def hard_phi(literals, k = 3): literals = shuffle(literals) n = len(literals) L = [prop.And(literals[i*n/k:(i+1)*n/k]) for i in range(n/k)] L.append(literals[0]) print L return prop.Or(L).cnf()
def random_phi(length = 10, vars = 10): V = ["phi"+str(i) for i in range(vars)] phi = [] i = 0 while i < length: tmp = [] if i >= length-3: break k = rnd.randint(3, length-i-1) for j in range(k): tmp_var = V[rnd.randint(0, vars-1)] # P(negacija formula) = 1/2? if rnd.randint(0, 5) <= 3: tmp.append(prop.Not(tmp_var) if rnd.randint(0,1) == 0 else tmp_var) # E[dolzina] = k * 3/5 else: break i = i+1 if len(prop.Or(tmp).l) > 0: phi.append(prop.Or(tmp)) # Protislovja spustimo (zato vrnemo formulo dolzine kvecjemu length :-) i = i+1 return prop.And(phi)
def hadamard2sat(n):
assert n % 2 == 0
l = []
#naredi matrikco za lazje mislit
ma3ka = {}
for i in range(n):
for j in range(n):
ma3ka[(i, j)] = "v" + str(j) + "s" + str(i)
#vjsi = vrstica j, stolpec i
print ma3ka
xOri = []
#nardimo xOre vrstic za vsak stolpec.
for i in range(n - 1):
for j in range(n):
xOri.append(
prop.Or([
prop.And([ma3ka[(i, j)],
prop.Not(ma3ka[(i + 1, j)])]),
prop.And([prop.Not(ma3ka[(i, j)]), ma3ka[(i + 1, j)]])
]))
#vse mozne true - false kombinacije
a = list(itertools.combinations(xOri,
len(xOri) / 2))
#generiramo mozne stolpce
stolpec = []
for j in range(len(a)):
stolpec.append(
prop.And([x if x in a[j] else prop.Not(x) for x in xOri]))
#nardimo or moznih stolpcev
l.append(prop.Or(stolpec))
xOri = []
#vrnemo koncno formulo
return prop.And(l)
def test1():
## Sestavljanje in manipuliranje formul ##
# Sestavimo zelo enostavno formulo u \/ v
phi = prop.Or(["v", "u"])
## Uporaba SAT solverjev ##
# Test DPLL SAT solverja
print "*" * 80
print "Testiram DPLL SAT solver"
dpll_sat_test(phi)
# Test bruteforce SAT solverja
print "*" * 80
print "Testiram bruteforce SAT solver"
bf_sat_test(phi)
## Prevedbe ##
print "*" * 80
print "*" * 80
print "Testiram prevedbe"
# k-barvanje grafov
print "*" * 80
print "Barvanje grafov"
k = 2 # dvodelnost
G = (3, [(0, 1), (1, 2), (0, 2)])
phi = p.graph_coloring2sat(G, k)
print phi
# return
# Sudoku
print "*" * 80
print "Sudoku"
S = h.get_sudoku('sudoku01a.in')
phi = p.sudoku2sat(S)
print phi
x = sat.sat(phi.cnf())
print S
print p.solveSudoku(S, x)
# return
# Hadamard
print "*" * 80
print "Hadamard"
phi = hadamard(4)
print phi
# return
# Erdosev problem diskrepance
print "*" * 80
print "Erdosev problem diskrepance"
phi = edp(1, 5)
print phi
def hadamard(n):
#hadamardove matrike obstajajo le za sode n
assert n % 2 == 0
if (n % 2 != 0):
return prop.Fls()
l = []
#predpostavka - prva vrstica so vsi True
l.append(prop.And(["v0s%d" % i for i in range(n)]))
#generiramo mozne kombinacije, tako da je n/2 clenov True in n/2 False
a = list(itertools.combinations(range(n), n / 2))
#print a;
#print a;
#gneriraj vse mozne vrstice | v moz[i] so spravljenje vse mogoce kombinacije spremenljivk za vrstico i
moz = [[]]
for i in range(1, n):
moz.append([])
for j in range(len(a)):
moz[i].append(
prop.And([
prop.Not("v" + str(i) + "s%d" % k) if
(k not in a[j]) else "v" + str(i) + "s%d" % k
for k in range(n)
]))
#print moz;
#generiraj vse mozne matrike, katerih vrstice bi ustrezale Hadamardovi matriki
#b = list(itertools.combinations(range(1,n*len(a)), n-1)); #iz vseh moznih vrstic moramo izbrati n-1, ker prvo ze imamo
#print b;
moz2 = []
b = list(itertools.combinations(range(len(moz[1])), n - 1))
for j in range(1): #range(len(b)):
moz2.append(
prop.And(
[l[0],
prop.And([moz[i][b[j][i - 1]] for i in range(1, n)])]))
print moz2
return prop.Or(moz2)
"""
def graph_coloring2sat(G, k):
# G[0] naj bo stevilo povezav
# G[1] naj bo seznam parov vozlics, a.k.a, seznan pobexzav
# k > 0 je stevilo barv
assert k > 0, "Premalo barv"
l = []
# vsako vozlisce ima vsaj eno barvo
l.append(
prop.And([
prop.Or(["v%dc%d" % (i, j) for j in range(k)]) for i in range(G[0])
]))
# pari krajisc so razlicnih barv
for (u, v) in G[1]:
l.append(
prop.And([
prop.And([
prop.Not(
prop.And([
"v" + str(u) + "c" + str(c),
"v" + str(v) + "c" + str(c)
])) for c in range(k)
])
]))
# vsako vozlisce je kvecjemu ene barve
for v in range(G[0]):
for i in range(k):
for j in range(i + 1, k):
l.append(
prop.Not(
prop.And([
"v" + str(v) + "c" + str(i),
"v" + str(v) + "c" + str(j)
])))
phi = prop.And(l)
#print phi
return phi
def sudoku2sat(s):
V = 81
#stevilo kvadratkov (vozlisc)
k = 9
#stevilo barv
l = []
#list logicnih formul
# sudoku je oznacen:
#
# 1 2 3 4 5 6 7 8 9
# 10 11 12 13 14 15 16 17 18
# ...
#
# pretvori sudoku s v 1d seznam
novS = []
for i in range(k):
novS = novS + s[i]
s = novS
# print s;
povVrst = [] # povezani kvadratki v vrsticah
povStolp = [] # povezani kvadratki v stolpcih
povKvadr = [] # povezani kvadratki v 3x3 kvadratih
# Konstriramo graf, 9-barvljiv <==> Sudoku resljiv
# povezemo kvadratke v vrsticah
for k in range(0, 9):
for i in range(1, 10):
for j in range(i + 1, 10):
povVrst.append((i + k * 9, j + k * 9))
#p ovezemo kvadratke v stolpcih
for i in range(1, 10): # index stolpca v k-ti vrstici
for k in range(0, 9): # index vrstice
for j in range(k + 1,
9): # stoplci, ki se niso povezavi s k-tim stolpcem
povStolp.append((i + 9 * k, i + 9 * j))
#povezemo kvadratke znotraj 3x3 kvadratov
for i in range(0, 3):
for j in range(1, 4):
povKvadr.append(((j * 3 - 1) + (9 * (3 * i + 1) - 10),
(j * 3 - 1) + (9 * (3 * i + 1)) + 10))
# Doda povezavo, ki je manjkala
povKvadr.append(((j * 3 - 1) + (9 * (3 * i + 1)),
(j * 3 - 1) + (9 * (3 * i + 1)) + 10))
povKvadr.append(((j * 3 - 1) + (9 * (3 * i + 1)),
(j * 3 - 1) + (9 * (3 * i + 1)) - 10))
povKvadr.append(((j * 3 - 1) + (9 * (3 * i + 1) - 8),
(j * 3 - 1) + (9 * (3 * i + 1)) + 8))
# Doda povezav, ki jo manjkala
povKvadr.append(((j * 3 - 1) + (9 * (3 * i + 1)),
(j * 3 - 1) + (9 * (3 * i + 1)) + 8))
povKvadr.append(((j * 3 - 1) + (9 * (3 * i + 1)),
(j * 3 - 1) + (9 * (3 * i + 1)) - 8))
povKvadr.append(((j * 3 - 1) + (9 * (3 * i + 1) - 1),
(j * 3 - 1) + (9 * (3 * i + 1)) - 9)) # /
povKvadr.append(((j * 3 - 1) + (9 * (3 * i + 1) + 1),
(j * 3 - 1) + (9 * (3 * i + 1)) - 9)) # \
povKvadr.append(((j * 3 - 1) + (9 * (3 * i + 1) - 1),
(j * 3 - 1) + (9 * (3 * i + 1)) + 9)) # /
povKvadr.append(((j * 3 - 1) + (9 * (3 * i + 1) + 1),
(j * 3 - 1) + (9 * (3 * i + 1)) + 9)) # \
povezave = povVrst + povStolp + povKvadr
# vsako vozlisce ima vsaj eno barvo
# vozlisca v imajo stevilko v zacetnem sudoku-ju pobarvamo z znano barvo
#l.append(prop.And([prop.Or(["v%dc%d" % (i, j) for j in range(k)]) for i in range(G[0])]))
barveVoz = []
for i in range(1, V + 1):
if (
s[i - 1] == None
): #ce je None, je lahko ubistvu kjerekoli barve, ce ne upostevamo pogojev
barveVoz.append(
prop.Or(["v%dc%d" % (i, j) for j in range(1, 9 + 1)]))
else:
#print str(i)+" "+s[i-1];
barveVoz.append("v%dc%d" % (i, int(s[i - 1])))
#ce ni None, ima ze tocno doloceno barvo...
barveVoz = prop.And(barveVoz)
l.append(barveVoz)
barveKraj = []
# pari krajisc so razlicnih barv
for (u, v) in povezave:
if (s[u - 1] != None):
barveKraj.append(prop.Not("v" + str(v) + "c" + s[u - 1]))
elif (s[v - 1] != None):
barveKraj.append(prop.Not("v" + str(u) + "c" + s[v - 1]))
else:
for c in range(1, 9 + 1):
barveKraj.append(
prop.Not(
prop.And([
"v" + str(u) + "c" + str(c),
"v" + str(v) + "c" + str(c)
])))
l.append(prop.And(barveKraj))
# vsako vozlisce je kvecjemu ene barve
for v in range(1, V + 1):
if (s[v - 1] == None):
for i in range(1, 9 + 1):
for j in range(i + 1, 9 + 1):
l.append(
prop.Not(
prop.And([
"v" + str(v) + "c" + str(i),
"v" + str(v) + "c" + str(j)
])))
return prop.And(l)
j = n = len(L)
while True:
R.append(L[1:])
j = n - 1
# NOTE: Changing `L[j] == 1' to `L[j] == mi' will generate [m1] x [m2] x ... x [mn], where [mi] denotes {1,2, ..., mi}
while j > 0 and L[j] == 1:
L[j] = 0
j = j - 1
L[j] = L[j] + 1
if j == 0: break
return R
# tries all possible assignments
def brute_force(phi):
# TODO: Get a list of variables from phi; then feed the list to comb() to get all possible assignments
for asg in comb([0, 0, 0, 0, 0, 0, 0, 0]):
print asg
# print { i : asg[i] for i in range(len(asg)) }
# Vstopna tocka
if __name__ == "__main__":
phi = prop.And([("a"), "a",
prop.Or(prop.Not("b"), "d"), "c",
prop.Or([prop.Not("b"), "a"])]).cnf()
print phi
print satBruteForce(phi)
print sat(phi)
print sat3(phi)
def edp2sat(fname):
L = open(fname).read().split('\n')
fmt = L[0]
return prop.And([
prop.Or([clean(c) for c in clause.split()[:-1]]) for clause in L[1:-1]
])
def parse_output(fname = 'new.cnf'):
L = open(fname).read().split('\n')
fmt = L[0]
return prop.And([prop.Or([clean(c) for c in clause.split()[:-1]]) for clause in L[1:-1]])
def rnd_cnf(literals, k = 3): literals = shuffle(literals) n = len(literals) return prop.And([prop.Or(literals[i*n/k:(i+1)*n/k]) for i in range(n/k)])
def sudoku(s):
V = 81; #stevilo kvadratkov (vozlisc)
k = 9; #stevilo barv
l = []; #list logicnih formul
#sudoku je oznacen:
"""
1 2 3 4 5 6 7 8 9
10 11 12 13 14 15 16 17 18
...
"""
#pretvori sudoku s v 1d seznam
novS = [];
for i in range(k):
novS = novS + s[i];
s = novS;
#print s;
povVrst = [] #povezani kvadratki v vrsticah
povStolp = [] #povezani kvadratki v stolpcih
povKvadr = [] #povezani kvadratki v 3x3 kvadratih
"""
for i in range(1, 82):
if (i % 9 != 0 and i != 81):
povVrst.append((i,i+1)); #delamo povezave vrstic
if (i < 73):
povStolp.append((i,9+i)); #delamo povezave stolpcev
"""
# Konstriramo graf, 9-barvljiv <==> Sudoku resljiv
#povezemo kvadratke v vrsticah
for k in range(0,9):
for i in range(1,10):
for j in range(i+1, 10):
povVrst.append((i+k*9, j+k*9));
#povezemo kvadratke v stolpcih
for i in range(1,10): # index stolpca v k-ti vrstici
for k in range(0,9): # index vrstice
for j in range(k+1, 9): # stoplci, ki se niso povezavi s k-tim stolpcem
povStolp.append((i+9*k, i+9*j));
#print povVrst;
#print povStolp;
#povezemo kvadratke znotraj 3x3 kvadratov
for i in range(0,3):
for j in range(1,4):
#povKvadr.append(((j*3-1)+(9*(3*i+1)), (j*3-1)+(9*(3*i+1))+1));
#povKvadr.append(((j*3-1)+(9*(3*i+1)), (j*3-1)+(9*(3*i+1))-1));
#povKvadr.append(((j*3-1)+(9*(3*i+1)), (j*3-1)+(9*(3*i+1))+9));
#povKvadr.append(((j*3-1)+(9*(3*i+1)), (j*3-1)+(9*(3*i+1))-9));
povKvadr.append(((j*3-1)+(9*(3*i+1)-10), (j*3-1)+(9*(3*i+1))+10)); # Doda povezavo, ki je manjkala
povKvadr.append(((j*3-1)+(9*(3*i+1)), (j*3-1)+(9*(3*i+1))+10));
povKvadr.append(((j*3-1)+(9*(3*i+1)), (j*3-1)+(9*(3*i+1))-10));
povKvadr.append(((j*3-1)+(9*(3*i+1)-8), (j*3-1)+(9*(3*i+1))+8)); # Doda povezav, ki jo manjkala
povKvadr.append(((j*3-1)+(9*(3*i+1)), (j*3-1)+(9*(3*i+1))+8));
povKvadr.append(((j*3-1)+(9*(3*i+1)), (j*3-1)+(9*(3*i+1))-8));
povKvadr.append(((j*3-1)+(9*(3*i+1)-1), (j*3-1)+(9*(3*i+1))-9)) # /
povKvadr.append(((j*3-1)+(9*(3*i+1)+1), (j*3-1)+(9*(3*i+1))-9)) # \
povKvadr.append(((j*3-1)+(9*(3*i+1)-1), (j*3-1)+(9*(3*i+1))+9)) # /
povKvadr.append(((j*3-1)+(9*(3*i+1)+1), (j*3-1)+(9*(3*i+1))+9)) # \
#print povKvadr;
povezave = povVrst + povStolp + povKvadr;
#povezave = povVrst + povKvadr;
#povezave = list(set(povezave)); #odstranimo duplikate povezav
#print povezave
# vsako vozlisce ima vsaj eno barvo
# vozlisca v imajo stevilko v zacetnem sudoku-ju pobarvamo z znano barvo
#l.append(prop.And([prop.Or(["v%dc%d" % (i, j) for j in range(k)]) for i in range(G[0])]))
barveVoz = [];
for i in range(1, V+1):
if(s[i-1] == None): #ce je None, je lahko ubistvu kjerekoli barve, ce ne upostevamo pogojev
barveVoz.append(prop.Or(["v%dc%d" % (i, j) for j in range(1, 9+1)]));
else:
#print str(i)+" "+s[i-1];
barveVoz.append("v%dc%d" % (i, int(s[i-1]))); #ce ni None, ima ze tocno doloceno barvo...
barveVoz = prop.And(barveVoz);
l.append(barveVoz);
#print barveVoz;
barveKraj = [];
# pari krajisc so razlicnih barv
for (u, v) in povezave:
if(s[u-1] != None):
#barveKraj.append(prop.Not(prop.And(["v"+str(u)+"c"+s[u-1], "v"+str(v)+"c"+s[u-1]])));
barveKraj.append(prop.Not("v"+str(v)+"c"+s[u-1]));
elif(s[v-1] != None):
#barveKraj.append(prop.Not(prop.And(["v"+str(u)+"c"+s[v-1], "v"+str(v)+"c"+s[v-1]])));
barveKraj.append(prop.Not("v"+str(u)+"c"+s[v-1]));
else:
for c in range(1, 9+1):
barveKraj.append(prop.Not(prop.And(["v"+str(u)+"c"+str(c), "v"+str(v)+"c"+str(c)])));
#l.append(prop.And([prop.And([prop.Not(prop.And(["v"+str(u)+"c"+str(c), "v"+str(v)+"c"+str(c)])) for c in range(k)])]))
#print barveKraj;
l.append(prop.And(barveKraj));
#print l;
# print "BARVEEEEE", k
# vsako vozlisce je kvecjemu ene barve
for v in range(1, V+1):
if(s[v-1] == None):
for i in range(1, 9+1):
for j in range(i+1, 9+1):
l.append(prop.Not(prop.And(["v"+str(v)+"c"+str(i), "v"+str(v)+"c"+str(j)])))
l = prop.And(l);
#print l;
return l;