mirror of
https://github.com/carlospolop/hacktricks
synced 2024-11-26 06:30:37 +00:00
198 lines
7.3 KiB
Markdown
198 lines
7.3 KiB
Markdown
<details>
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<summary><strong>Aprende hacking en AWS de cero a héroe con</strong> <a href="https://training.hacktricks.xyz/courses/arte"><strong>htARTE (HackTricks AWS Red Team Expert)</strong></a><strong>!</strong></summary>
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Otras formas de apoyar a HackTricks:
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* Si quieres ver tu **empresa anunciada en HackTricks** o **descargar HackTricks en PDF**, consulta los [**PLANES DE SUSCRIPCIÓN**](https://github.com/sponsors/carlospolop)!
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* Consigue el [**merchandising oficial de PEASS & HackTricks**](https://peass.creator-spring.com)
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* Descubre [**La Familia PEASS**](https://opensea.io/collection/the-peass-family), nuestra colección de [**NFTs**](https://opensea.io/collection/the-peass-family) exclusivos
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* **Únete al** 💬 [**grupo de Discord**](https://discord.gg/hRep4RUj7f) o al [**grupo de telegram**](https://t.me/peass) o **sigue**me en **Twitter** 🐦 [**@carlospolopm**](https://twitter.com/carlospolopm)**.**
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* **Comparte tus trucos de hacking enviando PRs a los repositorios de GitHub** [**HackTricks**](https://github.com/carlospolop/hacktricks) y [**HackTricks Cloud**](https://github.com/carlospolop/hacktricks-cloud).
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</details>
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Muy básicamente, esta herramienta nos ayudará a encontrar valores para variables que necesitan satisfacer algunas condiciones y calcularlos a mano sería muy molesto. Por lo tanto, puedes indicarle a Z3 las condiciones que las variables necesitan satisfacer y encontrará algunos valores (si es posible).
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# Operaciones Básicas
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## Booleanos/And/Or/Not
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```python
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#pip3 install z3-solver
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from z3 import *
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s = Solver() #The solver will be given the conditions
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x = Bool("x") #Declare the symbos x, y and z
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y = Bool("y")
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z = Bool("z")
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# (x or y or !z) and y
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s.add(And(Or(x,y,Not(z)),y))
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s.check() #If response is "sat" then the model is satifable, if "unsat" something is wrong
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print(s.model()) #Print valid values to satisfy the model
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```
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## Enteros/Simplificar/Reales
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```python
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from z3 import *
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x = Int('x')
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y = Int('y')
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#Simplify a "complex" ecuation
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print(simplify(And(x + 1 >= 3, x**2 + x**2 + y**2 + 2 >= 5)))
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#And(x >= 2, 2*x**2 + y**2 >= 3)
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#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.)
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#so you can get the decimals you need from the solution
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r1 = Real('r1')
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r2 = Real('r2')
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#Solve the ecuation
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print(solve(r1**2 + r2**2 == 3, r1**3 == 2))
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#Solve the ecuation with 30 decimals
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set_option(precision=30)
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print(solve(r1**2 + r2**2 == 3, r1**3 == 2))
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```
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## Impresión del Modelo
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```python
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from z3 import *
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x, y, z = Reals('x y z')
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s = Solver()
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s.add(x > 1, y > 1, x + y > 3, z - x < 10)
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s.check()
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m = s.model()
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print ("x = %s" % m[x])
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for d in m.decls():
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print("%s = %s" % (d.name(), m[d]))
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```
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# Aritmética de Máquina
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Las CPU modernas y los lenguajes de programación principales utilizan aritmética sobre **vectores de bits de tamaño fijo**. La aritmética de máquina está disponible en Z3Py como **Bit-Vectors**.
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```python
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from z3 import *
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x = BitVec('x', 16) #Bit vector variable "x" of length 16 bit
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y = BitVec('y', 16)
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e = BitVecVal(10, 16) #Bit vector with value 10 of length 16bits
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a = BitVecVal(-1, 16)
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b = BitVecVal(65535, 16)
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print(simplify(a == b)) #This is True!
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a = BitVecVal(-1, 32)
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b = BitVecVal(65535, 32)
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print(simplify(a == b)) #This is False
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```
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## Números con signo/sin signo
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Z3 ofrece versiones especiales firmadas de operaciones aritméticas donde importa si el **vector de bits se trata como con signo o sin signo**. En Z3Py, los operadores **<, <=, >, >=, /, % y >>** corresponden a las versiones **con signo**. Los operadores **sin signo** correspondientes son **ULT, ULE, UGT, UGE, UDiv, URem y LShR.**
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```python
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from z3 import *
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# Create to bit-vectors of size 32
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x, y = BitVecs('x y', 32)
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solve(x + y == 2, x > 0, y > 0)
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# Bit-wise operators
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# & bit-wise and
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# | bit-wise or
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# ~ bit-wise not
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solve(x & y == ~y)
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solve(x < 0)
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# using unsigned version of <
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solve(ULT(x, 0))
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```
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## Funciones
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**Funciones interpretadas** como la aritmética donde la **función +** tiene una **interpretación estándar fija** (suma dos números). Las **funciones no interpretadas** y las constantes son **máximamente flexibles**; permiten **cualquier interpretación** que sea **consistente** con las **restricciones** sobre la función o constante.
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Ejemplo: f aplicada dos veces a x resulta en x de nuevo, pero f aplicada una vez a x es diferente de x.
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```python
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from z3 import *
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x = Int('x')
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y = Int('y')
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f = Function('f', IntSort(), IntSort())
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s = Solver()
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s.add(f(f(x)) == x, f(x) == y, x != y)
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s.check()
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m = s.model()
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print("f(f(x)) =", m.evaluate(f(f(x))))
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print("f(x) =", m.evaluate(f(x)))
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print(m.evaluate(f(2)))
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s.add(f(x) == 4) #Find the value that generates 4 as response
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s.check()
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print(m.model())
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```
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# Ejemplos
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## Solucionador de Sudoku
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```python
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# 9x9 matrix of integer variables
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X = [ [ Int("x_%s_%s" % (i+1, j+1)) for j in range(9) ]
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for i in range(9) ]
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# each cell contains a value in {1, ..., 9}
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cells_c = [ And(1 <= X[i][j], X[i][j] <= 9)
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for i in range(9) for j in range(9) ]
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# each row contains a digit at most once
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rows_c = [ Distinct(X[i]) for i in range(9) ]
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# each column contains a digit at most once
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cols_c = [ Distinct([ X[i][j] for i in range(9) ])
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for j in range(9) ]
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# each 3x3 square contains a digit at most once
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sq_c = [ Distinct([ X[3*i0 + i][3*j0 + j]
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for i in range(3) for j in range(3) ])
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for i0 in range(3) for j0 in range(3) ]
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sudoku_c = cells_c + rows_c + cols_c + sq_c
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# sudoku instance, we use '0' for empty cells
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instance = ((0,0,0,0,9,4,0,3,0),
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(0,0,0,5,1,0,0,0,7),
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(0,8,9,0,0,0,0,4,0),
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(0,0,0,0,0,0,2,0,8),
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(0,6,0,2,0,1,0,5,0),
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(1,0,2,0,0,0,0,0,0),
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(0,7,0,0,0,0,5,2,0),
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(9,0,0,0,6,5,0,0,0),
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(0,4,0,9,7,0,0,0,0))
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instance_c = [ If(instance[i][j] == 0,
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True,
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X[i][j] == instance[i][j])
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for i in range(9) for j in range(9) ]
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s = Solver()
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s.add(sudoku_c + instance_c)
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if s.check() == sat:
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m = s.model()
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r = [ [ m.evaluate(X[i][j]) for j in range(9) ]
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for i in range(9) ]
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print_matrix(r)
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else:
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print "failed to solve"
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```
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# Referencias
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* [https://ericpony.github.io/z3py-tutorial/guide-examples.htm](https://ericpony.github.io/z3py-tutorial/guide-examples.htm)
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<details>
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<summary><strong>Aprende hacking en AWS de cero a héroe con</strong> <a href="https://training.hacktricks.xyz/courses/arte"><strong>htARTE (HackTricks AWS Red Team Expert)</strong></a><strong>!</strong></summary>
|
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|
|
Otras formas de apoyar a HackTricks:
|
|
|
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* Si quieres ver a tu **empresa anunciada en HackTricks** o **descargar HackTricks en PDF** revisa los [**PLANES DE SUSCRIPCIÓN**](https://github.com/sponsors/carlospolop)!
|
|
* Consigue el [**merchandising oficial de PEASS & HackTricks**](https://peass.creator-spring.com)
|
|
* Descubre [**La Familia PEASS**](https://opensea.io/collection/the-peass-family), nuestra colección de [**NFTs**](https://opensea.io/collection/the-peass-family) exclusivos
|
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* **Únete al** 💬 [**grupo de Discord**](https://discord.gg/hRep4RUj7f) o al [**grupo de telegram**](https://t.me/peass) o **sígueme** en **Twitter** 🐦 [**@carlospolopm**](https://twitter.com/carlospolopm)**.**
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* **Comparte tus trucos de hacking enviando PRs a los repositorios de github** [**HackTricks**](https://github.com/carlospolop/hacktricks) y [**HackTricks Cloud**](https://github.com/carlospolop/hacktricks-cloud).
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</details>
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