dpll | Davis–Putnam–Logemann–Loveland algorithm | Learning library
kandi X-RAY | dpll Summary
kandi X-RAY | dpll Summary
An implementation of the Davis-Putnam-Logemann-Loveland (DPLL) algorithm for solving the CNF-SAT problem. This algorithm decides the satisfiability of propositional logic formulae in conjunctive normal form (CNF) - i.e. an AND of ORs.
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QUESTION
Backtracking search is a well-known problem-solving technique, that recurs through all possible combinations of variable assignments in search of a valid solution. The general algorithm is abstracted into a concise higher-order function: https://en.wikipedia.org/wiki/Backtracking
Some problems require partial backtracking, that is, they have a mixture of don't-know non-determinism (have a choice to make, that matters, if you get it wrong you have to backtrack) and don't-care non-determinism (have a choice to make, that doesn't matter, maybe it matters for how long it takes you to find the solution, but not for the correctness thereof, you don't have to backtrack).
Consider for example the Boolean satisfiability problem that can be solved with the DPLL algorithm. If you try to represent that with the general backtracking algorithm, the result will not only recur through all 2^N variable assignments (which is sadly necessary in the general case), but all N! orders of trying the variables (completely unnecessary and hopelessly inefficient).
Is there a general algorithm for partial backtracking? A concise higher-order function that takes function parameters for both don't-know and don't-care choices?
...ANSWER
Answered 2020-Sep-07 at 02:34If I understand you correctly, you’re asking about symmetry-breaking in tree search. In the specific example you gave, all permutations of the list of variable assignments are equivalent.
Symmetries are going to be domain-specific. So is the more-general technique of pruning the search tree, by short-circuiting and backtracking eagerly. There are a few symmetry-breaking techniques I’ve used that generalize.
One is to search the problem space in a canonical order. If the branch that sets variable 10 only tries variables 11, 12 and up, not variables 9, 8 or 7, it won’t search any permutation of the same solution. It will only test solutions that are unique up to permutation. (In the specific case of SAT-solving, this might rule out an optimal search order—although you could re-order the variables arbitrarily.)
Another is to make a test that only one distinct solution of any equivalence class will pass, ideally one that can be checked near the top of the search tree. The classic example of this is, in the 8-queens problem, checking whether the queen on the row you look at first is on the left or the right side of the chessboard. Any solution where she’s on the right is a mirror-image of one other solution where she’s on the left, so you can cut the search space in half. (You can actually do better than this with that problem.) If you only need to test for satisfiability, you can get by with a filter that merely guarantees that, if any solution exists, at least one solution will pass.
If you have enough memory, you might also store a set of branches that have already been searched, and then check whether a branch that you are considering whether to search is equivalent to one already in the set. This would be more practical for a search space with a huge number of symmetries than one with a huge number of solutions unique up to symmetry.
QUESTION
I recently work on the DPLL problem and I find an interesting situation. I cannot describe my question very well, but here is an example to show it.
The first one is a recursive function without class.
...ANSWER
Answered 2020-Jul-17 at 19:50In the first example you do my_list+[variable1], which creates a new list that is the concatenation of my_list and [variable1]. This is also referred to as an "out-of-place" add. This creates a copy of the list which is passed down to the next invocation of the function.
In the second example you are appending to the list instead (an "in-place" add). There is no copy created in this case, and there is only one list object used in the entire run of test([1, 2, 3, 4])
QUESTION
I am using Runtime.getRuntime().exec() method for running 'optimathsat.exe' file. My code is like
...ANSWER
Answered 2018-May-02 at 05:02You are using cmd for 2 things:
- Setting the current directory
- Redirecting input and output
Rather than using cmd for that, you should use Java's ProcessBuilder.
QUESTION
I wanna implement the DPLL algorithm. Therefore i have to remove all occurrencies of a variable in a list of other variables, e.g. deleting neg(X) out of [neg(X), pos(X), neg(Y), pos(Y)] should return [pos(X), neg(Y), pos(Y)].
I've tried some built-in predicates like exclude/3 or delete/3 but all left me with asuming X = Y and a result [pos(X), pos(Y)], with all neg(_) removed, but I only want neg(X) removed and not neg(Y). Is this somehow possible?
ANSWER
Answered 2020-Jan-31 at 16:31From the Logtalk library list object:
QUESTION
I'm implementing DPLL algorithm that counts the number of visited nodes. I managed to implement DPLL that doesn't count visited nodes but I can't think of any solutions to the problem of counting. The main problem is that as the algorithm finds satisfying valuation and returns True, the recursion rolls up and returns counter from the moment the recursion started. In any imperative language I would just use global variable and increment it as soon as function is invoked, but it is not the case in Haskell.
The code I pasted here does not represent my attempts to solve the counting problem, it is just my solution without it. I tried to use tuples such as (True,Int) but it will always return integer value from the moment the recursion started.
This is my implementation where (Node -> Variable) is a heuristic function, Sentence is list of clauses in CNF to be satisfied, [Variable] is a list of Literals not assigned and Model is just a truth valuation.
...ANSWER
Answered 2017-Mar-13 at 21:48Basically what you described as your solution in imperative language can be modeled by passing around a counting variable, adding the variable to the result at the time you return it (the bottom of recursion that reaches satisfiable assignment), i.e. for a function a -> b you would create a new function a -> Int -> (b, Int). The Int argument is the current state of the counter, the result is enriched with the updated state of the counter.
This can further be re-expressed elegantly using the state monad. A very nice tutorial on haskell in general and state monad is here. Basically the transformation of a -> b to a -> Int -> (b, Int) can be seen as transformation of a -> b into a -> State Int b by simply given a nicer name to the function Int -> (b, Int). There is a very nice blog that explains where these nice abstractions come from in a very accessible way.
QUESTION
I am writing a SAT solver and I started implementing a DPLL algorithm. I understand the algorithm and how it works, I also implemented a variation of it, but what bothers me is the next thing.
...ANSWER
Answered 2017-Nov-02 at 17:22Φ is a consistent set of literals when it contains only literals and a literal and its negation does not appear in Φ. The DPLL algorithm, through the pure and unit rules, gradually converts the list of clauses into a list of literals that satisfy all the original clauses. The algorithm is done when that happens (Φ is a consistent set of literals) or when it runs out of literal assignments to try and the topmost DPLL call returns false.
QUESTION
DPLL SAT solvers typically apply a Phase Saving heuristic. The idea is to remember the last assignment of each variable and use it first in branching.
To experiment with the effects of branching polarity and phase saving, I tried several SAT solvers and modified the phase settings. All are Windows 64-bit ports, run in single-threaded mode. I always used the same example input of moderate complexity (5619 variables, 11261 clauses, in the solution 4% of all variable are true, 96% false).
The resulting run-times are listed below:
It might be just (bad) luck, but the differences are remarkably big. It is a special surprise, that MiniSat outperformed all modern solvers for my example.
My question:
...Are there any explanations for the differences?
Best practices for polarity and phase saving?
ANSWER
Answered 2018-Oct-30 at 15:51Nothing conclusive can be deduced from your test. DPLL and solvers based on it are known to exhibit heavy-tailed behavior depending on initial search conditions. This means the same solver can have both short and long runtimes on the same instance depending on factors like when random restarts occur. Search times across different solvers can vary wildly depending on (for example) how they choose decision variables, even without the added complications of phase saving and random restarts.
QUESTION
I just learned about DPLL(T) and the DPLL algorithm in relation to SMT solvers. I have seen z3 referenced in a few places regarding SMT solvers as well.
Wondering what the z3 uses for its algorithms at a high level for implementing the SMT solving. If it is DPLL algorithms, a variation, something custom, a bunch of things, etc. Hoping to learn about the different types of algorithms a modern SMT solver uses.
...ANSWER
Answered 2018-Jul-19 at 22:36SMT solvers come from a long line of research in automated reasoning, both in computer-based theorem-proving communities and in traditional mathematical logic. It's impossible to summarize all the algorithms/research in a stack-overflow answer. However, the book http://www.decision-procedures.org/ is an excellent read, and have many references that can help you guide into the literature. (The first edition is already 10 years old, but I see now that they have a second edition that came out in 2016.)
QUESTION
So I'm debugging my DPLL implementation and it's not quite working right so I step through the code line by line in the debugger, it gets to a return statement but the thing is it doesn't return, it just keeps on executing the same function. WTF I thought, am I really seeing this? So I looked at the dissasembly and sure enough one of the return statements jumps to the wrong place. Never have I seen VS generate incorrect code so I'm wondering if I screwed up somewhere but I can't find anything. The jump is incorrect even when compiling with all optimizations off.
This illustrates whats going on.
...ANSWER
Answered 2017-Mar-03 at 18:56The source interleaving is wrong. This is the correct place you want to look at:
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