text: """
  Numerical Root Finding and Optimization
  
  AUTHOR:
  
  - William Stein (2007): initial version
  - Nathann Cohen (2008) : Bin Packing
  
  
  Functions and Methods
  ----------------------
  """
  from six.moves import range
  
  from sage.modules.free_module_element import vector
  from sage.rings.real_double import RDF
  
  
  def find_root(f, a, b, xtol=10e-13, rtol=2.0**-50, maxiter=100, full_output=False):
      r"""
      Numerically find a root of ``f`` on the closed interval `[a,b]`
      (or `[b,a]`) if possible, where ``f`` is a function in the one variable.
      Note: this function only works in fixed (machine) precision, it is not
      possible to get arbitrary precision approximations with it.
  
      INPUT:
  
      - ``f`` -- a function of one variable or symbolic equality
  
      - ``a``, ``b`` -- endpoints of the interval
  
      - ``xtol``, ``rtol`` -- the routine converges when a root is known
        to lie within ``xtol`` of the value return. Should be `\geq 0`.
        The routine modifies this to take into account the relative precision
        of doubles. By default, rtol is ``4*numpy.finfo(float).eps``, the
        minimum allowed value for ``scipy.optimize.brentq``, which is what
        this method uses underneath. This value is equal to ``2.0**-50`` for
        IEEE-754 double precision floats as used by Python.
  
      - ``maxiter`` -- integer; if convergence is not achieved in
        ``maxiter`` iterations, an error is raised. Must be `\geq 0`.
  
      - ``full_output`` -- bool (default: ``False``), if ``True``, also return
        object that contains information about convergence.
  
  
      EXAMPLES:
  
      An example involving an algebraic polynomial function::
  
          sage: R.<x> = QQ[]
          sage: f = (x+17)*(x-3)*(x-1/8)^3
          sage: find_root(f, 0,4)
          2.999999999999995
          sage: find_root(f, 0,1)  # abs tol 1e-6 (note -- precision of answer isn't very good on some machines)
          0.124999
          sage: find_root(f, -20,-10)
          -17.0
  
      In Pomerance's book on primes he asserts that the famous Riemann
      Hypothesis is equivalent to the statement that the function `f(x)`
      defined below is positive for all `x \geq 2.01`::
  
          sage: def f(x):
          ....:     return sqrt(x) * log(x) - abs(Li(x) - prime_pi(x))
  
      We find where `f` equals, i.e., what value that is slightly smaller
      than `2.01` that could have been used in the formulation of the Riemann
      Hypothesis::
  
          sage: find_root(f, 2, 4, rtol=0.0001)
          2.0082...
  
      This agrees with the plot::
  
          sage: plot(f,2,2.01)
          Graphics object consisting of 1 graphics primitive
  
      The following example was added due to :trac:`4942` and demonstrates that
      the function need not be defined at the endpoints::
  
          sage: find_root(x^2*log(x,2)-1,0, 2)  # abs tol 1e-6
          1.41421356237   
          
      The following is an example, again from :trac:`4942` where Brent's method
      fails. Currently no other method is implemented, but at least we 
      acknowledge the fact that the algorithm fails::
  
          sage: find_root(1/(x-1)+1,0, 2) 
          0.0 
          sage: find_root(1/(x-1)+1,0.00001, 2)
          Traceback (most recent call last):
          ...
          NotImplementedError: Brent's method failed to find a zero for f on the interval
  
      An example of a function which evaluates to NaN on the entire interval::
  
          sage: f(x) = 0.0 / max(0, x)
          sage: find_root(f, -1, 0)
          Traceback (most recent call last):
          ...
          RuntimeError: f appears to have no zero on the interval
  
      """
      try:
          return f.find_root(a=a,b=b,xtol=xtol,rtol=rtol,maxiter=maxiter,full_output=full_output)
      except AttributeError:
          pass
      a = float(a); b = float(b)
      if a > b:
          a, b = b, a
      left = f(a)
      right = f(b)
     
      if left > 0 and right > 0:
          # Refine further -- try to find a point where this
          # function is negative in the interval
          val, s = find_local_minimum(f, a, b)
          if val > 0:
              if val < rtol:
                  if full_output:
                      return s, "No extra data"
                  else:
                      return s
              raise RuntimeError("f appears to have no zero on the interval")
          # If we found such an s, then we just instead find
          # a root between left and s or s and right.
          a = s   # arbitrary choice -- maybe should try both and take one that works?
  
      elif left < 0 and right < 0:
          # Refine further
          val, s = find_local_maximum(f, a, b)
          if val < 0:
              if abs(val) < rtol:
                  if full_output:
                      return s, "No extra data"
                  else:
                      return s
              raise RuntimeError("f appears to have no zero on the interval")
          a = s
  
      # Fixing :trac:`4942` - if the answer on any of the endpoints is NaN, 
      # we restrict to looking between minimum and maximum values in the segment
      # Note - this could be used in all cases, but it requires some more 
      # computation
  
      if (left != left) or (right != right):
          minval, s_1 = find_local_minimum(f, a, b)
          maxval, s_2 = find_local_maximum(f, a, b)
          if ((minval > 0) or (maxval < 0) or 
             (minval != minval) or (maxval != maxval)):
              raise RuntimeError("f appears to have no zero on the interval")
          a = min(s_1, s_2)
          b = max(s_1, s_2)
  
      import scipy.optimize
      brentqRes = scipy.optimize.brentq(f, a, b,
                                   full_output=full_output, xtol=xtol, rtol=rtol, maxiter=maxiter)
      # A check following :trac:`4942`, to ensure we actually found a root
      # Maybe should use a different tolerance here?
      # The idea is to take roughly the derivative and multiply by estimated
      # value of the root
      root = 0
      if full_output:
          root = brentqRes[0]
      else:
          root = brentqRes
      if abs(f(root)) > max(abs(root * rtol * (right - left) / (b - a)), 1e-6):
          raise NotImplementedError("Brent's method failed to find a zero for f on the interval")
      return brentqRes
  
  def find_local_maximum(f, a, b, tol=1.48e-08, maxfun=500):
      """
      Numerically find a local maximum of the expression `f` on the interval
      `[a,b]` (or `[b,a]`) along with the point at which the maximum is attained.
  
      Note that this function only finds a *local* maximum, and not the
      global maximum on that interval -- see the examples with
      :func:`find_local_maximum`.
  
      See the documentation for :func:`find_local_maximum` for more
      details and possible workarounds for finding the global minimum on
      an interval.
  
      EXAMPLES::
  
          sage: f = lambda x: x*cos(x)
          sage: find_local_maximum(f, 0, 5)
          (0.561096338191..., 0.8603335890...)
          sage: find_local_maximum(f, 0, 5, tol=0.1, maxfun=10)
          (0.561090323458..., 0.857926501456...)
          sage: find_local_maximum(8*e^(-x)*sin(x) - 1, 0, 7)
          (1.579175535558..., 0.7853981...)
      """
      try:
          return f.find_local_maximum(a=a, b=b, tol=tol, maxfun=maxfun)
      except AttributeError:
          pass
      minval, x = find_local_minimum(lambda z: -f(z), a=a, b=b, tol=tol, maxfun=maxfun)
      return -minval, x
  
  def find_local_minimum(f, a, b, tol=1.48e-08, maxfun=500):
      """
      Numerically find a local minimum of the expression ``f`` on the
      interval `[a,b]` (or `[b,a]`) and the point at which it attains that
      minimum.  Note that ``f`` must be a function of (at most) one
      variable.
  
      Note that this function only finds a *local* minimum, and not the
      global minimum on that interval -- see the examples below.
  
      INPUT:
  
      - ``f`` -- a function of at most one variable.
  
      - ``a``, ``b`` -- endpoints of interval on which to minimize self.
  
      - ``tol`` -- the convergence tolerance
  
      - ``maxfun`` -- maximum function evaluations
  
  
      OUTPUT:
  
      - ``minval`` -- (float) the minimum value that self takes on in the
        interval `[a,b]`
  
      - ``x`` -- (float) the point at which self takes on the minimum value
  
  
      EXAMPLES::
  
          sage: f = lambda x: x*cos(x)
          sage: find_local_minimum(f, 1, 5)
          (-3.28837139559..., 3.4256184695...)
          sage: find_local_minimum(f, 1, 5, tol=1e-3)
          (-3.28837136189098..., 3.42575079030572...)
          sage: find_local_minimum(f, 1, 5, tol=1e-2, maxfun=10)
          (-3.28837084598..., 3.4250840220...)
          sage: show(plot(f, 0, 20))
          sage: find_local_minimum(f, 1, 15)
          (-9.4772942594..., 9.5293344109...)
  
      Only local minima are found; if you enlarge the interval, the
      returned minimum may be *larger*! See :trac:`2607`.
  
      ::
  
          sage: f(x) = -x*sin(x^2)
          sage: find_local_minimum(f, -2.5, -1)
          (-2.182769784677722, -2.1945027498534686)
  
      Enlarging the interval returns a larger minimum::
  
          sage: find_local_minimum(f, -2.5, 2)
          (-1.3076194129914434, 1.3552111405712108)
  
      One work-around is to plot the function and grab the minimum from
      that, although the plotting code does not necessarily do careful
      numerics (observe the small number of decimal places that we
      actually test)::
  
          sage: plot(f, (x,-2.5, -1)).ymin()
          -2.1827...
          sage: plot(f, (x,-2.5, 2)).ymin()
          -2.1827...
  
      ALGORITHM:
  
      Uses `scipy.optimize.fminbound
      <http://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.fminbound.html>`_
      which uses Brent's method.
  
  
      AUTHOR:
  
      - William Stein (2007-12-07)
      """
      try:
          return f.find_local_minimum(a=a, b=b, tol=tol, maxfun=maxfun)
      except AttributeError:
          pass
      a = float(a)
      b = float(b)
      import scipy.optimize
      xmin, fval, iter, funcalls = scipy.optimize.fminbound(f, a, b, full_output=1, xtol=tol, maxfun=maxfun)
      return fval, xmin
  
  
  def minimize(func, x0, gradient=None, hessian=None, algorithm="default",
               verbose=False, **args):
      r"""
      This function is an interface to a variety of algorithms for computing
      the minimum of a function of several variables.
  
      INPUT:
  
      - ``func`` -- Either a symbolic function or a Python function whose
        argument is a tuple with `n` components
  
      - ``x0`` -- Initial point for finding minimum.
  
      - ``gradient`` -- Optional gradient function. This will be computed
        automatically for symbolic functions.  For Python functions, it allows
        the use of algorithms requiring derivatives.  It should accept a
        tuple of arguments and return a NumPy array containing the partial
        derivatives at that point.
  
      - ``hessian`` --  Optional hessian function. This will be computed
        automatically for symbolic functions. For Python functions, it allows
        the use of algorithms requiring derivatives. It should accept a tuple
        of arguments and return a NumPy array containing the second partial
        derivatives of the function.
  
      - ``algorithm`` -- String specifying algorithm to use. Options are
        ``'default'`` (for Python functions, the simplex method is the default)
        (for symbolic functions bfgs is the default):
  
         - ``'simplex'`` -- using the downhill simplex algorithm
  
         - ``'powell'`` -- use the modified Powell algorithm
  
         - ``'bfgs'`` -- (Broyden-Fletcher-Goldfarb-Shanno) requires gradient
  
         - ``'cg'`` -- (conjugate-gradient) requires gradient
  
         - ``'ncg'`` -- (newton-conjugate gradient) requires gradient and hessian
  
      - ``verbose`` -- (optional, default: False) print convergence message
  
      .. NOTE::
  
          For additional information on the algorithms implemented in this function,
          consult SciPy's `documentation on optimization and root
          finding <https://docs.scipy.org/doc/scipy/reference/optimize.html>`_
  
      EXAMPLES:
  
      Minimize a fourth order polynomial in three variables (see the
      :wikipedia:`Rosenbrock_function`)::
  
          sage: vars = var('x y z')
          sage: f = 100*(y-x^2)^2+(1-x)^2+100*(z-y^2)^2+(1-y)^2
          sage: minimize(f, [.1,.3,.4]) # abs tol 1e-6
          (1.0, 1.0, 1.0)
  
      Try the newton-conjugate gradient method; the gradient and hessian are 
      computed automatically::
  
          sage: minimize(f, [.1, .3, .4], algorithm="ncg") # abs tol 1e-6
          (1.0, 1.0, 1.0)
  
      We get additional convergence information with the `verbose` option::
  
          sage: minimize(f, [.1, .3, .4], algorithm="ncg", verbose=True)
          Optimization terminated successfully.
          ...
          (0.9999999..., 0.999999..., 0.999999...)
  
      Same example with just Python functions::
  
          sage: def rosen(x): # The Rosenbrock function
          ....:    return sum(100.0r*(x[1r:]-x[:-1r]**2.0r)**2.0r + (1r-x[:-1r])**2.0r)
          sage: minimize(rosen, [.1,.3,.4]) # abs tol 3e-5
          (1.0, 1.0, 1.0)
  
      Same example with a pure Python function and a Python function to
      compute the gradient::
  
          sage: def rosen(x): # The Rosenbrock function
          ....:    return sum(100.0r*(x[1r:]-x[:-1r]**2.0r)**2.0r + (1r-x[:-1r])**2.0r)
          sage: import numpy
          sage: from numpy import zeros
          sage: def rosen_der(x):
          ....:    xm = x[1r:-1r]
          ....:    xm_m1 = x[:-2r]
          ....:    xm_p1 = x[2r:]
          ....:    der = zeros(x.shape, dtype=float)
          ....:    der[1r:-1r] = 200r*(xm-xm_m1**2r) - 400r*(xm_p1 - xm**2r)*xm - 2r*(1r-xm)
          ....:    der[0] = -400r*x[0r]*(x[1r]-x[0r]**2r) - 2r*(1r-x[0])
          ....:    der[-1] = 200r*(x[-1r]-x[-2r]**2r)
          ....:    return der
          sage: minimize(rosen, [.1,.3,.4], gradient=rosen_der, algorithm="bfgs") # abs tol 1e-6
          (1.0, 1.0, 1.0)
      """
      from sage.symbolic.expression import Expression
      from sage.ext.fast_eval import fast_callable
      import numpy
      from scipy import optimize
      if isinstance(func, Expression):
          var_list=func.variables()
          var_names = [str(_) for _ in var_list]
          fast_f=fast_callable(func, vars=var_names, domain=float)
          f=lambda p: fast_f(*p)
          gradient_list=func.gradient()
          fast_gradient_functions=[fast_callable(gradient_list[i], vars=var_names, domain=float)  for i in range(len(gradient_list))]
          gradient=lambda p: numpy.array([ a(*p) for a in fast_gradient_functions])
      else:
          f=func
  
      if algorithm=="default":
          if gradient is None:
              min = optimize.fmin(f, [float(_) for _ in x0], disp=verbose, **args)
          else:
              min= optimize.fmin_bfgs(f, [float(_) for _ in x0],fprime=gradient, disp=verbose, **args)
      else:
          if algorithm=="simplex":
              min= optimize.fmin(f, [float(_) for _ in x0], disp=verbose, **args)
          elif algorithm=="bfgs":
              min= optimize.fmin_bfgs(f, [float(_) for _ in x0], fprime=gradient, disp=verbose, **args)
          elif algorithm=="cg":
              min= optimize.fmin_cg(f, [float(_) for _ in x0], fprime=gradient, disp=verbose, **args)
          elif algorithm=="powell":
              min= optimize.fmin_powell(f, [float(_) for _ in x0], disp=verbose, **args)
          elif algorithm=="ncg":
              if isinstance(func, Expression):
                  hess=func.hessian()
                  hess_fast= [ [fast_callable(a, vars=var_names, domain=float) for a in row] for row in hess]
                  hessian=lambda p: [[a(*p) for a in row] for row in hess_fast]
                  hessian_p=lambda p,v: scipy.dot(numpy.array(hessian(p)),v)
                  min = optimize.fmin_ncg(f, [float(_) for _ in x0], fprime=gradient, \
                        fhess=hessian, fhess_p=hessian_p, disp=verbose, **args)
      return vector(RDF, min)
  
  def minimize_constrained(func,cons,x0,gradient=None,algorithm='default', **args):
      r"""
      Minimize a function with constraints.
  
  
      INPUT:
  
      - ``func`` -- Either a symbolic function, or a Python function whose
        argument is a tuple with n components
  
      - ``cons`` -- constraints. This should be either a function or list of
        functions that must be positive. Alternatively, the constraints can
        be specified as a list of intervals that define the region we are
        minimizing in. If the constraints are specified as functions, the
        functions should be functions of a tuple with `n` components
        (assuming `n` variables). If the constraints are specified as a list
        of intervals and there are no constraints for a given variable, that
        component can be (``None``, ``None``).
  
      - ``x0`` -- Initial point for finding minimum
  
      - ``algorithm`` -- Optional, specify the algorithm to use:
  
        - ``'default'``  -- default choices
  
        - ``'l-bfgs-b'`` -- only effective if you specify bound constraints.
          See [ZBN1997]_.
  
      - ``gradient`` -- Optional gradient function. This will be computed
        automatically for symbolic functions. This is only used when the
        constraints are specified as a list of intervals.
  
  
      EXAMPLES:
  
      Let us maximize `x + y - 50` subject to the following constraints:
      `50x + 24y \leq 2400`, `30x + 33y \leq 2100`, `x \geq 45`,
      and `y \geq 5`::
  
          sage: y = var('y')
          sage: f = lambda p: -p[0]-p[1]+50
          sage: c_1 = lambda p: p[0]-45
          sage: c_2 = lambda p: p[1]-5
          sage: c_3 = lambda p: -50*p[0]-24*p[1]+2400
          sage: c_4 = lambda p: -30*p[0]-33*p[1]+2100
          sage: a = minimize_constrained(f,[c_1,c_2,c_3,c_4],[2,3])
          sage: a
          (45.0, 6.25...)
  
      Let's find a minimum of `\sin(xy)`::
  
          sage: x,y = var('x y')
          sage: f = sin(x*y)
          sage: minimize_constrained(f, [(None,None),(4,10)],[5,5])
          (4.8..., 4.8...)
  
      Check if L-BFGS-B finds the same minimum::
  
          sage: minimize_constrained(f, [(None,None),(4,10)],[5,5], algorithm='l-bfgs-b')
          (4.7..., 4.9...)
  
      Rosenbrock function (see the :wikipedia:`Rosenbrock_function`)::
  
          sage: from scipy.optimize import rosen, rosen_der
          sage: minimize_constrained(rosen, [(-50,-10),(5,10)],[1,1],gradient=rosen_der,algorithm='l-bfgs-b')
          (-10.0, 10.0)
          sage: minimize_constrained(rosen, [(-50,-10),(5,10)],[1,1],algorithm='l-bfgs-b')
          (-10.0, 10.0)
  
      TESTS:
  
      Check if :trac:`6592` is fixed::
  
          sage: x, y = var('x y')
          sage: f = (100 - x) + (1000 - y)
          sage: c = x + y - 479 # > 0
          sage: minimize_constrained(f, [c], [100, 300])
          (805.985..., 1005.985...)
          sage: minimize_constrained(f, c, [100, 300])
          (805.985..., 1005.985...)
      """
      from sage.symbolic.expression import Expression
      import numpy
      from scipy import optimize
      function_type = type(lambda x,y: x+y)
  
      if isinstance(func, Expression):
          var_list = func.variables()
          var_names = [str(_) for _ in var_list]
          fast_f = func._fast_float_(*var_names)
          f = lambda p: fast_f(*p)
          gradient_list = func.gradient()
          fast_gradient_functions = [gi._fast_float_(*var_names) for gi in gradient_list]
          gradient = lambda p: numpy.array([ a(*p) for a in fast_gradient_functions])
          if isinstance(cons, Expression):
              fast_cons = cons._fast_float_(*var_names)
              cons = lambda p: numpy.array([fast_cons(*p)])
          elif isinstance(cons, list) and isinstance(cons[0], Expression):
              fast_cons = [ci._fast_float_(*var_names) for ci in cons]
              cons = lambda p: numpy.array([a(*p) for a in fast_cons])
      else:
          f = func
  
      if isinstance(cons,list):
          if isinstance(cons[0], tuple) or isinstance(cons[0], list) or cons[0] is None:
              if gradient is not None:
                  if algorithm == 'l-bfgs-b':
                      min = optimize.fmin_l_bfgs_b(f, x0, gradient, bounds=cons, iprint=-1, **args)[0]
                  else:
                      min = optimize.fmin_tnc(f, x0, gradient, bounds=cons, messages=0, **args)[0]
              else:
                  if algorithm == 'l-bfgs-b':
                      min = optimize.fmin_l_bfgs_b(f, x0, approx_grad=True, bounds=cons, iprint=-1, **args)[0]
                  else:
                      min = optimize.fmin_tnc(f, x0, approx_grad=True, bounds=cons, messages=0, **args)[0]
          elif isinstance(cons[0], function_type) or isinstance(cons[0], Expression):
              min = optimize.fmin_cobyla(f, x0, cons, **args)
      elif isinstance(cons, function_type) or isinstance(cons, Expression):
          min = optimize.fmin_cobyla(f, x0, cons, **args)
      return vector(RDF, min)
  
  
  def linear_program(c, G, h, A=None, b=None, solver=None):
      r"""
      Solve the dual linear programs:
  
      - Minimize  `c'x` subject to `Gx + s = h`, `Ax = b`, and `s \geq 0` where
        `'` denotes transpose.
  
      - Maximize  `-h'z - b'y` subject to `G'z + A'y + c = 0` and `z \geq 0`.
  
  
      INPUT:
  
      - ``c`` -- a vector
  
      - ``G`` -- a matrix
  
      - ``h`` -- a vector
  
      - ``A`` -- a matrix
  
      - ``b`` --- a vector
  
      - ``solver`` (optional) --- solver to use. If None, the cvxopt's lp-solver
                                  is used. If it is 'glpk', then glpk's solver
                                  is used.
  
      These can be over any field that can be turned into a floating point
      number.
  
  
      OUTPUT:
  
      A dictionary ``sol`` with keys ``x``, ``s``, ``y``, ``z`` corresponding
      to the variables above:
  
      - ``sol['x']`` -- the solution to the linear program
  
      - ``sol['s']`` -- the slack variables for the solution
  
      - ``sol['z']``, ``sol['y']`` -- solutions to the dual program
  
  
      EXAMPLES:
  
      First, we minimize `-4x_1 - 5x_2` subject to `2x_1 + x_2 \leq 3`,
      `x_1 +  2x_2 \leq 3`, `x_1 \geq 0`, and `x_2 \geq 0`::
  
          sage: c=vector(RDF,[-4,-5])
          sage: G=matrix(RDF,[[2,1],[1,2],[-1,0],[0,-1]])
          sage: h=vector(RDF,[3,3,0,0])
          sage: sol=linear_program(c,G,h)
          sage: sol['x']
          (0.999..., 1.000...)
  
      Here we solve the same problem with 'glpk' interface to 'cvxopt'::
  
          sage: sol=linear_program(c,G,h,solver='glpk')
          GLPK Simplex Optimizer...
          ...
          OPTIMAL LP SOLUTION FOUND
          sage: sol['x']
          (1.0, 1.0)
  
      Next, we maximize `x+y-50` subject to `50x + 24y \leq 2400`,
      `30x + 33y \leq 2100`, `x \geq 45`, and `y \geq 5`::
  
          sage: v=vector([-1.0,-1.0,-1.0])
          sage: m=matrix([[50.0,24.0,0.0],[30.0,33.0,0.0],[-1.0,0.0,0.0],[0.0,-1.0,0.0],[0.0,0.0,1.0],[0.0,0.0,-1.0]])
          sage: h=vector([2400.0,2100.0,-45.0,-5.0,1.0,-1.0])
          sage: sol=linear_program(v,m,h)
          sage: sol['x']
          (45.000000..., 6.2499999..., 1.00000000...)
          sage: sol=linear_program(v,m,h,solver='glpk')
          GLPK Simplex Optimizer...
          OPTIMAL LP SOLUTION FOUND
          sage: sol['x']
          (45.0..., 6.25..., 1.0...)
      """
      from cvxopt.base import matrix as m
      from cvxopt import solvers
      solvers.options['show_progress']=False
      if solver=='glpk':
          from cvxopt import glpk
          glpk.options['LPX_K_MSGLEV'] = 0
      c_=m(c.base_extend(RDF).numpy())
      G_=m(G.base_extend(RDF).numpy())
      h_=m(h.base_extend(RDF).numpy())
      if A is not None and b is not None:
          A_=m(A.base_extend(RDF).numpy())
          b_=m(b.base_extend(RDF).numpy())
          sol=solvers.lp(c_,G_,h_,A_,b_,solver=solver)
      else:
          sol=solvers.lp(c_,G_,h_,solver=solver)
      status=sol['status']
      if status != 'optimal':
         return  {'primal objective':None,'x':None,'s':None,'y':None,
                'z':None,'status':status}
      x=vector(RDF,list(sol['x']))
      s=vector(RDF,list(sol['s']))
      y=vector(RDF,list(sol['y']))
      z=vector(RDF,list(sol['z']))
      return  {'primal objective':sol['primal objective'],'x':x,'s':s,'y':y,
                 'z':z,'status':status}
  
  
  def find_fit(data, model, initial_guess = None, parameters = None, variables = None, solution_dict = False):
      r"""
      Finds numerical estimates for the parameters of the function model to
      give a best fit to data.
  
  
      INPUT:
  
      - ``data`` -- A two dimensional table of floating point numbers of the
        form `[[x_{1,1}, x_{1,2}, \ldots, x_{1,k}, f_1],
        [x_{2,1}, x_{2,2}, \ldots, x_{2,k}, f_2],
        \ldots,
        [x_{n,1}, x_{n,2}, \ldots, x_{n,k}, f_n]]` given as either a list of
        lists, matrix, or numpy array.
  
      - ``model`` -- Either a symbolic expression, symbolic function, or a
        Python function. ``model`` has to be a function of the variables
        `(x_1, x_2, \ldots, x_k)` and free parameters
        `(a_1, a_2, \ldots, a_l)`.
  
      - ``initial_guess`` -- (default: ``None``) Initial estimate for the
        parameters `(a_1, a_2, \ldots, a_l)`, given as either a list, tuple,
        vector or numpy array. If ``None``, the default estimate for each
        parameter is `1`.
  
      - ``parameters`` -- (default: ``None``) A list of the parameters
        `(a_1, a_2, \ldots, a_l)`. If model is a symbolic function it is
        ignored, and the free parameters of the symbolic function are used.
  
      - ``variables`` -- (default: ``None``) A list of the variables
        `(x_1, x_2, \ldots, x_k)`. If model is a symbolic function it is
        ignored, and the variables of the symbolic function are used.
  
      - ``solution_dict`` -- (default: ``False``) if ``True``, return the
        solution as a dictionary rather than an equation.
  
  
      EXAMPLES:
  
      First we create some data points of a sine function with some random
      perturbations::
  
          sage: data = [(i, 1.2 * sin(0.5*i-0.2) + 0.1 * normalvariate(0, 1)) for i in xsrange(0, 4*pi, 0.2)]
          sage: var('a, b, c, x')
          (a, b, c, x)
  
      We define a function with free parameters `a`, `b` and `c`::
  
          sage: model(x) = a * sin(b * x - c)
  
      We search for the parameters that give the best fit to the data::
  
          sage: find_fit(data, model)
          [a == 1.21..., b == 0.49..., c == 0.19...]
  
      We can also use a Python function for the model::
  
          sage: def f(x, a, b, c): return a * sin(b * x - c)
          sage: fit = find_fit(data, f, parameters = [a, b, c], variables = [x], solution_dict = True)
          sage: fit[a], fit[b], fit[c]
          (1.21..., 0.49..., 0.19...)
  
      We search for a formula for the `n`-th prime number::
  
          sage: dataprime = [(i, nth_prime(i)) for i in range(1, 5000, 100)]
          sage: find_fit(dataprime, a * x * log(b * x), parameters = [a, b], variables = [x])
          [a == 1.11..., b == 1.24...]
  
  
      ALGORITHM:
  
      Uses ``scipy.optimize.leastsq`` which in turn uses MINPACK's lmdif and
      lmder algorithms.
      """
      import numpy
  
      if not isinstance(data, numpy.ndarray):
          try:
              data = numpy.array(data, dtype = float)
          except (ValueError, TypeError):
              raise TypeError("data has to be a list of lists, a matrix, or a numpy array")
      elif data.dtype == object:
          raise ValueError("the entries of data have to be of type float")
  
      if data.ndim != 2:
          raise ValueError("data has to be a two dimensional table of floating point numbers")
  
      from sage.symbolic.expression import Expression
  
      if isinstance(model, Expression):
          if variables is None:
              variables = list(model.arguments())
          if parameters is None:
              parameters = list(model.variables())
              for v in variables:
                  parameters.remove(v)
  
      if data.shape[1] != len(variables) + 1:
          raise ValueError("each row of data needs %d entries, only %d entries given" % (len(variables) + 1, data.shape[1]))
  
      if parameters is None or len(parameters) == 0 or \
         variables is None or len(variables) == 0:
          raise ValueError("no variables given")
  
      if initial_guess is None:
          initial_guess = len(parameters) * [1]
  
      if not isinstance(initial_guess, numpy.ndarray):
          try:
              initial_guess = numpy.array(initial_guess, dtype = float)
          except (ValueError, TypeError):
              raise TypeError("initial_guess has to be a list, tuple, or numpy array")
      elif initial_guess.dtype == object:
          raise ValueError("the entries of initial_guess have to be of type float")
  
      if len(initial_guess) != len(parameters):
          raise ValueError("length of initial_guess does not coincide with the number of parameters")
  
      if isinstance(model, Expression):
          var_list = variables + parameters
          var_names = [str(_) for _ in var_list]
          func = model._fast_float_(*var_names)
      else:
          func = model
  
      def function(x_data, params):
          result = numpy.zeros(len(x_data))
          for row in range(len(x_data)):
              fparams = numpy.hstack((x_data[row], params)).tolist()
              result[row] = func(*fparams)
          return result
  
      def error_function(params, x_data, y_data):
          result = numpy.zeros(len(x_data))
          for row in range(len(x_data)):
              fparams = x_data[row].tolist() + params.tolist()
              result[row] = func(*fparams)
          return result - y_data
  
      x_data = data[:, 0:len(variables)]
      y_data = data[:, -1]
  
      from scipy.optimize import leastsq
      estimated_params, d = leastsq(error_function, initial_guess, args = (x_data, y_data))
  
      if isinstance(estimated_params, float):
          estimated_params = [estimated_params]
      else:
          estimated_params = estimated_params.tolist()
  
      if solution_dict:
         dict = {}
         for item in zip(parameters, estimated_params):
             dict[item[0]] = item[1]
         return dict
  
      return [item[0] == item[1] for item in zip(parameters, estimated_params)]
  
  def binpacking(items, maximum=1, k=None, solver=None, verbose=0):
      r"""
      Solve the bin packing problem.
  
      The Bin Packing problem is the following :
  
      Given a list of items of weights `p_i` and a real value `k`, what is the
      least number of bins such that all the items can be packed in the bins,
      while ensuring that the sum of the weights of the items packed in each bin
      is at most `k` ?
  
      For more informations, see :wikipedia:`Bin_packing_problem`.
  
      Two versions of this problem are solved by this algorithm :
  
      - Is it possible to put the given items in `k` bins ?
      - What is the assignment of items using the least number of bins with
        the given list of items ?
  
      INPUT:
  
      - ``items`` -- list or dict; either a list of real values (the items'
        weight), or a dictionary associating to each item its weight.
  
      - ``maximum`` -- (default: 1); the maximal size of a bin
  
      - ``k`` -- integer (default: ``None``); Number of bins
  
        - When set to an integer value, the function returns a partition of the
          items into `k` bins if possible, and raises an exception otherwise.
  
        - When set to ``None``, the function returns a partition of the items
          using the least possible number of bins.
  
      - ``solver`` -- (default: ``None``); Specify a Linear Program (LP) solver to
        be used. If set to ``None``, the default one is used. For more information
        on LP solvers and which default solver is used, see the method
        :meth:`~sage.numerical.mip.MixedIntegerLinearProgram.solve` of the class
        :class:`~sage.numerical.mip.MixedIntegerLinearProgram`.
  
      - ``verbose`` -- integer (default: ``0``); sets the level of verbosity. Set
        to 0 by default, which means quiet.
  
      OUTPUT:
  
      A list of lists, each member corresponding to a bin and containing either
      the list of the weights inside it when ``items`` is a list of items' weight,
      or the list of items inside it when ``items`` is a dictionary. If there is
      no solution, an exception is raised (this can only happen when ``k`` is
      specified or if ``maximum`` is less than the weight of one item).
  
      EXAMPLES:
  
      Trying to find the minimum amount of boxes for 5 items of weights
      `1/5, 1/4, 2/3, 3/4, 5/7`::
  
          sage: from sage.numerical.optimize import binpacking
          sage: values = [1/5, 1/3, 2/3, 3/4, 5/7]
          sage: bins = binpacking(values)
          sage: len(bins)
          3
  
      Checking the bins are of correct size ::
  
          sage: all(sum(b) <= 1 for b in bins)
          True
  
      Checking every item is in a bin ::
  
          sage: b1, b2, b3 = bins
          sage: all((v in b1 or v in b2 or v in b3) for v in values)
          True
  
      And only in one bin ::
  
          sage: sum(len(b) for b in bins) == len(values)
          True
  
      One way to use only three boxes (which is best possible) is to put
      `1/5 + 3/4` together in a box, `1/3+2/3` in another, and `5/7`
      by itself in the third one.
  
      Of course, we can also check that there is no solution using only two boxes ::
  
          sage: from sage.numerical.optimize import binpacking
          sage: binpacking([0.2,0.3,0.8,0.9], k=2)
          Traceback (most recent call last):
          ...
          ValueError: this problem has no solution !
  
      We can also provide a dictionary keyed by items and associating to each item
      its weight. Then, the bins contain the name of the items inside it ::
  
          sage: values = {'a':1/5, 'b':1/3, 'c':2/3, 'd':3/4, 'e':5/7}
          sage: bins = binpacking(values)
          sage: set(flatten(bins)) == set(values.keys())
          True
  
      TESTS:
  
      Wrong type for parameter items::
  
          sage: binpacking(set())
          Traceback (most recent call last):
          ...
          TypeError: parameter items must be a list or a dictionary.
      """
      if isinstance(items, list):
          weight = {i:w for i,w in enumerate(items)}
      elif isinstance(items, dict):
          weight = items
      else:
          raise TypeError("parameter items must be a list or a dictionary.")
  
      if max(weight.values()) > maximum:
          raise ValueError("this problem has no solution !")
  
      if k is None:
          from sage.functions.other import ceil
          k = ceil(sum(weight.values())/maximum)
          while True:
              from sage.numerical.mip import MIPSolverException
              try:
                  return binpacking(items, k=k, maximum=maximum, solver=solver, verbose=verbose)
              except MIPSolverException:
                  k = k + 1
  
      from sage.numerical.mip import MixedIntegerLinearProgram, MIPSolverException
      p = MixedIntegerLinearProgram(solver=solver)
  
      # Boolean variable indicating whether the ith element belongs to box b
      box = p.new_variable(binary=True)
  
      # Capacity constraint of each bin
      for b in range(k):
          p.add_constraint(p.sum(weight[i]*box[i,b] for i in weight) <= maximum)
  
      # Each item is assigned exactly one bin
      for i in weight:
          p.add_constraint(p.sum(box[i,b] for b in range(k)) == 1)
  
      try:
          p.solve(log=verbose)
      except MIPSolverException:
          raise ValueError("this problem has no solution !")
  
      box = p.get_values(box)
  
      boxes = [[] for i in range(k)]
  
      for i,b in box:
          if box[i,b] == 1:
              boxes[b].append(weight[i] if isinstance(items, list) else i)
  
      return boxes
  