Comment on Who's up for a challenge?

e0qdk@kbin.social ⁨9⁩ ⁨months⁩ ago

• ExLisper@linux.community ⁨9⁩ ⁨months⁩ ago

  Yep, that’s correct. I never heard about Z3 and I did it by reverting all the operation. It takes couple of seconds of computer time to solve my way but it took me closer to 7h to figure it out. 1h is impressive.

  There are actually two possible solutions because some bits are lost when generating numbers. Can Z3 account for lost bits? Did it come up with just one solution?

  source

  • e0qdk@kbin.social ⁨9⁩ ⁨months⁩ ago

    Can Z3 account for lost bits? Did it come up with just one solution?

    It gave me just one solution the way I asked for it. With additional constraints added to exclude the original solution, it also gives me a second solution -- but the solution it produces is peculiar to my implementation and does not match your implementation. If you implemented exactly how the bits are supposed to end up in the result, you could probably find any other solutions that exist correctly, but I just did it in a quick and dirty way.

    This is (with a little clean up) what my code looked like:

    ::: spoiler solver code
    #!/usr/bin/env python3

    import z3
    
    rand1 = 0.38203435111790895
    rand2 = 0.5012949781958014
    rand3 = 0.5278898433316499
    rand4 = 0.5114834443666041
    
    def xoshiro128ss(a,b,c,d):
        t = 0xFFFFFFFF & (b << 9)
        r = 0xFFFFFFFF & (b * 5)
        r = 0xFFFFFFFF & ((r << 7 | r >> 25) * 9)
        c = 0xFFFFFFFF & (c ^ a)
        d = 0xFFFFFFFF & (d ^ b)
        b = 0xFFFFFFFF & (b ^ c)
        a = 0xFFFFFFFF & (a ^ d)
        c = 0xFFFFFFFF & (c ^ t)
        d = 0xFFFFFFFF & (d << 11 | d >> 21)
        return r, (a, b, c, d)
    
    a,b,c,d = z3.BitVecs("a b c d", 64)
    nodiv_rand1, state = xoshiro128ss(a,b,c,d)
    nodiv_rand2, state = xoshiro128ss(*state)
    nodiv_rand3, state = xoshiro128ss(*state)
    nodiv_rand4, state = xoshiro128ss(*state)
    
    z3.solve(a >= 0, b >= 0, c >= 0, d >= 0, 
      nodiv_rand1 == int(rand1*4294967296),
      nodiv_rand2 == int(rand2*4294967296),
      nodiv_rand3 == int(rand3*4294967296),
      nodiv_rand4 == int(rand4*4294967296)
      )
    

    :::

    I never heard about Z3

    If you're not familiar with SMT solvers, they are a useful tool to have in your toolbox. Here are some links that may be of interest:

    • https://en.wikipedia.org/wiki/Satisfiability_modulo_theories
    • https://en.wikipedia.org/wiki/Z3_Theorem_Prover

    source

    • ExLisper@linux.community ⁨9⁩ ⁨months⁩ ago

      a >= 0, b >= 0, c >= 0, d >= 0

      I think that’s the issue, in the second possible solution one of the parameters is negative :)

      This looks great, I didn’t even know it’s possible to solve it this way. I’m glad someone dedicated some time to it. Let’s see if anyone will try solving it in other way.

      source