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175 lines
5.4 KiB
Markdown
175 lines
5.4 KiB
Markdown
{{#include ../../banners/hacktricks-training.md}}
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非常に基本的に、このツールは、いくつかの条件を満たす必要がある変数の値を見つけるのに役立ちます。手作業で計算するのは非常に面倒です。したがって、Z3に変数が満たす必要のある条件を示すと、可能であればいくつかの値を見つけてくれます。
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**いくつかのテキストと例は[https://ericpony.github.io/z3py-tutorial/guide-examples.htm](https://ericpony.github.io/z3py-tutorial/guide-examples.htm)から抽出されています**
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# 基本操作
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## ブール値/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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## 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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## モデルの印刷
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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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# マシン算術
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現代のCPUと主流のプログラミング言語は、**固定サイズのビットベクター**に対する算術を使用します。マシン算術はZ3Pyで**ビットベクター**として利用可能です。
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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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## 符号付き/符号なしの数
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Z3は、**ビットベクターが符号付きまたは符号なしとして扱われるかどうか**が重要な場合に、特別な符号付きの算術演算を提供します。Z3Pyでは、演算子**<, <=, >, >=, /, % および >>**は**符号付き**バージョンに対応しています。対応する**符号なし**演算子は**ULT, ULE, UGT, UGE, UDiv, URem および LShR.**です。
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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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## 関数
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**解釈された関数**は、**関数 +**が**固定された標準解釈**を持つ算術のようなもので（2つの数を加算します）、**解釈されていない関数**と定数は**最大限の柔軟性**を持ちます。これらは、関数や定数に対する**制約**と**一貫性**のある**任意の解釈**を許可します。
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例：fをxに2回適用すると再びxになりますが、fをxに1回適用するとxとは異なります。
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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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# 例
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## 数独ソルバー
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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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## 参考文献
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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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{{#include ../../banners/hacktricks-training.md}}
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