--- 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) ```