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Author Topic: Self-referential quiz  (Read 2429 times)

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infangthief

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Self-referential quiz
« on: April 08, 2019, 05:09:44 am »

Here is a quiz a devised a couple of years ago.
I've seen enough references on this site to logic and paradoxes that I think this might go down well with some of you.

There are six questions. The tricky thing is that the questions are about your answers (or, in the case of question 6, about all possible 6-tuples of answers).

There may, or may not, be a way to answer all six questions correctly. The aim is to find a 6-tuple which maximises the number of correct answers.

Each answer must be a decimal integer (in simplest form).

Now for the questions:

1. How many times does the digit ‘1’ occur in your answers?

2. Which question number has your (strictly) lowest answer?

3. What is double the median of your answers?

4. What is the range of your answers?

5. What is the result of subtracting your answer to Q1 from your answer to Q2?

6. How many ways are there of answering all six questions correctly?

Good luck.
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faust

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Re: Self-referential quiz
« Reply #1 on: April 08, 2019, 06:27:50 am »

There is a way to answer at least   6  of the questions correctly.

Suppose there is not. Then the correct answer to 6 is 0. Consider the answer set (2, 6, 10, 10, 4, 0). The answers to the questions 1-5 are correct. But with our assumption, this means that that this set forms a way of answering all questions correctly! Contradiction to our assumption.

I guess the difficult part is to figure out how many sets of correct answers there are.


Alternatively:

The answer set (2,6,10,9,4,1) answers 1-5 correctly. If no other solutions exist, then this is a solution. Otherwise, well, other solutions exist.
« Last Edit: April 08, 2019, 06:41:15 am by faust »
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faust

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Re: Self-referential quiz
« Reply #2 on: April 08, 2019, 06:33:55 am »

I assume the formulation of 2 is supposed to imply that there is no correct answer if 2 answers are tied for lowest?
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infangthief

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Re: Self-referential quiz
« Reply #3 on: April 08, 2019, 06:45:34 am »

I assume the formulation of 2 is supposed to imply that there is no correct answer if 2 answers are tied for lowest?
Yes. If your 6-tuple of answers has two or more answers tied for lowest then your answer to question 2 is incorrect, whatever it is.
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faust

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Re: Self-referential quiz
« Reply #4 on: April 08, 2019, 07:09:33 am »

The best limiting factor are possible answers to 2.

The answer to 2 can never be 1: If it were, then the answer to 1 would have to be at least 1, i.e. not lower than the answer to 2.
The answer to 2 can never be 2: The answer to 5 is always smaller or equal than the answer to 2.
The answer to 2 can never be 3: Since that would imply that the median is smaller than every element of the list.
The answer to 2 can never be 4: If it were, then the range would be less than 4. But it also has to be at least 2 (if it's 1, then between answers 2 and 4 alone we already have a range of 3). So every other answer is at least 3. That means that there are at least 3 occurrences of the digit 1. But these could only occur in double-digit numbers, contradiction to the range being less than 4.

Thus the only viable answers for 2 are 5 or 6.
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ghostofmars

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Re: Self-referential quiz
« Reply #5 on: April 08, 2019, 07:40:47 am »

How do you deal with infinity in this puzzle?

Suppose there are N possible solutions, where the numbers N and 2N + 5 contain at most 5 digits '1' combined. Now I can answer
5          (1. How many times does the digit ‘1’ occur in your answers?)
5          (2. Which question number has your (strictly) lowest answer?)
2N + 5 (3. What is double the median of your answers?)
Any number larger than 2N+5 that leads to the correct amount of '1's (4. What is the range of your answers?)
0          (5. What is the result of subtracting your answer to Q1 from your answer to Q2?)
N          (6. How many ways are there of answering all six questions correctly?)
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ghostofmars

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Re: Self-referential quiz
« Reply #6 on: April 08, 2019, 07:42:25 am »

It should be N+5 not 2N + 5.
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infangthief

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Re: Self-referential quiz
« Reply #7 on: April 08, 2019, 07:59:54 am »

How do you deal with infinity in this puzzle?
Each answer must be a decimal integer, hence must be finite.
I think you're asking something a bit more than that, but I don't think I want to answer more than that at the moment!
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faust

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Re: Self-referential quiz
« Reply #8 on: April 08, 2019, 08:29:15 am »

How do you deal with infinity in this puzzle?

Suppose there are N possible solutions, where the numbers N and 2N + 5 contain at most 5 digits '1' combined. Now I can answer
5          (1. How many times does the digit ‘1’ occur in your answers?)
5          (2. Which question number has your (strictly) lowest answer?)
N + 5 (3. What is double the median of your answers?)
Any number larger than N+5 that leads to the correct amount of '1's (4. What is the range of your answers?)
0          (5. What is the result of subtracting your answer to Q1 from your answer to Q2?)
N          (6. How many ways are there of answering all six questions correctly?)


I think this implies that there is no correct answer to 6. (You have to walk through the case where N and N+5 contain more than 5 1s, but it should be possible to get a contradiction there as well).
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ghostofmars

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Re: Self-referential quiz
« Reply #9 on: April 09, 2019, 02:22:17 am »

How do you deal with infinity in this puzzle?
Each answer must be a decimal integer, hence must be finite.
I think you're asking something a bit more than that, but I don't think I want to answer more than that at the moment!
I guess the question is: If a certain type of answers leads to infinitely many solutions, should we discard this type of solution because they would lead to question 6 being wrong?
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infangthief

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Re: Self-referential quiz
« Reply #10 on: April 09, 2019, 08:40:50 am »

It is probably time to come clean and say that I don't believe there is much more to this puzzle than what has already been said by faust and ghostofmars.

In a couple of days I'll post my own thoughts on the puzzle. And I'll be interested and open to your interpretations too.
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faust

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Re: Self-referential quiz
« Reply #11 on: April 09, 2019, 08:51:45 am »

It is probably time to come clean and say that I don't believe there is much more to this puzzle than what has already been said by faust and ghostofmars.

In a couple of days I'll post my own thoughts on the puzzle. And I'll be interested and open to your interpretations too.
It is mainly a more intricate version of the barber paradox, right? The possible solutions do not form a set under ZFC, therefore we cannot assign a cardinality.
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MiX

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Re: Self-referential quiz
« Reply #12 on: April 09, 2019, 09:03:45 am »

How do you deal with infinity in this puzzle?

Suppose there are N possible solutions, where the numbers N and 2N + 5 contain at most 5 digits '1' combined. Now I can answer
5          (1. How many times does the digit ‘1’ occur in your answers?)
5          (2. Which question number has your (strictly) lowest answer?)
2N + 5 (3. What is double the median of your answers?)
Any number larger than 2N+5 that leads to the correct amount of '1's (4. What is the range of your answers?)
0          (5. What is the result of subtracting your answer to Q1 from your answer to Q2?)
N          (6. How many ways are there of answering all six questions correctly?)


If you suppose there's 0 correct answers, then that answer isn't correct, because you don't answer 2: so the answer (5, 5, 10, 1111, 0, 0) seems to be the best we can take, since if N's more than 0 then your reasoning suggets there's more than N correct answers. Am I right? Just doing it intuitively.
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ghostofmars

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Re: Self-referential quiz
« Reply #13 on: April 09, 2019, 10:46:22 am »

If you suppose there's 0 correct answers, then that answer isn't correct, because you don't answer 2: so the answer (5, 5, 10, 1111, 0, 0) seems to be the best we can take, since if N's more than 0 then your reasoning suggets there's more than N correct answers. Am I right? Just doing it intuitively.
Yeah, you need to modify the answers a bit once N < 5, because the median is not formed by N and 5 anymore. But the tuple (5, 6, 11, *, 1, 0) would also work. In general, because the answer to question 4 is directly related to itself if there is at least one 0 in any of the other questions, you can vary that number to produce infinitely many answers.

I think you would still get a paradox, if you modified 4 to ask only about the range of the other questions, but I'm not sure.
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infangthief

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Re: Self-referential quiz
« Reply #14 on: April 11, 2019, 04:28:32 am »

It is mainly a more intricate version of the barber paradox, right? The possible solutions do not form a set under ZFC, therefore we cannot assign a cardinality.
Pretty much, yes. I am not familiar enough with ZF to say it myself, but that sounds plausible.
Deciding whether the answer to question 6 is zero or non-zero ends up being equivalent to answering "Is the answer to this question 'no'?" with a 'yes' or a 'no'.

When I devised the questions, my first aim was to find a set of questions 1-5 such that the correct answer to question 6 would be '0', unparadoxically.
I tried out this set of questions, and found it didn't meet that aim; '0' cannot be correct without leading to a paradox.
Then I found the 'solution' (2,6,10,9,4,1) and thought the process of finding it was interesting enough to set this quiz to my work colleagues. Anyway, life is only so long, and I didn't pursue my first aim further. So long as there weren't any other 'solutions' with a '1' at the end, then everything would be ok. Basically everything that faust got in his first reply.
It was a few days later, when I set out to prove the solution, that I came across 6-tuples similar to the ones ghostofmars found. (5,5,10,X,0,1) was the critical one which disproved my first solution attempt, and then we quickly see that nothing else works either.
So we're in paradox land.

However, in spite of the paradox, I think the stated aim is still valid:
The aim is to find a 6-tuple which maximises the number of correct answers.
The number of correct answers is maximised by any 6-tuple which answers Q1-Q5 correctly. I think.

Some ideas for developing this further if I (or anyone else) gets round to it:
- If Q4 is changed to exclude its own answer from the range, then I think (2,6,10,9,4,1) becomes valid. But that seems a clumsy thing to do in a quiz which is meant to be self-referential.
- With different Q1-Q5, there is probably an interesting quiz which has '0' as the correct answer to Q6, as per my original aim.
- With different Q1-Q5, it may be possible to end up with potential solutions (a): (a1,a2,a3,a4,a5,1); (b): (b1,b2,b3,b4,b5,2) and (c): (c1,c2,c3,c4,c5,2). In that case either (a) is correct, or (b) and (c) are correct. That seems similar to deciding how to answer "Is the answer to this question 'yes'?".
- Just leave off questions like Q6 and have fun with other sets of questions which refer to the answers.
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