[AuRomin]

The way that I would do it is:

1 -> 4
2 -> 3
3 -> 1
4 -> 5
5 -> 6
etc.

There are infinitely many solutions using the above archetype.

This hits all (natural) numbers without repeating, all inputs have unique solutions, and every input has one output (i.e., no abs. inequalities)

BTW, isn't rule 1 redundant because of rule 2?

[Drlee]

I don't do math. I do statistics. Math just fucks them up. It up. I mean maths just fuck it (the statistic) up. Them up.

Fuck this.

[Saeko]

The way that I would do it is:

1 -> 4
2 -> 3
3 -> 1
4 -> 5
5 -> 6
etc.

There are infinitely many solutions using the above archetype.

This hits all (natural) numbers without repeating, all inputs have unique solutions, and every input has one output (i.e., no abs. inequalities)

I assume that by "etc." you mean

6 -> 7
7 -> 8
...?

But if so, then where does 2 appear in your list of answers?

Because if 2 appears as an answer to a problem beyond 5, say 13 for example, you will have the cycle 13 -> 2 -> 3 -> 1 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 10 -> 11 -> 12 -> 13.

BTW, isn't rule 1 redundant because of rule 2?

Yes, I think I originally meant to say that every number appears at least once in the list of answers, which, when combined with rule 1 would imply that every number appears exactly once. But then I just said that last part to avoid confusion.

[AuRomin]

Thank you, Saeko. Right before i wrote the post on how I would answer it, I had halfway written a post on how it was impossible, then I thought of that. I guess I shouldn't rethink posts on a whim.

Now I have to write out why it is not possible. (in my thinking)

for whatever number is started at, it must: a) not compute to itself (as to not make a loop), and b) be an output eventually(for rule 2).

This naturally means that there is a loop at some point to reach back to the starting input(which leads to an absence of another number or repeats), or that the starting input is never an output (which is against the rules). Of course, you could do an odd (I like to imagine it as a parabola) thing like going through the evens to 2, then do the odds, but that has no valid starting input that fulfills the rules.

Theoretically, you might be able to say that the pattern is f(x) = x+1, and this goes on until you reach a number for which infinity is a factor, then change the pattern to f(x) = x/infinity (to get one and not repeat).

This only works if you like to think it does, though. This is sort of a grey area in math. Of course, it would never reasonably work (you don't ever reach 'all numbers', much less divide by them), and division of this kind isn't supported if only natural numbers are allowed.

[Saeko]

^ Mmm... nope. This puzzle definitely has a solution. Infinitely many, in fact.

[Saeko]

Assuming all natural numbers as the set of numbers in which every number appears exactly once, you would have to start at infinity in order to reach all even numbers. Infinity is not a number and therefore cannot be used in your function. I don't see how this is a valid answer.

Please inform me of how or if I'm wrong.

Sorry I didn't see this post.

My solution is the function f(x) = x + 2 for x odd, f(x) = x - 2 for x even and x > 2, f(x) = 1 if x = 2. This function never takes infinity as an argument.

[AuRomin]

How would you do it? Your answer in the opening post cannot be correct, so what is the answer? I will admit that I am not very good with proofs for a negative, but I believe that I have done so to the fullest necessary extent in my previous post.

EDIT: Disregard this. Yet to read the post you did while i wrote this, regarding my previous post.

[AuRomin]

What I meant is that, wherever you start, you never get to the even numbers above it. The only way to hit all evens as well as all odds is to start at the "last" even number, hence the necessity of using infinity as an argument. Maybe I am missing something, but I can't see how you get to 12 if start at 10.

I hope this clarifies what I was saying.

[Saeko]

What I meant is that, wherever you start, you never get to the even numbers above it. The only way to hit all evens as well as all odds is to start at the "last" even number, hence the necessity of using infinity as an argument. Maybe I am missing something, but I can't see how you get to 12 if start at 10.

I hope this clarifies what I was saying.

Read the rules very carefully. The required solution is a function, and not a dynamical system. Unlike dynamical systems, functions have no "starting points". It is not at all necessary to "hit" every natural number along the way from some starting point, it is only necessary to avoid hitting the same natural number more than once.

[AuRomin]

Well, that's where I went wrong. But, with it only needing to be a function, why not f(x) = x?

[Saeko]

Well, that's where I went wrong. But, with it only needing to be a function, why not f(x) = x?

Infinitely many loops.

[AuRomin]

I have to disagree with this one (possibly on the premise of my ignorance).

If you take f(x)={for x%2 = 0 and x>2, x-2; for x%2 != 0, x+2; for x = 2, x-1}, you need to start at a point to get the original x. If there is no starting point, then worrying about loops is irrelevant, but as you have said, loops are an issue because you do take f(g) = f(x) next, then f(h) = f(g), and repeat it. This indicates (very strongly) a starting point. The starting x to be evaluated must be a realized number (i.e., not simply the largest even number (which doesn't exist)). My point is that any number you start with is too low because it exists.

[Saeko]

I have to disagree with this one (possibly on the premise of my ignorance).

If you take f(x)={for x%2 = 0 and x>2, x-2; for x%2 != 0, x+2; for x = 2, x-1}, you need to start at a point to get the original x. If there is no starting point, then worrying about loops is irrelevant, but as you have said, loops are an issue because you do take f(g) = f(x) next, then f(h) = f(g), and repeat it. This indicates (very strongly) a starting point. The starting x to be evaluated must be a realized number (i.e., not simply the largest even number (which doesn't exist)). My point is that any number you start with is too low because it exists.

You don't need to "get the original x". It is only necessary that there exist a y such that f(y) = x.

[AuRomin]

Alright. I finally concede defeat.

Is there any other function archetype that would work?

[Saeko]

Alright. I finally concede defeat.

Is there any other function archetype that would work?

What are you asking me for? Go find them.

[AuRomin]

Just wondering if you were hiding some from us.

My first guess an inverse of your archetype. A negative parabola, if evens are negative and odds are positive, if you can visualize my crude ramblings. I'm thinking this might not work as it would have an undefined vertex.

EDIT: I'm correct. This would't work.

[Saeko]

Just wondering if you were hiding some from us.

My first guess an inverse of your archetype. A negative parabola, if evens are negative and odds are positive, if you can visualize my crude ramblings. I'm thinking this might not work as it would have an undefined vertex.

It's easy to see how to get solutions more interesting than just inverses and compositions of my original. Simply:

1) divide up the naturals into two or more partitions. For example, prime numbers with 1 and composite numbers.

2) Define a basepoint for each set. In this example, 1 for the first and 4 for the second.

3) Define a successor or predecessor relation for each set. For the first set, S(x) = smallest prime greater than x, and for the second set P(x) = greatest composite smaller than x

4) Link the basepoints. For this example, P(4) = 1.

[AuRomin]

I wasn't trying to simply mutate your function; I wanted a very different function style, and what you propose seems very similar to your original function.

I do believe that the partitioning is crucial, though.

[AuRomin]

Using the logic in one of the previous posts in which i tried to disprove you, you could make a proof that your archetype of function is the only one. I suppose any amount of axes is possible for this problem as well.