package math.matrix;

import java.io.Serializable;

import math.Maths;

/**
 * Singular Value Decomposition.
 * <p>
 * For an m-by-n matrix A with m &ge; 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;<sub>k</sub> = S<sub>kk</sub>, are ordered so that
 * &sigma;<sub>1</sub> &ge; &sigma;<sub>2</sub> &ge; ... &ge; &sigma;<sub>n</sub>.
 * <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.
 * @author Daniel Weschke
 */
public class SingularValueDecomposition implements Serializable {

   /**
	 * UID
	 */
	private static final long serialVersionUID = -4816488865370991490L;

/**
    * Array for internal storage of U.
    * @serial internal storage of U.
    */
   private double[][] U;

   /**
    * Array for internal storage of V.
    * @serial internal storage of V.
    */
   private double[][] V;
   
   /**
    * Array for internal storage of singular values.
    * @serial internal storage of singular values.
    */
   private double[] s;

   /**
    * Row dimensions.
    * @serial row dimension.
    */
   private int m;

   /**
    * Column dimensions.
    * @serial column dimension.
    */
   private int n;

/* ------------------------
   Constructor
 * ------------------------ */

   /**
    * Construct the singular value decomposition Structure to access U, S and V.
    * @param Arg rectangular matrix
    */
   public SingularValueDecomposition(Matrix Arg){
      int i, j, k;
      // Derived from LINPACK code.
      // Initialize.
      double[][] A = Arg.getCopy();
      m = Arg.getM();
      n = Arg.getN();
      /* TODO: Apparently the failing cases are only a proper subset of (m<n), 
	   * so let's not throw error.  Correct fix to come later?
       * if (m<n) {
	   * throw new IllegalArgumentException("SVD only works for m >= n"); }
       */
      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));
      for(k=0; k<Math.max(nct,nrt); k++){
         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(i=k; i<m; i++)
               s[k] = Maths.hypot(s[k],A[i][k]);
            if(s[k] != 0.0){
               if(A[k][k] < 0.0)
                  s[k] = -s[k];
               for(i=k; i<m; i++)
                  A[i][k] /= s[k];
               A[k][k] += 1.0;
            }
            s[k] = -s[k];
         }
         for(j=k+1; j<n; j++){
            if((k < nct) & (s[k] != 0.0)){
               // Apply the transformation.
               double t = 0;
               for(i=k; i<m; i++)
                  t += A[i][k]*A[i][j];
               t = -t/A[k][k];
               for(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(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(i=k+1; i<n; i++)
               e[k] = Maths.hypot(e[k],e[i]);
            if(e[k] != 0.0){
               if(e[k+1] < 0.0)
                  e[k] = -e[k];
               for(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(i=k+1; i<m; i++) 
                  work[i] = 0.0;
               for(j=k+1; j<n; j++)
                  for(i=k+1; i<m; i++)
                     work[i] += e[j]*A[i][j];
               for(j=k+1; j<n; j++){
                  double t = -e[j]/e[k+1];
                  for(i=k+1; i<m; i++)
                     A[i][j] += t*work[i];
               }
            }
            if(wantv){
               // Place the transformation in V for subsequent
               // back multiplication.
               for(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){
         for(j=nct; j<nu; j++){
            for(i=0; i<m; i++)
               U[i][j] = 0.0;
            U[j][j] = 1.0;
         }
         for(k=nct-1; k>=0; k--){
            if(s[k] != 0.0) {
               for(j=k+1; j<nu; j++){
                  double t = 0;
                  for(i=k; i < m; i++)
                     t += U[i][k]*U[i][j];
                  t = -t/U[k][k];
                  for(i=k; i<m; i++)
                     U[i][j] += t*U[i][k];
               }
               for(i=k; i < m; i++)
                  U[i][k] = -U[i][k];
               U[k][k] = 1.0 + U[k][k];
               for(i=0; i<k-1; i++)
                  U[i][k] = 0.0;
            } else {
               for(i=0; i < m; i++)
                  U[i][k] = 0.0;
               U[k][k] = 1.0;
            }
         }
      }
      // If required, generate V.
      if(wantv)
         for(k=n-1; k>=0; k--){
            if((k < nrt) & (e[k] != 0.0))
               for(j=k+1; j<nu; j++){
                  double t = 0;
                  for(i=k+1; i < n; i++)
                     t += V[i][k]*V[i][j];
                  t = -t/V[k+1][k];
                  for(i=k+1; i<n; i++)
                     V[i][j] += t*V[i][k];
               }
            for(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);
      double tiny = Math.pow(2.0,-966.0);
      while(p > 0){
         int 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]) <=
                  tiny + 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]) <= tiny + 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(j=p-2; j>=k; j--){
                  double t = Maths.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(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(j=k; j<p; j++){
                  double t = Maths.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(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(j=k; j<p-1; j++){
                  double t = Maths.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(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 = Maths.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(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(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(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(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;
         }
      }
   }

   /**
    * @return the left singular vectors U
    */
   public Matrix getU(){
      return new Matrix(U);
   }

   /**
    * @return the right singular vectors V
    */
   public Matrix getV(){
      return new Matrix(V);
   }

   /**
    * @return the one-dimensional array of singular values, diagonal of S.
    */
   public double[] getSingularValues(){
      return s;
   }

   /**
    * @return the diagonal matrix of singular values S
    */
   public Matrix getS(){
      int i, j;
      Matrix S = new Matrix(n,n);
      for(i=0; i<n; i++){
         for(j=0; j<n; j++)
            S.set(i,j, 0.0);
         S.set(i,i, this.s[i]);
      }
      return S;
   }

   /**
    * 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 i, r = 0;
      for(i=0; i<s.length; i++)
         if(s[i] > tol)
            r++;
      return r;
   }
}
