Yo bet, new math vid

1:40 How is that curly brace } so perfect??

@carbrickscity

9 ай бұрын

lol

Turing machines can only compute computable numbers, so you certainly can't encode the Rayo function in a Turing machine. BB itself is only uncomputable because we're diagonalizing over Turing machines and you can't predict which ones produce a number at all (halting problem).

@007Rincewind

9 ай бұрын

A SuperBusyBeaver would probably solve this that can evaluate if a BB halts or not, (but not itself,). Since it only need to evalate if FOST strings terminate and not itself. A SuperBusyBeaver if probably impossible to build in real life. but would be stronger than Rayo.

@MarkBettner-fi2ec

9 ай бұрын

Don't know how it would be possible to decode non computable functions that don't even fit on the FGH.

The number of digits of n is just log(n)+1. For a massive power tower like 3↑↑↑3, a singe log is basically neglectable

Thanks for another awesome vid!

@carbrickscity

9 ай бұрын

Thanks.

My request is successful! Thank you for the video! 真係多謝你啊！

Do a video on BMS aka Bashicu Matrix System.

@AXIOMATICLIMIT

9 ай бұрын

good idea

Very good video

You can crate BB in about a 10^10 symbols whit fost

I think hes argument is somewhat solid. howevere since a BB would need to loop over all possible Rayo strings to find the strongest one a BB would end up in a non halting state and therefore fail to halt and give a number. This raises the intresting question that the Super Busy Beaver function (having a halting oracle) would probably surpase Rayo number since it can aviod running non halting Rayo Strings. Why is SBB suddenly stronger than Rayo(n), since a lot of research saying the opposite, I think has to do with the definition of infinte tape. At a certain N not that great Tapelenght becomes a substitute for Symbols. By having a primenumber generator you can for example segment the memory into infinite number of infinte long tapes etc (think of hilbert hotel for method and indexing). A quick seeing this is that Loaders number is the largest number that can fit in 512 bytes(here the limitiation is the 512 bytes), Rayo is the largest number that can fit in 10^100 bytes. But SBB is the larget number that can fit in (N) bytes where N is going to Infinity.

@OnlySkeep

9 ай бұрын

But sorry if you say that it is the largest number you can write within N Bytes, then how can there be the largest finite number if N goes to infinity

@carbrickscity

9 ай бұрын

Well if you are comparing BB(n) to Rayo(10^100) then sure eventually BB(n) would be bigger. But that's not a fair comparison, you have to compare BB(n) to Rayo(n), in which case Rayo(n) should beat BB(n).

@007Rincewind

9 ай бұрын

The statemachine is not infinite, and the starting state is all Zeros, it will be impossible to reach the end of the tape without any infinte loop and it is necessary for a loop to halt for a final value to be produced. @@OnlySkeep

@OnlySkeep

9 ай бұрын

@@007Rincewind Oh, thx!

yo thanks for clearing up the misconception on my last comment and sorry for being wrong

@carbrickscity

9 ай бұрын

It's not just you, but many people.

damm bruh 10^3 is indeed 4digits but 3^3 isn't 4 digits, is 2, you need to convert tower of 3s to 10s

So which do you think is bigger: Rayo's Number or BusyBeaver(TREE(3)) ?

@carbrickscity

9 ай бұрын

From Googology wiki, Rayo.

@TheGamingG810

7 ай бұрын

What's bigger? Rayo or BB(BB(BB(...(TREE(3))...) (Busy Beaver is iterated BB(TREE(3)) times)

@carbrickscity

7 ай бұрын

Rayo. Anything BB doesn't complete with Rayo at all.

@garvsharma4471

4 ай бұрын

Comparing Rayo(n) and BB(n), For any n greater than 7339 Rayo will always be bigger than BB. That's just saying if we put any value in both functions, considering that value is greater than 7339 rayo will always win. for any value smaller than that BB is greater than Rayo. here's the link to the full video - kzread.info/dash/bejne/X36bsLGFkrfHkrQ.html

That’s enough pentation for a day

@carbrickscity

9 ай бұрын

Never get enough lol

if a function is uncomputable, it cannot be computed by a turing machine

hi

Yay a new video pls let there be a new googology video be quick my English might be wrong

Funny....

@carbrickscity

9 ай бұрын

Don't know what you mean.

People think 3^^^3 has 3.6trillion digits because they view it as (3^3)^3)...^3 with 7.6trillion 3's as opposed to 3^(3^(...(3^3) with 7.6trillion 3's.

@carbrickscity

9 ай бұрын

Yes people may think it works from bottom to top but I don't think that's the main reason people think it has 3.6 trillion digits. For some reason they were thinking 3^^^3 = 3^^4. They just think adding another arrow means adding another 3 to the tower, which is a vast underestimation of how arrow works.

@nickronca1562

9 ай бұрын

@@carbrickscity I realize since reading your comment that going bottom to top is equal to 3^^5 not 3^^4

@carbrickscity

9 ай бұрын

I don't know whether going from bottom to top is 3^^5. But power tower works from top to bottom.

@nickronca1562

9 ай бұрын

@@carbrickscity It is because (3^3)^3)...^3 with 7.6trillion 3's = 3^(3^7.6trillion) using an extension of the (a^b)^c = a^(b*c) formula. Technically speaking, it will be 3^(3^(3^27 - 1)), so the whole thing is cube root of 3^^5 since the exponent is reduced a third, but that's basically the same thing.

@carbrickscity

9 ай бұрын

I see what you mean. I suppose it's 3^(3^7.6 trillion), which is 3^^5. I guess I learned something.

10↑↑↑↑↑↑↑↑↑↑10 and 10↑↑↑↑↑↑↑↑↑↑↑↑10÷10^^^10 are the same?

@carbrickscity

7 ай бұрын

No of course not.

@mattmarshall4142

4 ай бұрын

10 with 11 arrows 10 divided by 10 with 10 arrows 10 would almost be indistinguishable from 10 with 11 arrows 10. The hyper operation sequence is unbelievably powerful.

uwutation

@carbrickscity

9 ай бұрын

Don't know what you mean.