main
Matthew Butterick 10 years ago
parent 0caa63af76
commit 2b8dd131fa

@ -3,7 +3,7 @@
;; Adapted from work by Peter Norvig
;; http://aima-python.googlecode.com/svn/trunk/csp.py
(require racket/list racket/bool racket/contract)
(require racket/list racket/bool racket/contract racket/class)
(require "utils.rkt" "search.rkt")
#|
@ -49,8 +49,9 @@ class CSP(search.Problem):
|#
(define csp% (class problem%))
(define (init csp vars domains neighbors constraints)
(define (?init csp vars domains neighbors constraints)
;; Construct a CSP problem. If vars is empty, it becomes domains.keys().
(define vars (if (null? vars) (hash-keys domains) vars))
(hash-set*! csp 'vars vars 'domains domains
@ -161,4 +162,328 @@ class CSP(search.Problem):
return [var for var in self.vars
if self.nconflicts(var, current[var], current) > 0]
#______________________________________________________________________________
# CSP Backtracking Search
def backtracking_search(csp, mcv=False, lcv=False, fc=False, mac=False):
"""Set up to do recursive backtracking search. Allow the following options:
mcv - If true, use Most Constrained Variable Heuristic
lcv - If true, use Least Constraining Value Heuristic
fc - If true, use Forward Checking
mac - If true, use Maintaining Arc Consistency. [Fig. 5.3]
>>> backtracking_search(australia)
{'WA': 'B', 'Q': 'B', 'T': 'B', 'V': 'B', 'SA': 'G', 'NT': 'R', 'NSW': 'R'}
"""
if fc or mac:
csp.curr_domains, csp.pruned = {}, {}
for v in csp.vars:
csp.curr_domains[v] = csp.domains[v][:]
csp.pruned[v] = []
update(csp, mcv=mcv, lcv=lcv, fc=fc, mac=mac)
return recursive_backtracking({}, csp)
def recursive_backtracking(assignment, csp):
"""Search for a consistent assignment for the csp.
Each recursive call chooses a variable, and considers values for it."""
if len(assignment) == len(csp.vars):
return assignment
var = select_unassigned_variable(assignment, csp)
for val in order_domain_values(var, assignment, csp):
if csp.fc or csp.nconflicts(var, val, assignment) == 0:
csp.assign(var, val, assignment)
result = recursive_backtracking(assignment, csp)
if result is not None:
return result
csp.unassign(var, assignment)
return None
def select_unassigned_variable(assignment, csp):
"Select the variable to work on next. Find"
if csp.mcv: # Most Constrained Variable
unassigned = [v for v in csp.vars if v not in assignment]
return argmin_random_tie(unassigned,
lambda var: -num_legal_values(csp, var, assignment))
else: # First unassigned variable
for v in csp.vars:
if v not in assignment:
return v
def order_domain_values(var, assignment, csp):
"Decide what order to consider the domain variables."
if csp.curr_domains:
domain = csp.curr_domains[var]
else:
domain = csp.domains[var][:]
if csp.lcv:
# If LCV is specified, consider values with fewer conflicts first
key = lambda val: csp.nconflicts(var, val, assignment)
domain.sort(lambda(x,y): cmp(key(x), key(y)))
while domain:
yield domain.pop()
def num_legal_values(csp, var, assignment):
if csp.curr_domains:
return len(csp.curr_domains[var])
else:
return count_if(lambda val: csp.nconflicts(var, val, assignment) == 0,
csp.domains[var])
#______________________________________________________________________________
# Constraint Propagation with AC-3
def AC3(csp, queue=None):
"""[Fig. 5.7]"""
if queue == None:
queue = [(Xi, Xk) for Xi in csp.vars for Xk in csp.neighbors[Xi]]
while queue:
(Xi, Xj) = queue.pop()
if remove_inconsistent_values(csp, Xi, Xj):
for Xk in csp.neighbors[Xi]:
queue.append((Xk, Xi))
def remove_inconsistent_values(csp, Xi, Xj):
"Return true if we remove a value."
removed = False
for x in csp.curr_domains[Xi][:]:
# If Xi=x conflicts with Xj=y for every possible y, eliminate Xi=x
if every(lambda y: not csp.constraints(Xi, x, Xj, y),
csp.curr_domains[Xj]):
csp.curr_domains[Xi].remove(x)
removed = True
return removed
#______________________________________________________________________________
# Min-conflicts hillclimbing search for CSPs
def min_conflicts(csp, max_steps=1000000):
"""Solve a CSP by stochastic hillclimbing on the number of conflicts."""
# Generate a complete assignement for all vars (probably with conflicts)
current = {}; csp.current = current
for var in csp.vars:
val = min_conflicts_value(csp, var, current)
csp.assign(var, val, current)
# Now repeapedly choose a random conflicted variable and change it
for i in range(max_steps):
conflicted = csp.conflicted_vars(current)
if not conflicted:
return current
var = random.choice(conflicted)
val = min_conflicts_value(csp, var, current)
csp.assign(var, val, current)
return None
def min_conflicts_value(csp, var, current):
"""Return the value that will give var the least number of conflicts.
If there is a tie, choose at random."""
return argmin_random_tie(csp.domains[var],
lambda val: csp.nconflicts(var, val, current))
#______________________________________________________________________________
# Map-Coloring Problems
class UniversalDict:
"""A universal dict maps any key to the same value. We use it here
as the domains dict for CSPs in which all vars have the same domain.
>>> d = UniversalDict(42)
>>> d['life']
42
"""
def __init__(self, value): self.value = value
def __getitem__(self, key): return self.value
def __repr__(self): return '{Any: %r}' % self.value
def different_values_constraint(A, a, B, b):
"A constraint saying two neighboring variables must differ in value."
return a != b
def MapColoringCSP(colors, neighbors):
"""Make a CSP for the problem of coloring a map with different colors
for any two adjacent regions. Arguments are a list of colors, and a
dict of {region: [neighbor,...]} entries. This dict may also be
specified as a string of the form defined by parse_neighbors"""
if isinstance(neighbors, str):
neighbors = parse_neighbors(neighbors)
return CSP(neighbors.keys(), UniversalDict(colors), neighbors,
different_values_constraint)
def parse_neighbors(neighbors, vars=[]):
"""Convert a string of the form 'X: Y Z; Y: Z' into a dict mapping
regions to neighbors. The syntax is a region name followed by a ':'
followed by zero or more region names, followed by ';', repeated for
each region name. If you say 'X: Y' you don't need 'Y: X'.
>>> parse_neighbors('X: Y Z; Y: Z')
{'Y': ['X', 'Z'], 'X': ['Y', 'Z'], 'Z': ['X', 'Y']}
"""
dict = DefaultDict([])
for var in vars:
dict[var] = []
specs = [spec.split(':') for spec in neighbors.split(';')]
for (A, Aneighbors) in specs:
A = A.strip();
dict.setdefault(A, [])
for B in Aneighbors.split():
dict[A].append(B)
dict[B].append(A)
return dict
australia = MapColoringCSP(list('RGB'),
'SA: WA NT Q NSW V; NT: WA Q; NSW: Q V; T: ')
usa = MapColoringCSP(list('RGBY'),
"""WA: OR ID; OR: ID NV CA; CA: NV AZ; NV: ID UT AZ; ID: MT WY UT;
UT: WY CO AZ; MT: ND SD WY; WY: SD NE CO; CO: NE KA OK NM; NM: OK TX;
ND: MN SD; SD: MN IA NE; NE: IA MO KA; KA: MO OK; OK: MO AR TX;
TX: AR LA; MN: WI IA; IA: WI IL MO; MO: IL KY TN AR; AR: MS TN LA;
LA: MS; WI: MI IL; IL: IN; IN: KY; MS: TN AL; AL: TN GA FL; MI: OH;
OH: PA WV KY; KY: WV VA TN; TN: VA NC GA; GA: NC SC FL;
PA: NY NJ DE MD WV; WV: MD VA; VA: MD DC NC; NC: SC; NY: VT MA CA NJ;
NJ: DE; DE: MD; MD: DC; VT: NH MA; MA: NH RI CT; CT: RI; ME: NH;
HI: ; AK: """)
#______________________________________________________________________________
# n-Queens Problem
def queen_constraint(A, a, B, b):
"""Constraint is satisfied (true) if A, B are really the same variable,
or if they are not in the same row, down diagonal, or up diagonal."""
return A == B or (a != b and A + a != B + b and A - a != B - b)
class NQueensCSP(CSP):
"""Make a CSP for the nQueens problem for search with min_conflicts.
Suitable for large n, it uses only data structures of size O(n).
Think of placing queens one per column, from left to right.
That means position (x, y) represents (var, val) in the CSP.
The main structures are three arrays to count queens that could conflict:
rows[i] Number of queens in the ith row (i.e val == i)
downs[i] Number of queens in the \ diagonal
such that their (x, y) coordinates sum to i
ups[i] Number of queens in the / diagonal
such that their (x, y) coordinates have x-y+n-1 = i
We increment/decrement these counts each time a queen is placed/moved from
a row/diagonal. So moving is O(1), as is nconflicts. But choosing
a variable, and a best value for the variable, are each O(n).
If you want, you can keep track of conflicted vars, then variable
selection will also be O(1).
>>> len(backtracking_search(NQueensCSP(8)))
8
>>> len(min_conflicts(NQueensCSP(8)))
8
"""
def __init__(self, n):
"""Initialize data structures for n Queens."""
CSP.__init__(self, range(n), UniversalDict(range(n)),
UniversalDict(range(n)), queen_constraint)
update(self, rows=[0]*n, ups=[0]*(2*n - 1), downs=[0]*(2*n - 1))
def nconflicts(self, var, val, assignment):
"""The number of conflicts, as recorded with each assignment.
Count conflicts in row and in up, down diagonals. If there
is a queen there, it can't conflict with itself, so subtract 3."""
n = len(self.vars)
c = self.rows[val] + self.downs[var+val] + self.ups[var-val+n-1]
if assignment.get(var, None) == val:
c -= 3
return c
def assign(self, var, val, assignment):
"Assign var, and keep track of conflicts."
oldval = assignment.get(var, None)
if val != oldval:
if oldval is not None: # Remove old val if there was one
self.record_conflict(assignment, var, oldval, -1)
self.record_conflict(assignment, var, val, +1)
CSP.assign(self, var, val, assignment)
def unassign(self, var, assignment):
"Remove var from assignment (if it is there) and track conflicts."
if var in assignment:
self.record_conflict(assignment, var, assignment[var], -1)
CSP.unassign(self, var, assignment)
def record_conflict(self, assignment, var, val, delta):
"Record conflicts caused by addition or deletion of a Queen."
n = len(self.vars)
self.rows[val] += delta
self.downs[var + val] += delta
self.ups[var - val + n - 1] += delta
def display(self, assignment):
"Print the queens and the nconflicts values (for debugging)."
n = len(self.vars)
for val in range(n):
for var in range(n):
if assignment.get(var,'') == val: ch ='Q'
elif (var+val) % 2 == 0: ch = '.'
else: ch = '-'
print ch,
print ' ',
for var in range(n):
if assignment.get(var,'') == val: ch ='*'
else: ch = ' '
print str(self.nconflicts(var, val, assignment))+ch,
print
#______________________________________________________________________________
# The Zebra Puzzle
def Zebra():
"Return an instance of the Zebra Puzzle."
Colors = 'Red Yellow Blue Green Ivory'.split()
Pets = 'Dog Fox Snails Horse Zebra'.split()
Drinks = 'OJ Tea Coffee Milk Water'.split()
Countries = 'Englishman Spaniard Norwegian Ukranian Japanese'.split()
Smokes = 'Kools Chesterfields Winston LuckyStrike Parliaments'.split()
vars = Colors + Pets + Drinks + Countries + Smokes
domains = {}
for var in vars:
domains[var] = range(1, 6)
domains['Norwegian'] = [1]
domains['Milk'] = [3]
neighbors = parse_neighbors("""Englishman: Red;
Spaniard: Dog; Kools: Yellow; Chesterfields: Fox;
Norwegian: Blue; Winston: Snails; LuckyStrike: OJ;
Ukranian: Tea; Japanese: Parliaments; Kools: Horse;
Coffee: Green; Green: Ivory""", vars)
for type in [Colors, Pets, Drinks, Countries, Smokes]:
for A in type:
for B in type:
if A != B:
if B not in neighbors[A]: neighbors[A].append(B)
if A not in neighbors[B]: neighbors[B].append(A)
def zebra_constraint(A, a, B, b, recurse=0):
same = (a == b)
next_to = abs(a - b) == 1
if A == 'Englishman' and B == 'Red': return same
if A == 'Spaniard' and B == 'Dog': return same
if A == 'Chesterfields' and B == 'Fox': return next_to
if A == 'Norwegian' and B == 'Blue': return next_to
if A == 'Kools' and B == 'Yellow': return same
if A == 'Winston' and B == 'Snails': return same
if A == 'LuckyStrike' and B == 'OJ': return same
if A == 'Ukranian' and B == 'Tea': return same
if A == 'Japanese' and B == 'Parliaments': return same
if A == 'Kools' and B == 'Horse': return next_to
if A == 'Coffee' and B == 'Green': return same
if A == 'Green' and B == 'Ivory': return (a - 1) == b
if recurse == 0: return zebra_constraint(B, b, A, a, 1)
if ((A in Colors and B in Colors) or
(A in Pets and B in Pets) or
(A in Drinks and B in Drinks) or
(A in Countries and B in Countries) or
(A in Smokes and B in Smokes)): return not same
raise 'error'
return CSP(vars, domains, neighbors, zebra_constraint)
def solve_zebra(algorithm=min_conflicts, **args):
z = Zebra()
ans = algorithm(z, **args)
for h in range(1, 6):
print 'House', h,
for (var, val) in ans.items():
if val == h: print var,
print
return ans['Zebra'], ans['Water'], z.nassigns, ans,
|#

Loading…
Cancel
Save