1 /****************************************************************************
2 *
3 * ViSP, open source Visual Servoing Platform software.
4 * Copyright (C) 2005 - 2019 by Inria. All rights reserved.
5 *
6 * This software is free software; you can redistribute it and/or modify
7 * it under the terms of the GNU General Public License as published by
8 * the Free Software Foundation; either version 2 of the License, or
9 * (at your option) any later version.
10 * See the file LICENSE.txt at the root directory of this source
11 * distribution for additional information about the GNU GPL.
12 *
13 * For using ViSP with software that can not be combined with the GNU
14 * GPL, please contact Inria about acquiring a ViSP Professional
15 * Edition License.
16 *
17 * See http://visp.inria.fr for more information.
18 *
19 * This software was developed at:
20 * Inria Rennes - Bretagne Atlantique
21 * Campus Universitaire de Beaulieu
22 * 35042 Rennes Cedex
23 * France
24 *
25 * If you have questions regarding the use of this file, please contact
26 * Inria at visp@inria.fr
27 *
28 * This file is provided AS IS with NO WARRANTY OF ANY KIND, INCLUDING THE
29 * WARRANTY OF DESIGN, MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE.
30 *
31 * Description:
32 * Homography estimation.
33 *
34 * Authors:
35 * Eric Marchand
36 *
37 *****************************************************************************/
38
39 /*!
40 \file vpHomographyDLT.cpp
41
42 This file implements the fonctions related with the homography
43 estimation using the DLT algorithm
44 */
45
46 #include <visp3/core/vpMatrix.h>
47 #include <visp3/core/vpMatrixException.h>
48 #include <visp3/vision/vpHomography.h>
49
50 #include <cmath> // std::fabs
51 #include <limits> // numeric_limits
52
53 #ifndef DOXYGEN_SHOULD_SKIP_THIS
54
HartleyNormalization(const std::vector<double> & x,const std::vector<double> & y,std::vector<double> & xn,std::vector<double> & yn,double & xg,double & yg,double & coef)55 void vpHomography::HartleyNormalization(const std::vector<double> &x, const std::vector<double> &y,
56 std::vector<double> &xn, std::vector<double> &yn, double &xg, double &yg,
57 double &coef)
58 {
59 if (x.size() != y.size())
60 throw(vpException(vpException::dimensionError, "Hartley normalization require that x and y vector "
61 "have the same dimension"));
62
63 unsigned int n = (unsigned int)x.size();
64 if (xn.size() != n)
65 xn.resize(n);
66 if (yn.size() != n)
67 yn.resize(n);
68
69 xg = 0;
70 yg = 0;
71
72 for (unsigned int i = 0; i < n; i++) {
73 xg += x[i];
74 yg += y[i];
75 }
76 xg /= n;
77 yg /= n;
78
79 // Changement d'origine : le centre de gravite doit correspondre
80 // a l'origine des coordonnees
81 double distance = 0;
82 for (unsigned int i = 0; i < n; i++) {
83 double xni = x[i] - xg;
84 double yni = y[i] - yg;
85 xn[i] = xni;
86 yn[i] = yni;
87 distance += sqrt(vpMath::sqr(xni) + vpMath::sqr(yni));
88 } // for translation sur tous les points
89
90 // Changement d'echelle
91 distance /= n;
92 // calcul du coef de changement d'echelle
93 // if(distance ==0)
94 if (std::fabs(distance) <= std::numeric_limits<double>::epsilon())
95 coef = 1;
96 else
97 coef = sqrt(2.0) / distance;
98
99 for (unsigned int i = 0; i < n; i++) {
100 xn[i] *= coef;
101 yn[i] *= coef;
102 }
103 }
104
HartleyNormalization(unsigned int n,const double * x,const double * y,double * xn,double * yn,double & xg,double & yg,double & coef)105 void vpHomography::HartleyNormalization(unsigned int n, const double *x, const double *y, double *xn, double *yn,
106 double &xg, double &yg, double &coef)
107 {
108 unsigned int i;
109 xg = 0;
110 yg = 0;
111
112 for (i = 0; i < n; i++) {
113 xg += x[i];
114 yg += y[i];
115 }
116 xg /= n;
117 yg /= n;
118
119 // Changement d'origine : le centre de gravite doit correspondre
120 // a l'origine des coordonnees
121 double distance = 0;
122 for (i = 0; i < n; i++) {
123 double xni = x[i] - xg;
124 double yni = y[i] - yg;
125 xn[i] = xni;
126 yn[i] = yni;
127 distance += sqrt(vpMath::sqr(xni) + vpMath::sqr(yni));
128 } // for translation sur tous les points
129
130 // Changement d'echelle
131 distance /= n;
132 // calcul du coef de changement d'echelle
133 // if(distance ==0)
134 if (std::fabs(distance) <= std::numeric_limits<double>::epsilon())
135 coef = 1;
136 else
137 coef = sqrt(2.0) / distance;
138
139 for (i = 0; i < n; i++) {
140 xn[i] *= coef;
141 yn[i] *= coef;
142 }
143 }
144
145 //---------------------------------------------------------------------------------------
146
HartleyDenormalization(vpHomography & aHbn,vpHomography & aHb,double xg1,double yg1,double coef1,double xg2,double yg2,double coef2)147 void vpHomography::HartleyDenormalization(vpHomography &aHbn, vpHomography &aHb, double xg1, double yg1, double coef1,
148 double xg2, double yg2, double coef2)
149 {
150
151 // calcul des transformations a appliquer sur M_norm pour obtenir M
152 // en fonction des deux normalisations effectuees au debut sur
153 // les points: aHb = T2^ aHbn T1
154 vpMatrix T1(3, 3);
155 vpMatrix T2(3, 3);
156 vpMatrix T2T(3, 3);
157
158 T1.eye();
159 T2.eye();
160 T2T.eye();
161
162 T1[0][0] = T1[1][1] = coef1;
163 T1[0][2] = -coef1 * xg1;
164 T1[1][2] = -coef1 * yg1;
165
166 T2[0][0] = T2[1][1] = coef2;
167 T2[0][2] = -coef2 * xg2;
168 T2[1][2] = -coef2 * yg2;
169
170 T2T = T2.pseudoInverse(1e-16);
171
172 vpMatrix aHbn_(3, 3);
173 for (unsigned int i = 0; i < 3; i++)
174 for (unsigned int j = 0; j < 3; j++)
175 aHbn_[i][j] = aHbn[i][j];
176
177 vpMatrix maHb = T2T * aHbn_ * T1;
178
179 for (unsigned int i = 0; i < 3; i++)
180 for (unsigned int j = 0; j < 3; j++)
181 aHb[i][j] = maHb[i][j];
182 }
183
184 #endif // #ifndef DOXYGEN_SHOULD_SKIP_THIS
185
186 /*!
187 From couples of matched points \f$^a{\bf p}=(x_a,y_a,1)\f$ in image a
188 and \f$^b{\bf p}=(x_b,y_b,1)\f$ in image b with homogeneous coordinates,
189 computes the homography matrix by resolving \f$^a{\bf p} = ^a{\bf H}_b\;
190 ^b{\bf p}\f$ using the DLT (Direct Linear Transform) algorithm.
191
192 At least 4 couples of points are needed.
193
194 To do so, we use the DLT algorithm on the data,
195 ie we resolve the linear system by SDV : \f$\bf{Ah} =0\f$ where
196 \f$\bf{h}\f$ is the vector with the terms of \f$^a{\bf H}_b\f$ and
197 \f$\mathbf{A}\f$ depends on the points coordinates.
198
199 For each point, in homogeneous coordinates we have:
200 \f[
201 ^a{\bf p} = ^a{\bf H}_b\; ^b{\bf p}
202 \f]
203 which is equivalent to:
204 \f[
205 ^a{\bf p} \times {^a{\bf H}_b \; ^b{\bf p}} =0
206 \f]
207 If we note \f$\mathbf{h}_j^T\f$ the \f$j^{\textrm{th}}\f$ line of
208 \f$^a{\bf H}_b\f$, we can write: \f[ ^a{\bf H}_b \; ^b{\bf p} = \left(
209 \begin{array}{c}\mathbf{h}_1^T \;^b{\bf p} \\\mathbf{h}_2^T \; ^b{\bf p}
210 \\\mathbf{h}_3^T \;^b{\bf p} \end{array}\right) \f]
211
212 Setting \f$^a{\bf p}=(x_{a},y_{a},w_{a})\f$, the cross product can be
213 rewritten by: \f[ ^a{\bf p} \times ^a{\bf H}_b \; ^b{\bf p} =\left(
214 \begin{array}{c}y_{a}\mathbf{h}_3^T \; ^b{\bf p}-w_{a}\mathbf{h}_2^T \;
215 ^b{\bf p} \\w_{a}\mathbf{h}_1^T \; ^b{\bf p} -x_{a}\mathbf{h}_3^T \; ^b{\bf
216 p} \\x_{a}\mathbf{h}_2^T \; ^b{\bf p}- y_{a}\mathbf{h}_1^T \; ^b{\bf
217 p}\end{array}\right) \f]
218
219 \f[
220 \underbrace{\left( \begin{array}{ccc}\mathbf{0}^T & -w_{a} \; ^b{\bf p}^T
221 & y_{a} \; ^b{\bf p}^T \\ w_{a}
222 \; ^b{\bf p}^T&\mathbf{0}^T & -x_{a} \; ^b{\bf p}^T \\
223 -y_{a} \; ^b{\bf p}^T & x_{a} \; ^b{\bf p}^T &
224 \mathbf{0}^T\end{array}\right)}_{\mathbf{A}_i (3\times 9)}
225 \underbrace{\left( \begin{array}{c}\mathbf{h}_{1}^{T} \\
226 \mathbf{h}_{2}^{T}\\\mathbf{h}_{3}^{T}\end{array}\right)}_{\mathbf{h}
227 (9\times 1)}=0 \f]
228
229 leading to an homogeneous system to be solved:
230 \f$\mathbf{A}\mathbf{h}=0\f$ with \f$\mathbf{A}=\left(\mathbf{A}_1^T, ...,
231 \mathbf{A}_i^T, ..., \mathbf{A}_n^T \right)^T\f$.
232
233 It can be solved using an SVD decomposition:
234 \f[\bf A = UDV^T \f]
235 <b>h</b> is the column of <b>V</b> associated with the smalest singular
236 value of <b>A
237 </b>
238
239 \param xb, yb : Coordinates vector of matched points in image b. These
240 coordinates are expressed in meters. \param xa, ya : Coordinates vector of
241 matched points in image a. These coordinates are expressed in meters. \param
242 aHb : Estimated homography that relies the transformation from image a to
243 image b. \param normalization : When set to true, the coordinates of the
244 points are normalized. The normalization carried out is the one preconized
245 by Hartley.
246
247 \exception vpMatrixException::rankDeficient : When the rank of the matrix
248 that should be 8 is deficient.
249 */
DLT(const std::vector<double> & xb,const std::vector<double> & yb,const std::vector<double> & xa,const std::vector<double> & ya,vpHomography & aHb,bool normalization)250 void vpHomography::DLT(const std::vector<double> &xb, const std::vector<double> &yb, const std::vector<double> &xa,
251 const std::vector<double> &ya, vpHomography &aHb, bool normalization)
252 {
253 unsigned int n = (unsigned int)xb.size();
254 if (yb.size() != n || xa.size() != n || ya.size() != n)
255 throw(vpException(vpException::dimensionError, "Bad dimension for DLT homography estimation"));
256
257 // 4 point are required
258 if (n < 4)
259 throw(vpException(vpException::fatalError, "There must be at least 4 matched points"));
260
261 try {
262 std::vector<double> xan, yan, xbn, ybn;
263
264 double xg1 = 0., yg1 = 0., coef1 = 0., xg2 = 0., yg2 = 0., coef2 = 0.;
265
266 vpHomography aHbn;
267
268 if (normalization) {
269 vpHomography::HartleyNormalization(xb, yb, xbn, ybn, xg1, yg1, coef1);
270 vpHomography::HartleyNormalization(xa, ya, xan, yan, xg2, yg2, coef2);
271 } else {
272 xbn = xb;
273 ybn = yb;
274 xan = xa;
275 yan = ya;
276 }
277
278 vpMatrix A(2 * n, 9);
279 vpColVector h(9);
280 vpColVector D(9);
281 vpMatrix V(9, 9);
282
283 // We need here to compute the SVD on a (n*2)*9 matrix (where n is
284 // the number of points). if n == 4, the matrix has more columns
285 // than rows. This kind of matrix is not supported by GSL for
286 // SVD. The solution is to add an extra line with zeros
287 if (n == 4)
288 A.resize(2 * n + 1, 9);
289
290 // build matrix A
291 for (unsigned int i = 0; i < n; i++) {
292 A[2 * i][0] = 0;
293 A[2 * i][1] = 0;
294 A[2 * i][2] = 0;
295 A[2 * i][3] = -xbn[i];
296 A[2 * i][4] = -ybn[i];
297 A[2 * i][5] = -1;
298 A[2 * i][6] = xbn[i] * yan[i];
299 A[2 * i][7] = ybn[i] * yan[i];
300 A[2 * i][8] = yan[i];
301
302 A[2 * i + 1][0] = xbn[i];
303 A[2 * i + 1][1] = ybn[i];
304 A[2 * i + 1][2] = 1;
305 A[2 * i + 1][3] = 0;
306 A[2 * i + 1][4] = 0;
307 A[2 * i + 1][5] = 0;
308 A[2 * i + 1][6] = -xbn[i] * xan[i];
309 A[2 * i + 1][7] = -ybn[i] * xan[i];
310 A[2 * i + 1][8] = -xan[i];
311 }
312
313 // Add an extra line with zero.
314 if (n == 4) {
315 for (unsigned int i = 0; i < 9; i++) {
316 A[2 * n][i] = 0;
317 }
318 }
319
320 // solve Ah = 0
321 // SVD Decomposition A = UDV^T (destructive wrt A)
322 A.svd(D, V);
323
324 // on en profite pour effectuer un controle sur le rang de la matrice :
325 // pas plus de 2 valeurs singulieres quasi=0
326 int rank = 0;
327 for (unsigned int i = 0; i < 9; i++)
328 if (D[i] > 1e-7)
329 rank++;
330 if (rank < 7) {
331 throw(vpMatrixException(vpMatrixException::rankDeficient, "Matrix rank %d is deficient (should be 8)", rank));
332 }
333 // h = is the column of V associated with the smallest singular value of A
334
335 // since we are not sure that the svd implemented sort the
336 // singular value... we seek for the smallest
337 double smallestSv = 1e30;
338 unsigned int indexSmallestSv = 0;
339 for (unsigned int i = 0; i < 9; i++)
340 if ((D[i] < smallestSv)) {
341 smallestSv = D[i];
342 indexSmallestSv = i;
343 }
344
345 h = V.getCol(indexSmallestSv);
346
347 // build the homography
348 for (unsigned int i = 0; i < 3; i++) {
349 for (unsigned int j = 0; j < 3; j++)
350 aHbn[i][j] = h[3 * i + j];
351 }
352
353 if (normalization) {
354 // H after denormalization
355 vpHomography::HartleyDenormalization(aHbn, aHb, xg1, yg1, coef1, xg2, yg2, coef2);
356 } else {
357 aHb = aHbn;
358 }
359
360 } catch (vpMatrixException &me) {
361 vpTRACE("Matrix Exception ");
362 throw(me);
363 } catch (const vpException &me) {
364 vpERROR_TRACE("caught another error");
365 std::cout << std::endl << me << std::endl;
366 throw(me);
367 }
368 }
369