#include "tutil.h"

#include "tmathutil.h"

#include "tconvert.h"

#include <iterator></iterator>

using TConsts::epsilon;

TMathException::TMathException(std::string msg) : m_msg(::to_wstring(msg)) {}

namespace {

//-------------------------------------------------------------------------

//! maximum order for a polynomial

const int MAX_ORDER = 12;

//! smallest relative error we want

const double RELERROR = 1.0e-14;

//! max power of 10 we wish to search to

const int MAXPOW = 32;

//! max number of iterations

const int MAXIT = 800;

//! a coefficient smaller than SMALL_ENOUGH is considered to be zero (0.0).

const double SMALL_ENOUGH = 1.0e-12;

//-------------------------------------------------------------------------

inline int getEl(int i, int j, int n) { return (j - 1) + (i - 1) * n; }

//-------------------------------------------------------------------------

//! structure type for representing a polynomial

typedef struct {

  int ord;

  double coef[MAX_ORDER + 1];

} poly;

//-------------------------------------------------------------------------

//  compute number of root in (min,max)

double evalpoly(int ord, double *coef, double x);

int numchanges(int np, poly *sseq, double a);

int buildsturm(int ord, poly *sseq);

int modrf(int ord, double *coef, double a, double b, double *val);

//-------------------------------------------------------------------------

void convert(const std::vector<double> &v, poly &p) {</double>

  assert((int)v.size() <= MAX_ORDER);

  if ((int)v.size() > MAX_ORDER) return;

  p.ord = v.size() - 1;

  std::copy(v.begin(), v.end(), p.coef);

}

//-------------------------------------------------------------------------

void convert(const poly &p, std::vector<double> &v) {</double>

  v.resize(p.ord);

  std::copy(p.coef, p.coef + p.ord, v.begin());

}

//-------------------------------------------------------------------------

/*

  * modp

  *

  *  calculates the modulus of u(x) / v(x) leaving it in r, it

  *  returns 0 if r(x) is a constant.

  *  note: this function assumes the leading coefficient of v

  *  is 1 or -1

  */

int modp(poly *u, poly *v, poly *r) {

  int k, j;

  double *nr, *end, *uc;

  nr  = r->coef;

  end = &u->coef[u->ord];

  uc                      = u->coef;

  while (uc <= end) *nr++ = *uc++;

  if (v->coef[v->ord] < 0.0) {

    for (k = u->ord - v->ord - 1; k >= 0; k -= 2) r->coef[k] = -r->coef[k];

    for (k = u->ord - v->ord; k >= 0; k--)

      for (j       = v->ord + k - 1; j >= k; j--)

        r->coef[j] = -r->coef[j] - r->coef[v->ord + k] * v->coef[j - k];

  } else {

    for (k = u->ord - v->ord; k >= 0; k--)

      for (j = v->ord + k - 1; j >= k; j--)

        r->coef[j] -= r->coef[v->ord + k] * v->coef[j - k];

  }

  k = v->ord - 1;

  while (k >= 0 && fabs(r->coef[k]) < SMALL_ENOUGH) {

    r->coef[k] = 0.0;

    k--;

  }

  r->ord = (k < 0) ? 0 : k;

  return (r->ord);

}

//-------------------------------------------------------------------------

/*

  * buildsturm

  *

  *  build up a sturm sequence for a polynomial in sseq, returning

  * the number of polynomials in the sequence

  */

int buildsturm(int ord, poly *sseq) {

  int i;

  double f, *fp, *fc;

  poly *sp;

  sseq[0].ord = ord;

  sseq[1].ord = ord - 1;

  //  calculate the derivative and normalise the leadingcoefficient.

  f  = fabs(sseq[0].coef[ord] * ord);

  fp = sseq[1].coef;

  fc = sseq[0].coef + 1;

  for (i = 1; i <= ord; i++) *fp++ = *fc++ * i / f;

  // construct the rest of the Sturm sequence

  for (sp = sseq + 2; modp(sp - 2, sp - 1, sp); sp++) {

    //  reverse the sign and normalise

    f = -fabs(sp->coef[sp->ord]);

    for (fp = &sp->coef[sp->ord]; fp >= sp->coef; fp--) *fp /= f;

  }

  sp->coef[0] = -sp->coef[0];  // reverse the sign

  return (sp - sseq);

}

//-------------------------------------------------------------------------

/*

  return the number of distinct real roots of the polynomial

  described in sseq

  */

int numroots(int np, poly *sseq, int &atneg, int &atpos) {

  int atposinf = 0, atneginf = 0;

  poly *s;

  double f, lf;

  // change at positive infinity

  lf = sseq[0].coef[sseq[0].ord];

  for (s = sseq + 1; s <= sseq + np; ++s) {

    f = s->coef[s->ord];

    if (0.0 == lf || lf * f < 0.0) ++atposinf;

    lf = f;

  }

  // change at negative infinity

  if (sseq[0].ord & 1)

    lf = -sseq[0].coef[sseq[0].ord];

  else

    lf = sseq[0].coef[sseq[0].ord];

  for (s = sseq + 1; s <= sseq + np; ++s) {

    if (s->ord & 1)

      f = -s->coef[s->ord];

    else

      f = s->coef[s->ord];

    if (0.0 == lf || lf * f < 0.0) ++atneginf;

    lf = f;

  }

  atneg = atneginf;

  atpos = atposinf;

  return atneginf - atposinf;

}

//-------------------------------------------------------------------------

/*

  * numchanges

  *

  * return the number of sign changes in the Sturm sequence in

  * sseq at the value a.

  */

int numchanges(int np, poly *sseq, double a) {

  int changes;

  double f, lf;

  poly *s;

  changes = 0;

  lf = evalpoly(sseq[0].ord, sseq[0].coef, a);

  for (s = sseq + 1; s <= sseq + np; s++) {

    f = evalpoly(s->ord, s->coef, a);

    if (lf == 0.0 || lf * f < 0) changes++;

    lf = f;

  }

  return (changes);

}

//-------------------------------------------------------------------------

/*

  * sbisect

  *

  * uses a bisection based on the sturm sequence for the polynomial

  * described in sseq to isolate intervals in which roots occur,

  * the roots are returned in the roots array in order of magnitude.

  */

void sbisect(int np, poly *sseq, double min, double max, int atmin, int atmax,

             double *roots) {

  double mid;

  int n1 = 0, n2 = 0, its, atmid;

  if ((atmin - atmax) == 1) {

    // first try a less expensive technique.

    if (modrf(sseq->ord, sseq->coef, min, max, &roots[0])) return;

    /*

* if we get here we have to evaluate the root the hard

* way by using the Sturm sequence.

*/

    for (its = 0; its < MAXIT; its++) {

      mid = (min + max) / 2;

      atmid = numchanges(np, sseq, mid);

      if (fabs(mid) > RELERROR) {

        if (fabs((max - min) / mid) < RELERROR) {

          roots[0] = mid;

          return;

        }

      } else if (fabs(max - min) < RELERROR) {

        roots[0] = mid;

        return;

      }

      if ((atmin - atmid) == 0)

        min = mid;

      else

        max = mid;

    }

    if (its == MAXIT) {

      /*

fprintf(stderr, "sbisect: overflow min %f max %f\

diff %e nroot %d n1 %d n2 %d\n",

min, max, max - min, nroot, n1, n2);

*/

      roots[0] = mid;

    }

    return;

  }

  /*

* more than one root in the interval, we have to bisect...

*/

  for (its = 0; its < MAXIT; its++) {

    mid = (min + max) / 2;

    atmid = numchanges(np, sseq, mid);

    n1 = atmin - atmid;

    n2 = atmid - atmax;

    if (n1 != 0 && n2 != 0) {

      sbisect(np, sseq, min, mid, atmin, atmid, roots);

      sbisect(np, sseq, mid, max, atmid, atmax, &roots[n1]);

      break;

    }

    if (n1 == 0)

      min = mid;

    else

      max = mid;

  }

  if (its == MAXIT) {

    /*

fprintf(stderr, "sbisect: roots too close together\n");

fprintf(stderr, "sbisect: overflow min %f max %f diff %e\

nroot %d n1 %d n2 %d\n",

min, max, max - min, nroot, n1, n2);

*/

    for (n1 = atmax; n1 < atmin; n1++) roots[n1 - atmax] = mid;

  }

}

//-------------------------------------------------------------------------

/*

  * evalpoly

  *

  * evaluate polynomial defined in coef returning its value.

    */

double evalpoly(int ord, double *coef, double x) {

  double *fp, f;

  fp = &coef[ord];

  f  = *fp;

  for (fp--; fp >= coef; fp--) f = x * f + *fp;

  return (f);

}

//-------------------------------------------------------------------------

/*

    * modrf

    *

    * uses the modified regula-falsi method to evaluate the root

    * in interval [a,b] of the polynomial described in coef. The

    * root is returned is returned in *val. The routine returns zero

    * if it can't converge.

    */

int modrf(int ord, double *coef, double a, double b, double *val) {

  int its;

  double fa, fb, x, fx, lfx;

  double *fp, *scoef, *ecoef;

  scoef = coef;

  ecoef = &coef[ord];

  fb = fa = *ecoef;

  for (fp = ecoef - 1; fp >= scoef; fp--) {

    fa = a * fa + *fp;

    fb = b * fb + *fp;

  }

  //  if there is no sign difference the method won't work

  if (fa * fb > 0.0) return (0);

  if (fabs(fa) < RELERROR) {

    *val = a;

    return (1);

  }

  if (fabs(fb) < RELERROR) {

    *val = b;

    return (1);

  }

  lfx = fa;

  for (its = 0; its < MAXIT; its++) {

    x = (fb * a - fa * b) / (fb - fa);

    fx = *ecoef;

    for (fp = ecoef - 1; fp >= scoef; fp--) fx = x * fx + *fp;

    if (fabs(x) > RELERROR) {

      if (fabs(fx / x) < RELERROR) {

        *val = x;

        return (1);

      }

    } else if (fabs(fx) < RELERROR) {

      *val = x;

      return (1);

    }

    if ((fa * fx) < 0) {

      b  = x;

      fb = fx;

      if ((lfx * fx) > 0) fa /= 2;

    } else {

      a  = x;

      fa = fx;

      if ((lfx * fx) > 0) fb /= 2;

    }

    lfx = fx;

  }

  // fprintf(stderr, "modrf overflow %f %f %f\n", a, b, fx);

  return (0);

}

//-------------------------------------------------------------------------

/*!

    a x^2 + b x + c

      Remark:

      poly[0] = c

      poly[1] = b

      poly[2] = a

    */

int rootForQuadraticEquation(const std::vector<double> &v,</double>

                             std::vector<double> &sol) {</double>

  double q, delta;

  /*

if( isAlmostZero(v[2]))

{

if ( isAlmostZero(v[1]) )

return -1;

  sol.push_back(-v[0]/v[1]);

  return 1;

  }

*/

  if (isAlmostZero(v[1])) {

    q = -v[0] / v[2];

    if (q < 0)

      return 0;

    else if (isAlmostZero(q)) {

      sol.push_back(0);

      return 1;

    }

    q = sqrt(q);

    sol.push_back(-q);

    sol.push_back(q);

    return 2;

  }

  delta = sq(v[1]) - 4.0 * v[0] * v[2];

  if (delta < 0.0) return 0;

  assert(v[2] != 0);

  if (isAlmostZero(delta)) {

    sol.push_back(-v[1] / (v[2] * 2.0));

    return 1;

  }

  q = -0.5 * (v[1] + tsign(v[1]) * sqrt(delta));

  assert(q != 0);

  sol.push_back(v[0] / q);

  sol.push_back(q / v[2]);

  return 2;

}

//-----------------------------------------------------------------------------

/*

    a x^3+b x^2 + c x + d

      Remark:

      poly[0] = d

      poly[1] = c

      poly[2] = b

      poly[3] = a

    */

int rootForCubicEquation(const std::vector<double> &p,</double>

                         std::vector<double> &sol) {</double>

  /*

if( isAlmostZero(p[3]) )

return rootForQuadraticEquation(p,sol);

*/

  if (isAlmostZero(p[0])) {

    int numberOfSol;

    std::vector<double> redPol(3);</double>

    redPol[0] = p[1];

    redPol[1] = p[2];

    redPol[2] = p[3];

    numberOfSol = rootForQuadraticEquation(redPol, sol);

    for (int i = 0; i < numberOfSol; ++i)

      if (0.0 == sol[i]) return numberOfSol;

    // altrimenti devo contare la soluzione nulla

    ++numberOfSol;

    sol.push_back(0);

    return numberOfSol;

  }

  double inv_v3 = 1.0 / p[3], a = p[2] * inv_v3, b = p[1] * inv_v3,

         c = p[0] * inv_v3;

  static const double inv_3  = 1.0 / 3.0;

  static const double inv_9  = 1.0 / 9.0;

  static const double inv_54 = 1.0 / 54.0;

  double Q = (sq(a) - 3.0 * b) * inv_9,

         R = (2.0 * sq(a) * a - 9.0 * a * b + 27.0 * c) * inv_54;

  double R_2 = sq(R), Q_3 = sq(Q) * Q;

  if (R_2 < Q_3) {

    double Q_sqrt = sqrt(Q);

    double theta  = acos(R / (Q * Q_sqrt));

    sol.push_back(-2 * Q_sqrt * cos(theta * inv_3) - a * inv_3);

    sol.push_back(-2 * Q_sqrt * cos((theta - M_2PI) * inv_3) - a * inv_3);

    sol.push_back(-2 * Q_sqrt * cos((theta + M_2PI) * inv_3) - a * inv_3);

    std::sort(sol.begin(), sol.end());

    return 3;

  }

  double A = -tsign(R) * pow((fabs(R) + sqrt(R_2 - Q_3)), inv_3);

  double B = A != 0 ? Q / A : 0;

  sol.push_back((A + B) - a * inv_3);

  return 1;

}

//-----------------------------------------------------------------------------

int rootForGreaterThanThreeEquation(const std::vector<double> &p,</double>

                                    std::vector<double> &sol) {</double>

  poly sseq[MAX_ORDER];

  convert(p, sseq[0]);

  int np = buildsturm(p.size() - 1, sseq);

  int atmin, atmax;

  int nroot = numroots(np, sseq, atmin, atmax);

  if (nroot == 0) return 0;

  double minVal = -1.0;

  UINT i = 0;

  int nchanges = numchanges(np, sseq, minVal);

  for (i = 0; nchanges != atmin && i != (UINT)MAXPOW; ++i) {

    minVal *= 10.0;

    nchanges = numchanges(np, sseq, minVal);

  }

  if (nchanges != atmin) {

    atmin = nchanges;

  }

  double maxVal = 1.0;

  nchanges      = numchanges(np, sseq, maxVal);

  for (i = 0; nchanges != atmax && i != (UINT)MAXPOW; ++i) {

    maxVal *= 10.0;

    nchanges = numchanges(np, sseq, maxVal);

  }

  if (nchanges != atmax) {

    atmax = nchanges;

  }

  nroot = atmin - atmax;

  assert(nroot > 0);

  poly outPoly;

  outPoly.ord = nroot;

  sbisect(np, sseq, minVal, maxVal, atmin, atmax, outPoly.coef);

  convert(outPoly, sol);

  return nroot < 0 ? -1 : nroot;

}

//-----------------------------------------------------------------------------

}  // end of unnamed namespace

//-----------------------------------------------------------------------------

void tLUDecomposition(double *a, int n, int *indx, double &d) {

  int i, imax, j, k;

  double big, dum, sum, temp;

  std::vector<double> vv(n);</double>

  d = 1.0;

  for (i = 1; i <= n; i++) {

    big = 0.0;

    for (j = 1; j <= n; j++)

      if ((temp = fabs(a[getEl(i, j, n)])) > big) big = temp;

    if (big == 0.0)

      throw TMathException("Singular matrix in routine tLUDecomposition\n");

    vv[i - 1] = 1.0 / big;

  }

  for (j = 1; j <= n; j++) {

    for (i = 1; i < j; i++) {

      sum = a[getEl(i, j, n)];

      for (k = 1; k < i; k++) sum -= a[getEl(i, k, n)] * a[getEl(k, j, n)];

      a[getEl(i, j, n)] = sum;

    }

    big = 0.0;

    for (i = j; i <= n; i++) {

      sum = a[getEl(i, j, n)];

      for (k = 1; k < j; k++) sum -= a[getEl(i, k, n)] * a[getEl(k, j, n)];

      a[getEl(i, j, n)] = sum;

      if ((dum = vv[i - 1] * fabs(sum)) >= big) {

        big  = dum;

        imax = i;

      }

    }

    if (j != imax) {

      for (k = 1; k <= n; k++) {

        dum = a[getEl(imax, k, n)];

        a[getEl(imax, k, n)] = a[getEl(j, k, n)];

        a[getEl(j, k, n)]    = dum;

      }

      d            = -(d);

      vv[imax - 1] = vv[j - 1];

    }

    indx[j - 1] = imax;

    if (a[getEl(j, j, n)] == 0.0) a[getEl(j, j, n)] = TConsts::epsilon;

    if (j != n) {

      dum = 1.0 / (a[getEl(j, j, n)]);

      for (i = j + 1; i <= n; i++) a[getEl(i, j, n)] *= dum;

    }

  }

}

//-----------------------------------------------------------------------------

void tbackSubstitution(double *a, int n, int *indx, double *b) {

  int i, ii = 0, ip, j;

  double sum;

  for (i = 1; i <= n; i++) {

    ip        = indx[i - 1];

    sum       = b[ip - 1];

    b[ip - 1] = b[i - 1];

    if (ii)

      for (j = ii; j <= i - 1; j++) sum -= a[getEl(i, j, n)] * b[j - 1];

    else if (sum)

      ii     = i;

    b[i - 1] = sum;

  }

  for (i = n; i >= 1; i--) {

    sum = b[i - 1];

    for (j   = i + 1; j <= n; j++) sum -= a[getEl(i, j, n)] * b[j - 1];

    b[i - 1] = sum / a[getEl(i, i, n)];

  }

}

//-----------------------------------------------------------------------------

double tdet(double *LUa, int n, double d) {

  for (int i = 1; i <= n; ++i) d *= LUa[getEl(i, i, n)];

  return d;

}

//-----------------------------------------------------------------------------

double tdet(double *a, int n) {

  double d;

  std::vector<int> indx(n);</int>

  tLUDecomposition(a, n, &indx[0], d);

  for (int i = 1; i <= n; ++i) d *= a[getEl(i, i, n)];

  return d;

}

//-----------------------------------------------------------------------------

void tsolveSistem(double *a, int n, double *res) {

  double d;

  std::vector<int> indx(n);</int>

  tLUDecomposition(a, n, &indx[0], d);

  assert(tdet(a, n, d) != 0);

  /*

if( isAlmostZero(tdet(a, n, d)) )

throw TMathException("Singular matrix in routine tLUDecomposition\n");

*/

  tbackSubstitution(a, n, &indx[0], res);

}

//-----------------------------------------------------------------------------

int rootFinding(const std::vector<double> &in_poly, std::vector<double> &sol) {</double></double>

  // per ora risolvo solo i polinomi di grado al piu' pari a 3

  assert((int)in_poly.size() <= MAX_ORDER);

  if (in_poly.empty() || (int)in_poly.size() > MAX_ORDER) return -1;

  std::vector<double> p;</double>

  std::copy(in_poly.begin(), in_poly.end(), std::back_inserter(p));

  // eat zero in poly

  while (!p.empty() && isAlmostZero(p.back())) p.pop_back();

  sol.clear();

  while (!p.empty() && p.front() == 0) {

    sol.push_back(0.0);

    p.erase(p.begin());  // se i coefficienti bassi sono zero, ci sono soluzioni

                         // pari a 0.0. le metto,

  }                      // e abbasso il grado del polinomio(piu' veloce)

  switch (p.size()) {

  case 0:

    if (sol.empty()) return INFINITE_SOLUTIONS;

    break;

  case 1:  // no solutions

    break;

  case 2:

    sol.push_back(-p[0] / p[1]);

    break;

  case 3:

    rootForQuadraticEquation(p, sol);

    break;

  case 4:

    rootForCubicEquation(p, sol);

    break;

  default:

    rootForGreaterThanThreeEquation(p, sol);

  }

  return sol.size();

}

//-----------------------------------------------------------------------------

int solveEquation2(std::complex<double> *roots, double a, double b, double c) {</double>

  if (isAlmostZero(a)) {

    if (isAlmostZero(b)) return 0;

    roots[0] = -c/b;

    return 1;

  }

  double D = b*b - a*c*4;

  double k = 0.5/a;

  std::complex<double> d = D < 0</double>

                         ? std::complex<double>(0, sqrt(-D))</double>

                         : std::complex<double>(sqrt(D));</double>

  roots[0] = (-b - d)*k;

  roots[1] = (-b + d)*k;

  return 2;

}

//-----------------------------------------------------------------------------

int solveEquation3(std::complex<double> *roots, double a, double b, double c, double d) {</double>

  if (isAlmostZero(a))

    return solveEquation2(roots, b, c, d);

  // x = y - b/(3*a)

  // y*y*y + p*y + q = 0

  double p = (3*a*c - b*b)/(3*a*a);

  double q = (27*a*a*d - 9*a*b*c + 2*b*b*b)/(27*a*a*a);

  double Q = p*p*p/27 + q*q/4;

  std::complex<double> Qs = Q < 0</double>

                          ? std::complex<double>(0, sqrt(-Q))</double>

                          : std::complex<double>(sqrt(Q));</double>

  std::complex<double> A = pow(-q/2 + Qs, 1.0/3);</double>

  std::complex<double> B = pow(-q/2 - Qs, 1.0/3);</double>

  // choose complimentary B for A (A*B must be equal -p/3)

  std::complex<double> rot(-0.5, sqrt(3.0)/2);</double>

  if (!isAlmostZero(A*B + p/3)) B *= rot;

  if (!isAlmostZero(A*B + p/3)) B *= rot;

  std::complex<double> Y = (A - B)*std::complex<double>(0, sqrt(3.0));</double></double>

  std::complex<double> y0 = A + B;</double>

  std::complex<double> y1 = (-y0 - Y)/2.0;</double>

  std::complex<double> y2 = (-y0 + Y)/2.0;</double>

  double dd = b/(3*a);

  roots[0] = y0 - dd;

  roots[1] = y1 - dd;

  roots[2] = y2 - dd;

  return 3;

}

//-----------------------------------------------------------------------------

int solveEquation4(std::complex<double> *roots, double a, double b, double c, double d, double e) {</double>

  if (isAlmostZero(a))

    return solveEquation3(roots, b, c, d, e);

  // x = y - b/(4*a)

  // y^4 + p*y^2 + q*y + r = 0

  double dd = b/(4*a);

  double p = (8*a*c - 3*b*b)/(8*a*a);

  double q = (8*a*a*d - 4*a*b*c + b*b*b)/(8*a*a*a);

  double r = (256*a*a*a*e - 64*a*a*b*d + 16*a*b*b*c - 3*b*b*b*b)/(256*a*a*a*a);

  if (isAlmostZero(q)) {

    // biquadratic equation

    // y^4 + p*y^2 + r = 0

    // handling this special case will give us more accurate results

    std::complex<double> y[2];</double>

    solveEquation2(y, 1, p, r);

    y[0] = sqrt(y[0]);

    y[1] = sqrt(y[1]);

    roots[0] = -y[0] - dd;

    roots[1] =  y[0] - dd;

    roots[2] = -y[1] - dd;

    roots[3] =  y[1] - dd;

    return 4;

  }

  // solve cubic equation

  // z*z*z + (p/2)*z*z + ((p*p - 4*r)/16)*z - q*q/64 = 0

  double pp = p/2;

  double qq = (p*p - 4*r)/16;

  double rr = -q*q/64;

  std::complex<double> z[3];</double>

  solveEquation3(z, 1, pp, qq, rr);

  z[0] = sqrt(z[0]);

  z[1] = sqrt(z[1]);

  z[2] = sqrt(z[2]);

  // we need to find signs combination where following is valid:

  // (+-z0)*(+-z1)*(+-z2) = -q/8

  // magnitudes are always equals, we need to check signs only

  std::complex<double> zzz = z[0]*z[1]*z[2];</double>

  if ((zzz.real() > 0) == (q > 0))

    z[0] = -z[0];

  roots[0] =  z[0] - z[1] - z[2] - dd;

  roots[1] = -z[0] + z[1] - z[2] - dd;

  roots[2] = -z[0] - z[1] + z[2] - dd;

  roots[3] =  z[0] + z[1] + z[2] - dd;

  return 4;

}

//-----------------------------------------------------------------------------

/*

*/

int numberOfRootsInInterval(int order, const double *polyH, double min,

                            double max) {

  poly sseq[MAX_ORDER];