georef.c

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/****************************************************************************
 *
 * MODULE:       imagery library
 * AUTHOR(S):    Original author(s) name(s) unknown - written by CERL
 *               Written By: Brian J. Buckley
 *
 *               At: The Center for Remote Sensing
 *               Michigan State University
 *
 * PURPOSE:      Image processing library
 * SPDX-FileCopyrightText: 1999, 2005 GRASS Development Team
 * SPDX-License-Identifier: GPL-2.0-or-later
 *
 *****************************************************************************/
 
/*
 *  Written: 12/19/91
 *
 *  Last Update: 12/26/91 Brian J. Buckley
 *  Last Update:  1/24/92 Brian J. Buckley
 *  Added printout of trnfile. Triggered by BDEBUG.
 *  Last Update:  1/27/92 Brian J. Buckley
 *  Fixed bug so that only the active control points were used.
 *
 */
 
#include <stdlib.h>
#include <math.h>
#include <grass/gis.h>
#include <grass/imagery.h>
#include <signal.h>
 
/* STRUCTURE FOR USE INTERNALLY WITH THESE FUNCTIONS.  THESE FUNCTIONS EXPECT
   SQUARE MATRICES SO ONLY ONE VARIABLE IS GIVEN (N) FOR THE MATRIX SIZE */
 
struct MATRIX {
    int n; /* SIZE OF THIS MATRIX (N x N) */
    double *v;
};
 
/* CALCULATE OFFSET INTO ARRAY BASED ON R/C */
 
#define M(row, col) m->v[(((row) - 1) * (m->n)) + (col) - 1]
 
#define MSUCCESS    1  /* SUCCESS */
#define MNPTERR     0  /* NOT ENOUGH POINTS */
#define MUNSOLVABLE -1 /* NOT SOLVABLE */
#define MMEMERR     -2 /* NOT ENOUGH MEMORY */
#define MPARMERR    -3 /* PARAMETER ERROR */
#define MINTERR     -4 /* INTERNAL ERROR */
 
#define MAXORDER    3 /* HIGHEST SUPPORTED ORDER OF TRANSFORMATION */
 
/***********************************************************************
 
  FUNCTION PROTOTYPES FOR STATIC (INTERNAL) FUNCTIONS
 
************************************************************************/
 
static int calccoef(struct Control_Points *, double *, double *, int);
static int calcls(struct Control_Points *, struct MATRIX *, double *, double *,
                  double *, double *);
static int exactdet(struct Control_Points *, struct MATRIX *, double *,
                    double *, double *, double *);
static int solvemat(struct MATRIX *, double *, double *, double *, double *);
static double term(int, double, double);
 
/***********************************************************************
 
  TRANSFORM A SINGLE COORDINATE PAIR.
 
************************************************************************/
 
int I_georef(double e1,  /* EASTING TO BE TRANSFORMED */
             double n1,  /* NORTHING TO BE TRANSFORMED */
             double *e,  /* EASTING, TRANSFORMED */
             double *n,  /* NORTHING, TRANSFORMED */
             double E[], /* EASTING COEFFICIENTS */
             double N[], /* NORTHING COEFFICIENTS */
             int order   /* ORDER OF TRANSFORMATION TO BE PERFORMED, MUST MATCH
                            THE   ORDER USED TO CALCULATE THE COEFFICIENTS */
)
{
    double e3, e2n, en2, n3, e2, en, n2;
 
    switch (order) {
    case 1:
        *e = E[0] + E[1] * e1 + E[2] * n1;
        *n = N[0] + N[1] * e1 + N[2] * n1;
        break;
 
    case 2:
        e2 = e1 * e1;
        n2 = n1 * n1;
        en = e1 * n1;
 
        *e = E[0] + E[1] * e1 + E[2] * n1 + E[3] * e2 + E[4] * en + E[5] * n2;
        *n = N[0] + N[1] * e1 + N[2] * n1 + N[3] * e2 + N[4] * en + N[5] * n2;
        break;
 
    case 3:
        e2 = e1 * e1;
        en = e1 * n1;
        n2 = n1 * n1;
        e3 = e1 * e2;
        e2n = e2 * n1;
        en2 = e1 * n2;
        n3 = n1 * n2;
 
        *e = E[0] + E[1] * e1 + E[2] * n1 + E[3] * e2 + E[4] * en + E[5] * n2 +
             E[6] * e3 + E[7] * e2n + E[8] * en2 + E[9] * n3;
        *n = N[0] + N[1] * e1 + N[2] * n1 + N[3] * e2 + N[4] * en + N[5] * n2 +
             N[6] * e3 + N[7] * e2n + N[8] * en2 + N[9] * n3;
        break;
 
    default:
        return MPARMERR;
    }
 
    return MSUCCESS;
}
 
/***********************************************************************
 
  COMPUTE THE FORWARD AND BACKWARD GEOREFFERENCING COEFFICIENTS
  BASED ON A SET OF CONTROL POINTS
 
************************************************************************/
 
int I_compute_georef_equations(struct Control_Points *cp, double E12[],
                               double N12[], double E21[], double N21[],
                               int order)
{
    double *tempptr;
    int status;
 
    if (order < 1 || order > MAXORDER)
        return MPARMERR;
 
    /* CALCULATE THE FORWARD TRANSFORMATION COEFFICIENTS */
 
    status = calccoef(cp, E12, N12, order);
 
    if (status != MSUCCESS)
        return status;
 
    /* SWITCH THE 1 AND 2 EASTING AND NORTHING ARRAYS */
 
    tempptr = cp->e1;
    cp->e1 = cp->e2;
    cp->e2 = tempptr;
    tempptr = cp->n1;
    cp->n1 = cp->n2;
    cp->n2 = tempptr;
 
    /* CALCULATE THE BACKWARD TRANSFORMATION COEFFICIENTS */
 
    status = calccoef(cp, E21, N21, order);
 
    /* SWITCH THE 1 AND 2 EASTING AND NORTHING ARRAYS BACK */
 
    tempptr = cp->e1;
    cp->e1 = cp->e2;
    cp->e2 = tempptr;
    tempptr = cp->n1;
    cp->n1 = cp->n2;
    cp->n2 = tempptr;
 
    return status;
}
 
/***********************************************************************
 
  COMPUTE THE GEOREFFERENCING COEFFICIENTS
  BASED ON A SET OF CONTROL POINTS
 
************************************************************************/
 
static int calccoef(struct Control_Points *cp, double E[], double N[],
                    int order)
{
    struct MATRIX m;
    double *a;
    double *b;
    int numactive; /* NUMBER OF ACTIVE CONTROL POINTS */
    int status, i;
 
    /* CALCULATE THE NUMBER OF VALID CONTROL POINTS */
 
    for (i = numactive = 0; i < cp->count; i++) {
        if (cp->status[i] > 0)
            numactive++;
    }
 
    /* CALCULATE THE MINIMUM NUMBER OF CONTROL POINTS NEEDED TO DETERMINE
       A TRANSFORMATION OF THIS ORDER */
 
    m.n = ((order + 1) * (order + 2)) / 2;
 
    if (numactive < m.n)
        return MNPTERR;
 
    /* INITIALIZE MATRIX */
 
    m.v = G_calloc(m.n * m.n, sizeof(double));
    a = G_calloc(m.n, sizeof(double));
    b = G_calloc(m.n, sizeof(double));
 
    if (numactive == m.n)
        status = exactdet(cp, &m, a, b, E, N);
    else
        status = calcls(cp, &m, a, b, E, N);
 
    G_free(m.v);
    G_free(a);
    G_free(b);
 
    return status;
}
 
/***********************************************************************
 
  CALCULATE THE TRANSFORMATION COEFFICIENTS WITH EXACTLY THE MINIMUM
  NUMBER OF CONTROL POINTS REQUIRED FOR THIS TRANSFORMATION.
 
************************************************************************/
 
static int exactdet(struct Control_Points *cp, struct MATRIX *m, double a[],
                    double b[], double E[], /* EASTING COEFFICIENTS */
                    double N[]              /* NORTHING COEFFICIENTS */
)
{
    int pntnow, currow, j;
 
    currow = 1;
    for (pntnow = 0; pntnow < cp->count; pntnow++) {
        if (cp->status[pntnow] > 0) {
            /* POPULATE MATRIX M */
 
            for (j = 1; j <= m->n; j++)
                M(currow, j) = term(j, cp->e1[pntnow], cp->n1[pntnow]);
 
            /* POPULATE MATRIX A AND B */
 
            a[currow - 1] = cp->e2[pntnow];
            b[currow - 1] = cp->n2[pntnow];
 
            currow++;
        }
    }
 
    if (currow - 1 != m->n)
        return MINTERR;
 
    return solvemat(m, a, b, E, N);
}
 
/***********************************************************************
 
  CALCULATE THE TRANSFORMATION COEFFICIENTS WITH MORE THAN THE MINIMUM
  NUMBER OF CONTROL POINTS REQUIRED FOR THIS TRANSFORMATION.  THIS
  ROUTINE USES THE LEAST SQUARES METHOD TO COMPUTE THE COEFFICIENTS.
 
************************************************************************/
 
static int calcls(struct Control_Points *cp, struct MATRIX *m, double a[],
                  double b[], double E[], /* EASTING COEFFICIENTS */
                  double N[]              /* NORTHING COEFFICIENTS */
)
{
    int i, j, n, numactive = 0;
 
    /* INITIALIZE THE UPPER HALF OF THE MATRIX AND THE TWO COLUMN VECTORS */
 
    for (i = 1; i <= m->n; i++) {
        for (j = i; j <= m->n; j++)
            M(i, j) = 0.0;
        a[i - 1] = b[i - 1] = 0.0;
    }
 
    /* SUM THE UPPER HALF OF THE MATRIX AND THE COLUMN VECTORS ACCORDING TO
       THE LEAST SQUARES METHOD OF SOLVING OVER DETERMINED SYSTEMS */
 
    for (n = 0; n < cp->count; n++) {
        if (cp->status[n] > 0) {
            numactive++;
            for (i = 1; i <= m->n; i++) {
                for (j = i; j <= m->n; j++)
                    M(i, j) += term(i, cp->e1[n], cp->n1[n]) *
                               term(j, cp->e1[n], cp->n1[n]);
 
                a[i - 1] += cp->e2[n] * term(i, cp->e1[n], cp->n1[n]);
                b[i - 1] += cp->n2[n] * term(i, cp->e1[n], cp->n1[n]);
            }
        }
    }
 
    if (numactive <= m->n)
        return MINTERR;
 
    /* TRANSPOSE VALUES IN UPPER HALF OF M TO OTHER HALF */
 
    for (i = 2; i <= m->n; i++)
        for (j = 1; j < i; j++)
            M(i, j) = M(j, i);
 
    return solvemat(m, a, b, E, N);
}
 
/***********************************************************************
 
  CALCULATE THE X/Y TERM BASED ON THE TERM NUMBER
 
  ORDER\TERM   1    2    3    4    5    6    7    8    9   10
  1            e0n0 e1n0 e0n1
  2            e0n0 e1n0 e0n1 e2n0 e1n1 e0n2
  3            e0n0 e1n0 e0n1 e2n0 e1n1 e0n2 e3n0 e2n1 e1n2 e0n3
 
************************************************************************/
 
static double term(int term, double e, double n)
{
    switch (term) {
    case 1:
        return 1.0;
    case 2:
        return e;
    case 3:
        return n;
    case 4:
        return e * e;
    case 5:
        return e * n;
    case 6:
        return n * n;
    case 7:
        return e * e * e;
    case 8:
        return e * e * n;
    case 9:
        return e * n * n;
    case 10:
        return n * n * n;
    }
 
    return 0.0;
}
 
/***********************************************************************
 
  SOLVE FOR THE 'E' AND 'N' COEFFICIENTS BY USING A SOMEWHAT MODIFIED
  GAUSSIAN ELIMINATION METHOD.
 
  | M11 M12 ... M1n | | E0   |   | a0   |
  | M21 M22 ... M2n | | E1   | = | a1   |
  |  .   .   .   .  | | .    |   | .    |
  | Mn1 Mn2 ... Mnn | | En-1 |   | an-1 |
 
  and
 
  | M11 M12 ... M1n | | N0   |   | b0   |
  | M21 M22 ... M2n | | N1   | = | b1   |
  |  .   .   .   .  | | .    |   | .    |
  | Mn1 Mn2 ... Mnn | | Nn-1 |   | bn-1 |
 
************************************************************************/
 
static int solvemat(struct MATRIX *m, double a[], double b[], double E[],
                    double N[])
{
    int i, j, i2, j2, imark;
    double factor, temp;
    double pivot; /* ACTUAL VALUE OF THE LARGEST PIVOT CANDIDATE */
 
    for (i = 1; i <= m->n; i++) {
        j = i;
 
        /* find row with largest magnitude value for pivot value */
 
        pivot = M(i, j);
        imark = i;
        for (i2 = i + 1; i2 <= m->n; i2++) {
            temp = fabs(M(i2, j));
            if (temp > fabs(pivot)) {
                pivot = M(i2, j);
                imark = i2;
            }
        }
 
        /* if the pivot is very small then the points are nearly co-linear */
        /* co-linear points result in an undefined matrix, and nearly */
        /* co-linear points results in a solution with rounding error */
 
        if (pivot == 0.0)
            return MUNSOLVABLE;
 
        /* if row with highest pivot is not the current row, switch them */
 
        if (imark != i) {
            for (j2 = 1; j2 <= m->n; j2++) {
                temp = M(imark, j2);
                M(imark, j2) = M(i, j2);
                M(i, j2) = temp;
            }
 
            temp = a[imark - 1];
            a[imark - 1] = a[i - 1];
            a[i - 1] = temp;
 
            temp = b[imark - 1];
            b[imark - 1] = b[i - 1];
            b[i - 1] = temp;
        }
 
        /* compute zeros above and below the pivot, and compute
           values for the rest of the row as well */
 
        for (i2 = 1; i2 <= m->n; i2++) {
            if (i2 != i) {
                factor = M(i2, j) / pivot;
                for (j2 = j; j2 <= m->n; j2++)
                    M(i2, j2) -= factor * M(i, j2);
                a[i2 - 1] -= factor * a[i - 1];
                b[i2 - 1] -= factor * b[i - 1];
            }
        }
    }
 
    /* SINCE ALL OTHER VALUES IN THE MATRIX ARE ZERO NOW, CALCULATE THE
       COEFFICIENTS BY DIVIDING THE COLUMN VECTORS BY THE DIAGONAL VALUES. */
 
    for (i = 1; i <= m->n; i++) {
        E[i - 1] = a[i - 1] / M(i, i);
        N[i - 1] = b[i - 1] / M(i, i);
    }
 
    return MSUCCESS;
}