# ECOSolveR Examples

#### 2023-05-15

## \(L_1\) 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.

    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.

    sprandn <- function(nrow, ncol, density) {
        items <- ceiling(nrow * ncol * density)
        matrix(c(rnorm(items),
                 rep(0, nrow * ncol - items)),
               nrow = nrow)
    }

    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.

    ## 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.

    names(z)

    ## [1] "x"          "y"          "s"          "z"          "infostring"
    ## [6] "retcodes"   "summary"    "timing"

    z$infostring

    ## [1] "Optimal solution found"

Extract the solution.

    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)

    ## [1] "x reconstructed with 1 non-zero entries"