Reduce the time for multiple modles

Question

It get Multiple Models But It takes hours of time.So Kindly suggest me to reduce the time to get all models.how to get all possible solution for Satisfy equation in less time? Is there any function in z3python for getting all possible solutions in number.

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))

score 0 · Answer 1 · answered Mar 04 '17 at 23:40

0

I ran your program unmodified and it took a mere 0.5 seconds to complete. And I don't really have a very fast machine either. So, I'm curious what sort of a machine you're running this on to see hours of runtime?

answered Mar 04 '17 at 23:40

alias

28,120
2
23
40

8.5 rounds of code? It's impossible to know exactly what that's referring to. If you could post, or link to, something people can actually look at that exhibits the problem, you might get better feedback. – alias Mar 06 '17 at 15:40
I am implementing the code for simon block cipher in z3py for fixing plaintext and cipher text and retrieving possible keys.The above is for 2 rounds of simon.while implementing for more rounds it takes more time its like days and after 7 rounds it may not give keys. Here is paper [link] ( https://eprint.iacr.org/2013/404.pdf ) – praveen gundaram Mar 08 '17 at 05:01
An SMT solver would be extremely unlikely to figure out the keys from a pair of pt-ct. That would go against cryptographic security of such algorithms. Essentially, you'd be forcing z3 down to what's necessarily a brute-force search; no better than simply trying all possible keys. – alias Mar 08 '17 at 16:48

Reduce the time for multiple modles

1 Answers1