edges in $C$ are among those induced by $R$, so that $r\geq c=\binom{k}{2}-d$. Putting these inequalities together yields $\binom{k}{2}-d\leq r\leq\binom{k-d}{2}$. However, since we assumed $k>4$, it holds that $\binom{k}{2}-d>\binom{k-d}{2}$, yielding a contradiction. Consequently, every circuit $C$ of $A$ must satisfy $\lvert{C}\rvert\geq\binom{k}{2}$. ∎

The smallest size (cardinality) of a circuit can be expressed as

Clearly, there exists a circuit of size at most $k$ if and only if the spark is at most $k$. Hence, Theorem 1 immediately yields the following.

The idea of reducing from the clique problem to prove Theorem 1 is due to Larry Stockmeyer and appears in Theorem 3.3.6 of (see also ). However, uses generic matrices that, in fact, represent transversal matroids (of bipartite graphs) and therefore have certain properties needed in the proof. The entries of these generic matrices are not specified, and to date there is no known deterministic way to do so such that the matrix represents a transversal matroid. We replaced the corresponding machinery by our explicit matrix construction and the arguments using Lemmas 1, 2 and 3 to become independent of transversal matroid representations and work directly on vector matroids. Note that the proof of Theorem 1 also shows NP-completeness of the problem to decide whether $\bm{A}$ has a circuit $C$ with equal to $k$; see also Example 1.

The results above are related to, but different from, the following.

Theorem 1 in shows that for an $m\times n$ matrix $\bm{A}$, it is NP-complete to decide whether $\bm{A}$ has an $m\times m$ submatrix with zero determinant. This implies NP-completeness of deciding whether $\spark(\bm{A})\leq k$ for the special case $k=m$. This restriction of $k$ to the row number $m$ of $\bm{A}$ could in principle be removed by appending all-zero rows, but one would then no longer be in the interesting case where the matrix has full (row) rank. Our proof admits spark values other than the row number for full-rank matrices; however, the row and column numbers in the reduction depend on the instance.

In contrast to the results above, for graphic matroids, the girth can be computed in polynomial time .

The paper proves NP-hardness of computing the girth of the binary matroid, i.e., a vector matroid over $\mathds{F}_{2}$. This, however, does not imply NP-hardness over the field of rational or real numbers, and the proof cannot be extended accordingly. Similarly, in it was shown that, over $\mathds{F}_{2}$, it is NP-complete to decide whether there exists a vector with $k$ nonzero entries in the nullspace of a matrix. However, while the proof for this result can straightforwardly be extended to the rational case, it does not imply hardness of computing the spark either: Since in , there is no lower bound on the spark (such as we provide in the last paragraph of the proof of Theorem 1), a situation as in Example 1 is not explicitly avoided there. (Note also that it was already remarked in that the problem to decide whether an ($\mathds{F}_{2}$-)nullspace vector with at most $k$ nonzeros exists is not covered by their proof.)

We also have the following result, which shows another relation between minimum cardinality circuits and the task of finding sparsest solutions to underdetermined linear systems.

Given a matrix $\bm{B}$, a specific column $\bm{b}$ of $\bm{B}$, and a positive integer $k$, the problem of deciding whether there exists a circuit of $\bm{B}$ of size $k$ which contains $\bm{b}$ is NP-complete in the strong sense. Consequently, it is strongly NP-hard to determine the minimum cardinality of circuits that contain a specific column $\bm{b}$ of $\bm{B}$.

Denote by $\bm{A}$ the matrix $\bm{B}$ without the column $\bm{b}$. Then it is easy to see that $\bm{B}$ has a circuit of size $k$ that contains $\bm{b}$ if and only if $\bm{A}\bm{x}=\bm{b}$ has a solution with $k-1$ nonzero entries. Deciding the latter is well-known to be NP-complete in the strong sense (it amounts to the decision version of ( P 0 )), see [MP5] in . ∎

As mentioned in , one can reduce spark computations to ( P 0 ) as follows: For each column of $\bm{A}\in\mathds{Q}^{m\times n}$ in turn, add a new row with a $1$ in this column and $0$ elsewhere. The right hand side $\bm{b}$ is the $(m+1)$-th unit vector. Now solve each such ( P 0 ) problem, and take the solution with smallest support. (Note that this is a (Turing-)reduction, using Theorem 1 to show NP-hardness of ( P 0 ).) Interestingly, we do not know an easy reduction for the reverse direction.

Let us now briefly consider full spark frames. An $m\times n$ matrix $\bm{A}$ with full rank $m$ ($m\leq n$) is said to be full spark if $\spark(\bm{A})=m+1$, i.e., every submatrix consisting of at most $m$ columns of $\bm{A}$ has full rank. In , it was shown that testing a matrix for this property is hard for NP under randomized reductions. In fact, the following stronger result holds:

Given a rational matrix $\bm{A}$, deciding whether $\bm{A}$ is a full spark frame is coNP-complete.

We can assume w.l.o.g. that $\bm{A}\in\mathds{Q}^{m\times n}$ with rank $m\leq n$. Thus, $\bm{A}$ is full spark if $\spark(\bm{A})=m+1$. Clearly, $\spark(\bm{A})=m+1$ holds if and only if the question whether $\bm{A}$ has a singular $m\times m$ submatrix has a negative answer. Since the latter decision problem is NP-complete by (or [17, Proposition 4] and the results in ), deciding whether $\bm{A}$ is a full spark frame is NP-hard. Moreover, this problem is contained in coNP, since the “no” answer can be certified in polynomial time by specifying a singular (square) submatrix and because singularity can be verified in polynomial time. ∎

Above, and in the related complexity results to follow, we show that the decision problem under consideration has a negative answer if and only if a known NP-complete problem has a positive answer. Since, by definition, the complementary problem of an NP-complete problem is coNP-complete, the respective hardness results follow—see also . Both NP- and coNP-completeness imply that no polynomial time algorithm exists unless P$=$NP (or equivalently P$=$coNP). Since a problem is NP-hard if and only if it is coNP-hard (every problem in coNP can be Turing-reduced to it; cf [47, Ch. 15.7]), we use the term NP-hard throughout the article.

In the next sections we shall see how we can deduce NP-hardness of computing restricted isometry or nullspace constants from the above results. (In particular, the extension to $k<m$ provided by Theorem 1, cf. Remark 2, will allow for making statements about RIP or NSP orders other than the row number.)

III NP-hardness of computing the restricted isometry constant

Recall that for a positive integer $k$, $\bm{A}$ satisfies the RIP of order $k$ with constant $\delta_{k}$ if (1) holds. Given $\bm{A}$ and $k$, the smallest such constant is the RIC $\underline{\delta}_{k}$. Note that (1) holds for $\delta_{k}=0$ if and only if $\bm{A}$ is orthogonal, and that $\delta_{k}<0$ is impossible.

The suspicions about computational intractability of the RIP are based on the observation that a brute-force method would have to inspect all submatrices induced by column subsets of sizes up to $k$. Of course, by itself, this does not generally rule out the possible existence of an efficient algorithm. However, we show below that (given $\bm{A}$ and $k$) deciding whether there exists a constant $\delta_{k}<1$ such that (1) holds is coNP-complete; consequently, computing the RIC is NP-hard. Moreover, we show that RIP certification (for a given $\delta_{k}\in(0,1)$) is NP-hard. Thus, unless P$=$NP, there exists no polynomial time algorithm for any of these problems (cf. Remark 4).

We will need the following technical result.

First, observe that the largest singular value of $\bm{A}$, $\sigma_{\max}(\bm{A})$, satisfies

By the singular value interlacing theorem (see, e.g., ), $\sigma_{\max}(\bm{A})$ is an upper bound for the largest singular value of every submatrix of $\bm{A}$. Thus, the above lemma essentially shows that by scaling the matrix $\bm{A}$, one can focus on the lower part of the RIP (1) (a similar argument has been derived independently in ). This leads to the following complexity result.

Given a matrix $\bm{A}\in\mathds{Q}^{m\times n}$ and a positive integer $k$, the problem to decide whether there exists some rational constant $\delta_{k}<1$ such that $\bm{A}$ satisfies the RIP of order $k$ with constant $\delta_{k}$ is coNP-complete.

implies that we must have $\delta_{k}\geq 1$.

For the converse, assume that there does not exist $\delta_{k}<1$ for which (1) holds. By Lemma 4, the upper part of (1) is always satisfied. Consequently, there must exist a vector $\hat{\bm{x}}$ with $1\leq\lVert{\hat{\bm{x}}}\rVert_{0}\leq k$ such that the lower part is tight, i.e.,

Usually, one is interested in the smallest constant $\delta_{k}$ for which (1) holds, i.e., the restricted isometry constant (RIC)

Recall that $\delta<0$ is impossible, resulting in the condition $\delta\geq 0$ in (5). We immediately obtain the following complexity result.

For a given matrix $\bm{A}\in\mathds{Q}^{m\times n}$ and positive integer $k$, it is NP-hard to compute the RIC $\underline{\delta}_{k}$.

In this section, we show that the RIP certification problem, i.e., deciding whether a matrix $\bm{A}$ satisfies the RIP with given order $k$ and given constant $\delta_{k}\in(0,1)$, is (co)NP-hard. The main arguments used in the proofs of the following Lemma and Theorem have been independently derived by us and the authors of .

Given a matrix $\bm{A}\in\mathds{Q}^{m\times n}$ and a positive integer $k\leq n$, if $\spark(\bm{A})>k$, there exists a constant $\varepsilon>0$ with encoding length polynomially bounded by that of $\bm{A}$ such that $\lVert{\bm{A}\bm{x}}\rVert_{2}^{2}\geq\varepsilon\,\lVert{\bm{x}}\rVert_{2}^{2}$ for all $x$ with $\lVert{\bm{x}}\rVert_{0}\leq k$.

Assume without loss of generality that $\bm{A}$ has only integer entries (this can always be achieved by scaling with the least common denominator of the matrix entries, which influences $\varepsilon$ by a polynomial factor only).

Define $\alpha$ as in Lemma 4. Note that $\spark(\bm{A})>k$ implies that every submatrix $\bm{A}_{S}$ with $S\subseteq\{1,\dots,n\}$, $\lvert{S}\rvert\leq k$, has linearly independent columns. Consider an arbitrary such $S$. Then, $\bm{A}_{S}^{\top}\bm{A}_{S}$ is positive definite, so its smallest eigenvalue fulfills $\lambda_{\min}(\bm{A}_{S}^{\top}\bm{A}_{S})>0$, and also $\det(\bm{A}_{S}^{\top}\bm{A}_{S})>0$. Moreover, since the absolute values of entries of $\bm{A}$ are integers in $\{0,1,\dots,\alpha\}$, the entries of $\bm{A}_{S}^{\top}\bm{A}_{S}$ are also integral and lie in $\{0,1,\dots,m\,\alpha^{2}\}$. Therefore, it must in fact hold that $\det(\bm{A}_{S}^{\top}\bm{A}_{S})\geq 1$ and $\lambda_{\max}(\bm{A}_{S}^{\top}\bm{A}_{S})\geq 1$. It follows that

Since $S$ was arbitrary, $\lVert{\bm{A}\bm{x}}\rVert_{2}^{2}\geq\lambda_{\min}(\bm{A}_{S}^{\top}\bm{A}_{S})\lVert{\bm{x}}\rVert_{2}^{2}\geq\varepsilon\lVert{\bm{x}}\rVert_{2}^{2}$ for all $\bm{x}$ with support $S\subseteq\{1,\dots,n\}$, $\lvert{S}\rvert\leq k$. Moreover, the encoding length of $\alpha$, and therefore that of $\varepsilon$, is clearly polynomially bounded by the encoding length of $\bm{A}$, which completes the proof. ∎

Given a matrix $\bm{A}\in\mathds{Q}^{m\times n}$, a positive integer $k$, and some constant $\delta_{k}\in(0,1)$, it is NP-hard to decide whether $\bm{A}$ satisfies the RIP of order $k$ with constant $\delta_{k}$.

It is an open question whether the problem in Theorem 3 is in coNP.

Clearly, Theorem 3 leads to another easy proof for Corollary 4 (and Corollary 5 below): computing the (lower asymmetric) RIC would also decide the RIP certification problem. On the other hand, our proof of Theorem 3 essentially reduces the RIP certification problem to the setting of Theorem 2, which therefore can be seen as the core RIP hardness result (by establishing the direct link to spark computations); see also Remark 9 below.

III-B Asymmetric restricted isometry constants

It has been remarked in that the symmetric nature of the RIP can be overly restrictive. In particular, the influence of the upper inequality in (5) is often stronger, although the lower inequality is more important in the context of sparse recovery. For instance, the often stated condition $\underline{\delta}_{2k}<1$ for uniqueness of $k$-sparse solutions (see, e.g., ) should in fact read $\underline{\delta}_{2k}^{L}<1$, where

is the lower asymmetric restricted isometry constant (see also ). Correspondingly, the upper asymmetric RIC is

The central argument in the proof of Theorem 2 in fact shows the following:

Given a matrix $\bm{A}\in\mathds{Q}^{m\times n}$ and a positive integer $k$, it is NP-hard to compute $\underline{\delta}_{k}^{L}$.

Moreover, the next result settles the computational complexity of computing the upper asymmetric RIC $\underline{\delta}^{U}_{k}$.

Given a matrix $\bm{A}\in\mathds{Q}^{m\times n}$, a positive integer $k\leq m$ and a parameter $\delta>0$, it is NP-hard in the strong sense to decide whether $\underline{\delta}^{U}_{k}<\delta$, even in the square case $m=n$. Consequently, it is strongly NP-hard to compute $\underline{\delta}_{k}^{U}$.

To prove this theorem, we need some auxiliary results.

Let $G=(V,E)$ be a simple undirected graph with $n=\lvert{V}\rvert$ and let $\bm{A}_{G}$ be its $n\times n$ adjacency matrix, i.e., $(\bm{A}_{G})_{ij}=1$ if $\{i,j\}\in E$ and $0$ otherwise. Denote by $K_{n}$ the complete graph with $n$ vertices.

If $G=K_{n}$, i.e., $G$ is a clique, then $\bm{A}_{G}$ has eigenvalues $-1$ and $n-1$ with respective multiplicities $n-1$ and $1$.

Removing an edge from $G$ does not increase $\lambda_{\max}(\bm{A}_{G})$. In fact, if $G$ is connected, this strictly decreases $\lambda_{\max}(\bm{A}_{G})$.

If $G=K_{n}\setminus e$, i.e., a clique with one edge removed, then the largest eigenvalue of $\bm{A}_{G}$ is $(n-3+\sqrt{n^{2}+2n-7})/2$.

The first two statements can be found in, or deduced easily from, [9, Ch. 1.4.1 and Prop. 3.1.1], respectively. The third result is a special case of [33, Theorem 1]. ∎

Lemma 6 shows that, in a graph $G=(V,E)$ with $\lvert{V}\rvert=n$, the largest eigenvalue of the adjacency matrix of any induced subgraph with $k\in\{2,\dots,n\}$ vertices is either $k-1$ (if and only if the subgraph is a $k$-clique) or at most $(k-3+\sqrt{k^{2}+2k-7})/2$.

Given a matrix $\bm{H}\in\mathds{Q}^{n\times n}$, a positive integer $k\leq n$ and a parameter $\lambda>0$, it is coNP-complete in the strong sense to decide whether $\lambda_{\max}^{(k)}<\lambda$, where

Consequently, solving the sparse principal component analysis (Sparse PCA) problem is strongly NP-hard.

We reduce from the $k$-Clique Problem. Let $G=(V,E)$ be a simple undirected graph with $n$ vertices (w.l.o.g., $n\geq 2$). From $G$, construct its $n\times n$ adjacency matrix $\bm{H}\coloneqq\bm{A}_{G}$. By the previous Lemma, $G$ contains a $k$-clique if and only if $\lambda_{\max}^{(k)}(\bm{H})=k-1\eqqcolon\lambda$. (Note that $\lambda_{\max}^{(k)}\leq k-1$ always holds by construction.) Hence, the Sparse PCA decision problem has a negative answer for the instance $(\bm{H},k,\lambda)$ if and only if the $k$-Clique Problem has a positive answer. Since the latter is NP-complete in the strong sense, and because all numbers appearing in the constructed Sparse PCA instance, and their encoding lengths, are polynomially bounded by $n$, the former is strongly (co)NP-hard.

Moreover, consider a “no” instance of the Sparse PCA decision problem. Then, as we just saw, there is a $k$-clique $S$ in $G$, and it is easily verified that $\hat{\bm{x}}$ with $\hat{\bm{x}}_{i}=1/\sqrt{k}$ for $i\in S$, and zeros everywhere else, achieves $\hat{\bm{x}}^{\top}\bm{H}\hat{x}=\lambda_{\max}^{(k)}(\bm{H})=\lambda$. Scaling (9) by $k$, we see that equivalently,

The Sparse PCA problem (see, e.g., ) is often mentioned to be (NP-)hard, but we could not locate a rigorous proof of this fact. In [11, Section 6], the authors sketch a reduction from the $k$-Clique Problem but do not give the details; the central spectral argument mentioned there, however, is exactly what we exploit in the above proof.

We will extend the proof of Proposition 1 to show Theorem 4 by suitably approximating the Cholesky decomposition of a matrix very similar to the adjacency matrix; the following technical result will be useful for this extension.

With deg$(v)$ denoting the degree of a vertex $v$ of $G$, and $\lambda_{i}(\bm{A}_{G})$, $i=1,\dots,n$, the eigenvalues of $\bm{A}_{G}$,

Thus, it is easy to see that $\bm{H}=\bm{A}_{G}+n^{2}\bm{I}$ is (symmetric) positive definite; in particular, the eigenvalues of $\bm{H}$ obey $\lambda_{i}(\bm{H})=\lambda_{i}(\bm{A}_{G})+n^{2}$ for all $i$. Then, $\bm{H}$ has a unique Cholesky factorization $\bm{H}=\bm{L}\bm{D}\bm{L}^{\top}$, and $\bm{L}\in\mathds{Q}^{n\times n}$ (unit lower triangular) and $\bm{D}\in\mathds{Q}^{n\times n}$ (diagonal) can be obtained by Gaussian elimination; see, e.g., [37, Section 4.9.2] or [40, Section 4.2.3].

Let $\bm{H}^{(0)}\coloneqq\bm{H}$ and let $\bm{H}^{(k)}=(h^{(k)}_{ij})$ be the matrix obtained from $\bm{H}$ after $k$ iterations of (symmetric) Gaussian elimination. There are $n-1$ such iterations, and each matrix $\bm{H}^{(k)}$ has block structure with $\diag(d_{1},\dots,d_{k})$ in the upper left part and a symmetric matrix in the lower right block. We show by induction that for all $k=1,\dots,n$, and all $i,j\in\{0,\dots,n-k\}$,

Clearly, by construction of $\bm{H}$, $h^{(0)}_{ii}=n^{2}$ and $h^{(0)}_{ij}\in\{0,1\}$ for all $i,j\in\{1,\dots,n\}$, $i\neq j$, so (10) holds true for $k=1$. Suppose (10) holds for some $k\in\{1,\dots,n-1\}$, i.e., throughout the first $k-1$ iterations of the Gaussian elimination process. Performing the $k$-th iteration, we obtain $h^{(k)}_{ik}=h^{(k)}_{ki}=0$ for all $i>k$, $h^{(k)}_{kk}=h^{(k-1)}_{kk}$, and

Thus, in particular, by symmetry of $\bm{H}^{(k-1)}$, for any $i\in\{1,\dots,n-k\}$,

Applying the induction hypothesis (10) to (12) yields the first interval inclusion (note that $h^{(k-1)}_{kk}>0$, so $h^{(k)}_{k+i,k+i}\leq h^{(k-1)}_{k+i,k+i}$):

Similarly, for the off-diagonal entries $h^{(k)}_{k+i,k+j}$ with $i,j\in\{1,\dots,n-k\}$, $i\neq j$, from (11) and (10) we obtain

which shows the second interval inclusion and concludes the induction.

Given an instance $(G,k)$ for the $k$-Clique Problem, we construct $\bm{H}=\bm{A}_{G}+n^{2}\bm{I}$ from the graph’s adjacency matrix $\bm{A}_{G}$ (w.l.o.g., $n\geq 2$). From the proof of Proposition 1, recall that $G$ has no $k$-clique if and only if $\lambda_{\max}^{(k)}(\bm{A}_{G})<k-1$, or equivalently $\lambda_{\max}^{(k)}(\bm{H})<n^{2}+k-1$ (cf. the beginning of the proof of Lemma 7). Let $\bm{D}$ and $\bm{L}$ be the Cholesky factors of $\bm{H}$, i.e., $\bm{H}=\bm{L}\bm{D}\bm{L}^{\top}$. Letting $\bm{D}^{1/2}\coloneqq\diag(\sqrt{d_{1}},\dots,\sqrt{d_{n}})$, observe that the upper asymmetric RIC for the matrix $\bm{A}^{\prime}\coloneqq\bm{D}^{1/2}\bm{L}^{\top}$ can be written as

Consequently, $G$ has a $k$-clique $S$ if and only if $\bm{H}$ has a $k\times k$ submatrix $\bm{H}_{SS}$ with largest eigenvalue $n^{2}+k-1$, i.e., $\underline{\delta}^{U}_{k}(\bm{A}^{\prime})=n^{2}+k-2$. Moreover, by Lemma 6, $\lambda_{\max}(\bm{H}_{TT})\leq n^{2}+(k-3+\sqrt{k^{2}+2k-7})/2<n^{2}+k-1$ for any incomplete induced subgraph of $G$ with vertex set $T$, $\lvert{T}\rvert=k$. However, while $\bm{L}$ and $\bm{D}$ are rational, $\bm{D}^{1/2}$ can contain irrational entries, so we cannot directly use $\bm{A}^{\prime}$ as the input matrix for the upper asymmetric RIC decision problem. The remainder of this proof shows that we can replace $\bm{D}^{1/2}$ by a rational approximation to within an accuracy that still allows us to distinguish between eigenvalues associated to $k$-cliques and those belonging to incomplete induced subgraphs.

Therefore (cf., e.g., [40, Corollary 8.1.6]), we have for all $S\subseteq\{1,\dots,n\}$ that

Consequently, if $S$ is a $k$-clique, we have

whereas for any $T\subseteq\{1,\dots,n\}$ with $\lvert{T}\rvert=k$ which induces no clique, it holds that

If $G$ has a $k$-clique, then by (14), $\underline{\delta}^{U}_{k}(\bm{A})\geq\delta$, and if not, $\underline{\delta}^{U}_{k}(\bm{A})\leq n^{2}+(k-3+\sqrt{k^{2}+2k-7})/2+2\cdot 10^{-p}\cdot n^{3}-1$ by (15). In fact, our choice of $p$ implies

which shows that $\underline{\delta}^{U}_{k}(\bm{A})<\delta$ if no $k$-clique is contained in $G$. Therefore, $G$ has a $k$-clique if and only if $\underline{\delta}^{U}_{k}(\bm{A})\geq\delta$, i.e., the upper asymmetric RIC decision problem under consideration has a negative answer.

Thus, since the Clique Problem is well-known to be NP-complete in the strong sense, and because our polynomial reduction in fact preserves boundedness of the numbers within $\mathcal{O}(\poly(n))$, the upper asymmetric RIC decision problem is (co)NP-hard in the strong sense. This completes the proof of Theorem 4. ∎

In fact, observe that for the matrix $\bm{A}$ constructed in the proof of Theorem 4, the upper asymmetric RIC is always larger than the lower asymmetric RIC, whence the former therefore coincides with the (symmetric) RIC $\underline{\delta}_{k}$ of $\bm{A}$. Thus, the following result holds true, which slightly strengthens Corollary 4.

Given a matrix $A\in\mathds{Q}^{m\times n}$ and a positive integer $k$, it is NP-hard in the strong sense to compute the RIC $\underline{\delta}_{k}$, even in the square case $m=n$.

Theorem 2 (which yielded Corollary 4) is of interest in its own right, as it reveals, for instance, the intrinsic relation between the $k$-sparse solution uniqueness conditions $2k<\spark(\bm{A})$ and $\underline{\delta}_{2k}<1$ (or $\underline{\delta}_{2k}^{L}<1$, respectively, see Corollary 5); consequently, verifying uniqueness via these conditions is NP-hard because deciding whether $\spark(\bm{A})\leq k$ is.

We can easily extend the construction from the proof of Theorem 4 to cover the non-square case $m<n$ as well. The idea is as follows: For instance, let $\bm{B}$ be the matrix from Example 1, and replace $\bm{A}$ in the above proof by the block diagonal matrix $\hat{\bm{A}}$ which has $\bm{A}$ in the first block and $\bm{B}$ in the second. This matrix has dimensions $(n+3)\times(n+4)$, and since the eigenvalues of $\bm{B}^{\top}\bm{B}$ are contained in $[0,5)$, it is easy to see that the relation between eigenvalues of symmetrically chosen submatrices and cliques is the same for $\hat{\bm{A}}^{\top}\hat{\bm{A}}$ as for $\bm{A}^{\top}\bm{A}$ itself (note that $n^{2}+(k-3+\sqrt{k^{2}+2k-7})/2-2\cdot 10^{-p}\cdot n^{3}\geq 5$ whenever $n\geq 3$, $k\geq 2$, which can of course be assumed without loss of generality). A similar idea is exploited in the proof of [49, Theorem 6].

The rational certificate for the Sparse PCA problem (see the proof of Proposition 1) cannot be employed to verify the “no” answer of the upper asymmetric RIC decision problem from Theorem 4 in polynomial time; containment in coNP of this latter problem thus remains an open question.

IV NP-hardness of computing the nullspace constant

Recall that $\bm{A}\in\mathds{Q}^{m\times n}$ satisfies the NSP of order $k$ with constant $\alpha_{k}$ if (2) holds for all $\bm{x}\in\mathds{R}^{n}$ with $\bm{A}\bm{x}=0$. As for the RIP, one is typically interested in the smallest constant $\underline{\alpha}_{k}$, the nullspace constant (NSC), such that the NSP of order $k$ holds with this constant. Thus, the NSC is given by

Sparse recovery by ( P 1 ) is ensured if and only if $\underline{\alpha}_{k}<1/2$. However, the following results show that computing $\underline{\alpha}_{k}$ is a challenging problem.

Given a matrix $\bm{A}\in\mathds{Q}^{m\times n}$ and a positive integer $k$, the problem to decide whether $\bm{A}$ satisfies the NSP of order $k$ with some constant $\alpha_{k}<1$ is coNP-complete.

First of all, note that the NSP (2) is equivalent to the condition that

holds for all $S\subseteq\{1,\dots,n\}$, $\lvert{S}\rvert\leq k$, and $\bm{x}\in\mathds{R}^{n}$ with $\bm{A}\bm{x}=0$. Clearly, (2) and (18) always hold for some $\alpha_{k}\in$.

To show hardness, we claim that the matrix $\bm{A}$ has a circuit of size at most $k$ if and only if there does not exist any $\alpha_{k}<1$ such that (2) holds. Since the former problem is NP-complete by Theorem 1, this completes the proof.

Assume $\bm{A}$ has a circuit of size at most $k$. Then there exists a vector $\bm{x}$ in the nullspace of $\bm{A}$ with $1\leq\lVert{\bm{x}}\rVert_{0}\leq k$. It follows that $\lVert{\bm{x}}\rVert_{k,1}=\lVert{\bm{x}}\rVert_{1}$. Thus, (2) implies $\alpha_{k}\geq 1$. Since, trivially, $\alpha_{k}\leq 1$, we conclude that $\alpha_{k}=1$.

Conversely, assume that there exists no $\alpha_{k}<1$ such that (2) holds for $\bm{A}$ and $k$. This implies that there is a vector $\bm{x}$ with $\bm{A}\bm{x}=0$ and $1\leq\lVert{\bm{x}}\rVert_{0}\leq k$ such that $\lVert{\bm{x}}\rVert_{k,1}=\lVert{\bm{x}}\rVert_{1}$, because otherwise $\alpha_{k}<1$ would be possible. But this means that the support of $\bm{x}$ contains a circuit of $\bm{A}$ of size at most $k$, which shows the claim. ∎

Given a matrix $\bm{A}\in\mathds{Q}^{m\times n}$ and a positive integer $k$, it is NP-hard to compute the nullspace constant $\underline{\alpha}_{k}$.

V Concluding remarks

Instead of the intractable RIP, NSP, or spark, the weaker but efficiently computable mutual coherence is sometimes used. It can be shown that the mutual coherence yields bounds on the RIC, NSC, and the spark; see, for instance, . Thus, imposing certain conditions involving the mutual coherence of a matrix can yield uniqueness and recoverability (by basis pursuit or other heuristics), see, e.g., . However, the sparsity levels for which the mutual coherence can guarantee recoverability are quite often too small to be of practical use. This emphasizes the importance of other concepts.

An interesting question for future research is whether it is hard to approximate the constants associated with the RIP or NSP in polynomial time to within some factor, or whether spark and NSC computations are also strongly NP-hard (as is RIC computation, see Corollary 6). A first step in this direction was taken in , where inapproximability of RIP parameters is shown under certain less common complexity assumptions (interestingly, also making use of Cholesky decompositions of certain matrices related to a type of adjacency matrix for random graphs), see also ; strong inapproximability results for ( P 0 ) appear in .

Moreover, (co)NP-hardness does not necessarily exclude the possibility of practically efficient algorithms. So far, to the best of our knowledge, the focus has been laid largely on relaxations or heuristics. In , it was shown that one may sometimes do better than exhaustive search to certify the RIP, making use of the nondecreasing monotonicity of $\underline{\delta}_{k}$ with growing $k$. A “sandwiching” procedure to compute the NSC $\underline{\alpha}_{k}$ (exactly) was very recently proposed in and empirically demonstrated to be faster than brute force. However, neither method can guarantee a running time improvement with respect to simple enumeration. Moreover, Corollary 6 shows that, in general, no pseudo-polynomial time algorithm (i.e., a method with running time polynomially bounded by the input size and the largest occuring numerical value) can exist for computing the RIC $\underline{\delta}_{k}$, unless P$=$NP. More work on exact algorithms could shed more light on the behavior of the RIP and NSP.

Acknowledgment

We thank the anonymous referees for their constructive comments and for spotting an error in an earlier version of this paper.

References

References