hacktricks/reversing/reversing-tools-basic-methods/satisfiability-modulo-theories-smt-z3.md

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Veoma jednostavno, ovaj alat će nam pomoći da pronađemo vrednosti za promenljive koje treba da zadovolje određene uslove, a ručno izračunavanje bi bilo veoma dosadno. Stoga, možete Z3 ukazati na uslove koje promenljive treba da zadovolje i on će pronaći neke vrednosti (ako je moguće).

**Neki tekstovi i primeri su preuzeti sa [https://ericpony.github.io/z3py-tutorial/guide-examples.htm](https://ericpony.github.io/z3py-tutorial/guide-examples.htm)**

# Osnovne Operacije

## Booleovi/And/Or/Not
```python
#pip3 install z3-solver
from z3 import *
s = Solver() #The solver will be given the conditions

x = Bool("x") #Declare the symbos x, y and z
y = Bool("y")
z = Bool("z")

# (x or y or !z) and y
s.add(And(Or(x,y,Not(z)),y))
s.check() #If response is "sat" then the model is satifable, if "unsat" something is wrong
print(s.model()) #Print valid values to satisfy the model
```
## Ints/Simplify/Reals
```python
from z3 import *

x = Int('x')
y = Int('y')
#Simplify a "complex" ecuation
print(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 solution
r1 = Real('r1')
r2 = Real('r2')
#Solve the ecuation
print(solve(r1**2 + r2**2 == 3, r1**3 == 2))
#Solve the ecuation with 30 decimals
set_option(precision=30)
print(solve(r1**2 + r2**2 == 3, r1**3 == 2))
```
## Štampanje modela
```python
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]))
```
# Mašinska Aritmetika

Moderni CPU-i i mainstream programski jezici koriste aritmetiku nad **fiksno velikim bit-vektorima**. Mašinska aritmetika je dostupna u Z3Py kao **Bit-Vektori**.
```python
from z3 import *

x = BitVec('x', 16) #Bit vector variable "x" of length 16 bit
y = BitVec('y', 16)

e = BitVecVal(10, 16) #Bit vector with value 10 of length 16bits
a = 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
```
## Potpisani/Ne potpisani brojevi

Z3 pruža posebne potpisane verzije aritmetičkih operacija gde je važno da li se **bit-vektor tretira kao potpisan ili ne potpisan**. U Z3Py, operatori **<, <=, >, >=, /, % i >>** odgovaraju **potpisanim** verzijama. Odgovarajući **ne potpisani** operatori su **ULT, ULE, UGT, UGE, UDiv, URem i LShR.**
```python
from z3 import *

# Create to bit-vectors of size 32
x, y = BitVecs('x y', 32)
solve(x + y == 2, x > 0, y > 0)

# Bit-wise operators
# & bit-wise and
# | bit-wise or
# ~ bit-wise not
solve(x & y == ~y)
solve(x < 0)

# using unsigned version of <
solve(ULT(x, 0))
```
## Functions

**Interpretirane funkcije** kao što su aritmetičke gde **funkcija +** ima **fiksnu standardnu interpretaciju** (sabira dva broja). **Neinterpretirane funkcije** i konstante su **maksimalno fleksibilne**; omogućavaju **bilo koju interpretaciju** koja je **dosledna** sa **ograničenjima** nad funkcijom ili konstantom.

Primer: f primenjena dva puta na x rezultira ponovo u x, ali f primenjena jednom na x je drugačija od x.
```python
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 response
s.check()
print(m.model())
```
# Primeri

## Rešavač Sudokua
```python
# 9x9 matrix of integer variables
X = [ [ Int("x_%s_%s" % (i+1, j+1)) for j in range(9) ]
for i in range(9) ]

# each cell contains a value in {1, ..., 9}
cells_c  = [ And(1 <= X[i][j], X[i][j] <= 9)
for i in range(9) for j in range(9) ]

# each row contains a digit at most once
rows_c   = [ Distinct(X[i]) for i in range(9) ]

# each column contains a digit at most once
cols_c   = [ Distinct([ X[i][j] for i in range(9) ])
for j in range(9) ]

# each 3x3 square contains a digit at most once
sq_c     = [ Distinct([ X[3*i0 + i][3*j0 + j]
for i in range(3) for j in range(3) ])
for i0 in range(3) for j0 in range(3) ]

sudoku_c = cells_c + rows_c + cols_c + sq_c

# sudoku instance, we use '0' for empty cells
instance = ((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 in range(9) for j in range(9) ]

s = Solver()
s.add(sudoku_c + instance_c)
if s.check() == sat:
m = s.model()
r = [ [ m.evaluate(X[i][j]) for j in range(9) ]
for i in range(9) ]
print_matrix(r)
else:
print "failed to solve"
```
## Reference

* [https://ericpony.github.io/z3py-tutorial/guide-examples.htm](https://ericpony.github.io/z3py-tutorial/guide-examples.htm)


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