@El'endiaStarman you up for a game now?

Yeah, sure.

Lemme grab dinner.

Okay, now I'm getting interested in the small cases...

Who wins on a 3x3 board where the minimum area per zone is 2?

Does the game differ significantly when the total number of cells is odd?

It makes sense that it would since the no-more-than-half restriction means that there's a free cell that will never be zoned, so it can "move around", so to speak.

I think the chances of a 3x3 board having two squares open at the end is super unlikely

I think it's a tie nearly every game

First player can definitely force a tie.

XXX 11X 112
XXX 11X 112
XXX XXX 22X

XXX X1X X1X
XXX X1X X1X
XXX XXX 222

It's a win for the first player!

No wait, second player can take two cells. Hmmm...

Okay, perfect play by both definitely leads to a tie in all cases.

If one or the other player zones three cells, they lose.

Oooh: I just thought of a really good rule: if the game ends in a tie, the player that had the last turn wins

It completely removes the "take half" strategy

Hmm, very true.

That turns the n=3, m=2 case into a win for the second player, I believe.

Unless you allow them to have 5 squares

So, limit it to n^2//2+1?

That's a win for the first player.

For odd square boards

(n^2+1)//2 then is the formula. Works for even and odd boards.

Yeah, I think so too. P1 plays 3, P2 must play 3, P1 takes 2

Back on the topic of quantum vibrations... what do you consider the proper way to discretize a continuous "string" into an array of numbers?

As in, each number represents the displacement of a particular point on the string, the points being evenly spaced.

The question is... how much of a gap should there be between the first/last data points and the end of the string?

Any reason you wouldn't want to make the first/last data points be the ends of the string?

Right now, I have it so that if each point is L distance away, the first/last data points are L/2 away from the end.

Which was actually easier when it came to classical strings.

Or maybe this is actually a non-issue in the greater scheme of things.

(arccos((a+c)/2-b+1)/(2*pi)) ^ 2 = rotation speed for point b, given consecutive points a,b,c on the string

took me a long time to figure that out

(arccos((a+c)/(2*b))/(2*pi)) ^ 2

maybe right now?

For the Schrodinger wave function on a string?

yes, I'm trying to get the thing to rotate properly

@NathanMerrill: Another potential variant on your game is to allow N-ominoes. Winning would then mean isolating 1-cell spaces.

Yeah, but that is harder to write a submission for

PSA: quantum physics is hard

Lol, really?

@flawr: I too found the perfect murder part hilarious. Batman dies, but none of the assassins killed him. :P

That shows that we don't need the axiom of choice to get nonsense in maths=)

Infinity is enough XD

@NathanMerrill: Regarding your zoning game, it seems to me like you could perhaps apply a strategy-stealing argument to...figure something out. I'm not sure it's symmetrical enough for that though.

@flawr I was actually thinking at the end of the video that finite numbers are "rigid", in a sense. Once you hit infinity, though, they become "fluid".

Also, I saw Vsauce's brachistochrone video before you linked to it. :P (And it made me pretty happy that they showed some of 3Blue1Brown's video on the same subject.)

btw: Is this supposed to be the new go to room for the interesting stuff here on ppcg?:)

^ a magic square of areas.

how cool is that

carresmagiques.blogspot.ch/2017/01/…

@flawr Woahhhhh, epic cool

@flawr it's science, math, and programming. What isn't interesting about that?

@El'endiaStarman I'm not sure what you mean

@flawr That's so cool.

@NathanMerrill In the game we played, I could've won fairly easily. I think the reason is that you zoned a sizable chunk, which allowed me to isolate some cells and then (if I had realized it) take most of the remaining area at once. I think that the more area you take in one zone, the more vulnerable you are to this strategy. If that's the case, then it's better to zone small areas.

I agree.

You want to prevent your opponent from getting multiple turns

Yep. IIRC that was the reasoning behind me zoning 6 cells after you zoned 4; it guaranteed that you only got one turn.

Strategy idea: always zone one unit at a time, unless an opportunity is presented that allows you to win the game by zoning the remaining area

Minimum area is 4

But you can make a 7 square and still be safe

Then instead zone 2x2 squares instead

Also, I think I might increase the minimum to 5

@NathanMerrill Yes. Doing a 2x3 though allowed me to isolate a 3-cell region on my next turn.

Actually, doing a large region on your first turn in the center still works I think

How large are we talking?

They get 2 turns in a row

Or maybe zone a 399x399 region on your first turn. That'd make the game interesting

Maximum number of squares is 200

For one player, that is. Board is 20x20.

I thought board was 400x400

And if we have a tie-is-a-win-for-the-last-player rule, then zoning half the board in one go is a losing move.

Already in the spec

@KritixiLithos Well, it could be. You still wouldn't be able to zone a 399x399 region in one go since that's more than half the board.

Let me read the rules again...

Hm: what if players take turns as long as one is under 2x of the other?

Hmm, curious idea. How would that affect the gameplay?

Like, if you have 16 squares, I'd get repeated turns until I had 8 or more squares

Having more squares is a definite advantage, because you are closer to the "grab remaining area"

Hmm... interesting, that'd change my prior strategy

But it gives your opponent more room to play without giving you 2 turns in a row

I think I like that change. I'm also going to increase the minimum size to 5

room topic changed to 1100101110110101: For discussion about programming, math, and science, and related topics. Not about PPCG (the site itself) or golfing. (no tags)

What is that supposed to be?

˵ in binary. I don't get it either. :P

(I'm guessing it's just a random 16-bit binary number.)

@El'endiaStarman ... Em?

Maybe that's what it means

Right now I'm having "fun" implementing complex Arccos.

@KritixiLithos If so, then that means this room is about music too!

Surprisingly appropriate given Phi's work on vibrating strings...

confirmed: imaginary numbers are magic

@PhiNotPi :)

I just recently explained the difficulties of that on SO

Ah no, it was atanh.

@PhiNotPi What are you doing anyway?

@flawr Right now I'm trying to simulate the quantum wavefunction on a 1D "string".

Sounds awesome, but I have no idea what you're talking about=)

@flawr Do you know anything about "classical" vibrating string and stuff?

You mean as in classical mechanics?

x'' = -a x etc?

yeah that... well the quantum version is also a wave equation which means it's not too different to simulate, except for the whole "it's a different equation" thing.

for starters, let W(x,t) be the value of the wavefunction at a particular place (x) and time (t)

this is 1D for now... then the equation is

dW/dt = i * k * d^2W/dx^2

where i is imaginary and k is some constant (depending on the particle)

Hm if you remove the i it almost looks like the heat equation, doesn't it?

yeah

right now I'm trying to convert that differential equation into polar coordinates

my condolences...

is it even possible? I don't know

It is just 1d, why bothering with polar coordinates?

Or in what way would polar coordinates even make sense?

I'm trying to express W (a complex number) in polar coordinates instead of a+bi

is t real or complex too?

real

Isn't that just r=sqrt(a^2+b^2) and theta=atan2(b,a)?

@El'endiaStarman I think he wants the polar form of the differential equation?

using the relations you just noted, can you express d^2/dx^2 W(t,a,b) with just W(a,r,theta)

@PhiNotPi Can't you just use the polar form of the laplace operator? We're basically in 2D

wikimedia.org/api/rest_v1/media/math/render/svg/…

@KritixiLithos @El'endiaStarman I was hoping you guys would have figured out the purpose of my last title, so I changed it as another clue

Wait, Nonsquare rectangles on an n X n board that's your koth

1100
1011
1011
0101

@KritixiLithos nope. It was inspiration for my koth, not the other way around

Uhhh....dominoes?

30232311 is the base-4 equivalent and CBB5 is the base-16 equivalent of 1100101110110101.

So I still have no idea. :P

1100101110110101 to binary is 11111010001000100100111111111000101001011110010101

which has a length of 50 characters: 22 0s , the rest 1s

I've been thinking... since the four-colour theorem is referring to planar graphs, which are just triangulations of 2-spheres, then it should should be possible to state an n-dimensional variant of the four-colour theorem using triangulations of n-spheres. I wonder then, if it's possible to, for any dimension number n, find a lowerbound for the numbers of colours c required for colouring an n-sphere.

I've been playing around with MagicaVoxel for some rudimentary insight, but I believe that the complexity of cubes is insufficient for ascertaining the lowerbound, if any.

You can also try to generalize the classical result to other geni.

It probably has already been done.

probably, but I haven't seen any (free) online papers covering it after some search

herea rea some: en.wikipedia.org/wiki/Four_color_theorem#Generalizations

@ConorO'Brien you'd first have to make a few definitions first

a n-sphere is an n-dimensional manifold, right?

> There is no obvious extension of the coloring result to three-dimensional solid regions.

That might be why it's hard to find results.

@flawr oo didn't see this. I shouldn't doubt wikipedia

@flawr I believe so yes

I mean even for a non-planar graph it doesn't make any sense to talk about a coloring problem a priori :)

Wait, is a 2-sphere like a disc or the surface of a ball?

I have no idea. Aren't they equivalent?

Uh...I'm not sure.

@ConorO'Brien nope

oh, right--the first is bounded and the second wraps around, right?

A n-sphere can be embedded in a n+1-dimensional space

S^n = {x in R^(n+1) | ||x|| = 1 }

So a 2-sphere is the surface of a ball, got it.

just checking: a ball is a 3-sphere

no, a 2-sphere

I'm using "ball" here in the colloquial sense. Like a basketball.

oh

It does also have a specific meaning in mathematics, as do all ordinary words... :P

which is really not confusing at all >_< "A ball is the surface of a 'ball'"

Y'know, I actually think that "ball" does mean the solid region.

Since it can be used in definitions of continuity.

Or maybe I'm thinking of connectedness.

Well ball is generally as soon as you have a metric space

So a real 0-sphere comprises just two points?

@flawr Yeah, the math class(es) I took that used that term most often was Real Variables I/II, if I recall correctly.

@flawr I believe so, yes. -1 and 1.

So I'm not gonna say "... up to sign" anymore but only "...up to a factor in S0" XD

@flawr You're like the chemist who asks for H20 in an attempt to assassinate his friend.

(The intent is that the "friend" orders some "H20 too", i.e., H2O2.)

...I just noticed that I typed 'H20' instead of 'H2O' twice...

H20 woud surely be an interesting molecule too=)

@flawr You'd probably get a Nobel Prize for it.

for what reason?

@ConorO'Brien Because it is probably very impossible=)

If you managed to get 20 hydrogen atoms to bond together into a molecule, you deserve a big prize.

ahh. Apparently, Wolfram thinks its very possible :P

I just noticed I'm the third highest ranking user with only 2 gold badges.

And at the same time the lowest ranking user above 20k

@flawr Fourth, actually. feersum, DavidC, and Howard.

Ah right=)

WHAT.

It's mind-bending to think that it's likely that there is some range [k..k+n) that has more primes in it than [2..n+1].

Cool thread=)

@El'endiaStarman I would actually have guessed that there are ranges k...k+n with more primes than 2...n+1

When you think of the twin primes, triple primes ... sexy primes usw.

The greater the m in those m-tuplet primes, the further away you get from the origin.