How to retrieve the optimization problem class - drake

Is there any way to show what class of optimization problem we have? E.g. SOCP, SDP, or totally nonconvex?
Or do we just need to try a bunch of solvers and see when they fail? Is SNOPT the only one that supports nonconvex optimization?

You can refer to GetProgramType which returns if the program is LP, QP, SDP, etc.
Is SNOPT the only one that supports nonconvex optimization?
No, we also have IPOPT and NLOpt for non-convex nonlinear optimization.

Or do we just need to try a bunch of solvers and see when they fail?
You can use ChooseBestSolver to find the best-suited solver to the program.
from pydrake.solvers import mathematicalprogram as mp
# Make a program.
prog = mp.MathematicalProgram()
x = prog.NewContinuousVariables(2, "x")
prog.AddLinearConstraint(x[0] + x[1] == 0)
prog.AddLinearConstraint(2*x[0] - x[1] == 1)
# Find the best solver.
solver_id = mp.ChooseBestSolver(prog)
solver = mp.MakeSolver(solver_id)
assert solver.solver_id().name() == "Linear system"
# Solve.
result = solver.Solve(prog)
There is also a mp.Solve function that does all of that at once.
Is SNOPT the only one that supports nonconvex optimization?
The documentation has the full list of solvers. At the moment SNOPT, Ipopt, NLopt are the available non-linear solvers.

Related

(Semi-decidable) combination of first-order theories is possible in Z3, but what about an actual semantic/signature-wise combination?

Disclaimer: This is a rather theoretical question, but think it fits here; in case not, let me know an alternative :)
Z3 seems expressive
Recently, I realized I can specify this type of formulae in Z3:
Exists x,y::Integer s.t. [Exists i::Integer s.t. (0<=i<|seq|) & (avg(seq)+t<seq[i])] & (y<Length(seq)) & (y<x)
Here is the code (in Python):
from z3 import *
#Average function
IntSeqSort = SeqSort(IntSort())
sumArray = RecFunction('sumArray', IntSeqSort, IntSort())
sumArrayArg = FreshConst(IntSeqSort)
RecAddDefinition( sumArray
, [sumArrayArg]
, If(Length(sumArrayArg) == 0
, 0
, sumArrayArg[0] + sumArray(SubSeq(sumArrayArg, 1, Length(sumArrayArg) - 1))
)
)
def avgArray(arr):
return ToReal(sumArray(arr)) / ToReal(Length(arr))
###The specification
t = Int('t')
y = Int('y')
x = Int('x')
i = Int('i') #Has to be declared, even if it is only used in the Existential
seq = Const('seq', SeqSort(IntSort()))
avg_seq = avgArray(seq)
phi_0 = And(2<t, t<10)
phi_1 = And(0 <= i, i< Length(seq))
phi_2 = (t+avg_seq<seq[i])
big_vee = And([phi_0, phi_1, phi_2])
phi = Exists(i, big_vee)
phi_3 = (y<Length(seq))
phi_4 = (y>x)
union = And([big_vee, phi_3, phi_4])
phiTotal = Exists([x,y], union)
s = Solver()
s.add(phiTotal)
print(s.check())
#s.model()
solve(phiTotal) #prettier display
We can see how it outputs sat and outputs models.
But...
However, even if this expressivity is useful (at least for me), there is something I am missing: formalization.
I mean, I am combining first-order theories that have different signature and semantics: a sequence-like theory, an integer arithmetic theory and also a (uninterpreted?) function avg. Thus, I would like to combine these theories with a Nelson-Oppen-like procedure, but this procedure only works with quantifier-free fragments.
I mean, I guess this combined theory is semi-decidable (because of quantifiers and because of sequences), but can we formalize it? In case yes, I would like to (in a correct way) combine these theories, but I have no idea how.
An exercise (as an orientation)
Thus, in order to understand this, I proposed a simpler exercise to myself: take the decidable array property fragment (What's decidable about arrays?http://theory.stanford.edu/~arbrad/papers/arrays.pdf), which has a particular set of formulae and signature.
Now, suppose I want to add an avg function to it. How can I do it?
Do I have to somehow combine the array property fragment with some kind of recursive function theory and an integer theory? How would I do this? Note that these theories involve quantifiers.
Do I have to first combine these theories and then create a decision procedure for the combined theory? How?
Maybe it suffices with creating a decision procedure within the array property fragment?
Or maybe it suffices with a syntactic adding to the signature?
Also, is the theory array property fragment with an avg function still decidable?
A non-answer answer: Start by reading Chapter 10 of https://www.decision-procedures.org/toc/
A short answer: Unless your theory supports quantifier-elimination, SMT solvers won't have a decision-procedure. Assuming they all admit quantifier-elimitation, then you can use Nelson-Oppen. Adding functions like avg etc. do not add significantly to expressive power: They are definitions that are "unfolded" as needed, and so long as you don't need induction, they're more or less conveniences. (This is a very simplified account, of course. In practice, you'll most likely need induction for any interesting property.)
If these are your concerns, it's probably best to move to more expressive systems than push-button solvers. Start looking at Lean: Sure, it's not push-button, but it provides a very nice framework for general purpose theorem proving: https://leanprover.github.io
Even longer answer is possible, but stack-overflow isn't the right forum for it. You're now looking into the the theory of decision procedures and theorem-proving, something that won't fit into any answer in any internet-based forum. (Though https://cstheory.stackexchange.com might be better, if you want to give it a try.)

How to improve binary search based optimization in Z3py

I am trying to optimize with Z3py an instance of Set Covering Problem (SCP41) based on minimize.
The results are the following:
Using
(1) I know that Z3 supports optimization (https://rise4fun.com/Z3/tutorial/optimization). Many times I get to the optimum in SCP41 and others instances, a few do not.
(2) I understand that if I use the Z3py API without the optimization module I would have to do the typical sequential search described in (Minimum and maximum values of integer variable) by #Leonardo de Moura. It never gives me results.
My approach
(3) I have tried to improve the sequential search approach by implementing a binary search similar to how it explains #Philippe in (Does Z3 have support for optimization problems), when I run my algorithm it waits and I do not get any result.
I understand that the binary search should be faster and work in this case? I also know that the instance SCP41 is something big and that many restrictions are generated and it becomes extremely combinatorial, this is my full code (Code large instance) and this is my binary search it:
def min(F, Z, M, LB, UB, C):
i = 0
s = Solver()
s.add(F[0])
s.add(F[1])
s.add(F[2])
s.add(F[3])
s.add(F[4])
s.add(F[5])
r = s.check()
if r == sat:
UB = s.model()[Z]
while int(str(LB)) <= int(str(UB)):
C = int(( int(str(LB)) + int(str(UB)) / 2))
s.push()
s.add( Z > LB, Z <= C)
r = s.check()
if r==sat:
UB = Z
return s.model()
elif r==unsat:
LB = C
s.pop()
i = i + 1
if (i > M):
raise Z3Exception("maximum not found, maximum number of iterations was reached")
return unsat
And, this is another instance (Code short instance) that I used in initial tests and it worked well in any case.
What is incorrect binary search or some concept of Z3 not applied correctly?
regards,
Alex
I don't think your problem is to do with minimization itself. If you put a print r after r = s.check() in your program, you see that z3 simply struggles to return a result. So your loop doesn't even execute once.
It's impossible to read through your program since it's really large! But I see a ton of things of the form:
Or(X250 == 0, X500 == 1)
This suggests your variables X250 X500 etc. (and there's a ton of them) are actually booleans, not integers. If that is indeed true, you should absolutely stick to booleans. Solving integer constraints is significantly harder than solving pure boolean constraints, and when you use integers to model booleans like this, the underlying solver simply explores the search space that's just unreachable.
If this is indeed the case, i.e., if you're using Int values to model booleans, I'd strongly recommend modelling your problem to get rid of the Int values and just use booleans. If you come up with a "small" instance of the problem, we can help with modeling.
If you truly do need Int values (which might very well be the case), then I'd say your problem is simply too difficult for an SMT solver to deal with efficiently. You might be better off using some other system that is tuned for such optimization problems.

Z3 getting last valid model

I using Z3 C++ api to find a satisfiable formula that is minimal with respect to some boolean variables (let us call them b0,...,bn) being true.
I have a formula that includes boolean variables b0,...,bn and I want to find some satisfiable formula where I have the least number of b0,...,bn set to true.
I do this by initially finding a subset of b0,...,bn that can be assigned to true and satisfy my formula, and I incrementally ask the solver to find smaller subsets (i.e. where one of these boolean variables is flipped to false).
I find my local minimum when I cannot find a smaller subset, i.e. I get a unsat result from the Z3. At this point, I would like to access the last valid model.
Is that possible? Does Z3 modify the model when a call to "check" is unsat?
If so, how can I do this using the C++ api?
Many thanks in advance,
You can retrieve a model if the solver returns "sat". The model refers to the state of the solver, so if you add assertions, the state changes and models are no longer valid until you check satisfiability and it returns sat.
So you can retrieve a model every time the solver returns SAT, and then discharge all but the last model.
As Nikolaj mentioned, you need to keep track of models after each call that results in sat and return the last one when you get an unsat if you follow the strategy you outlined.
However, there might be another alternative that avoids repeated calls altogether. Instead of a satisfaction problem, you can cast your problem as an optimization one. You mentioned you have control variables b0, b1, .. bn such that you want to minimize the number of them getting set to true for a satisfying model. Create a metric that counts the number of ones in these variables. Something like:
metric = (if b0 then 1 else 0)
+ (if b1 then 1 else 0)
+ ...
+ (if bn then 1 else 0)
Then use Z3's optimization routines to minimize metric. I believe this will provide you with the solution you are looking for in one call only.
Some helpful references:
Here's the Z3 optimization tutorial: http://rise4fun.com/z3opt/tutorialcontent/guide.
This example, in particular, talks about soft-constraints, and might quite be applicable in your case as well: http://rise4fun.com/z3opt/tutorialcontent/guide#h25.
Here's the C++ API reference for the optimizer: http://z3prover.github.io/api/html/classz3_1_1optimize.html.

Use Z3 to determine difficulty of quantifier elimination for BV-queries

I'm currently using the Z3 C++ API for solving queries over bitvectors. Some queries may contain an existential quantifier at the top level.
Often times the quantifier elimination is simple and can be performed by Z3 quickly. However, in those cases where the quantifier elimination falls back to enumerating thousands of feasible solutions I'd like to abort this tactic and handle the query myself in some other way.
I've tried wrapping the 'qe'-tactic with a 'try-for'-tactic, hoping that if quantifier elimination fails (in say 100ms) I'd know that I'd better handle the query in some other way. Unfortunately, the 'try-for'-tactic fails to cancel the quantifier elimination (for any time bound).
In an old post a similar issue is discussed and the 'smt' tactic is being blamed for being not responsive. Does the same reasoning apply to the 'qe' tactic? The same post indicates that 'future' versions should be more responsive though. Is there any way or heuristic to determine whether quantifier elimination would take long (besides running the solver in a separate thread and killing it on timeout)?
I've attached a minimal example so you can try it yourselves:
z3::context ctx;
z3::expr bv1 = ctx.bv_const("bv1", 10);
z3::expr bv2 = ctx.bv_const("bv2", 10);
z3::goal goal(ctx);
goal.add(z3::exists(bv1, bv1 != bv2));
z3::tactic t = z3::try_for(z3::tactic(ctx,"qe"), 100);
auto res = t.apply(goal);
std::cout << res << std::endl;
Thanks!
The timeout cancellations have to be checked periodically by the tactic that is running.
We basically have to ensure that the code checks for cancellations and does not descend into a long running loop without checking. You can probably identify the code segment that fails to check for cancellation by running your code in a debugger, break and then determine which procedures it is in. Then file a bug on GitHub to have cancellation flag checked in the place that will help.
Overall, the quantifier elimination tactic is currently fairly simplistic when it comes to bit-vectors so it would be better to avoid qe for all but simple cases.

Does Z3 have support for optimization problems

I saw in a previous post from last August that Z3 did not support optimizations.
However it also stated that the developers are planning to add such support.
I could not find anything in the source to suggest this has happened.
Can anyone tell me if my assumption that there is no support is correct or was it added but I somehow missed it?
Thanks,
Omer
If your optimization has an integer valued objective function, one approach that works reasonably well is to run a binary search for the optimal value. Suppose you're solving the set of constraints C(x,y,z), maximizing the objective function f(x,y,z).
Find an arbitrary solution (x0, y0, z0) to C(x,y,z).
Compute f0 = f(x0, y0, z0). This will be your first lower bound.
As long as you don't know any upper-bound on the objective value, try to solve the constraints C(x,y,z) ∧ f(x,y,z) > 2 * L, where L is your best lower bound (initially, f0, then whatever you found that was better).
Once you have both an upper and a lower bound, apply binary search: solve C(x,y,z) ∧ 2 * f(x,y,z) > (U - L). If the formula is satisfiable, you can compute a new lower bound using the model. If it is unsatisfiable, (U - L) / 2 is a new upper-bound.
Step 3. will not terminate if your problem does not admit a maximum, so you may want to bound it if you are not sure it does.
You should of course use push and pop to solve the succession of problems incrementally. You'll additionally need the ability to extract models for intermediate steps and to evaluate f on them.
We have used this approach in our work on Kaplan with reasonable success.
Z3 currently does not support optimization. This is on the TODO list, but it has not been implemented yet. The following slide decks describe the approach that will be used in Z3:
Exact nonlinear optimization on demand
Computation in Real Closed Infinitesimal and Transcendental Extensions of the Rationals
The library for computing with infinitesimals has already been implemented, and is available in the unstable (work-in-progress) branch, and online at rise4fun.

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