def newton_iterate(evaluate_fn, s, t):
    r"""Perform a Newton iteration.

    In this function, we assume that :math:`s` and :math:`t` are nonzero,
    this makes convergence easier to detect since "relative error" at
    ``0.0`` is not a useful measure.

    There are several tolerance / threshold quantities used below:

    * :math:`10` (:attr:`MAX_NEWTON_ITERATIONS`) iterations will be done before
      "giving up". This is based on the assumption that we are already starting
      near a root, so quadratic convergence should terminate quickly.
    * :math:`\tau = \frac{1}{4}` is used as the boundary between linear
      and superlinear convergence. So if the current error
      :math:`\|p_{n + 1} - p_n\|` is not smaller than :math:`\tau` times
      the previous error :math:`\|p_n - p_{n - 1}\|`, then convergence
      is considered to be linear at that point.
    * :math:`\frac{2}{3}` of all iterations must be converging linearly
      for convergence to be stopped (and moved to the next regime). This
      will only be checked after 4 or more updates have occurred.
    * :math:`\tau = 2^{-42}` (:attr:`NEWTON_ERROR_RATIO`) is used to
      determine that an update is sufficiently small to stop iterating. So if
      the error :math:`\|p_{n + 1} - p_n\|` smaller than :math:`\tau` times
      size of the term being updated :math:`\|p_n\|`, then we
      exit with the "correct" answer.

    It is assumed that ``evaluate_fn`` will use a Jacobian return value of
    :data:`None` to indicate that :math:`F(s, t)` is exactly ``0.0``. We
    **assume** that if the function evaluates to exactly ``0.0``, then we are
    at a solution. It is possible however, that badly parameterized curves
    can evaluate to exactly ``0.0`` for inputs that are relatively far away
    from a solution (see issue #21).

    Args:
        evaluate_fn (Callable[Tuple[float, float], tuple]): A callable
            which takes :math:`s` and :math:`t` and produces an evaluated
            function value and the Jacobian matrix.
        s (float): The (first) parameter where the iteration will start.
        t (float): The (second) parameter where the iteration will start.

    Returns:
        Tuple[bool, float, float]: The triple of

        * Flag indicating if the iteration converged.
        * The current :math:`s` value when the iteration stopped.
        * The current :math:`t` value when the iteration stopped.
    """
    # Several quantities will be tracked throughout the iteration:
    # * norm_update_prev: ||p{n}   - p{n-1}|| = ||dp{n-1}||
    # * norm_update     : ||p{n+1} - p{n}  || = ||dp{n}  ||
    # * linear_updates  : This is a count on the number of times that
    #                     ``dp{n}`` "looks like" ``dp{n-1}`` (i.e.
    #                     is within a constant factor of it).
    norm_update_prev = None
    norm_update = None
    linear_updates = 0  # Track the number of "linear" updates.
    current_s = s
    current_t = t
    for index in range(MAX_NEWTON_ITERATIONS):
        jacobian, func_val = evaluate_fn(current_s, current_t)
        if jacobian is None:
            return True, current_s, current_t

        singular, delta_s, delta_t = _py_helpers.solve2x2(
            jacobian, func_val[:, 0]
        )
        if singular:
            break

        norm_update_prev = norm_update
        norm_update = np.linalg.norm([delta_s, delta_t], ord=2)
        # If ||p{n} - p{n-1}|| > 0.25 ||p{n-1} - p{n-2}||, then that means
        # our convergence is acting linear at the current step.
        if index > 0 and norm_update > 0.25 * norm_update_prev:
            linear_updates += 1
        # If ``>=2/3`` of the updates have been linear, we are near a
        # non-simple root. (Make sure at least 5 updates have occurred.)
        if index >= 4 and 3 * linear_updates >= 2 * index:
            break

        # Determine the norm of the "old" solution before updating.
        norm_soln = np.linalg.norm([current_s, current_t], ord=2)
        current_s -= delta_s
        current_t -= delta_t
        if norm_update < NEWTON_ERROR_RATIO * norm_soln:
            return True, current_s, current_t

    return False, current_s, current_t
Exemple #2
0
    def _call_function_under_test(lhs, rhs):
        from bezier.hazmat import helpers

        return helpers.solve2x2(lhs, rhs)
def newton_refine(s, nodes1, t, nodes2):
    r"""Apply one step of 2D Newton's method.

    .. note::

       There is also a Fortran implementation of this function, which
       will be used if it can be built.

    We want to use Newton's method on the function

    .. math::

       F(s, t) = B_1(s) - B_2(t)

    to refine :math:`\left(s_{\ast}, t_{\ast}\right)`. Using this,
    and the Jacobian :math:`DF`, we "solve"

    .. math::

       \left[\begin{array}{c}
           0 \\ 0 \end{array}\right] \approx
           F\left(s_{\ast} + \Delta s, t_{\ast} + \Delta t\right) \approx
           F\left(s_{\ast}, t_{\ast}\right) +
           \left[\begin{array}{c c}
               B_1'\left(s_{\ast}\right) &
               - B_2'\left(t_{\ast}\right) \end{array}\right]
           \left[\begin{array}{c}
               \Delta s \\ \Delta t \end{array}\right]

    and refine with the component updates :math:`\Delta s` and
    :math:`\Delta t`.

    .. note::

       This implementation assumes the curves live in
       :math:`\mathbf{R}^2`.

    For example, the curves

    .. math::

        \begin{align*}
        B_1(s) &= \left[\begin{array}{c} 0 \\ 0 \end{array}\right] (1 - s)^2
            + \left[\begin{array}{c} 2 \\ 4 \end{array}\right] 2s(1 - s)
            + \left[\begin{array}{c} 4 \\ 0 \end{array}\right] s^2 \\
        B_2(t) &= \left[\begin{array}{c} 2 \\ 0 \end{array}\right] (1 - t)
            + \left[\begin{array}{c} 0 \\ 3 \end{array}\right] t
        \end{align*}

    intersect at the point
    :math:`B_1\left(\frac{1}{4}\right) = B_2\left(\frac{1}{2}\right) =
    \frac{1}{2} \left[\begin{array}{c} 2 \\ 3 \end{array}\right]`.

    However, starting from the wrong point we have

    .. math::

        \begin{align*}
        F\left(\frac{3}{8}, \frac{1}{4}\right) &= \frac{1}{8}
            \left[\begin{array}{c} 0 \\ 9 \end{array}\right] \\
        DF\left(\frac{3}{8}, \frac{1}{4}\right) &=
            \left[\begin{array}{c c}
            4 & 2 \\ 2 & -3 \end{array}\right] \\
        \Longrightarrow \left[\begin{array}{c} \Delta s \\ \Delta t
            \end{array}\right] &= \frac{9}{64} \left[\begin{array}{c}
            -1 \\ 2 \end{array}\right].
        \end{align*}

    .. image:: ../images/newton_refine1.png
       :align: center

    .. testsetup:: newton-refine1, newton-refine2, newton-refine3

       import numpy as np
       import bezier
       from bezier._py_intersection_helpers import newton_refine

       machine_eps = np.finfo(np.float64).eps

       def realroots(*coeffs):
           all_roots = np.roots(coeffs)
           return all_roots[np.where(all_roots.imag == 0.0)].real

    .. doctest:: newton-refine1

       >>> nodes1 = np.asfortranarray([
       ...     [0.0, 2.0, 4.0],
       ...     [0.0, 4.0, 0.0],
       ... ])
       >>> nodes2 = np.asfortranarray([
       ...     [2.0, 0.0],
       ...     [0.0, 3.0],
       ... ])
       >>> s, t = 0.375, 0.25
       >>> new_s, new_t = newton_refine(s, nodes1, t, nodes2)
       >>> 64.0 * (new_s - s)
       -9.0
       >>> 64.0 * (new_t - t)
       18.0

    .. testcleanup:: newton-refine1

       import make_images
       curve1 = bezier.Curve(nodes1, degree=2)
       curve2 = bezier.Curve(nodes2, degree=1)
       make_images.newton_refine1(s, new_s, curve1, t, new_t, curve2)

    For "typical" curves, we converge to a solution quadratically.
    This means that the number of correct digits doubles every
    iteration (until machine precision is reached).

    .. image:: ../images/newton_refine2.png
       :align: center

    .. doctest:: newton-refine2

       >>> nodes1 = np.asfortranarray([
       ...     [0.0, 0.25,  0.5, 0.75, 1.0],
       ...     [0.0, 2.0 , -2.0, 2.0 , 0.0],
       ... ])
       >>> nodes2 = np.asfortranarray([
       ...     [0.0, 0.25, 0.5, 0.75, 1.0],
       ...     [1.0, 0.5 , 0.5, 0.5 , 0.0],
       ... ])
       >>> # The expected intersection is the only real root of
       >>> # 28 s^3 - 30 s^2 + 9 s - 1.
       >>> expected, = realroots(28, -30, 9, -1)
       >>> s_vals = [0.625, None, None, None, None]
       >>> t = 0.625
       >>> np.log2(abs(expected - s_vals[0]))
       -4.399...
       >>> s_vals[1], t = newton_refine(s_vals[0], nodes1, t, nodes2)
       >>> np.log2(abs(expected - s_vals[1]))
       -7.901...
       >>> s_vals[2], t = newton_refine(s_vals[1], nodes1, t, nodes2)
       >>> np.log2(abs(expected - s_vals[2]))
       -16.010...
       >>> s_vals[3], t = newton_refine(s_vals[2], nodes1, t, nodes2)
       >>> np.log2(abs(expected - s_vals[3]))
       -32.110...
       >>> s_vals[4], t = newton_refine(s_vals[3], nodes1, t, nodes2)
       >>> np.allclose(s_vals[4], expected, rtol=6 * machine_eps, atol=0.0)
       True

    .. testcleanup:: newton-refine2

       import make_images
       curve1 = bezier.Curve(nodes1, degree=4)
       curve2 = bezier.Curve(nodes2, degree=4)
       make_images.newton_refine2(s_vals, curve1, curve2)

    However, when the intersection occurs at a point of tangency,
    the convergence becomes linear. This means that the number of
    correct digits added each iteration is roughly constant.

    .. image:: ../images/newton_refine3.png
       :align: center

    .. doctest:: newton-refine3

       >>> nodes1 = np.asfortranarray([
       ...     [0.0, 0.5, 1.0],
       ...     [0.0, 1.0, 0.0],
       ... ])
       >>> nodes2 = np.asfortranarray([
       ...     [0.0, 1.0],
       ...     [0.5, 0.5],
       ... ])
       >>> expected = 0.5
       >>> s_vals = [0.375, None, None, None, None, None]
       >>> t = 0.375
       >>> np.log2(abs(expected - s_vals[0]))
       -3.0
       >>> s_vals[1], t = newton_refine(s_vals[0], nodes1, t, nodes2)
       >>> np.log2(abs(expected - s_vals[1]))
       -4.0
       >>> s_vals[2], t = newton_refine(s_vals[1], nodes1, t, nodes2)
       >>> np.log2(abs(expected - s_vals[2]))
       -5.0
       >>> s_vals[3], t = newton_refine(s_vals[2], nodes1, t, nodes2)
       >>> np.log2(abs(expected - s_vals[3]))
       -6.0
       >>> s_vals[4], t = newton_refine(s_vals[3], nodes1, t, nodes2)
       >>> np.log2(abs(expected - s_vals[4]))
       -7.0
       >>> s_vals[5], t = newton_refine(s_vals[4], nodes1, t, nodes2)
       >>> np.log2(abs(expected - s_vals[5]))
       -8.0

    .. testcleanup:: newton-refine3

       import make_images
       curve1 = bezier.Curve(nodes1, degree=2)
       curve2 = bezier.Curve(nodes2, degree=1)
       make_images.newton_refine3(s_vals, curve1, curve2)

    Unfortunately, the process terminates with an error that is not close
    to machine precision :math:`\varepsilon` when
    :math:`\Delta s = \Delta t = 0`.

    .. testsetup:: newton-refine3-continued

       import numpy as np
       import bezier
       from bezier._py_intersection_helpers import newton_refine

       nodes1 = np.asfortranarray([
           [0.0, 0.5, 1.0],
           [0.0, 1.0, 0.0],
       ])
       nodes2 = np.asfortranarray([
           [0.0, 1.0],
           [0.5, 0.5],
       ])

    .. doctest:: newton-refine3-continued

       >>> s1 = t1 = 0.5 - 0.5**27
       >>> np.log2(0.5 - s1)
       -27.0
       >>> s2, t2 = newton_refine(s1, nodes1, t1, nodes2)
       >>> s2 == t2
       True
       >>> np.log2(0.5 - s2)
       -28.0
       >>> s3, t3 = newton_refine(s2, nodes1, t2, nodes2)
       >>> s3 == t3 == s2
       True

    Due to round-off near the point of tangency, the final error
    resembles :math:`\sqrt{\varepsilon}` rather than machine
    precision as expected.

    .. note::

       The following is not implemented in this function. It's just
       an exploration on how the shortcomings might be addressed.

    However, this can be overcome. At the point of tangency, we want
    :math:`B_1'(s) \parallel B_2'(t)`. This can be checked numerically via

    .. math::

        B_1'(s) \times B_2'(t) = 0.

    For the last example (the one that converges linearly), this is

    .. math::

        0 = \left[\begin{array}{c} 1 \\ 2 - 4s \end{array}\right] \times
            \left[\begin{array}{c} 1 \\ 0 \end{array}\right] = 4 s - 2.

    With this, we can modify Newton's method to find a zero of the
    over-determined system

    .. math::

        G(s, t) = \left[\begin{array}{c} B_0(s) - B_1(t) \\
            B_1'(s) \times B_2'(t) \end{array}\right] =
            \left[\begin{array}{c} s - t \\ 2 s (1 - s) - \frac{1}{2} \\
            4 s - 2\end{array}\right].

    Since :math:`DG` is :math:`3 \times 2`, we can't invert it. However,
    we can find a least-squares solution:

    .. math::

        \left(DG^T DG\right) \left[\begin{array}{c}
            \Delta s \\ \Delta t \end{array}\right] = -DG^T G.

    This only works if :math:`DG` has full rank. In this case, it does
    since the submatrix containing the first and last rows has rank two:

    .. math::

        DG = \left[\begin{array}{c c} 1 & -1 \\
            2 - 4 s & 0 \\
            4 & 0 \end{array}\right].

    Though this avoids a singular system, the normal equations have a
    condition number that is the square of the condition number of the matrix.

    Starting from :math:`s = t = \frac{3}{8}` as above:

    .. testsetup:: newton-refine4

       import numpy as np
       from bezier.hazmat import helpers

       def modified_update(s, t):
           minus_G = np.asfortranarray([
               [t - s],
               [0.5 - 2.0 * s * (1.0 - s)],
               [2.0 - 4.0 * s],
           ])
           DG = np.asfortranarray([
               [1.0, -1.0],
               [2.0 - 4.0 * s, 0.0],
               [4.0, 0.0],
           ])
           DG_t = np.asfortranarray(DG.T)

           LHS = helpers.matrix_product(DG_t, DG)
           RHS = helpers.matrix_product(DG_t, minus_G)
           delta_params = np.linalg.solve(LHS, RHS)
           delta_s, delta_t = delta_params.flatten()
           return s + delta_s, t + delta_t

    .. doctest:: newton-refine4

       >>> s0, t0 = 0.375, 0.375
       >>> np.log2(0.5 - s0)
       -3.0
       >>> s1, t1 = modified_update(s0, t0)
       >>> s1 == t1
       True
       >>> 1040.0 * s1
       519.0
       >>> np.log2(0.5 - s1)
       -10.022...
       >>> s2, t2 = modified_update(s1, t1)
       >>> s2 == t2
       True
       >>> np.log2(0.5 - s2)
       -31.067...
       >>> s3, t3 = modified_update(s2, t2)
       >>> s3 == t3 == 0.5
       True

    Args:
        s (float): Parameter of a near-intersection along the first curve.
        nodes1 (numpy.ndarray): Nodes of first curve forming intersection.
        t (float): Parameter of a near-intersection along the second curve.
        nodes2 (numpy.ndarray): Nodes of second curve forming intersection.

    Returns:
        Tuple[float, float]: The refined parameters from a single Newton
        step.

    Raises:
        ValueError: If the Jacobian is singular at ``(s, t)``.
    """
    # NOTE: We form -F(s, t) since we want to solve -DF^{-1} F(s, t).
    func_val = _py_curve_helpers.evaluate_multi(
        nodes2, np.asfortranarray([t])
    ) - _py_curve_helpers.evaluate_multi(nodes1, np.asfortranarray([s]))
    if np.all(func_val == 0.0):
        # No refinement is needed.
        return s, t

    # NOTE: This assumes the curves are 2D.
    jac_mat = np.empty((2, 2), order="F")
    jac_mat[:, :1] = _py_curve_helpers.evaluate_hodograph(s, nodes1)
    jac_mat[:, 1:] = -_py_curve_helpers.evaluate_hodograph(t, nodes2)
    # Solve the system.
    singular, delta_s, delta_t = _py_helpers.solve2x2(jac_mat, func_val[:, 0])
    if singular:
        raise ValueError("Jacobian is singular.")

    return s + delta_s, t + delta_t