content: package com.actelion.research.calc;
  
  
     /** Singular Value Decomposition.
     <P>
     For an m-by-n matrix A with m >= n, the singular value decomposition is
     an m-by-n orthogonal matrix U, an n-by-n diagonal matrix S, and
     an n-by-n orthogonal matrix V so that A = U*S*V'.
     <P>
     The singular values, sigma[k] = S[k][k], are ordered so that
     sigma[0] >= sigma[1] >= ... >= sigma[n-1].
     <P>
     The singular value decompostion always exists, so the constructor will
     never fail.  The matrix condition number and the effective numerical
     rank can be computed from this decomposition.
     */
  
  public class SingularValueDecomposition implements java.io.Serializable {
      static final long serialVersionUID = 0x20110324;
  
  /* ------------------------
     Class variables
   * ------------------------ */
  
     /** Arrays for internal storage of U and V.
     @serial internal storage of U.
     @serial internal storage of V.
     */
     private double[][] U, V;
  
     /** Array for internal storage of singular values.
     @serial internal storage of singular values.
     */
     private double[] s;
  
     /** Row and column dimensions.
     @serial row dimension.
     @serial column dimension.
     */
     private int m, n;
  
  /* ------------------------
     Constructor
   * ------------------------ */
  
     /** Construct the singular value decomposition
     @param A    Rectangular matrix
     @return     Structure to access U, S and V.
     */
  
     public SingularValueDecomposition (double[][] A, ProgressListener progressListener, ThreadMaster threadMaster) {
  
  	  // Derived from LINPACK code.
  	  // Initialize.
  	  m = A.length;
  	  n = A[0].length;
  	  int nu = Math.min(m,n);
  	  s = new double [Math.min(m+1,n)];
  	  U = new double [m][nu];
  	  V = new double [n][n];
  	  double[] e = new double [n];
  	  double[] work = new double [m];
  	  boolean wantu = true;
  	  boolean wantv = true;
  
  	  // Reduce A to bidiagonal form, storing the diagonal elements
  	  // in s and the super-diagonal elements in e.
  
  	  int nct = Math.min(m-1,n);
  	  int nrt = Math.max(0,Math.min(n-2,m));
  
  	  if (progressListener != null)
  		progressListener.startProgress("Initializing SVD...", 0, Math.max(nct,nrt));
  
  	  for (int k = 0; k < Math.max(nct,nrt); k++) {
  		if (progressListener != null)
  		  progressListener.updateProgress(k);
  		if (threadMaster != null && threadMaster.threadMustDie())
  		  return;
  
  		 if (k < nct) {
  
  			// Compute the transformation for the k-th column and
  			// place the k-th diagonal in s[k].
  			// Compute 2-norm of k-th column without under/overflow.
  			s[k] = 0;
  			for (int i = k; i < m; i++) {
  			   s[k] = hypot(s[k],A[i][k]);
  			}
  			if (s[k] != 0.0) {
  			   if (A[k][k] < 0.0) {
  				  s[k] = -s[k];
  			   }
  			   for (int i = k; i < m; i++) {
  				  A[i][k] /= s[k];
  			   }
  			   A[k][k] += 1.0;
  			}
  			s[k] = -s[k];
  		 }
  		 for (int j = k+1; j < n; j++) {
  			if ((k < nct) & (s[k] != 0.0))  {
  
  			// Apply the transformation.
  
  			   double t = 0;
  			   for (int i = k; i < m; i++) {
  				  t += A[i][k]*A[i][j];
  			   }
  			   t = -t/A[k][k];
  			   for (int i = k; i < m; i++) {
  				  A[i][j] += t*A[i][k];
  			   }
  			}
  
  			// Place the k-th row of A into e for the
  			// subsequent calculation of the row transformation.
  
  			e[j] = A[k][j];
  		 }
  		 if (wantu & (k < nct)) {
  
  			// Place the transformation in U for subsequent back
  			// multiplication.
  
  			for (int i = k; i < m; i++) {
  			   U[i][k] = A[i][k];
  			}
  		 }
  		 if (k < nrt) {
  
  			// Compute the k-th row transformation and place the
  			// k-th super-diagonal in e[k].
  			// Compute 2-norm without under/overflow.
  			e[k] = 0;
  			for (int i = k+1; i < n; i++) {
  			   e[k] = hypot(e[k],e[i]);
  			}
  			if (e[k] != 0.0) {
  			   if (e[k+1] < 0.0) {
  				  e[k] = -e[k];
  			   }
  			   for (int i = k+1; i < n; i++) {
  				  e[i] /= e[k];
  			   }
  			   e[k+1] += 1.0;
  			}
  			e[k] = -e[k];
  			if ((k+1 < m) & (e[k] != 0.0)) {
  
  			// Apply the transformation.
  
  			   for (int i = k+1; i < m; i++) {
  				  work[i] = 0.0;
  			   }
  			   for (int j = k+1; j < n; j++) {
  				  for (int i = k+1; i < m; i++) {
  					 work[i] += e[j]*A[i][j];
  				  }
  			   }
  			   for (int j = k+1; j < n; j++) {
  				  double t = -e[j]/e[k+1];
  				  for (int i = k+1; i < m; i++) {
  					 A[i][j] += t*work[i];
  				  }
  			   }
  			}
  			if (wantv) {
  
  			// Place the transformation in V for subsequent
  			// back multiplication.
  
  			   for (int i = k+1; i < n; i++) {
  				  V[i][k] = e[i];
  			   }
  			}
  		 }
  	  }
  
  	  // Set up the final bidiagonal matrix or order p.
  
  	  int p = Math.min(n,m+1);
  	  if (nct < n) {
  		 s[nct] = A[nct][nct];
  	  }
  	  if (m < p) {
  		 s[p-1] = 0.0;
  	  }
  	  if (nrt+1 < p) {
  		 e[nrt] = A[nrt][p-1];
  	  }
  	  e[p-1] = 0.0;
  
  	  // If required, generate U.
  
  	  if (wantu) {
  		if (progressListener != null)
  		  progressListener.startProgress("generating eigenvalues...", 0, nct);
  
  		  for (int j = nct; j < nu; j++) {
  			for (int i = 0; i < m; i++) {
  			   U[i][j] = 0.0;
  			}
  			U[j][j] = 1.0;
  		 }
  		 for (int k = nct-1; k >= 0; k--) {
  			if (progressListener != null)
  			  progressListener.updateProgress(nct-k);
  			if (threadMaster != null && threadMaster.threadMustDie())
  			  return;
  
  			if (s[k] != 0.0) {
  			   for (int j = k+1; j < nu; j++) {
  				  double t = 0;
  				  for (int i = k; i < m; i++) {
  					 t += U[i][k]*U[i][j];
  				  }
  				  t = -t/U[k][k];
  				  for (int i = k; i < m; i++) {
  					 U[i][j] += t*U[i][k];
  				  }
  			   }
  			   for (int i = k; i < m; i++ ) {
  				  U[i][k] = -U[i][k];
  			   }
  			   U[k][k] = 1.0 + U[k][k];
  			   for (int i = 0; i < k-1; i++) {
  				  U[i][k] = 0.0;
  			   }
  			} else {
  			   for (int i = 0; i < m; i++) {
  				  U[i][k] = 0.0;
  			   }
  			   U[k][k] = 1.0;
  			}
  		 }
  	  }
  
  	  // If required, generate V.
  
  	  if (wantv) {
  		if (progressListener != null)
  		  progressListener.startProgress("generating eigenvectors...", 0, n);
  
  		 for (int k = n-1; k >= 0; k--) {
  			if (progressListener != null)
  			  progressListener.updateProgress(n-k);
  			if (threadMaster != null && threadMaster.threadMustDie())
  			  return;
  
  			if ((k < nrt) & (e[k] != 0.0)) {
  			   for (int j = k+1; j < nu; j++) {
  				  double t = 0;
  				  for (int i = k+1; i < n; i++) {
  					 t += V[i][k]*V[i][j];
  				  }
  				  t = -t/V[k+1][k];
  				  for (int i = k+1; i < n; i++) {
  					 V[i][j] += t*V[i][k];
  				  }
  			   }
  			}
  			for (int i = 0; i < n; i++) {
  			   V[i][k] = 0.0;
  			}
  			V[k][k] = 1.0;
  		 }
  	  }
  
  	  // Main iteration loop for the singular values.
  
  	  int pp = p-1;
  	  int iter = 0;
  	  double eps = Math.pow(2.0,-52.0);
  
  	  if (progressListener != null)
  		progressListener.startProgress("locating negligible elements...", 0, pp+1);
  
  	  while (p > 0) {
  		if (progressListener != null)
  		  progressListener.updateProgress(pp-p);
  		if (threadMaster != null && threadMaster.threadMustDie())
  		  return;
  
  		 int k,kase;
  
  		 // Here is where a test for too many iterations would go.
  
  		 // This section of the program inspects for
  		 // negligible elements in the s and e arrays.  On
  		 // completion the variables kase and k are set as follows.
  
  		 // kase = 1     if s(p) and e[k-1] are negligible and k<p
  		 // kase = 2     if s(k) is negligible and k<p
  		 // kase = 3     if e[k-1] is negligible, k<p, and
  		 //              s(k), ..., s(p) are not negligible (qr step).
  		 // kase = 4     if e(p-1) is negligible (convergence).
  
  		 for (k = p-2; k >= -1; k--) {
  			if (k == -1) {
  			   break;
  			}
  			if (Math.abs(e[k]) <= eps*(Math.abs(s[k]) + Math.abs(s[k+1]))) {
  			   e[k] = 0.0;
  			   break;
  			}
  		 }
  		 if (k == p-2) {
  			kase = 4;
  		 } else {
  			int ks;
  			for (ks = p-1; ks >= k; ks--) {
  			   if (ks == k) {
  				  break;
  			   }
  			   double t = (ks != p ? Math.abs(e[ks]) : 0.) +
  						  (ks != k+1 ? Math.abs(e[ks-1]) : 0.);
  			   if (Math.abs(s[ks]) <= eps*t)  {
  				  s[ks] = 0.0;
  				  break;
  			   }
  			}
  			if (ks == k) {
  			   kase = 3;
  			} else if (ks == p-1) {
  			   kase = 1;
  			} else {
  			   kase = 2;
  			   k = ks;
  			}
  		 }
  		 k++;
  
  		 // Perform the task indicated by kase.
  
  		 switch (kase) {
  
  			// Deflate negligible s(p).
  
  			case 1: {
  			   double f = e[p-2];
  			   e[p-2] = 0.0;
  			   for (int j = p-2; j >= k; j--) {
  				  double t = hypot(s[j],f);
  				  double cs = s[j]/t;
  				  double sn = f/t;
  				  s[j] = t;
  				  if (j != k) {
  					 f = -sn*e[j-1];
  					 e[j-1] = cs*e[j-1];
  				  }
  				  if (wantv) {
  					 for (int i = 0; i < n; i++) {
  						t = cs*V[i][j] + sn*V[i][p-1];
  						V[i][p-1] = -sn*V[i][j] + cs*V[i][p-1];
  						V[i][j] = t;
  					 }
  				  }
  			   }
  			}
  			break;
  
  			// Split at negligible s(k).
  
  			case 2: {
  			   double f = e[k-1];
  			   e[k-1] = 0.0;
  			   for (int j = k; j < p; j++) {
  				  double t = hypot(s[j],f);
  				  double cs = s[j]/t;
  				  double sn = f/t;
  				  s[j] = t;
  				  f = -sn*e[j];
  				  e[j] = cs*e[j];
  				  if (wantu) {
  					 for (int i = 0; i < m; i++) {
  						t = cs*U[i][j] + sn*U[i][k-1];
  						U[i][k-1] = -sn*U[i][j] + cs*U[i][k-1];
  						U[i][j] = t;
  					 }
  				  }
  			   }
  			}
  			break;
  
  			// Perform one qr step.
  
  			case 3: {
  
  			   // Calculate the shift.
  
  			   double scale = Math.max(Math.max(Math.max(Math.max(
  					   Math.abs(s[p-1]),Math.abs(s[p-2])),Math.abs(e[p-2])),
  					   Math.abs(s[k])),Math.abs(e[k]));
  			   double sp = s[p-1]/scale;
  			   double spm1 = s[p-2]/scale;
  			   double epm1 = e[p-2]/scale;
  			   double sk = s[k]/scale;
  			   double ek = e[k]/scale;
  			   double b = ((spm1 + sp)*(spm1 - sp) + epm1*epm1)/2.0;
  			   double c = (sp*epm1)*(sp*epm1);
  			   double shift = 0.0;
  			   if ((b != 0.0) | (c != 0.0)) {
  				  shift = Math.sqrt(b*b + c);
  				  if (b < 0.0) {
  					 shift = -shift;
  				  }
  				  shift = c/(b + shift);
  			   }
  			   double f = (sk + sp)*(sk - sp) + shift;
  			   double g = sk*ek;
  
  			   // Chase zeros.
  
  			   for (int j = k; j < p-1; j++) {
  				  double t = hypot(f,g);
  				  double cs = f/t;
  				  double sn = g/t;
  				  if (j != k) {
  					 e[j-1] = t;
  				  }
  				  f = cs*s[j] + sn*e[j];
  				  e[j] = cs*e[j] - sn*s[j];
  				  g = sn*s[j+1];
  				  s[j+1] = cs*s[j+1];
  				  if (wantv) {
  					 for (int i = 0; i < n; i++) {
  						t = cs*V[i][j] + sn*V[i][j+1];
  						V[i][j+1] = -sn*V[i][j] + cs*V[i][j+1];
  						V[i][j] = t;
  					 }
  				  }
  				  t = hypot(f,g);
  				  cs = f/t;
  				  sn = g/t;
  				  s[j] = t;
  				  f = cs*e[j] + sn*s[j+1];
  				  s[j+1] = -sn*e[j] + cs*s[j+1];
  				  g = sn*e[j+1];
  				  e[j+1] = cs*e[j+1];
  				  if (wantu && (j < m-1)) {
  					 for (int i = 0; i < m; i++) {
  						t = cs*U[i][j] + sn*U[i][j+1];
  						U[i][j+1] = -sn*U[i][j] + cs*U[i][j+1];
  						U[i][j] = t;
  					 }
  				  }
  			   }
  			   e[p-2] = f;
  			   iter = iter + 1;
  			}
  			break;
  
  			// Convergence.
  
  			case 4: {
  
  			   // Make the singular values positive.
  
  			   if (s[k] <= 0.0) {
  				  s[k] = (s[k] < 0.0 ? -s[k] : 0.0);
  				  if (wantv) {
  					 for (int i = 0; i <= pp; i++) {
  						V[i][k] = -V[i][k];
  					 }
  				  }
  			   }
  
  			   // Order the singular values.
  
  			   while (k < pp) {
  				  if (s[k] >= s[k+1]) {
  					 break;
  				  }
  				  double t = s[k];
  				  s[k] = s[k+1];
  				  s[k+1] = t;
  				  if (wantv && (k < n-1)) {
  					 for (int i = 0; i < n; i++) {
  						t = V[i][k+1]; V[i][k+1] = V[i][k]; V[i][k] = t;
  					 }
  				  }
  				  if (wantu && (k < m-1)) {
  					 for (int i = 0; i < m; i++) {
  						t = U[i][k+1]; U[i][k+1] = U[i][k]; U[i][k] = t;
  					 }
  				  }
  				  k++;
  			   }
  			   iter = 0;
  			   p--;
  			}
  			break;
  		 }
  	  }
     }
  
  /* ------------------------
     Public Methods
   * ------------------------ */
  
     /** Return the left singular vectors
     @return     U
     */
  
     public double[][] getU () {
  //	   return new Matrix(U);
  	   return U;
     }
  
     /** Return the right singular vectors
     @return     V
     */
  
     public double[][] getV () {
  //	   return new Matrix(V,n,n);
  	   return V;
     }
  
     /** Return the one-dimensional array of singular values
     @return     diagonal of S.
     */
  
     public double[] getSingularValues () {
  	  return s;
     }
  
     /** Return the diagonal matrix of singular values
     @return     S
     */
  
  /* public Matrix getS () {
  	  Matrix X = new Matrix(n,n);
  	  for (int i = 0; i < n; i++) {
  		 for (int j = 0; j < n; j++) {
  			X.set(i,j,0.0);
  		 }
  		 X.set(i,i,this.s[i]);
  	  }
  	  return X;
     }*/
  
     /** Two norm
     @return     max(S)
     */
  
     public double norm2 () {
  	  return s[0];
     }
  
     /** Two norm condition number
     @return     max(S)/min(S)
     */
  
     public double cond () {
  	  return s[0]/s[Math.min(m,n)-1];
     }
  
     /** Effective numerical matrix rank
     @return     Number of nonnegligible singular values.
     */
  
     public int rank () {
  	  double eps = Math.pow(2.0,-52.0);
  	  double tol = Math.max(m,n)*s[0]*eps;
  	  int r = 0;
  	  for (int i = 0; i < s.length; i++) {
  		 if (s[i] > tol) {
  			r++;
  		 }
  	  }
  	  return r;
     }
  
     public static double hypot(double a, double b) {
        double r;
        if (Math.abs(a) > Math.abs(b)) {
           r = b/a;
           r = Math.abs(a)*Math.sqrt(1+r*r);
        } else if (b != 0) {
           r = a/b;
           r = Math.abs(b)*Math.sqrt(1+r*r);
        } else {
           r = 0.0;
        }
        return r;
     }
  }
  