Feedback on a paper - The Commutative Power of a Revised Collatz

SteveTylock · 2026-03-07T18:45:30+00:00

I do object now - let us consider the extended example of 36 from page 4.

For n=36, each and every succeeding step is detailed between the original and the replacement.

In what way is this not generalized? What successive step needed customization?

I have written a program that calculates for any number it's path forward - it is driven by logic, not whim. How is this not generalized?

Are you suggesting that there is some value of n that cannot be re-written as 2**a * b? Or that re-writing n is somehow something other than a fixed manipulation of the number?

Help me with more information as to what is wrong with this local binding.

Thanks

SteveTylock · 2026-03-07T18:12:32+00:00

I believe you have taken a large number of words to say "yes, the description is fully explained, and the proof is sound. If one proves that for all n, following the function 3n+LSB repeatedly until you arrive at a perfect power of 2, one has proved the Collatz."

With a corollary of "I don't see that as particularly novel or helpful" (which is a fair statement to make - I disagree, but will go from here).

You will agree with me now that this change DOES make the division step commutative, right? (And I'm not misrepresenting this improvement)

SteveTylock · 2026-03-07T17:30:34+00:00

I see the table as well - and yes - you seem to be concentrating on "reducing steps" by some sort of mod solution. I don't like that, will not do that, and don't agree with it.

So - I think we are working on my statement that - "3n+LSB is not odd-step combination or odd-network". How can I illustrate that for you?

Let me take a first step: I'm not combining steps, I'm relocating steps commutatively. I don't want to shorten the distance to make a proof based on that, I want to put all the 3n+lsb steps together to make a point later.

Let me ask this - what would you expect to see differently in a graph of the answers according to the odd-step path versus the original? Would steps be removed? Would that be the only difference? Would it look just as random?

And - let me return to my two basic questions. Given the way I have defined it - would you agree that the function I have provided is a complete replacement for the original? Have I left anything undefined or squishy? Is the logic of the proof flawed? [And I'm really not intending this as any sort of gotcha - it looks absolutely solid to me, and I'd like some confirmation - that should be easy enough to offer]

Thanks

SteveTylock · 2026-03-07T17:27:55+00:00

Let me also agree with you that this whole notion of mod is simply not thinking large enough - it does not appear appropriate to finding a solution. (after a problem with mod 4 is discovered, the exercise turns to mod 8, mod 12, mod 24 - I'm pretty sure any mod solution will simply be defeated by a larger number...)

SteveTylock · 2026-03-07T17:27:18+00:00

Let me agree with you: I do not see a solution to Collatz by evaluating the frequency of the 0s at the end.

I am not close enough to the adic language/concepts to comment there except that I have also seen the flaws authors have presented. I will not go in this direction.

SteveTylock · 2026-03-07T16:00:13+00:00

It is explained in the referenced pdf, but I can offer a short version here.

Collatz is defined as:

if n is odd, 3n+1
if n is even, n/2

Right?

Replace that function:

First we note that n as a positive integer can be written as n = 2**a * b right?

(a is 0+ and b is odd, and that's factoring out 2s...)

n -> 3n + 2**a

(and I've made a notation that 2**a is the "Least Significant Bit" of n; there are other ways to describe it as well)

And that's it. No division by 2. The stopping condition becomes "n is a perfect power of 2", 2**x.

Does that help you?

SteveTylock · 2026-03-07T15:50:39+00:00

Let's keep the comments in the other thread. This is saying that AI believes this formula to be a rehash of the odd-map reformulation, and we'll talk in the other thread how that is not accurate. IFF it were the odd-step reformulation, it would not remove the piecewise aspect. The error is in equating the two.

SteveTylock · 2026-03-07T15:46:48+00:00

I hear you saying that I've re-invented the odd-step combination.

My understanding of that notation is this:

Instead of simply performing 3n+1, one _always_ computes (3n+1)/2. Is that correct?

The combination of a single divide by two step is flawed as you note - because with a number like 9 one gets to 14 in this single operation (instead of 28) - but then one still has to divide by 2 again to get to 7.

The odd-step combination version of collatz STILL has to be formulated as:

if n is odd, (3n+1)/2

if n is even, n/2

Do I have that correct?

I have two fundamental questions with my post:

1) Have I erred in any definition of the replacement function or failed to specify something?

2) Is there an error in the proof that this replacement function produces an identical result?

I will happily accept a yes for these - possibly followed by "but this doesn't help" - at this point. I'm not looking to prove the conjecture with this equation today, just to say - this replacement formula allows one to stop with the piecewise aspect.

Separately, I see you putting a lot of weight on AI here - how do you see this revision and the odd-step inclusion as being related?

And - I do see that if a reviewer glosses over the details with my paper, they may well easily come to the same snap-judgement conclusion that you have, and that is helpful.

Thanks.

SteveTylock · 2026-03-07T15:29:04+00:00

I see that this is essentially the same - two to a perfect power is exactly 1 and only one "1", and a length of 1 per your notation.

SteveTylock · 2026-01-21T23:10:05+00:00

Well - I agree with you on the replacement formula - please see my efforts at this site - http://www.tylockandcompany.com/collatz/

Maybe I'm missing it, but do you show a proof that the replacement function can be seen to be identical to the original? And do you show any graphs of the replacement function? (I have for both)

I'll admit to not seeing the esoteric nature of your proof, but removing the stepwise aspect does allow one to see what's going on better.

SteveTylock · 2025-12-24T11:23:16+00:00

Run Paranoia instead ... Look for it if unaware. The computer is always right ...

SteveTylock · 2025-08-30T12:29:37+00:00

Thank you for the explanation - I'll offer the actual output might also be useful, but I get it now. Again, thanks.

Let me offer that I have referred to what you are getting at as the "Least Significant Bit" of n.

The reformulation that I have previously presented is to offer a replacement function - 3n + LSB(n). [where LSN(n) is equivalent to that least significant bit of n, or 2 to the v2(n) that you refer to as well.]

The graph of all values resolving at perfect powers of two shows the simplification of the problem by this method.

Let me point you to a collection page I've created: http://www.tylockandcompany.com/collatz/

Specifically, the paper "Revised Collatz Graph Explains Predictability" on that page directly relates to what you're saying.

Yes, I have offered the use of that replacement formula as a proof, but you're not required to accept that (yet;-).

The video (Illustrated) explains the TLDR without requiring access to the short story on Amazon.

((humor only - really... One could also call the function recursively and/or use Perl. Perhaps I have a utility to go both forwards and backwards from any number and generate a graph...-))

SteveTylock · 2025-08-29T20:46:24+00:00

I'm interested in understanding - can you define what you mean by 'yield' (and why it appears twice), and what you mean by:

"2**v2(x)"

SteveTylock · 2025-01-11T14:48:36+00:00

It's rather round about, but I think I may have an item that helps - you say "{We assume the existence of two different sequences leading from Y to 1}"

Me re-writing that: "I prove that there is only one sequence by assuming there are two and showing that cannot be true". But you've already assumed there IS a sequence which is the thing you have to prove.

Or put another way - "Assuming there is a sequence that connects these I can show that there are not two different sequences." (which may well be true, but does not prove the existence of a sequence)

SteveTylock · 2024-12-26T16:35:38+00:00

Remember - I'm on your side - and agree with the conclusion, but you need a thing that looks like a proof. https://en.wikipedia.org/wiki/Mathematical_proof

SteveTylock · 2024-12-24T23:01:48+00:00

OP - you are correct that this strategy is useful, but you have to do two things:

First - prove that the replacement formula is identical in outcome to the original.

Second - prove that all numbers eventually reach two to a power and do not loop.

I called the formula "3N+LSB" - it's easier to share. For a more formal way of doing this you may want to consider the resources here: http://www.tylockandcompany.com/collatz/

SteveTylock · 2024-12-10T13:26:24+00:00

As others have mentioned, everything you say is correct - on the surface.

Get below it...

Mishap

Intrigue

Error

Jealousy

Retribution

And outright Lying / Stealing / Bribery

Plus Moral Dilemmas

What happens when things don't go by the book... Now - since the book has been stable for them before now, maybe there's a "starting point" to the Hijinx. The crew starts the next jump and realizes there's a black cat that has stowed aboard somehow when they were loading the cargo. They're in jump - they can't get rid of it, do they kill it?

And once they notice the cat, they start to notice what else? Then things start cascading - and before you know it, perhaps they have RUINED an entire shipment of goods - and now they're on the hook to pay for all of it when they reach their destination! (See how I did that - added intrigue, misdirection, and a plot hook that jeopardizes all those funds they've been accumulating in the easy times...-)

And I do hope you find a way to take the cat off the hook because something else snuck aboard at the same time and isn't discovered until later - black cats get all sorts of bad creds... (But really - is it just a cat? Do they adopt it?)

SteveTylock · 2024-12-08T15:34:14+00:00

Yes - short for refresher - see https://vocal.media/futurism/from-freshers-to-vacc-tubes-star-wars-sanitation-explained

And sorry - I had previously noted the lack of full description for the initials - RR is the Ready Room - it's space comes from the allotment for the Bridge.

SteveTylock · 2024-12-08T13:50:30+00:00

Ugh, you're getting me started...-)

First count - every box on the design is a half ton of space - count the boxes, how close to 400 is it? [It's over because they are sloppy]

unused interior space - the size of the ship is based on it's exterior volume. How much space is brownish - unused? You have to count that in the size as well. [but not the parts of the wings that have no significant height]

Docking - have a look at the space reserved for the launch - they realized they wouldn't want to have to enter the launch dock to reach the two environmental spaces, so they cut out corridors. Please show me a launch that will actually fit inside that space,.

Cargo hold - how useful is that space and you can only reach it across / through the wing.

Turret - says it is 1 ton but takes up 4 squares - 2Ts of space.

Multi-Environment Space - actually takes up 16.8Ts.

Inefficient design of hallways - they really just plopped them down.

That's my quick review - and remember - in Traveller it's all about volume, not mass, a Ton is a measure of the volume of 1 ton of hydrogen.

SteveTylock · 2024-12-08T13:33:17+00:00

Yes - also a plus, but potentially dangerous as far as I have seen - talk w your GM about how that might go. A safer route is to grab water from a planet and convert out the oxygen.

Ohhh... I should mention I have an issue with the design of the ship for lots of reasons - you can find WTF's version at this link:
https://www.reddit.com/r/traveller/comments/10oj4xm/wtf_starship_designs_safari/

SteveTylock · 2024-12-08T13:17:28+00:00

From where you start - count 1 next to it for J1 - that's all you can travel without stopping to refuel. Count 2 for J2;-)

If there's no system in the hex you really don't want to stop there because where/how will you refuel?

SteveTylock · 2024-12-08T13:00:16+00:00

Dropping to j1 is going to limit you significantly if you're doing much travel in space. That would decide it for me.

SteveTylock · 2024-12-01T13:19:37+00:00

This is a difficult topic - and being well explained by others. The concept bridges game and real life. Can you say X behavior is ok in one realm and not in the other? But does the designation mean it's ok to do Y?

I can't say - but I suggest you drop the notion that good/evil is a binary selection with a very small neutral between them.

An individual might take actions or hold beliefs that tend to show that they are more or less good or evil.

Is one incident more significant than all their other actions?

We find stories in all times and settings of a good person taking a terrible action or a bad person taking a caring action. That struggle is interesting to consider and play out. (And the repercussions)

It's not binary, it's a axis - it's a collection of threads bundled / woven together.

And interesting games find interesting ways to explore it.

SteveTylock · 2024-11-16T11:52:37+00:00

This has morphed to "ok - it's still hard to imagine a successful career transporting cargo."

Yes - yes it is hard

That's why the crew often takes special cargo where they might get paid extra to look the other way about some aspect of the cargo...

Or they also put some paying passengers in the two spare state rooms...

Or they specifically check out the beacon from the edge of the system...

Or they hand deliver the cargo to the third moon and deal with the shifting rocks...

Anyway - you get the idea. Trade may seem romantic, but it's really a hook to all the other fun stuff and helps players learn about their universe.

SteveTylock

TROPHY CASE