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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:

def empirical_sinkhorn_divergence(X_s, X_t, reg, a=None, b=None, metric='sqeuclidean', numIterMax=10000, stopThr=1e-9, verbose=False, log=False, **kwargs):
    '''
    Compute the sinkhorn divergence loss from empirical data

    The function solves the following optimization problems and return the
    sinkhorn divergence :math:`S`:

    .. math::

        W &= \min_\gamma <\gamma,M>_F + reg\cdot\Omega(\gamma)

        W_a &= \min_{\gamma_a} <\gamma_a,M_a>_F + reg\cdot\Omega(\gamma_a)

        W_b &= \min_{\gamma_b} <\gamma_b,M_b>_F + reg\cdot\Omega(\gamma_b)

        S &= W - 1/2 * (W_a + W_b)

    .. math::
        s.t. \gamma 1 = a

             \gamma^T 1= b

             \gamma\geq 0

             \gamma_a 1 = a

             \gamma_a^T 1= a

             \gamma_a\geq 0

             \gamma_b 1 = b

             \gamma_b^T 1= b

             \gamma_b\geq 0
    where :

    - :math:`M` (resp. :math:`M_a, M_b`) is the (ns,nt) metric cost matrix (resp (ns, ns) and (nt, nt))
    - :math:`\Omega` is the entropic regularization term :math:`\Omega(\gamma)=\sum_{i,j} \gamma_{i,j}\log(\gamma_{i,j})`
    - :math:`a` and :math:`b` are source and target weights (sum to 1)


    Parameters
    ----------
    X_s : np.ndarray (ns, d)
        samples in the source domain
    X_t : np.ndarray (nt, d)
        samples in the target domain
    reg : float
        Regularization term >0
    a : np.ndarray (ns,)
        samples weights in the source domain
    b : np.ndarray (nt,)
        samples weights in the target domain
    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


    Returns
    -------
    gamma : (ns x nt) ndarray
        Regularized optimal transportation matrix for the given parameters
    log : dict
        log dictionary return only if log==True in parameters

    Examples
    --------

    >>> n_s = 2
    >>> n_t = 4
    >>> reg = 0.1
    >>> X_s = np.reshape(np.arange(n_s), (n_s, 1))
    >>> X_t = np.reshape(np.arange(0, n_t), (n_t, 1))
    >>> emp_sinkhorn_div = empirical_sinkhorn_divergence(X_s, X_t, reg)
    >>> print(emp_sinkhorn_div)
    >>> [2.99977435]


    References
    ----------

    .. [23] Aude Genevay, Gabriel Peyré, Marco Cuturi, Learning Generative Models with Sinkhorn Divergences,  Proceedings of the Twenty-First International Conference on Artficial Intelligence and Statistics, (AISTATS) 21, 2018
    '''
    if log:
        sinkhorn_loss_ab, log_ab = empirical_sinkhorn2(X_s, X_t, reg, a, b, metric=metric, numIterMax=numIterMax, stopThr=1e-9, verbose=verbose, log=log, **kwargs)

        sinkhorn_loss_a, log_a = empirical_sinkhorn2(X_s, X_s, reg, a, b, metric=metric, numIterMax=numIterMax, stopThr=1e-9, verbose=verbose, log=log, **kwargs)

        sinkhorn_loss_b, log_b = empirical_sinkhorn2(X_t, X_t, reg, a, b, metric=metric, numIterMax=numIterMax, stopThr=1e-9, verbose=verbose, log=log, **kwargs)

        sinkhorn_div = sinkhorn_loss_ab - 1 / 2 * (sinkhorn_loss_a + sinkhorn_loss_b)

        log = {}
        log['sinkhorn_loss_ab'] = sinkhorn_loss_ab
        log['sinkhorn_loss_a'] = sinkhorn_loss_a
        log['sinkhorn_loss_b'] = sinkhorn_loss_b
        log['log_sinkhorn_ab'] = log_ab
        log['log_sinkhorn_a'] = log_a
        log['log_sinkhorn_b'] = log_b

        return max(0, sinkhorn_div), log

    else:
        sinkhorn_loss_ab = empirical_sinkhorn2(X_s, X_t, reg, a, b, metric=metric, numIterMax=numIterMax, stopThr=1e-9, verbose=verbose, log=log, **kwargs)

        sinkhorn_loss_a = empirical_sinkhorn2(X_s, X_s, reg, a, b, metric=metric, numIterMax=numIterMax, stopThr=1e-9, verbose=verbose, log=log, **kwargs)

        sinkhorn_loss_b = empirical_sinkhorn2(X_t, X_t, reg, a, b, metric=metric, numIterMax=numIterMax, stopThr=1e-9, verbose=verbose, log=log, **kwargs)

        sinkhorn_div = sinkhorn_loss_ab - 1 / 2 * (sinkhorn_loss_a + sinkhorn_loss_b)
        return max(0, sinkhorn_div)