Now let us consider the solutions of the combinatorial pairs. the weights are the square of the corresponding determinants,

Transcription

1 Algebraic Geodesy and Geoinformatics PART I METHODS 7 Gauss- Jacobi Combinatorial Algorithm 7-1 Linear model Another technique to solve overdetermined system is proposed by Gauss and Jacobi. The Gauss-Jacobi combinatorial solution can be employed originally for linear regression problem. This means, if we have more independent equations, m than variables, n, so m > n, then the solution - in least-squares sense - can be achieved by solving the m n combinatorial subsets of m equations, and then weighting these solutions properly. This method can be extended to nonlinear case using linearization. A key point of this extention, is that we can solve nonlinear subsets without a priori information about the solutions, which would be necessary for the traditional linearization. The determination of weights through linearization takes place after the solutions of these nonlinear subsets. Therefore Gauss- Jacobi method can be considered as a gobal method First, let us illustrate this method for linear system. We consider a system with m = 2 unknowns, (x, y) and with n = 3 equations, (eq1, eq2, eq3). Clear@"Global *"D eq1 = a1 x + b1 y - y1 0; eq2 = a2 x + b2 y - y2 0; eq3 = a3 x + b3 y - y3 0; We shall show that in case of linear system, the result of the least square solution and that of the weighted combinatorial solution are the same. The least square solution can be computed in the following way: - the matrix form of the system, 8b, A< = eq2, eq3<, 8x, y<d; MatrixForm@AD a1 b1 a2 b2 a3 b3 b = - b; MatrixForm@bD y1 y2 y3 - then the least square solution is

2 2 GaussJacobi_07.nb xy Simplify PseudoInverse A.b, Element a1, a2, a3, b1, b2, b3, y1, y2, y3, Reals ; MatrixForm xy a3 b1 b3 y1 b2 b3 y2 b1 2 y3 b2 2 y3 a1 b2 2 y1 b3 2 y1 b1 b2 y2 b1 b3 y3 a2 b1 b2 y1 b1 2 y2 b3 b3 y2 b2 y3 a3 2 b1 2 b2 2 2 a1 a3 b1 b3 2 a2 b2 a1 b1 a3 b3 a2 2 b1 2 b3 2 a1 2 b2 2 b3 2 a3 2 b1 y1 b2 y2 a1 a3 b3 y1 b1 y3 a2 a1 b2 y1 a1 b1 y2 a3 b3 y2 a3 b2 y3 a2 2 b1 y1 b3 y3 a1 2 b2 y2 b3 y3 a3 2 b1 2 b2 2 2 a1 a3 b1 b3 2 a2 b2 a1 b1 a3 b3 a2 2 b1 2 b3 2 a1 2 b2 2 b3 2 Now let us consider the solutions of the combinatorial pairs A12 CoefficientArrays eq1, eq2, x, y 2 ; xy12 LinearSolve A12, y1, y2 ; MatrixForm xy12 b2 y1 b1 y2 a2 b1 a1 b2 a2 y1 a1 y2 a2 b1 a1 b2 A13 CoefficientArrays eq1, eq3, x, y 2 ; xy13 LinearSolve A13, y1, y3 ; MatrixForm xy13 b3 y1 b1 y3 a3 b1 a1 b3 a3 y1 a1 y3 a3 b1 a1 b3 A23 CoefficientArrays eq2, eq3, x, y 2 ; xy23 LinearSolve A23, y2, y3 ; MatrixForm xy23 b3 y2 b2 y3 a3 b2 a2 b3 a3 y2 a2 y3 a3 b2 a2 b3 the weights are the square of the corresponding determinants, Π12 a1 b2 a2 b1 2 ; Π13 a1 b3 a3 b1 2 ; Π23 a2 b3 a3 b2 2 ; leading to the weighted combinatorical solution for variable x as xc Π12 xy12 1 Π13 xy13 1 Π23 xy23 1 Π12 Π13 Π23 Simplify a2 b1 a1 b2 b2 y1 b1 y2 a3 b1 a1 b3 b3 y1 b1 y3 a3 b2 a2 b3 b3 y2 b2 y3 a2 b1 a1 b2 2 a3 b1 a1 b3 2 a3 b2 a2 b3 2 Similarly, the solution for y is given by yc Π12 xy12 2 Π13 xy13 2 Π23 xy23 2 Π12 Π13 Π23 Simplify a3 2 b1 y1 b2 y2 a1 a3 b3 y1 b1 y3 a2 a1 b2 y1 a1 b1 y2 a3 b3 y2 a3 b2 y3 a2 2 b1 y1 b3 y3 a1 2 b2 y2 b3 y3 a3 2 b1 2 b2 2 2 a1 a3 b1 b3 2 a2 b2 a1 b1 a3 b3 a2 2 b1 2 b3 2 a1 2 b2 2 b3 2 This Gauss-Jacobi combinatorial solution and the least square solution are the same, indeed FullSimplify xc, yc xy 0, 0

3 GaussJacobi_07.nb 7-2 Nonlinear model Arithmetical Average of the Combinatorial Solution In order to employ this result for a nonlinear model, this model should be linearized, but where? To solve this problem the following algorithm may be used, improving the solution step by step, - Compute the combinatorial solutions of the nonlinear model via resultants or Groebner basis, alternatively with global numerical methods, like homotopy, - Compute the arithmetical average of these solutions, - Consider this arithmetical average as the point, where the Jacobian should be evaluated and employ weighting technique, - Linearize the nonlinear system at this weighted result, and solve it as a linear least square problem. Let us illustrate this technique with the combinatorial solution of the 2D positioning problem. As we have seen in Section , this problem can be solved easily via ALESS using Groebner basis. Considering 4 points with known coordinates, P i (x i,y i ) and their distances (t i ) from a point P 0 with unknown coordinates (x 0,y 0 ), which should be computed. The values are presented in the following table, Table 7.1 Data for 2D Positioning problem The coordinates and the distances, x ; x ; x ; x ; y ; y ; y ; y ; t ; t ; t ; t ; The lists of variables, X x1, x2, x3, x4 ; Y y1, y2, y3, y4 ; T t1, t2, t3, t4 ; The observational equations are f1 x1 x0 ^2 y1 y0 ^2 t1 2 ; f2 x2 x0 ^2 y2 y0 ^2 t2 2 ; f3 x3 x0 ^2 y3 y0 ^2 t3 2 ; f4 x4 x0 ^2 y4 y0 ^2 t4 2 ; So we have 4 equations and 2 unknowns, m 4; n 2; The number of the combinations,

4 4 GaussJacobi_07.nb mn Binomial m, n 6 The combinatorial pairs, R Subsets Range m, n 1, 2, 1, 3, 1, 4, 2, 3, 2, 4, 3, 4 The corresponding data set is in form of {x i,y i, t i, x j,y j, t j } datac Map X 1, Y 1, T 1, X 2, Y 2, T 2 &, R , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , The solutions of a general pair can be computed via reduced Groebner basis. The system, F xi x0 ^2 yi y0 ^2 ti 2, xj x0 ^2 yj y0 ^2 tj 2 ; The second order polynomial for the x0 coordinate, px0 GroebnerBasis F, x0, y0, y0 ti 4 2 ti 2 tj 2 tj 4 4 ti 2 x0 xi 4 tj 2 x0 xi 2 ti 2 xi 2 2 tj 2 xi 2 4 x0 2 xi 2 4 x0 xi 3 xi 4 4 ti 2 x0 xj 4 tj 2 x0 xj 8 x0 2 xi xj 4 x0 xi 2 xj 2 ti 2 xj 2 2 tj 2 xj 2 4 x0 2 xj 2 4 x0 xi xj 2 2 xi 2 xj 2 4 x0 xj 3 xj 4 2 ti 2 yi 2 2 tj 2 yi 2 4 x0 2 yi 2 4 x0 xi yi 2 2 xi 2 yi 2 4 x0 xj yi 2 2 xj 2 yi 2 yi 4 4 ti 2 yi yj 4 tj 2 yi yj 8 x0 2 yi yj 8 x0 xi yi yj 4 xi 2 yi yj 8 x0 xj yi yj 4 xj 2 yi yj 4 yi 3 yj 2 ti 2 yj 2 2 tj 2 yj 2 4 x0 2 yj 2 4 x0 xi yj 2 2 xi 2 yj 2 4 x0 xj yj 2 2 xj 2 yj 2 6 yi 2 yj 2 4 yi yj 3 yj 4 its solution in general, sol. Solve a2 2 a1 a0 0, a1 a12 4 a0 a2 2 a2, a1 a12 4 a0 a2 2 a2 It means we have two solutions. The coefficients A0 Coefficient px0, x0, 0 ti 4 2 ti 2 tj 2 tj 4 2 ti 2 xi 2 2 tj 2 xi 2 xi 4 2 ti 2 xj 2 2 tj 2 xj 2 2 xi 2 xj 2 xj 4 2 ti 2 yi 2 2 tj 2 yi 2 2 xi 2 yi 2 2 xj 2 yi 2 yi 4 4 ti 2 yi yj 4 tj 2 yi yj 4 xi 2 yi yj 4 xj 2 yi yj 4 yi 3 yj 2 ti 2 yj 2 2 tj 2 yj 2 2 xi 2 yj 2 2 xj 2 yj 2 6 yi 2 yj 2 4 yi yj 3 yj 4 A1 Coefficient px0, x0, 1 4 ti 2 xi 4 tj 2 xi 4 xi 3 4 ti 2 xj 4 tj 2 xj 4 xi 2 xj 4 xi xj 2 4 xj 3 4 xi yi 2 4 xj yi 2 8 xi yi yj 8 xj yi yj 4 xi yj 2 4 xj yj 2 A2 Coefficient px0, x0, 2 4 xi 2 8 xi xj 4 xj 2 4 yi 2 8 yi yj 4 yj 2

5 GaussJacobi_07.nb The function which computes the x0 coordinates for a pair {i, j} X0 xi_, yi_, ti_, xj_, yj_, tj_ sol. a0 A0, a1 A1, a2 A2 Flatten; For example in case of pair {1, 2} X0 datac , The values of all of the x0 coordinates are, x0c Map X0 &, datac , , , , , , , , , , , The same computations can be done for y0, namely py0 GroebnerBasis F, x0, y0, x0 ti 4 2 ti 2 tj 2 tj 4 2 ti 2 xi 2 2 tj 2 xi 2 xi 4 4 ti 2 xi xj 4 tj 2 xi xj 4 xi 3 xj 2 ti 2 xj 2 2 tj 2 xj 2 6 xi 2 xj 2 4 xi xj 3 xj 4 4 xi 2 y0 2 8 xi xj y0 2 4 xj 2 y0 2 4 ti 2 y0 yi 4 tj 2 y0 yi 4 xi 2 y0 yi 8 xi xj y0 yi 4 xj 2 y0 yi 2 ti 2 yi 2 2 tj 2 yi 2 2 xi 2 yi 2 4 xi xj yi 2 2 xj 2 yi 2 4 y0 2 yi 2 4 y0 yi 3 yi 4 4 ti 2 y0 yj 4 tj 2 y0 yj 4 xi 2 y0 yj 8 xi xj y0 yj 4 xj 2 y0 yj 8 y0 2 yi yj 4 y0 yi 2 yj 2 ti 2 yj 2 2 tj 2 yj 2 2 xi 2 yj 2 4 xi xj yj 2 2 xj 2 yj 2 4 y0 2 yj 2 4 y0 yi yj 2 2 yi 2 yj 2 4 y0 yj 3 yj 4 B0 Coefficient py0, y0, 0 ti 4 2 ti 2 tj 2 tj 4 2 ti 2 xi 2 2 tj 2 xi 2 xi 4 4 ti 2 xi xj 4 tj 2 xi xj 4 xi 3 xj 2 ti 2 xj 2 2 tj 2 xj 2 6 xi 2 xj 2 4 xi xj 3 xj 4 2 ti 2 yi 2 2 tj 2 yi 2 2 xi 2 yi 2 4 xi xj yi 2 2 xj 2 yi 2 yi 4 2 ti 2 yj 2 2 tj 2 yj 2 2 xi 2 yj 2 4 xi xj yj 2 2 xj 2 yj 2 2 yi 2 yj 2 yj 4 B1 Coefficient py0, y0, 1 4 ti 2 yi 4 tj 2 yi 4 xi 2 yi 8 xi xj yi 4 xj 2 yi 4 yi 3 4 ti 2 yj 4 tj 2 yj 4 xi 2 yj 8 xi xj yj 4 xj 2 yj 4 yi 2 yj 4 yi yj 2 4 yj 3 B2 Coefficient py0, y0, 2 4 xi 2 8 xi xj 4 xj 2 4 yi 2 8 yi yj 4 yj 2 Y0 xi_, yi_, ti_, xj_, yj_, tj_ sol. a0 B0, a1 B1, a2 B2 Flatten; y0c Map Y0 &, datac , , , , , , , , , , , In order to decide, which combination is the admissible solution, we should define the least square objective, obj Apply Plus, Map ^2 &, f1, f2, f3, f x y x y x y x y0 2 2 For example let us consider the first pair. The x0 values

6 6 GaussJacobi_07.nb x0c , The y0 values, y0c , The possible combinations, xyc Tuples x0c 1, y0c , , , , , , , Let us compute the values of the objective function, objc Map obj. x0 1, y0 2 &, xyc , , , This means that the third combination is the best, since it results smallest residual. In general way, xyc First Flatten Position objc, Min objc , Let us generalize this selection technique for all points. The combinations, xyc MapThread Tuples 1, 2 &, x0c, y0c , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , The objective functions, objc Map Map obj. x0 1, y0 2 &, &, xyc , , , , , , , , , , , , , , , , , , , , , , , since the positions of the minimums are Map First Flatten Position, Min &, objc 3, 1, 3, 1, 2, 2 The proper solutions of the combinatorial subsets, x0y0 MapThread xyc 1 2 &, Range mn, Map First Flatten Position, Min &, objc , , , , , , , , , , ,

7 GaussJacobi_07.nb Let us compute the arithmetical average, x0a, y0a Map Mean &, Transpose x0y , NumberForm, , Weighted Combinatorial Solution As we have seen in the previous chapter, the solution with high precision is, {x , y } therefore one may improve this combinatorial solution by employing weighting. The weighting for nonlinear system can be carried out like in case of the linear system. Let us consider the j-th equation of a linear model, For example in our case n = 2, then the j-th observation equation is, The linearized form of the j-th nonlinear model equation at (Η 1,...,Η n ), In our case j = 1,..., m, where m = 4. For example, in case of n = 2, the j-th equation, or As we have seen in case of the linear model (n = 2 and m = 4) the weight of the combinatorial solution {1, 2} was In similar way the weight of the nonlinear model for the solution (x 12 ), where Η x 12. In our case, the partial derivatives,

8 8 GaussJacobi_07.nb f1x D f1, x x0 f1y D f1, y y0 f2x D f2, x x0 f2y D f2, y y0 f3x D f3, x x0 f3y D f3, y y0 f4x D f4, x x0 f4y D f4, y y0 The selected combinatorial solutions of the subsets in rule form, x0y0s Map x0 1, y0 2 &, x0y0 x , y , x , y , x , y , x , y , x , y , x , y Then the weights, Π12 f1x f2y f2x f1y. x0y0s Π13 f1x f3y f3x f1y. x0y0s Π14 f1x f4y f4x f1y. x0y0s Π23 f2x f3y f3x f2y. x0y0s Π24 f2x f4y f4x f2y. x0y0s Π34 f3x f4y f4x f3y. x0y0s Employing these weights, the weighted solutions, {x0c, y0c} are, Πw Π12, Π13, Π14, Π23, Π24, Π34 ;

9 GaussJacobi_07.nb xd Transpose x0y , , , , , yd Transpose x0y , , , , , Πw.xd x0c Apply Plus, Πw NumberForm, Πw.yd y0c Apply Plus, Πw NumberForm, Additional Linearization This result can be further refined, if we linearize our nonlinear system at this point {x0c, y0c}, and solve it as a linear least square problem. The linearized equations q1 Series f1, x0, x0c, 1, y0, y0c, 1 Normal Expand x y0 q2 Series f2, x0, x0c, 1, y0, y0c, 1 Normal Expand x y0 q3 Series f3, x0, x0c, 1, y0, y0c, 1 Normal Expand x y0 q4 Series f4, x0, x0c, 1, y0, y0c, 1 Normal Expand x y0 In matrix form, where and Clear b b, A CoefficientArrays q1, q2, q3, q4, x0, y0 ; b b; MatrixForm A

10 10 GaussJacobi_07.nb MatrixForm b The pseudo inverse solution, PseudoInverse A.b , NumberForm, , Here we summarized the results of the different approaches, Table 7.2 Results of Gauss-Jacobi solution and its improvement The improvement can be clearly seen. 7-3 Examples GPS 4 - Point Problem This problem has been solved in Section with linear homotopy. Now, we solve the problem in a different way, using computer algebra, namely Dixon resultant. The reason is, that this result will be utilized in case of N- point problem, in the next example, using Gauss-Jacobi method. Employing our equations, e i = 0, i = 1...4, where Clear x1, x2, x3, x4, b e1 x1 a 0 2 x2 b 0 2 x3 c 0 2 x4 d 0 2 ; e2 x1 a 1 2 x2 b 1 2 x3 c 1 2 x4 d 1 2 ; e3 x1 a 2 2 x2 b 2 2 x3 c 2 2 x4 d 2 2 ; e4 x1 a 3 2 x2 b 3 2 x3 c 3 2 x4 d 3 2 ; First, this system of polynomial equations, will be transformed into a system of linear equations and a quadratic equation. Let us expand and multiply by minus one, and arranged the original equations, eqsl e1l, e2l, e3l, e4l Map Sort Expand &, e1, e2, e3, e4 x1 2 x2 2 x3 2 x4 2 2 x1 a 0 a x2 b 0 b x3 c 0 c x4 d 0 d 0 2, x1 2 x2 2 x3 2 x4 2 2 x1 a 1 a x2 b 1 b x3 c 1 c x4 d 1 d 1 2, x1 2 x2 2 x3 2 x4 2 2 x1 a 2 a x2 b 2 b x3 c 2 c x4 d 2 d 2 2, x1 2 x2 2 x3 2 x4 2 2 x1 a 3 a x2 b 3 b x3 c 3 c x4 d 3 d 3 2

11 GaussJacobi_07.nb Subtract the fourth equation from the other three ones, we get q Table eqsl i eqsl 4, i, 1, 3 Simplify 2 x1 a 0 a x1 a 3 a x2 b 0 b x2 b 3 b x3 c 0 c x3 c 3 c x4 d 0 d x4 d 3 d 3 2, 2 x1 a 1 a x1 a 3 a x2 b 1 b x2 b 3 b x3 c 1 c x3 c 3 c x4 d 1 d x4 d 3 d 3 2, 2 x1 a 2 a x1 a 3 a x2 b 2 b x2 b 3 b x3 c 2 c x3 c 3 c x4 d 2 d x4 d 3 d 3 2 This is a system of three linear equations, and they can be written as, g 1 a 0,3 x1 b 0,3 x2 c 0,3 x3 d 3,0 x4 e 0,3 ; g 2 g , 0 1 x1 a 1,3 x2 b 1,3 x3 c 1,3 x4 d 3,1 e 1,3 g 3 g , 0 2 x1 a 2,3 x2 b 2,3 x3 c 2,3 x4 d 3,2 e 2,3 The coefficients a i,3, b i,3, c i,3, d 3,i, e i,3, i , can be determined as, coeffs0 Table Coefficient q i 1, x1, Coefficient q i 1, x2, Coefficient q i 1, x3, Coefficient q i 1, x4 Factor, i, 0, 2 ; which are the coefficients of the variables x 1, x 2, x 3, x 4. The constant part is, coeffs1 Table q i coeffs0 i. x1, x2, x3, x4 Simplify, i, 1, 3 ; Therefore, all of the coefficients are, coeffs Table Union coeffs0 i, coeffs1 i, i, 1, 3 2 a 0 a 3, 2 b 0 b 3, 2 c 0 c 3, 2 d 0 d 3, a 0 2 a 3 2 b 0 2 b 3 2 c 0 2 c 3 2 d 0 2 d 3 2, 2 a 1 a 3, 2 b 1 b 3, 2 c 1 c 3, 2 d 1 d 3, a 1 2 a 3 2 b 1 2 b 3 2 c 1 2 c 3 2 d 1 2 d 3 2, 2 a 2 a 3, 2 b 2 b 3, 2 c 2 c 3, 2 d 2 d 3, a 2 2 a 3 2 b 2 2 b 3 2 c 2 2 c 3 2 d 2 2 d 3 2 Let us assign these coefficients to the linear system, coeffsn Flatten Table Inner 1 2 &, a i,3, b i,3, c i,3, d 3,i, e i,3, coeffs i 1, List, i, 0, 2 a 0,3 2 a 0 a 3, b 0,3 2 b 0 b 3, c 0,3 2 c 0 c 3, d 3,0 2 d 0 d 3, e 0,3 a 0 2 a 3 2 b 0 2 b 3 2 c 0 2 c 3 2 d 0 2 d 3 2, a 1,3 2 a 1 a 3, b 1,3 2 b 1 b 3, c 1,3 2 c 1 c 3, d 3,1 2 d 1 d 3, e 1,3 a 1 2 a 3 2 b 1 2 b 3 2 c 1 2 c 3 2 d 1 2 d 3 2, a 2,3 2 a 2 a 3, b 2,3 2 b 2 b 3, c 2,3 2 c 2 c 3, d 3,2 2 d 2 d 3, e 2,3 a 2 2 a 3 2 b 2 2 b 3 2 c 2 2 c 3 2 d 2 2 d 3 2 In addition, we take one of the nonlinear equations, let say the fourth one, e 4 x1 a 3 2 x2 b 3 2 x3 c 3 2 x4 d 3 2 ; Now, we shall solve the linear system for the variables x 1, x 2, x 3, with x 4 as parameter. It means, the relations x 1 = g(x 4 ), x 2 = g(x 4 ) and x 3 = g(x 4 ) will be computed. To do that, different methods can be employed. Here we shall employ Dixon resultant, Resultant Dixon Now we can solve the original system, eliminating the variables x 2 and x 3 to get x 1 = g(x 4 ),

12 12 GaussJacobi_07.nb drx1 DixonResultant g 1, g 2, g 3, x2, x3, u2, u3 x1 a 2,3 b 1,3 c 0,3 x1 a 1,3 b 2,3 c 0,3 x1 a 2,3 b 0,3 c 1,3 x1 a 0,3 b 2,3 c 1,3 x1 a 1,3 b 0,3 c 2,3 x1 a 0,3 b 1,3 c 2,3 x4 b 2,3 c 1,3 d 3,0 x4 b 1,3 c 2,3 d 3,0 x4 b 2,3 c 0,3 d 3,1 x4 b 0,3 c 2,3 d 3,1 x4 b 1,3 c 0,3 d 3,2 x4 b 0,3 c 1,3 d 3,2 b 2,3 c 1,3 e 0,3 b 1,3 c 2,3 e 0,3 b 2,3 c 0,3 e 1,3 b 0,3 c 2,3 e 1,3 b 1,3 c 0,3 e 2,3 b 0,3 c 1,3 e 2,3 This is a linear expression containing only x 1 and x 4, Exponent drx1, x1, x2, x3, x4 1, 0, 0, 1 Then the solution for x 1 = g(x 4 ) solx1 Solve drx1 0, x1 x1 x4 b 2,3 c 1,3 d 3,0 x4 b 1,3 c 2,3 d 3,0 x4 b 2,3 c 0,3 d 3,1 x4 b 0,3 c 2,3 d 3,1 x4 b 1,3 c 0,3 d 3,2 x4 b 0,3 c 1,3 d 3,2 b 2,3 c 1,3 e 0,3 b 1,3 c 2,3 e 0,3 b 2,3 c 0,3 e 1,3 b 0,3 c 2,3 e 1,3 b 1,3 c 0,3 e 2,3 b 0,3 c 1,3 e 2,3 a 2,3 b 1,3 c 0,3 a 1,3 b 2,3 c 0,3 a 2,3 b 0,3 c 1,3 a 0,3 b 2,3 c 1,3 a 1,3 b 0,3 c 2,3 a 0,3 b 1,3 c 2,3 Similarly, for the two additional variables, x 2 = g(x 4 ) and x 3 = g(x 4 ), and drx2 DixonResultant g 1, g 2, g 3, x1, x3, u1, u3 x2 a 2,3 b 1,3 c 0,3 x2 a 1,3 b 2,3 c 0,3 x2 a 2,3 b 0,3 c 1,3 x2 a 0,3 b 2,3 c 1,3 x2 a 1,3 b 0,3 c 2,3 x2 a 0,3 b 1,3 c 2,3 x4 a 2,3 c 1,3 d 3,0 x4 a 1,3 c 2,3 d 3,0 x4 a 2,3 c 0,3 d 3,1 x4 a 0,3 c 2,3 d 3,1 x4 a 1,3 c 0,3 d 3,2 x4 a 0,3 c 1,3 d 3,2 a 2,3 c 1,3 e 0,3 a 1,3 c 2,3 e 0,3 a 2,3 c 0,3 e 1,3 a 0,3 c 2,3 e 1,3 a 1,3 c 0,3 e 2,3 a 0,3 c 1,3 e 2,3 Exponent drx2, x1, x2, x3, x4 0, 1, 0, 1 solx2 Solve drx2 0, x2 x2 x4 a 2,3 c 1,3 d 3,0 x4 a 1,3 c 2,3 d 3,0 x4 a 2,3 c 0,3 d 3,1 x4 a 0,3 c 2,3 d 3,1 x4 a 1,3 c 0,3 d 3,2 x4 a 0,3 c 1,3 d 3,2 a 2,3 c 1,3 e 0,3 a 1,3 c 2,3 e 0,3 a 2,3 c 0,3 e 1,3 a 0,3 c 2,3 e 1,3 a 1,3 c 0,3 e 2,3 a 0,3 c 1,3 e 2,3 a 2,3 b 1,3 c 0,3 a 1,3 b 2,3 c 0,3 a 2,3 b 0,3 c 1,3 a 0,3 b 2,3 c 1,3 a 1,3 b 0,3 c 2,3 a 0,3 b 1,3 c 2,3 drx3 DixonResultant g 1, g 2, g 3, x1, x2, u1, u2 x3 a 2,3 b 1,3 c 0,3 x3 a 1,3 b 2,3 c 0,3 x3 a 2,3 b 0,3 c 1,3 x3 a 0,3 b 2,3 c 1,3 x3 a 1,3 b 0,3 c 2,3 x3 a 0,3 b 1,3 c 2,3 x4 a 2,3 b 1,3 d 3,0 x4 a 1,3 b 2,3 d 3,0 x4 a 2,3 b 0,3 d 3,1 x4 a 0,3 b 2,3 d 3,1 x4 a 1,3 b 0,3 d 3,2 x4 a 0,3 b 1,3 d 3,2 a 2,3 b 1,3 e 0,3 a 1,3 b 2,3 e 0,3 a 2,3 b 0,3 e 1,3 a 0,3 b 2,3 e 1,3 a 1,3 b 0,3 e 2,3 a 0,3 b 1,3 e 2,3 Exponent drx3, x1, x2, x3, x4 0, 0, 1, 1 solx3 Solve drx3 0, x3 x3 x4 a 2,3 b 1,3 d 3,0 x4 a 1,3 b 2,3 d 3,0 x4 a 2,3 b 0,3 d 3,1 x4 a 0,3 b 2,3 d 3,1 x4 a 1,3 b 0,3 d 3,2 x4 a 0,3 b 1,3 d 3,2 a 2,3 b 1,3 e 0,3 a 1,3 b 2,3 e 0,3 a 2,3 b 0,3 e 1,3 a 0,3 b 2,3 e 1,3 a 1,3 b 0,3 e 2,3 a 0,3 b 1,3 e 2,3 a 2,3 b 1,3 c 0,3 a 1,3 b 2,3 c 0,3 a 2,3 b 0,3 c 1,3 a 0,3 b 2,3 c 1,3 a 1,3 b 0,3 c 2,3 a 0,3 b 1,3 c 2,3 After substitution of these results into the nonlinear equation e 4, we get a quadratic equation for x 4, G e 4. solx1 1, 1, solx2 1, 1, solx3 1, 1 ;

13 GaussJacobi_07.nb Exponent G, x4, List 0, 1, 2 The coefficents of the quadratic equation are, h2d Coefficient G, x4 2 ; h1d Coefficient G, x4 ; h0d Simplify G h2d x4 2 h1d x4 ; Indeed, this is a quadratic equation, D h0d, x GPS N - Point Problem Let us now apply Gauss- Jacobi combinatorial solution. First, the subsets should be determined. In our case we have six satellites, which means six equations, and four unknowns. For illustration, we have the following numerical values, datan a , a , a , a , a , a , b , b , b , b , b , b , c , c , c , c , c , c , d , d , d , d , d , d ; m 6; n 4; The number of the subsets mn Binomial m, n 15 These subsets are, qs Partition Map 1 &, Flatten Subsets Range m, n, n 0, 1, 2, 3, 0, 1, 2, 4, 0, 1, 2, 5, 0, 1, 3, 4, 0, 1, 3, 5, 0, 1, 4, 5, 0, 2, 3, 4, 0, 2, 3, 5, 0, 2, 4, 5, 0, 3, 4, 5, 1, 2, 3, 4, 1, 2, 3, 5, 1, 2, 4, 5, 1, 3, 4, 5, 2, 3, 4, 5 The values of the indices start from zero in correspondence with the indices of the coefficients of the equations. Now, we shall utilize the symbolic solution of the GPS 4-points problem, namely the expressions of the coefficients of the quadratic equation (h 2, h 1, h 0 ). Therefore, we construct a new data list, datap similar to datan,which assignes the proper values to the coefficients of the equations of the 15 subsets, datap Table Map Select datan, MemberQ qs i, 1, 2 &. 1 0, 2 1, 3 2, 4 3 &, qs i, i, 1, mn ; For example, the fourth subset is indexed as qs 4 0, 1, 3, 4

14 14 GaussJacobi_07.nb and it has the proper data assignments, datap 4 a , a , a , a , b , b , b , b , c , c , c , c , d , d , d , d Now, we can employ the symbolic expressions of the coefficients of the quadratic equation for x 4, (h 2, h 1, h 0 ), which were developed in the previous example. These coefficients can be evaluated for all of the 15 combinatorial subsets, (H 2, H 1, H 0 ), H2 Map h2d. coeffsn. Flatten &, datap ; H1 Map h1d. coeffsn. Flatten &, datap ; H0 Map h0d. coeffsn. Flatten &, datap ; It is useful to display these coeffients, H210 Transpose H2, H1, H0 ; H210c Map. datan &, H210 ; TableForm SetPrecision H210c, mn This table indicates that the 10-th combination has a poor geometry, which fact can be also detected by computing its PDOP (Position Dilution of Precision). Then the 15 quadratic equations can be solved for x 4, X4 Map x4. Solve 1 x4 ^ 2 2 x4 3 0, x4 1, 1 &, H210c ; These values of x 4 can be substituted into the symbolic relations x 1 = g(x 4 ), x 2 = g(x 4 ) and x 3 = g(x 4 ) developed for GPS 4 - points problem, X1 MapThread x1. solx1 1, 1. coeffsn. Flatten 1. x4 2 &, datap, X4 ; X2 MapThread x2. solx2 1, 1. coeffsn. Flatten 1. x4 2 &, datap, X4 ; X3 MapThread x3. solx3 1, 1. coeffsn. Flatten 1. x4 2 &, datap, X4 ; Let us display these solutions for (x 1, x 2, x 3, x 4 ),

15 GaussJacobi_07.nb X Transpose X1, X2, X3, X4 ; TableForm SetPrecision X, In order to compute the weights of these solutions, one has to compute the square of the 15 Jacobi determinants. Each has size of 4 4, because of four equations and four variables in each subset. Starting with the general form of the i-th equation, e x1 a i 2 x2 b i 2 x3 c i 2 x4 d i 2 ; The partial derivatives are, de D e, x1, D e, x2, D e, x3, D e, x4 2 x1 a i, 2 x2 b i, 2 x3 c i, 2 x4 d i The numerical values of these partial derivatives will be computed at the corresponding combinatorial solutions, see the table above (variable X). Therefore, the weights, Π j, the square of the 15 Jacobi determinants are, Πs Table Map Det de. i 1, de. i 2, de. i 3, de. i 4. datan 2 &, qs j. x1 X j, 1, x2 X j, 2, x3 X j, 3, x4 X j, 4, j, 1, 15 ; The sum of these weights are sπ, sπs Apply Plus, Πs Then the weighted solution of the variable x i is X1s, X2s, X3s, X4s Map Πs. &, X1, X2, X3, X4 sπs; SetPrecision X1s, X2s, X3s, X4s, , , , which is very close to the direct numerical solution. This direct numerical solution can be computed as follows:

16 16 GaussJacobi_07.nb - the objective to be minimized is the sum of the residiums of the equations f Apply Plus, Table e 2. datan, i, 0, 5 Simplify x x x x x x x x x x x x x x x x x x x x x x x x4 2 2 To find the global minimum, the built-in function NMinimize is more promising than FindMinimum designed to search local minimum, or soln NMinimize f, x1, x2, x3, x , x , x , x , x SetPrecision soln 2, 10 x , x , x , x However, if we employ the norm of the distance error, instead of the residium of the equations, namely en d i x1 a i 2 x2 b i 2 x3 c i 2 x4 2 ; and then the objective is,

17 GaussJacobi_07.nb fn Apply Plus, Table en. datan, i, 0, x x x3 2 x x x x3 2 2 x x x x3 2 x x x x3 2 2 x x x x3 2 x x x x3 2 x4 The optimum will be somewhat different, solnn NMinimize fn, x1, x2, x3, x , x , x , x , x or SetPrecision solnn 2, 10 x , x , x , x We shall deal with problem in Chapter 11, in Part II Parallel Computing Similar to linear homotopy, the computation time of Gauss-Jacobi combinatorial method can be also naturally reduced by parallel computing! The solution of the equations of the different subsets can be achieved independently and therefore simultaneously. Concerning GPS N -point problem, the computation via parallel technique required more time than in normal way. AbsoluteTiming X4 Map x4. Solve 1 x4 ^ 2 2 x4 3 0, x4 1, 1 &, H210c ; 0., Null

18 18 GaussJacobi_07.nb AbsoluteTiming X4 ParallelMap x4. Solve 1 x4 ^ 2 2 x4 3 0, x4 1, 1 &, H210c ; , Null The explanation for this controversial fact is, that the computation time originally is already very short and using parallel mode, the communication time between the cores dominate and it needs more time than the net computation itself. However, when the normal computation time is longer, the parallel method can reduce the running time considerably, see e.g. Section