[Haskell-cafe] Haskell DPLL
Houdini
andressocrates90 at yahoo.com
Wed Feb 2 18:08:54 CET 2011
One :Thank you Carsten Schultz,Daniel Fischer and all the other for your
help.
Two:After my last post I wrote some function that should help me in the
future,but I need some help with the followint as I'm tired and have to fit
in a schedule.I know it is long so I'll try and explain it as short and
consice as I can.
The DPLL procedure has two main stages: a simplification stage and a search
stage. In the simplification
stage, functions are applied to the formula to assign truth values to
certain propositional variables. The
simplifications are made because if there is a satisfying assignment for the
simplified formula then it is also
a satisfying assignment for the original formula. This reduces the need for
search which can take a long
time. The search stage is performed when no more simplifications can be
made. In this stage a literal is
chosen and is assigned true or false leading to two new branches in the
search space.
I wrote some function wich should be helpfull,but I have to fit in a
schedule and I'm getting tired I need some additional help if possible.
module Algorithm where
import System.Random
import Data.Maybe
import Data.List
type Atom = String
type Literal = (Bool,Atom)
type Clause = [Literal]
type Formula = [Clause]
type Model = [(Atom, Bool)]
type Node = (Formula, ([Atom], Model))
-- This function takess a Clause and return the set of Atoms of that
Clause.
atomsClause :: Clause -> [Atom]
-- This function takes a Formula returns the set of Atoms of a Formula
atoms :: Formula -> [Atom]
-- This function returns True if the given Literal can be found within
-- the Clause.
isLiteral :: Literal -> Clause -> Bool
-- this function takes a Model and an Atom and flip the truthvalue of
-- the atom in the model
flipSymbol :: Model -> Atom -> Model -- is this ok?
Additional functions that I wrote:
remove :: (Eq a) )a ->[a] ->[a]
-This function removes an item from a list.
neg :: Literal->Literal
-This function flips a literal (ie. from P to :P and from :P to P).
falseClause :: Model -> Clause -> Bool
-This function takes a Model and a Clause and returns True
if the clause is unsatisfied by the model or False otherwise.
falseClauses :: Formula -> Model -> [Clause]
-This function takes a Formula and a Model and returns the list of
clauses of the formula that are not satisfied.
assignModel :: Model -> Formula -> Formula
-This function applies the assign function for all the assignments of a
given model.
checkFormula :: Formula -> Maybe Bool This function checks whether a
formula can be decided to be satisfiable or unsatisfiable based on the
effects of the assign function.
satisfies :: Model -> Formula -. Bool This function checks whether a
model satisfies a formula. This is done with the combination of the
assignModel and checkFormula functions.
--Where do I need help:
removeTautologies :: Formula->Formula
This function should output a simplified formula if tautologies
can be found in one or more clauses in the input
Notes: If in a clause, a literal and its negation are found, it means that
the clause will be true, regardless of the value
finally assigned to that propositional variable. Consider the following
example:
(A v B v -A) ^ (B v C v A)
The first clause contains the literals A and -A. This means that the clause
will always be true, in which case
it can be simplify the whole set to simply (B v C v A) (the second clause
alone)
pureLiteralDeletion :: Formula->Formula
This function is suppose to implement a simplification step that assumes as
true any atom in a formula that appears exclusively in a positive or
negative form (not both). Consider the formula:
(P v Q v R) ^ (P v Q v -R) ^ (-Q v R)
Note that in this formula P is present but -P is not. Using Pure Literal
Deletion it can be assumed that the value of P will be True thus
simplifying the formula to (-Q v R). If the literal were false then the
literal would simply be deleted from the clauses it appears in. In that case
any satisfying model for the resulting formula would also be a satisfying
model for the formula when we assume that the literal is true. Hence this
simplification is sound in that if there is a solution to the simplified
formula then there is a solution to the original formula.
propagateUnits :: Formula->Formula
If a clause in a propositional formula contains only one literal, then that
literal must be true (so that the particular clause can be satisfied). When
this happens,we can remove the unit clauses (the ones that contain only one
literal), all the clauses where the literal appears and also, from the
remaining clauses, we can delete the negation of the literal (because if P
is true, -P will be false).For example, in the formula (P v Q v R) ^ (-P v Q
v -R) ^ (P) we have one unit clause (the third clause(P) ). Because this one
has to be true for the whole formula to be true we assign True to P and try
to find
a satisfying assignment for the remaining formula. Finally because -P cannot
be true (given the assigned value of P) then the second clause is reduced by
eliminating the symbol -P . This simplification results in the revised
formula (Q v -R).
The resulting simplification can create other unit clauses. For example in
the formula (-P v Q) ^ (P) is simplified to (Q) when the unit clause (P) is
propagated. This makes (Q) a unit clause which can now also be simplified to
give a satisfying assignment to the formula. The function should apply unit
propagation until it can no longer make any further simplifications.
Note that if both P and -P are unit clauses then the formula is
unsatisfiable. In this case the function should return a formula with an
empty clause in it to indicate that the formula could not be satisfied.
update :: Node -> [Node]
The update function should take in a Node and return a list of the Nodes
that result from assigning True to an unassigned atom in one case and False
in the other (ie. a case
split). So the list returned should have two nodes as elements. One node
should contain the formula with an atom assigned True and the model updated
with this assignment, and the other should contain the formula with the atom
assigned False and the model updated to show this. The lists of unassigned
atoms of each node should also be updated accordingly. This function should
use your implemented assign function to make the assignments. It should also
use the chooseAtom function provided to select the literal to assign.
search :: (Node -> [Node]) -> [Node] -> Int -> (Bool, Int)
The search function should perform a backtracking search. The function takes
the update function as input and uses it to generate nodes in the search
space. The search function also takes in a list which has one element, the
initial node consisting of the formula along with an initial model. It
should generate nodes using the update function and check nodes using the
given check function. If a node is unsatisfiable then it should
abandon that branch of the search. If a node is satisfiable then a
satisfying assignment has been found and so it should return True. If a node
is neither satisfiable or unsatisfiable then it should generate new nodes
from this node. If all possible branches of the search space have been
tried, i.e. the list of nodes to try has become the empty list, then it
should return False since all possible assignments have been tried. The
search function also has an Int argument which should be an integer that
tracks how many
calls to the update function have been made. The search function should
return a pair consisting of the truth value of the formula and the number of
calls to update made.
Thank you(because ,I'm anticipating your help),and until I get a reply I
will continue to try an squize my exhausted brain.
--
View this message in context: http://haskell.1045720.n5.nabble.com/Haskell-DPLL-tp3368123p3368123.html
Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.
More information about the Haskell-Cafe
mailing list