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---
title: "ECOSolveR Examples"
date: '`r Sys.Date()`'
output:
html_document:
fig_caption: yes
theme: cerulean
toc: yes
toc_depth: 2
vignette: >
%\VignetteIndexEntry{ECOSolveR Examples}
%\VignetteEngine{knitr::rmarkdown}
\usepackage[utf8]{inputenc}
---
```{r echo=F}
### get knitr just the way we like it
knitr::opts_chunk$set(
message = FALSE,
warning = FALSE,
error = FALSE,
tidy = FALSE,
cache = FALSE
)
```
## $L_1$ minimization (Linear Programming)
We solve the following problem that arises for example in sparse
signal reconstruction problems such as compressed sensing:
$$
\mbox{minimize } ||x||_1 \mbox{ ($L_1$) }\\
\mbox{subject to } Ax = b
$$
with $x\in R^n$, $A \in R^{m \times n}$ and $m\leq n.$ Reformulate the
problem expressing the $L_1$ norm of $x$ as follows
$$
x \leq u\\
-x \leq u\\
$$
where $u\in R^n$ and we minimize the sum of $u$. The reformulated
problem using the stacked variables
$$
z = \begin{pmatrix}x\\u\end{pmatrix}
$$
is now
$$
\mbox{minimize } c^{\top}z\\
\mbox{subject to } \tilde{A}x = b \mbox{ (LP) }\\
Gx \leq h
$$
where the inequality is with respective to the positive orthant.
Here is the R code that generates a random instance of this problem
and solves it.
```{r}
library(ECOSolveR)
library(Matrix)
set.seed(182391)
n <- 1000L
m <- 10L
density <- 0.01
c <- c(rep(0.0, n), rep(1.0, n))
```
First, a function to generate random sparse matrices with normal
entries.
```{r}
sprandn <- function(nrow, ncol, density) {
items <- ceiling(nrow * ncol * density)
matrix(c(rnorm(items),
rep(0, nrow * ncol - items)),
nrow = nrow)
}
```
```{r}
A <- sprandn(m, n, density)
Atilde <- Matrix(cbind(A, matrix(rep(0.0, m * n), nrow = m)), sparse = TRUE)
b <- rnorm(m)
I <- diag(n)
G <- rbind(cbind(I, -I),
cbind(-I, -I))
G <- as(G, "dgCMatrix")
h <- rep(0.0, 2L * n)
dims <- list(l = 2L * n, q = NULL, e = 0L)
```
Note how ECOS expects sparse matrices, not ordinary matrices.
```{r}
## Solve the problem
z <- ECOS_csolve(c = c, G = G, h = h, dims = dims, A = Atilde, b = b)
```
We check that the solution was found.
```{r}
names(z)
z$infostring
```
Extract the solution.
```{r}
x <- z$x[1:n]
u <- z$x[(n+1):(2*n)]
nnzx = sum(abs(x) > 1e-8)
sprintf("x reconstructed with %d non-zero entries", nnzx / length(x) * 100)
```
|
