Actually, the problem as stated in this video is solvable... If you take a finite set of programs, you can always create a program that says true or false whether the input program is flawed or not. The number of programs which size is < 1To is finite. Therefore, you could even write a program that checks all the programs < 1 To. It would take a long time to write though XD .
Though there is indeed an algorithm that solves the halting problem for the finite collection on the disk, I don't think anyone could write it, since they'd have to hard-code the solution. They could, actually, but only buy guessing correctly. However, saying that "it would take a long time to write" seems to indicate that it's figure out-able, which it's not (i.e. the solution is not uniform in the finite set).
@eabod I think I understand. So if the program you are testing on your finite inputs halts, then hooray. But if the program is taking forever, you could look at the loop conditions and the current input state and determine if the loop will quit or not. Is that right?
@stoobers1 No, I'm saying there's a program that "guesses" correctly, since only finitely many guesses are needed. E.g. if there is only one program, P1, on the disk, then either P1 halts or it doesn't, so two guessing algorithms are sufficient; one that outputs "yes" (no matter what) and one that outputs "no". One of these algorithms is correct. Similarly if there are n programs on the disk, then 2^n guessing algorithms (on for each combination of "yes" and "no") are sufficient.
@stoobers1 Notice that what I'm suggesting would be impossible if there were infinitely many programs on the disk, since an algorithm cannot hold infinitely much hard-coded information.
Eve is such a bitch.
ClericAgent 8 months ago 3
Actually, the problem as stated in this video is solvable... If you take a finite set of programs, you can always create a program that says true or false whether the input program is flawed or not. The number of programs which size is < 1To is finite. Therefore, you could even write a program that checks all the programs < 1 To. It would take a long time to write though XD .
boumbh 11 months ago
Though there is indeed an algorithm that solves the halting problem for the finite collection on the disk, I don't think anyone could write it, since they'd have to hard-code the solution. They could, actually, but only buy guessing correctly. However, saying that "it would take a long time to write" seems to indicate that it's figure out-able, which it's not (i.e. the solution is not uniform in the finite set).
eabod 2 days ago
Why does bob have a Mohawk
Punk rockers aren't educated enough to ponder this sort of thing
and why is eve shirtless that dirt whore
manrayer88 11 months ago
There is indeed an algorithm which, given input "n", outputs "yes" if the nth program on the disk halts and "no" otherwise. Bob just can't find it.
eabod 1 year ago
@eabod Prove it.
stoobers1 4 days ago
@stoobers1 The disk has finite memory, hence the collection of (indices of) programs on the disk is finite. Every finite set is computable. QED
eabod 3 days ago
@eabod I think I understand. So if the program you are testing on your finite inputs halts, then hooray. But if the program is taking forever, you could look at the loop conditions and the current input state and determine if the loop will quit or not. Is that right?
stoobers1 3 days ago
Comment removed
eabod 2 days ago
@stoobers1 No, I'm saying there's a program that "guesses" correctly, since only finitely many guesses are needed. E.g. if there is only one program, P1, on the disk, then either P1 halts or it doesn't, so two guessing algorithms are sufficient; one that outputs "yes" (no matter what) and one that outputs "no". One of these algorithms is correct. Similarly if there are n programs on the disk, then 2^n guessing algorithms (on for each combination of "yes" and "no") are sufficient.
eabod 2 days ago
@stoobers1 Notice that what I'm suggesting would be impossible if there were infinitely many programs on the disk, since an algorithm cannot hold infinitely much hard-coded information.
eabod 2 days ago
i must know!
fractal420 2 years ago
That's the kind of reaction I like to hear :D
kjlg74 2 years ago
The light at the end...invoking presents of a spiritual being from another dimensional plain? Hmmmm....what will happen next?
rgman66 2 years ago
It's a gripping cliffhanger, isn't it? ;D
kjlg74 2 years ago
OMG what is going to happen next?
1GOD1JESUS 2 years ago
It's anyone's guess...
kjlg74 2 years ago