mirror of
https://github.com/carlospolop/hacktricks
synced 2024-12-18 17:16:10 +00:00
202 lines
7.4 KiB
Markdown
202 lines
7.4 KiB
Markdown
{% hint style="success" %}
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Learn & practice AWS Hacking:<img src="/.gitbook/assets/arte.png" alt="" data-size="line">[**HackTricks Training AWS Red Team Expert (ARTE)**](https://training.hacktricks.xyz/courses/arte)<img src="/.gitbook/assets/arte.png" alt="" data-size="line">\
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Learn & practice GCP Hacking: <img src="/.gitbook/assets/grte.png" alt="" data-size="line">[**HackTricks Training GCP Red Team Expert (GRTE)**<img src="/.gitbook/assets/grte.png" alt="" data-size="line">](https://training.hacktricks.xyz/courses/grte)
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<details>
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<summary>Support HackTricks</summary>
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* Check the [**subscription plans**](https://github.com/sponsors/carlospolop)!
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* **Join the** 💬 [**Discord group**](https://discord.gg/hRep4RUj7f) or the [**telegram group**](https://t.me/peass) or **follow** us on **Twitter** 🐦 [**@hacktricks\_live**](https://twitter.com/hacktricks\_live)**.**
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* **Share hacking tricks by submitting PRs to the** [**HackTricks**](https://github.com/carlospolop/hacktricks) and [**HackTricks Cloud**](https://github.com/carlospolop/hacktricks-cloud) github repos.
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</details>
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{% endhint %}
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Basicamente, esta ferramenta nos ajudará a encontrar valores para variáveis que precisam satisfazer algumas condições e calculá-los manualmente será muito irritante. Portanto, você pode indicar ao Z3 as condições que as variáveis precisam satisfazer e ele encontrará alguns valores (se possível).
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**Alguns textos e exemplos são extraídos de [https://ericpony.github.io/z3py-tutorial/guide-examples.htm](https://ericpony.github.io/z3py-tutorial/guide-examples.htm)**
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# Operações Básicas
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## Booleanos/E/OU/Não
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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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## Ints/Simplify/Reals
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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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## Impressão de 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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CPUs modernas e linguagens de programação convencionais usam aritmética sobre **vetores de bits de tamanho fixo**. A aritmética de máquina está disponível no Z3Py como **Vetores de Bits**.
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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 Assinados/Não Assinados
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Z3 fornece versões especiais assinadas de operações aritméticas onde faz diferença se o **bit-vector é tratado como assinado ou não assinado**. No Z3Py, os operadores **<, <=, >, >=, /, % e >>** correspondem às versões **assinadas**. Os operadores **não assinados** correspondentes são **ULT, ULE, UGT, UGE, UDiv, URem e 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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## Funções
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**Funções interpretadas** como aritmética onde a **função +** tem uma **interpretação padrão fixa** (ela soma dois números). **Funções não interpretadas** e constantes são **maximamente flexíveis**; elas permitem **qualquer interpretação** que seja **consistente** com as **restrições** sobre a função ou constante.
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Exemplo: f aplicada duas vezes a x resulta em x novamente, mas f aplicada uma vez a x é 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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# Exemplos
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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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## Referências
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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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{% hint style="success" %}
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Aprenda e pratique Hacking AWS:<img src="/.gitbook/assets/arte.png" alt="" data-size="line">[**HackTricks Training AWS Red Team Expert (ARTE)**](https://training.hacktricks.xyz/courses/arte)<img src="/.gitbook/assets/arte.png" alt="" data-size="line">\
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Aprenda e pratique Hacking GCP: <img src="/.gitbook/assets/grte.png" alt="" data-size="line">[**HackTricks Training GCP Red Team Expert (GRTE)**<img src="/.gitbook/assets/grte.png" alt="" data-size="line">](https://training.hacktricks.xyz/courses/grte)
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<details>
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<summary>Support HackTricks</summary>
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* Confira os [**planos de assinatura**](https://github.com/sponsors/carlospolop)!
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* **Junte-se ao** 💬 [**grupo do Discord**](https://discord.gg/hRep4RUj7f) ou ao [**grupo do telegram**](https://t.me/peass) ou **siga**-nos no **Twitter** 🐦 [**@hacktricks\_live**](https://twitter.com/hacktricks\_live)**.**
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* **Compartilhe truques de hacking enviando PRs para o** [**HackTricks**](https://github.com/carlospolop/hacktricks) e [**HackTricks Cloud**](https://github.com/carlospolop/hacktricks-cloud) repositórios do github.
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</details>
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{% endhint %}
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