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abs_normal qp_interior: Example and Test

Problem
Our original problem is $$\R{minimize} \; | u - 1| \; \R{w.r.t} \; u \in \B{R}$$ We reformulate this as the following problem $$\begin{array}{rlr} \R{minimize} & v & \R{w.r.t} \; (u,v) \in \B{R}^2 \\ \R{subject \; to} & u - 1 \leq v \\ & 1 - u \leq v \end{array}$$ This is equivalent to $$\begin{array}{rlr} \R{minimize} & (0, 1) \cdot (u, v)^T & \R{w.r.t} \; (u,v) \in \B{R}^2 \\ \R{subject \; to} & \left( \begin{array}{cc} 1 & -1 \\ -1 & -1 \end{array} \right) \left( \begin{array}{c} u \\ v \end{array} \right) + \left( \begin{array}{c} -1 \\ 1 \end{array} \right) \leq 0 \end{array}$$ which is in the form expected by qp_interior .

Source

# include <limits>
# include "qp_interior.hpp"

bool qp_interior(void)
{     bool ok = true;
//
size_t n = 2;
size_t m = 2;
vector C(m*n), c(m), G(n*n), g(n), xin(n), xout(n), yout(m), sout(m);
C[ 0 * n + 0 ] =  1.0; // C(0,0)
C[ 0 * n + 1 ] = -1.0; // C(0,1)
C[ 1 * n + 0 ] = -1.0; // C(1,0)
C[ 1 * n + 1 ] = -1.0; // C(1,1)
//
c[0]           = -1.0;
c[1]           =  1.0;
//
g[0]           =  0.0;
g[1]           =  1.0;
//
// G = 0
for(size_t i = 0; i < n * n; i++)
G[i] = 0.0;
//
// If (u, v) = (0,2), C * (u, v) + c = (-2,-2)^T + (1,-1)^T < 0
// Hence (0, 2) is feasible.
xin[0] = 0.0;
xin[1] = 2.0;
//
double epsilon = 99.0 * std::numeric_limits<double>::epsilon();
size_t maxitr  = 10;
size_t level   = 0;
//
}