Question

它得到了多个模型但需要花费数小时的时间。所以请建议我缩短获得所有模型的时间。如何在更短的时间内获得所有可能的Satisfy方程解？ z3python中是否有任何函数可以获得所有可能的解决方案。

from z3 import *
x0,x1,x2,x3,x4,x5=BitVecs('x0 x1 x2 x3 x4 x5',32)
y0,y1,y2,y3,y4,y5=BitVecs('y0 y1 y2 y3 y4 y5',32)
k0,k1,k2,k3,k4=BitVecs('k0 k1 k2 k3 k4',32)
c0,c1,c2=BitVecs('c0 c1 c2',32)
s = Solver()
s.add(x0==0x656b696c)
s.add(y0==0x20646e75)
s.add(x5==0xcf9919c3)
s.add(y5==0xf776ba96)
s.add(x1==simplify((RotateLeft(x0,1)&RotateLeft(x0,8))^(RotateLeft(x0,2))^y0^k0))
s.add(y1==x0)

s.add(x2==simplify((RotateLeft(x1,1)&RotateLeft(x1,8))^(RotateLeft(x1,2))^y1^k1))
s.add(y2==x1)

s.add(x3==simplify((RotateLeft(x2,1)&RotateLeft(x2,8))^(RotateLeft(x2,2))^y2^k2))
s.add(y3==x2)

s.add(x4==simplify((RotateLeft(x3,1)&RotateLeft(x3,8))^(RotateLeft(x3,2))^y3^k3))
s.add(y4==x3)

s.add(x5==simplify((RotateLeft(x4,1)&RotateLeft(x4,8))^(RotateLeft(x4,2))^y4^k4))
s.add(y5==x4)

s.add(c1==0)
s.add(c2==1)
s.add(k3==(RotateRight(RotateRight(k2,3),1)^(RotateRight(k2,3)^k0))^c0^c1)
s.add(k4==(RotateRight(RotateRight(k3,3),1)^(RotateRight(k3,3)^k1))^c0^c1)

count = 1
while s.check() == sat:
    if (count > 10):
        break
    print 'The count is:', count
    count=count + 1
    print 'x1=',hex(s.model()[x1].as_long()),'y1=',hex(s.model()[y1].as_long()),'k0=',hex(s.model()[k0].as_long()),"\n "
    print 'x2=',hex(s.model()[x2].as_long()),'y2=',hex(s.model()[y2].as_long()),'k1=',hex(s.model()[k1].as_long()),"\n "
    print 'x3=',hex(s.model()[x3].as_long()),'y3=',hex(s.model()[y3].as_long()),'k2=',hex(s.model()[k2].as_long()),"\n "
    print 'x4=',hex(s.model()[x4].as_long()),'y4=',hex(s.model()[y4].as_long()),'k3=',hex(s.model()[k3].as_long()),"\n "
    print 'x5=',hex(s.model()[x5].as_long()),'y5=',hex(s.model()[y5].as_long()),'k4=',hex(s.model()[k4].as_long()),"\n "
    m = s.model()
    block = []
    for d in m:
        c = d()
        block.append(c != m[d])
    s.add(Or(block))

Answer 1

我没有修改你的程序，只用了0.5秒就完成了。我也没有一台非常快的机器。所以，我很好奇你正在运行什么样的机器来查看运行时间？

减少多个模块的时间

1 个答案: