Source code for par_trans.utils.expm_multiply_np

"""Compute the action of the matrix exponential. This module is taken
from scipy, the one difference is we add the option use_frag_31 to see if the
more difficult algorithm requiring the estimation of a higher norm makes much
of a difference. This more difficult estimate is one of the reasons the algorithm
is still not adapted for JAX. We use this module to show it is sufficient to use only the 1-norm in our case.
"""
from warnings import warn

import numpy as np
import numpy.linalg as la

import scipy.linalg
import scipy.sparse.linalg
from scipy.linalg._decomp_qr import qr
from scipy.sparse._sputils import is_pydata_spmatrix
from scipy.sparse.linalg import aslinearoperator
from scipy.sparse.linalg._interface import IdentityOperator
from scipy.sparse.linalg._onenormest import onenormest

__all__ = ['expm_multiply']


def traceest(A, m3, seed=None):
    """Estimate `np.trace(A)` using `3*m3` matrix-vector products.

    The result is not deterministic.

    Parameters
    ----------
    A : LinearOperator
        Linear operator whose trace will be estimated. Has to be square.
    m3 : int
        Number of matrix-vector products divided by 3 used to estimate the
        trace.
    seed : optional
        Seed for `numpy.random.default_rng`.
        Can be provided to obtain deterministic results.

    Returns
    -------
    trace : LinearOperator.dtype
        Estimate of the trace

    Notes
    -----
    This is the Hutch++ algorithm given in [1]_.

    References
    ----------
    .. [1] Meyer, Raphael A., Cameron Musco, Christopher Musco, and David P.
       Woodruff. "Hutch++: Optimal Stochastic Trace Estimation." In Symposium
       on Simplicity in Algorithms (SOSA), pp. 142-155. Society for Industrial
       and Applied Mathematics, 2021
       https://doi.org/10.1137/1.9781611976496.16

    """
    rng = np.random.default_rng(seed)
    if len(A.shape) != 2 or A.shape[-1] != A.shape[-2]:
        raise ValueError("Expected A to be like a square matrix.")
    n = A.shape[-1]
    S = rng.choice([-1.0, +1.0], [n, m3])
    Q, _ = qr(A.matmat(S), overwrite_a=True, mode='economic')
    trQAQ = np.trace(Q.conj().T @ A.matmat(Q))
    G = rng.choice([-1, +1], [n, m3])
    right = G - Q@(Q.conj().T @ G)
    trGAG = np.trace(right.conj().T @ A.matmat(right))
    return trQAQ + trGAG/m3


def _ident_like(A):
    # A compatibility function which should eventually disappear.
    if scipy.sparse.issparse(A):
        # Creates a sparse matrix in dia format
        out = scipy.sparse.eye(A.shape[0], A.shape[1], dtype=A.dtype)
        if isinstance(A, scipy.sparse.spmatrix):
            return out.asformat(A.format)
        return scipy.sparse.dia_array(out).asformat(A.format)
    elif is_pydata_spmatrix(A):
        import sparse
        return sparse.eye(A.shape[0], A.shape[1], dtype=A.dtype)
    elif isinstance(A, scipy.sparse.linalg.LinearOperator):
        return IdentityOperator(A.shape, dtype=A.dtype)
    else:
        return np.eye(A.shape[0], A.shape[1], dtype=A.dtype)


[docs] def expm_multiply(A, B, start=None, stop=None, num=None, endpoint=None, traceA=None, use_frag_31=True): """ Compute the action of the matrix exponential of A on B, using the algorithm described in [1]_ and [2]_ . Parameters ---------- A : transposable linear operator The operator whose exponential is of interest. B : ndarray The matrix or vector to be multiplied by the matrix exponential of A. start : scalar, optional The starting time point of the sequence. stop : scalar, optional The end time point of the sequence, unless `endpoint` is set to False. In that case, the sequence consists of all but the last of ``num + 1`` evenly spaced time points, so that `stop` is excluded. Note that the step size changes when `endpoint` is False. num : int, optional Number of time points to use. endpoint : bool, optional If True, `stop` is the last time point. Otherwise, it is not included. traceA : scalar, optional Trace of `A`. If not given the trace is estimated for linear operators, or calculated exactly for sparse matrices. It is used to precondition `A`, thus an approximate trace is acceptable. For linear operators, `traceA` should be provided to ensure performance as the estimation is not guaranteed to be reliable for all cases. use_frag_31 : bool, optional Indicates if we use high_p or not. Returns ------- expm_A_B : ndarray The result of the action :math:`e^{t_k A} B`. Warns ----- UserWarning If `A` is a linear operator and ``traceA=None`` (default). Notes ----- The optional arguments defining the sequence of evenly spaced time points are compatible with the arguments of `numpy.linspace`. The output ndarray shape is somewhat complicated so I explain it here. The ndim of the output could be either 1, 2, or 3. It would be 1 if you are computing the expm action on a single vector at a single time point. It would be 2 if you are computing the expm action on a vector at multiple time points, or if you are computing the expm action on a matrix at a single time point. It would be 3 if you want the action on a matrix with multiple columns at multiple time points. If multiple time points are requested, expm_A_B[0] will always be the action of the expm at the first time point, regardless of whether the action is on a vector or a matrix. References ---------- .. [1] Awad H. Al-Mohy and Nicholas J. Higham (2011) "Computing the Action of the Matrix Exponential, with an Application to Exponential Integrators." SIAM Journal on Scientific Computing, 33 (2). pp. 488-511. ISSN 1064-8275 http://eprints.ma.man.ac.uk/1591/ .. [2] Nicholas J. Higham and Awad H. Al-Mohy (2010) "Computing Matrix Functions." Acta Numerica, 19. 159-208. ISSN 0962-4929 http://eprints.ma.man.ac.uk/1451/ Examples -------- >>> import numpy as np >>> from scipy.sparse import csc_matrix >>> from scipy.sparse.linalg import expm, expm_multiply >>> A = csc_matrix([[1, 0], [0, 1]]) >>> A.toarray() array([[1, 0], [0, 1]], dtype=int64) >>> B = np.array([np.exp(-1.), np.exp(-2.)]) >>> B array([ 0.36787944, 0.13533528]) >>> expm_multiply(A, B, start=1, stop=2, num=3, endpoint=True) array([[ 1. , 0.36787944], [ 1.64872127, 0.60653066], [ 2.71828183, 1. ]]) >>> expm(A).dot(B) # Verify 1st timestep array([ 1. , 0.36787944]) >>> expm(1.5*A).dot(B) # Verify 2nd timestep array([ 1.64872127, 0.60653066]) >>> expm(2*A).dot(B) # Verify 3rd timestep array([ 2.71828183, 1. ]) """ if all(arg is None for arg in (start, stop, num, endpoint)): X = _expm_multiply_simple(A, B, traceA=traceA, use_frag_31=use_frag_31) else: X, status = _expm_multiply_interval(A, B, start, stop, num, endpoint, traceA=traceA) return X
def _expm_multiply_simple(A, B, t=1.0, traceA=None, balance=False, use_frag_31=True): """ Compute the action of the matrix exponential at a single time point. Parameters ---------- A : transposable linear operator The operator whose exponential is of interest. B : ndarray The matrix to be multiplied by the matrix exponential of A. t : float A time point. traceA : scalar, optional Trace of `A`. If not given the trace is estimated for linear operators, or calculated exactly for sparse matrices. It is used to precondition `A`, thus an approximate trace is acceptable balance : bool Indicates whether or not to apply balancing. Returns ------- F : ndarray :math:`e^{t A} B` Notes ----- This is algorithm (3.2) in Al-Mohy and Higham (2011). """ if balance: raise NotImplementedError if len(A.shape) != 2 or A.shape[0] != A.shape[1]: raise ValueError('expected A to be like a square matrix') if A.shape[1] != B.shape[0]: raise ValueError(f'shapes of matrices A {A.shape} and B {B.shape}' ' are incompatible') ident = _ident_like(A) is_linear_operator = isinstance(A, scipy.sparse.linalg.LinearOperator) n = A.shape[0] if len(B.shape) == 1: n0 = 1 elif len(B.shape) == 2: n0 = B.shape[1] else: raise ValueError('expected B to be like a matrix or a vector') u_d = 2**-53 tol = u_d if traceA is None: if is_linear_operator: warn("Trace of LinearOperator not available, it will be estimated." " Provide `traceA` to ensure performance.", stacklevel=3) # m3=1 is bit arbitrary choice, a more accurate trace (larger m3) might # speed up exponential calculation, but trace estimation is more costly traceA = traceest(A, m3=1) if is_linear_operator else A.trace() mu = traceA / float(n) A = A - mu * ident A_1_norm = onenormest(A) if is_linear_operator else la.norm(A, 1) if t*A_1_norm == 0: m_star, s = 0, 1 else: ell = 2 norm_info = LazyOperatorNormInfo(t*A, A_1_norm=t*A_1_norm, ell=ell) m_star, s = _fragment_3_1(norm_info, n0, tol, ell=ell, use_frag_31=use_frag_31) return _expm_multiply_simple_core(A, B, t, mu, m_star, s, tol, balance) def _expm_multiply_simple_core(A, B, t, mu, m_star, s, tol=None, balance=False): """ A helper function. """ if balance: raise NotImplementedError if tol is None: u_d = 2 ** -53 tol = u_d F = B eta = np.exp(t*mu / float(s)) for i in range(s): c1 = la.norm(B) for j in range(m_star): coeff = t / float(s*(j+1)) B = coeff * A.dot(B) c2 = la.norm(B) F = F + B if c1 + c2 <= tol * la.norm(F): break c1 = c2 F = eta * F B = F return F # This table helps to compute bounds. # They seem to have been difficult to calculate, involving symbolic # manipulation of equations, followed by numerical root finding. _theta = { # The first 30 values are from table A.3 of Computing Matrix Functions. 1: 2.29e-16, 2: 2.58e-8, 3: 1.39e-5, 4: 3.40e-4, 5: 2.40e-3, 6: 9.07e-3, 7: 2.38e-2, 8: 5.00e-2, 9: 8.96e-2, 10: 1.44e-1, # 11 11: 2.14e-1, 12: 3.00e-1, 13: 4.00e-1, 14: 5.14e-1, 15: 6.41e-1, 16: 7.81e-1, 17: 9.31e-1, 18: 1.09, 19: 1.26, 20: 1.44, # 21 21: 1.62, 22: 1.82, 23: 2.01, 24: 2.22, 25: 2.43, 26: 2.64, 27: 2.86, 28: 3.08, 29: 3.31, 30: 3.54, # The rest are from table 3.1 of # Computing the Action of the Matrix Exponential. 35: 4.7, 40: 6.0, 45: 7.2, 50: 8.5, 55: 9.9, } def _onenormest_matrix_power(A, p, t=2, itmax=5, compute_v=False, compute_w=False): """ Efficiently estimate the 1-norm of A^p. Parameters ---------- A : ndarray Matrix whose 1-norm of a power is to be computed. p : int Non-negative integer power. t : int, optional A positive parameter controlling the tradeoff between accuracy versus time and memory usage. Larger values take longer and use more memory but give more accurate output. itmax : int, optional Use at most this many iterations. compute_v : bool, optional Request a norm-maximizing linear operator input vector if True. compute_w : bool, optional Request a norm-maximizing linear operator output vector if True. Returns ------- est : float An underestimate of the 1-norm of the sparse matrix. v : ndarray, optional The vector such that ||Av||_1 == est*||v||_1. It can be thought of as an input to the linear operator that gives an output with particularly large norm. w : ndarray, optional The vector Av which has relatively large 1-norm. It can be thought of as an output of the linear operator that is relatively large in norm compared to the input. """ #XXX Eventually turn this into an API function in the _onenormest module, #XXX and remove its underscore, #XXX but wait until expm_multiply goes into scipy. from scipy.sparse.linalg._onenormest import onenormest return onenormest(aslinearoperator(A) ** p) class LazyOperatorNormInfo: """ Information about an operator is lazily computed. The information includes the exact 1-norm of the operator, in addition to estimates of 1-norms of powers of the operator. This uses the notation of Computing the Action (2011). This class is specialized enough to probably not be of general interest outside of this module. """ def __init__(self, A, A_1_norm=None, ell=2, scale=1): """ Provide the operator and some norm-related information. Parameters ---------- A : linear operator The operator of interest. A_1_norm : float, optional The exact 1-norm of A. ell : int, optional A technical parameter controlling norm estimation quality. scale : int, optional If specified, return the norms of scale*A instead of A. """ self._A = A self._A_1_norm = A_1_norm self._ell = ell self._d = {} self._scale = scale def set_scale(self,scale): """ Set the scale parameter. """ self._scale = scale def onenorm(self): """ Compute the exact 1-norm. """ if self._A_1_norm is None: self._A_1_norm = la.norm(self._A, 1) return self._scale*self._A_1_norm def d(self, p): """ Lazily estimate :math:`d_p(A) ~= || A^p ||^(1/p)` where :math:`||.||` is the 1-norm. """ if p not in self._d: est = _onenormest_matrix_power(self._A, p, self._ell) self._d[p] = est ** (1.0 / p) return self._scale*self._d[p] def alpha(self, p): """ Lazily compute max(d(p), d(p+1)). """ return max(self.d(p), self.d(p+1)) def _compute_cost_div_m(m, p, norm_info): """ A helper function for computing bounds. This is equation (3.10). It measures cost in terms of the number of required matrix products. Parameters ---------- m : int A valid key of _theta. p : int A matrix power. norm_info : LazyOperatorNormInfo Information about 1-norms of related operators. Returns ------- cost_div_m : int Required number of matrix products divided by m. """ return int(np.ceil(norm_info.alpha(p) / _theta[m])) def _compute_p_max(m_max): """ Compute the largest positive integer p such that p*(p-1) <= m_max + 1. Do this in a slightly dumb way, but safe and not too slow. Parameters ---------- m_max : int A count related to bounds. """ sqrt_m_max = np.sqrt(m_max) p_low = int(np.floor(sqrt_m_max)) p_high = int(np.ceil(sqrt_m_max + 1)) return max(p for p in range(p_low, p_high+1) if p*(p-1) <= m_max + 1) def _fragment_3_1(norm_info, n0, tol, m_max=55, ell=2, use_frag_31=True): """ A helper function for the _expm_multiply_* functions. Parameters ---------- norm_info : LazyOperatorNormInfo Information about norms of certain linear operators of interest. n0 : int Number of columns in the _expm_multiply_* B matrix. tol : float Expected to be :math:`2^{-24}` for single precision or :math:`2^{-53}` for double precision. m_max : int A value related to a bound. ell : int The number of columns used in the 1-norm approximation. This is usually taken to be small, maybe between 1 and 5. Returns ------- best_m : int Related to bounds for error control. best_s : int Amount of scaling. Notes ----- This is code fragment (3.1) in Al-Mohy and Higham (2011). The discussion of default values for m_max and ell is given between the definitions of equation (3.11) and the definition of equation (3.12). """ if ell < 1: raise ValueError('expected ell to be a positive integer') best_m = None best_s = None if (not use_frag_31) or _condition_3_13(norm_info.onenorm(), n0, m_max, ell): for m, theta in _theta.items(): s = int(np.ceil(norm_info.onenorm() / theta)) if best_m is None or m * s < best_m * best_s: best_m = m best_s = s else: # Equation (3.11). for p in range(2, _compute_p_max(m_max) + 1): for m in range(p*(p-1)-1, m_max+1): if m in _theta: s = _compute_cost_div_m(m, p, norm_info) if best_m is None or m * s < best_m * best_s: best_m = m best_s = s best_s = max(best_s, 1) return best_m, best_s def _condition_3_13(A_1_norm, n0, m_max, ell): """ A helper function for the _expm_multiply_* functions. Parameters ---------- A_1_norm : float The precomputed 1-norm of A. n0 : int Number of columns in the _expm_multiply_* B matrix. m_max : int A value related to a bound. ell : int The number of columns used in the 1-norm approximation. This is usually taken to be small, maybe between 1 and 5. Returns ------- value : bool Indicates whether or not the condition has been met. Notes ----- This is condition (3.13) in Al-Mohy and Higham (2011). """ # This is the rhs of equation (3.12). p_max = _compute_p_max(m_max) a = 2 * ell * p_max * (p_max + 3) # Evaluate the condition (3.13). b = _theta[m_max] / float(n0 * m_max) return A_1_norm <= a * b def _expm_multiply_interval(A, B, start=None, stop=None, num=None, endpoint=None, traceA=None, balance=False, status_only=False): """ Compute the action of the matrix exponential at multiple time points. Parameters ---------- A : transposable linear operator The operator whose exponential is of interest. B : ndarray The matrix to be multiplied by the matrix exponential of A. start : scalar, optional The starting time point of the sequence. stop : scalar, optional The end time point of the sequence, unless `endpoint` is set to False. In that case, the sequence consists of all but the last of ``num + 1`` evenly spaced time points, so that `stop` is excluded. Note that the step size changes when `endpoint` is False. num : int, optional Number of time points to use. traceA : scalar, optional Trace of `A`. If not given the trace is estimated for linear operators, or calculated exactly for sparse matrices. It is used to precondition `A`, thus an approximate trace is acceptable endpoint : bool, optional If True, `stop` is the last time point. Otherwise, it is not included. balance : bool Indicates whether or not to apply balancing. status_only : bool A flag that is set to True for some debugging and testing operations. Returns ------- F : ndarray :math:`e^{t_k A} B` status : int An integer status for testing and debugging. Notes ----- This is algorithm (5.2) in Al-Mohy and Higham (2011). There seems to be a typo, where line 15 of the algorithm should be moved to line 6.5 (between lines 6 and 7). """ if balance: raise NotImplementedError if len(A.shape) != 2 or A.shape[0] != A.shape[1]: raise ValueError('expected A to be like a square matrix') if A.shape[1] != B.shape[0]: raise ValueError(f'shapes of matrices A {A.shape} and B {B.shape}' ' are incompatible') ident = _ident_like(A) is_linear_operator = isinstance(A, scipy.sparse.linalg.LinearOperator) n = A.shape[0] if len(B.shape) == 1: n0 = 1 elif len(B.shape) == 2: n0 = B.shape[1] else: raise ValueError('expected B to be like a matrix or a vector') u_d = 2**-53 tol = u_d if traceA is None: if is_linear_operator: warn("Trace of LinearOperator not available, it will be estimated." " Provide `traceA` to ensure performance.", stacklevel=3) # m3=5 is bit arbitrary choice, a more accurate trace (larger m3) might # speed up exponential calculation, but trace estimation is also costly # an educated guess would need to consider the number of time points traceA = traceest(A, m3=5) if is_linear_operator else A.trace() mu = traceA / float(n) # Get the linspace samples, attempting to preserve the linspace defaults. linspace_kwargs = {'retstep': True} if num is not None: linspace_kwargs['num'] = num if endpoint is not None: linspace_kwargs['endpoint'] = endpoint samples, step = np.linspace(start, stop, **linspace_kwargs) # Convert the linspace output to the notation used by the publication. nsamples = len(samples) if nsamples < 2: raise ValueError('at least two time points are required') q = nsamples - 1 h = step t_0 = samples[0] t_q = samples[q] # Define the output ndarray. # Use an ndim=3 shape, such that the last two indices # are the ones that may be involved in level 3 BLAS operations. X_shape = (nsamples,) + B.shape X = np.empty(X_shape, dtype=np.result_type(A.dtype, B.dtype, float)) t = t_q - t_0 A = A - mu * ident A_1_norm = onenormest(A) if is_linear_operator else la.norm(A, 1) ell = 2 norm_info = LazyOperatorNormInfo(t*A, A_1_norm=t*A_1_norm, ell=ell) if t*A_1_norm == 0: m_star, s = 0, 1 else: m_star, s = _fragment_3_1(norm_info, n0, tol, ell=ell) # Compute the expm action up to the initial time point. X[0] = _expm_multiply_simple_core(A, B, t_0, mu, m_star, s) # Compute the expm action at the rest of the time points. if q <= s: if status_only: return 0 else: return _expm_multiply_interval_core_0(A, X, h, mu, q, norm_info, tol, ell,n0) elif not (q % s): if status_only: return 1 else: return _expm_multiply_interval_core_1(A, X, h, mu, m_star, s, q, tol) elif (q % s): if status_only: return 2 else: return _expm_multiply_interval_core_2(A, X, h, mu, m_star, s, q, tol) else: raise Exception('internal error') def _expm_multiply_interval_core_0(A, X, h, mu, q, norm_info, tol, ell, n0): """ A helper function, for the case q <= s. """ # Compute the new values of m_star and s which should be applied # over intervals of size t/q if norm_info.onenorm() == 0: m_star, s = 0, 1 else: norm_info.set_scale(1./q) m_star, s = _fragment_3_1(norm_info, n0, tol, ell=ell) norm_info.set_scale(1) for k in range(q): X[k+1] = _expm_multiply_simple_core(A, X[k], h, mu, m_star, s) return X, 0 def _expm_multiply_interval_core_1(A, X, h, mu, m_star, s, q, tol): """ A helper function, for the case q > s and q % s == 0. """ d = q // s input_shape = X.shape[1:] K_shape = (m_star + 1, ) + input_shape K = np.empty(K_shape, dtype=X.dtype) for i in range(s): Z = X[i*d] K[0] = Z high_p = 0 for k in range(1, d+1): F = K[0] c1 = la.norm(F) for p in range(1, m_star+1): if p > high_p: K[p] = h * A.dot(K[p-1]) / float(p) coeff = float(pow(k, p)) F = F + coeff * K[p] inf_norm_K_p_1 = la.norm(K[p]) c2 = coeff * inf_norm_K_p_1 if c1 + c2 <= tol * la.norm(F): break c1 = c2 X[k + i*d] = np.exp(k*h*mu) * F return X, 1 def _expm_multiply_interval_core_2(A, X, h, mu, m_star, s, q, tol): """ A helper function, for the case q > s and q % s > 0. """ d = q // s j = q // d r = q - d * j input_shape = X.shape[1:] K_shape = (m_star + 1, ) + input_shape K = np.empty(K_shape, dtype=X.dtype) for i in range(j + 1): Z = X[i*d] K[0] = Z high_p = 0 if i < j: effective_d = d else: effective_d = r for k in range(1, effective_d+1): F = K[0] c1 = la.norm(F) for p in range(1, m_star+1): if p == high_p + 1: K[p] = h * A.dot(K[p-1]) / float(p) high_p = p coeff = float(pow(k, p)) F = F + coeff * K[p] inf_norm_K_p_1 = la.norm(K[p]) c2 = coeff * inf_norm_K_p_1 if c1 + c2 <= tol * la.norm(F): break c1 = c2 X[k + i*d] = np.exp(k*h*mu) * F return X, 2