SingularValueDecomposition.java

package Jama;
import Jama.util.*;

   

Singular Value Decomposition.

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 = USV'.

The singular values, sigma[k] = S[k][k], are ordered so that sigma[0] >= sigma[1] >= ... >= sigma[n-1].

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 {

/* ------------------------
   Class variables
 * ------------------------ */

   

Arrays for internal storage of U and V.

Serial:  internal storage of U., 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., column dimension.


   private int m, n;

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

   

Construct the singular value decomposition

Returns:  Structure to access U, S and V.
Parameters: 
- Rectangular matrix



   public SingularValueDecomposition (Matrix Arg) {

      // Derived from LINPACK code.
      // Initialize.
      double[][] A = Arg.getArrayCopy();
      m = Arg.getRowDimension();
      n = Arg.getColumnDimension();

      /* 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("Jama 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 (int 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 (int 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 (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] = Maths.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) {
         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 (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) {
         for (int k = n-1; k >= 0; k--) {
            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);
      double tiny = Math.pow(2.0,-966.0);
      while (p > 0) {
         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]) <=
                  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 (int 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 (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 = 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 (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 = 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 (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 = 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 (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

Returns:  U



   public Matrix getU () {
      return new Matrix(U,m,Math.min(m+1,n));
   }

   

Return the right singular vectors

Returns:  V



   public Matrix getV () {
      return new Matrix(V,n,n);
   }

   

Return the one-dimensional array of singular values

Returns:  diagonal of S.



   public double[] getSingularValues () {
      return s;
   }

   

Return the diagonal matrix of singular values

Returns:  S



   public Matrix getS () {
      Matrix X = new Matrix(n,n);
      double[][] S = X.getArray();
      for (int i = 0; i < n; i++) {
         for (int j = 0; j < n; j++) {
            S[i][j] = 0.0;
         }
         S[i][i] = this.s[i];
      }
      return X;
   }

   

Two norm

Returns:  max(S)



   public double norm2 () {
      return s[0];
   }

   

Two norm condition number

Returns:  max(S)/min(S)



   public double cond () {
      return s[0]/s[Math.min(m,n)-1];
   }

   

Effective numerical matrix rank

Returns:  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;
   }
}


This file was generated on Sun Jun 13 10:31:45 PDT 2010 from file SingularValueDecomposition.java
by the ilog.language.tools.Hilite Java tool written by Hassan Aït-Kaci