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SVD Hanging?

Last post 09-28-2009 12:40 by tahansen. 4 replies.
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  • 11-15-2008 0:07

    SVD Hanging?

    I've been incorporating Math.NET Iridium into a little hobby project of mine, and I seem to have run into a bug. In the code, I have several matrices, and compute the SVD for each. The first two compute just fine, but when I hit the third matrix, the call to myMatrix.SingularValueDecomposition just gets stuck in an infinite loop. Also, trying to compute the condition number of the same matrix also causes the program to hang.

     

    Any ideas? I can provide the offending matrix if you'd like, but it is 120x120 so I didn't want to post it here without asking...

     

    P.S. -- As a feature request...how about adding methods to the Matrix and Vector classes to parse/output a Matlab-format string (e.g. [[1 2 3; 4 5 6; 7 8 9;]])?

    Filed under: ,
  • 11-15-2008 0:20 In reply to

    Re: SVD Hanging?

    Sorry, I should have also mentioned that I know there's not a problem with the matrix itself, since I can put it into MATLAB and compute the SVD just fine. However, I'm trying to do the whole project in .NET, so that doesn't help me ;)

  • 11-21-2008 17:18 In reply to

    Re: SVD Hanging?

    Seems there are two issues here:

    • The algorithm doesn't converge (nicely or at all) as it should. Matlab apparently converges.
    • The algorithm doesn't stop even though it does not converge.

    I noticed that although iterations are counted, this count is never checked in the current revision (there's actually a comment where the check should be placed), that's why it hangs instead of throwing an exception. I'll have to check that.

    Can you attach the failing matrix? So I can analyze the issue and build a unit test around it - if there are no data privacy/IP concerns.

  • 12-02-2008 6:07 In reply to

    Re: SVD Hanging?

    It's happening on several different matrices. I'm using the SVD to do some analysis on graphs; the matrix I'm including below is the Laplace matrix of a disconnected graph (specifically, a weighted digraph with 120 vertices). I've also included the singular values of the matrix as calculated by Matlab. NOTE: If you want to put the matrix into Matlab yourself, the whole thing needs to be on one line (you may have to remove the newline characters in notepad or similar first).

    Thanks for all your hard work! Perhaps one day I will be able to contribute some code back to the project...

     

    >> testmatrix = [34,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-29,0,0,0,0,0,0,0,0,0,0,0,0,0,0; 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0; 0,0,32,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-48,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0; 0,0,0,52,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-26,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0; 0,0,0,0,19,0,0,0,0,0,0,0,0,-31,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0; 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0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-34,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,27,0,0,0,0,0,0,0,0,0; 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,18,0,0,0,0,0,0,0,-53; 0,0,0,0,0,0,0,0,0,0,0,-26,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,24,0,0,0,0,0,0,0; 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,32,0,0,0,0,-46,0; 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-28,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,41,0,0,0,0,0; 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,44,0,-31,0,0; 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-33,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,32,0,0,0; 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-44,0,31,0,0; 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-32,0,0,0,0,46,0; 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-18,0,0,0,0,0,0,0,53; ];

    >> svd(testmatrix)

    ans =

      102.8591
       98.0000
       96.7471
       87.1436
       86.6487
       86.3134
       84.8646
       84.0238
       83.1986
       82.2192
       81.5843
       81.2158
       80.8084
       80.3243
       79.6994
       79.2465
       79.1581
       79.0569
       77.3692
       76.1183
       75.3127
       75.1665
       74.9667
       74.9667
       74.6726
       74.0945
       73.3348
       72.9932
       72.8011
       72.4431
       71.8471
       71.3863
       71.0634
       71.0634
       70.2282
       70.2140
       67.0075
       65.0077
       64.5600
       63.3877
       63.1981
       62.8013
       61.4003
       55.7136
       51.4198
       50.0400
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
        0.0000
             0
             0
             0
            
            
            


     

  • 09-28-2009 12:40 In reply to

    Re: SVD Hanging?

    I have the same problem with the code, and I have changed the code so it is checking for numbers of iter and the hypot is also changed to a new function "givens" which is used for the four cases.

    I have also changed the constructor interface so you should speify if you want U and Y , but this is just done to make the algorithm faster.

     

    I have included the updated code below

    #region Math.NET Iridium (LGPL) by Vermorel + Contributors
    // Math.NET Iridium, part of the Math.NET Project
    // http://mathnet.opensourcedotnet.info
    //
    // Copyright (c) 2004-2008, Joannes Vermorel, http://www.vermorel.com
    //
    // Contribution: The MathWorks and NIST [2000]
    //
    // This program is free software; you can redistribute it and/or modify
    // it under the terms of the GNU Lesser General Public License as published
    // by the Free Software Foundation; either version 2 of the License, or
    // (at your option) any later version.
    //
    // This program is distributed in the hope that it will be useful,
    // but WITHOUT ANY WARRANTY; without even the implied warranty of
    // MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
    // GNU Lesser General Public License for more details.
    //
    // You should have received a copy of the GNU Lesser General Public
    // License along with this program; if not, write to the Free Software
    // Foundation, Inc., 675 Mass Ave, Cambridge, MA 02139, USA.
    #endregion

    using System;
    using System.Diagnostics;

    namespace MathNet.Numerics.LinearAlgebra
    {
      /// <summary>Singular Value Decomposition.</summary>
      /// <remarks>
      /// <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>
      ///
      /// <p>The singular values, sigma[k] = S[k, k], are ordered so that
      /// sigma[0] >= sigma[1] >= ... >= sigma[n-1].</p>
      ///
      /// <p>The singular value decomposition always exists, so the constructor will
      /// never fail.  The matrix condition number and the effective numerical
      /// rank can be computed from this decomposition.</p>
      /// </remarks>
      [Serializable]
      public class SingularValueDecomposition
      {
      
        Matrix _u;
        Matrix _v;

        /// <summary>Array for internal storage of singular values.</summary>
        Vector _singular;

        /// <summary>Row dimensions.</summary>
        int m;

        /// <summary>Column dimensions.</summary>
        int n;

        /// <summary>Indicates whether all the results provided by the
        /// method or properties should be transposed.</summary>
        /// <remarks>
        /// (vermorel) The initial implementation was assuming that
        /// m &gt;= n, but in fact, it is easy to handle the case m &lt; n
        /// by transposing all the results.
        /// </remarks>
        bool transpose;

        OnDemandComputation<Matrix> _diagonalSingularValuesOnDemand;
        OnDemandComputation<int> _rankOnDemand;

        /// <summary>Construct the singular value decomposition.</summary>
        /// <remarks>Provides access to U, S and V.</remarks>
        /// <param name="Arg">Rectangular matrix</param>
        public
        SingularValueDecomposition(
            Matrix Arg,
            bool wantu,
            bool wantv
            )
        {
          transpose = (Arg.RowCount < Arg.ColumnCount);

          // Derived from LINPACK code.
          // Initialize.
          double[[ A;
          if (transpose)
          {
            // copy of internal data, independent of Arg
            A = Matrix.Transpose(Arg).GetArray();
            m = Arg.ColumnCount;
            n = Arg.RowCount;
          }
          else
          {
            A = Arg.CopyToJaggedArray();
            m = Arg.RowCount;
            n = Arg.ColumnCount;
          }

          int nu = Math.Min(m, n);
          double[ s = new double[Math.Min(m + 1, n)];
          double[[ U = Matrix.CreateMatrixData(m, nu);
          double[[ V = Matrix.CreateMatrixData(n, n);

          double[ e = new double[n];
          double[ work = new double[m];


          // 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] = Fn.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] = Fn.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 = Number.PositiveRelativeAccuracy;
          double tiny = double.Epsilon / eps;
          double snorm = 0;
          for (int ii = 0; ii < nu; ii++)
          {
            snorm = Math.Max(snorm, Math.Max(Math.Abs(s[ii]), Math.Abs(e[ii])));
          }
          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;
              }
              double test0 = Math.Abs(s[k]) + Math.Abs(s[k + 1]);
              double ztest0 = Math.Abs(e[k]);         
              if ((ztest0 <= eps * test0) || (ztest0 <= tiny)
                || (iter > 20 && ztest0 <= eps*snorm))
              {
                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]) : 0.0) + (ks != k + 1 ? Math.Abs(e[ks - 1]) : 0.0);
                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, cs, sn;
                    givens(s[j], f, out t, out cs, out sn);
                    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, cs, sn;
                    givens(s[j], f, out t, out cs, out sn);
                    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, cs, sn;
                    givens( f,g, out t, out cs, out sn);

                    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;
                      }
                    }

                    givens(f, g, out t, out cs, out sn);
                    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;
            }
          }

          // (vermorel) transposing the results if needed
          if (transpose)
          {
            // swaping U and V
            double[[ T = V;
            V = U;
            U = T;
          }

          _u = new Matrix(U);
          _v = new Matrix(V);
          _singular = new Vector(s);

          InitOnDemandComputations();
        }

        /// <summary>Gets the one-dimensional array of singular values.</summary>
        /// <returns>diagonal of S.</returns>
        public Vector SingularValues
        {
          get { return _singular; }
        }

        /// <summary>Get the diagonal matrix of singular values.</summary>
        public Matrix S
        {
          get
          {
            // TODO: bad name for this property
            return _diagonalSingularValuesOnDemand.Compute();
          }
        }

        /// <summary>Gets the left singular vectors (U matrix).</summary>
        public Matrix LeftSingularVectors
        {
          get { return _u; }
        }

        /// <summary>Gets the right singular vectors (V matrix).</summary>
        public Matrix RightSingularVectors
        {
          get { return _v; }
        }

        /// <summary>Two norm.</summary>
        /// <returns>max(S)</returns>
        public
        double
        Norm2()
        {
          // TODO (cdr, 2008-03-11): Change to property
          return _singular[0];
        }

        /// <summary>Two norm condition number.</summary>
        /// <returns>max(S)/min(S)</returns>
        public
        double
        Condition()
        {
          // TODO (cdr, 2008-03-11): Change to property
          return _singular[0] / _singular[Math.Min(m, n) - 1];
        }

        /// <summary>Effective numerical matrix rank - Number of nonnegligible singular values.</summary>
        public
        int
        Rank()
        {
          // TODO (cdr, 2008-03-11): Change to property
          return _rankOnDemand.Compute();
        }

        void
        InitOnDemandComputations()
        {
          _diagonalSingularValuesOnDemand = new OnDemandComputation<Matrix>(ComputeDiagonalSingularValues);
          _rankOnDemand = new OnDemandComputation<int>(ComputeRank);
        }

        Matrix
        ComputeDiagonalSingularValues()
        {
          return Matrix.Diagonal(_singular);
        }

        int
        ComputeRank()
        {
          double tol = Math.Max(m, n) * _singular[0] * Number.PositiveRelativeAccuracy;
          int r = 0;

          for (int i = 0; i < _singular.Length; i++)
          {
            if (_singular[i] > tol)
            {
              r++;
            }
          }

          return r;
        }
        private double dlapy2(double a, double b)
        {
          b = Math.Abs(b);
          a = Math.Abs(a);
          double y = a;
          if (b > a || double.IsNaN(b))
          {
            y = b;
            b = a;
          }
          if (!(Number.AlmostZero(y) || double.IsInfinity(y)))
          {
            double r = b / y;
            y = y * Math.Sqrt(1 + r * r);
          }
          return y;
        }

        private void givens(double a, double b, out double r, out double c, out double s)
        {
          r = dlapy2(a, b);
          if (r == 0)
          {
            c = 1;
            s = 0;
          }
          else
          {
            double rho;
            if (Math.Abs(a) > Math.Abs(b))
              rho = a;
            else
              rho = b;

            if (rho < 0)
              r = -r;

            c = a / r;
            s = b / r;
          }
        }
      }
    }
     

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