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Davis Putnam Satisfiability Algorithm | false |
The Davis Putnam (DP) algorithm is the first resolution based algorithm created in 1960 that checks whether or not a given formula is satisfiable. That is, there exists an assignment of propositions that can make the formula true.
They later made a refinement of this algorithm in 1962 called the "Davis Putnam Logemann Loveland" (DPLL). In fact, the refinement is so widely used that it is difficult to find a clear document describing the earlier edition. This page will attempt to describe the origial DP algorithm. To obtain the DPLL algorithm replace the resolution rule with the splitting rule.
The core idea of the algorithm is to transform the formula to a simpler but equisatisfiable one.
This algorithm takes a list of clauses, which is considered to be in CNF.
For example for the following list of clauses: [[a, b, c], [d, e]]
we get the following formula (a + b + c)(d + e)
The algorithm has three rules associated with it:
Unit Propagation
If there is a clause that only contains one literal p
then:
- Remove clauses containing
p
- Remove all instances of
Not(p)
def _unit_propagation(x: List[Clause]) -> List[Clause]:
"""
Applies unit propagation.
If there is a clause with one literal p then
(1) Remove clauses containing p
(2) Remove all instances of Not(p)
"""
# Find all unit clauses
single_literals = (c[0] for c in x if len(c) == 1)
# Only focus on one this rule application as
# [[p], [Not(p)]] is an edge case
single_literal = next(single_literals, None)
# If there are no unit clauses, move onto the next rule.
if single_literal is None:
return x
# (1) Remove clauses containing the unit clause
new_x = [c for c in x if all((l != single_literal for l in c))]
# (2) Remove all instances of Not(p)
negated_single_literal = negate(single_literal)
new_x = [[l for l in c if l != negated_single_literal] for c in new_x]
return new_x
Affirmative-Negative
Remove clauses that contains literals that only occur positively in the list of clauses or negatively.
def _affirmative_negative(x: List[Clause]) -> List[Clause]:
"""
Remove clauses containing a literal
that only occurs positively or negatively
"""
literal_counts = _literal_counts(x)
positive_literals, negative_literals = _pos_neg_lits(literal_counts)
# Collect literals that are strictly positive or negative
# but not both
strict_literals = positive_literals ^ negative_literals
# Only consider one strict literal
strict_literal = next(iter(strict_literals), None)
# If there are no strict literals,
# move on to the next rule
if strict_literal is None:
return x
# Remove clauses that contain the strict literal
return [c for c in x
if all(
(l != strict_literal and negate(l) != strict_literal
for l in c)
)]
DP Resolution
- Find a literal
p
that occurs both positively and negatively in the list of clauses.- As a heuristic, choosing a
p
that is the least common to minimize the state size.
- As a heuristic, choosing a
- Create a list
C
containing clauses wherep
occurs positively; removingp
from each clause. - Create a list
D
containing clauses wherep
occurs negatively; removingNot(p)
from each clause. - Create a list
X
that contains clauses that do not containp
orNot(p)
in them. - Return
X + C + D
as the new list of clauses.
def _dp_resolution(x: List[Clause]) -> List[Clause]:
"""
Take a literal that occurs positvely and negatively
and combine clauses that occur positively and
negatively with p.
"""
# Find literals that occur both positively and negatively
literal_counts = _literal_counts(x)
positive_literals, negative_literals = _pos_neg_lits(literal_counts)
polar_literals = positive_literals & negative_literals
# If there are no such literals, then move
# onto the next rule.
if len(polar_literals) == 0:
return x
# Count the number of occurances of each of the polar
# literals and choose the least common literal p as
# a heuristic.
polar_literal_counts = dict(filter(lambda y: y[0] in polar_literals, literal_counts.items()))
p, _ = min(polar_literal_counts.items(), key=itemgetter(1))
# Create a list C containing clauses where p occurs
# positively; removing p from the clause
C = [[l for l in c if l != p] for c in x if p in c]
# Create a list D containing clauses where p occurs
# negatively; removing Not(p) from the clause
D = [[l for l in c if l != negate(p)] for c in x if negate(p) in c]
# Combine C and D
new_x: List[Clause] = []
for ci in C:
for di in D:
# Or(ci, di) removing duplicate literals
cdi = list(set(ci + di))
# No point in adding a clause that already exists.
if cdi not in new_x:
new_x.append(cdi)
# Clauses that don't contain p or not p in it
x0 = [c for c in x if p not in c and negate(p) not in c]
return new_x + x0
Conclusion
The complete DP algorithm is a recursive one with two base cases:
- If the length of the list of clauses is zero, then it is satisfiable
- If
[]
is in the list of clauses, then it is not satisfiable.
Then it applies the first rule that matches out of the three:
def davis_putnam(x: List[Clause]) -> bool:s
"""
Davis-Putnam procedure for deciding
satisfiability from CNF clauses.
This is called recursively and it applies
the first rule that matches: Unit Propagation,
Affirmative-Negative, Resolution.
"""
# Base Cases
if len(x) == 0:
return True
if [] in x:
return False
# (1) Unit Propagation
x_new = _unit_propagation(x)
if x_new != x:
return davis_putnam(x_new)
# (2) Affirmative-Negative
x_new = _affirmative_negative(x)
if x_new != x:
return davis_putnam(x_new)
# (3) DP Resolution
x_new = _dp_resolution(x)
return davis_putnam(x_new)