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# -*- coding: utf-8 -*-
"""
Bregman projections for regularized OT with GPU
"""
import numpy as np
import cudamat
def sinkhorn(a, b, M_GPU, reg, numItermax=1000, stopThr=1e-9, verbose=False,
log=False, returnAsGPU=False):
"""
Solve the entropic regularization optimal transport problem on GPU
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,)
samples in the target domain
M_GPU : cudamat.CUDAMatrix (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
returnAsGPU : bool, optional
return the OT matrix as a cudamat.CUDAMatrix
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
"""
# init data
Nini = len(a)
Nfin = len(b)
if log:
log = {'err': []}
# we assume that no distances are null except those of the diagonal of
# distances
u = (np.ones(Nini)/Nini).reshape((Nini, 1))
u_GPU = cudamat.CUDAMatrix(u)
a_GPU = cudamat.CUDAMatrix(a.reshape((Nini, 1)))
ones_GPU = cudamat.empty(u_GPU.shape).assign(1)
v = (np.ones(Nfin)/Nfin).reshape((Nfin, 1))
v_GPU = cudamat.CUDAMatrix(v)
b_GPU = cudamat.CUDAMatrix(b.reshape((Nfin, 1)))
M_GPU.divide(-reg)
K_GPU = cudamat.exp(M_GPU)
ones_GPU.divide(a_GPU, target=a_GPU)
Kp_GPU = cudamat.empty(K_GPU.shape)
K_GPU.mult_by_col(a_GPU, target=Kp_GPU)
tmp_GPU = cudamat.empty(K_GPU.shape)
cpt = 0
err = 1
while (err > stopThr and cpt < numItermax):
uprev_GPU = u_GPU.copy()
vprev_GPU = v_GPU.copy()
KtransposeU_GPU = K_GPU.transpose().dot(u_GPU)
b_GPU.divide(KtransposeU_GPU, target=v_GPU)
ones_GPU.divide(Kp_GPU.dot(v_GPU), target=u_GPU)
if (np.any(KtransposeU_GPU.asarray() == 0) or
not u_GPU.allfinite() or not v_GPU.allfinite()):
# we have reached the machine precision
# come back to previous solution and quit loop
print('Warning: numerical errors at iteration', cpt)
u_GPU = uprev_GPU.copy()
v_GPU = vprev_GPU.copy()
break
if cpt % 10 == 0:
# we can speed up the process by checking for the error only all
# the 10th iterations
K_GPU.mult_by_col(u_GPU, target=tmp_GPU)
tmp_GPU.mult_by_row(v_GPU.transpose(), target=tmp_GPU)
bcopy_GPU = b_GPU.copy().transpose()
bcopy_GPU.add_sums(tmp_GPU, axis=0, beta=-1)
err = bcopy_GPU.euclid_norm()**2
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 += 1
if log:
log['u'] = u_GPU.asarray()
log['v'] = v_GPU.asarray()
K_GPU.mult_by_col(u_GPU, target=K_GPU)
K_GPU.mult_by_row(v_GPU.transpose(), target=K_GPU)
if returnAsGPU:
res = K_GPU
else:
res = K_GPU.asarray()
if log:
return res, log
else:
return res
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