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

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Kwa msingi, chombo hiki kitatusaidia kupata thamani za mabadiliko ambayo yanahitaji kutimiza masharti fulani na kuhesabu kwa mkono kutakuwa na usumbufu mkubwa. Hivyo, unaweza kuonyesha kwa Z3 masharti ambayo mabadiliko yanahitaji kutimiza na itapata baadhi ya thamani (ikiwa inawezekana).

**Baadhi ya maandiko na mifano yameondolewa kutoka [https://ericpony.github.io/z3py-tutorial/guide-examples.htm](https://ericpony.github.io/z3py-tutorial/guide-examples.htm)**

# Basic Operations

## Booleans/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))
```
## Kuchapisha Mfano
```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]))
```
# Hesabu ya Mashine

CPUs za kisasa na lugha za programu za kawaida hutumia hesabu juu ya **bit-vectors zenye ukubwa wa kudumu**. Hesabu ya mashine inapatikana katika Z3Py kama **Bit-Vectors**.
```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
```
## Signed/Unsigned Numbers

Z3 inatoa toleo maalum la operesheni za kihesabu ambapo ina umuhimu ikiwa **bit-vector inachukuliwa kama signed au unsigned**. Katika Z3Py, waendeshaji **<, <=, >, >=, /, % na >>** wanalingana na toleo la **signed**. Waendeshaji wa **unsigned** husika ni **ULT, ULE, UGT, UGE, UDiv, URem na 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

**Kazi za tafsiri** kama za hesabu ambapo **kazi +** ina **tafsiri ya kawaida isiyobadilika** (inaongeza nambari mbili). **Kazi zisizo na tafsiri** na constants ni **za kiwango cha juu cha kubadilika**; zinaruhusu **tafsiri yoyote** ambayo ni **sawa** na **vizuizi** juu ya kazi au constant.

Mfano: f inatumika mara mbili kwa x inarudi x tena, lakini f inatumika mara moja kwa x ni tofauti na 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())
```
# Mifano

## Mchambuzi wa Sudoku
```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"
```
## References

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


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