Baie basies, hierdie hulpmiddel sal ons help om waardes vir veranderlikes te vind wat aan sekere voorwaardes moet voldoen, en om dit met die hand te bereken sal baie irriterend wees. Daarom kan jy vir Z3 die voorwaardes aandui waaraan die veranderlikes moet voldoen en dit sal 'n paar waardes vind (indien moontlik).
#pip3 install z3-solverfrom z3 import*s =Solver()#The solver will be given the conditionsx =Bool("x")#Declare the symbos x, y and zy =Bool("y")z =Bool("z")# (x or y or !z) and ys.add(And(Or(x,y,Not(z)),y))s.check()#If response is "sat" then the model is satifable, if "unsat" something is wrongprint(s.model())#Print valid values to satisfy the model
Ints/Simplify/Reals
from z3 import*x =Int('x')y =Int('y')#Simplify a "complex" ecuationprint(simplify(And(x +1>=3, x**2+ x**2+ y**2+2>=5)))#And(x >= 2, 2*x**2 + y**2 >= 3)#Note that Z3 is capable to treat irrational numbers (An irrational algebraic number is a root of a polynomial with integer coefficients. Internally, Z3 represents all these numbers precisely.)#so you can get the decimals you need from the solutionr1 =Real('r1')r2 =Real('r2')#Solve the ecuationprint(solve(r1**2+ r2**2==3, r1**3==2))#Solve the ecuation with 30 decimalsset_option(precision=30)print(solve(r1**2+ r2**2==3, r1**3==2))
Druk Model
from z3 import*x, y, z =Reals('x y z')s =Solver()s.add(x >1, y >1, x + y >3, z - x <10)s.check()m = s.model()print ("x = %s"% m[x])for d in m.decls():print("%s = %s"% (d.name(), m[d]))
Masjien Aritmetiek
Moderne CPU's en hoofstroom programmeertale gebruik aritmetiek oor vaste-grootte bit-vektore. Masjien aritmetiek is beskikbaar in Z3Py as Bit-Vektore.
from z3 import*x =BitVec('x', 16)#Bit vector variable "x" of length 16 bity =BitVec('y', 16)e =BitVecVal(10, 16)#Bit vector with value 10 of length 16bitsa =BitVecVal(-1, 16)b =BitVecVal(65535, 16)print(simplify(a == b))#This is True!a =BitVecVal(-1, 32)b =BitVecVal(65535, 32)print(simplify(a == b))#This is False
Getekende/Ongetekende Getalle
Z3 bied spesiale getekende weergawes van wiskundige operasies waar dit 'n verskil maak of die bit-vectore as getekend of ongetekend behandel word. In Z3Py, die operateurs <, <=, >, >=, /, % en >> stem ooreen met die getekende weergawes. Die ooreenstemmende ongetekende operateurs is ULT, ULE, UGT, UGE, UDiv, URem en LShR.
from z3 import*# Create to bit-vectors of size 32x, y =BitVecs('x y', 32)solve(x + y ==2, x >0, y >0)# Bit-wise operators# & bit-wise and# | bit-wise or# ~ bit-wise notsolve(x & y ==~y)solve(x <0)# using unsigned version of <solve(ULT(x, 0))
Funksies
Geïnterpreteerde funksies soos aritmetika waar die funksie + 'n vaste standaardinterpretasie het (dit voeg twee getalle by). Nie-geïnterpreteerde funksies en konstantes is maksimaal buigsaam; hulle laat enige interpretasie toe wat konsekwent is met die beperkings oor die funksie of konstante.
Voorbeeld: f wat twee keer op x toegepas word, lei tot x weer, maar f wat een keer op x toegepas word, is anders as x.
from z3 import*x =Int('x')y =Int('y')f =Function('f', IntSort(), IntSort())s =Solver()s.add(f(f(x)) == x, f(x) == y, x != y)s.check()m = s.model()print("f(f(x)) =", m.evaluate(f(f(x))))print("f(x) =", m.evaluate(f(x)))print(m.evaluate(f(2)))s.add(f(x) ==4)#Find the value that generates 4 as responses.check()print(m.model())
Voorbeelde
Sudoku-oplosser
# 9x9 matrix of integer variablesX = [ [ Int("x_%s_%s"% (i+1, j+1))for j inrange(9) ]for i inrange(9) ]# each cell contains a value in {1, ..., 9}cells_c = [ And(1<= X[i][j], X[i][j] <=9)for i inrange(9)for j inrange(9) ]# each row contains a digit at most oncerows_c = [ Distinct(X[i])for i inrange(9) ]# each column contains a digit at most oncecols_c = [ Distinct([ X[i][j] for i inrange(9) ])for j inrange(9) ]# each 3x3 square contains a digit at most oncesq_c = [ Distinct([ X[3*i0 + i][3*j0 + j]for i inrange(3) for j inrange(3) ])for i0 inrange(3)for j0 inrange(3) ]sudoku_c = cells_c + rows_c + cols_c + sq_c# sudoku instance, we use '0' for empty cellsinstance = ((0,0,0,0,9,4,0,3,0),(0,0,0,5,1,0,0,0,7),(0,8,9,0,0,0,0,4,0),(0,0,0,0,0,0,2,0,8),(0,6,0,2,0,1,0,5,0),(1,0,2,0,0,0,0,0,0),(0,7,0,0,0,0,5,2,0),(9,0,0,0,6,5,0,0,0),(0,4,0,9,7,0,0,0,0))instance_c = [ If(instance[i][j] ==0,True,X[i][j] == instance[i][j])for i inrange(9)for j inrange(9) ]s =Solver()s.add(sudoku_c + instance_c)if s.check()== sat:m = s.model()r = [ [ m.evaluate(X[i][j])for j inrange(9) ]for i inrange(9) ]print_matrix(r)else:print"failed to solve"
