I am trying to set "mosek_param" settings, but, am getting errors. For instance, for the following case
MSK_IPAR_INTPNT_SOLVE_FORM
Controls whether the primal or the dual problem is solved.
Default:
"FREE"
Accepted:
"FREE", "PRIMAL", "DUAL"
Example:
param.MSK_IPAR_INTPNT_SOLVE_FORM = 'MSK_SOLVE_FREE'
Groups:
Interior-point method
from https://docs.mosek.com/9.0/toolbox/parameters.html --> I tried
prob.solve(solver=MOSEK,
mosek_params={mosek.iparam.intpnt_solve_form: mosek.solve.primal}, # mosek.iparam.presolve_use:mosek.presolvemode.off
verbose=True)
but, run into errors .... the commented part works.
When I was working in Matlab --> using
cvx_solver_settings('MSK_IPAR_INTPNT_SOLVE_FORM','MSK_SOLVE_PRIMAL')
worked well for me. But, doesn't work in the present case. Also, I was able to set precision as follows
cvx_precision low
but, cannot do so now. Is there another way to do both of these in cvxpy? thank you.
PS: this question has also been posted in the CVXPY forum --> https://groups.google.com/forum/#!topic/cvxpy/MEAewGMlqjI
Below is an example
# Solves a bounded least-squares problem.
import mosek
from cvxpy import *
import numpy
# Problem data.
m = 10
n = 5
numpy.random.seed(1)
A = numpy.random.randn(m, n)
b = numpy.random.randn(m)
# Construct the problem
x = Variable(n)
objective = Minimize(sum_squares(A*x - b))
constraints = [0 <= x, x <= 1]
prob = Problem(objective, constraints)
prob.solve(solver=MOSEK,
mosek_params={mosek.iparam.intpnt_solve_form: mosek.solve.primal}, # mosek.iparam.presolve_use:mosek.presolvemode.off
verbose=True)
which gives me the error. I also tried using 'MSK_IPAR_INTPNT_SOLVE_FORM' = 'MSK_SOLVE_PRIMAL' but to no avail.
Look at
https://docs.mosek.com/9.1/pythonapi/parameters.html#mosek.iparam.intpnt_solve_form
The correct form is
mosek.solveform.primal
Related
I'm still learning to use caret package through The caret Package by Max Kuhn and got stuck in 16.2 Partial Least Squares Discriminant Analysis section while trying to predict using the plsBayesFit model through predict(plsBayesFit, head(testing), type = "prob") as shown in the book as well.
The data used is data(Sonar) from mlbench package, with the data being split as:
inTrain <- createDataPartition(Sonar$Class, p = 2/3, list = FALSE)
sonarTrain <- Sonar[ inTrain, -ncol(Sonar)]
sonarTest <- Sonar[-inTrain, -ncol(Sonar)]
trainClass <- Sonar[ inTrain, "Class"]
testClass <- Sonar[-inTrain, "Class"]
and then preprocessed as follows:
centerScale <- preProcess(sonarTrain)
centerScale
training <- predict(centerScale, sonarTrain)
testing <- predict(centerScale, sonarTest)
after this the model is trained using plsBayesFit <- plsda(training, trainClass, ncomp = 20, probMethod = "Bayes"), followed by predicted using predict(plsBayesFit, head(testing), type = "prob").
When I'm trying to do this I get the following error:
Error in predict.NaiveBayes(object$probModel[[ncomp[i]]], as.data.frame(tmpPred[, : Not all variable names used in object found in newdata
I've checked both the training and testing sets to check for any missing variable but there isn't any. I've also tried to predict using the 2.7.1 version of pls package which was used to render the book at that time but that too is giving me same error. What's happening?
I've tried to replicate your problem using different models, as I have encountered this error as well, but I failed; and caret seems to behave differently now from when I used it.
In any case stumbled upon this Github-issues here, and it seems like that there is a specific problem with the klaR-package. So my guess is that this is simply a bug - and nothing that can be readily fixed here!
I want to find the range of valid values that a variable can have, given some constraints. Eg,
x = Int('x')
s = Solver()
s.add(x >= 1)
s.add(x < 5+2)
Is there some way that I can get z3 to print 1..6 for this variable?
I tried using the following, but range() applies only to declarations.
print("x.range():", x.range()) # this does not work
Note: 1. This question seems to ask the same, but I did not understand its answers, and I am looking for python answer.
in reply to #Malte: I am not looking for all the answers, I just want to simplify multiple constraints in to a valid range. If constraints on both sides of the variable cannot be merged, then at least only on one side as is mentioned in above mentioned question.
This question comes up occasionally, and the answer isn't very trivial, unfortunately. It really depends on what your constraints are and exactly what you are trying to do. See:
Is it possible to get a legit range info when using a SMT constraint with Z3
And
(Sub)optimal way to get a legit range info when using a SMT constraint with Z3
Essentially, the problem is too difficult (and I'd say not even well defined) if you have multiple variables. If you have exactly one variable, you can use the optimizer to some extent, assuming the variable is indeed bounded. In case you have multiple variables, one idea might be to fix all but one to satisfying constants, and compute the range of that last variable based on the constant assignment to the others. But again, it depends on what you're really trying to achieve.
Please take a look at the above two answers and see if it helps you. If not, please show us what you tried: Stack-overflow works the best when you post some code and see how it can be improved/fixed.
As a SAT/SMT solver, Z3 "only" needs to find a single model (satisfying assignment) to show that a formula is satisfiable. Finding all models is therefore not directly supported.
The question comes up regularly, though, and the solution is to repeatedly find and then block (assume in negated form) models until no further model can be found. For example, for your snippet of code:
x = Int('x')
s = Solver()
s.add(x >= 1)
s.add(x < 5+2)
result = s.check()
while result == sat:
m = s.model()
print("Model: ", m)
v_x = m.eval(x, model_completion=True)
s.add(x != v_x)
result = s.check()
print(result, "--> no further models")
Executing the script yields the solution you asked for, albeit in a less concise form:
Model: [x = 1]
Model: [x = 2]
Model: [x = 3]
Model: [x = 4]
Model: [x = 5]
Model: [x = 6]
unsat --> no further models
In general,
you would have iterate over all variables (here: just x)
model completion is necessary for variables whose value doesn't affect satisfiability; since any value will do, they won't be explicit in the model
Related questions whose answers provide additional details:
(Z3Py) checking all solutions for equation
Why Z3Py does not provide all possible solutions
Getting all solutions of a boolean expression in Z3Py never ends
Drake has an interface where you can give it a generic function as a constraint and it can set up the nonlinearly-constrained mathematical program automatically (as long as it supports AutoDiff). I have a situation where my constraint does not support AutoDiff (the constraint function conducts a line search to approximate the maximum value of some function), but I have a closed-form expression for the gradient of the constraint. In my case, the math works out so that it's difficult to find a point on this function, but once you have that point it's easy to linearize around it.
I know many optimization libraries will allow you to provide your own analytical gradient when available; can you do this with Drake's MathematicalProgram as well? I could not find mention of it in the MathematicalProgram class documentation.
Any help is appreciated!
It's definitely possible, but I admit we haven't provided helper functions that make it pretty yet. Please let me know if/how this helps; I will plan to tidy it up and add it as an example or code snippet that we can reference in drake.
Consider the following code:
from pydrake.all import AutoDiffXd, MathematicalProgram, Solve
prog = MathematicalProgram()
x = prog.NewContinuousVariables(1, 'x')
def cost(x):
return (x[0]-1.)*(x[0]-1.)
def constraint(x):
if isinstance(x[0], AutoDiffXd):
print(x[0].value())
print(x[0].derivatives())
return x
cost_binding = prog.AddCost(cost, vars=x)
constraint_binding = prog.AddConstraint(
constraint, lb=[0.], ub=[2.], vars=x)
result = Solve(prog)
When we register the cost or constraint with MathematicalProgram in this way, we are allowing that it can get called with either x being a float, or x being an AutoDiffXd -- which is simply a wrapping of Eigen's AutoDiffScalar (with dynamically allocated derivatives of type double). The snippet above shows you roughly how it works -- every scalar value has a vector of (partial) derivatives associated with it. On entry to the function, you are passed x with the derivatives of x set to dx/dx (which will be 1 or zero).
Your job is to return a value, call it y, with the value set to the value of your cost/constraint, and the derivatives set to dy/dx. Normally, all of this happens magically for you. But it sounds like you get to do it yourself.
Here's a very simple code snippet that, I hope, gets you started:
from pydrake.all import AutoDiffXd, MathematicalProgram, Solve
prog = MathematicalProgram()
x = prog.NewContinuousVariables(1, 'x')
def cost(x):
return (x[0]-1.)*(x[0]-1.)
def constraint(x):
if isinstance(x[0], AutoDiffXd):
y = AutoDiffXd(2*x[0].value(), 2*x[0].derivatives())
return [y]
return 2*x
cost_binding = prog.AddCost(cost, vars=x)
constraint_binding = prog.AddConstraint(
constraint, lb=[0.], ub=[2.], vars=x)
result = Solve(prog)
Let me know?
For solving a SAT-problem I decided to use the Z3-solver from Microsoft and Python 3. The aim is to take a long model (up to 500,000 features) and find all possible solutions. To find them I want to add the first solution S1 to the initial equation and exclude S1 it and so forth. I will do it using a while-loop.
Solving a SAT-problem is important for me, because I wanna analyse feature models.
But I'm facing a problem with adding sth to the initial equation. I will share a minimal example:
# Import statements
import sys
sys.path.insert(0,'/.../z3/bin')
from z3 import * # https://github.com/Z3Prover/z3/wiki
def main():
'''
Solves after transformation a given boolean equation by using the Z3-Solver from Microsoft.
'''
fd = dict()
fd['_r'] = Bool('_r')
fd['_r_1'] = Bool('_r_1')
equation = '''And(fd.get("_r"),Or(Not(fd.get("_r")),fd.get("_r_1")))'''
# Solve the equation
s = Solver()
exec('s.add(' + equation + ')')
s.check()
print(s.model())
###################################
# Successfull until here.
###################################
s.add(Or(fd['_r'] != bool(s.model()[fd.get('_r')])))
# s.add(Or(fd['_r'] != False))
s.check()
print(s.model())
if __name__=='__main__':
main()
The first coded line after # Successfull... throws an z3types.Z3Exception: model is not available error. So i tried the line above adding simply false to the model. That works just fine.
I'm stucked here. I believe the error is easy to solve, but I don't see the solution. Does one of you? Thanks!
Models become available only after s.check() returns 'sat'.
The model maps Boolean propositions to {True, False} and
generally maps constants and functions to fixed values.
The requirement is that the model provides an interpretation
that satisfies the formula that is added to the solver 's'.
We don't know whether the solver state is satisfiable before
we have called 's.check()'.
Suppose you want to say:
s.add(Or(fd['_r'] != bool(s.model()[fd.get('_r')])))
meaning that in the model that satisfies the constraint should
have the property that if '_r' is true under the model, then
fd['_r'] != True, and if '_r' is false under the model, then
fd['_r'] != False. This is equivalent to saying
fd['_r'] != '_r'. So it isn't really necessary to access the value
of '_r' in whatever model may evaluate '_r' in order to say something
about the evaluation of it.
I basically want Sympy to generate latex code \frac{x-1}{3} = y but whenever I ask it generate the Tex component of things Sympy always returns \frac{x}{3} - \frac{1}{3} .
How do I avoid splitting up the equations, and assign an equals operator to another variable.
I have not attempted to add the "y =" part to the code yet as I wanted to clarify the fraction situation first, but since I have had to come cap in hand to stack exchange I thought I would ask both questions. I have been through every tutorial page I could find but to no avail.
Any help would be much appreciated although I would ask you keep it relatively simple !!!
Thanks in advance .
import sympy
from sympy import *
x = Symbol("x")
a = (x-Integer(1))
b = (3)
c = a/b
print(latex(c))
The issue comes from sympy automatically expanding (x-1)/3 into x/3-1/3. So one solution is to ask sympy to factor this back:
In [18]: c = a/b; c
Out[18]: x/3 - 1/3
In [19]: d = c.factor(); d
Out[19]: (x - 1)/3
In [20]: print(latex(d))
\frac{1}{3} \left(x - 1\right)