Thread for Non-CA Academic Questions

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Re: Thread for Non-CA Academic Questions

Post by fluffykitty » November 14th, 2019, 6:28 pm

Still not sure why you haven't joined the Googology Discord yet btw
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Re: Thread for Non-CA Academic Questions

Post by Moosey » November 14th, 2019, 7:17 pm

fluffykitty wrote:
November 14th, 2019, 6:28 pm
Still not sure why you haven't joined the Googology Discord yet btw
I don't have a discord account
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Re: Thread for Non-CA Academic Questions

Post by Moosey » November 15th, 2019, 6:25 pm

fluffykitty wrote:
November 14th, 2019, 6:28 pm
Still not sure why you haven't joined the Googology Discord yet btw
Hold it wait, I just joined. To confirm, I am Moooosey#7155.
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Re: Thread for Non-CA Academic Questions

Post by Moosey » November 15th, 2019, 8:02 pm

Are there any online SKIO calculus websites? I've already seen a SKI calculus one.
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Re: Thread for Non-CA Academic Questions

Post by Moosey » November 27th, 2019, 1:35 pm

what happens if we extend bf with a handful of extra commands:
* = mark/remove marker: add marker to data at pointer
( = check for marker; if none is present, skip to corresponding )
) = check for marker; if one is present, go back to corresponding (
Note: ( and ) are like [ and ] but they do it with markers instead of checking for data
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Re: Thread for Non-CA Academic Questions

Post by Moosey » November 29th, 2019, 4:44 pm

is it possible to run hex13 (https://gitlab.com/apgoucher/hex13) on mac?

already did git clone, which worked
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Re: Thread for Non-CA Academic Questions

Post by gameoflifemaniac » December 4th, 2019, 3:01 am

Can someone explain Bowers Exploding Array Function?
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Re: Thread for Non-CA Academic Questions

Post by Ian07 » December 4th, 2019, 7:19 am

gameoflifemaniac wrote:
December 4th, 2019, 3:01 am
Can someone explain Bowers Exploding Array Function?
There's an explanation of it here on Googology Wiki which I really liked, though it is a bit long.

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Re: Thread for Non-CA Academic Questions

Post by Moosey » December 7th, 2019, 2:46 pm

I found this, and I'm on mac OS X:
http://www.srm.org.uk/downloads.html
what would I need to put into terminal in order to compile and then run the CBlocksSource stuff?
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Re: Thread for Non-CA Academic Questions

Post by Moosey » December 14th, 2019, 7:04 pm

There's been some discussion of this on Discord:

Code: Select all

#C both these solutions were found by BlinkerSpawn
x = 100, y = 99, rule = LifeHistory
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18.FC.3B2.CF$18.F2.C2.3CF$18.F3CA.D2.F$18.F3.2A3DF$18.F4.A3.F$18.10F!
#C [[ VIEWONLY ]]
So, the CPP (Crappy packing problem) asks for the least dense way to fill an object with N-ominoes for some N such that you cannot place any more N-ominoes. ECPP (Eventual CPP) asks for the same thing, except that it's possible to place one more N-omino but no more
So far, for the 8x8 grid CPP with tetrominoes and 8x8 grid tetromino-ECPP have solutions (both by Blinkerspawn)

So here's a question: Suppose that E2CPP asks for that, but you can place an N-omino such that it brings you to an ECPP position, and never into any position which eventually leads to ECPP, E3CPP asks for that, but you can place an N-omino such that it brings you to an E2CPP position, and never into any position which eventually leads to E2CPP, etc.
Does anyone have solutions for E2CPP, and E3CPP with tetrominoes on an 8x8 grid?
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Re: Thread for Non-CA Academic Questions

Post by pcallahan » January 30th, 2020, 10:22 pm

I'm not sure this is academic, but it's non-CA, at least directly. I'll post it as a puzzle in case someone else wants to think of a solution.

We have a 4x4 grid of values (or you could generalize it, but I am interested in n=4). Assigning variable names, it looks like this:

Code: Select all

a b c d
e f g h
i j k l
m n o p
We can make a domino tiling of this grid. I.e., we use 8 dominos to cover all the grid cell such that each domino covers two adjacent cells and dominoes never overlap (there are apparently 36 ways to do this, but that's not the question).

Suppose we make a domino tiling like this and we swap each adjacent pair. Symbolically, if you did it with all horizontal dominoes, you get this series of swaps: a↔b, c↔d, e↔f, g↔h, i↔j, k↔l, m↔n, o↔p. That's one possible permutation of the values, but each domino tiling defines a different one.

So the first question is: What permutations of these 16 values can you obtain just by applying a series of domino-tiling permutations? Can you obtain all permutations? Just, e.g, even permutations? There is no limit on the number of domino tilings you have to apply in succession but you could come up with a lower bound of log_36 16! (right?) assuming you could get every permutation.

Now I'll reveal that I'm actually only interested in cases where the values are 0 or 1. In this case, a swap of identical values doesn't change anything. But you still have to cover all the grid values at each step.

And further, I'm really only interested in whether you can start with an arbitrary set of 0s and 1s and get a set that is monotonic in rows in columns, i.e. like

Code: Select all

0 0 0 0
0 0 1 1
1 1 1 1
1 1 1 1
And yes, this is very remotely connected to a CGOL thing I'm working on but I wonder if anyone knows the answer. (Not that I don't like puzzles but if someone else likes this one, I am really more interested in the answer than the challenge.)

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Re: Thread for Non-CA Academic Questions

Post by pcallahan » January 31st, 2020, 10:40 am

pcallahan wrote:
January 30th, 2020, 10:22 pm
Suppose we make a domino tiling like this and we swap each adjacent pair. Symbolically, if you did it with all horizontal dominoes, you get this series of swaps: a↔b, c↔d, e↔f, g↔h, i↔j, k↔l, m↔n, o↔p. That's one possible permutation of the values, but each domino tiling defines a different one.

So the first question is: What permutations of these 16 values can you obtain just by applying a series of domino-tiling permutations? Can you obtain all permutations?
If anyone else cares, I think the answer is yes within parity constraints. The main trick is to use the dominoes to move values around cyclic paths. E.g., consider these two successive domino tilings (a domino is represented by two adjacent cells with the same capital letter).

Code: Select all

A A E E
D B F F
D B G G
C C H H
and

Code: Select all

A B E E
A B F F
D C G G
D C H H
The A, B, C, D dominoes move the values around the cycle over two steps while the E, F, G, H dominoes swap the same value back and forth. (Actually, the cycle is more complex than I pictured in my head; every other value moves either two steps clockwise or counterclockwise.)

It's still a puzzle to create a specific permutation and you are limited by parity unless I'm mistaken. E.g. let the grid be a checkerboard of white and black squares. Any value on a white square is on a black square after an odd number of steps, and back on a white square after an even number of steps.

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Re: Thread for Non-CA Academic Questions

Post by gameoflifemaniac » February 11th, 2020, 4:13 pm

What's the general formula for the area of a regular n-gon with side length 1? I know, not very academic question
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Re: Thread for Non-CA Academic Questions

Post by pcallahan » February 11th, 2020, 8:00 pm

gameoflifemaniac wrote:
February 11th, 2020, 4:13 pm
What's the general formula for the area of a regular n-gon with side length 1? I know, not very academic question
You can look that up pretty easily (I checked). If you want to work it out yourself, think of the n-gon as consisting of n isosceles triangles with a base length 1 and angle 360°/n. Divide those into right triangles and use the fact that the short leg (1/2 length) to long leg ratio is tan(180°/n). The area of the isosceles triangle is the product of short and long legs, and you multiply that by n for the total area.

Let's see, for a square, n=4, 180/4=45. Each isosceles triangle has area (1/2)(1/2) and that times 4 is 1. For a hexagon, n=6, 180/6=30. The tangent is 1/sqrt(3) I think, so total area is 6*(1/2)(sqrt(3)/2) and so on.

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Re: Thread for Non-CA Academic Questions

Post by Moosey » February 13th, 2020, 3:07 pm

How incredibly unstable would 1,1,3,3,5,5-hexaazido-2,2,4,4,6,6-hexanitrocyclohexane be?
Would it be more or less stable compared to 1,2,3,4,5,6-hexaazido-1,2,3,4,5,6-hexanitrocyclohexane?
And how would those compare to 1,1,2,2,3,3-hexaazido-4,4,5,5,6,6-hexanitrocyclohexane? (And am I right in thinking that these are more unstable than even, say, 1-diazidocarbamoyl-5-azidotetrazole?)

And are those standard nomenclature?
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Re: Thread for Non-CA Academic Questions

Post by Moosey » February 13th, 2020, 3:32 pm

A more serious question: how would I name the following group?

Code: Select all

   NO2
    |
NO2-C-NO2
    |
    O
    |
As in, like, say

1,2,3,4,5-pentaazido-6,7,8,9,10-penta(group)-1,2,3,4,5,6,7,8,9,10-decanitrodecane

I'm gonna guess it's something in parentheses, maybe (trinitromethoxy) ?

EDIT: apparently yes.
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Re: Thread for Non-CA Academic Questions

Post by GUYTU6J » February 24th, 2020, 11:45 pm

The base of natural logarithm e has many characteristicss/defnitions, e.g.
e₁=lim(x→∞)(1+1/x)^x
e₂=∑(n=0 to ∞)(1/n!)
e₃ is the unique solution x for ∫(t=1 to x)(1/t)dt = 1
e₄ is the unique number a such that f(x)=a^x is equal to its own derivative and has unit tangent slope at (0,f(0))
How to prove their equivalency, i.e. e₁=e₂=e₃=e₄?
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Re: Thread for Non-CA Academic Questions

Post by PkmnQ » February 25th, 2020, 8:08 am

GUYTU6J wrote:
February 24th, 2020, 11:45 pm
The base of natural logarithm e has many characteristicss/defnitions, e.g.
e₁=lim(x→∞)(1+1/x)^x
e₂=∑(n=0 to ∞)(1/n!)
e₃ is the unique solution x for ∫(t=1 to x)(1/t)dt = 1
e₄ is the unique number a such that f(x)=a^x is equal to its own derivative and has unit tangent slope at (0,f(0))
How to prove their equivalency, i.e. e₁=e₂=e₃=e₄?
I have no idea what I’m doing, but I’m just gonna experiment.
I’ll start with e2
Notice that each factorial is a factor of the next.
I’ll ignore the first 1/0! because it’s just mess.
That means you can do this:
1/1 + 1/2 + 1/6 = 6/6 + 3/6 + 1/6
If you sum up everything you get this sequence: 1/1, 3/2, 10/6, 41/24
It evaluates to this:
n -> Infinity
Σ[k=0->n-1](n^k)/n!
How to XSS:

Code: Select all

Function(‘a’+’lert(1)’)()

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Re: Thread for Non-CA Academic Questions

Post by toroidalet » February 25th, 2020, 7:57 pm

In order to prove that en=em, you would use whichever property is used in the proof that en=e (some more complex relations may use multiple properties, in which case you would want to prove those en's to be equivalent first).
For example, to prove that e3=e4, you would find that d/dx(ln4(x))=1/x, and so by the fundamental theorem of calculus ∫(t=1 to e4)(1/t)dt=ln4(e4)-ln4(1)=1-0=1 (the e3 condition), and so e4 is the same as e3.
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Re: Thread for Non-CA Academic Questions

Post by GUYTU6J » February 27th, 2020, 1:52 am

Can we not introduce natural logarithm?
e₅ is the unique number a that a^(∫(t=1 to x)(1/t)dt) = x
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Re: Thread for Non-CA Academic Questions

Post by BlinkerSpawn » February 28th, 2020, 11:57 pm

GUYTU6J wrote:
February 24th, 2020, 11:45 pm
The base of natural logarithm e has many characteristics/defnitions, e.g.
e₁=lim(x→∞)(1+1/x)^x
e₂=∑(n=0 to ∞)(1/n!)
e₃ is the unique solution x for ∫(t=1 to x)(1/t)dt = 1
e₄ is the unique number a such that f(x)=a^x is equal to its own derivative and has unit tangent slope at (0,f(0))
How to prove their equivalency, i.e. e₁=e₂=e₃=e₄?
e1 -> e4: This is the equation for an interest rate of 100% applied x times per unit of time. Letting the width of each compounding period become infinitesimally small, the rate of change becomes instantaneous (a derivative) while retaining its rate of 100%/period, and so the derivative is equal to the original function.
e4 -> e2: Starting with e^0 = 1, repeated application of "equals its own derivative" provides the power series e^x = ∑(n=0 to ∞)(x^n/n!); let x=1.
e4 -> e3: Let y = ln x. Then x = e^y, and implicit differentiation yields 1 = e^y dy/dx, and dy/dx = e^-y = 1/x.
Therefore, ln x is the antiderivative of 1/x, and the needed upper bound for the integral is the unique t such that ln t - ln 1 = 1. But 1 = e^0, so ln t = 1 and t = e.
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Re: Thread for Non-CA Academic Questions

Post by Entity Valkyrie 2 » February 29th, 2020, 6:50 am

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Re: Thread for Non-CA Academic Questions

Post by GUYTU6J » February 29th, 2020, 11:00 am

BlinkerSpawn wrote:
February 28th, 2020, 11:57 pm
GUYTU6J wrote:
February 24th, 2020, 11:45 pm
The base of natural logarithm e has many characteristics/defnitions, e.g.
e₁=lim(x→∞)(1+1/x)^x
e₂=∑(n=0 to ∞)(1/n!)
e₃ is the unique solution x for ∫(t=1 to x)(1/t)dt = 1
e₄ is the unique number a such that f(x)=a^x is equal to its own derivative and has unit tangent slope at (0,f(0))
How to prove their equivalency, i.e. e₁=e₂=e₃=e₄?
e1 -> e4: This is the equation for an interest rate of 100% applied x times per unit of time. Letting the width of each compounding period become infinitesimally small, the rate of change becomes instantaneous (a derivative) while retaining its rate of 100%/period, and so the derivative is equal to the original function.
...
I've figured out a formal proof for e₁→e₁=e₄. Define a function f(x) = lim(t→∞)(1+x/t)^t, then f(1) = e₁ and
f(x) = lim(t→∞)(1+x/t)^t = lim(t→∞)(1+1/(t/x))^((t/x)·x) = (lim(t/x→∞)(1+1/(t/x))^(t/x))^x = (f(1))^x = e₁^x
f(x) is equal to its own derivative because
(lim(t→∞)(1+(x/t))^t)' = lim(t→∞)(t·(1+(x/t))^(t-1)·(1/t)) = lim(t→∞)(1+(x/t))^(t-1) = lim(t→∞)(1+(x/t))^t
f(0) = 1, so the tangent slope at (0,f(0)) is indeed 1.
Therefore e₁=e₄. The remaining part about e₄→e₄=e₂ and e₄=e₃ is what you've said, and e₅ is a corollary from e₄=e₃.
---
What if we start from e₂ or e₃ and prove they are equal to e₁? (Their uniqueness is easy to see, btw)
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Re: Thread for Non-CA Academic Questions

Post by Moosey » March 5th, 2020, 3:22 pm

Can someone explain the symbols \models, \prec, and V_k found in pages like these?
https://web.archive.org/web/20190727152 ... fo/Worldly
https://web.archive.org/web/20190727125 ... Reflecting
They're blocking my understanding of worldlies and stuff
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Re: Thread for Non-CA Academic Questions

Post by BlinkerSpawn » March 6th, 2020, 4:26 pm

It looks like a "model" is essentially a set large enough that logical axioms and formulae can be interpreted and manipulated as subsets of said set.
Someone better-trained in set theory than myself will have to refine that.
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