@flawr I understand the basics of topology (how shapes are equivalent across deformation). However, that operation doesn't make sense: How am I getting two new paths? Am I making a figure 8 or something?

and if I cut two circles in half, I can certainly glue them back together to make a new loop if I wanted to

@NathanMerrill my explanation was probably not the best: let's say you have each a parametrization of two such closed paths, one is f:[0,1] -> {topological space}, the other one is g:[0,1] -> {topological space}, both must obviously be continuous. And as they are closed we have f(0) = f(1) and g(0) = g(1)

under deformation we can assume that f(0) = f(1) = g(0) = g(1)

when we "add" those two paths we get a new parametrization by using (f+g)(x) = f(2*x) for x in [0,1/2], and g(2*x-1) for x in (1/2,1]

@flawr what does "=" mean?

oh, equality of course, we're talking maths:)

does it make sense?

that's not strictly true. We're in topology. If we simply define it to be "The same spot in space", then that doesn't work because I can move any shape to any position, so any two points are equal

I'm asusming that f(0) is a point on the path of f

@NathanMerrill that is right

@NathanMerrill I don't understand what you mean

well, take a torus. Any given point I can match to any other point

no matter the path I chose

and the same is true for spheres

so f(0) = f(1) for toruses and g(0) = g(1) for spheres

so, I still have no idea how one pairs up to the integers, and the other pairs up to 0

er, it was circles and a "filled in circle"

platter?

that's a filled in circle

Ah I forgot to mention: You do usually first choose one point on the space, and then define all the loops such that they start and end there.

@NathanMerrill disk?:)

disk :)

@flawr how does that work with a double-figure-8?

there's no 1 point that contains all loops

@NathanMerrill you can chose any point (and one can show that you still get the same group, no matter what point you chose)

ok. So I choose a point on a circle. I can pick any other point between 0 and 1 that represents another spot on that circle

I'm cool with that

however, that only works because a circle doesn't have a thick border

if it was a circle that had a border with a thickness, suddenly, the number .5 doesn't represent a unique point on the circle

@NathanMerrill but you can continuously deform any loop that goes -let say clockwise- around the circle such that 0.5 is mapped to a certain point

doesn't that mean that any number (except for 1 and 0) maps to any point?

because I deform all I want

anyways, back to the problem: I can get behind mapping loops to a range between 0 and 1

@NathanMerrill not if we choose a concrete representant of the class of such loops

and I can get behind the fact that discs don't have loops

there are still two more holes that I see: That 0-1 maps to the integers (because there are more floats than integers), and that "no loop" maps to 0

@NathanMerrill well it does, but they are all equivalent to the trivial constant "null" loop that maps [0,1] to the very same point

why are we considering that a loop?

loops need to have a hole

@NathanMerrill the start and the end point is the same, and it is continuous

(this trivial loop will later become the "identity" element of the group)

so, a loop is defined as a list of continuous points that eventually touch itself

which makes all loops on the disk equivalent.

not not a set, it is a functin that maps [0,1] to the topological space

right, not a set :)

or actually not [0,1], but [0,1] where we identify 0=1

anyways, a circle has two loops then?

the null loop and the actual loop?

@NathanMerrill no, even more: we can also go aroudn the circle twice

this is where the distinction betwen sets / function is important:

oooh, clever.

if we consider a loop just as a set, we cannot count how many times it goes around (and in what direction)

right, that's why I edited it to a list

though, a function is more appropriate

I get the concept of ordering the points

so, -1 represents counter clockwise, while 1 represents clockwise?

exactly:)

so, when we add loops, we aren't adding the shapes together

we're adding the loops

that's what I missed

right! well that was also not mentioned in the video, but it is important

whenever we take a loop and try to find all other equivalent loops on the circle, we find that they have the number of times it goes around the circle in common.

doesn't that mean that a figure 8 can't be represented by integers?

you mean as a topological space?

let me be more specific

lets say I have a double-figure eight (3 circles connected to each other)

there's no 1 point that they all share

therefore, it doesn't make sense to add any two loops

because not every loop touches every other loop

so, you need a more complicated representation than integers

@NathanMerrill that is true, but it is not for the reason you mentioned

@NathanMerrill you can choose any point, let me draw an example:

here we choose the red point as our fundamental point. (we consider the black figure of 8 as our topological space)

here the blue and the green loop are in the same equivalence class

as you can continuously deform them to one another

but as representants we choose loops that go through the red point

it turns out the fundamental group of the figure of 8 is already relatively complicated

oh, because if 1 is a loop around one side, then what integer do you pick for a loop around the other.

its impossible because it can't be possible to add/subtract to it

bah, the grammar of that sentence is hard.

hehe:)

"it needs to be impossible to add or subtract to it"

couldn't you represent it as a pair of integers?

no it is even more complicated than that

because you need to keep track in which order you go anround which side (and which direction)

so let a / b be loops around the left circle in clockwise/counterclockwise direction, and x / y be loops around the right circle in clockwise/counterclockwise direction

then we can represent the group as strings over the alphabet {a,b,x,y}

right so, a list of those

I was going to say a list of integers from 0 to 4

but alphabet is much cleaner

where we can cancel ab or ba and also xy and yx

so xabb is the same as xb

oh, that's really complicated

the cancellation makes things 20 times worse

and the "addition" is just concatenation

right

@NathanMerrill oddly specific number XD

I actually use that number all the time for "X times worse"

yeah?? I only ever heard some powers of ten for that use :)

now consider the torus (i.e. the surface of a donut), any guess what the fundamental group might be?

(it is less complicated)

so, a loop around the top cancels out a loop around the girth

wait...this is weird, because a loop around the girth can cancel out a loop around the top

so, they are equivalent across cancellation, but can't be deformed into each other

that doesn't make sense

I'm not sure this is right, what do you mean by loop around the "top"?

the part of a donut that touches the table

and around the girth is "through the hole"?

yeah

no they do not cancel out

but they commute

ok...tell me why this doesn't work

Just found a nice gif that explains why. source (includes spoilers)

that's not the operation I'm considering

pick a point that both loops go through.

oooh, I figured it out

ok, now that I'm thinking about commute, I understand that gif

but two loops around the top don't commute to two loops around the girth

yes they do:)

let's say a is a loop clockwise around the girth, and b is a loop clockwise around the top

then assuming ab==ba (they commute) we can infer aabb == abab == baab == baba == bbaa

right, I agree that you can reorder, but you can't do aa == bb

ah right, they are different!

oh, I was using the wrong term with "commute"

sorry, I meant that you can't convert two loops across the top to to loops around the girth

then a torus is just a pair of integers

yeah I agree

@NathanMerrill 1000 20 points to gryffindor!

XD

lol

is a 4-d torus represented by a triplet of integers?

aka, can I generalize and say "all circle-shapes in (N)D is represented by a N-1 tuple"?

@NathanMerrill one would first have to define what you mean by a 4d torus, but you can generalize a usual 2-torus to n-torus by just taking the cartesian product of a circle (S1) with itself.

So the n-torus is topologically speaking defined as S1 x S1 x ... x S1 (n factors)

and then we have indeed that the n-torus can be represented by n-tuples of integers.

but this is just the abstract topological view, I'm not sure if or how such a torus could be embedded in some real higher dimensional space R^n

yeah, well, I have a really hard time visualizing in 4D

even after watching 3b1b video about it

"Let n be the number of dimensions, and then let n go to 9." <- Simple!

@NathanMerrill maybe this analogy: sometimes in 2d games we have a rectangle as our map, and if you leave through the left side you enter through the right (same for top and bottom), this is the very same (homeomorphic) to a 2-torus (hence torroidal topology)

so you could imagine a 3-torus as a cube (including the inside)

where you "identify" opposite sides

if you use a dice as a model: if you leave through the side 1, you enter though the side 6.

live chess!

Sounds like watching paint dry. (I'm kidding.)

hey, magnus live right now

he's in a tough position

bah, just as I said that, they switched

this reminds me:

I wanted to make a challenge around detecting the strength of a position given only one side

aka, you only know the white pieces

and you give it a score, and we run your function twice and compare the scores, and use that composite score to guess win/loss

@NathanMerrill I'm enjoying having this in the corner of a monitor and glancing at the boards from time to time.

Mamedyarov (2804) vs Adhiban (2655) is the game they're showing most now and it's fairly interesting. Adhiban is really short on time (less than a minute to move, though I think it's +15 seconds for each move), but it looks to me like he has an ultimately stronger set of pieces. A queen, two rooks, and three pawns vs a queen, one rook, one bishop, and four pawns.

It's making me think about producing a visualization that colors squares according to the relative strength of pieces attacking that square. I wonder if it might be possible to take e.g. Lichess' chess website and stick it in an iframe to make building on it with custom JavaScript easier.

So let's say that some buddies and I want to win HQ Trivia by pooling our resources. What's the optimal strategy?

Some background (even though I've never personally played HQ Trivia). The goal of the game is to answer 12 multiple-choice questions correctly in a row, with each question having 3 answer choices. There is a strict/short time limit per question (10 seconds so no time for google or talking). No fee for entry, but if you win you get like $20 or something (depending on the number of winners).

It's a live mobile game with 1 or 2 tournaments daily. So each phone = 1 chance to win.

I think the best strategy is to perform some kind of pooling, where if we have X = 20 people and Y <= 20 phones (because phones are eliminated over time), we take the votes of all 20 people and use it to proportionally allocate the remaining phones among the 3 answer choices.

This increases our ability to win because the amount of knowledge being pooled together doesn't decrease over time.. a person whose own personal phone has been eliminated still gets to contribute to choosing later answers.

This will have to be almost fully automated, since there's no practical way for the players to communicate during the competition, so there will have to be some mechanism for many people to push buttons and also for the software to use those votes to make the actual answer selections on people's phones.

But that's beside the point.

I'm more interested in the mathematics of this strategy (since this is a pretty uneconomical way to earn a dollar)

Live stream of TATA Steel Chess just ended. Mamedyarov won the game in the end.