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abs_min_linear: Example and Test

Purpose
The function $f : \B{R}^3 \rightarrow \B{R}$ defined by $$\begin{array}{rcl} f( x_0, x_1 ) & = & | d_0 - x_0 | + | d_1 - x_0 | + | d_2 - x_0 | \\ & + & | d_3 - x_1 | + | d_4 - x_1 | + | d_5 - x_1 | \\ \end{array}$$ is affine, except for its absolute value terms. For this case, the abs_normal approximation should be equal to the function itself. In addition, the function is convex and abs_min_linear should find its global minimizer. The minimizer of this function is $x_0 = \R{median}( d_0, d_1, d_2 )$ and $x_1 = \R{median}( d_3, d_4, d_5 )$

Source

# include "abs_min_linear.hpp"

namespace {
const CPPAD_TESTVECTOR(double)& x ,
const CPPAD_TESTVECTOR(double)& u )
{     size_t n = x.size();
size_t s = u.size();
CPPAD_TESTVECTOR(double) xu(n + s);
for(size_t j = 0; j < n; j++)
xu[j] = x[j];
for(size_t j = 0; j < s; j++)
xu[n + j] = u[j];
return xu;
}
}
bool abs_min_linear(void)
{     bool ok = true;
//
//
//
size_t dpx   = 3;          // number of data points per x variable
size_t level = 0;          // level of tracing
size_t n     = 2;          // size of x
size_t m     = 1;          // size of y
size_t s     = dpx * n;    // number of data points and absolute values
// data points
d_vector  data(s);
for(size_t i = 0; i < s; i++)
data[i] = double(s - i) + 5.0 - double(i % 2) / 2.0;
//
// record the function f(x)
for(size_t j = 0; j < n; j++)
ad_x[j] = double(j + 1);
AD<double> sum = 0.0;
for(size_t j = 0; j < n; j++)
for(size_t k = 0; k < dpx; k++)
sum += abs( data[j * dpx + k] - ad_x[j] );

// create its abs_normal representation in g, a
f.abs_normal_fun(g, a);

// check dimension of domain and range space for g
ok &= g.Domain() == n + s;
ok &= g.Range()  == m + s;

// check dimension of domain and range space for a
ok &= a.Domain() == n;
ok &= a.Range()  == s;

// --------------------------------------------------------------------
// Choose a point x_hat
d_vector x_hat(n);
for(size_t j = 0; j < n; j++)
x_hat[j] = double(0.0);

// value of a_hat = a(x_hat)
d_vector a_hat = a.Forward(0, x_hat);

// (x_hat, a_hat)
d_vector xu_hat = join(x_hat, a_hat);

// value of g[ x_hat, a_hat ]
d_vector g_hat = g.Forward(0, xu_hat);

// Jacobian of g[ x_hat, a_hat ]
d_vector g_jac = g.Jacobian(xu_hat);

// trust region bound (make large enough to include solutuion)
d_vector bound(n);
for(size_t j = 0; j < n; j++)
bound[j] = 10.0;

// convergence criteria
d_vector epsilon(2);
double eps99 = 99.0 * std::numeric_limits<double>::epsilon();
epsilon[0]   = eps99;
epsilon[1]   = eps99;

// maximum number of iterations
s_vector maxitr(2);
maxitr[0] = 10; // maximum number of abs_min_linear iterations
maxitr[1] = 35; // maximum number of qp_interior iterations

// minimize the approxiamtion for f, which is equal to f because
// f is affine, except for absolute value terms
d_vector delta_x(n);
level, n, m, s, g_hat, g_jac, bound, epsilon, maxitr, delta_x
);

// number of data points per variable is odd
ok &= dpx % 2 == 1;

// check that the solution is the median of the corresponding data
for(size_t j = 0; j < n; j++)
{     // data[j * dpx + 0] , ... , data[j * dpx + dpx - 1] corresponds to x[j]
// the median of this data has index j * dpx + dpx / 2
size_t j_median = j * dpx + (dpx / 2);
//
ok &= CppAD::NearEqual( delta_x[j], data[j_median], eps99, eps99 );
}

return ok;
}`

Input File: example/abs_normal/abs_min_linear.cpp