This paper is concerned with linear time-invariant control systems (LTI systems). Such a system in dimension is specified by a transition matrix and a set of controls that is a linear subspace of . The evolution of the system is described by the recurrence , where for all .
An important class of decision problems concerning control systems involves reachability [6, 7, 4]. Given an LTI system with control set , an initial state , and target state , the point-to-point reachability problem asks whether there exists and controls such that . In presenting this problem many authors fix either the initial state or target state to be . The former case is often referred to as reachability (steering the system from the origin to a given state) while the latter case if often referred to as controllability (steering a given state to the origin). In this paper we will speak uniformly of reachability and we will consider both point-to-point problems and point-to-set problems.
The point-to-point reachability problem for LTI systems was shown to be decidable in polynomial time by reduction to Kannan and Lipton’s Orbit Problem  (the techniques used in this reduction are standard in control theory, and have appeared, in particular, in ). The central theme of the present paper is to consider a more general formulation of the linear control problem in which the set of controls is defined by a boolean combination of linear inequalities. We show that in this setting decidability is much more delicate. Our main results are as follows:
When is the union of finitely many affine subspaces, the point-to-point reachability problem for LTI systems is undecidable.
When is the union of two affine subspaces, the point-to-point reachability problem is as hard as the Skolem Problem for linear recurrence sequences.
When is a bounded convex polytope, the point-to-point reachability problem is as hard as the Positivity Problem for linear recurrence sequences.
When, for all eigenvaluesof , it holds that and that is a root of unity, and when is a closed and bounded convex polytope containing as an interior point, then the problem of reaching a given convex polytope starting from is decidable.
Skolem’s Problem asks whether a given integer linear recurrence sequence has a zero term, while the Positivity Problem asks whether all terms are positive. The decidability of both problems has been open since at least the 1970s [23, 22, 28, 11]. To date, decidability of Skolem’s Problem is known only for recurrences of order at most 4 [31, 20] and decidability of the Positivity Problem is known only for recurrences of order at most 5 . The hardness results in this paper suggest that deciding reachability in LTI systems for any class of control sets more general than linear subspaces will prove a very challenging problem. Note however that the reachability problem is straightforwardly semi-decidable, as reachability in steps for each fixed
is easily reduced to solving a linear program.
As one might expect, our positive decidability result (Item 4) requires fairly strong hypotheses. Intuitively, the requirement that lie in the interior of ensures that we can control in every direction, while the requirement that the spectral radius of be less than one is equivalent to the requirement that the system without input be asymptotically stable (also called Schur stable).222This is a strictly stronger condition than Bounded-Input Bounded-Output stability. The remaining spectral assumption on can be related to notions in the theory of self-affine fractals (see below).
Related Work. From the point of view of decidability it is well understood that control problems are hard even for mild generalisations of linear systems . For example, point-to-point reachability is undecidable for piecewise linear systems [3, 5, 19] and for saturated linear systems . On the other hand, for linear systems it is often considered that the main controllability problems are efficiently decidable. The results of this paper illustrate that this view crucially depends on the assumption that the set on controls is a linear subspace. However such an assumption does not allow to express many natural requirements, e.g., that the set of controls be bounded.
Control of LTI systems (both in the continuous-time and discrete-time settings) under more general constraints on the set of controls has been extensively studied [8, 26, 27, 16, 14, 15, 29, 10, 24, 13, 9, 1, 32]. Although it is very common to consider systems with saturated inputs, we do not consider the problem of controller design and thus saturated inputs reduce to having inputs in the unit hypercube in our case. LTI systems with convex input constraints have been considered in the past (see the references above) but we are not aware of any complete characterisation of the reachable set in this case, except for conical constraints.
There is a clear relationship between the reachability problem with bounded convex control sets and self-affine fractals. For LTI systems with spectral radius , the closure of the set of states reachable from is the convex hull of the (unique) self-affine fractal that satisfies the set equation , where denotes the set of extreme points of . We are aware of several results [18, 30] on the computability of the convex hulls of such fractals (and more general types of fractals). However those results only apply to the case when the convex hull is a polytope and usually only in dimension . The requirements on the spectrum of under Item 4 above are related to the so-called fractal of unity of .
2 Undecidability and Hardness
In this section we give evidence for the hardness of the reachability problem. We show undecidability if the set of controls is a finite union of affine subspaces and give a reduction from the Positivity Problem in case the set of controls is a bounded convex polytope. The case where the set of controls is a union of two affine subspaces is treated in Appendix B.
The goal of this subsection is to prove the following result. The reachability problem for LTI systems whose sets of controls are finite unions of affine subspaces is undecidable.
We prove Theorem 2.1 by reduction from the vector reachability problem for invertible matrices: given invertible matrices
and vectors, do there exist non-zero integers such that . The undecidability of this problem is folklore, however we provide a proof in the appendix for the reader’s convenience. The key idea underlying the reduction of this problem to the reachability problem for LTI systems is to form an LTI whose transition matrix incorporates , and to provide a set of controls that can be used to simulate the successive application of powers of , , etc, by repeated application of . A subtle technical point here is to make the reduction robust with respect to the different orders in which the controls can be applied.
Proof of Theorem 2.1.
We reduce the vector reachability problem for invertible matrices to the reachability problem for LTI systems.
Let be invertible matrices and . From these data we define an LTI system in dimension . We consider states of this LTI system to be -tuples comprising vectors in and a vector in . We write each component of such a tuple in bold.
Matrix is a block diagonal matrix of dimension , given by
For let be the -th coordinate vector. For we define the following affine subspaces of :
We now define the set of admissible controls to be
Notice that each control is determined by a single vector and the effect of the control is to subtract from one block within the global state and add to another block. Finally, we define an initial state , and target state .
This completes the definition of the LTI system. We now argue that is reachable from if and only if there exist such that .
For the “if” direction, suppose that such exist. Starting from , choose the first control to be
leading to the state
From there, use as control for steps, leading to
Choose the next control to be
Continuing this construction in the natural way we obtain a sequence of controls from to .
We now proceed to showing the converse. Suppose that is reachable from . In order to reach the last block of (the -dimensional vector of all ones) it must be that:
a control from was used exactly once,
for all , a control from was used exactly once,
a control from was used exactly once.
Let be the vectors in that respectively determine these controls.
Since the first block of is the zero vector in we must have . This in turn implies that for some . By similar reasoning, for all we have for some and for some . Finally, we have since the penultimate block in equals . Altogether we have that , which concludes the proof.
As a final remark, note that the preceding argument does not assume any particular order in which the controls are applied. In particular, the controls need not be applied in the “natural” order , etc. ∎
2.2 Positivity Hardness for Convex Control Sets
In this section we show that if the set of controls is a convex polytope then the LTI reachability problem is as hard as the Positivity Problem for linear recurrences. Instead of reducing from the Positivity Problem directly, we give a reduction from the Markov Reachability Problem
: given a column-stochastic matrix, determine whether there exists such that . A reduction from the Positivity Problem to the Markov Reachability Problem has been given in . There is a reduction from the Positivity Problem to the reachability problem for LTI systems whose sets of controls are compact convex polytopes (with rational vertices).
We give a polynomial-time reduction from the Markov Reachability Problem to the problem at hand. Given a column-stochastic matrix , we define an LTI system comprising a matrix and a compact convex polytope
The initial state is and target state .
We argue that is reachable from if and only if there exists such that .
First, suppose that there exists such that . Consider the sequence of controls and This sequence steers to .
On the other hand, suppose that there exists a sequence of controls controlling to . Since the -nd and -rd coordinates of are equal, noting that the matrix erases coordinate but not , it follows that is the only non-zero control, that is, . Therefore at time the state is , and the only way to reach in the remaining step is to take (otherwise one of the first coordinates will be non-zero). But this is only possible if . This concludes the proof. ∎
3 Decidability of Reachability for Simple LTI Systems
Define an LTI system to be simple if:
the set of controls is a bounded convex polytope that contains in its relative interior (i.e. for some open ball around );
the spectral radius of is less than one;
every non-zero eigenvalue of is such that is a root of unity.
The rationale behind those assumptions will be explained in the following paragraphs.
We show decidability of the following problem: given a simple LTI system in dimension and a bounded convex polytope , determine whether is reachable from . We call this the reachability problem for simple LTI systems. The set of states reachable from is . Thus the reachability problem for simple LTI systems is equivalent to asking whether meets .
It is clear that the reachability problem for LTI systems is semi-decidable. Fixing , the problem of whether meets a given polytope can straightforwardly be cast as a linear program. Iterating over all we thus have a semi-decision procedure for reachability.
In the rest of this section we describe a semi-decision procedure for non-reachability. The idea is to use a hyperplane that separatesand as a certificate of non-reachability, as illustrated on Figure 1. The key technical step here is to show that it suffices to consider hyperplanes whose normal vectors have algebraic entries. To establish this we consider the cone of all hyperplanes that separate and and show that the extremal elements of this cone are algebraic. This reasoning heavily relies on the spectral assumptions about the matrix in the definition of simple LTI systems.
The main difficulty in certifying non-reachability is that the convex set is difficult to describe in general. In particular, it can have infinitely many faces, even though the control polyhedron only has finitely many faces, as illustrated in Figure 2. This is, however, not the only difficulty. Indeed, one might get the impression from Figure 2 that the boundary of consists solely of (possibly countably many) facets, that is faces of dimension . If that were true then we could write333Recall that any closed convex set can be written as the intersection of its closed supporting half-spaces. the closure of as the intersection of the supporting half-spaces of its facets. Since any facet of a simple LTI systems has an algebraic supporting hyperplane, this would immediately give us a description of the reachable set. Unfortunately, this is not always the case—there are cases, as illustrated in Figure 3, where part of the boundary does not belong to any facet, but rather to some lower-dimensional faces. Consequently, we need to include in the intersection of supporting half-spaces some directions for lower dimensional faces. However such faces, by definition, do not usually have a unique supporting hyperplane and it is not, in fact, clear that they admit a supporting hyperplane with algebraic coefficients. The main technical result of this section (Proposition 3) shows that it is the case.
The following proposition identifies some simplifying assumptions that can be made without loss of generality for analysing reachability. See Appendix C for a proof.
The reachability problem for a simple LTI system can be reduced to the special case in which it is assumed that all eigenvalues of are real and strictly positive and that is full dimensional.
In the rest of this section we will assume that all simple LTI systems are such that is full dimensional and all eigenvalues of are strictly positive.
Given a simple LTI system , is a convex open subset of whose closure is .
Since it holds that is an increasing collection of convex sets and hence is convex.
To see that is open, consider a typical element . Now is full dimensional and contains in its interior. Moreover since is invertible we have that is also full dimensional and contains in its interior. Hence lies in the interior of , which is contained in the interior of .
From the fact that the spectral radius of is strictly less than one it easily follows that is dense in , so it remains to observe that is closed. But is a fixed point of the contractive self map on the metric space of all bounded subsets of under the Hausdorff metric. Such a self-map has a unique fixed point. Moreover, since the collection of compact subsets of is complete under the Hausdorff metric and is preserved by , we conclude that is compact and thus closed. ∎
The following is a (version of a) classical result of convex analysis: [Theorem of the Separating Hyperplane] Let and be compact convex subsets of . Then if and only if there exists and such that for all and for all .
It is straightforward to verify that is a topologically closed cone in . Moreover by the assumption that is full dimensional we have that is a pointed cone, that is, for all if both then . See Figure 1 for graphical representation of the cone.
Given and a closed set , define
For fixed , note that is maximised for if and only if each individual inner product is maximised. In other words,
Given , define
Then is a vector subspace of —indeed is the unique translation of the affine hull of that contains the origin.
The vector space has a basis of rational vectors. Furthermore, for any .
Let be such that
for all . Such an exists since the right-hand side of (2) forms an increasing family of subspaces of as and such a sequence must eventually stabilize. We claim that
From the claim it follows that has a basis of rational vectors since is bounded polytope with rational vertices for .
It remains to prove the claimed equality (3). The left-to-right inclusion follows directly from Equations (1) and (2). For the right-to-left inclusion, it suffices to note that for all . Indeed, suppose that for some and . Define for by , , and for . Then
Let the eigenvalues of matrix be . Then there is a collection of bilinear forms with algebraic coefficients such that for all we have
By the Jordan–Chevalley decomposition, we have where is diagonal, is nilpotent, and commute, and all matrices have algebraic coefficients. Moreover we can write for appropriate idempotent diagonal matrices . Then for all we have
where is defined in (3). This concludes the proof because each is clearly bilinear with algebraic coefficients. ∎
Given , define to be the linear subspace of comprising all such that for all and every bilinear form as in (4) if then . It is clear that that has a basis of vectors all of whose entries are algebraic numbers. See Figure 4 for a geometrical intuition of this notion.
We say that a sequence of real numbers is positive if for all . We moreover say that is ultimately positive if there exists such that for all .
Suppose that are such that . Then for all , if the sequence is positive (resp. ultimately positive) then there exists such that the sequence