Reconstruction in One Dimension from Unlabeled Euclidean Lengths

Abstract

Let G be a 3-connected ordered graph with n vertices and m edges. Let $\textbf{p}$ be a randomly chosen mapping of these n vertices to the integer range $\{1, 2,3, \ldots , 2^b\}$ for $b\ge m^2$. Let $\ell$ be the vector of m Euclidean lengths of G’s edges under $\textbf{p}$. In this paper, we show that, with high probability over $\textbf{p}$, we can efficiently reconstruct both G and $\textbf{p}$ from $\ell$. This reconstruction problem is NP-HARD in the worst case, even if both G and $\ell$ are given. We also show that our results stand in the presence of small amounts of error in $\ell$, and in the real setting, with sufficiently accurate length measurements. Our method combines lattice reduction, which has previously been used to solve random subset sum problems, with an algorithm of Seymour that can efficiently reconstruct an ordered graph given an independence oracle for its matroid.

1 Introduction

Let $\textbf{p}$ be a configuration of n points $p_i$ on the real line. Let G be an ordered graph with n vertices, numbered $\{1,\ldots , n\}$ (briefly, [n]) and m edges, indexed by [m]. Having edge indices is very useful when applying linear algebraic methods to graphs. We denote the edge with endpoints i and j as ij or $\{i,j\}$. For each edge ij in G, we denote its Euclidean length as $\ell _{ij}(\textbf{p}):= |p_i-p_j|$. We denote by $\ell =\ell (\textbf{p})$, the m-vector of all of these measured lengths. In a labeled reconstruction problem, we are given G and $\ell$, and we are asked to reconstruct $\textbf{p}$, up to an unknowable translation and reflection. Since the graph G and the vector $\ell$ are ordered, this means we are given the bijection between lengths and edges. In an unlabeled reconstruction problem, the input is the vector $\ell$, and the task is to reconstruct G and $\textbf{p}$. As in the unlabeled setting, $\textbf{p}$ can be reconstructed only up to a translation and reflection. Since G is ordered, reconstructing it requires determining a bijection between the edges of G and the entries of $\ell$. We note that the number of entries of $\ell$ determines m, but not n, and that n is not in the input. That problems of this type are difficult to solve in the worst case is classical. Saxe [30] showed in 1979 that the labeled version of the problem in one dimension is strongly NP-hard.

Two and three-dimensional labeled and unlabeled reconstruction problems often arise in applications such as sensor network localization and molecular shape determination. These problems are even harder than the one-dimensional setting studied here. But due to the practical importance of these applications, many heuristics have been studied, especially in the labeled setting. See [8, 9] for some surveys of this topic. There do exist polynomial time algorithms that have guarantees of success, but only on restricted classes of graphs and configurations. For example, labeled problems can be solved with semidefinite programming if the unknown configuration $\textbf{p}$ and graph G correspond to a “universally rigid framework” $(G,\textbf{p})$ [33]. Labeled problems in dimension d can be solved with greedy algorithms if the underlying graph happens to be “d-trilateral”. This means that the graph has an edge set that allows one to localize one or more $K_{d+2}$ subgraphs, and then iteratively glue vertices onto already-localized subgraphs using $d+1$ edges, (and unambiguously glue together already localized subgraphs) in a way to cover the entire graph. There is also a huge literature on the related problem of low-rank matrix completion; see [28] for a survey of this topic. Typically, those results assume randomness in the choice of matrix sample locations, roughly corresponding to the graph G in our setting.

Unlabeled problems are much harder, and there are fewer approaches known to be viable. In [4], the authors prove that for all dimensions d and configurations $\textbf{p}$ outside of a Zariski-closed subset of d-dimensional configurations, there is a unique way to associate the values in $\ell$ to the edges of the complete graph $K_n$. Thus, when all pairwise distances are given, an exhaustive combinatorial search (exponential time) must correctly reconstruct such configurations. For d-trilateral graphs, a backtracking approach called LIGA is presented in [21] and a greedy approach called TRIBOND is presented in [12]. TRIBOND is shown to run in $O(m^k)$ time, where k is a fixed exponent for each dimension. The work described by these groups has spearheaded the research in unlabeled reconstruction. In [2], the authors define a property for a measured length vector $\ell$ they call “strongly generic”. They then say that the TRIBOND approach must succeed if $\ell$ is strongly generic. Notably, no sufficient condition is provided for any graph G to have even a single strongly generic measured length vector. In [15] it is shown that for $d \ge 2$, when the underlying graph is d-trilateral then the TRIBOND algorithm, given exact measurements of a d-dimensional configuration $\textbf{p}$, must correctly reconstruct $\textbf{p}$, provided that $\textbf{p}$ is not in an exceptional, Zariski-closed subset of all d-dimensional configurations. When $d=1$ the authors show that the graph must be 2-trilateral (instead of 1-trilateral) in order for a greedy trilateration method to work.

The unlabeled reconstruction problem in the one-dimensional integer setting also goes by names such as “the turnpike problem” and the “partial digest problem”. Theoretical work on these problems typically uses worst case analysis and often only applies to the complete graph (e.g. [5, 23, 32]).

In this paper we present an algorithm that can provably solve the one-dimensional unlabeled (and thus, also the labeled) reconstruction problems in polynomial time, for each fixed ordered graph, with high probability over randomly chosen $\textbf{p}$, as long as we have sufficiently accurate length measurements. We only require that G is 2-connected for the labeled case and 3-connected for the unlabeled case. As we will see below, these conditions are necessary for any method to work.

The algorithm is based on combining the LLL lattice reduction algorithm [13, 22, 24] together with the graphic matroid reconstruction algorithm of Seymour [31]. The limiting factor of our algorithm in practice is its reliance on very precise measurements of the edge lengths $\ell$. But the existence of an efficient algorithm with these theoretical guarantees was surprising to us. Even though the required precision grows with the number of edges, we give an explicit bound. This is in contrast with theoretical results about generic configurations often found in rigidity theory, which do not apply to configurations with integer coordinates or inexactly-measured lengths.

1.1 Connectivity Assumption

Clearly we can never detect a translation or reflection of $\textbf{p}$ in the line, so we will only be interested in recovery of the configuration up to such congruence. In the labeled setting, if i is a cut vertex of G, then G is the union of two graphs $H_1$ and $H_2$ with at least two vertices each, so that the vertex set of $H_1$ and $H_2$ intersect on exactly $\{i\}$. Hence, we can reflect the positions of the vertices in $H_1$ across $p_i$ without changing $\ell$. Such a flip is invisible to our input data. To avoid such partial reflections, we must assume that G is 2-connected.

The question of whether it is possible to recover $\textbf{p}$, a configuration in $\mathbb {R}^d$, uniquely up to isometry from the the labeled length measurements is known in rigidity theory as the question of whether the framework $(G,\textbf{p})$ is globally rigid in d-dimensions. If, for generic $\textbf{p}$, $(G,\textbf{p})$ is always globally rigid in dimension d, then G is called generically globally rigid in dimension d [7, 16]. It is a folklore result that G is generically globally rigid in dimension one if and only if it is 2-connected.

In higher dimensions, $(d+1)$-connectivity is necessary for a graph to be generically globally rigid [19], but it is not sufficient [6]. In dimension two, there is a combinatorial characterization of generically globally rigid graphs [20]. In dimensions $d\ge 3$, there is no known combinatorial characterization of generically globally rigid graphs, but there is a characterization that can be checked with a randomized algorithm that uses linear algebra [16].

Fig. 1

The graphs, a and d are related by a 2-swap. Note that the edge lengths in one dimension are unchanged under a 2-swap

In the unlabeled setting, we can never detect a renumbering of the points in $\textbf{p}$ along with a corresponding vertex renumbering in G, so we will be unconcerned with such isomorphic outputs. More formally, let $G = (V,E)$ be an ordered graph with vertex set [n] and with an ordering of the edges given by a bijection $\sigma : E\rightarrow [m]$. If $H = (V,E')$ is another ordered graph with the same vertex set, such that there is a bijection $\pi : V\rightarrow V$ such that $ij\in E$ if and only if $\pi (i)\pi (j)\in E'$ and the edges of H are ordered via a bijection $\tau : E'\rightarrow [m]$ so that $\sigma (ij) = \tau (\pi (i)\pi (j))$, then G and H are isomorphic as ordered graphs, and it is impossible to detect whether a measurement vector $\ell$ arose from G or H.

Without labeling information, if G has a cut vertex i, then, decomposing G into $H_1$ and $H_2$ as above, we cannot hope to distinguish measurements from G and those from a graph made from identifying $H_1$ and $H_2$ along some other vertex besides i, because besides reflecting the positions of vertices in $H_2$ across i, we can also translate them. Even if G is 2-connected but not 3-connected, the unlabeled problem does not have a unique solution. Let $\{i,j\}$ be a cut pair and decompose G into graphs $H_1$ and $H_2$ so that their vertex sets intersect exactly on $\{i,j\}$. Any set of length measurements of G also arises from the graph obtained by identifying the copy of i in $H_1$ with the copy of j in $H_2$ and vice versa. This operation, illustrated in Fig. 1, is called a 2-swap. Graphs that are related by a 2-swap are not isomorphic, but they have the same graphic matroids. To avoid ambiguities such swapping, we must assume, at least, that G is 3-connected. There are a number of subtleties in how to even define an appropriate notion of generic global rigidity in the unlabeled setting. These are discussed in [15, 18].

1.2 Integer Setting

Moving to an algorithmic setting, we will now assume that the points are placed at integer locations in the set [B], with $B=2^b$ for some positive integer b. For the labeled problem, we will want an algorithm that takes as input G and an integer vector $\ell \in [B]^m$, and outputs $\textbf{p}$ (up to congruence). For the unlabeled problem, we will want an algorithm that takes as input the integer vector $\ell$, and outputs G (up to vertex relabeling) and $\textbf{p}$ (up to congruence).

Even the labeled problem is strongly NP-HARD by a result of Saxe [30]. Interestingly, one of his reductions is from SUBSET-SUM. This is a problem that, while hard in the worst case, can be efficiently solved, using the Lenstra–Lenstra–Lovász (LLL) lattice reduction algorithm [24], for most instances that have sufficiently large integers [22]. Indeed, we will take cue from this insight and use LLL to solve both the labeled and unlabeled reconstruction problems.

We say that a sequence, $A_m$, of events in a probability space occurs with high probability as $m \rightarrow \infty$ (WHP) if $\Pr [A_m] = 1 - o(1)$. Here is our basic result.

Theorem 1.1

Let G be a 3-connected (resp. 2-connected) graph with m edges. Let $b \ge m^2$. Let each $p_i$ be chosen independently and uniformly at random from $[2^b]$. Then there is a polynomial time algorithm that succeeds, WHP, on the unlabeled (resp. labeled) reconstruction problems.

The success probability, the convergence rate and the minimal required b can be estimated more carefully (see Proposition 3.9 and Remark 3.10 below). In any case, what we can prove seems to be pessimistic. Experiments (see below) show that our algorithm succeeds with very high probability on randomly chosen $\textbf{p}$ even with significantly smaller b. In part, this is because the LLL algorithm very often outperforms its worst case behavior [27] on random input lattices. On the other hand, lattice reduction is a brittle operation, and, for fixed m, it will fail ungracefully once b becomes too small.

1.3 Errors in Input

To move towards a numerical setting, we need to differentiate between the geometric data in $\textbf{p}$ and the combinatorial data that records the order of the points of $\textbf{p}$ along the line. Since we only have access to the measurements $\ell$, the ordering data will be represented by an orientation $\sigma _\textbf{p}$ of a graph G produced by the algorithm; the orientation of an edge indicates the ordering of its endpoints on the line using the configuration $\textbf{p}$. To model error or noise, we consider the case where there is $\{-1,0,1\}$ error, added adversarially, to each of the integer measurements $\ell _{ij}$. Relative to our scale of $2^{m^2}$, a $\{-1,0,1\}$ error is very small.

Theorem 1.2

Let G be an ordered graph with m edges. For the labeled setting, assume G is 2-connected. For the unlabeled setting, assume G is 3-connected. Let $b \ge m^2$. Let each $p_i$ be chosen independently and uniformly at random from $[2^b]$. Assume that $\{-1,0,1\}$ noise has been added adversarially to the $\ell _{ij}$. Then, in both the labeled and unlabeled settings, there is a polynomial time algorithm that, WHP, correctly outputs G and $\sigma _\textbf{p}$, up to reversing the orientation of every edge in $\sigma _\textbf{p}$.

The bound on b given here can be improved slightly, at the cost of a more complicated statement. Remark 3.10 describes such an improvement. In Sect. 4 we discuss ways in which our algorithm can fail.

Once we have the combinatorial data of G and $\sigma _\textbf{p}$, it is easy to reconstruct $\textbf{p}$, up to congruence, when $\ell$ has no error. We can simply use a spanning tree of G to lay out the $p_i$ one at a time. If there are errors, then it is unlikely that there is any configuration corresponding to the observed measurements. However, we can apply a linear least squares algorithm to get an estimate of $\textbf{p}$ with small error.

Theorem 1.3

Let G be a graph with m edges, and suppose that we are given measurements $\ell$ of an unknown configuration $\textbf{p}$, with adversarial noise in the model of Theorem 1.2, and an orientation $\sigma _\textbf{p}$ of G that is consistent with the ordering of the points in $\textbf{p}$ on the line.

Then we can efficiently compute a vector $\textbf{p}^* \in \mathbb {Q}^n$ so that, there is a vector $\textbf{q}= \pm \textbf{p}+ t\textbf{1}$ with

$$\begin{aligned} \Vert \textbf{p}^*-\textbf{q}\Vert \le n^{2.5}\sqrt{m} \end{aligned}$$

where $\pm \textbf{p}$ represents either $\textbf{p}$ or the configuration $-\textbf{p}$ obtained from negating each of the points $p_i$ in $\textbf{p}$ and $t\textbf{1}$ represents translation of $\textbf{p}$ by an unknown rational number t.

Since we are working with signals at the scale of $B=2^{m^2}$, this is a very small error indeed. We show in Remark 3.11 shows that when we have noise of larger magnitude than 1, the best approach is to round and truncate the input in such a way that the errors on each length measurement are in the set $\{-1,0,1\}$, albeit with a smaller b. As long as this new, smaller, b is sufficiently large, then our theorem will still apply. Informally, the important thing is not the number of error bits, but the number of signal bits. Additionally, Remark 3.12 shows that with sufficiently large m, we can tolerate any constant error maintained in our representation, or even a number of error bits that is bounded by a function that is o(m).

1.4 Real Setting

Theorem 1.2 can be used in the setting where $\textbf{p}$ is real valued. We suppose that each $p_i$ is chosen independently from the uniform distribution on the closed interval [1, B], which determines the underlying $\ell _{ij}$. We also suppose that each $\ell _{ij}$ is corrupted by an adversarial error $\varepsilon _{ij}$ with $-\frac{1}{2} \le \varepsilon _{ij} \le \frac{1}{2}$, so that the input data is

$$\begin{aligned} \ell _{ij} + \varepsilon _{ij}. \end{aligned}$$

Let $\ell '_{ij}:= |\textrm{nint}(p_i)-\textrm{nint}(p_j)|$, where $\textrm{nint}$ is the nearest integer. Under these assumptions we have

$$\begin{aligned} |\ell '_{ij} - \textrm{nint}(\ell _{ij}+\varepsilon _{ij}) |\le 1. \end{aligned}$$

As a result, the rounded data, $\textrm{nint}(\ell _{ij}+\varepsilon _{ij})$, will be a $\{-1,0,1\}$ noisy version of the lengths obtained from the rounded configuration, $\textrm{nint}(\textbf{p})$. Hence, our Theorem 1.2 applies to this integer input. If we instead start with a real configuration $\textbf{p}$ with points lying in the closed interval [0, 1], scaling by B and replaying the above construction shows that we can tolerate adversarial error on the order of $\frac{1}{2^{m^2}}$.

If we are given data with higher $\varepsilon$ values, we can simply divide by an appropriate constant so that, afterwards, $-\frac{1}{2} \le \varepsilon _{ij} \le \frac{1}{2}$, and then proceed with a smaller effective b and B. As long as b is large enough, Theorem 1.2 will still apply. In light of Remark 3.11, this approach is strictly better than keeping around additional error bits. Though in light of Remark 3.12, our theorem can tolerate o(m) extra error bits.

1.5 Basic Ideas

Consider traversing a cycle C in G, which is an ordered list of distinct vertices (except for the first and last, which are equal)

$$\begin{aligned} v_1, v_2, \ldots , v_k =v_1 \end{aligned}$$

so that the edges $v_{i}v_{i+1}$ and $v_{k-1}v_1$ are in G. As we traverse C, we consider the configuration $\textbf{p}$. Some of the edges go to the right ($p_{v_{i+1}} > p_{v_i}$) and some to the left ($p_{v_{i+1}} < p_{v_i}$). If we sum up the Euclidean lengths of these edges, with a $+1$ coefficient for edges that go to the right and and a $-1$ coefficient for edges that go to the left, the sum is zero, since the cycle traversal starts and ends at $p_{v_1}$. Thus, the cycle C determines an integer relation on the coordinates of the length vector $\ell$. The integer coefficients in this relation are all “small”, (the actual bounds we use will become apparent in Sect. 3.2) in fact only 0, $-1$, and 1, and, hence, so is the norm of the coefficient vector for such a cycle relation.

Let us call $\sigma _\textbf{p}$ the ordering of the vertices along the line. Repeating the argument for each cycle in G defines more small integer relations on the coordinates of $\ell$. We define $S^{\sigma _\textbf{p}}$ to be the linear span, in $\mathbb {R}^m$, of all the coefficient vectors of integer relations defined by cycles in G. The space $S^{\sigma _\textbf{p}}$ contains all the small integer relations that are forced on $\ell$ because it arises from measuring $\textbf{p}$. Heuristically, if $\textbf{p}$ is is selected uniformly at random, then $\ell$ should not satisfy any other small integer relation, because $\ell$ still behaves otherwise randomly.

In Sect. 3.2, we make this heuristic reasoning precise. What we show is that, if $\textbf{p}$ is selected uniformly at random from configurations of n integer points in [0, B], then, WHP any integer relation that is satisfied by the coordinates of $\ell$ and has norm below a cutoff value is in $S^{\sigma _\textbf{p}}$. This gives us a way to find a basis for $S^{\sigma _\textbf{p}}$, provided that we can find a basis for an appropriate lattice with small enough basis vectors.

Finding a lattice basis consisting of vectors that are as short as possible is an NP-hard problem. However, the cutoff we need is large enough so that any LLL reduced basis for the lattice we construct will, WHP, allow us to find a basis for $S^{\sigma _\textbf{p}}$. Our analysis builds on work of Lagarias and Odylyzko [22] and Frieze [13] on random SUBSET-SUM problems. The main technical difference between our problem and random SUBSET-SUM instances is that $\ell$ is obtained from $\textbf{p}$ via a non-linear mapping. We handle this issue using a counting argument that extends those of Frieze [13]. Because the LLL algorithm [24] can find a reduced basis in polynomial time, this step of the algorithm is polynomial.

Once we recover $S^{\sigma _{\textbf{p}}}$, our next task is to use it to recover the ordered graph G. For this, we will use a polynomial time algorithm of Seymour [31] that constructs an unknown graph using an independence oracle for its graphic matroid. We can implement the oracle in polynomial time using the basis for $S^{\sigma _\textbf{p}}$ found in the previous step, so we can recover G in polynomial time. To complete the combinatorial steps, we need to recover the orientation $\sigma _\textbf{p}$. Using $S^{\sigma _\textbf{p}}$, we can compute a consistent orientation for the edges of any cycle in G. We then find a globally consistent orientation, which must be $\pm \sigma _{\textbf{p}}$, using an iterative procedure based on an ear decomposition of G. In this step, we use 2-connectivity of G, because otherwise an ear decomposition will not exist.

Finally, we can use G, $\sigma _{\textbf{p}}$ and $\ell$ to determine the positions of each of the $p_i$. In the noiseless case, this can be done by traversing a spanning tree of G and laying out the points. If there is noise, we solve a least squares problem to approximate $\textbf{p}$.

Algorithm 1

The full Algorithm

The methods in this paper do not directly generalize to higher dimensions. In fact, each of our steps poses a challenging problem once the dimension becomes at least two. There is no direct analogue to the $\pm 1$ cycle relations, and instead, we would have to find integer polynomial relations on the measured lengths. To this end, we have no known non-trivial bound on either the degree or coefficient-sizes of these polynomials. Even if it were possible to recover the polynomials, and then, somehow, using a (so-far unknown) algorithmic version of [18], the unknown graph G, the remaining labeled graph realization problem is difficult in practice.

1.6 Related LLL Work

Recently in [14], a similar approach has been carefully applied to recover an unknown discrete signal from multiple randomly chosen large integer linear measurements. As described in [14], an algorithm based on LLL lattice reduction can even work in the presence of adversarially chosen error on the input, provided that the error is sufficiently small. Theorem 2.1 of [14] provides a detailed analysis of the relationship between the various parameters.

In other related recent work, [1] has shown that the LLL method can be used to efficiently solve the correspondence retrieval problem and the phase retrieval problem with no noise. Song et al. [34] have shown how LLL can be used to efficiently learn single periodic neurons in the low noise regime (this problem also generalizes phase retrieval). Work in [10, 38] shows how a number of other problems in machine learning, such as learning mixtures of Gaussians, can be solved by LLL, provided that the noise is sufficiently small. The main difference between our setting and those of [1, 10, 14, 22, 34, 38], is that in our case, we are looking for a number of linearly independent small relations in one random data vector, instead of a single small relation in one or more random data vectors.

In [14, 22], the data that satisfies a small integer relation ($\ell$ in our setting) is directly generated by a random process. In our work, as in the work of [1, 10, 34, 38], the input data is obtained indirectly from the random process. And in particular, in this paper, the process going from the randomly chosen configuration $\textbf{p}$ to the length measurements $\ell$ involves an absolute value operation.

In [10, 14], the authors show that when one has O(m) random data vectors that satisfy a common small integer relation, then LLL can be successfully applied with only around m bits of reliable data, instead of the $m^2$ bits that we need here. For our reconstruction problem, we only have one underlying random configuration $\textbf{p}$, so we cannot exploit any additional linear relations to get a quantitative improvement.

In our random model, each $p_i$ is chosen independently from a uniform distribution. In rigidity applications, this is a natural assumption to make. Presumably, similar results could hold under some other sampling assumptions for the points $p_i$. In particular, the works in [1, 10, 34, 38] consider a Gaussian random process. What appears to be key is that the sampling distribution possesses some sufficient anti-concentration properties [10, 38].

2 Graph Theoretic Preliminaries

First we set up some graph-theoretic background.

2.1 Graphs, Orientations, and Measurements

In this paper, $G = (V,E)$ will denote a simple, undirected ordered graph with n ordered vertices and m ordered edges. These orderings will allow us to index various matrix rows and columns associated with edges and vertices.

Definition 2.1

Let $G = (V,E)$ be an undirected simple graph with n vertices and m edges. We denote undirected edges by $\{i,j\}$, with $i, j\in V$.

An orientation $\sigma$ of G is a map that assigns each undirected edge $\{i,j\}$ of G to exactly one of the directed edges (i, j) or (j, i); i.e., $\sigma (\{i,j\})\in \{(i,j),(j,i)\}$. Informally, $\sigma$ assigns one endpoint of each edge to be the tail and the other to be the head.

An oriented graph $G^\sigma$ is a directed graph obtained from an undirected graph G and an orientation $\sigma$ of G by replacing each undirected edge $\{i,j\}$ of G with $\sigma (\{i,j\})$.

We are usually interested in a pair $(G,\textbf{p})$ where G has n vertices and $\textbf{p}$ is a configuration of n points $(p_1, \ldots , p_n)$ on the line. Such a pair has a natural orientation of G associated with it.

Definition 2.2

Let $(G,\textbf{p})$ denote a graph with n vertices and $\textbf{p}$ a configuration of n points. Define the configuration orientation $\sigma _\textbf{p}$ by

$$\begin{aligned} \sigma _\textbf{p}(\{i,j\}) = {\left\{ \begin{array}{ll} (i,j) &{} \hbox { if}\ p_i < p_j \\ (j,i) &{}\hbox { if}\ p_i \ge p_j \end{array}\right. } \qquad (\text {for all} \{i,j\}\in E) \end{aligned}$$

We say that an orientation $\sigma$ of G is vertex consistent if $\sigma = \sigma _\textbf{p}$ for some configuration $\textbf{p}$.

There are $2^m$ possible orientations, but, when $m > n$, many fewer are vertex consistent.

Lemma 2.3

Let G be a graph with n vertices. There are at most n! vertex consistent orientations of G.

Proof

The orientation $\sigma _\textbf{p}$ is determined by the permutation of $\{1, \ldots , n\}$ induced by the ordering of the $p_i$ on the line. $\square$

We now introduce some specialized facts about a standard linear representation of a graphic matroid over $\mathbb {R}$. We start by associating two vector configurations with a graph G.

Definition 2.4

Let G be a graph with n vertices and m edges. For $i\ne j$, both in V, a signed edge incidence vector $e_{(i,j)}$ has n coordinates indexed by V; all the entries are zero except for the ith, which is $-1$ and the jth, which is $+1$. (So that $e_{(i,j)} = - e_{(j,i)}$.)

Given an ordered graph G and an orientation $\sigma$ of G, the signed edge incidence matrix, $M^\sigma (G)$, has as its rows the vectors $e_{\sigma (\{i,j\})}$, over all of the edges $\{i,j\}$ of G.

Let $\rho = v_1, v_2, \ldots , v_k = v_1$ be a cycle in G. The signed cycle vector $\textbf{w}_{\rho ,\sigma }$ of $\rho$ in $G^\sigma$ has m coordinates indexed by E. Coordinates corresponding to edges not traversed by $\rho$ are zero. The coordinate corresponding to the edge $\{v_{i}, v_{i+1}\}$ is 1 if $\sigma (\{v_{i}, v_{i+1}\}) = (v_{i}, v_{i+1})$ and otherwise it is $-1$.

If we have fixed some $\textbf{p}$, we can shorten $\textbf{w}_{\rho ,\sigma _{\textbf{p}}}$ to $\textbf{w}_{\rho }$ (with no explicit orientation written).

It is easy to see that a signed cycle vector $\textbf{w}_{\rho ,\sigma _{\textbf{p}}}$ must be in the kernel of the transpose of $M^\sigma (G)$. Now we shift focus to an associated linear space.

Definition 2.5

Let G be a graph with n vertices and m edges and $\sigma$ an orientation. An $\mathbb {R}$-cycle is a vector in the kernel of the transpose of $M^\sigma (G)$. The $\mathbb {R}$-cycle space (cycle space, for short) $S^\sigma \subseteq \mathbb {R}^m$ is the subspace of all $\mathbb {R}$-cycles. A basis of $S^\sigma$ is induced by graph cycles if it contains only signed cycle vectors.

For our intended application, it is important that $S^\sigma$ have a basis comprising integer vectors of small norm. It is a classical fact that this is the case.

Proposition 2.6

(See e.g, [29, Chapter 5]) Let G be a connected ordered graph with n vertices and m edges and $\sigma$ an orientation of G. The $\mathbb {R}$-cycle space $S^\sigma$, has dimension $c:= m - n + 1$. Moreover, it is spanned by the signed cycle vectors in $G^\sigma$. In particular, $S^\sigma$ has a basis induced by graph cycles.

If we denote by $-\sigma$ the orientation of G obtained by reversing every edge in $G^\sigma$, it is easy to see that $S^\sigma = S^{-\sigma }$. We now describe a family of linear maps related to edge length measurements.

Definition 2.7

Let G be an ordered graph with n vertices and $\sigma$ an orientation of G. We define the oriented measurement map $\ell _\sigma : \mathbb {R}^n \rightarrow \mathbb {R}^m$ by

$$\begin{aligned} \ell _\sigma (\textbf{p})_{\{i,j\}}:= {\left\{ \begin{array}{ll} p_j - p_i &{} \hbox { if}\ \sigma (\{i,j\}) = (i,j) \\ p_i - p_j &{} \hbox { if}\ \sigma (\{i,j\}) = (j,i) \end{array}\right. } \end{aligned}$$

For any choice of $\sigma$, $\ell _\sigma$ is a linear map. We also define the measurement map $\ell$ by

$$\begin{aligned} \ell (\textbf{p})_{\{i,j\}} = |p_j - p_i| \end{aligned}$$

The measurement map is only piecewise linear.

Lemma 2.8

Let $(G,\textbf{p})$ be an ordered graph with n vertices and $\textbf{p}$ a configuration of n points. Let $\rho$ be a cycle of G. Let $\textbf{w}_{\rho }$ be its signed cycle vector in $G^{\sigma _\textbf{p}}$. Then

$$\begin{aligned} {\textbf{w}}^\top _\rho \ell (\textbf{p}) = 0 \end{aligned}$$

Proof

Unraveling definitions, we see that $\ell (\textbf{p}) = \ell _{\sigma _\textbf{p}}(\textbf{p})$, and that, in the standard bases for the domain $\mathbb {R}^n$ and codomain $\mathbb {R}^m$, respectively, the matrix of the linear transformation $\ell _{\sigma _\textbf{p}}$ is $M^{\sigma _p}(G)$. We then compute

$$\begin{aligned} {\textbf{w}}^\top _\rho \ell (\textbf{p}) = {\textbf{w}}^\top _\rho \ell _{\sigma _\textbf{p}}(\textbf{p}) = {\textbf{w}}^\top _\rho M^{\sigma _p}(G) \textbf{p}= 0 \end{aligned}$$

because ${\textbf{w}}^\top _\rho$ is in the cokernel of $M^{\sigma _p}(G)$ by definition. $\square$

3 The Algorithm

In this section we describe the algorithm in detail and develop the tools required for its analysis. We will focus on the harder, unlabeled setting, and afterwards describe the various simplifications that can be applied in the labeled case.

3.1 The Sampling Model

The algorithm has the following parameters. We have a precision $b\in \mathbb {N}$. There is fixed ordered graph G with n vertices and m edges. Our algorithm takes as input $\ell = \ell (\textbf{p}) + \varepsilon$, where $\textbf{p}$ is selected uniformly at random among configurations of n points with coordinates in $[2^b]$ and $\varepsilon$ is an error vector in $\{-1,0,1\}^m$.

The algorithm does not have access to n, only m (via the number of coordinates in $\ell$). All the probability statements are taken with respect to $\textbf{p}$, which is not part of the input. This complicates the probabilistic analysis a little bit relative to [13, 22], but much remains similar.

3.2 Reconstruct $S^{\sigma _\textbf{p}}$

In this section, we will show how, WHP, to reconstruct $S^{\sigma _\textbf{p}}$ from $\ell$. Our main tool will be the LLL algorithm [24], as used by [22], and analyzed by [13].

Definition 3.1

For any vector $\textbf{r}$ in $\mathbb {Z}^{m}$ define the lattice ${{\,\textrm{Lat}\,}}(\textbf{r})$ in $\mathbb {Z}^{m+1}$ as the integer span of the columns of the following $(m+1)$-by-m lattice generating matrix

$$\begin{aligned} L(\textbf{r}):= \begin{pmatrix} I_m \\ {\textbf{r}}^\top \end{pmatrix} \end{aligned}$$

where $I_m$ is the $m\times m$ identity matrix.

Remark 3.2

Frieze [13] multiplies the last row of $L(\textbf{r})$ by a large constant. We do not do that, so that we can easily deal with noise. In [17], we elaborate on this issue and show how, in the original context of SUBSET-SUM, Frieze’s analysis goes through without the use of the large constant.

In our algorithm we will work over the lattice ${{\,\textrm{Lat}\,}}(\ell )$, where $\ell$ is the edge length measurements of the unknown, configuration $\textbf{p}$, plus adervarially chosen noise. Figure 2 summarizes the algorithm in pseudo-code.

Algorithm 2

The ComputeRelations Algorithm

If $\textbf{x}$ is a vector in $\mathbb {R}^m$ and $a\in \mathbb {R}$ a real number, we denote by $\begin{pmatrix} \textbf{x}\\ a \end{pmatrix}$ the vector in $\mathbb {R}^{m+1}$ that has $\textbf{x}$ as its first m coordinates and a as its $(m+1)$st. We call $\textbf{x}$ the truncation of $\begin{pmatrix} \textbf{x}\\ a \end{pmatrix}$. Now we can describe the algorithm to reconstruct $S^{\sigma _\textbf{p}}$ in more detail.

• Given, $\ell$, we let m be the number of coordinates in $\ell$. This is the number of edges of G. The parameter m determines a threshold value $\sqrt{2m}2^{m/2}$.

• We then use the LLL algorithm to compute a reduced basis $\Lambda$ for the lattice ${{\,\textrm{Lat}\,}}(\ell )$.

• Finally, we truncate every vector in $\Lambda$ of norm at most the threshold value $\sqrt{2m}2^{m/2}$ and return these as the basis for $S^{\sigma _\textbf{p}}$.

The rest of the section contains the analysis of this algorithm. Fix an ordered graph G, a configuration $\textbf{p}$ and an error vector $\varepsilon$, which determine the observed measurements $\ell = \ell (\textbf{p}) + \varepsilon$. Because $\ell$ arises from measuring an integer configuration on the line, ${{\,\textrm{Lat}\,}}(\ell )$ will deterministically contain some short vectors associated with generators of the $\mathbb {R}$-cycle space $S^{\sigma _\textbf{p}}$.

Definition 3.3

Let G be an ordered graph with n vertices and m edges, $\varepsilon \in \mathbb {Z}^m$, and an orientation $\sigma$ of G be fixed. Let $S^\sigma$ be the $\mathbb {R}$-cycle space of $G^\sigma$. We define

$$\begin{aligned} {\hat{S}}^\sigma := \left\{ \begin{pmatrix} \textbf{v}\\ {\varepsilon }^\top \textbf{v}\end{pmatrix}: \textbf{v}\in S^\sigma \right\} \end{aligned}$$

The subspace ${\hat{S}}^\sigma \subseteq \mathbb {R}^{m+1}$ is the image of $S^\sigma$ under an injective linear transformation.

Lemma 3.4

Let G be a connected ordered graph with n vertices and m edges, $\sigma$ an orientation of G, $\textbf{p}$ a configuration of n points, and $\varepsilon \in \{-1,0,1\}^m$ an error vector. Then ${\hat{S}}^\sigma$ is contained in the linear span of a set of vectors in ${{\,\textrm{Lat}\,}}(\ell )$, each of norm at most $\sqrt{2m}$.

Proof

Let $\textbf{w}_\rho$ be the signed cycle vector of a cycle $\rho$ in the oriented graph $G^{\sigma _\textbf{p}}$. Define $e_\rho = {\textbf{w}_\rho }^\top \ell$. From Lemma 2.8, we have

$$\begin{aligned} e_\rho = {\textbf{w}_\rho }^\top \ell = {\textbf{w}_\rho }^\top (\ell (\textbf{p}) + \varepsilon ) = {\textbf{w}_\rho }^\top \varepsilon \end{aligned}$$

We then define ${\hat{\textbf{w}}}_\rho := \begin{pmatrix} \textbf{w}_\rho \\ e_\rho \end{pmatrix}$. For any $\textbf{r}\in \mathbb {Z}^m$, the column span of the lattice generating matrix $L(\textbf{r})$ is exactly vectors of the form

$$\begin{aligned} {\hat{\textbf{x}}} = \begin{pmatrix} \textbf{x}\\ {\textbf{r}}^\top \textbf{x}\end{pmatrix}. \end{aligned}$$

(1)

The vector ${\hat{\textbf{w}}}_\rho$ is in ${{\,\textrm{Lat}\,}}(\ell )$, because it is integral and is in the column span of $L(\ell )$.

From Proposition 2.6, there is a basis $\textbf{w}_{\rho _1}, \ldots , \textbf{w}_{\rho _c}$ of signed cycle vectors for $S^{\sigma _\textbf{p}}$, where $c:=m-n+1$. As the images under the injective linear map of Definition 3.3, the vectors ${\hat{\textbf{w}}}_{\rho _1}, \ldots , {\hat{\textbf{w}}}_{\rho _c}$ are linearly independent, and, by the discussion above, in ${{\,\textrm{Lat}\,}}(\ell )$. Because the vectors $\textbf{w}_{\rho _i}$ and $\varepsilon$ are $\pm 1$ vectors, we get, for each $1\le i\le c$, the estimate

$$\begin{aligned} \Vert {\hat{\textbf{w}}}_\rho \Vert ^2_2 = \Vert \textbf{w}_\rho \Vert ^2_2 + |e_\rho |^2 \le 2m \end{aligned}$$

It now follows that ${{\,\textrm{Lat}\,}}(\ell )$ contains c linearly independent vectors of norm at most $\sqrt{2m}$ that are contained in ${\hat{S}}^\sigma$. As ${\hat{S}}^\sigma$ is c-dimensional, these must be a spanning set. $\square$

Now we proceed with the probabilistic analysis of our algorithm. The idea, which is already present in Frieze’s work [13], is that if $\ell$ (instead of $\textbf{p}$) has been selected uniformly at random, then ${{\,\textrm{Lat}\,}}(\ell )$ would, WHP, not contain any vector of length $\sqrt{2m}2^{m/2}$. The point of Lemma 3.4 is that, because $\ell$ is a (corrupted) measurement of a configuration $\textbf{p}$, ${{\,\textrm{Lat}\,}}(\ell )$ must contain some “planted” short vectors of norm at most $\sqrt{2m}$ that span the linear space ${\hat{S}}^\sigma$. We won’t be able to recover these efficiently, so the next step of the analysis is to show that, WHP (in $\textbf{p}$), every vector of norm at most $\sqrt{2m}2^{m/2}$ in ${{\,\textrm{Lat}\,}}(\ell )$ lies in ${\hat{S}}^\sigma$. Informally, ${{\,\textrm{Lat}\,}}(\ell )$ is still “random enough” looking that we can spot the vectors in it that arise from cycle relations.

Definition 3.5

We call any vector of norm at most $\sqrt{2m}2^{m/2}$ medium sized.

We defer the proof of the following key estimate to Sect. 3.3.1.

Proposition 3.6

Let G be a connected graph with m edges and n vertices, $\varepsilon \in \{-1,0,1\}^m$, and ${\hat{\textbf{x}}}\in \mathbb {Z}^{m+1}$. If $\textbf{p}$ is selected uniformly at random from $[B]^n$, then the probability that ${\hat{\textbf{x}}}$ is in ${{\,\textrm{Lat}\,}}(\ell (\textbf{p})+\varepsilon )$ but ${\hat{\textbf{x}}} \not \in {\hat{S}}^{\sigma _\textbf{p}}$ is at most $\frac{2^n n!}{B}$.

Using the key estimate, we prove the result mentioned in the informal discussion above.

Lemma 3.7

Let G be a connected graph with m edges and n vertices. If $\textbf{p}$ is selected uniformly at random from $[B]^n$, then, with probability at least

$$\begin{aligned} 1 - \frac{2^{m^2/2} \;\; 2^{O(m\log m)}}{B} \end{aligned}$$

for every $\varepsilon \in \{-1,0,1\}^m$, every medium sized vector in ${{\,\textrm{Lat}\,}}(\ell (\textbf{p}) + \varepsilon )$ is in ${\hat{S}}^{\sigma _\textbf{p}}$.

Proof

For the moment, fix $\varepsilon \in \{-1,0,1\}$. Set $K=(2\,m)^{1/2}2^{m/2}$. If ${\hat{\textbf{x}}}\in \mathbb {Z}^{m+1}$ is a medium sized integer vector, then certainly all its coordinates have magnitude at most K. Hence, there are at most $(2K+1)^{m+1} \le (3K)^{m+1}$ medium sized integer vectors. We bound this latter quantity as follows

$$\begin{aligned} \left( 3(2m)^{1/2}2^{m/2} \right) ^{m+1} \le 2^{\frac{m(m+1)}{2}} 2^{\frac{m+1}{2}} 3^{(m+1)} m^{\frac{m+1}{2}} = 2^{m^2/2}2^{O(m\log m)} \end{aligned}$$

With $\varepsilon$ fixed, Proposition 3.6 then implies that, if $\textbf{p}$ is selected uniformly at random from $[B]^n$, the probability that there is some medium sized ${\hat{\textbf{x}}}$ so that ${\hat{\textbf{x}}}\in {{\,\textrm{Lat}\,}}(\ell (\textbf{p}) + \varepsilon )$ and ${\hat{\textbf{x}}}\notin {\hat{S}}^{\sigma _\textbf{p}}$ is at most

$$\begin{aligned} \frac{2^n n!}{B}2^{m^2/2}2^{O(m\log m)} \end{aligned}$$

Finally, we let $\varepsilon$ vary over $3^m$ choices to conclude that the statement fails with probability at most

$$\begin{aligned} 3^m\frac{2^n n!}{B}2^{m^2/2}2^{O(m\log m)} \end{aligned}$$

Because G is connected it has at least $n-1$ edges, so we can absorb the term $3^m2^nn!$ into the $2^{O(m\log m)}$. The final failure probability is

$$\begin{aligned} \frac{2^{m^2/2} \;\; 2^{O(m\log m)}}{B} \end{aligned}$$

as claimed. $\square$

Now picking B to make the probability of success high is routine.

Lemma 3.8

Let $b\ge (\frac{1}{2}+\delta )m^2$ for fixed $\delta > 0$. For m sufficiently large (depending on $\delta$), if $\textbf{p}$ is selected uniformly at random from $[2^b]^n$, with probability at least $1 - \frac{1}{2^{\delta m^2/2}}$, for every $\varepsilon \in \{-1,0,1\}^n$, if ${\hat{\textbf{x}}}$ is medium sized and ${\hat{\textbf{x}}} \in {{\,\textrm{Lat}\,}}(\ell (\textbf{p})+\varepsilon )$, then ${\hat{\textbf{x}}}\in {\hat{S}}^{\sigma _\textbf{p}}$.

Proof

For sufficiently large m, the quantity $2^{m^2/2}2^{O(m\log m)}$ is bounded by $2^{(\frac{1}{2} + \frac{\delta }{2})m^2}$. The probability of the event in the statement fails is, by Lemma 3.7, is then at most

$$\begin{aligned} \frac{2^{(\frac{1}{2} + \frac{\delta }{2})m^2}}{2^{(\frac{1}{2}+\delta )m^2}} = \frac{1}{2^{\delta m^2 /2}} \end{aligned}$$

$\square$

At this point, we can analyze the algorithm.

Proposition 3.9

For all $\delta > 0$, there is an $M\in \mathbb {N}$ and a polynomial time algorithm that correctly computes, with probability at least $1 - 2^{-\delta m^2/2}$, for any connected ordered graph G with n vertices and $m\ge M$ edges, and any noise vector $\varepsilon \in \{-1,0,1\}^m$, a basis for the space $S^{\sigma _p}$ from $\ell = \ell (\textbf{p}) + \varepsilon$. The probability is with respect to $\textbf{p}$ chosen uniformly at random among configurations of n integer points in [B], where $B = 2^b$ and $b \ge (\frac{1}{2} + \delta )m^2$.

Proof

Let us recall that, given $\ell$, the algorithm first computes an LLL reduced basis $\Lambda$, of ${{\,\textrm{Lat}\,}}(\ell )$ and then returns the truncation of every medium sized vector in $\Lambda$.

By Lemma 3.4, ${{\,\textrm{Lat}\,}}(\ell )$ contains at least c linearly independent vectors of norm at most $\sqrt{2m}$. As these vectors can be extended to a lattice basis, the approximation guarantee of an LLL reduced basis [24, Proposition 1.12] implies that $\Lambda$ must contain at least c vectors of norm at most $\sqrt{2m}2^{(m-1)/2}$. Let us call these ${\hat{\textbf{v}}}_1, \ldots , {\hat{\textbf{v}}}_c$.

Since $b \ge (\frac{1}{2} + \delta )m^2$, by Lemma 3.8, there is an $M > 0$ so that, if $m > M$, with probability at least $1 - 2^{-\delta m^2/2}$, all the vectors ${\hat{\textbf{v}}}_i$ are in the c-dimensional linear space ${\hat{S}}^{\sigma _\textbf{p}}$. As the ${\hat{\textbf{v}}}_i$ are linearly independent, they are a basis for ${\hat{S}}^{\sigma _\textbf{p}}$. The algorithm returns the truncated vectors $\textbf{v}_1, \ldots , \textbf{v}_c$. By construction, the $\textbf{v}_i$ are in $S^{\sigma _\textbf{p}}$. They must also be linearly independent. Otherwise the ${\hat{\textbf{v}}}_i$ would be linearly dependent as the image of a dependent set under the linear transformation of Definition 3.3. Hence, the algorithm returns a basis for $S^{\sigma _\textbf{p}}$ with probability at least $1 - 2^{-\delta m^2/2}$.

The LLL algorithm runs in polynomial time [24], and all the other steps are evidently polynomial, so the algorithm to compute a basis of $S^{\sigma _\textbf{p}}$ runs in polynomial time. $\square$

We note, for later, than the polynomial running time of the algorithm to find a basis for $S^{\sigma _p}$ implies that the coordinates of the vectors it returns have a number of bits that is polynomially bounded in m.

3.3 More Error/Fewer Bits

In the following remarks, we touch on a number of noise related issues.

Remark 3.10

Lemma 3.8, Proposition 3.9, and our main Theorems 1.1 and 1.2 can be generalized somewhat. In particular we can set $b\ge (\frac{1}{2}+\delta )m^2-f(m)$ where $f(m)=o(m^{2})$. In this case, the result can still be shown to hold for sufficiently large values of m (which now depends on f as well as $\delta$).

Remark 3.11

Suppose that instead of errors in $\{-1,0,1\}$, we had integer errors in $\{-(2^k-1),\ldots ,2^k-1\}$, for some k. The estimate on the size of the small vectors in Lemma 3.4 increases somewhat to $\sqrt{2(2^k - 1)m}$. This means, that in the proof of Lemma 3.7, we need to increase the quantity K to $\sqrt{2(2^k - 1)m}2^{m/2}$ and replace the $3^m$ choices of $\varepsilon$ with $(2^{k+1}-1)^m$. This second change raises the failure probability in the conclusion of Lemma 3.7 by a factor of $(\frac{2^{k+1}-1}{3})^m$.

Alternatively, by dividing by $2^{k+1}$, we could model the resulting rational numbers as a real valued input, as in Sect. 1.4, with errors bounded in magnitude by 1/2. Then we can round the data, putting us back to the $\{-1,0,1\}$ error model, but on integers in the interval $\{1,\ldots ,2^{b-k-1}\}$. Then in the proof of of Lemma 3.7, the $B=2^b$ term in the denominator would become $2^{b-k-1}$. This would raise the probability of failure instead by only $2^{k+1}$.

Since $(\frac{2^{k+1}-1}{3})^m$ is bigger than $2^{k+1}$, we see, that if we have integer errors in $\{-(2^k-1),\ldots ,2^k-1\}$, for some k, we are much better off rounding off the lowest $k+1$ bits and treating this as $\{-1,0,1\}$ error, than keeping around the noise! This observation is in line with the conclusions of [14].

Remark 3.12

If we do have errors of magnitude bounded by some constant K, the proof of Lemma 3.7 can be modified to accommodate the larger error. Likewise Lemma 3.8 and its proof still hold but the sufficiently large values required for m will need to increase. Indeed, as long as $\log (K)=o(m)$ Lemma 3.8 can still be shown to hold. For this generalization, the statement of Lemma 3.7 needs to have the $O(m\log m)$ term replaced by $o(m^{2})$. This is in line with [14, Theorem 2.1], where the role of error is tracked more carefully than we are doing here.

3.3.1 Proof of Proposition 3.6

Now we develop the proof of Proposition 3.6. In this section, we fix a connected ordered graph G with n vertices and m edges, as well as an error vector $\varepsilon \in \{-1,0,1\}^m$ and any vector ${\hat{\textbf{x}}}\in \mathbb {Z}^{m+1}$. Let $B\in \mathbb {N}$ be given, and define

$$\begin{aligned} \mathcal {S}({\hat{\textbf{x}}},\varepsilon ) = \left\{ \textbf{p}\in [B]^n: {\hat{\textbf{x}}}\in {{\,\textrm{Lat}\,}}(\ell (\textbf{p}) + \varepsilon ) \text {and} {\hat{\textbf{x}}}\notin {\hat{S}}^{\sigma _\textbf{p}}\right\} \end{aligned}$$

The Proposition asserts that

$$\begin{aligned} |\mathcal {S}({\hat{\textbf{x}}},\varepsilon )| \le 2^n n! B^{n-1} \end{aligned}$$

(2)

Establishing (2) is complicated by the fact that the measurement map $\ell$ is non-linear. To deal with this, we first define, for each orientation $\sigma$ of G, the set

$$\begin{aligned} \mathcal {S}({\hat{\textbf{x}}},\varepsilon ,\sigma ) = \left\{ \textbf{p}\in \mathcal {S}({\hat{\textbf{x}}},\varepsilon ): \sigma _\textbf{p}= \sigma \right\} \end{aligned}$$

And we have

$$\begin{aligned} |\mathcal {S}({\hat{\textbf{x}}},\varepsilon )| = \sum _{\sigma } |\mathcal {S}({\hat{\textbf{x}}},\varepsilon ,\sigma )| \end{aligned}$$

(3)

where the sum ranges over all vertex consistent orientations of G. By Lemma 2.3, there are at most n! vertex consistent orientations.

Next we fix a $\sigma$ so that $\mathcal {S}({\hat{\textbf{x}}},\varepsilon ,\sigma )$ is non-empty. For any $\textbf{p}$ in $\mathcal {S}({\hat{\textbf{x}}},\varepsilon ,\sigma )$, we know that $\ell (\textbf{p}) = \ell _{\sigma }(\textbf{p})$. The map $\ell _\sigma (\textbf{p})$ is linear. Since G is connected, Proposition 2.6 implies that the rank of $\ell _{\sigma }$ is $n-1$, so it has a one-dimensional kernel. It follows that the fiber $\ell ^{-1}_\sigma (\ell _\sigma (\textbf{p}))$ consists only of translations of $\textbf{p}$. We observe that there are at most B of these where the points $p_i$ remain in the range [B].

Meanwhile, because $\textbf{p}\in \mathcal {S}({\hat{\textbf{x}}},\varepsilon ,\sigma )$, the measurement $\ell (\textbf{p}) + \varepsilon$ must be in the set

$$\begin{aligned} \mathcal {B}({\hat{\textbf{x}}},\sigma )= & {} \{ \textbf{r}\in \{-1, \ldots , B\}^m: {\hat{\textbf{x}}}\in {{\,\textrm{Lat}\,}}(\textbf{r}), {\hat{\textbf{x}}}\notin {\hat{S}}^{\sigma }, \\{} & {} \text {and} {\hat{S}}^\sigma \text {is in the linear span of} {{\,\textrm{Lat}\,}}(\textbf{r}) \} \end{aligned}$$

where we have used Lemma 3.4 to ensure that ${{\,\textrm{Lat}\,}}(\ell (\textbf{p}) + \varepsilon )$ contains a basis for ${\hat{S}}^\sigma$. From the observations above, we obtain

$$\begin{aligned} |\mathcal {S}({\hat{\textbf{x}}},\varepsilon ,\sigma )| \le B |\mathcal {B}({\hat{\textbf{x}}},\sigma )| \end{aligned}$$

(4)

The next task is to bound the size of a non-empty $\mathcal {B}({\hat{\textbf{x}}},\sigma )$. We do this by showing that such a $\mathcal {B}({\hat{\textbf{x}}},\sigma )$ is contained in an $(n-2)$-dimensional affine subspace of $\mathbb {R}^m$. Our main tool will be the following simple fact.

Lemma 3.13

Let A be a d-dimensional affine subset of $\mathbb {R}^{m}$. Then

$$\begin{aligned} |\{-1,0,\ldots , B\}^m \cap A| \le (B + 2)^d \le (2B)^d \end{aligned}$$

Proof

For convenience, set $C^m = \{-1,0,\ldots , B\}^m$ and define $C^d = \{-1,0,\ldots , B\}^d$. Without loss of generality, the projection of A onto the first d coordinates of $\mathbb {R}^m$ by forgetting the last $m-d$ coordinates is bijective (if not, one can pick a different coordinate subspace), and in particular injective. This affine map sends points in $A\cap C^m$ to points in $C^d$. It follows that $|A\cap C^m| \le |C^d| = (B+2)^d$. Since $B\ge 2$, then $B + 2 \le 2B$. $\square$

Returning to the proof of the Proposition, we let $\{{\hat{\textbf{v}}}_1,\ldots , {\hat{\textbf{v}}}_c\}$ be some fixed basis for ${\hat{S}}^{\sigma }$; recall that, because G is connected the dimension of ${\hat{S}}^\sigma$ is $c = m - n + 1$. Each of these basis vectors is of the form ${\hat{\textbf{v}}}_i =: \begin{pmatrix} \textbf{v}_i \\ f_i \end{pmatrix}$ for some $\textbf{v}_i$ and $f_i$. Let us record for later the following linear system with variable $\textbf{r}\in \mathbb {R}^m$:

$$\begin{aligned} \begin{pmatrix} {\textbf{v}}^\top _1 \\ \vdots \\ {\textbf{v}}^\top _c \end{pmatrix} \textbf{r}= \begin{pmatrix} f_1 \\ \vdots \\ f_c \end{pmatrix} \end{aligned}$$

(5)

Since the $\textbf{v}_i$ are linearly independent, the rank of this system is c. (For linear independence recall that if the $\textbf{v}_i$ were dependent, then the ${\hat{\textbf{v}}}_i$ would be have to be linearly dependent as the image of a dependent set under the linear transformation of Definition 3.3.)

Now suppose that for some fixed $\textbf{r}\in \mathbb {R}^m$, $\textbf{r}\in \mathcal {B}({\hat{\textbf{x}}},\sigma )$. This implies that ${{\,\textrm{Lat}\,}}(\textbf{r})$ contains a basis ${\hat{\textbf{w}}}_1, \ldots , {\hat{\textbf{w}}}_c$ of ${\hat{S}}^\sigma$. Each of these basis vectors is of the form ${\hat{\textbf{w}}}_i = \begin{pmatrix} \textbf{w}_i \\ e_i \end{pmatrix}$, and as each ${\hat{\textbf{w}}}_i$ is in ${{\,\textrm{Lat}\,}}(\textbf{r})$, $\textbf{r}$ satisfies (recalling Eq. (1)) the following linear system:

$$\begin{aligned} \begin{pmatrix} {\textbf{w}}^\top _1 \\ \vdots \\ {\textbf{w}}^\top _c \end{pmatrix} \textbf{r}= \begin{pmatrix} e_1 \\ \vdots \\ e_c \end{pmatrix} \end{aligned}$$

(6)

Meanwhile, putting the ${\hat{\textbf{v}}}_i$ and ${\hat{\textbf{w}}}_j$ as the columns of $(m+1)\times c$ matrices, we can find an invertible $c\times c$ matrix W so that

$$\begin{aligned} \begin{pmatrix} {\hat{\textbf{v}}}_1&\cdots&{\hat{\textbf{v}}}_c \end{pmatrix} = \begin{pmatrix} {\hat{\textbf{w}}}_1&\cdots&{\hat{\textbf{w}}}_c \end{pmatrix} W \end{aligned}$$

it now follows that $\textbf{r}$ must also satisfy this next linear system, which which is equivalent to (6)

$$\begin{aligned} {W}^\top \begin{pmatrix} {\textbf{w}}^\top _1 \\ \vdots \\ {\textbf{w}}^\top _c \end{pmatrix} \textbf{r}= {W}^\top \begin{pmatrix} e_1 \\ \vdots \\ e_c \end{pmatrix} \end{aligned}$$

But this system is exactly (5). So we see that $\textbf{r}$ satisfies (5) whether or not the ${\hat{\textbf{v}}}_i$ are in ${{\,\textrm{Lat}\,}}(\textbf{r})$.

Up to this point, we have not yet used the other hypothesis coming from our assumption that $\textbf{r}\in \mathcal {B}({\hat{\textbf{x}}},\sigma )$: that ${\hat{\textbf{x}}}\in {{\,\textrm{Lat}\,}}(\textbf{r})$ and ${\hat{\textbf{x}}}\notin {\hat{S}}^\sigma$. Now we will. Let ${\hat{\textbf{x}}} = \begin{pmatrix} \textbf{x}\\ g \end{pmatrix}$. Because ${\hat{\textbf{x}}}\in {{\,\textrm{Lat}\,}}(\textbf{r})$, $\textbf{r}$ satisfies the linear equation

$$\begin{aligned} {\textbf{x}}^\top \textbf{r}= g \end{aligned}$$

and, because $\mathcal {B}({\hat{\textbf{x}}},\sigma )$ is not empty, we have ${\hat{\textbf{x}}}\notin {\hat{S}}^\sigma$, so the rank of the the linear system that adds this equation to (5), namely

$$\begin{aligned} \begin{pmatrix} {\textbf{v}}^\top _1 \\ \vdots \\ {\textbf{v}}^\top _c\\ {\textbf{x}}^\top \end{pmatrix} \textbf{r}= \begin{pmatrix} f_1 \\ \vdots \\ f_c \\ g \end{pmatrix} \end{aligned}$$

(7)

must be $c+1$. Because we assume that $\mathcal {B}({\hat{\textbf{x}}},\sigma )$ is non-empty, this final system (7) is consistent, and has, as its solution space, an affine subspace A of dimension $m - (c - 1) = n -2$. Since (7) depends only on the data defining $\mathcal {B}({\hat{\textbf{x}}},\sigma )$, we conclude that

$$\begin{aligned} \mathcal {B}({\hat{\textbf{x}}},\sigma )\subseteq A \cap \{-1, \ldots , B\}^m \end{aligned}$$

And now using Lemma 3.13 we get

$$\begin{aligned} |\mathcal {B}({\hat{\textbf{x}}},\sigma )| \le (2B)^{n-2} \end{aligned}$$

(8)

a Finally, plugging (8) into (4), we get

$$\begin{aligned} |\mathcal {S}({\hat{\textbf{x}}},\varepsilon ,\sigma )| \le 2^n B^{n-1} \end{aligned}$$

whenever the l.h.s. is non-zero. Plugging this estimate into (3), and summing over the at most n! non-zero terms, we obtain (2), and the proposition is proved. $\square$

3.4 Reconstruct G

Next we show how to reconstruct G from $S^{\sigma _\textbf{p}}$. Our main tool will be an algorithm by Seymour [31]. First we recall some standard facts about graphic matroids. A standard reference for matroids is [29]; the material in this section can be found in [29, Chapter 5].

Definition 3.14

Let $G = (V,E)$ be an ordered graph. The graphic matroid of G is the matroid on the ground set E that has, as its independent sets, the subsets of E corresponding to acyclic subgraphs of G.

If G is connected and has n vertices, the rank of its graphic matroid is $n-1$. The following lemma is due to Whitney [36].

Lemma 3.15

Let $G^\sigma$ be any orientation of an ordered graph $G = (V,E)$. Then a subset $E'\subseteq E$ is independent in the graphic matroid of G if and only if $S^\sigma \subseteq \mathbb {R}^m$ contains no non-zero vector with support contained in $E'$.

Definition 3.16

Let $\mathcal {M}$ be a matroid defined on a ground set E. An independence oracle for $\mathcal {M}$ takes as its input a subset $E'$ of E and outputs “yes” or “no” depending on whether $E'$ is independent in $\mathcal {M}$.

Since G has n vertices and is connected, the dimension of $S^\sigma$ is $m - n + 1$. Given an $m\times (m - n + 1)$ matrix A that has as its columns a basis of $S^\sigma$, we can check whether a set $E'\subseteq E$ is independent follows. Let $A'$ be the matrix obtained from A by discarding the rows corresponding to $E'$. If $A'$ has a non-zero vector $\textbf{v}$ in its null space then $A\textbf{v}$ has zeros in coordinates corresponding to edges outside of $E'$. Since $\textbf{v}$ is non-zero and A has independent columns, $A\textbf{v}$ is non-zero, so, by Lemma 3.15, $E'$ is dependent. On the other hand, if $S^\sigma$ contains a non-zero vector $\textbf{w}$ with support in $E'$, there is a $\textbf{v}$ so that $\textbf{w}= A\textbf{v}$, and this $\textbf{v}$ must be in the null space of $A'$.

It now follows that, if every entry of A has a number of bits polynomially bounded in n and m, we can check independence of $E'$ in polynomial time by forming $A'$ and computing its rank with Gaussian elimination. This gives us a polynomial time independence oracle for the graphic matroid of G.

A striking result of Seymour [31], which builds on work of Tutte [35] (see also [3]) is that an independence oracle can be used to efficiently determine a graph G from its matroid.

Theorem 3.17

([31]) Let $\mathcal {M}$ be a matroid over an ordered ground set. There is an algorithm that uses an independence oracle to determine whether $\mathcal {M}$ is the graphic matroid of some ordered graph G. If so, the algorithm outputs some such G. If the independence oracle is polynomial time, so is the whole algorithm.

Since the matroid of an ordered graph G depends only on whether each subset of edges is a cycle or not, if G and H are isomorphic as ordered graphs, they will have the same matroid and so Seymour’s algorithm can only return an ordered graph H that is isomorphic to G. A foundational result of Whitney [37] says that if an (ordered) graph G is 3-connected, then G is determined up to (ordered) graph isomorphism by its graphic matroid.

Corollary 3.18

If G is 3-connected, then Seymour’s algorithm outputs an ordered graph isomorphic to G, given an independence oracle for the graphic matroid of G.

Which for our purposes, together with Lemma 3.15 gives us:

Proposition 3.19

Let G be a 3-connected ordered graph with n vertices and m edges. Let $\sigma$ be an orientation and A be a $m\times c$ matrix with column span $S^\sigma$. Suppose that every entry in A has a bit complexity that is polynomial in n and m. Then there is an algorithm, with running time polynomial in n and m that correctly computes G from A, up to isomorphism.

3.5 Reconstructing $\sigma _\textbf{p}$

Next we show how to reconstruct $\sigma _\textbf{p}$ from G and $S^{\sigma _\textbf{p}}$.

Lemma 3.20

Let $G^\sigma$ be any orientation of an ordered graph $G = (V,E)$. Let $E'$ be the edge set of a cycle $\rho$ in G. Then any non-zero vector in $S^\sigma$ with support contained in coordinates corresponding to $E'$ is a scale of the signed cycle vector $\textbf{w}_{\rho ,\sigma }$.

Proof

As a cycle, every proper subset of $E'$ must be independent in the graphic matroid of G. Hence, by Lemma 3.15, any non-zero vector in $S^\sigma$ that is supported on coordinates corresponding to $E'$ must have no zero entries. If there are two such vectors that are not scalar multiples of each other, then an appropriate linear combination would be non-zero and have smaller support. The resulting contradiction says that, up to scale there is a single vector in $S^\sigma$ supported on $E'$; since $\textbf{w}_{\rho , \sigma }$ is such a vector, we are done. $\square$

If we have an $m\times (m - n + 1)$ matrix A with column span $S^\sigma$, we can find the vector $\pm \textbf{w}_{\rho , \sigma }$ from the edge set $E'$ of $\rho$ by finding a non-zero vector $\textbf{v}$ in the null space of the matrix $A'$ obtained by discarding the rows of A corresponding to $E'$. We then have $\alpha \textbf{w}_{\rho , \sigma } = A\textbf{v}$, for some scalar $\alpha \in \mathbb {R}$. Lemma 3.20 implies that $\textbf{v}$ is unique up to scale, and then we can scale $A\textbf{v}$ to obtain a vector with only entries $\pm 1$, which must be $\pm \textbf{w}_{\rho , \sigma }$.

In the algorithm to recover $\sigma _\textbf{p}$, we will use the following graph-theoretic notion.

Definition 3.21

Let G be a graph. An ear decomposition of G is a sequence of cycles

$$\begin{aligned} \rho _1, \ldots , \rho _k \end{aligned}$$

such that, for each $2\le j\le k$

$$\begin{aligned} \rho _j \setminus (\rho _1 \cup \cdots \cup \rho _{j-1}) \end{aligned}$$

is a path, and the edge set of G is the union of the cycles $\rho _i$.

A graph is 2-connected, if and only if it has an ear decomposition [11, Proposition 3.1.1], and an ear decomposition can be computed in polynomial time (e.g., [25]).

Proposition 3.22

Let $G = (V,E)$ be a 2-connected ordered graph, and A an $m\times (m - n + 1)$ matrix whose column span, $S^\sigma$, is the $\mathbb {R}$-cycle space of an unknown orientation of G. Also assume that the entries of A have a number of bits that is bounded polynomially in m. Then we can find $\sigma$ from this data in polynomial time, up to a global flipping of the entire orientation.

Proof

Lemma 3.20 allows us to construct, an orientation of any cycle in G that is consistent with $\sigma$ or the orientation $-\sigma$, in which all the edges are reversed. To reconstruct $\sigma$ (or $-\sigma$), we will iteratively orient the whole graph, working one cycle at a time, in a way that makes sure that all the cycles are oriented consistently with each other, and thus globally correct up to a single sign.

We first compute an ear decomposition

$$\begin{aligned} \rho _1, \ldots , \rho _k \end{aligned}$$

Using the matrix A and Lemma 3.20, we find a signed cycle vector $\textbf{w}_1 = \pm \textbf{w}_{\rho _1,\sigma }$. Now, for each $j = 2, \ldots , k$, we use A and Lemma 3.20 to find a vector $\textbf{w}_j = \pm \textbf{w}_{\rho _j,\sigma }$ so that the signs in coordinates corresponding to edges in $\rho _j\cap (\cup _{i=1}^{j-1} \rho _i)$ agree with the signs in $\textbf{w}_1, \ldots , \textbf{w}_{j-1}$. The defining property of an ear decompostion is that

$$\begin{aligned} \rho _j\setminus (\cup _{i=1}^{j-1} \rho _i) \end{aligned}$$

is a path. Because $\rho _j$ is a cycle, this implies that

$$\begin{aligned} \rho _j\cap (\cup _{i=1}^{j-1} \rho _i) \end{aligned}$$

is non-empty. Hence, exactly one of the two possible orientations of $\rho _j$ consistent with $\sigma$ or $-\sigma$ will also be consistent with the previous choices of the algorithm. At the end of the process, the coordinate corresponding to every edge is either $\pm 1$ in at least one of the vectors $\textbf{w}_j$, and the signs will be the same whenever non-zero. Hence, we may read off the orientation $\sigma$, up to an unknown sign.

Each of the steps is polynomial time. As discussed above, an ear decomposition can be computed in polynomial time. The number of cycles in the ear decomposition is at most m, so there are at most m calls to the polynomial time independence oracle obtained from A. All of the other steps are evidently polynomial, and so the algorithm is too. $\square$

Taken together, Propositions 3.9 (with $\delta$ set to 1/2), 3.19, and 3.22, constitute a proof of our Theorem 1.2.

3.6 Reconstruct $\textbf{p}$

Now armed with $\ell$, G and $\sigma _\textbf{p}$ (up to a global flip) we can reconstruct $\textbf{p}$, up to translation and reflection.

In the noiseless setting, we can simply traverse a spanning tree of G and lay out each of the vertices in a greedy manner. This step completes the proof of Theorem 1.1.

In the setting with noise, we can set up a least-squares problem, finding the $\textbf{p}$ that minimizes the squared error in the length measurements, as follows. Suppose G and $\sigma _\textbf{p}$ have been successfully reconstructed. Without loss of generality, assume that we have chosen the global sign of $\sigma _\textbf{p}$ correctly as well. Then we can set up the linear system

$$\begin{aligned} M \textbf{x}= \ell , \end{aligned}$$

where M is the signed edge incidence matrix $M^{\sigma _\textbf{p}}(G)$.

The matrix M has a kernel spanned by the $\textbf{1}$ (all ones) vector, which we now want to mod out. To do this, let Q be an n-by-$(n-1)$ matrix whose columns form an orthonormal basis for $\textbf{1}^\perp$. We now consider the linear system

$$\begin{aligned} M Q\textbf{y}= \ell \end{aligned}$$

To solve for the least squares solution, we set up the normal equations

$$\begin{aligned} {Q}^\top {M}^\top M Q\textbf{y}= {Q}^\top {M}^\top \ell \end{aligned}$$

with solution

$$\begin{aligned} \textbf{y}^* = \left( {Q}^\top {M}^\top M Q\right) ^{-1} \left( {Q}^\top {M}^\top \right) \ell \end{aligned}$$

We then output the configuration $\textbf{p}^*:=Q \textbf{y}^*$.

We now want to compare this output to $\textbf{q}:=Q{Q}^\top \textbf{p}$, which is the configuration $\textbf{p}$ translated to have its center of mass at the origin. Thus we want to bound

$$\begin{aligned} ||\textbf{p}^*-\textbf{q}||=||Q\textbf{y}^*-Q{Q}^\top \textbf{p}||=||\textbf{y}^*-{Q}^\top \textbf{p}||. \end{aligned}$$

Let us define the correct lengths as $\ell ':= M \textbf{p}= (M Q)({Q}^\top \textbf{p})$ and the residual $\textbf{r}:= \ell - \ell '$. Then

$$\begin{aligned} \textbf{y}^*= & {} \left( {Q}^\top {M}^\top M Q\right) ^{-1} \left( {Q}^\top {M}^\top \right) [\ell ' + \textbf{r}] \\= & {} {Q}^\top \textbf{p}+ \left( {Q}^\top {M}^\top M Q\right) ^{-1} \left( {Q}^\top {M}^\top \right) \textbf{r}\end{aligned}$$

and, hence,

$$\begin{aligned} \textbf{y}^* - {Q}^\top \textbf{p}= \left( {Q}^\top {M}^\top M Q\right) ^{-1} \left( {Q}^\top {M}^\top \right) \textbf{r}. \end{aligned}$$

Under our noise assumption, the norm of $\textbf{r}$ is at most $\sqrt{m}$.

Recognizing ${M}^\top M$ as the unweighted Laplacian matrix of the graph G, we see its smallest non-zero eigenvalue is at least $\frac{1}{n^2}$ [26, Eq. (6.10)] and the largest singular value of M is at most $\sqrt{n}$ [26, Theorem 2.2(c)]. The only effect of Q on the spectral decomposition of ${M}^\top M$ is to remove the 0 eigenvalue. Likewise, its only effect on M is to remove the 0 singular value. Putting things together, we see that, up to translation and reflection on $\textbf{p}$, the norm of $\textbf{p}^*-\textbf{q}$ is at most $n^{2.5}\sqrt{m}$. This completes the proof of Theorem 1.3.

3.7 Labeled Setting

In the labeled setting the algorithm is even easier. In particular, we can skip the step of reconstructing G from $S^{\sigma _\textbf{p}}$. (That is the only step that depends on 3-connectivity.)

Alternatively, since we have access to G, we can also apply a more expensive, but conceptually more simple algorithm. Instead of using LLL as in Sect. 3.2 to recover $S^{\sigma _\textbf{p}}$ all at once, we can compute an ear decomposition of G (as described in Sect. 3.5), and then extract the orientation of the edges by solving a single SUBSET-SUM problem (using LLL) separately for each cycle. (These are not independent SUBSET-SUM problems, but there are at most n of them, so the failure probability is still exponentially small in m.) Then we continue with the computation as described in Sect. 3.5.

4 Detecting Failures

Theorem 1.1 provides only a high probability guarantee of reconstructing G, in the unlabeled setting, and $\textbf{p}$, in the labelled and unlabelled settings. We now discuss where the algorithm can fail, and some tests to detected failure.

The first observation is that the only step requiring a probabilistic analysis is the computation of $S^{\sigma _\textbf{p}}$ using an LLL reduced basis for ${{\,\textrm{Lat}\,}}(\ell )$. After this, the rest of the steps are deterministic and, if $S^{\sigma _\textbf{p}}$ is correctly computed, will compute G and then $\textbf{p}$. Hence, every failure of the algorithm arises from failure of the LLL step to correctly compute $S^{\sigma _\textbf{p}}$. Moreover, since the truncations of the medium sized vectors in $\Lambda$ span a subspace S of $\mathbb {R}^m$ that contains $S^{\sigma _\textbf{p}}$, failure occurs iff $S\supsetneq S^{\sigma _\textbf{p}}$.

One example of how this can happen is when the pair $(G,\textbf{p})$ is not “globally rigid”. This means there is another configuration $\textbf{q}$ so that $\ell = \ell (\textbf{p}) = \ell (\textbf{q})$, but $\textbf{q}$ is not related to $\textbf{p}$ by translation or reflection. Since we assume that G is 3-connected, there is some cycle in G that is oriented differently in $\sigma _\textbf{p}$ and $\sigma _\textbf{q}$. Since $\ell$ satisfies the cycle relations from both $G^{\sigma _\textbf{p}}$ and $G^{\sigma _q}$, the lattice reduction will find more than c medium-sized relations. We don’t know that failures of global rigidity are the only way that the lattice reduction step can fail.

4.1 Labeled Setting

In labeled problems, the algorithm has access to n, and thus also c. Since the computation of $S^{\sigma _\textbf{p}}$ succeeds if and only if $\Lambda$ has exactly c medium sized vectors in it, the algorithm can stop and report failure if there are more than c medium sized vectors in $\Lambda$.

4.2 Unlabeled Setting

In the unlabeled setting, the algorithm does not have access to c, and so there is a possibility that it reports an incorrect result. Here we describe some, to our knowledge non-exhaustive, tests that can detect a failure to correctly compute a basis for $S^{\sigma _\textbf{p}}$.

Seymour’s algorithm may report that the matroid represented by the linear space S obtained in the lattice reduction step is not graphic. Since $S^{\sigma _\textbf{p}}$ does represent a graphic matroid, whenever Seymour’s algorithm fails, $S\ne S^{\sigma _\textbf{p}}$.

Even if an incorrectly computed S represents a graphic matroid, of a graph H necessarily not isomorphic to G, the orientation-finding step may fail. If H is not 2-connected, then the failure will occur when that step tries to find an ear-decomposition. If H is 2-connected, the algorithm may find an orientation $H^\tau$ of H that is not a DAG, and, hence, $\tau$ is not vertex consistent.

Finally, if the previous tests all pass, we can check that the output configuration $\textbf{q}$ has length measurements consistent with $\ell$. If there is no noise, the check is exact: we can verify that each edge of the output graph G has, in $\ell (\textbf{q})$, the same length as the corresponding entry of $\ell$. If not, the algorithm has failed. When there is noise, the test is not exact, but a residual error in the measured lengths larger than that allowed by Theorem 1.3 indicates the algorithm has failed.

While we expect these tests are useful in practice, the possibility that all of them pass for an incorrectly computed $S^{\sigma _\textbf{p}}$ remains a (low probability) possibility.

5 Denser Graphs

The limiting factor of our algorithm is its need for highly accurate measurements. Here we describe an option for certain denser graphs, which can require many fewer bits of accuracy.

Definition 5.1

We say that G has a k-basis, for some k, and orientation $\sigma$, the $\mathbb {R}$-cycle space of $G^\sigma$ has a cycle induced basis $\{\textbf{w}_{\rho ,\sigma }\}$, where each of the cycles $\rho$ has at most k edges.

If we assume that G has a k-basis, then, given a noisy measurement $\ell = \ell (\textbf{p}) + \varepsilon$, with $\varepsilon \in \{-1,0,1\}^m$ we can compute a basis for ${{\,\textrm{Lat}\,}}(\ell )$ by enumerating every vector

$$\begin{aligned} {\hat{\textbf{w}}} = \begin{pmatrix} \textbf{w}\\ e \end{pmatrix} \end{aligned}$$

where $\textbf{w}$ has at most k non-zero entries, each of which is $\pm 1$ and $-k\le e\le k$ and checking whether it is in ${{\,\textrm{Lat}\,}}(\ell )$ using the lattice generating matrix $L(\ell )$. Assuming that G has a k-basis, a maximal linearly-independent subset of the vectors of this form will span a subspace of ${{\,\textrm{Lat}\,}}(\ell )$ that contains ${\hat{S}}^{\sigma _\textbf{p}}$. A similar analysis to that in Sect. 3.2, shows that the span of the vectors returned by the brute-force procedure is, with high probability over a uniformly selected $\textbf{p}$, a basis for ${\hat{S}}^{\sigma _\textbf{p}}$. Once we have recovered ${\hat{S}}^{\sigma _\textbf{p}}$, we can proceed as if we had used LLL.

The complexity of the brute-force approach to find a k-basis for $S^\sigma$ is exponential in k, but polynomial in m. If k is fixed and independent of m, it is polynomial, albeit much slower than LLL for even modest k. The advantage of the brute-force procedure is that, because it solves the shortest-vector problem for ${{\,\textrm{Lat}\,}}(\ell )$ exactly, we can take b to be any quantity that is $\omega (m\log m)$, since we no longer have to account for the approximation factor in the LLL algorithm.

6 Experiments

We implemented the LLL step in our algorithm to explore how it performs on different graph families. We looked at three family of graphs: cycles; nearly 3-regular graphs (with all vertices degree 3 except for one which has degree 4 or 5); and complete graphs.

Fig. 2

Number of bits required for the LLL step to recover the cycle space with probability 0.9 vs. number of vertices. Log–Log on right

For a fixed number of vertices (and hence edges), we used our implementation to search for the number of bits necessary for the probability (over random point sets) of the LLL step correctly identifying the cycle space of the graph to be at least 0.9. (Here the configurations were integer, and the lengths were measured with $\{-1,0,1\}$ error.) For the nearly 3-regular graphs, we tried 5 different random ensembles and used the maximum number of bits required for each size. These experiments are “optimistic”, since they use knowledge of the number of vertices to avoid using a threshold value to identify small vectors in the reduced lattice basis.

In these results, we see that for all 3 families of graphs, the bit requirement grows at about the rate $n^{1.5}$ (see the log–log plot of Fig. 2). More bits are needed for the cycle graphs than the nearly 3-regular graphs. This is perhaps due to the fact that the nearly 3-regular graphs have basis cycles, with each cycle much smaller than n. For the complete graphs, where $m=O(n^2)$, this gives us growth that is somewhat sub-linear in m. All of this behaviour is significantly better than the pessimistic $m^2$ bounds of our theory. On the other hand, even for moderate sized graphs, the required bit accuracy quickly becomes higher than we would be able to get from a physical measurement system.