I'm trying to minimize a variable, but z3 takes to long in order to give me a solution.
And I would like to know if it's possible to get a solution when timeout gets triggered.
If yes how can i do that?
Thx in advance!
If by "solution" you mean the latest approximation of the optimal value, then you may be able to retrieve it, provided that the optimization algorithm being used finds any intermediate solution along the way. (Some optimization algorithms --like, e.g., maxres-- don't find any intermediate solution).
Example:
import z3
o = z3.Optimize()
o.add(...very hard problem...)
cf = z3.Int('cf')
o.add(cf = ...)
obj = o.minimize(cf)
o.set(timeout=...)
res = o.check()
print(res)
print(obj.upper())
Even when res = unknown because of a timeout, the objective instance contains the latest approximation of the optimum value found by z3 before the timeout.
Unfortunately, I am not sure whether it is also possible to retrieve the corresponding sub-optimal model with o.model() (or any other method).
For OptiMathSAT, I show how to retrieve the latest approximation of the optimum value and the corresponding model in the unit-test timeout.py.
Related
I want to know which given string is random.
For example, If there was given words (A)"protectsky" and (B)"ptctpkysui", A human would recognize (A) as a meaningful string and (B) as a meaningless string.
In this way, is there any way for the computer to differentiate between any given string in the same way as above?
Machine learning, dictionary matching, entropy calculation, etc.. any method or idea is fine.
It is also okay if the result is a probability value or a specific threshold value.
If there is a better way, please let me know.
This is far from a perfect solution, but it may help you.
You can use the spelling correction suggestions from hunspell. If there are no suggestions, it may be "meaningless". If there are suggestions, it is possibly meaningful. Using your examples:
library(hunspell)
TestWords = c("protectsky", "ptctpkysui")
hunspell_suggest(TestWords)
[[1]]
[1] "protect sky" "protect-sky" "protectorate"
[[2]]
character(0)
You could go a little further, by requiring that the suggestions be not too far from the original.
library(stringdist)
Possible = unlist(hunspell_suggest("protectsky"))
stringdist(Possible, "protectsky")
[1] 1 1 5
Since "protectorate" isn't very close to "protectsky" we would reject it as a possibility. But "protect sky" and "protect-sky"are only at distant of 1, it is likely that they are a "meaningful" version of the test word.
We are using Mosek solver via its Cvxpy interface.
We deal with large-scale optimization problem on a regular-basis and sometimes the runtime is very high. So, we specify a upper limit on runtime using Mosek's mosek.dparam.optimizer_max_time parameter.
In those cases, the pain-point is that we get no solution.
Is it possible to get the suboptimal/best found solution so far?
If Mosek did not find any feasible integer solution within the time limit then there is nothing to return, so you get nothing.
If Mosek found some feasible integer solution then CVXPY should return it with solution status s.OPTIMAL_INACCURATE, judging from a quick look at the code.
So the question is what does it say in the log output and what happens at the end of the optimization when CVXPY is processing the answer from the solver.
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.
So in Lua it's common knowledge that you can use math.randomseed but it's also obvious that math.random sets the seed as well (calling it twice does not return the same result), what does it set it to, and how can I keep track of it, and if it's impossible, please explain why that is so.
This is not a Lua question, but general question on how some RNG algorithm works.
First, Lua don't have their own RNG - they just output you (slightly mangled) value from RNG of underlying C library. Most RNG implementations do not reveal you their inner state, but sometimes you can caclulate it yourself.
For example when you use Lua on Windows, you'll be using LCG-based RNG from MS C library. The numbers you get is a slice of seed, not full value. There are two ways you can deal with that:
If you know how many times you called random, you can just take initial seed value, feed it to your copy of the same algorithm with same constants that are hardcoded in MS library and get exact value of seed.
If you don't, but you can be sure that nobody interferes in between your two calls to random, you can get two generated numbers, and reverse LCG algorithm by shifting bits back to their place. This will leave you with several missing bits (with one more bit thanks to Lua mangling) that you will need to simply bruteforce - just reiterate over all missing bits until your copy of algorithm produces exactly same two "random" numbers you've recorded before. That will be current seed stored inside library's RNG as well. Well programmed solution in Lua can bruteforce this in about 0.2-0.5s on somewhat dated PC - I did it past. Here's example on Crypto.SE talking about this task in more details: Predicting values from a Linear Congruential Generator.
First approach can be used with any other RNG algorithm that doesn't use any real entropy, second with most RNGs that don't mask too much bits in slice to make bruteforcing unreasonable.
Real answer though is: you don't need to keep track of seed at all. What you want is probably something else.
If you set a seed all numbers math.random() generates are pseudo-random (This is always the case as the system will generate a seed by itself).
math.randomseed(4)
print(math.random())
print(math.random())
math.randomseed(4)
print(math.random())
Outputs
0.50827539156303
0.75454387490399
0.50827539156303
So if you reset the seed to the same value you can predict all values that are going to come up to the maximum number of consecutive values that you already generated using that seed.
What the seed does not do is keep the output of math.random() the same. It would be the same if you kept resetting it to the same value.
An analogy as an example
Imagine the random number is an integer between 0 and 9 (instead of a double between 0 and 1).
math.random() could traverse pi's decimals from an arbitrary starting position (default could be system time).
What you do when you use set.seed() is (not literally, this is an analogy as mentioned) set the starting decimals of where in pi you are going to retrieve your numbers.
If you now reset the seed to the same starting position the numbers are going to be the same as the last time you reset the starting position.
You will know the numbers of to the last call, after that you can't be certain anymore.
Firstly , For those of your who dont know - Anytime Algorithm is an algorithm that get as input the amount of time it can run and it should give the best solution it can on that time.
Weighted A* is the same as A* with one diffrence in the f function :
(where g is the path cost upto node , and h is the heuristic to the end of path until reaching a goal)
Original = f(node) = g(node) + h(node)
Weighted = f(node) = (1-w)g(node) +h(node)
My anytime algorithm runs Weighted A* with decaring weight from 1 to 0.5 until it reaches the time limit.
My problem is that most of the time , it takes alot time until this it reaches a solution , and if given somthing like 10 seconds it usaully doesnt find solution while other algorithms like anytime beam finds one in 0.0001 seconds.
Any ideas what to do?
If I were you I'd throw the unbounded heuristic away. Admissible heuristics are much better in that given a weight value for a solution you've found, you can say that it is at most 1/weight times the length of an optimal solution.
A big problem when implementing A* derivatives is the data structures. When I implemented a bidirectional search, just changing from array lists to a combination of hash augmented priority queues and array lists on demand, cut the runtime cost by three orders of magnitude - literally.
The main problem is that most of the papers only give pseudo-code for the algorithm using set logic - it's up to you to actually figure out how to represent the sets in your code. Don't be afraid of using multiple ADTs for a single list, i.e. your open list. I'm not 100% sure on Anytime Weighted A*, I've done other derivatives such as Anytime Dynamic A* and Anytime Repairing A*, not AWA* though.
Another issue is when you set the g-value too low, sometimes it can take far longer to find any solution that it would if it were a higher g-value. A common pitfall is forgetting to check your closed list for duplicate states, thus ending up in a (infinite if your g-value gets reduced to 0) loop. I'd try starting with something reasonably higher than 0 if you're getting quick results with a beam search.
Some pseudo-code would likely help here! Anyhow these are just my thoughts on the matter, you may have solved it already - if so good on you :)
Beam search is not complete since it prunes unfavorable states whereas A* search is complete. Depending on what problem you are solving, if incompleteness does not prevent you from finding a solution (usually many correct paths exist from origin to destination), then go for Beam search, otherwise, stay with AWA*. However, you can always run both in parallel if there are sufficient hardware resources.