Spectrally Optimized Pointset Configurations

Abstract

The search for optimal configurations of pointsets, the most notable examples being the problems of Kepler and Thompson, have an extremely rich history with diverse applications in physics, chemistry, communication theory, and scientific computing. In this paper, we introduce and study a new optimality criteria for pointset configurations. Namely, we consider a certain weighted graph associated with a pointset configuration and seek configurations that minimize certain spectral properties of the adjacency matrix or graph Laplacian defined on this graph, subject to geometric constraints on the pointset configuration. This problem can be motivated by solar cell design and swarming models, and we consider several spectral functions with interesting interpretations such as spectral radius, algebraic connectivity, effective resistance, and condition number. We prove that the regular simplex extremizes several spectral invariants on the sphere. We also consider pointset configurations on flat tori via (i) the analogous problem on lattices and (ii) through a variety of computational experiments. For many of the objectives considered (but not all), the triangular lattice is extremal.

Appendices

Appendix A: Sensitivity Analysis of Spectral Quantities

In Sects. 2.1 and 2.2, we introduced the weighted adjacency matrix and graph Laplacian associated to a pointset configuration, discussed some spectral properties of the matrices, and recalled a variety of spectral quantities that are of interest in various applications. In this appendix, we discuss the sensitivity of these spectral quantities with respect to changes in the pointset configuration.

We say that a function $J :{\mathbb {R}}^n \rightarrow {\mathbb {R}}$ is a symmetric if $J(x) = J([x])$, where $[\cdot ]:{\mathbb {R}}^n \rightarrow {\mathbb {R}}^n$ rearranges the components of a vector in nondecreasing order. In other words, the value of J is invariant to permuting the components of the argument. Recall that we have defined $\mathbf{S}^n$ to be the set of real symmetric $n \times n$ matrices, and let $\lambda :\mathbf{S}^n \rightarrow {\mathbb {R}}^n$ be the vector containing the ordered eigenvalues of a symmetric matrix. We say that the composition $J\circ \lambda$ is a spectral function if J is a symmetric function. Roughly speaking, $J\circ \lambda$ is convex and differentiable when J is convex and differentiable [12].

Recall that the weighted adjacency matrix associated with a Euclidean pointset $\{{{\mathbf {x}}}_i \}_{i\in [n]}$ is defined as in (2). For the spectral function $J\circ \lambda$, we consider the composition

$$\begin{aligned} J \circ \lambda \circ W^{{\mathbf {x}}}. \end{aligned}$$

The following proposition shows how the composite function, $J \circ \lambda \circ W^{{\mathbf {x}}}$, changes as we move a single point in the configuration, say ${{\mathbf {x}}}_i \in {\mathbb {R}}^d$.

Proposition A.1

Let J be differentiable at $\lambda (W^{{\mathbf {x}}})$. The gradient of the objective function, $J \circ \lambda \circ W^{{\mathbf {x}}}:{\mathbb {R}}^{N\times 2} \rightarrow {\mathbb {R}}$, with respect to ${{\mathbf {x}}}_i$ is given by

$$\begin{aligned} \nabla _{{{\mathbf {x}}}_i} (J \circ \lambda \circ W^{{\mathbf {x}}}) = 4 \sum _k \left( U \ \mathrm {diag} \nabla J (\lambda ) \ U^t \right) _{ik} f'( d^2({{\mathbf {x}}}_i, {{\mathbf {x}}}_k) ) \ ({{\mathbf {x}}}_ i - {{\mathbf {x}}}_k) \end{aligned}$$

for any diagonalizing matrix $U\in O^n$ satisfying $W^{{\mathbf {x}}}= U \ \mathrm {diag}\ \lambda (W^{{\mathbf {x}}}) \ U^t$.

Proof

The proof is an exercise in the chain rule. When J is differentiable at $\lambda (A)$, the composition $J\circ \lambda :\mathbf{S}^N \rightarrow {\mathbb {R}}$ is Fréchet differentiable with derivative

$$\begin{aligned} \nabla (J\circ \lambda ) (A) = U \ \mathrm {diag} \nabla J (\lambda ) \ U^t \end{aligned}$$

for any diagonalizing matrix $U\in O^n$ satisfying $A = U \ \mathrm {diag}\ \lambda (A) \ U^t$. See, for example, [12, Corollary 5.2.5]. For simplicity, we write $V = U \ \mathrm {diag} \nabla J (\lambda ) \ U^t$. Thus, we compute

$$\begin{aligned} \nabla _{{{\mathbf {x}}}_i}( J \circ \lambda \circ W^{{\mathbf {x}}})&= \langle V , \nabla _{{{\mathbf {x}}}_i} W^{{\mathbf {x}}}\rangle _F \end{aligned}$$

(30a)

$$\begin{aligned}&= \sum _{jk} V_{jk} \nabla _{{{\mathbf {x}}}_i} W_{jk}^{{\mathbf {x}}}\end{aligned}$$

(30b)

$$\begin{aligned}&= \sum _{jk} V_{jk} \left( \delta _{ij} \nabla _{{{\mathbf {x}}}_i} W_{ik}^{{\mathbf {x}}}+ \delta _{ik} \nabla _{{{\mathbf {x}}}_i} W_{ji}^{{\mathbf {x}}}\right) \end{aligned}$$

(30c)

$$\begin{aligned}&= \sum _{k} V_{ik} \ \nabla _{{{\mathbf {x}}}_i} W_{ik}^{{\mathbf {x}}}+ \sum _{j} V_{ji} \ \nabla _{{{\mathbf {x}}}_i} W_{ji}^{{\mathbf {x}}}\end{aligned}$$

(30d)

$$\begin{aligned}&= 2 \sum _{k} V_{ik} \ \nabla _{{{\mathbf {x}}}_i} W_{ik}^{{\mathbf {x}}}. \end{aligned}$$

(30e)

In (30a), $\langle \cdot , \cdot \rangle _F$ denotes the Frobenius inner product. In (30c), $\delta _{ij}$ denotes the Kronecker delta function. In (30e), we used the symmetry of V and W. We compute

$$\begin{aligned} \nabla _{{{\mathbf {x}}}_i} W_{ik}^{{\mathbf {x}}}= f'( d^2({{\mathbf {x}}}_i, {{\mathbf {x}}}_k) ) \ \nabla _{{{\mathbf {x}}}_i} d^2({{\mathbf {x}}}_i, {{\mathbf {x}}}_k) = 2 f'( d^2({{\mathbf {x}}}_i, {{\mathbf {x}}}_k) ) \ ({{\mathbf {x}}}_ i - {{\mathbf {x}}}_k), \end{aligned}$$

(31)

from which the result follows. $\square$

Example

Consider the spectral function $J(\lambda ) = \sum _i \lambda _i^2$. In this case, $U \ \mathrm {diag} \nabla J (\lambda ) \ U^t = 2 W^{{\mathbf {x}}}$. Thus from Proposition A.1 we can check that

$$\begin{aligned} \nabla _{{{\mathbf {x}}}_i} (J \circ \lambda \circ W^{{\mathbf {x}}}) = 2 \langle W^{{\mathbf {x}}}, \nabla _{{{\mathbf {x}}}_i} W^{{\mathbf {x}}}\rangle _F = \nabla _{{{\mathbf {x}}}_i} \Vert W^{{\mathbf {x}}}\Vert _F^2 . \end{aligned}$$

This is a trivial identity since $\Vert W^{{\mathbf {x}}}\Vert _F^2 = \mathrm {tr} [ (W^{{\mathbf {x}}})^2 ] = \sum _i \lambda _i(W^{{\mathbf {x}}})^2$.

As in (3), the graph Laplacian associated with a pointset $\{{{\mathbf {x}}}_i \}_{i\in [n]}$ is given by $L^{{\mathbf {x}}}= D^{{\mathbf {x}}}- W^{{\mathbf {x}}}$. For the spectral function $J\circ \lambda$, we consider the composition

$$\begin{aligned} J \circ \lambda \circ L^{{\mathbf {x}}}. \end{aligned}$$

The following proposition shows how this composite function, $J \circ \lambda \circ L^{{\mathbf {x}}}$, changes as we move a single point in the configuration, say ${{\mathbf {x}}}_i \in {\mathbb {R}}^n$.

Proposition A.2

Let J be differentiable at $\lambda (L^{{\mathbf {x}}})$. The gradient of the objective function, $J \circ \lambda \circ L^{{\mathbf {x}}}:{\mathbb {R}}^{N\times 2} \rightarrow {\mathbb {R}}$, with respect to ${{\mathbf {x}}}_i$ is given by

$$\begin{aligned} \nabla _{{{\mathbf {x}}}_i} (J \circ \lambda \circ L^{{\mathbf {x}}}) = 2 \sum _k(V_{kk} - 2 V_{ik} + V_{ii}) \ f'( d^2({{\mathbf {x}}}_i, {{\mathbf {x}}}_k) ) \ ({{\mathbf {x}}}_ i - {{\mathbf {x}}}_k), \end{aligned}$$

(32)

where $V = \left( U \ \mathrm {diag} \nabla J (\lambda ) \ U^t \right) _{ik}$ for any diagonalizing matrix $U\in O^n$ satisfying $L^{{\mathbf {x}}}= U \ \mathrm {diag}\ \lambda (L^{{\mathbf {x}}}) \ U^t$. The gradient can alternatively be expressed using the arc-vertex incidence matrix decomposition (7),

$$\begin{aligned} \nabla _{{{\mathbf {x}}}_i} (J \circ \lambda \circ L^{{\mathbf {x}}}) = 2 \sum _{k = (i,j)} g_k \ f'( d^2({{\mathbf {x}}}_i, {{\mathbf {x}}}_j) ) \ ({{\mathbf {x}}}_ i - {{\mathbf {x}}}_j), \end{aligned}$$

(33)

where $g = \sum _{j=1}^n (Bu_j)^2 \big ( \nabla J(\lambda ) \big )_j$.

Proof

As in the proof of Proposition A.1, when J be differentiable at $\lambda (L^{{\mathbf {x}}})$, the composition $J\circ \lambda :\mathbf{S}^n \rightarrow {\mathbb {R}}$ is Fréchet differentiable with derivative

$$\begin{aligned} (J\circ \lambda )' (L^{{\mathbf {x}}}) = U \ \mathrm {diag} \nabla J (\lambda ) \ U^t \end{aligned}$$

for any diagonalizing matrix $U\in O^n$ satisfying $L^{{\mathbf {x}}}= U \ \mathrm {diag} \lambda (L^{{\mathbf {x}}}) \ U^t$. Writing $V = U \ \mathrm {diag} \nabla J (\lambda ) \ U^t$, it follows that

$$\begin{aligned} \nabla _{{{\mathbf {x}}}_i}( J \circ \lambda \circ L^{{\mathbf {x}}})&= \langle V , \nabla _{{{\mathbf {x}}}_i} (D^{{\mathbf {x}}}- W^{{\mathbf {x}}}) \rangle _F \nonumber \\&= \sum _k V_{kk} \nabla _{{{\mathbf {x}}}_i} D_{kk}^{{\mathbf {x}}}- 2 \sum _k V_{ik} \nabla _{{{\mathbf {x}}}_i} W_{ik}^{{\mathbf {x}}}, \end{aligned}$$

(34)

where we used (30e). From

$$\begin{aligned} \nabla _{{{\mathbf {x}}}_i} D_{kk}^{{\mathbf {x}}}= \nabla _{{{\mathbf {x}}}_i} W_{ik}^{{\mathbf {x}}}+ \delta _{ik} \sum _j \nabla _{{{\mathbf {x}}}_i} W_{ij}^{{\mathbf {x}}}, \end{aligned}$$

we then have that

$$\begin{aligned} \nabla _{{{\mathbf {x}}}_i}( J \circ \lambda \circ L^{{\mathbf {x}}})&= \sum _k V_{kk} \nabla _{{{\mathbf {x}}}_i} W_{ik}^{{\mathbf {x}}}+ \sum _k V_{kk} \delta _{ik} \sum _j \nabla _{{{\mathbf {x}}}_i} W_{ij}^{{\mathbf {x}}}- 2 \sum _k V_{ik} \nabla _{{{\mathbf {x}}}_i} W_{ik}^{{\mathbf {x}}}\\&= \sum _k V_{kk} \nabla _{{{\mathbf {x}}}_i} W_{ik}^{{\mathbf {x}}}+ V_{ii} \sum _j \nabla _{{{\mathbf {x}}}_i} W_{ij}^{{\mathbf {x}}}- 2 \sum _k V_{ik} \nabla _{{{\mathbf {x}}}_i} W_{ik}^{{\mathbf {x}}}\\&= \sum _k (V_{kk} - 2 V_{ik} + V_{ii}) \nabla _{{{\mathbf {x}}}_i} W_{ik}^{{\mathbf {x}}}. \end{aligned}$$

Equation (32) now follows from (31).

From (34), we could also write

$$\begin{aligned} \nabla _{{{\mathbf {x}}}_i} (J\circ \lambda \circ L)&= \langle V, \nabla _{{{\mathbf {x}}}_i} (B^t \text {diag}(w^{{\mathbf {x}}}) B) \rangle _F \nonumber \\&= \langle V, B^t \text {diag}( \nabla _{{{\mathbf {x}}}_i} w) B \rangle _F \nonumber \\&= \langle (BU) \ \mathrm {diag} \nabla J (\lambda ) \ (BU)^t, \text {diag}( \nabla _{{{\mathbf {x}}}_i} w^{{\mathbf {x}}}) \rangle _F \nonumber \\&= \Big \langle g , \nabla _{{{\mathbf {x}}}_i} w^{{\mathbf {x}}}\Big \rangle , \end{aligned}$$

(35)

where

$$\begin{aligned} g = \text {diag} \left( (BU) \ \mathrm {diag} \nabla J (\lambda ) \ (BU)^t \right) . \end{aligned}$$

Thus, $g\in {\mathbb {R}}^{\left( {\begin{array}{c}n\\ 2\end{array}}\right) }$ has entries given by

$$\begin{aligned} g_k = \sum _{j=1}^n (BU)_{k,j}^2 \big ( \nabla J(\lambda ) \big )_j. \end{aligned}$$

Finally, we compute

$$\begin{aligned} \nabla _{{{\mathbf {x}}}_i} w_k = {\left\{ \begin{array}{ll} 2 f'( d^2({{\mathbf {x}}}_i, {{\mathbf {x}}}_j) ) \ ({{\mathbf {x}}}_ i - {{\mathbf {x}}}_j), &{} k = (i,j), \\ 0, &{} \text {otherwise}, \end{array}\right. } \end{aligned}$$

which gives (33). $\square$

Example

Consider the spectral function $J(\lambda ) = \sum _i \lambda _i$ so that the objective function is simply $\text {tr} L^{{\mathbf {x}}}$. In this case, $V= \text {Id}$, and

$$\begin{aligned} \nabla _{{{\mathbf {x}}}_i} (J \circ \lambda \circ L^{{\mathbf {x}}}) = 4 \sum _k f'( d^2({{\mathbf {x}}}_i, {{\mathbf {x}}}_k) ) \ ({{\mathbf {x}}}_ i - {{\mathbf {x}}}_k) = \nabla _{{{\mathbf {x}}}_i} \sum _{j,k} f( d^2({{\mathbf {x}}}_j, {{\mathbf {x}}}_k) ). \end{aligned}$$

Remark A.3

Equation (35) in the proof of Proposition A.2 implies that the variation of a spectral function of the graph Laplacian with respect to the edge weights is generally given by

$$\begin{aligned} \nabla _w [J \circ \lambda \circ (B^t \ \text {diag}(w) \ B) ] = \text {diag}[ (BU) \ \text {diag}\big (\nabla J(\lambda ) \big ) \ (BU)^t]. \end{aligned}$$

(36)

For example, the gradient of the trace, $\text {tr} L(w)$, is given by

$$\begin{aligned} \frac{\delta \text {tr} L}{\delta w} = \text {diag} (B B^t) = 2 \cdot 1_{\left( {\begin{array}{c}n\\ 2\end{array}}\right) }. \end{aligned}$$

The factor of two simply indicates that each edge weight effects the degree of two vertices. When $\lambda _2 \big ( L(w)\big )$ is a simple eigenvalue, the gradient is given by

$$\begin{aligned} \frac{\delta \lambda _2}{\delta w} = (B u_2)^2, \end{aligned}$$

which agrees with the formulas in [29, 30]. Finally, the gradient of the total effective resistance, $R_{tot} = \sum _j \lambda _j^{-1}$, is computed

$$\begin{aligned} \frac{\delta R_{tot}}{\delta w} = - \text {diag} \big ( B U \text {diag}(\lambda ^{-2} ) (BU)^t \big ) = - \text {diag}( B (L^\dag )^2 B^t), \end{aligned}$$

which can also be found in [31].

For a general spectral function J, the gradients computed in Propositions A.1 and A.2 can be used together with a quasi-Newton optimization method to efficiently search for a locally optimal pointset configuration; see Sect. 5.

Appendix B: Parameterization of Lattices

Let $B = [b_1,\ldots ,b_n]\in {\mathbb {R}}^{n\times n}$ have linearly independent columns. The lattice generated by the basis B is the set of integer linear combinations of the columns of B,

$$\begin{aligned} {\mathcal {L}}(B)=\{Bx:x \in {\mathbb {Z}}^n \}. \end{aligned}$$

Let B and C be two lattice bases. We recall that ${\mathcal {L}}(B)={\mathcal {L}}(C)$ if and only if there is a unimodular³ matrix U such that $B = CU$. Thus, there is a one-to-one correspondence between the unimodular $2\times 2$ matrices and the bases of a two-dimensional lattice.

We say that two lattices are isometric if there is a rigid transformation that maps one to the other. The following proposition parameterizes the space of two-dimensional, unit-volume lattices modulo isometry.

Proposition B.1

Every two-dimensional lattice with volume one is isometric to a lattice parameterized by the basis

$$\begin{aligned} \begin{pmatrix} \frac{1}{\sqrt{b}}&{}\frac{a}{\sqrt{b}} \\ 0 &{} \sqrt{b} \end{pmatrix}, \end{aligned}$$

where the parameters a and b are constrained to the set

$$\begin{aligned} U := \left\{ (a,b) \in {\mathbb {R}}^2 :b>0, \ a \in [ 0,1/2 ],\ \text {and} \ a^2 + b^2 \ge 1 \right\} . \end{aligned}$$

The set U defined in Proposition B.1 is illustrated in Fig. 8.

Fig. 8

The set U in Proposition B.1. Parameters (a, b) corresponding to square, triangular, rectangular, rhombic, and oblique lattices are also indicated

Proof

Consider an arbitrary lattice with unit volume. We first choose the basis vectors so that the angle between them is acute. After a suitable rotation and reflection, we can let the shorter basis vector (with length $\frac{1}{\sqrt{b}}$) be parallel to the x axis and the longer basis vector (with length $\sqrt{ \frac{a^2}{b} + b} = \sqrt{\frac{1}{b} \left( a^2 + b^2 \right) } \ge \sqrt{ \frac{1}{b}}$) lie in the first quadrant. Multiplying on the right by a unimodular matrix, $\begin{pmatrix} 1 &{} 1 \\ 0 &{} 1 \end{pmatrix}$, we compute

$$\begin{aligned} \begin{pmatrix} \frac{1}{\sqrt{b}} &{} \frac{a}{\sqrt{b}} \\ 0 &{} \sqrt{b} \end{pmatrix} \begin{pmatrix} 1 &{} 1 \\ 0 &{} 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{b}} &{} \frac{a+1}{\sqrt{b}} \\ 0 &{} \sqrt{b} \end{pmatrix} . \end{aligned}$$

Since this is equivalent to taking $a\mapsto a+1$, it follows that we can identify the lattices associated with the points (a, b) and $(a+1,b)$. Thus, we can restrict the parameter a to the interval $\left[ 0, 1/2 \right]$ by symmetry. For a complete picture of this restriction and how the symmetry naturally arises, see [33, Proposition 3.2 Figure 3]. $\square$