Set Theory Problem

There are at most aleph-zero disjunct 3D spheres in 3D space. And there are at least aleph-one disjunct 2D circles in every finite volume part of the 3D space?

The number of points in N dimensional space is always aleph-one. And you can also divide this space into aleph-one disjunct N-1 dimensional spheres.

Is there a way to divide 3D space into aleph-one cubes with no common volume? They may touch each other, but may not share some common volume.

If you can find one such space division, you brought down the ZF Set Theory.


Discussion there:


Recent 2 Problems

Were a part of the “scientific war”. Not so much a better known cultural war,  despite the fact one may see them as that, too!

Two corner stones for the Climate Change narrative are obviously shaky. Our planet was once warm, thanks to its faster rotation in the first place. And the locally rising sea is a bulshit as well.

These basic facts explains a lot!




Create 2314

This is how you create the number 7.

MakeIntVar A
Inc A
Inc A
Dec A


MakeIntVar A
MakeIntVar B
Inc A
Inc B
Shl A, A
Shl A, A

You can do all the basic operations + – * \ and you can shift right and left, and decrease or increase any variable, which you have to “MakeIntVar” first. By doing that, the variable is initialized to zero.

Now, make 2314 with the shortest possible algorithm.


And Another Physics Problem

The one I talked about before, but it is very obscure, even unknown on the whole World Wide Web.

We have two equal stars, A and B. Then we have a planet called X in the orbit around A, and a planet called Y around the star B.

Everything else equal, except for the rotation speed of both planets. Planet X rotates 20% faster than planet Y. The question is, which planet has a higher average surface temperature.

Edit: See discussion:


Chesslike Problem

Long ago, I invented a problem which solution looks like this:


The problem was to place a complete set of white chess pieces on the chessboard such that the number of bonds between them is maximal. Here you see that the king protects 8 neighbouring pieces and  the pawn in the top line covers no piece. The number of all protections is 53 and it can’t be done any better. (You may notice a pawn in the first line and that bishops are of the same color, but that’s fine.)

If you now add 40 more queens and rearrange all those 56 pieces, then there will be at most 338 bonds in total. We call that number B – the maximal bonding number for the set F. For all the white pieces it’s 53, for 2 knights only it’s 2, and so on.

We have a set F with bonding number B1. And a proper subset F’ with the bonding number B2. B1 is equal to B2. Which set is that?

In other words, for which set you can remove one or more pieces and the maximal bonding number remains the same as it was initially.

There is a trivial solution of any single piece and the empty set, but as always, we want a nontrivial solution.