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raoofha wrote: October 3rd, 2023, 12:45 pmas I wrote before the turing machine U halts if the given input halts and return the tape WITHOUT halting if the given input does not halt, so for all input you know that a given input halt or does not halt that's sound to me like a solution assuming I can find U

when does it return the tape?
how does it know when to return the tape? How does it know the input does or doesn't halt so it can determine whether to return the tape/etcetera??

olivia enessemir wrote: October 3rd, 2023, 1:00 pm when does it return the tape?
how does it know when to return the tape? How does it know the input does or doesn't halt so it can determine whether to return the tape/etcetera??

those are things that have to be discovered, but U can not solve the halting problem by having a "return the tape" state because that does not work (I can explain why if you are interested) it must have some kind of "underflow" situation to avoid creating a counter example for U

raoofha wrote: October 3rd, 2023, 1:22 pm U can not solve the halting problem by having a "return the tape" state because that does not work

yes, this is the point I'm making.
regardless, obviously if:

raoofha wrote: October 3rd, 2023, 12:45 pm for all input you know that a given input halt or does not halt that's sound to me like a solution

and you actually know that a given non-halting input doesn't halt after a finite amount of time, then if that can be done with a regular turing machine, you can construct a contradiction using the same construction that has already been mentioned many, many, many times in this thread.

olivia enessemir wrote: October 4th, 2023, 3:46 am and you actually know that a given non-halting input doesn't halt after a finite amount of time, then if that can be done with a regular turing machine, you can construct a contradiction using the same construction that has already been mentioned many, many, many times in this thread.

how do you create a contradiction exactly ? I'm curious to know
the contradiction only works for a turing machine that halts for all input not for U

raoofha wrote: October 4th, 2023, 5:50 am the contradiction only works for a turing machine that halts for all input not for U [i.e. U doesn't solve the halting problem for all turing machines]

... then U doesn't really solve the halting problem. Solving the halting problem for only some cases is possible but kind of pointless.

olivia enessemir wrote: October 4th, 2023, 12:39 pm ... then U doesn't really solve the halting problem. Solving the halting problem for only some cases is possible but kind of pointless.

as I wrote many times U(m) halts on m if m halts and return the tape WITHOUT halting if m does not halt
for me it is an interesting challenge to find U
if you or anybody else reading this think there is no turing machine U to be found I'm curious to know your reasons

Can we halt this thread?

hotdogPi wrote: October 4th, 2023, 2:32 pm Can we halt this thread?

yes