An old challenge

[YoshL]

Re: An old challenge

solution backpaged, assuming it's right Kappa

[Reincarnate]

Re: An old challenge

Maybe I am missing something, but IMO the question (may) still be malformed... not 100% sure yet.

"It can also predict exactly when this bug will occur, but can't avoid it. The bug inverts only the final answers, without affecting the values used to calculate the answers."

So no matter what you ask the computer, it will say false 50% of the time, and true 50% of the time. This is indistinguishable from having the computer's output being completely disconnected from the input.

And if we were able to extract any meaning from this, it may contradict a premise somewhere. It's sort of like Newcomb's Paradox -- if the psychic can indeed predict your every move, then probability goes out the window and it doesn't matter if you go by utility expectation or dominance principle anyway.

I mean, you could give it a statement with its own predicted bugginess in the statement itself, but I have to see if this contradicts any of the premises (e.g. some variant of "Is the truth value of X the same as the bugged form of this answer?").

[benguino]

Re: An old challenge

@reincarnate: Its not equivalent to the computer flipping a coin time. I think what the OP was trying to get at is that the computer knows ahead of time when its going to lie as if it did an infinite number of coin tosses and recorded them all before you were even there. And since that was done already, you can ask questions about that since its based on the computers previously knowledge.

Edit: I made a post of how the statement:

[Substatement A] xor you are currently bugged

Allows you to get the expected truth value of [substatement a] earlier in this thread. Don't know if you agree with that or find it useful.

[PaperclipGames]

Re: An old challenge

I'm working on a solution, I seem to be getting there. It doesn't seem that difficult, but I need to check that what I'm doing is correct. ^^

@Reincarnate: The computer knows beforehand that the answer he will give is going to be flipped or not (and this can affect the calculation).

[Reincarnate]

Re: An old challenge

But my point is that even if it knows it's going to flip the answer, the answer it gives is still going to be flipped.

The reason I likened this to Newcomb's Paradox is because it tries to get you to make actions even when you're already being predicted by-premise. Here, you're trying to get the computer to output a meaningful answer when the premise says the final answer will be flipped half the time. If the final answer output isn't flipped half the time, then the premise doesn't hold.

[YoshL]

Re: An old challenge

2 cases, final answer (not output) is true or false
2 cases, flipped or not flipped
true flipped - true xor flipped = T xor T = F -> flipped = T
true not-flipped - true xor not-flipped = T xor F = T

false flipped - false xor flipped = F xor T = T -> flipped = F
false not-flipped - false xor not-flipped = F xor F = F

[Reincarnate]

Re: An old challenge

Yeah I am retarded, just found a way and it turned out to be the same as what you guys came up with already.

Let "X xor 'there is a bug present'" be the same as the statement "If true=1 and false=0, then the binary sum sans carry of statement X and statement "'there is a bug present" is equal to 1"

Then let's say X is true.

Assume a bug is present. Then the statement is 1 xor 1 = 0, so false. Computer says true since there's a bug.
Assume a bug is not present. Then the statement is 1 xor 0 = 1, so true. Computer says true.

Now let's say X is false.

Assume a bug is present. Then the statement is 0 xor 1 = 1, so true. Computer says false since there's a bug.
Assume a bug is not present. Then the statement is 0 xor 0 = 0, so false. Computer says false.


So now we have a way to isolate the truth value of X in the computer's output.

[benguino]

Re: An old challenge

Glad we agree Anyways, off to work, I hope to see some good ideas when I get back.

[YoshL]

Re: An old challenge

basically, there's the true or false output, but you can also consider no answer to be an output too.

[PaperclipGames]

Re: An old challenge

Goal: to assign True, False, and Indeterminate answers to the three doors, regardless of bugginess.

Note: say (B xor bugged) yields True, then if it's bugged, (B xor True) is False, thus B is True. If it's not bugged, then B wins also.
Say (B xor bugged) yields False, then if it's bugged, (B xor True) is True, thus B is False. If it's not bugged, then (B xor False) = False implies B is false too.
Say (B xor bugged) yields no answer, then there must be an error in B.

In genearlized form: (statement XOR bugged) iff. statement; regardless of bugginess. This means that I can think of a statement that doesn't regard buginess, and then apply this XOR to it, to get rid of the bug. Thus, the current goal changes to "assign True, False, and Undefined answers to the three doors".

Let's try to, having doors A, B, C, labeled from left to right (if using actual variables in the machine isn't allowed, replace A/B/C in the actual statement with "the door that I have labeled A/B/C in my mind", and it still works), have a True answer link to A being awesome, False leading to B being awesome, and Indeterminate leading to C being awesome.

The statement "A is awesome" suffices for the first condition. Change the statement to fulfill the second statement: "A is awesome XOR (A is awesome XOR B is not awesome)". We still have A->True. If B wins, then we have "False XOR (False XOR False)" = "False XOR False" = False, thus B->False.

But what if C wins? Then we have "False XOR (False XOR True)" = "False XOR True" = True. But we only want A to be assigned to True. Thus, in case C wins, we need to create a paradox/error in the statement, so that the computer won't respond anymore if this happens.

Note that in any case, if A wins, then C loses (and vice versa). Thus, we wish to craft a statement such that if C wins, the statement (which would be True otherwise) implies that A wins also, since that would imply C loses again, resulting in an error. A common type of error is the construction "the statement to the right is false. the statement on the left is true". We can incorporate this into our statemtent. We add to our original statement "AND NOT (statement to the right)", where the right-hand statement is yet to be made.
The statement thus far is then "(A is awesome XOR (A is awesome XOR B is not awesome)) AND (NOT statementontheright)".

The first part of the AND results True if C wins, as we've already seen. Thus, we wish for the statementontheright to include C winning. Thus, the statement on the right will be "statementontheleft AND C is awesome".

Our total construction will thus be: "((A is awesome XOR (A is awesome XOR B is not awesome)) AND (NOT statementontheright)) XOR ( statementontheleft AND (C is awesome)", where statementontheright is the part after the bold XOR, and statemenetontheleft is the part before it.

Let's check again:
A wins: "((True XOR (True XOR True)) AND (NOT statementontheright)) XOR (statementontheleft AND False)"
= "(True AND (NOT statementontheright)) XOR (False)
= True. Phew.

B wins: "((False XOR (False XOR False)) AND (NOT statementontheright)) XOR (NOT(statementontheleft) AND False)
= "(False AND (statementontheright)) XOR (False)
= "(False AND False) XOR False"
= False. Phew.

C wins: "((False XOR (False XOR True)) AND (NOT statementontheright)) XOR (statementontheleft AND True)"
= "True AND (NOT statementontheright)) XOR (statementontheleft)"
= "NOT(statementontheright) XOR (statementontheleft)"
The left tells the right it's lying, whereas the right tells that the left is true.

If the left is True, then that directly implies the right is False. But if the right is False, then that means the left is also False. Contradiction.
If the left is False, then that directly implies the right is True. But if the right is True, then that means the left is also True. Contradiction.

Thus, it is impossible to determine whether either hand sides are true or false, and in particular, the XOR cannot be executed. Conclusion: If C wins, the computer will not give an answer.

To finalize the answer, we must incorporate the term we found in the very beginning to get rid of the bug. Thus, our final statement becomes:

(((A is awesome XOR (A is awesome XOR B is not awesome)) AND (NOT redstatement)) XOR (greenstatement AND (C is awesome))) XOR (You are bugged), where redstatement is everything red and greenstatement is everything green.

Please check my answer, rofl. I wrote down every step I took, and after a few errors I made and fixing them, I seem to have reached an answer.

[Dynam0]

Re: An old challenge

Lots of intense logical proofing here, I feel like I'm out of my league lol

Still going with:

Let the word "True" mean yes.
Let the word "False" mean that the bug has occurred, implying that the real answer is "True" which means yes.

Door 'x' is the awesome door.

**If the computer doesn't reply, it's because it can't declare that statement to be "False". I've effectively taken away it's ability to declare a statement false. Not sure if that breaks one of the rules since all I've done was changed the definition of the word false.

Pick other doors until the computer gives an answer. Any answer summarily implies 'yes' meaning that the computer will basically tell you when you've made a correct statement and thus picked the right door.

This method (if allowed) would work for any finite number of doors btw.

[PaperclipGames]

Re: An old challenge

Lots of intense logical proofing here, I feel like I'm out of my league lol

Still going with:

Let the word "True" mean yes.
Let the word "False" mean that the bug has occurred, implying that the real answer is "True" which means yes.

Door 'x' is the awesome door.

**If the computer doesn't reply, it's because it can't declare that statement to be "False". I've effectively taken away it's ability to declare a statement false. Not sure if that breaks one of the rules since all I've done was changed the definition of the word false.

Pick other doors until the computer gives an answer. Any answer summarily implies 'yes' meaning that the computer will basically tell you when you've made a correct statement and thus picked the right door.

This method (if allowed) would work for any finite number of doors btw.

This would result in a 1/3 chance of the computer saying either True or False, and a 2/3 chance of it saying nothing. Therefore, no guarantees. Your method doesn't work.

You can only say one statement, pal.

[Dynam0]

Re: An old challenge

Only one "answerable" statement, yes.

I can make as many unanswerable statements as I want; read the rules again.

[benguino]

Re: An old challenge

Nice answer Alex! It seems to flow all right with me, definitely agree with the whole "xor you are bugged" thing at least since I posted my finding of that already in this thread. I didn't think of trying to create a paradox like that, very clever. I'll try to go through the proof by hand with pencil and paper when I get home if you'd like a second verification.

[Zaevod]

Re: An old challenge

Please check my answer, rofl. I wrote down every step I took, and after a few errors I made and fixing them, I seem to have reached an answer.

I will, in a moment.

Edit: I read it. It was beautiful. Good job! 8D

I'll post my own solution later, just to provide an alternative. And remember this whenever you get trapped in a chamber with 3 doors and a bugged magical computer!