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""" Numerical Root Finding and Optimization
AUTHOR:
- William Stein (2007): initial version - Nathann Cohen (2008) : Bin Packing
Functions and Methods ---------------------- """
""" 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) # 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 """ a, b = b, a # Refine further -- try to find a point where this # function is negative in the interval 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.
# Refine further 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")
full_output=full_output, xtol=xtol, rtol=rtol, maxiter=maxiter)
""" 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...) """
""" 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) """
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) """ else:
else: else: min= optimize.fmin(f, [float(_) for _ in x0], disp=verbose, **args) min= optimize.fmin_cg(f, [float(_) for _ in x0], fprime=gradient, disp=verbose, **args) min= optimize.fmin_powell(f, [float(_) for _ in x0], disp=verbose, **args) fhess=hessian, fhess_p=hessian_p, disp=verbose, **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...) """
else:
else: else: else: min = optimize.fmin_tnc(f, x0, approx_grad=True, bounds=cons, messages=0, **args)[0]
""" Solves 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...) """ 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: return {'primal objective':None,'x':None,'s':None,'y':None, 'z':None,'status':status} 'z':z,'status':status}
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. """
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")
raise ValueError("data has to be a two dimensional table of floating point numbers")
raise ValueError("each row of data needs %d entries, only %d entries given" % (len(variables) + 1, data.shape[1]))
variables is None or len(variables) == 0: raise ValueError("no variables given")
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")
raise ValueError("length of initial_guess does not coincide with the number of parameters")
else:
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
estimated_params = [estimated_params] else:
r""" Solves 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 put in the bins, while keeping sure that each bin contains a weight of at most `K` ?
For more informations, see :wikipedia:`Bin_packing_problem`
Two version of this problem are solved by this algorithm : * Is it possible to put the given items in `L` bins ? * What is the assignment of items using the least number of bins with the given list of items ?
INPUT:
- ``items`` -- A list of real values (the items' weight)
- ``maximum`` -- The maximal size of a bin
- ``k`` -- 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 number possible of bins.
OUTPUT:
A list of lists, each member corresponding to a box and containing the list of the weights inside it. If there is no solution, an exception is raised (this can only happen when ``k`` is specified or if ``maximum`` is less that the size 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
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 ! """
raise ValueError("This problem has no solution !")
except MIPSolverException: k = k + 1
# Boolean variable indicating whether # the i th element belongs to box b
# Each bin contains at most max
# Each item is assigned exactly one bin
|