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[MagmaMcFry]

Alright. Here you go. The multiplicative group F₂¹⁸ has a generator x satisfying x¹⁸ = x⁷+1. Using this generator, and the F₂-basis {1,x,x², ..., x¹⁷} (this is always a basis when x is a generator), we get the following graph:

Spoiler (click to show/hide)

Clearly the next activation state is always the current activation state multiplied by x.

[Reelya]

Hey I was wondering if any of you math genius' could help me out understanding some of the terminology in this. I'm really good at coding, don't know much advanced math's speak. Trying to implement quaternion splines according to this paper:

http://qspline.sourceforge.net/qspline.pdf

There are some steps I can get, other's are completely out of my field of knowledge e.g. the line

"Multiplying Equation (3) by q^_-1 we find that ... (equation 5)"

... exactly how is equation 5 derived from equation 3 by multiplying both sides by q^_-1? I have no reference for how this works.

[frostshotgg]

You can treat q like a variable, mostly.

They flipped it across the = sign, to get 1/2qw = qdot, then multiplied by q^-1 which is the same as dividing by q, so you get 1/2w = qdot * q^-1. Then it multiplies by 2 to get rid of the 1/2 and solve for w.

[Helgoland]

And the lambda thing comes from (1), I think.

[TheBiggerFish]

So @MagmaMcFry and @Ipsil, did anything end up happening of that?  I have no idea what you're doing, but I'm intrigued.

[TheBiggerFish]

Aww, darn.

Would've been cool to be watching on the sidelines when a new mathy thing happened.

[MagmaMcFry]

Well, I did manage to turn the Collatz conjecture into a finite-state machine operating on a string in base 3. Not sure what good that does, though, towards proving it.

Well, now you've basically reduced the Collatz conjecture to an instance of the halting problem, so all we gotta do now is solve the halting problem!

[da_nang]

Well, I did manage to turn the Collatz conjecture into a finite-state machine operating on a string in base 3. Not sure what good that does, though, towards proving it.

Well, now you've basically reduced the Collatz conjecture to an instance of the halting problem, so all we gotta do now is solve the halting problem!

And since it's a finite-state machine, that problem's decidable.

[TheBiggerFish]

Well, I did manage to turn the Collatz conjecture into a finite-state machine operating on a string in base 3. Not sure what good that does, though, towards proving it.

Well, now you've basically reduced the Collatz conjecture to an instance of the halting problem, so all we gotta do now is solve the halting problem!

And since it's a finite-state machine, that problem's decidable.

Waaaaaaait.
What does that mean?

[da_nang]

A decidable problem is a problem where, using our current axioms, we can (theoretically) either prove or disprove the problem in a rigorous, consistent manner.

An undecidable problem is one where our current axioms just aren't enough. We could in theory get contradictory results and they'd both be correct!

The halting problem for the general Turing machine is one such undecidable problem.

[MagmaMcFry]

Well, I did manage to turn the Collatz conjecture into a finite-state machine operating on a string in base 3. Not sure what good that does, though, towards proving it.

Well, now you've basically reduced the Collatz conjecture to an instance of the halting problem, so all we gotta do now is solve the halting problem!

And since it's a finite-state machine, that problem's decidable.

Wait no, the algorithm that always halts iff the Collatz conjecture is true is not the finite-state machine itself, but the algorithm of applying that finite-state machine repeatedly to the input number until you reach 1.

[TheBiggerFish]

Well, I did manage to turn the Collatz conjecture into a finite-state machine operating on a string in base 3. Not sure what good that does, though, towards proving it.

Well, now you've basically reduced the Collatz conjecture to an instance of the halting problem, so all we gotta do now is solve the halting problem!

And since it's a finite-state machine, that problem's decidable.

Wait no, the algorithm that always halts iff the Collatz conjecture is true is not the finite-state machine itself, but the algorithm of applying that finite-state machine repeatedly to the input number until you reach 1.

...Bother.

[Reelya]

Well Conway proved that the Collatz problem has no non-trivial cycles < 400 in length and later Lagarias proved that there are no trivial cycles < 275000. My hunch is that this is the likeliest avenue for solving the thing. If you could construct a proof by induction that the minimum non-trivial solution was always of greater length than some value 'n' then you could effectively prove that there are no cycles of finite length. It would need to be a computer-constructed proof that could take 'n' and always construct the proof needed.

IDK whether that would rule out a cycle of infinite length however, but you might prove that as a corollary.

[TheBiggerFish]

How could you possibly have a cycle of infinite length?

Doesn't that mean it comes back around?

[TheBiggerFish]

So it's not a cycle.  As in, it is not cyclical, it just keeps going.

Yes.

Yes.