3 ms·
Movfuscator only works because x86's MOV is like the Swiss Army knife of move instructions. It doesn't just copy bits from one location to another, but has para
by SilasX 6y ago
Movfuscator only works because x86's MOV is like the Swiss Army knife of move instructions. It doesn't just copy bits from one location to another, but has parameters for conditionality and arithmetic, which is what allows it to be so general. That still doesn't address the parent's confusion (that I share) of how you get Turing-completeness just from addition and multiplication.
- dlubarov 6y agoTrue, but there are simpler (abstract) machines that also achieve Turing completeness: https://en.wikipedia.org/wiki/One-instruction_set_computer https://en.wikipedia.org/wiki/One-instruction_set_computer > That still doesn't address the parent's confusion (that I share) of how you get Turing-completeness just from addition and multiplication. I wouldn't say arithmetic circuits are Turing complete since they can only perform bounded computations, but we could say that they're functionally complete. It is known that boolean AND and OR gates are functionally complete, and we can reduce them to multiplication and addition in any finite field: x and y = x * y x or y = ¬(¬x and ¬y) = 1 - (1 - x)(1 - y)
- giomasce 6y agoWhat does "functionally complete" means for you?
- dlubarov 6y agoInformally, I mean reducible from time-bounded Turing machines. Given a Turing machine TM and a time bound T, a (binary or arithmetic) circuit can be constructed to simulate TM for T steps. Or in other words, a circuit can be constructed to solve any particular instance of FNTIME(T).
- rrobukef 6y agoYou get bounded-Turing completeness. This is sufficient for all practical applications as we live in a finite universe.