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

<details>

<summary><strong>从零开始学习AWS黑客技术，成为专家</strong> <a href="https://training.hacktricks.xyz/courses/arte"><strong>htARTE（HackTricks AWS红队专家）</strong></a><strong>！</strong></summary>

其他支持HackTricks的方式：

* 如果您想看到您的**公司在HackTricks中做广告**或**下载PDF格式的HackTricks**，请查看[**订阅计划**](https://github.com/sponsors/carlospolop)!
* 获取[**官方PEASS & HackTricks周边产品**](https://peass.creator-spring.com)
* 探索[**PEASS家族**](https://opensea.io/collection/the-peass-family)，我们的独家[**NFTs**](https://opensea.io/collection/the-peass-family)
* **加入** 💬 [**Discord群**](https://discord.gg/hRep4RUj7f) 或 [**电报群**](https://t.me/peass) 或 **关注**我们的**Twitter** 🐦 [**@carlospolopm**](https://twitter.com/hacktricks_live)**。**
* 通过向[**HackTricks**](https://github.com/carlospolop/hacktricks)和[**HackTricks Cloud**](https://github.com/carlospolop/hacktricks-cloud) github仓库提交PR来分享您的黑客技巧。

</details>


非常基本地，这个工具将帮助我们找到需要满足一些条件的变量的值，手动计算将会很烦人。因此，您可以告诉Z3变量需要满足的条件，它将找到一些值（如果可能的话）。

**一些文本和示例摘自[https://ericpony.github.io/z3py-tutorial/guide-examples.htm](https://ericpony.github.io/z3py-tutorial/guide-examples.htm)**

# 基本操作

## 布尔值/与/或/非
```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
```
## 整数/简化/实数
```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))
```
## 打印模型
```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]))
```
# 机器算术

现代 CPU 和主流编程语言使用固定大小比特向量进行算术运算。在 Z3Py 中，可以使用**比特向量**来进行机器算术。
```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
```
## 有符号/无符号数字

Z3提供了特殊的有符号版本的算术操作，在这些操作中，**位向量被视为有符号或无符号**会产生不同的结果。在Z3Py中，操作符**<, <=, >, >=, /, % 和 >>**对应于**有符号**版本。相应的**无符号**操作符是**ULT, ULE, UGT, UGE, UDiv, URem 和 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))
```
## 函数

**解释函数**，如算术，其中**函数 +**具有**固定的标准解释**（它将两个数字相加）。**未解释函数**和常量具有**最大的灵活性**；它们允许**与函数或常量上的约束一致的任何解释**。

示例：f两次应用于x会再次得到x，但f应用一次于x与x不同。
```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())
```
# 例子

## 数独求解器
```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"
```
## 参考资料

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


<details>

<summary><strong>从零开始学习AWS黑客技术，成为专家</strong> <a href="https://training.hacktricks.xyz/courses/arte"><strong>htARTE（HackTricks AWS Red Team Expert）</strong></a><strong>！</strong></summary>

支持HackTricks的其他方式：

* 如果您想在HackTricks中看到您的**公司广告**或**下载PDF格式的HackTricks**，请查看[**订阅计划**](https://github.com/sponsors/carlospolop)!
* 获取[**官方PEASS & HackTricks周边产品**](https://peass.creator-spring.com)
* 探索我们的独家[**NFTs**](https://opensea.io/collection/the-peass-family)收藏品[**The PEASS Family**](https://opensea.io/collection/the-peass-family)
* **加入** 💬 [**Discord群**](https://discord.gg/hRep4RUj7f) 或 [**电报群**](https://t.me/peass) 或在**Twitter**上关注我们 🐦 [**@carlospolopm**](https://twitter.com/hacktricks_live)**。**
* 通过向[**HackTricks**](https://github.com/carlospolop/hacktricks)和[**HackTricks Cloud**](https://github.com/carlospolop/hacktricks-cloud) github仓库提交PR来分享您的黑客技巧。 

</details>