Tian

https://toot.kif.rocks/@tian

Meine erste KIF war in Bielefeld

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Profilbild: Kolorierte Blindzeichnung von Sanne (npub1r2p…xz6j)

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npub1lqzfls55t8epshmq7lrqkwqhxz8p8gtvx7pw6hj5w5l6s7lye2nq6ptgt7

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tian_at_toot.kif.rocks@momostr.pink

Profile Code

nprofile1qqs0spylc229nusctas003st8qtnprsn59kr0qhdte2820ag00jv4fspz3mhxue69uhhyetvv9ujuerpd46hxtnfduqs6amnwvaz7tmwdaejumr0dsrg6z29

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nostr.wine

relay.damus.io

nos.lol

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Last Notes

2026-07-09 15:27:15 UTC

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"Warum die Eilanträge erfolglos waren."
Spannende Frage, schade, dass dazu nichts im Artikel steht.
"Das Bundesverfassungsgericht hat seine Begründung zwar noch nicht veröffentlicht."
Das "zwar" lässt hoffen, dass danach im Artikel noch was konkretes kommt, aber nein.
Was soll das?

2026-07-04 13:37:20 UTC

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Zugegeben, die teilnehmenden Vermutungen würden nicht aus aller Welt stammen. Aber hätten sie nicht zumindest Bundesliga-Niveau?
Eigentlich ging es mir ja auch nur darum, dass die alleine schon hinreichend viele sind, um eine Meisterschaft durchzuführen.
https://www.erdosproblems.com/

2026-07-04 12:49:27 UTC

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Man könnte ja auch eine WM nur mit Erdős-Vermutungen machen…

2025-09-26 12:58:23 UTC

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The rules of go are nice and few.

2025-07-11 10:14:20 UTC

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die haben grade alle drei Wahlen von der heutigen Tagesordnung abgesetzt

2025-06-29 13:44:09 UTC

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Klingt binärlogisch so, als ob die metapersonalen Beziehungen eine Teilmenge der transpersonalen wären. Oder als ob die transpersonalen durch eine genauere Definition weiter eingeschränkt werden, so dass es drei disjunkte Mengen sind.
Oder wird es doch noch fuzzy-logisch, wie ich erst dachte?

2025-04-27 23:33:54 UTC

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Das Kontinuum (aka Zahlenstrahl) ist doch noch ganz anschaulich. Und das Einheitsintervall reicht ja auch schon. Aber der Unterschied zwischen einer Geraden und einer langen Geraden macht mich verrückt.
https://de.m.wikipedia.org/wiki/Lange_Gerade

2025-04-13 13:24:38 UTC

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Dreiste Korruption erfordert nicht, eine solche Trotteligkeit zu spielen. Das muss größtenteils echt sein.

2025-04-13 12:33:43 UTC

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Ich sehe nicht, welches Eigeninteresse eine durchschnittlich intelligente Person dazu bringt, so dumm zu wirken. Vielleicht überschätze ich den Durchschnitt aber auch.

2025-03-29 04:13:30 UTC

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boring computation like this:
hascpo := function(g)
local au, inner;
if IsAbelian(g) then return false; fi;
au := AutomorphismGroup(g);
inner := InnerAutomorphismsAutomorphismGroup(au);
return ForAny(au, p->not p in inner and ForAll(g,el->IsConjugate(g,el,el^p)));
end;

2025-03-29 03:08:40 UTC

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I used GAP to bruteforce the other smallest such group. It is SmallGroup(32,44) with StructureDescription
"(C2 x Q8) : C2".

2024-09-27 08:15:09 UTC

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Die Formulierung der GO ist nicht eindeutig. Man kann sie so lesen, dass auch im zweiten und in weiteren Wahlgängen nur die stärkste Fraktion ein Vorschlagsrecht hat. Der Alterspräsident hätte sich wohl diese "Rechtsansicht" zu eigen gemacht.

2024-09-22 14:55:34 UTC

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Hiermit bewiesen: wir haben mindestens k+1 Punkte mit gleichem parity type. Wähle einen. Die Mittelpunkte von diesem zu jedem der k anderen sind verschiedene weitere Punkte auf dem Gitter.
(Alle binomial(k+1,2) Mittelpunkte sind auf dem Gitter, aber wieviele davon sind garantiert unterschiedlich?)
Tatsächlich müssten es aber "viel mehr" blockierte Punkte sein.

2024-09-16 12:49:07 UTC

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I guess there is no easy way to turn your program into an android app?

2024-09-16 12:19:12 UTC

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I wonder if with 8 points there are actually at least three blocked. It's true in the "main branch" of the argument (midpoints of AM and MB) and in the case of at least three points with same parity type. But can we have four pairs of points with same p.t. such that the four midpoints are two different points, without blocking any other?

2024-09-15 09:43:08 UTC

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The argument doesn't depend on n. So if the final conclusion is not possible, an earlier assumption must fail.
If n=4, you won't find four pairs with different common parity type and same midpoint.
Maybe it's time to consider the four-coloring induced by parity types: take a chessboard pattern, and in each other row, replace black and white by red and blue.

2024-09-15 08:52:20 UTC

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Assume we have eight points. If at least three had the same parity type, that alone would give three different on-grid midpoints. So assume they form four pairs of points, each with a different common parity type. Each pair has an on-grid midpoint. Now assume these midpoints are all the same and call it M. Let A and B be the pair of points that has the same parity type as M. Then the midpoint of A and M is another on-grid point on the line AB.

2024-09-14 21:10:34 UTC

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You need to implement export, unless you prefer it like this:
[(18, 13), (0, 5), (16, 18), (2, 0), (17, 4), (1, 14), (13, 1)]
n: 19, points: 7, blocked: 1
(I knew I had seen such a thing before, but again took me some time to create)

2024-09-14 19:23:31 UTC

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To stop me wasting more time searching for a configuration with eight points and only one blocked, I have now proven that that is not possible.

2024-09-14 11:54:13 UTC

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yes, a/b gives a float, but a//b is integer division

2024-09-14 11:44:41 UTC

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Oh no, now I have a reason to boot my computer earlier today, in other words: nice!
But I really don't like
"if something return True else return False"
— that is just "something"!
Also: int(a/b) => a//b

2024-09-11 21:40:49 UTC

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Four in a quarter would allow 16. I think you can use halfes.
With 9 points, three must have the same parity type, and the three on-grid midpoints that we get from them are all different.

2024-09-11 21:06:10 UTC

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how about 1 blocked?
oxooooooo
ooooooxoo
xoooooooo
ooooooooo
ooooøoooo
ooooooooo
oooooooox
ooxoooooo
oooooooxo

2024-09-11 15:55:43 UTC

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n=13:
oxooooooooooo
oooooooooooxo
xoooooooooooo
oooooooooooox
ooooooooooooo
ooooooooooooo
ooooooooooooo
oøoooooooooøo
ooooooooooooo
ooooooooooooo
ooooooooooooo
ooooooooooxoo
ooxoooooooooo

2024-09-11 12:41:05 UTC

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you can add a line of 'o's to that example

2024-09-11 12:14:54 UTC

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oxooooooo
oooooooxo
xoooooooo
oooooooox
ooooooooo
oooooooøo
ooooooooo
ooooooxoo

2024-09-10 22:36:51 UTC

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Yes, that always works!

2024-09-10 17:27:58 UTC

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I was wondering the same. My immediate reaction: well sure—no, wait!
Do you have any results for small n>4?

2024-09-10 13:33:01 UTC

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wow…

2024-09-10 09:54:06 UTC

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I hope this is us doing math¹ and me contributing, not you having fun and me spoiling it.
¹ and having fun

2024-09-10 09:29:07 UTC

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Use integer coordinates in the obvious way. (The non-obvious details don't matter.) For a position at (x,y) call (x mod 2,y mod 2) its parity type. There are four different parity types, so five different positions have at least one pair of positions with the same parity type. Such a pair has an even difference in both the x- and the y-coordinate, so their midpoint has integer coordinates and therefore is another point on the grid.

2024-09-09 23:29:46 UTC

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Oh yes! Your writing about more than five made me believe you had an example with five. Four is good hint. My complicated proof is obsolete, oh well.

2024-09-09 23:03:19 UTC

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In that case: no OEIS entry for you. Sad!

2024-09-08 11:31:19 UTC

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oh, that is interesting. Damn!

2024-09-08 07:47:27 UTC

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We also have n=128, and the last proof should work for 128+delta for some small delta. Things can certainly be improved. For computer checking, the remaining grid sizes seem still a little too big to me.
What do you mean with the last question? k<n selected positions on an n x n grid? I have this in the final proof. Removing delta rows and columns from a hypothetical solution for n=64k+delta may remove 2*delta selected points, so I show there can't be 64k-delta of those on 64k x 64k.

2024-09-08 00:42:08 UTC

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Seems to be true for all n>=192: let n=64k+delta, delta<64. If there were a solution for n, we could remove the last delta rows and columns and get a 64k x 64k grid without blocked positions with at least 64k-delta>64k-64 selected positions. If we use the same numbers of subgrids as previously, we find that some corner 8k x 8k grid must contain more than ((64k-64)-24)/4=16k-22 selected positions, which is not possible if k>=3.

2024-09-07 22:43:02 UTC

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Proof for n=64, should work for other (arbitrary large) nice numbers and might work for other numbers with more bookkeeping:
Subdivide the grid into 4x4 subgrids of size 16x16. In each non-corner subgrid, at most one point can be selected, because thereis enough place (inthe whole grid) to repeat the step from one point to another in the same subgrid in at least one direction. There are 4*4-4=12 non-corner subgrids.
Next, subdivide the corner subgrids further into 2x2 grids of size 8x8.

2024-09-07 13:07:52 UTC

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In the 6 version, there is a blocked position between the x in the 1st and 3rd row

2024-09-07 11:53:26 UTC

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BTW, not trying to outsmart you, I genuinely thought there need to be restrictions and you have to tell me which. Now I think it's no-three-in-a-line or equivalently, extend the no blocked positions condition to occupied positions. It doesn't seem true for small n, though. Trivial cases (n=2) aside, there is
. . X .
X . . .
. . . X
. X . .

2024-09-07 10:52:56 UTC

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Are there restrictions on placing the n points? If they are all in one row, no other free grid point is blocked. (Or does "position" not mean grid point?)