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# -*- coding: utf-8 -*-
"""
Bregman projections for regularized OT with GPU
"""
# Author: Remi Flamary <remi.flamary@unice.fr>
# Leo Gautheron <https://github.com/aje>
#
# License: MIT License
import cupy as np # np used for matrix computation
import cupy as cp # cp used for cupy specific operations
from . import utils
def sinkhorn_knopp(a, b, M, reg, numItermax=1000, stopThr=1e-9,
verbose=False, log=False, to_numpy=True, **kwargs):
"""
Solve the entropic regularization optimal transport on GPU
If the input matrix are in numpy format, they will be uploaded to the
GPU first which can incur significant time overhead.
The function solves the following optimization problem:
.. math::
\gamma = arg\min_\gamma <\gamma,M>_F + reg\cdot\Omega(\gamma)
s.t. \gamma 1 = a
\gamma^T 1= b
\gamma\geq 0
where :
- M is the (ns,nt) metric cost matrix
- :math:`\Omega` is the entropic regularization term :math:`\Omega(\gamma)=\sum_{i,j} \gamma_{i,j}\log(\gamma_{i,j})`
- a and b are source and target weights (sum to 1)
The algorithm used for solving the problem is the Sinkhorn-Knopp matrix scaling algorithm as proposed in [2]_
Parameters
----------
a : np.ndarray (ns,)
samples weights in the source domain
b : np.ndarray (nt,) or np.ndarray (nt,nbb)
samples in the target domain, compute sinkhorn with multiple targets
and fixed M if b is a matrix (return OT loss + dual variables in log)
M : np.ndarray (ns,nt)
loss matrix
reg : float
Regularization term >0
numItermax : int, optional
Max number of iterations
stopThr : float, optional
Stop threshol on error (>0)
verbose : bool, optional
Print information along iterations
log : bool, optional
record log if True
to_numpy : boolean, optional (default True)
If true convert back the GPU array result to numpy format.
Returns
-------
gamma : (ns x nt) ndarray
Optimal transportation matrix for the given parameters
log : dict
log dictionary return only if log==True in parameters
Examples
--------
>>> import ot
>>> a=[.5,.5]
>>> b=[.5,.5]
>>> M=[[0.,1.],[1.,0.]]
>>> ot.sinkhorn(a,b,M,1)
array([[ 0.36552929, 0.13447071],
[ 0.13447071, 0.36552929]])
References
----------
.. [2] M. Cuturi, Sinkhorn Distances : Lightspeed Computation of Optimal Transport, Advances in Neural Information Processing Systems (NIPS) 26, 2013
See Also
--------
ot.lp.emd : Unregularized OT
ot.optim.cg : General regularized OT
"""
a = cp.asarray(a)
b = cp.asarray(b)
M = cp.asarray(M)
if len(a) == 0:
a = np.ones((M.shape[0],)) / M.shape[0]
if len(b) == 0:
b = np.ones((M.shape[1],)) / M.shape[1]
# init data
Nini = len(a)
Nfin = len(b)
if len(b.shape) > 1:
nbb = b.shape[1]
else:
nbb = 0
if log:
log = {'err': []}
# we assume that no distances are null except those of the diagonal of
# distances
if nbb:
u = np.ones((Nini, nbb)) / Nini
v = np.ones((Nfin, nbb)) / Nfin
else:
u = np.ones(Nini) / Nini
v = np.ones(Nfin) / Nfin
# print(reg)
# Next 3 lines equivalent to K= np.exp(-M/reg), but faster to compute
K = np.empty(M.shape, dtype=M.dtype)
np.divide(M, -reg, out=K)
np.exp(K, out=K)
# print(np.min(K))
tmp2 = np.empty(b.shape, dtype=M.dtype)
Kp = (1 / a).reshape(-1, 1) * K
cpt = 0
err = 1
while (err > stopThr and cpt < numItermax):
uprev = u
vprev = v
KtransposeU = np.dot(K.T, u)
v = np.divide(b, KtransposeU)
u = 1. / np.dot(Kp, v)
if (np.any(KtransposeU == 0) or
np.any(np.isnan(u)) or np.any(np.isnan(v)) or
np.any(np.isinf(u)) or np.any(np.isinf(v))):
# we have reached the machine precision
# come back to previous solution and quit loop
print('Warning: numerical errors at iteration', cpt)
u = uprev
v = vprev
break
if cpt % 10 == 0:
# we can speed up the process by checking for the error only all
# the 10th iterations
if nbb:
err = np.sum((u - uprev)**2) / np.sum((u)**2) + \
np.sum((v - vprev)**2) / np.sum((v)**2)
else:
# compute right marginal tmp2= (diag(u)Kdiag(v))^T1
tmp2 = np.sum(u[:, None] * K * v[None, :], 0)
#tmp2=np.einsum('i,ij,j->j', u, K, v)
err = np.linalg.norm(tmp2 - b)**2 # violation of marginal
if log:
log['err'].append(err)
if verbose:
if cpt % 200 == 0:
print(
'{:5s}|{:12s}'.format('It.', 'Err') + '\n' + '-' * 19)
print('{:5d}|{:8e}|'.format(cpt, err))
cpt = cpt + 1
if log:
log['u'] = u
log['v'] = v
if nbb: # return only loss
#res = np.einsum('ik,ij,jk,ij->k', u, K, v, M) (explodes cupy memory)
res = np.empty(nbb)
for i in range(nbb):
res[i] = np.sum(u[:, None, i] * (K * M) * v[None, :, i])
if to_numpy:
res = utils.to_np(res)
if log:
return res, log
else:
return res
else: # return OT matrix
res = u.reshape((-1, 1)) * K * v.reshape((1, -1))
if to_numpy:
res = utils.to_np(res)
if log:
return res, log
else:
return res
# define sinkhorn as sinkhorn_knopp
sinkhorn = sinkhorn_knopp
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