Multiview Attenuation Estimation and Correction

Abstract

Measuring attenuation coefficients is a fundamental problem that can be solved with diverse techniques such as X-ray or optical tomography and lidar. We propose a novel approach based on the observation of a sample from a few different angles. This principle can be used in existing devices such as lidar or various types of fluorescence microscopes. It is based on the resolution of a nonlinear inverse problem. We propose a specific computational approach to solve it and show the well-foundedness of the approach on simulated data. Some of the tools developed are of independent interest. In particular, we propose an efficient method to correct attenuation defects, new robust solvers for the lidar equation as well as new efficient algorithms to compute the proximal operator of the logsumexp function in dimension 2.

Appendix

1.1 Proof Proposition 1

Proof

The first item is obtained by direct inspection:

• for $\beta$ fixed, $\alpha \mapsto \left\langle \exp (-A_j\alpha ), \mathbbm {1}\right\rangle$ is the composition of a linear operator with a convex function; hence, it is convex. In addition, $\alpha \mapsto \left\langle A_j\alpha , \mathbbm {1}\right\rangle$ is a linear mapping.

• for $\alpha$ fixed, the first term in $\beta$ is linear and $\beta \mapsto \left\langle - \log (\beta ), \mathbbm {1}\right\rangle$ is convex.

Let us now focus on the second and third points. The function G can be rewritten as a sum of functions:

$$\begin{aligned} G(\alpha ,\beta ) = \sum _{i=1}^n g_{i}\left( (A_j\alpha [i])_{1\le j\le m},\beta [i]\right) , \end{aligned}$$

where $g_{i}:\mathbb {R}^m\times \mathbb {R}_+ \rightarrow \mathbb {R}$ is defined as follows:

$$\begin{aligned} (x,y) \mapsto g_{i}(x,y)&= \sum _{j=1}^m \exp (-x[j]) y \\&\quad +u_j[i]\left( x[j] - \log (y) \right) . \end{aligned}$$

To prove the convexity of G, it suffices to study the convexity of each function $g_{i}$. From now on, we skip the indices i to lighten the notation.

Let us analyze the eigenvalues of the Hessian $H_g$:

$$\begin{aligned} H_g(x,y)=\begin{pmatrix} \mathrm {diag}(y\exp (-x)) &{}\quad - \exp (-x)\\ - \exp (-x)^T &{}\quad \sum _{j=1}^m u_j/y^2 \end{pmatrix}. \end{aligned}$$

To study the positive semidefiniteness, let $(v,w)\in \mathbb {R}^{m+1}$, denote an arbitrary vector. We have:

$$\begin{aligned}&\left\langle \begin{pmatrix} v \\ w \end{pmatrix}, H_g(x,y)\begin{pmatrix} v \\ w \end{pmatrix} \right\rangle \\&\quad = \sum _{j=1}^m v[j]\exp (-x[j])\left( y v[j] -2w\right) + w^2\frac{\sum _{j=1}^m u_j}{y^2}. \end{aligned}$$

In the case $y>\frac{\sum _{j=1}^mu_j}{\sum _{j=1}^m\exp (-x[j])}$ and $w \ne 0$, we get that:

$$\begin{aligned}&\left\langle \begin{pmatrix} v \\ w \end{pmatrix}, H_g(x,y)\begin{pmatrix} v \\ w \end{pmatrix} \right\rangle \\&\quad < \sum _{j=1}^m v[j]\exp (-x[j])\left( y v[j]-2w\right) \\&\qquad + w^2\frac{\sum _{j=1}^m\exp (-x[j])}{y}\\&\quad = \sum _{j=1}^m \exp (-x[j])\left( y v[j]^2 -2wv[j] + \frac{w^2}{y}\right) \\&\quad = \sum _{j=1}^m \exp (-x[j])\left( v[j]\sqrt{y} -\frac{w}{\sqrt{y}}\right) ^2 \end{aligned}$$

where the last expression is equal to 0 for the particular choice $v[j] y =w$, for all $1\le j \le m$. This implies $\left\langle \begin{pmatrix} v \\ w \end{pmatrix}, H_g\begin{pmatrix} v \\ w \end{pmatrix} \right\rangle <0$, which shows that function g is not convex in $\mathbb {R}^n\times \mathbb {R}_+^n$ and proves the second item.

The same argument with $y\le \frac{\sum _{j=1}^mu_j}{\sum _{j=1}^m\exp (-x[j])}$ proves the third item. $\square$

1.2 Proof Proposition 2

Proof

With this specific choice, it is easy to check that the optimality conditions of problem (10) with respect to variable $\beta$ yield (14). By replacing this expression in (11), we obtain the optimization problem shown in equation (13).

Checking convexity of this problem can be done by simple inspection. The term $\langle u_j (A_j \alpha ),\mathbbm {1}\rangle$ is linear, hence convex. The term $\log \left( \sum _{j=1}^m \exp (-(A_j\alpha )[i])\right)$ is the composition of the convex logsumexp function with a linear operator; hence, it is convex. $\square$

1.3 Proximal Operator of logsumexp in Dimension 2

In this section, we propose a fast and accurate numerical algorithm based on Newton’s method to solve the following problem:

$$\begin{aligned} w&=\mathrm {prox}_{\gamma g_1}(z) \\&=\mathop {\mathrm {argmin}}_{x\in \mathbb {R}^{2n}} \frac{1}{2}\Vert x-z\Vert _2^2 \\&\quad +\gamma \sum _{i=1}^n \sum _{j=1}^2 u_j[i] \left[ x_j[i] + \log \left( \sum _{j=1}^2 \exp (-x_j[i])\right) \right] , \end{aligned}$$

where $z=\begin{pmatrix} z_1 \\ z_2 \end{pmatrix}$ and $x=\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$ are vectors in $\mathbb {R}^{2n}$. This problem may seem innocuous at first sight, but turns out to be quite a numerical challenge. The first observation is that it can be decomposed as nindependent problems of dimension 2 since:

$$\begin{aligned} w[i]&=\mathop {\mathrm {argmin}}_{(x_1,x_2)\in \mathbb {R}^{2}} \frac{1}{2}(x_j-z_j[i])_2^2 \nonumber \\&\quad + \gamma \sum _{j=1}^2 u_j[i] \left[ x_j + \log \left( \sum _{j=1}^2 \exp (-x_j)\right) \right] . \end{aligned}$$

(20)

To simplify the notation, we will skip the index i in what follows. The following proposition shows that our problem is equivalent to finding the proximal operator associated with the “logsumexp” function.

Proposition 6

Define the logsumexp function $\mathrm {lse}(x_1,x_2)=\log \left( \sum _{j=1}^2 \exp (x_j)\right)$. The solution of problem (20) coincides with the opposite of the proximal operator of $\mathrm {lse}$:

$$\begin{aligned} w[i]&= -\mathop {\mathrm {argmin}}_{(x_1,x_2)\in \mathbb {R}^2} a \mathrm {lse}(x_1,x_2)\nonumber \\&\quad + \frac{1}{2}((x_1-y_1)^2+(x_2-y_2)^2) \end{aligned}$$

(21)

$$\begin{aligned}&= -\mathrm {prox}_{a \mathrm {lse}} (y_1,y_2), \end{aligned}$$

(22)

where $a=\gamma (u_1+u_2)$ and $y_j = \gamma u_j - z_j$.

Proof

The first-order optimality conditions for problem (20) read

$$\begin{aligned} \left\{ \begin{array}{l} \gamma u_1 - \frac{\gamma (u_1+u_2)\exp (-x_1)}{\exp (-x_1)+\exp (-x_2)} + x_1 - z_1 = 0 \\ \gamma u_2 - \frac{\gamma (u_1+u_2)\exp (-x_2)}{\exp (-x_1)+\exp (-x_2)} + x_2 - z_2 = 0. \end{array}\right. \end{aligned}$$

(23)

By letting $a=\gamma (u_1+u_2)$ and $y_j = \gamma u_j - z_j$, this equation becomes

$$\begin{aligned} \left\{ \begin{array}{l} - \frac{a\exp (-x_1)}{\exp (-x_1)+\exp (-x_2)} + x_1 + y_1 = 0 \\ - \frac{a\exp (-x_2)}{\exp (-x_1)+\exp (-x_2)} + x_2 + y_2 = 0. \end{array}\right. \end{aligned}$$

(24)

It now suffices to make the change of variable $x'_i=-x_i$ to retrieve the optimality conditions of problem (22)

$$\begin{aligned} \left\{ \begin{array}{l} \frac{a\exp (x'_1)}{\exp (x'_1)+\exp (x'_2)} + x'_1 - y_1 = 0 \\ \frac{a\exp (x'_2)}{\exp (x'_1)+\exp (x'_2)} + x'_2 - y_2 = 0. \end{array}\right. \end{aligned}$$

(25)

$\square$

Remark 4

To the best of our knowledge, this is the first attempt to find a fast algorithm to evaluate the prox of logsumexp. This function is important in many regards. In particular, it is a $C^\infty$ approximation of the maximum value of a vector. In addition, its Fenchel conjugate coincides with the Shannon entropy restricted to the unit simplex. We refer to [14, §3.2] for some details. The algorithm that follows has potential applications outside the scope of this paper.

We now design a fast and accurate minimization algorithm for problem (22) or equivalently, a root-finding algorithm for problem (25). This algorithm differs depending on whether $y_1\ge y_2$ or $y_2\ge y_1$. We focus on the case $y_1\ge y_2$. The case $y_2\ge y_1$ can be handled by symmetry.

Let $\lambda = \frac{\exp (x'_1)}{\exp (x'_1)+\exp (x'_2)}$ and notice that

$$\begin{aligned} \frac{\exp (x'_2)}{\exp (x'_1)+\exp (x'_2)} =1-\lambda . \end{aligned}$$

Therefore, (25) becomes:

$$\begin{aligned} \left\{ \begin{array}{l} x'_1 = y_1 -a\lambda \\ x'_2 = y_2 - a(1-\lambda ). \end{array}\right. \end{aligned}$$

(26)

Hence,

$$\begin{aligned} \frac{1-\lambda }{\lambda } = \exp (x'_2-x'_1) = \exp (y_2-y_1-a)\exp (2a\lambda ). \end{aligned}$$

(27)

Taking the logarithm on each side yields¹:

$$\begin{aligned} \log (1-\lambda )-\log (\lambda ) = y_2-y_1-a + 2a\lambda . \end{aligned}$$

(28)

We are now facing the problem of finding the root $\lambda ^*$ of the following function:

$$\begin{aligned} f(\lambda ) = y_2-y_1-a + 2a\lambda - \log (1-\lambda )+\log (\lambda ). \end{aligned}$$

(29)

There are two important advantages for this approach compared to the direct resolution of (25). First, we have to solve a 1D problem instead of a 2D problem. More importantly, we directly constrain $x'$ to be of form $x'=y - a\delta$, where $\delta$ lives on the 2D simplex.

Let us collect a few properties of function f. First, we have:

$$\begin{aligned} f'(\lambda ) = 2a + \frac{1}{1-\lambda } + \frac{1}{\lambda } > 0,\quad \forall \lambda \in (0,1). \end{aligned}$$

(30)

Therefore, f is increasing on (0, 1). To use convergence results of Newton’s algorithm, we need to compute $f''$ as well:

$$\begin{aligned} f''(\lambda ) = -\frac{1}{\lambda ^2}+\frac{1}{(1-\lambda )^2}. \end{aligned}$$

(31)

Proposition 7

If $y_1\ge y_2$, then $x_1'\ge x_2'$ and

$$\begin{aligned}&\max \left( \frac{1}{2}, \frac{1}{1+\exp (y_2-y_1+a)}\right) \le \lambda ^* \nonumber \\&\quad \le \frac{1}{1+\exp (y_2-y_1)}. \end{aligned}$$

(32)

Proof

The first statement can be proven by contradiction. Assume that $x_2'> x_1'$, then Eq. (25) indicates that $y_2>y_1$.

For the second statement, it suffices to evaluate f at the extremities of the interval since $f'>0$. We get $f(1/2)=y_2-y_1\le 0$ and $f\left( \frac{1}{1+\exp (y_2-y_1)}\right) = -a + \frac{2a}{1 + \exp (y_2-y_1)} \ge 0$. $\square$

Proposition 8

Set $\lambda _0 = \frac{1}{1+\exp (y_2-y_1)}$. Then, the following Newton’s method

$$\begin{aligned} \lambda _{k+1} = \lambda _k - \frac{f(\lambda _k)}{f'(\lambda _k)} \end{aligned}$$

(33)

converges to the root $\lambda ^*$ of f, with a locally quadratic rate.

Fig. 9

Performance evaluation for Newton’s algorithm. Left: $\lambda ^*$ depending on a and $y_1-y_2$. Right: number of iterations of Newton’s method to reach machine precision

Proof

First notice that $f''(\lambda )\ge 0$ on the interval [1 / 2, 1). Hence, $f''$ is also positive on $I=[\lambda ^*,\lambda _0]$. This ensures that

$$\begin{aligned} \lambda _0 \ge \lambda _1 \ge \cdots \ge \lambda ^*. \end{aligned}$$

(34)

We prove this assertion by recurrence. Notice that $\lambda _0\ge \lambda ^*$ by Proposition 7. Now, assume that $\lambda _k\ge \lambda ^*$, then

$$\begin{aligned} f(\lambda _k) = f(\lambda ^*) + \int _{\lambda ^*}^{\lambda _k} f'(t)\,\mathrm{d}t \le f'(\lambda _k) (\lambda _k-\lambda ^*). \end{aligned}$$

(35)

Hence, $\lambda _k-\lambda ^* \ge \frac{f(\lambda _k)}{f'(\lambda _k)}$ and $\lambda _{k+1}\ge \lambda ^*$. In addition $\frac{f(\lambda _k)}{f'(\lambda _k)}\ge 0$ on I, so that $\lambda _{k+1}\ge \lambda _k$.

The sequence $(\lambda _k)_{k\in \mathbb {N}}$ is monotonically decreasing and bounded below; therefore it converges to some value $\lambda '\ge \lambda ^*$. Necessarily $\lambda '=\lambda ^*$, since for $\lambda '>\lambda ^*$, $\frac{f(\lambda ')}{f'(\lambda ')}>0$.

To prove the locally quadratic convergence rate, we just invoke the celebrated Newton–Kantorovich’s theorem [24, 25] that ensures local quadratic convergence if $f''$ is bounded in a neighborhood of the minimizer. $\square$

Finally, let us mention that computing $\lambda _0$ on a computer is a tricky due to underflow problems: in double precision the command $1+\exp (y_2-y_1)$ will return 1 for $y_2-y_1<-37\simeq \log (10^{-16})$. This may cause the algorithm to fail since f and its derivatives are undefined at $\lambda =1$. In practice, we therefore set $\lambda _0=1/(1+\exp (y_2-y_1))-10^{-16}$. Similarly, by bound (32), we get $\lambda ^*=1$ up to machine precision whenever $y_2-y_2-a<\log (10^{-16})$. Algorithm 2 summarizes all the ideas described in this paragraph.

An attentive reader may have remarked that the convergence of Newton’s algorithm depends only on the difference $y(1)-y(2)$ and a. A shift of y(1) and y(2) by the same value does not change Newton’s iteration. In Fig. 9, we show that the algorithm behaves very well for a wide range of parameters. For $y(1)-y(2)$ and a varying in the interval $[2^{-10},2^{20}]$, the algorithm never requires more than 18 iterations to reach machine precision and needs 2.8 iterations on average.