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Comment author: Thomas 17 July 2017 11:08:11AM 1 point [-]

Try this

Comment author: Gurkenglas 17 July 2017 07:43:26PM *  0 points [-]

The same can be said of unit masses at every whole negative number.

The arrow that points to the right is at the same place that the additional guest in Hilbert's Hotel goes. Such unintuitiveness is life when infinities/singularities such as the diverging forces acting on your points are involved.

Comment author: Thomas 12 July 2017 05:49:09AM 0 points [-]

50 moves rule is totally inappropriate in 4D. Let us dismiss that rule here, yes.

Comment author: Gurkenglas 12 July 2017 10:11:45PM *  0 points [-]

An upper bound is 17 queens: 16 threaten all 6^4 inner squares, then the 17th moves to the inner square closest to the king.

Edit: Nevermind, this amounts to the 17th queen checkmating the king on a 3D board with warp sides.

Comment author: Thomas 10 July 2017 06:35:56AM 1 point [-]

Kindly invited to solve this.

Comment author: Gurkenglas 11 July 2017 06:27:12PM *  1 point [-]

An upper bound is 27 queens, which can threaten all squares of a 3D chessboard hyperplane (and the two adjacent ones), which sweeps through the hypercube and smashes the king against a hyperwall. This assumes that the game doesn't draw after 50 turns.

Comment author: Thomas 05 July 2017 07:37:32AM 0 points [-]

Very well, congratulations again!

Perhaps a nonrecursive function would be faster.

Comment author: Gurkenglas 05 July 2017 02:36:31PM *  0 points [-]

Not really, the sequence grows quickly enough to outstrip the recursive overhead. To calculate the overhead, replace the * in f(i)*f(2n+1-i) with a +. Memoizing is of course trivial anyway, using memoFix.

Comment author: Thomas 03 July 2017 03:26:36PM 0 points [-]

It's something. But what are the codes? An algorithm to create them would suffice. A faster one is better, of course.

Comment author: Gurkenglas 03 July 2017 08:54:15PM *  1 point [-]

The same control flow generates them. In Haskell:

data T = N T T | L deriving Show
⠀
ts :: Int -> [T]
ts 1 = [L]
ts k | (n, 1) <- divMod k 2 = [N x y | i <- [1..n ], x <- ts i, y <- ts (k-i)]
ts k | (n, 0) <- divMod k 2 = [N x y | i <- [1..n-1], x <- ts i, y <- ts (k-i)]
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀++ [N x y | ys@(x:_) <- tails (ts n), y <- ys]
⠀
⠀
<Gurkenglas> > ts 4
<lambdabot> [N L (N L (N L L)),N (N L L) (N L L)]

(Beware, I had to use U+2800 to almost align the code block in spite of LW's software eating whitespace. Source here)

Edit: See also: oeis, where you enter an integer sequence and it tells you where people have seen it.

Comment author: Thomas 03 July 2017 08:38:51AM 1 point [-]
Comment author: Gurkenglas 03 July 2017 01:09:35PM *  1 point [-]

Your huffman codes with essential indifference are binary trees (each node has 0 or 2 children) with isomorphism.

Let f(n) be the number of trees with n leaves.

f(1)=1
f(2n+1)=sum from i=1 to n of f(i)*f(2n+1-i)
f(2n)=f(n)*(f(n)+1)/2 + sum from i=1 to n-1 of f(i)*f(2n-i)

Here's the first 26 numbers of such trees:

[1,1,1,2,3,6,11,23,46,98,207,451,983,2179,4850,10905, 24631,56011,127912,293547,676157,1563372,3626149,8436379,19680277]

Comment author: Thomas 20 June 2017 07:37:24PM 0 points [-]

Given aleph-one cubes with no common volume in 3D space

Unfortunately we don't know how to fill aleph-one cubes into 3D space in a way, that they don't overlap with 3D intersections.

We can easily do so with aleph-zero cubes, but have no known way to do it with aleph-one cubes.

Others are telling me, not to even try, because it's surely impossible and I will suffer some great misfortune, if I try.

I am not sure, if it's really impossible.You can surely fil N dimensional hyperspace with aleph-one N-1 dimensional hypercubes. That IS possible.

Now, what if we have countably infinite amount of space dimensions. How many (rational) points are there? And how many countably infinite dimensional hypercubes we can squizz there?

I don't know, but it's possible to calculate and to see how the ZF would handle that.

Comment author: Gurkenglas 21 June 2017 12:22:00AM *  0 points [-]

The first sentence of your post on protokol2020 is "There are at most aleph-zero disjunct 3D spheres in 3D space.", so I gave a way to make aleph-one spheres from aleph-one cubes, in order to disprove the possibility of aleph-one cubes.

Aleph-zero-dimensional space has aleph-one rationals. Note that the union of all finite-dimensional spaces (each embedded in the next as a slice) has aleph-zero rationals.

Science demands that you notice an anvil dropped on your head, and my heuristics are also saying you're turning into a math crank.

Then again, if in all spacetime there's one Jesus and a million madmen believing they're Jesus, would we rather that they all believe themselves madmen?

Comment author: Thomas 19 June 2017 08:03:22AM 1 point [-]
Comment author: Gurkenglas 20 June 2017 06:06:21PM 0 points [-]

Given aleph-one cubes with no common volume in 3D space, replacing each cube with the largest sphere that fits in it will give you aleph-one spheres with no common volume in 3D space.

Comment author: Viliam 20 June 2017 08:53:21AM 1 point [-]

what is the modern equivalent of learning to program as a kid in the 80s.

Cryptocurrency investment. Imagine how your kid's peers will be impressed to hear "when I was at elementary school, I put my pocket money in various altcoins, and... long story short, I am a billionaire now". :D

But maybe learning to program is the modern equivalent of learning to program. Just because there are many tablet games teaching kids how to build "a loop in a loop" programs from predefined blocks, doesn't mean that kids will bother to play the games, and will move to further stages of programming.

Comment author: Gurkenglas 20 June 2017 01:30:13PM *  0 points [-]

Markets are anti-inductive; why do you think there's future money lying on the street in buying some of many altcoins?

Comment author: MrMind 13 June 2017 08:00:49AM 1 point [-]

This is the non-central fallacy at its best.
Not as a defence, but seriously: who cares if you define yourself as a cultist or not? The important thing about a belief is the usual: how much reality does this belief allows you to grasp and manipulate?

Comment author: Gurkenglas 14 June 2017 08:57:06PM 1 point [-]

If he finds he doesn't want to be some things, and cultists are those things and also others, then it's fine epistemology to also try not to be those other things in case they're correlated.

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