コード例 #1
0
 /**
  * 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;
 }
コード例 #2
0
  /**
   * Construct the singular value decomposition Structure to access U, S and V.
   *
   * @param Arg 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;
      }
    }
  }