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I wasn't able to see my T.A. today but my friend asked him for the answer over AIM so when he responds I'll let you know. Hopefully it doesn't turn out to be a huge let down as I know DrPizza and other people actually put time into this.

Along with that, yes GooseMaster I'll punch my T.A. in the face if it's some trick question or has a solution you could never get.
 
Originally posted by: RedArmy
I wasn't able to see my T.A. today but my friend asked him for the answer over AIM so when he responds I'll let you know. Hopefully it doesn't turn out to be a huge let down as I know DrPizza and other people actually put time into this.

Along with that, yes GooseMaster I'll punch my T.A. in the face if it's some trick question or has a solution you could never get.

Make sure you make a flip book of it while drawing the scene as if the view point was rotating in 3D space...
 
Originally posted by: gsellis
"If the your number is 3, answer yes, but if it is 2, answer no."

Occum would be proud. 😉

That answer hit me about midway through calculus 4th period today.

edit: except it has to be phrased as a question.

But, I thought it was more enjoyable with yes, no, and the third choice: the problem has a solution, but it's unsolveable.

Oh, and in my case around the 90th post or so, use a set of ninety 25 digit even numbers. There are 2^90th different combinations (about 1.2*10^27), but the sum has to be less than 10^27. Or, have fun and use a million numbers such that the maximum number of digits in each number, times a million, is a little less than 2^millionth.

As far as computing time, suppose that you could compute sums at 3 GHz - that is, not just one operation, but, in the case of any subset of those 90 numbers, you can find the sum of the numbers in that subset in 1/3billionth of a second. In order to compute all of the possible sums, it would take roughly 30 BILLION years. (not that you'd need to do all of them, if you got a yes answer... with incredible luck, you might even find an answer of 0 in your lifetime) I don't want to break out the calculator and start doing logs in order to deal with a million numbers and how many digits in each one.
 
Originally posted by: b0mbrman
Originally posted by: mobobuff
Originally posted by: Goosemaster
If I square the number, is it less than, equal to, or greater than four.

Winnar.

How is that a yes or no question?

Is the number less than, equal to, or greater than two? Same thing, not yes or no.

I don't think a yes/no question can solve this. I mean, the answer has to eliminate 2 of the possibilities for you to know the number, so if "yes" would eliminate two possibilities, than "no" can only eliminate one possibility, and the other way around.

I don't think there is one yes/no question that'll guaruntee the answer.

The "which word has a lettercount equal to your number, yes or no" and "in the reciprocal of one less than the number equal to 1" are both very clever but assume a third response of "wtf?" to the question.

with a yes/no/wtf? question, it's very possible.
 
The question must have a yes or no answer, which (in a purely logical form) is a binary output, as 'no answer' is not an option. I could phrase the problem in a more general way that illustrates this. For example, "I am thinking of a, b, or c. Ask me one true/false question to determine which I am thinking of." In a pure true/false question, 'no answer' is not a valid output. It certainly is one potential output if the question is asked of humans. However, if I look at it in terms of, say, circuit-level logic, then the voltage will always be 'high' or 'low' - there is no third option. The interesting thought I had was that in software logic, it might be possible to make such a ternary judgment if the program threw an exception rather than simply true/false. However, this would only be possible if one could generate a question where one of the possibilities (a, b, or c) was neither true nor false.
 
Originally posted by: LordMorpheus
Originally posted by: b0mbrman
Originally posted by: mobobuff
Originally posted by: Goosemaster
If I square the number, is it less than, equal to, or greater than four.

Winnar.

How is that a yes or no question?

Is the number less than, equal to, or greater than two? Same thing, not yes or no.

I don't think a yes/no question can solve this. I mean, the answer has to eliminate 2 of the possibilities for you to know the number, so if "yes" would eliminate two possibilities, than "no" can only eliminate one possibility, and the other way around.

I don't think there is one yes/no question that'll guaruntee the answer.

The "which word has a lettercount equal to your number, yes or no" and "in the reciprocal of one less than the number equal to 1" are both very clever but assume a third response of "wtf?" to the question.

with a yes/no/wtf? question, it's very possible.

THE OP Added that yes/no bubcuss really late into the thread.
 
Originally posted by: DrPizza
But, I thought it was more enjoyable with yes, no, and the third choice: the problem has a solution, but it's unsolveable.
Your posts made me remember back to the college courses I took. Reminded me of the 4 color theorem:
http://en.wikipedia.org/wiki/Four_color_theorem

It's something where the five color theorem is a yes, the three color theorem is a no, and the four color theorem is a "most likely yes, but impossible to prove by hand."


There were another theorem that I very vaguely recall that was someone (not even a top mathematician in his time) who jotted down the solution to a complex mathematical problem in the margin of a random printed page, and it's something that so far has been impossible to prove, and also impossible to disprove.
 
This reminds me of something I read in a math book recently. It has to do with Turing Machines and whether math really exists or we made it all up, but the relevant chapter discussed a number that is easily defined, but we have no idea if it is even or odd.

It has to do with PI and whether there is a certain string (100 zero's in a row for example) and whether that occurs on an even or odd digit. Using that you can define another number that is certainly even or odd, we just have no idea which.

Fun stuff.

I agree that given that this is a yes or no question, then the person answering can answer only with "yes" or "no" and no answer is an answer in and of itself. Given that it is unsolvable. But riddles often use these "loopholes" so its probably something stupid like that.


Relevant web comic.
 
If we don't get to use Null answers, then I think we hit a paradox. Imagine I asked the correct question and it goes as follows.

If he responds Yes then he's thinking of A.
If he responds No then he's thinking of B.

Since there's no other response choices available, there's no way to conclude C. So assuming the correct question exists, he can only think of 2 of the 3 numbers and never a third. But since he's allowed to think about 3 of the 3 answers then we have to conclude that the question doesn't exist.



 
Originally posted by: TuxDave
If we don't get to use Null answers, then I think we hit a paradox. Imagine I asked the correct question and it goes as follows.

If he responds Yes then he's thinking of A.
If he responds No then he's thinking of B.

Since there's no other response choices available, there's no way to conclude C. So assuming the correct question exists, he can only think of 2 of the 3 numbers and never a third. But since he's allowed to think about 3 of the 3 answers then we have to conclude that the question doesn't exist.

what is a paradox though? can something that doesn't fit the bounds of our mathematical system be a pardox just because?




 
Originally posted by: Goosemaster
Originally posted by: TuxDave
If we don't get to use Null answers, then I think we hit a paradox. Imagine I asked the correct question and it goes as follows.

If he responds Yes then he's thinking of A.
If he responds No then he's thinking of B.

Since there's no other response choices available, there's no way to conclude C. So assuming the correct question exists, he can only think of 2 of the 3 numbers and never a third. But since he's allowed to think about 3 of the 3 answers then we have to conclude that the question doesn't exist.

what is a paradox though? can something that doesn't fit the bounds of our mathematical system be a pardox just because?

Since he only says Yes or No, you're only going to conclude A or B. And since we assume the question is correct to deduce his number, the person is never allowed to choose C. But since the premise of the question says he is allowed to choose C, that creates the paradox?

I guess we have to be formal with the assumptions and givens:

Given:
A person is allowed to think of A,B or C

Assume:
A question exists AND only results in a Yes/No and no nulls AND will correctly deduce his answer.

Deduce:
A person is not allowed to think of C. (contradicts given so assumption is wrong)

Conclusion:
A question does not exist or the question exists but has more or less than just a Yes/No answers or a question exists but doesn't deduce his answer.



 
umm no, NULL isn't a member of a set, it is the indication of an absense of members for that set.

EDIT: also it seems pretty trivial to me that no question can exist with only 2 answers which can choose between 3 possibilities.
 
Originally posted by: cubby1223
Originally posted by: DrPizza
But, I thought it was more enjoyable with yes, no, and the third choice: the problem has a solution, but it's unsolveable.
Your posts made me remember back to the college courses I took. Reminded me of the 4 color theorem:
http://en.wikipedia.org/wiki/Four_color_theorem

It's something where the five color theorem is a yes, the three color theorem is a no, and the four color theorem is a "most likely yes, but impossible to prove by hand."


There were another theorem that I very vaguely recall that was someone (not even a top mathematician in his time) who jotted down the solution to a complex mathematical problem in the margin of a random printed page, and it's something that so far has been impossible to prove, and also impossible to disprove.
Fermat's last theorem?

First off that's been proven. Second, he was one of the top mathematicians of his time.

Please get your facts straight.
 
Originally posted by: Kyteland
Originally posted by: cubby1223
Originally posted by: DrPizza
But, I thought it was more enjoyable with yes, no, and the third choice: the problem has a solution, but it's unsolveable.
Your posts made me remember back to the college courses I took. Reminded me of the 4 color theorem:
http://en.wikipedia.org/wiki/Four_color_theorem

It's something where the five color theorem is a yes, the three color theorem is a no, and the four color theorem is a "most likely yes, but impossible to prove by hand."


There were another theorem that I very vaguely recall that was someone (not even a top mathematician in his time) who jotted down the solution to a complex mathematical problem in the margin of a random printed page, and it's something that so far has been impossible to prove, and also impossible to disprove.
Fermat's last theorem?

First off that's been proven. Second, he was one of the top mathematicians of his time.

Please get your facts straight.
So which part of "very vaguely recall" did you not understand?

Fermat sounds right, though.
 
Originally posted by: 2Xtreme21
Originally posted by: TuxDave
Originally posted by: gsellis
"If the your number is 3, answer yes, but if it is 2, answer no."
Occum would be proud. 😉
I like it!
Still relies on the "inability to answer" to resolve the 3rd choice.
My version is:

If we assume 1 is "yes", 2 is "no", and 3 is "yes or no", is your number 1, 2 or 3?
 
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