Numerical Root Finding and Optimization¶
AUTHOR:
 William Stein (2007): initial version
 Nathann Cohen (2008) : Bin Packing
Functions and Methods¶

sage.numerical.optimize.
binpacking
(items, maximum=1, k=None, solver=None, verbose=0)¶ 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 article 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 bink
– 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 toNone
, the default one is used. For more information on LP solvers and which default solver is used, see the methodsolve()
of the classMixedIntegerLinearProgram
.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 whenitems
is a dictionary. If there is no solution, an exception is raised (this can only happen whenk
is specified or ifmaximum
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

sage.numerical.optimize.
find_fit
(data, model, initial_guess=None, parameters=None, variables=None, solution_dict=False)¶ 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. IfNone
, 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
) ifTrue
, 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*i0.2) + 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.

sage.numerical.optimize.
find_local_maximum
(f, a, b, tol=1.48e08, 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
find_local_maximum()
.See the documentation for
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...)

sage.numerical.optimize.
find_local_minimum
(f, a, b, tol=1.48e08, 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 thatf
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 tolerancemaxfun
– 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=1e3) (3.28837136189098..., 3.42575079030572...) sage: find_local_minimum(f, 1, 5, tol=1e2, 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 ticket #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 workaround 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 which uses Brent’s method.
AUTHOR:
 William Stein (20071207)

sage.numerical.optimize.
find_root
(f, a, b, xtol=1e12, rtol=8.881784197001252e16, maxiter=100, full_output=False)¶ Numerically find a root of
f
on the closed interval \([a,b]\) (or \([b,a]\)) if possible, wheref
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 equalitya
,b
– endpoints of the intervalxtol
,rtol
– the routine converges when a root is known to lie withinxtol
of the value return. Should be \(\geq 0\). The routine modifies this to take into account the relative precision of doubles. By default, rtol is4*numpy.finfo(float).eps
, the minimum allowed value forscipy.optimize.brentq
, which is what this method uses underneath. This value is equal to2.0**50
for IEEE754 double precision floats as used by Python.maxiter
– integer; if convergence is not achieved inmaxiter
iterations, an error is raised. Must be \(\geq 0\).full_output
– bool (default:False
), ifTrue
, also return object that contains information about convergence.
EXAMPLES:
An example involving an algebraic polynomial function:
sage: R.<x> = QQ[] sage: f = (x+17)*(x3)*(x1/8)^3 sage: find_root(f, 0,4) 2.999999999999995 sage: find_root(f, 0,1) # abs tol 1e6 (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 ticket #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 1e6 1.41421356237
The following is an example, again from trac ticket #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/(x1)+1,0, 2) 0.0 sage: find_root(1/(x1)+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

sage.numerical.optimize.
linear_program
(c, G, h, A=None, b=None, solver=None)¶ 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 vectorG
– a matrixh
– a vectorA
– a matrixb
— a vectorsolver
(optional) — solver to use. If None, the cvxopt’s lpsolver 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 keysx
,s
,y
,z
corresponding to the variables above:sol['x']
– the solution to the linear programsol['s']
– the slack variables for the solutionsol['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 \(x+y50\) 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...)

sage.numerical.optimize.
minimize
(func, x0, gradient=None, hessian=None, algorithm='default', verbose=False, **args)¶ 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 \(n\) componentsx0
– 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'
– (BroydenFletcherGoldfarbShanno) requires gradient'cg'
– (conjugategradient) requires gradient'ncg'
– (newtonconjugate 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
EXAMPLES:
Minimize a fourth order polynomial in three variables (see the Wikipedia article Rosenbrock_function):
sage: vars = var('x y z') sage: f = 100*(yx^2)^2+(1x)^2+100*(zy^2)^2+(1y)^2 sage: minimize(f, [.1,.3,.4]) # abs tol 1e6 (1.0, 1.0, 1.0)
Try the newtonconjugate gradient method; the gradient and hessian are computed automatically:
sage: minimize(f, [.1, .3, .4], algorithm="ncg") # abs tol 1e6 (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 + (1rx[:1r])**2.0r) sage: minimize(rosen, [.1,.3,.4]) # abs tol 3e5 (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 + (1rx[: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*(xmxm_m1**2r)  400r*(xm_p1  xm**2r)*xm  2r*(1rxm) ....: der[0] = 400r*x[0r]*(x[1r]x[0r]**2r)  2r*(1rx[0]) ....: der[1] = 200r*(x[1r]x[2r]**2r) ....: return der sage: minimize(rosen, [.1,.3,.4], gradient=rosen_der, algorithm="bfgs") # abs tol 1e6 (1.0, 1.0, 1.0)

sage.numerical.optimize.
minimize_constrained
(func, cons, x0, gradient=None, algorithm='default', **args)¶ Minimize a function with constraints.
INPUT:
func
– Either a symbolic function, or a Python function whose argument is a tuple with n componentscons
– 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 minimumalgorithm
– Optional, specify the algorithm to use:'default'
– default choices'lbfgsb'
– 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 LBFGSB finds the same minimum:
sage: minimize_constrained(f, [(None,None),(4,10)],[5,5], algorithm='lbfgsb') (4.7..., 4.9...)
Rosenbrock function (see the Wikipedia article Rosenbrock_function):
sage: from scipy.optimize import rosen, rosen_der sage: minimize_constrained(rosen, [(50,10),(5,10)],[1,1],gradient=rosen_der,algorithm='lbfgsb') (10.0, 10.0) sage: minimize_constrained(rosen, [(50,10),(5,10)],[1,1],algorithm='lbfgsb') (10.0, 10.0)