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1 y 4 u 1 y 4 d 1 y 4 l 1 y 4 r using barcode printing for none control to generate, create none image in none applications. Visal Basic .NET (3.8.4).

For gridpoi nts with measured values, on the other hand, a different (simple) equation is satis ed, (3.8.5) y0 D y0.

measured/ Note that these nonzero right-hand sides are what make an inhomogeneous, and therefore generally solvable, set of linear equations. We are not quite done, since we must provide special forms for the top, bottom, left, and right boundaries, and for the four corners. Homogeneous choices that embody natural boundary conditions (with no preferred function values) are y0 y0 y0 y0 y0 y0.

1 y 2 u 1 y 2 l 1 y 2 r 1 y 2 l 1 y 2 r 1 y 2 l 1 y 2 d 1 y 2 r 1 y 2 d 1 y 2 d 1 y 2 u 1 y 2 u D0 D0 D0 D0 D0 D0 (left and r none for none ight boundaries) (top and bottom boundaries) (top-left corner) (top-right corner) (bottom-left corner) (bottom-right corner) (3.8.6).

Since every gridpoint corresponds to exactly one of the equations in (3.8.4), (3.

8.5), or (3.8.

4), we have exactly as many equations as there are unknowns. If the grid is M by N , then there are MN of each. This can be quite a large number; but the equations are evidently very sparse.

We solve them by de ning a derived class from 2.7 s Linbcg base class. You can readily identify all the cases of equations (3.

8.4) (3.8.

6) in the code for atimes, below.. struct Lapl none for none ace_interp : Linbcg { Object for interpolating missing data in a matrix by solving Laplace s equation. Call constructor once, then solve one or more times (see text). MatDoub &mat; Int ii,jj; Int nn,iter; VecDoub b,y,mask; Laplace_interp(MatDoub_IO &matrix) : mat(matrix), ii(mat.

nrows()), jj(mat.ncols()), nn(ii*jj), iter(0), b(nn), y(nn), mask(nn) { Constructor. Values greater than 1.

e99 in the input matrix mat are deemed to be missing data. The matrix is not altered until solve is called. Int i,j,k; Doub vl = 0.

; interp laplace.h. 3. Interpolation and Extrapolation for (k=0;k& none none lt;nn;k++) { Fill the r.h.s.

vector, the initial guess, i = k/jj; and a mask of the missing data. j = k - i*jj; if (mat[i][j] < 1.e99) { b[k] = y[k] = vl = mat[i][j]; mask[k] = 1; } else { b[k] = 0.

; y[k] = vl; mask[k] = 0; } } } void asolve(VecDoub_I &b, VecDoub_O &x, const Int itrnsp); void atimes(VecDoub_I &x, VecDoub_O &r, const Int itrnsp); See de nitions below. These are the real algorithmic content. Doub solve(Doub tol=1.

e-6, Int itmax=-1) { Invoke Linbcg::solve with appropriate arguments. The default argument values will usually work, in which case this routine need be called only once. The original matrix mat is re lled with the interpolated solution.

Int i,j,k; Doub err; if (itmax <= 0) itmax = 2*MAX(ii,jj); Linbcg::solve(b,y,1,tol,itmax,iter,err); for (k=0,i=0;i<ii;i++) for (j=0;j<jj;j++) mat[i][j] = y[k++]; return err; } }; void Laplace_interp::asolve(VecDoub_I &b, VecDoub_O &x, const Int itrnsp) { Diagonal preconditioner. (Diagonal elements all unity.) Int i,n=b.

size(); for (i=0;i<n;i++) x[i] = b[i]; } void Laplace_interp::atimes(VecDoub_I &x, VecDoub_O &r, const Int itrnsp) { Sparse matrix, and matrix transpose, multiply. This routine embodies eqs. (3.

8.4), (3.8.

5), and (3.8.6).

Int i,j,k,n=r.size(),jjt,it; Doub del; for (k=0;k<n;k++) r[k] = 0.; for (k=0;k<n;k++) { i = k/jj; j = k - i*jj; if (mask[k]) { Measured point, eq.

(3.8.5).

r[k] += x[k]; } else if (i>0 && i<ii-1 && j>0 && j<jj-1) { Interior point, eq. (3.8.

4). if (itrnsp) { r[k] += x[k]; del = -0.25*x[k]; r[k-1] += del; r[k+1] += del; r[k-jj] += del; r[k+jj] += del; } else { r[k] = x[k] - 0.

25*(x[k-1]+x[k+1]+x[k+jj]+x[k-jj]); } } else if (i>0 && i<ii-1) { Left or right edge, eq. (3.8.

6). if (itrnsp) { r[k] += x[k]; del = -0.5*x[k]; r[k-jj] += del;.

3.8 Laplace Interpolation r[k+jj] += none none del; } else { r[k] = x[k] - 0.5*(x[k+jj]+x[k-jj]); } } else if (j>0 && j<jj-1) { if (itrnsp) { r[k] += x[k]; del = -0.5*x[k]; r[k-1] += del; r[k+1] += del; } else { r[k] = x[k] - 0.

5*(x[k+1]+x[k-1]); } } else { jjt = i==0 jj : -jj; it = j==0 1 : -1; if (itrnsp) { r[k] += x[k]; del = -0.5*x[k]; r[k+jjt] += del; r[k+it] += del; } else { r[k] = x[k] - 0.5*(x[k+jjt]+x[k+it]); } } } }.

Top or bott om edge, eq. (3.8.

6).. Corners, eq. (3.8.

6).. Usage is qu ite simple. Just ll a matrix with function values where they are known, and with 1.e99 where they are not; send the matrix to the constructor; and call the solve routine.

The missing values will be interpolated. The default arguments should serve for most cases..

Int m=...

,n none none =...

; MatDoub mat(m,n); ...

Laplace_interp mylaplace(mat); mylaplace.solve();. Quite decen t results are obtained for smooth functions on 300 300 matrices in which a random 10% of gridpoints have known function values, with 90% interpolated. However, since compute time scales as MN max .M; N / (that is, as the cube), this is not a method to use for much larger matrices, unless you break them up into overlapping tiles.

If you experience convergence dif culties, then you should call solve, with appropriate nondefault arguments, several times in succession, and look at the returned error estimate after each call returns..
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