Sujet : Re: No TM exists that can simulate all TM.
De : wyniijj5 (at) *nospam* gmail.com (wij)
Groupes : comp.theoryDate : 26. Oct 2024, 15:08:23
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <bedc2992ff0433fe5abfe8e572a75b1a44a67884.camel@gmail.com>
References : 1 2
User-Agent : Evolution 3.50.2 (3.50.2-1.fc39)
On Fri, 2024-10-25 at 18:17 -0400, Richard Damon wrote:
On 10/25/24 1:12 PM, wij wrote:
Proof: Simulating self is not possible (from all real programs, every one can verify).
This also implies UTM does not exist.
Did Turing made a mistake?
https://en.wikipedia.org/wiki/Universal_Turing_machine
The key point you are missing is that if the UTM emulating a program
like H^ (without the contrary nature at the end), so that we can get the
infinite chain of UTM emulating that UTM emulating that UTM ..., it just
ends up being a non-halting computation, and thus the non-halting
emulation of that computation is exactly right.
The problem is if you want it to be a UTM and also a decider, then you
run into the problem.
The UTM chain *IS* an infinite deep recursive simulation, which is just
like the infinite-recursion, which is non-halting.
When you change the UTM to be a decider (and thus no longer a UTM) it
finds it can't know the answer for the behavior of the UTM it was
previously.
Simulator::= A TM (utm) that performs the same function as its argument TM (f).
IOW, utm(f)=f (or utm(f,arg)=f(arg))
In case of self-simulation "utm(utm)" ... well, the argument utm has no argument
(empty tape?) to define its behavior. What is the outer utm supposed to 'simulate'?
How do you define 'simulator'? What exactly does UTM simulate?