def solve(instance, watchlist, assignment, d, verbose):
"""
Iteratively solve SAT by assigning to variables d, d+1, ..., n-1. Assumes
variables 0, ..., d-1 are assigned so far. A generator for all the
satisfying assignments is returned.
"""
# The state list wil keep track of what values for which variables
# we have tried so far. A value of 0 means nothing has been tried yet,
# a value of 1 means False has been tried but not True, 2 means True but
# not False, and 3 means both have been tried.
n = len(instance.variables)
state = [0] * n
while True:
if d == n:
yield assignment
d -= 1
continue
# Let's try assigning a value to v. Here would be the place to insert
# heuristics of which value to try first.
tried_something = False
for a in [0, 1]:
if (state[d] >> a) & 1 == 0:
if verbose:
print('Trying {} = {}'.format(instance.variables[d], a),
file=stderr)
tried_something = True
# Set the bit indicating a has been tried for d
state[d] |= 1 << a
assignment[d] = a
if not update_watchlist(instance, watchlist, d << 1 | a,
assignment, verbose):
assignment[d] = None
else:
d += 1
break
if not tried_something:
if d == 0:
# Can't backtrack further. No solutions.
return
else:
# Backtrack
state[d] = 0
assignment[d] = None
d -= 1
def recursive_solve(instance, watchlist, assignment, d, unit_propagation=False, verbose=False):
"""
Recursively solve SAT by assigning to variables d, d+1, ..., n-1. Assumes
variables 0, ..., d-1 are assigned so far. A generator for all the
satisfying assignments is returned.
"""
if d == len(instance.variables):
yield assignment
return
# if unit_propagation:
# if unit_pg(instance,
# watchlist,
# assignment,
# unit_propagation,
# verbose):
if unit_propagation:
pp_assignment = copy.deepcopy(assignment)
unit_pg(instance, watchlist, pp_assignment, unit_propagation, verbose)
# print(pp_assignment, assignment)
next_d = None
for i in range(len(pp_assignment)):
if pp_assignment[i] != assignment[i]:
next_d = i
# print(pp_assignment, assignment)
if next_d != None:
assignment = pp_assignment
d = next_d
# if not assignment[d] is None:
# return
for a in [0, 1]:
if verbose:
print('Trying {} = {}'.format(instance.variables[d], a),
file=stderr)
assignment[d] = a
if update_watchlist(instance,
watchlist,
variable_to_literal(d, a),
assignment,
unit_propagation,
verbose):
for a in recursive_solve(instance, watchlist, assignment, d+1, unit_propagation, verbose):
yield a
assignment[d] = None
def solve_iterative(instance, watchlist, assignment, d, verbose):
"""
- The state list wil keep track of what values for which variables were tried
- A value of 0 means nothing has been tried yet
- A value of 1 means False has been tried but not True
- A value of 2 means True has been tried but not False
- A value of 3 means True and False have been tried.
"""
n = len(instance.variables)
state = [0] * n
while True:
if d == n:
yield assignment
d -= 1
continue
# Let's try assigning a value to v. Here would be the place to insert
# heuristics of which value to try first.
tried_something = False
for a in [0, 1]:
if (state[d] >> a) & 1 == 0:
if verbose:
print('Trying {} = {}'.format(instance.variables[d], a),
file=stderr)
tried_something = True
# Set the bit indicating a has been tried for d
state[d] |= 1 << a
assignment[d] = a
if not update_watchlist(instance, watchlist, d << 1 | a,
assignment, verbose):
assignment[d] = None
else:
d += 1
break
if not tried_something:
if d == 0:
# Can't backtrack further. No solutions.
return
else:
# Backtrack
state[d] = 0
assignment[d] = None
d -= 1
def solve(instance, watchlist, assignment, d, verbose):
"""
Iteratively solve SAT by assigning to variables d, d+1, ..., n-1. Assumes
variables 0, ..., d-1 are assigned so far. A generator for all the
satisfying assignments is returned.
"""
# The state list wil keep track of what values for which variables
# we have tried so far. A value of 0 means nothing has been tried yet,
# a value of 1 means False has been tried but not True, 2 means True but
# not False, and 3 means both have been tried.
n = len(instance.variables)
state = [0] * n
while True:
if d == n:
yield assignment
d -= 1
continue
# Let's try assigning a value to v. Here would be the place to insert
# heuristics of which value to try first.
tried_something = False
for a in [0, 1]:
if (state[d] >> a) & 1 == 0:
if verbose:
print("Trying {} = {}".format(instance.variables[d], a), file=stderr)
tried_something = True
# Set the bit indicating a has been tried for d
state[d] |= 1 << a
assignment[d] = a
if not update_watchlist(instance, watchlist, d << 1 | a, assignment, verbose):
assignment[d] = None
else:
d += 1
break
if not tried_something:
if d == 0:
# Can't backtrack further. No solutions.
return
else:
# Backtrack
state[d] = 0
assignment[d] = None
d -= 1
def solve(instance, watchlist, assignment, d, verbose):
"""
Recursively solve SAT by assigning to variables d, d+1, ..., n-1. Assumes
variables 0, ..., d-1 are assigned so far. A generator for all the
satisfying assignments is returned.
"""
if d == len(instance.variables):
yield assignment
return
for a in [0, 1]:
if verbose:
print('Trying {} = {}'.format(instance.variables[d], a),
file=stderr)
assignment[d] = a
if update_watchlist(instance, watchlist, (d << 1) | a, assignment,
verbose):
for a in solve(instance, watchlist, assignment, d + 1, verbose):
yield a
assignment[d] = None
def solve(instance, watchlist, assignment, d, verbose):
"""
Recursively solve SAT by assigning to variables d, d+1, ..., n-1. Assumes
variables 0, ..., d-1 are assigned so far. A generator for all the
satisfying assignments is returned.
"""
if d == len(instance.variables):
yield assignment
return
for a in [0, 1]:
if verbose:
print('Trying {} = {}'.format(instance.variables[d], a),
file=stderr)
assignment[d] = a
if update_watchlist(instance,
watchlist,
(d << 1) | a,
assignment,
verbose):
for a in solve(instance, watchlist, assignment, d + 1, verbose):
yield a
assignment[d] = None
