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

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<details>
<summary><strong>从零开始学习AWS黑客技术</strong> <a href="https://training.hacktricks.xyz/courses/arte"><strong>htARTEHackTricks AWS红队专家</strong></a><strong></strong></summary>
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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
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# 基本操作
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## 布尔值/And/Or/Not
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```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
```
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## 整数/简化/实数
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```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))
```
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## 打印模型
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```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():
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print("%s = %s" % (d.name(), m[d]))
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```
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# 机器算术
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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。
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```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
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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中运算符**<, <=, >, >=, /, % 和 >>**对应于**有符号**版本。相应的**无符号**运算符是**ULT, ULE, UGT, UGE, UDiv, URem 和 LShR**。
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```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)
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# using unsigned version of <
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solve(ULT(x, 0))
```
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## 函数
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**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**。
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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。
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```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)))
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print(m.evaluate(f(2)))
s.add(f(x) == 4) #Find the value that generates 4 as response
s.check()
print(m.model())
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```
# 例子
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## 数独求解器
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```python
# 9x9 matrix of integer variables
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}
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
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) ])
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for j in range(9) ]
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# each 3x3 square contains a digit at most once
sq_c = [ Distinct([ X[3*i0 + i][3*j0 + j]
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for i in range(3) for j in range(3) ])
for i0 in range(3) for j0 in range(3) ]
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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),
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(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))
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instance_c = [ If(instance[i][j] == 0,
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True,
X[i][j] == instance[i][j])
for i in range(9) for j in range(9) ]
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s = Solver()
s.add(sudoku_c + instance_c)
if s.check() == sat:
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m = s.model()
r = [ [ m.evaluate(X[i][j]) for j in range(9) ]
for i in range(9) ]
print_matrix(r)
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else:
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print "failed to solve"
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```
## 参考资料
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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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2022-04-28 16:01:33 +00:00
<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>htARTEHackTricks AWS Red Team Expert</strong></a><strong></strong></summary>
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支持HackTricks的其他方式
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* 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)!
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* 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**我的**Twitter** 🐦 [**@carlospolopm**](https://twitter.com/carlospolopm)**。**
* 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。
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