Given a real number like Real('x') in Z3Py, how do I get its floor or ceiling? Equivalently, how do I round it? I see that to_int is a thing in the SMT specification, but I don't know how to call it from Python.
The function "ToInt" takes the floor.
from z3 import *
x = Real('x')
print ToInt(x)
print ToInt(RealVal("1/2"))
print simplify(ToInt(RealVal("1/2")))
Related
I am getting the error "dtw() got an unexpected keyword argument 'dist'" while I'm trying to calculate the dtw of 2 wav files. I can't figure out why or what to do to fix it. I am attaching the code below.
import librosa
import librosa.display
y1, sr1 = librosa.load('sample_data/Abir_Arshad_22.wav')
y2, sr2 = librosa.load('sample_data/Abir_Arshad_22.wav')
%pylab inline
subplot(1, 2, 1)
mfcc1 = librosa.feature.mfcc(y1, sr1)
librosa.display.specshow(mfcc1)
subplot(1, 2, 2)
mfcc2 = librosa.feature.mfcc(y2, sr2)
librosa.display.specshow(mfcc2)
from dtw import dtw
from numpy.linalg import norm
dist, cost, acc_cost, path = dtw(mfcc1.T, mfcc2.T, dist=lambda x, y: norm(x - y, ord=1))
print ('Normalized distance between the two sounds:', dist)
the error is occurring in the 2nd last line.
The error message is straight forward. Lets read the docs of the method you are calling:
https://dynamictimewarping.github.io/py-api/html/api/dtw.dtw.html#dtw.dtw
The dtw function has the following parameters:
Parameters x – query vector or local cost matrix
y – reference vector, unused if x given as cost matrix
dist_method – pointwise (local) distance function to use.
step_pattern – a stepPattern object describing the local warping steps
allowed with their cost (see [stepPattern()])
window_type – windowing function. Character: “none”, “itakura”,
“sakoechiba”, “slantedband”, or a function (see details).
open_begin,open_end – perform open-ended alignments
keep_internals – preserve the cumulative cost matrix, inputs, and
other internal structures
distance_only – only compute distance (no backtrack, faster)
You try to pass an argument named dist and that argument simply is not known.
Instead, removing that argument would solve the issue, such as
dist, cost, acc_cost, path = dtw(mfcc1.T, mfcc2.T)
I'm trying to obtain a numerical solution to the following integral:
1
The correct answer is -0.324 + 0.382i but as seen below I am not getting a numerical answer and would appreciate help with the Maxima syntax.
2
Perhaps related to why I am not getting a numerical output are two specific questions:
I read that e and i in Maxima need to be preceded by % in input but should these also appear as %e and %i as seen in the Maxima output?
Why is dy missing at the end of the integral in the Maxima output?
Thank you!
Looks to me like your input is okay, however, the function to compute approximations to integrals is named quad_qags. (There are actually several related functions. See ?? quad_ for more info.) Also, a wrinkle here is that the integrand is a complex-valued function (of a real variable), and quad_qags can only work on real-valued integrands, so we'll have to work around it. Here's how I would arrange it.
myintegrand: exp(%i*(1 + %i*y))/(1 + %i*y + 1/(1 + %i*y));
result_realpart: quad_qags (realpart (myintegrand), y, 0, 6);
result_imagpart: quad_qags (imagpart (myintegrand), y, 0, 6);
result: result_realpart[1] + %i*result_imagpart[1];
I get 0.3243496676292901*%i + 0.3820529930785175 as the final result. That's a little different from what you said; maybe a minus sign went missing? or there's a missing or extra factor of %i?
A quick approximation
0.1 * lsum (x, x, float (rectform (makelist (ev (myintegrand, y = k/10), k, 0, 60))));
seems to show the result from quad_qags is reasonable.
I am trying to formulate a trajectory optimization problem for a glider, where I want to maximize the average horisontal velocity. I have formulated the system as a drakesystem, and the state vector consists of the position and velocity.
Currently, I have something like the following:
dircol = DirectCollocation(
plant,
context,
num_time_samples=N,
minimum_timestep=min_dt,
maximum_timestep=max_dt,
)
... # other constraints etc
horisontal_pos = dircol.state()[0:2] # Only (x,y)
time = dircol.time()
dircol.AddFinalCost(-w.T.dot(horisontal_pos) / time)
where AddFinalCost() should replace all instances of state() and time() with the final values, as far as I understand from the documentation. min_dt is non-zero and w is a vector of linear weights.
However, I am getting the following error message
Expression (...) is not a polynomial. ParseCost does not support non-polynomial expression.
which makes me think that there is no way of adding the type of cost function that I am looking for. Is there anything that I am missing?
Thank you in advance!
When calling AddFinalCost(e) with e being a symbolic expression, we can only handle it when e is a polynomial function of the state (more precisely, either a quadratic function or a linear function). Hence the error you see complaining that the cost is not polynomial.
You could add the cost like this
def average_speed(v):
x = v[0]
time_steps = v[1:]
return v[0] / np.sum(time_steps)
h_vars = [dircol.timestep[i] for i in range(N-1)]
dircol.AddCost(average_speed, vars=[dircol.state(N-1)[0]] + h_vars)
which uses a function average_speed to evaluate the average speed. You could find example of doing this in https://github.com/RobotLocomotion/drake/blob/e5f3c3e5f7927ef675066d97d3afac55d3481305/bindings/pydrake/solvers/test/mathematicalprogram_test.py#L590
First, the cost function should be a scalar, but you a vector-valued horisontal_pos / time, which has two entries containing both position_x / dt and position_y / dt, namely a vector as the cost. You should instead provide a scalar valued cost.
Second, it is unclear to me why you divide time in the final cost. As far as I understand it, you want the final position to be close to the origin, so something like position_x² + position_y². The code can look like
dircol.AddFinalCost(horisontal_pos[0]**2 + horisontal_pos[1]**2)
If I understand correctly, cvxpy converts our high-level problem description to the standard canonical form before it is sent to a solver.
By the standard form I mean the form that can be used for the descent algorithms, so, for instance, it would convert all the absolute values in the objective to be a difference of two positive numbers with some new constraints, etc.
Wondering if its possible to see what the reduction looked like for a problem I specify in cvxpy?
For instance, lets say I have the following problem:
import numpy as np
import cvxpy as cp
x = cp.Variable(2)
L = np.asarray([[1,2],[2,3]])
P = L.T # L
constraints = []
constraints.append(x >= [-10, -10])
constraints.append(x <= [10, 10])
obj = cp.Minimize(cp.quad_form(x, P) - [1, 2] * x)
prob = cp.Problem(obj, constraints)
prob.solve(), prob.solver_stats.solver_name
(-0.24999999999999453, 'OSQP')
So, I would like to see the actual arguments (P, q, A, l, u) being sent to the OSQP solver https://github.com/oxfordcontrol/osqp-python/blob/master/module/interface.py#L278
Any help is greatly appreciated!
From looking at the documentation, it seems you can do this using the command get_problem_data as follows:
data, chain, inverse_data = prob.get_problem_data(prob.solver_stats.solver_name)
I have not tried it, and it says it output depends on the particular solver and the solver chain, but it may help you!
Let's assume a very simple constraint: solve(x > 0 && x < 5).
Can Z3 (or any other SMT solver, or any other automatic technique)
compute the minimum and maximum values of (integer) variable x that satisfies the given constraints?
In our case, the minimum is 1 and the maximum is 4.
Z3 has not support for optimizing (maximizing/minimizing) objective functions or variables.
We plan to add this kind of capability, but it will not happen this year.
In the current version, we can "optimize" an objective function by solving several problems where in each iteration we add additional constraints. We know we found the optimal when the problem becomes unsatisfiable. Here is a small Python script that illustrates the idea. The script maximizes the value of a variable X. For minimization, we just have to replace s.add(X > last_model[X]) with s.add(X < last_model[X]). This script is very naive, it performs a "linear search". It can be improved in many ways, but it demonstrates the basic idea.
You can also try the script online at: http://rise4fun.com/Z3Py/KI1
See the following related question: Determine upper/lower bound for variables in an arbitrary propositional formula
from z3 import *
# Given formula F, find the model the maximizes the value of X
# using at-most M iterations.
def max(F, X, M):
s = Solver()
s.add(F)
last_model = None
i = 0
while True:
r = s.check()
if r == unsat:
if last_model != None:
return last_model
else:
return unsat
if r == unknown:
raise Z3Exception("failed")
last_model = s.model()
s.add(X > last_model[X])
i = i + 1
if (i > M):
raise Z3Exception("maximum not found, maximum number of iterations was reached")
x, y = Ints('x y')
F = [x > 0, x < 10, x == 2*y]
print max(F, x, 10000)
As Leonardo pointed out, this was discussed in detail before: Determine upper/lower bound for variables in an arbitrary propositional formula. Also see: How to optimize a piece of code in Z3? (PI_NON_NESTED_ARITH_WEIGHT related).
To summarize, one can either use a quantified formula, or go iteratively. Unfortunately, these techniques are not equivalent:
Quantified approach needs no iteration, and can find global min/max in a single call to the solver; at least in theory. However, it does give rise to harder formulas. So, the backend solver can time-out, or simply return "unknown".
Iterative approach creates simple formulas for the backend solver to deal with, but it can loop forever if there's no optimal value; simplest example being trying to find the largest Int value. Quantified version can solve this problem nicely by quickly telling you that there is no such value, while the iterative version would go on indefinitely. This can be a problem if you don't know ahead of time that your constraints do have an optimal solution. (Needless to say, the "sufficient" iteration count is typically hard to guess, and might depend on random factors, like the seed used by the solver.)
Also keep in mind that if there is a custom optimization algorithm for the problem domain at hand, it's unlikely that a general purpose SMT solver can outperform it.
z3 now supports optimization.
from z3 import *
o = Optimize()
x = Int( 'x' )
o.add(And(x > 0, x < 5))
o.maximize(x)
print(o.check()) # prints sat
print(o.model()) # prints [x = 4]
This particular problem is an integer program.