package de.lmu.ifi.dbs.elki.math.linearalgebra;

/*
 This file is part of ELKI:
 Environment for Developing KDD-Applications Supported by Index-Structures

 Copyright (C) 2015
 Ludwig-Maximilians-Universität München
 Lehr- und Forschungseinheit für Datenbanksysteme
 ELKI Development Team

 This program is free software: you can redistribute it and/or modify
 it under the terms of the GNU Affero General Public License as published by
 the Free Software Foundation, either version 3 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 Affero General Public License for more details.

 You should have received a copy of the GNU Affero General Public License
 along with this program.  If not, see <http://www.gnu.org/licenses/>.
 */

import de.lmu.ifi.dbs.elki.math.MathUtil;

/**
 * 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 decomposition always exists, so the constructor will never
 * fail. The matrix condition number and the effective numerical rank can be
 * computed from this decomposition.
 * 
 * @author Arthur Zimek
 * @since 0.2
 *
 * @apiviz.uses Matrix - - transforms
 */
public class SingularValueDecomposition {
  private static final double EPS = Math.pow(2.0, -52.0);

  /**
   * 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;

  /**
   * Construct the singular value decomposition
   * 
   * @param Arg Rectangular matrix
   */
  public SingularValueDecomposition(Matrix Arg) {
    this(Arg.getArrayRef());
  }

  /**
   * Constructor.
   *
   * @param Arg Rectangular input matrix
   */
  public SingularValueDecomposition(double[][] Arg) {
    double[][] A = new Matrix(Arg).getArrayCopy();
    this.m = A.length;
    this.n = A[0].length;
    // Derived from LINPACK code.
    // Initialize.
    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] = MathUtil.fastHypot(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] = MathUtil.fastHypot(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 = EPS;
    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]) <= 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 = MathUtil.fastHypot(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 = MathUtil.fastHypot(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 = MathUtil.fastHypot(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 = MathUtil.fastHypot(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;
      }
    }
  }

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

  /**
   * Return the right singular vectors
   * 
   * @return V
   */
  public Matrix getV() {
    return new Matrix(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);
    double[][] S = X.getArrayRef();
    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
   * 
   * @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 non-negligible singular values.
   */
  public int rank() {
    double eps = EPS;
    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;
  }
}