Thread for NonCA Academic Questions

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Re: Thread for NonCA Academic Questions
Edit: VECTORPARTY is equivalent to TREE but restricted so every node has at most one child, so its maximum possible FGH level is the SVO (however, it is most likely sube0)
 testitemqlstudop
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Re: Thread for NonCA Academic Questions
No, cubeparty is finite.fluffykitty wrote:All versions of MATRIXPARTY(3) and CUBEPARTY(3) are infinite. The sequence isfollowed by extensions ofCode: Select all
2 1 1 1 1
(Element (x,y) is 1 iff x=y=0, x=y=max, or abs(xy)=1)Code: Select all
1 1 0 0 0 1 0 1 0 0 0 1 0 1 0 0 0 1 0 1 0 0 0 1 1
I said you can take the cube, CUT it, TRANSLATE it, and ROTATE it, and if after some cutting/rotating/translating one piece (one cut piece) matches a previous cube it's disallowed.
 Moosey
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Re: Thread for NonCA Academic Questions
I’m afraid that doesn’t change anything. It still is infinite, unless you can cut out of the middle, which you don’t allow.testitemqlstudop wrote:No, cubeparty is finite.fluffykitty wrote:All versions of MATRIXPARTY(3) and CUBEPARTY(3) are infinite. The sequence isfollowed by extensions ofCode: Select all
2 1 1 1 1
(Element (x,y) is 1 iff x=y=0, x=y=max, or abs(xy)=1)Code: Select all
1 1 0 0 0 1 0 1 0 0 0 1 0 1 0 0 0 1 0 1 0 0 0 1 1
I said you can take the cube, CUT it, TRANSLATE it, and ROTATE it, and if after some cutting/rotating/translating one piece (one cut piece) matches a previous cube it's disallowed.
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
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 testitemqlstudop
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Re: Thread for NonCA Academic Questions
 Moosey
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Re: Thread for NonCA Academic Questions
I mean remove the middle to have the outside, somehow turningtestitemqlstudop wrote:Yes you can cut out the middle through this process: cut front side, cut behind side, cut top, cut bottom, cut left, cut right, and then you have the middle.
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 a b c 
 d e f 
 g h i 
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 a c 
 g i 
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
Nanho walåt derwo esaato?
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Re: Thread for NonCA Academic Questions
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x = 20, y = 5, rule = //3
2B3A2.2BA.A4.2B2A$BAB2A2.BAB.A4.BABA$ABABA2.ABA.A4.AB2A$2ABAB11.3AB$
3A2B2.3A.B!
 A for awesome
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Re: Thread for NonCA Academic Questions
x₁=ηx
V*_η=c²√(Λη)
K=(Λu²)/2
Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt)
$$x_1=\eta x$$
$$V^*_\eta=c^2\sqrt{\Lambda\eta}$$
$$K=\frac{\Lambda u^2}2$$
$$P_a=1\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$
 Moosey
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Re: Thread for NonCA Academic Questions
I get that {x 1 anything} = x
And {x y} = {x y [any amount of 1s] } = x^y
But that does the other rule mean?
I’m also guessing that {anything 1} = {[that same anything]}, like on BAN (which I need an explanation on too)
EDIT:
After reading the introduction to BEAF article on the Googology Wiki, I have a less shaky idea of it (though it would be more intuitive once I read it without skimming or going off to find that ζ_0 is (it’s the first c such that c = ε_c))
Why do I use nested parentheses?!?
Also, if you want to have a more intuitive grasp on some of those ordinal things (Γ_0 in a nutshell! It’s the first c such that c=Φ_c(0), or, according to the googology wiki, {ω,ω,1,2} in ordinal BEAF), I highly recommend reading this series of blog posts that I found on the web by John Carlos Baez
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
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Re: Thread for NonCA Academic Questions
x₁=ηx
V*_η=c²√(Λη)
K=(Λu²)/2
Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt)
$$x_1=\eta x$$
$$V^*_\eta=c^2\sqrt{\Lambda\eta}$$
$$K=\frac{\Lambda u^2}2$$
$$P_a=1\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$

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Re: Thread for NonCA Academic Questions
 Moosey
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Re: Thread for NonCA Academic Questions
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
Nanho walåt derwo esaato?
Code: Select all
x = 4, y = 5, rule = B34e5e67c/S23ce45
4o2$b2o$o2bo$b2o!
[[ TRACK 0 7/325 ZOOM 16 THEME 0 ]]

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 A for awesome
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Re: Thread for NonCA Academic Questions
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3
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12
 That the second term in the sequence is not a minor of graphs containing isolated islands each consisting of a single label, but for which different islands can have different labels (as in the following example): Hence the sequence that would have generated SSCG(3) now generates a far larger number, although probably not approaching the order of SSCG(4) or even TREE(3).
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22222 111 2 2 2 1 1
And,  That taking the second term to be can actually lead to very long sequences consisting of vanillaSSCGtype graphs with a single substituted 2labeled node, similarly to the following:
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22
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3
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22
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2 2 2
, followed by the optimal SSCG(5) sequence with one arbitrary node in each graph labeled as 2 instead of 1 (for instance). This gives a phenomenally weak lower bound for LSSCG(3) as SSCG(SSCG(5)+4)+SSCG(5)+4. (Obviously the singlysubstituted version of SSCG(5) will almost certainly be much larger than the standard SSCG(5).)Code: Select all
2 2
Eliminating theterm and continuing with a doublysubstituted sequence yields a much stronger lower bound (possibly an actual approximate value) for LSSCG(3) on the order of SSCG(SSCG_s(SSCG_s(4, 2),1)), with SSCG_s(M, N) referring to the exactlyNtimessubstituted version of SSCG(M) with at most one substitution per island, but even determining any kind of bound on SSCG_s(4, 2) or whether or not it is greater than SSCG(4) is too much for my ability to visualize.Code: Select all
2 2
x₁=ηx
V*_η=c²√(Λη)
K=(Λu²)/2
Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt)
$$x_1=\eta x$$
$$V^*_\eta=c^2\sqrt{\Lambda\eta}$$
$$K=\frac{\Lambda u^2}2$$
$$P_a=1\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$

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Re: Thread for NonCA Academic Questions
 Moosey
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Re: Thread for NonCA Academic Questions
Oh. How do I get to ω^ω level?fluffykitty wrote:It's probably at most w^3.
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
Nanho walåt derwo esaato?
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Re: Thread for NonCA Academic Questions
 Moosey
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Re: Thread for NonCA Academic Questions
I choose the latter.fluffykitty wrote:You either need infinitely many arguments (which will lead you down the array notation path) or more complex methods.
What would I do?
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
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Re: Thread for NonCA Academic Questions
 Moosey
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How would I encode a level as a number?fluffykitty wrote:Well, you can do something like my f function, where you encode FGH levels as numbers, or do stuff like VECTORPARTY or TREE. I don't think there's any other way to get high FGH levels without doing one of those or using array notation methods like in BEAF.
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
Nanho walåt derwo esaato?
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Re: Thread for NonCA Academic Questions
 Moosey
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Re: Thread for NonCA Academic Questions
Okay, sow how would I turn that into a vgç_n(x,y,z) (very generalized ç)?fluffykitty wrote:One thing that may be useful is that p(a,b)=2^a*(2b+1) is different for all (nonnegative integer) values of a and b, so you can encode two numbers into one. Also, it never outputs 0, so you can do something different in that case. Using that, you can encode arbitrary binary trees into nonnegative integers.
Obviously I’d want something along the lines of a function which is f_n (x,y,z) = f_inverse p of its inputs (x,y,z) to expand back the inputs, but here I’m a little clueless.
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Re: Thread for NonCA Academic Questions
 Moosey
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Re: Thread for NonCA Academic Questions
fluffykitty wrote:One possible method of encoding ordinals as binary trees is to use a leaf node to represent 0 and a branch node to represent b+w^a, where a and b are its children. For example, 1=0+w^0 would be (()()) using a linear representation of trees. 2=1+w^0=(()(()())), w=0+w^1=((()())()), w+1=w+w^0=(((()())())()), etc. Using this representation, you can get a function that grows faster than any FGH level below e0 (=w^w^w^...). For levels of the form 0 or a+1, you can do what you normally do, and for other levels you can use fundamental sequences. The rules for fundamental sequences are (b+w^(a+1))[n+1]=((b+w^a)+w^(a+1))[n], (b+w^a)[n]=b+w^(a[n]) if the previous rule doesn't apply, and (b+w^a)[0]=b if the previous rules don't apply. The useful property of fundamental sequences is that for any ordinal a, an ordinal b is less than a if and only if it is less than a[n] for some n. If you set n=x, then the resulting function will eventually outgrow any instance with lower ordinal input for large enough x, since a[x] will eventually be greater than any ordinal less than a. A method like this one is used in my f, but with a much stronger ordinal notation based on ordinal collapsing functions.
So how would I use that to make a vgç function? Or would it be better just to define a new, simpler function?
Just to verify that I understand this, w^w^w=
((((()())())())())
If I typed it correctly.
I think I still need help in how to do all this.
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Re: Thread for NonCA Academic Questions
A good start would be to make a function for computing fundamental sequences in this tree representation. Also, if possible, simplifying functions is usually a good idea.Moosey wrote:fluffykitty wrote:trees
So how would I use that to make a vgç function? Or would it be better just to define a new, simpler function?
[quote="Moosey]
Just to verify that I understand this, w^w^w=
((((()())())())())
If I typed it correctly.
[/quote]
Yes.
 Moosey
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Re: Thread for NonCA Academic Questions
(ω^a)+b, ω^a >= b
b+(ω^a), ω^a < b
()= 0
Would be one way to do it, in a function notation.
I’m not sure I understand.fluffykitty wrote: A good start would be to make a function for computing fundamental sequences in this tree representation. Also, if possible, simplifying functions is usually a good idea.
If we do make a thing like this, how would one use the p(a,b) = (2^a)(2b+1) to encode an fgh level?
I get that presumably we’d need the inverse operation, which would make binary trees. Presumably, we would do this to all leaves with a value > 0, then iterate (a,b) to get the ordinal encoded by the integer.
That, I suppose, would let you encode an fgh level, but then what?
How does one turn the fgh level into a function without using the fgh itself?
Regardless, I guess we have a function which returns an ordinal given a finite number, sort of like a reverse ordinal collapsing function (but ordinal collapsing functions take uncountable values and make them countable rather than taking countable values and making them finite.)
Call it ord(n). It seems a little difficult to represent using function notation.
ord(0) =
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()
ord(1)=
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(()())
ord(2)=
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(ord(1), ord(0))
ord(3) =
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(ord(0), ord(1))
ord(4)=
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(ord(2), ord(0))
ord(5) =
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(ord(0),ord(2))
ord(6)=
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(ord(1), ord(1))
ord(7)=
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(ord(0),ord(3))
I think every ordinal, finite or infinite, but < ε_0, can be expressed this way.
ord(8)=
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(ord(3),ord(0))
Every finite number is, I think, mapped to a different tree (since every finite number has a unique prime factorization.)
I think every finite number c is represented exactly once (or perhaps 0 times) as ord(k), that is, c = ord(k) for only one k if c is finite.
This is because n = (0, n1) — thus no n = (0, nc) for any c un= 1 since nc+1 = (0, nc) and nc+1 = n iff c = 1
And no finite number n is (c, k) for any c > 0 and any k >= 0, otherwise n >= ω
Also only certain odd primes n make ord(n) finite (as well as 0 and 1 which are not primes)— namely, at least a subset such that, if n = 2c+1 c is prime (or 0 or 1) and satisfies the same property.
3 and 7 are some of these, I’m not sure there are any others. However, other nonprime numbers work, such as 15.
In general, are these just powers of 2, 1?
It seems that for k= (2^n)1, ord(k) = n
EDIT:
ord(9) =
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(ord(0),ord(4))
ord(10)=
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(ord(1),ord(2))
I think there’s a set, call it Š, of finite numbers and countable ordinals such that any value k in the set has exactly one corresponding finite n such that ord(n)=k. This is true for all finite numbers. How many countably infinite values are there in it though? 0? Infinitely many?
Every finite number maps to a unique tree, if that helps. Ergo:
If there is only a single way to represent a value as a binary tree in fluffykitty’s way, then that value is in š if you can find an integer that maps to that tree. That is do not need to worry about extra integers mapping to that tree.
I’m not sure whether every tree is mapped to by some integer, but I feel that there is a proof for that. Something along the lines of:
There is a countably infinite amount of binary trees, and there is a countably infinite amount of integers, thus every tree is mapped to.
This is not a good proof, since you can define a function which maps all integers to the same tree, but for our purposes I think some proof like that would work.
Anyways, now:
ord(11)=
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(ord(0),ord(5))
ord(12) =
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(ord(2)ord(1))
ord(13)=
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(ord(0),ord(6))
I feel the omega+3s will have three or four things in between.
ord(14)=
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(ord(1),ord(3))
Or not; maybe there’ll be 4 of them though
ord(15)=
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we all know what’s going on for a Mersenne notnecessarilyaprime
ord(16)=
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(ord(4),ord(0))
In general, ord(2^^n) = ω^^n
ord(17)=
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(ord(0),ord(8))
This is getting to be a whole lotta fun
ord(18)=
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(ord(1),ord(4))
ord(19) =
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(ord(0),ord(9))
Just one more
ord(20) =
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(ord(2),ord(2))
I think Š= Positive integers U limit ordinals, if I understand limit ordinals correctly.
Edit:
I think (w^w)+w, for instance, might be a limit ordinal, in which case Š ⊂ (Positive integers U limit ordinals)
Ord(currently largest known prime)=
82589933
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
Nanho walåt derwo esaato?
Code: Select all
x = 4, y = 5, rule = B34e5e67c/S23ce45
4o2$b2o$o2bo$b2o!
[[ TRACK 0 7/325 ZOOM 16 THEME 0 ]]