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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 solve(A, b, x0=None, tol=1e-5, maxiter=400, return_solver=False, existing_solver=None, verb=True, residuals=None):
 """Solve Ax=b. Solve the arbitrary system Ax=b with the best out-of-the box choice for a solver. The matrix A can be non-Hermitian, indefinite, Hermitian smoothed_aggregation_solver(..) are used to invert A. Parameters A : array, matrix, csr_matrix, bsr_matrix Matrix to invert, CSR or BSR format preferred for efficiency b : array Right hand side. x0 : array Initial guess (default random vector) tol : float Stopping criteria: relative residual r[k]/r[0] tolerance maxiter : int Stopping criteria: maximum number of allowable iterations return_solver : bool True: return the solver generated existing_solver : smoothed_aggregation_solver If instance of a multilevel solver, then existing_solver is used to invert A, thus saving time on setup cost. verb : bool If True, print verbose output during runtime residuals : list List to contain residual norms at each iteration. The preconditioned norm is used, namely ||r||_M = (M r, r)^(1/2) = (r, r)^(1/2) Returns ------- x : array Solution to Ax = b ml : multilevel_solver Optional return of the multilevel structure used for the solve Notes ----- is easy and efficient. Set "return_solver=True", and the return value will be a tuple, (x,ml), where ml is the solver used to invert A, and x is the "existing_solver=ml". Examples -------- 6.28e-06 """

    # Convert A to acceptable CSR/BSR format
    A = make_csr(A)

    # Generate solver if necessary
    if existing_solver is None:

        # Parameter dictionary for smoothed_aggregation_solver
        config = solver_configuration(A, B=None, verb=verb)
        # Generate solver
        existing_solver = solver(A, config)

    else:
        if existing_solver.levels[0].A.shape[0] != A.shape[0]:
            raise TypeError('Argument existing_solver must have level 0 matrix\
                             of same size as A')

    # Krylov acceleration depends on symmetry of A
    if existing_solver.levels[0].A.symmetry == 'hermitian':
        accel = 'cg'
    else:
        accel = 'gmres'

    # Initial guess
    if x0 is None:
        x0 = np.array(sp.rand(A.shape[0],), dtype=A.dtype)

    # Callback function to print iteration number
    if verb:
        iteration = np.zeros((1,))
        print("    maxiter = %d" % maxiter)

        def callback(x, iteration):
            iteration[0] = iteration[0] + 1
            print("    iteration %d" % iteration[0])

        def callback2(x):
            return callback(x, iteration)
    else:
        callback2 = None

    # Solve with accelerated Krylov method
    x = existing_solver.solve(b, x0=x0, accel=accel, tol=tol, maxiter=maxiter,
                              callback=callback2, residuals=residuals)

    if verb:
        r0 = b - A * x0
        rk = b - A * x
        M = existing_solver.aspreconditioner()
        nr0 = np.sqrt(np.inner(np.conjugate(M * r0), r0))
        nrk = np.sqrt(np.inner(np.conjugate(M * rk), rk))
        print("  Residuals ||r_k||_M, ||r_0||_M = %1.2e, %1.2e" % (nrk, nr0))
        if np.abs(nr0) > 1e-15:
            print("  Residual reduction ||r_k||_M/||r_0||_M = %1.2e"
                  % (nrk / nr0))

    if return_solver:
        return (x.reshape(b.shape), existing_solver)
    else:
        return x.reshape(b.shape)