Mathematics - 2017-11-25 (page 1 of 4)

Secret

01:06

Secret

01:17

> Theorem 1 (Zarach). It is consistent with ZFC- that ω1 exists but is singular and hence that a countable union of countable sets can be uncountable.
Proof. Consider the model W as constructed in the L ́evy collapse of אω above. Note that every cardinal אVn is collapsed in V [Gn ] and hence in W , but אVω remains a cardinal in every V [Gn ] and hence also in W . Thus, ω1W = אVω , which has cofinality ωinV andhenceinW,aswitnessedbythesequence אVn |n<ω ∈V ⊆W. So W satisfies that ω1 is singular. In particular, W satisfies that ω1 is a countable union of countable sets, as ω1W = {אVn | n ∈ ω}.

Pretty much saying that $\omega_1$ under the model W is the same as $\aleph_{\omega}$ under the model $V$

arxiv.org/pdf/1110.2430.pdf

NB ZFC^{-} is ZFC-P in the above arxiv

Secret

01:29

Some axiomatic systems based on ZF to be tested:

System 1: Extensionality, Regularity, Specification, Pairing, Union, Replacement, Infinity, Existence of an inaccessible cardinal, dependent choice (optional)

System 2: Extensionality, Regularity, Specification, Pairing, Union, Replacement, Infinity, (weaker and/or more generalised version of powers axiom to be figured out), dependent choice (optional)

System 3: Extensionality, Regularity, Specification, Pairing, Union, Replacement, Infinity, (weaker and/or more generalised version of powers axiom to be figured out), Existence of an inaccessible cardinal, dependent choice (optional)

Goal: To preserve as many notions of functional analysis and real analysis, quasiinfinite sets and allowing a richer construction of uncountable sets

Actually wait a minute, what does an inaccessible cardinal even mean without powerset...

Faust

02:53

morning

Secret

03:50

Afternoon nap dream: Reading a wikibook with a dark grey background. One of the pages talks about generalising to infinity and there's a paragraph saying something along the lines of there is no way to get more elements unless (forgot), followed by a set builder notation
$$\{f |f(k) \mapsto f(forgot)|(forgot) \mapsto (forgot) | (forgot)\}$$

The article then ends with "But what if there exists something that can compute every computing matter, that consciousness is (forgot) real"

I then click the link to the next page, which talks about something titled "Conscious Fir" and the 1st paragrap…

Of all my dreams, these afternoon dreams are the ones that are most potentially informative on how to solve problems. However they are so hard to recall thus most of the time, I only remember less than 1% of it

Aruka J

anyone know basic vector math / physics?

Secret

Depends on what type of vector/phyiscs

Aruka J

velocity, movement, 2d plane, applying force, changing directions

Secret

What is the concept in your problem you have trouble with. Remember SE prefers conceptual, not "how to do this problem" questions

Aruka J

Conceptually, if I know the current $x,y$ (position) of a moving object, as well as its velocity $V_x, V_y$ -- but I want the object to ultimately arrive at some point $x_t, y_t$

to get it to arrive at this point I am having trouble understanding how to change its velocities based on how hard i can strike the object

Secret

03:59

Since you mentioned it is a vector problem, the first step is to resolve the motion into x and y components. So you want to be able to given your initial position and velocity, compute the acceleration required to allow you to move some displacement to your final position.

Aruka J

the difficulty is the initial vx vy and the fact that I do not have unlimited force

I can't just immediately strike the object with the right force and angle to immediately change its vx vy to make it go towards the target point

sometimes i can if vx, vy is slow enough

but sometimes best i can do is try to get it as close as possible to the target but I don't know how to do this

Secret

Is your given velocity a function of time? If it is, you can simply integrate that expression for the velocity, plug in the initial velocity to obtain the arbitrary constant and then solve the final position

since velocity is the derivative of displacement and displacement is final position-initial position

Aruka J

no, i'm asking a different question than that

Secret

Does your object has a mass. If yes you might be able to use conservation of momentum to solve it.
...Actually I think I might need some context without the full problem. What kind of motion will that object be undergoing, is it standard scenarios like projectile motion, simple harmonic motion or something else?

Aruka J

mass is 1 for simplicity

imagine a billiard ball moving across a huge table

and at time $t=0$ you know the position and velocity of the ball

you want to send it towards the hole

you can strike it once with some maximum force F from any angle

score is determined by distance to hole at time $t=1$

maximum score is obviously hitting the ball directly into the hole but sometimes not possible

if the ball is already going very fast

and your max force is not enough to completely change the ball's course

but i am not sure how to solve for the cases where you can't 100% change the velocity so it's moving along the vector pointing towards the hole

Secret

04:18

Ok, so that means, choose a coordinate frame, you have some mass with initial position and velocities $(\mathbf{x}_i,\mathbf{v})$ and your hole has some position vector $\mathbf{x}_f$. There is a maximum to the force applied which means the magnitude of the acceleration vector has a maximum, and the angle of hitting determines the relative proportions of y and x components of the acceleration. So in other words, you want to solve for $a$ the following vector equation:

$$\mathbf{x}_f - \mathbf{x}_i=\int_0^1 \mathbf{v} + \mathbf{a}t dt$$

Actually, to do this more generally:

$$\text{Minimise} \left( \mathbf{x}_f + \mathbf{x}_i+\int_0^1 \mathbf{v} + \mathbf{a}t dt\right)$$

So if there is an exact solution, the final distance (given by the magnitude of displacement since all the motions are linear) of the ball from the hole is exactly zero, otherwise, it will be some finite value

Kevin Driscoll

04:55

@ArukaJ What matters is the impulse you can impart ot the object, not the maximum force. Given any finite force, if you can apply it for long enough you can make the velocity what ever you want.

So the question is, under what conditions can you not change the velocity enough so that the ball arrives at the desired point?

If you can arbitrarily change the direction of the force, then it doesn't matter that you can only apply a force of a fixed magnitude. Eventually you can always get it to the point you want.

Secret

I think the problem might be an undergraduate problem, thus the impulse is expected to be short for an elastic (billard ball) interaction

(and he does mentioned the object is a billard ball)

Kevin Driscoll

You still need some kind of impulse-scale though. Basically you're saying its acting like a delta-function force in Newton's 2nd law. A sudden 'kick'. But you need some dimension-ful coefficient to sit in front of that delta function, and that coefficient will be an impulse-scale.

Giving a maximum force doesn't help. Because if you apply any finite force over a shorter and shorter time scale, the impulse imparted goes to 0 as the timestep approaches 0

Aruka J

I don't know the right word for it, I have the ability to impart a new velocity to the object

Kevin Driscoll

But again, acceleration by itself doesn't help

Aruka J

thorugh vf = vi + a

Kevin Driscoll

05:01

that equation isn't dimensionally correct

v has units of m/s, a has units of m/s^2

Aruka J

it's for a game so the physics aren't going to be real-world here

i don't know the 100% correct terminology, i just know that i can pick a direction and force and i can change the velocity

for example if it's on the origin, not moving

Kevin Driscoll

Ok so the word youre looking for isnt force or acceleration, its impulse

Aruka J

and i strike it from the left with force 10

Kevin Driscoll

Impulse is a change in momentum

Aruka J

then i get vf = 0 + 10*1 = 10 velocity

towards the right

(mass is 1)

and in this case the position also updates to 10 as well (as opposed to what we'd expect with 1/2 at^2)

so a bit "simpler" in this regard

iirc impulse is F*t

Kevin Driscoll

05:03

Yea what you actually mean isnt that you strike with a force of 10N. You mean that you strike with a force, such that the impulse is 10 m/s

Aruka J

and since i'm interested in t=1 timescales it's equal to force

and mass is 1 so acceleration = f as well

Kevin Driscoll

You;re making lots of approximations

and not syaing that you're making them

and tis leading to your language being imprecise

Aruka J

i mentioned up front that mass is 1, and that i was interested in t=0 to t=1

only major simplification is how position is updated between the two times, and it's minor (doesn't really change the nature of the question)

Kevin Driscoll

Thats true, but if you actually apply a force of magnitude F over t=1timeunit, the position is changing over the whole time

Secret

@KevinDriscoll Yeah, that's how undergraduate billard ball type problems behave. I am not sure if these problems actually have an impulse scale in mind though, but they do basically act like delta distributions. I guess the only thing we can say is the impulse is some small but finite value short enough to minimic a kick

Aruka J

05:06

i understand that in practice the position is changing the whole time

Kevin Driscoll

Youre also making the approximation that 1 tiemunits is 'short enough' that you don't have to update the position during the application of the force

Aruka J

and that's normally where the vf = vi * t + 1/2 at^2 stuff comes in

Kevin Driscoll

and thats pretty important because it dignificantly changes the nature of the problem

Aruka J

but here i'm doing away with that aspect for simplcity's sake

Kevin Driscoll

So anyway, the point being that you have some maximum impulse you can apply to the object. And so in that case if you don't ahve enough impulse to actually make the object get to the required point, then you have to make the new velocity as close as possible to the velocity that would get it to the specified point

So if you draw the initial velocity and the velocity hat would make it pass through the specified point as vectors

Aruka J

05:10

yes

Kevin Driscoll

Then the 'best' you can do with just 1 'kick' will be to apply the maximum impulse if the direction of (velocity that would get to the specified point) - (initial velocity)

That gets you a final velocity vector that is as close as posible to the desired velocity

and since after this kick its just constant velocity motion, getting the velocity as clsoe as possible will get the position as clsoe as possible as well

Secret

hmm... I was not aware that one can do away with positions and simply work at the level of velocity vectors using impulses

Kevin Driscoll

its because you just have constant velocity motion after the 'kick'. Everything just goes in a straight line.

Aruka J

Kevin Driscoll

So as long as you get it moving along the line closest to the one you want, itll end up at the position as close as possible

Aruka J

05:15

I tossed this together to try to illustrate

ball's at origin, red vectors represent current motion (+ the horizontal and vertical decomposition)

black hole is the ideal destination point, blue vector is pointing from current position to hole

if i had unlimited impulse i could use paralellogram method to just figure out where to hit the ball and how hard and it'd go straight into the hole next time unit

Kevin Driscoll

Doesnt even have to be the next time unit

As long as you get the new velocity ot be along the blue vector, itll go into the hole eventually

because constant-velocity motion is just motion along a straight line

Aruka J

yes (my goal is to get it to that hole as fast as possible)

next time unit technically i can apply another hit, but this leads to this current subproblem of getting it to the hole each hit as close as possible

as this translates overall to getting it into the hole as fast as possible

Kevin Driscoll

Sure, you just want to get the new velocity to be along the blue vector, with as large a magnitude as possible

Aruka J

right

that is where i am struggling

since it also depends on magnitude of red relative to my max impulse ability

Kevin Driscoll

So then just take a vector that represents the maximum kick (AKA maximum impulse) you can apply, put its tail at the tip of the current velocity vector

and then move its tip until it touches the blue line

because $v_f = v_i + v_{kick}$

Aruka J

05:21

i tried that, e.g. i'll make a new picture here

Kevin Driscoll

The new velocity vector will point along the blue line

Aruka J

i think my problem was also the angling

Kevin Driscoll

Once yuve got the picture draw, you can use basic trig to say what the angle is

probably using some law of cosines or something

Ok, so the case where you do have enough impulse to get along the blue line is solved. You just carry out that procedure i mentioned, so let me think again about the case where you don'tahve enough

Aruka J

imgur is being difficult

so let's say that green vector is my max impulse

Kevin Driscoll

Oooooh, wait. The fact that you can do subsequent hits and want to find the mimumum total time actually might make this more complicated. I have to think more carefully

Aruka J

05:24

this is what i would do if I could, for sure, make the ball move along the blue line

(this is just the paralellogram method), my problem is what i do if i can't "reach the blue vector" with my green vector

Kevin Driscoll

Yea, no this is right for when you ahve enough impulse to move the velocity. because at each time step, you're maximizing the distance the ball travels toward the hole

So you're minimizing the total time, so thats fine

Aruka J

i am assuming that by constantly re-applying this same method each hit, it minimizes overall time

since it's minimizing distance to hole each time

i don't know how to prove that but it seems right to me(?)

Kevin Driscoll

It is right

Aruka J

but if the green vector can't reach the blue one, i'm at a loss

Kevin Driscoll

So to do it when you don't ahve enough, you know that eventually you want the ball velocity to be along the line connecting the current position and the hole.

Aruka J

05:27

i guess that's the best i can do

just hitting it back along the red

basically trying to slow it down along its current path repeatedly until i can make the green connect?

or something else?

Secret

If you want to do it in one kick but the green vector can never align with the blue one, then the next best option is to minimise their difference

Kevin Driscoll

Ya I agree with Secret, this was my original answer

Any velocity other then the one that minimizes the distance will lead to the ball being farther away fromt eh hole that the closest velocity in the next tiem step

and that extra distance menas itll take longer to get it there eventually

The problem is there is no fixed velocity you're minimziing with respect to

because the velocity can be ANY velocity along the blue line

AH! But any velocity that gets the ball to the hole in 1 time step is equivalent

So you can take the 'desired' velocity to be the minimum velocity that gets the ball to the hole in 1 time step

and then you want to new velocity to be as close to that velocity as possible

Aruka J

i may not always be able to do it in one kick, i can always do it in finite kicks but I'm generalizing to a single kick because whatever optimizes one kick will optimize the whole thing

by just doing it repeatedly

Kevin Driscoll

Yea we understand

Secret

Yeah, any sequence of velocities that can approach or get into the hole will be equivalent to one single velocity vector that is the optimal one because of how vector addition works

(only the initial and final value matters)

Aruka J

05:34

right

if i had unlimited impulse ability i can just one-shot the ball into the hole

Kevin Driscoll

Im not sure you udnerstand what Secret is saying

Aruka J

by just drawing a vector from the red vector tip to the hole and just applying that to the ball

?

Kevin Driscoll

Hes saying it doesnt matter what vector along the line connecting the ball and the hole you choose, as long as you apply the maximum impulse to the ball and get the new velocity to be as close as possible to ANY velocity along the blue line

Then youre doing the optimal thing

Aruka J

not sure i am following the logic there

it seems possible to just max-impulse it to any point along the vector infinitely but never hitting the hole

Kevin Driscoll

Draw a picture. Imagine the ball is going some direction and it will take at least 3 kicks to get it into the hole

And do the 3 kicks 2 different ways

Aruka J

05:38

he said any sequence of velocity sums that get to the hole is the same as one single vector due to how summation works, and that's fine / i agree with that

but we can do 5+5 = 10 but we can also do 1+1+1+1+1+1+1+1+1+1=10

former is more "optimal" than the latter

(to use an example)

Kevin Driscoll

1 second

Aruka J

yes we can "reorder" the kicks

vector sums are commutative

n! ways to arrange n kicks

Kevin Driscoll

I actually took Secret to be pointing out a stronger conclusion, but Im checking it now

Its kind of funny because the way you've setup your 'physics' if the ball starts at the origin, then the position and velocity vectors are always parallel

Secret

Kevin Driscoll

because you just move to the end of the velocity vector at each time step

Secret

05:46

I think I need to reread the chat transcript...

Aruka J

@Secret If you're saying the two sets of green arrows sum to the same thing, i agree/understand, but one of them is using more green arrows than the other, is what i mean

in my definition "optimal" means as few green arrows as possible

or are you saying something more than that?

Secret

Actually, I think I start to recall how billard ball games work. You have a billard ball that travel at some initial velocity in a straight line, you then hit it at some angle, the ball's final velocity after the kick is the sum of its initial velocity plus the velocity of the kick (picture updating..,)

Aruka J

yes, the velocities update in sum

Secret

So yes you can end up with a scenario where one kick is not sufficient to brought it to the hole if the ball travels too fast

Aruka J

yes

Secret

05:52

Let me update the picture first

Aruka J

in the event that green impulse is too small i almost wonder if this is the optimal decision:

Kevin Driscoll

It isn't but im struggling to articulate to you why

Aruka J

was the earlier approach wrong too?

Kevin Driscoll

Nope thats totally right

Aruka J

making the green vector connect to blue from tip of red

Kevin Driscoll

05:56

Notive youre doing something very different here

Aruka J

so as we make green smaller and smaller it gets closer and closer to the origin

as we drag it up the blue line

Kevin Driscoll

It works, but its not optimale

Aruka J

so what if green arrow = red?

in magnitude

Kevin Driscoll

If green = red then your proposal would have the ball stop after 1 time step, so you wouldnt get any clsoe ror farther away from teh hole

Secret

(Complicated diagram incoming!)

Kevin Driscoll

05:58

But I claim that if instead you aling the green arrow so that it points from the end of the red to the hole, then you'll actually get closer @ArukaJ

Aruka J

Secret

For the one kick case (I will be explaining it shortly)

Aruka J

so always aiming the green vector towards the intended destination no matter what?

with max impulse?

Kevin Driscoll

Depends onw hat you mean by aim

You dont aim the green arrow toward the hole from where the ball is

you aim the green vector toward the hole from where the end of the current velocity is

Secret

Basically, the blue vector is the displacement of the ball's current position to the hole's centre, the red vector is the initial velocity, the green vectors are the possible directions of impulses, and the dark red vectors are the final velocities after the kick. The green vectors are bounded by a dotted circle which means all the possible directions with maximum impulse is given by that circle. This particular scenario is produced so that the ball will never get into the hole in a single kick

Aruka J

06:00

i mean like:

Kevin Driscoll

Hold on I clearly need ot think more carefully about this

Secret

For all possible velocities after the kick, the trajectory of the ball is bounded by the cone shown. The dotted circle centred on the ball (the tail of the blue arrow) marks the t=1 point where the ball must stop. Therefore in order to get to the hole as close as possible, you need to compute:

Aruka J

are there just three green vectors for demonstration sake

(we could have just drawn one?)

trying to understand the diagram

Secret

yes. In reality there are uncountably many

Aruka J

ok

Kevin Driscoll

06:03

Yea here's the thing. After each time step, the ball always ends up at the point given by vi + vkick. So during each time tep you want to maximize the distance you travel toward the hole. So that means always pointing the green arrow toward the hole.

So yes, our original solution fo the case where you can reach the blue line in 1 time step actually is wrong.

Secret

To get to the hole as close as possible, compute Intersection p (t=1 circle, cone (which is a function of the circle of maximum impluse), circle centered at hole)

Kevin Driscoll

You always want to point the green vector from the end of the initial velocity toward the hole

Secret

Since the t=1 circle is fixed as a function of the final velocity, the intersection is unique for each direction. In particular, if the radius of the circle centered on the hole goes to zero, it corresponds to the case where you can one shot into the hole

Aruka J

I'm with you on everything in that diagram so far up until the t=1 part

let me re-read

Secret

(clearing clutter in diagram)

Kevin Driscoll

06:07

Notice in Secret;s diagram that for the circle bounding all of the maximum-impulse arrows there is 1 point thats closest to the hole

Aruka J

i don't undertstand the t=1 circle thing

you could have a green arrow pointing straight back at the red

Kevin Driscoll

That is the point where the circle intersects the line connecting the hole and the initial velocity vector

Aruka J

and the ball would be inside the dotted circle at t=1 would it not?

Kevin Driscoll

It would end up on the circle

Becuause the ball always travels to the end of the final velocity vector after each time tep

Secret

Aruka J

06:08

and in fact one of his green arrow / final velocity arrows are inside the circle

Kevin Driscoll

given how you've set up your physics

Secret

t=1 marks where the ball will end up after the kick 1 second later

Aruka J

unless he means the smaller dotted circle

as the t=1 circle

ok that must be it

Kevin Driscoll

The smaller circle shows you all possible maximum impulses

Aruka J

i'm still not following this big dotted circle

if the impulse is that small how are we saying the ball can be at any point around that big circle

Kevin Driscoll

06:10

Ya I dont see that either

In fact, the ball will end up at some point on the smaller circle

Aruka J

that's what i mean

wouldnt the small dotted circle represent all possible t=1 positions of the ball?

Secret

Ok let me try again:

Kevin Driscoll

Indeed it does

Aruka J

if we do nothing, the ball ends up at the tip of the red vector

Kevin Driscoll

and so you ant to pick the point on that smaller circle thats closest to the hole

Aruka J

06:11

if we apply a max kick in some direction, it goes to the small circle on the border

Kevin Driscoll

And that point is precisely the point where the circle intersects the line connecting the hole and the tip of the initial velocity vector

Aruka J

ahhhh

that's really interesting

Kevin Driscoll

which implied you want to make the green arrow point from the tip of the red arrow toward the hole

Aruka J

this whole time i was assuming we had to compensate for current velocity more complicatedly

Secret

Yes your understanding is right Kevin. Since all the arrows I have drawn are already velocities, everything happens between t=0,t=1 so the t=1 circle is redundant

Aruka J

06:15

@Secret wait are you saying in your diagram the green arrow is optimal?

i'm still not understanding those other circles

Secret

Now it's optimal

That is the direction you need to kick to get the ball closest to the hole

Aruka J

ok that makes more sense to me

Secret

As kevin said, the small circle marks all the positions at t=1

Kevin Driscoll

Im shocked I didnt see that that was the right answer earlier. It took sitting down with a pencil and paper and actually drawing it out

I did a whole one on my pad here tha took like 7 time steps ot get to the hole

Aruka J

i'm trying to expand his picture

and i'm ending up going in a spiral

is this right?

Secret

06:26

Does your table have friction: That is, does your ball come to a stop at t=1?

Kevin Driscoll

No its frictionless

Secret

ok

Aruka J

i can use friction though,, it would just decrease the size of the next velocity vector by some scalar amount

Kevin Driscoll

@ArukaJ I am not 100% sure, but I think this is right. I think the way youve drawn these particular velocities you will always overshoot the hole

Aruka J

so instead of mirroring my vector it would be slightly shorter

Kevin Driscoll

06:28

For example try to draw the same picutre, but instead of doing the thing we said was optimal

Aruka J

the picture above is using the "optimal" approach of always hitting towards the hole

Kevin Driscoll

Try and rotate the initial velocity so that it falls along the blue line. I think youll see you pass the hole before that can happen

Aruka J

i need to break out the paper and pencil

hard to use graphics

Kevin Driscoll

What I mean is, try drawing the green arrows so that they point toward the closest point on the blue line from where teh initial velocity is

and I thin youll STILL overshoot the hole

Aruka J

i think it also depends on how hard it's being hit

like we can always go sub-max, too

it's just we can't go more than max

Kevin Driscoll

06:30

Ya goign submax here though will definitely lead to shooting past the hole though

Aruka J

although submax would, yeah

same problem

Kevin Driscoll

I think, but am not 100% sure that actually you can ALWAYS hit it at a maximum, as long as the ball always goes in teh hole when it moves over it

Secret

Kevin Driscoll

If the ball were to skip over the hole if its goign too fast itd be a different story

Secret

This is some really shitty convergence speed...

(Note I have not take account riccochet of walls, so...)

Kevin Driscoll

06:32

Ya im sure theres some way of calculating how the number of steps scales with the velocity in multiples of the max or something

The worst-case scenario is you're headed directly away from the hole I imagine

Secret

I think intuitively, the first thing I will do if the ball travels too fast (> max impulse) is to hit it in the opposite direction so it slows down, then the problem becomes one where there is a one shot solution to the hole

Kevin Driscoll

That was Aruka's instinct too

and as a human thats probably how Id do it too

You could try that method with the drawing you already did. And hopefully if were right it takes more than 8 steps

Aruka J

oh secret already made something like this and more accurately LOL

Secret

(Faster strategy coming shortly)

Aruka J

but if i am always hitting towards the hole then I am spiraling into it with many moves

ill try our earlier strategy

Kevin Driscoll

06:40

Ya Im open to the idea that if youre going to overshoot this isnt the fastest

Its not 100% clear in my mind that you cant sacrifice some distance on one step to get more on the next

Secret

This is a two stage strategy:

Stage 1 (Initial velocity > Maximum impulse and there exists no one shot solution): Hit the ball so that it maximise the decrease in its velocity and also bring it closest to the hole's direction)

Stage 2: (One shot solution to the hole is found) Pick impulse that get to the hole

Aruka J

damn it you beat me to it again (I was trying to drop the impulse at 90 degree triangle to blue vector)

Kevin Driscoll

Ok ya so there is a thing about if youre going to overshoot, you want to go as close to the blue line as possible

There will still be velocities large enough though that even if you try this, youll stil overshoot

Ya doing it perpendicular to the blue vector isnt always going to be optimal. Sometimes to get it to not overshoot youll only have to go a little ways further toward the blue line from the locally optimal solution from earlier

Aruka J

how are you drawing these so quickly btw

what are you using?

Secret

power point

Aruka J

06:47

and the nice clean arrows and everything

Kevin Driscoll

I let my earlier thoughts about the single-step problem cloud my judgment

The 1-step answer is easy about whats optimal

The multi-step problem is much harder

I'd bet like $100 though that someone worked all this out, and probably 200+ years aho

Secret

Kevin Driscoll

ago*

Secret

The "hit it until it is slow enough, then one shot hole" strategy

only works without friction

Aruka J

in practice i do have some friction but it's applied at the end of the turn

Secret

06:50

Not sure which one is more optimal though

Aruka J

now i'm totally lost though as to what heuristic to use

Secret

There are many solution strategies, but it is the game mechanics that determines which strategy to use

Kevin Driscoll

If you can get the ball directly in the hole without overshooting by always applying to force 'toward the hole' then that will be faster than anything else

Aruka J

it's easy enough to get it to the hole if my impulse can one-shot it

Secret

If there's riccochet, the problem can become chaotic...

Kevin Driscoll

06:52

Depends on the shape of the boundary

Secret

Typical billard ball tables are rectangles

Kevin Driscoll

But the idea of chaotic doesnt really make sense when you can apply an impulse at each time step

So for a rectangle, the dynamics are integrable

Secret

I see

Kevin Driscoll

Any regular polyhedron is integrable.

The circle and the ellipse are also integrable I think

But if you cut the circle into 2 equal arclength halves, then glue the halves together using line segments

That system is chaotic

Aruka J

do you think making 90 degree triangles with the original blue line makes it optimal?

Kevin Driscoll

06:54

@ArukaJ Not always

For example, if the ball is only going to sliiiiiightly miss the hole using the 'force towar dhte hole' algorithm

then you only have to point the force sliiiiiightly away from the hole to get the optimal path

going 90 degrees to the blue line would be too much

Aruka J

ahh

Kevin Driscoll

So this idea occurs to me and Im not sure if its right

Aruka J

maybe this is a brachistochrone curve? googling around

Kevin Driscoll

ya I thought about it, it probably is

it cant be exactly that though

because this isnt a constant force

Aruka J

a choppy approximation thereof

Kevin Driscoll

06:57

But related tot he idea that occured to me, it WOULD be a brachistochrone if you used a constant force

so what you have is some piecewise-linear approximation to that

ya exactly

its a bit more complicated problem though because yo can change the direction of the force

so you could find the solution for a constant magnitude force where you can change the direction

it wont be exactly a brachistochrone

and then your solution here is a piecewise-linear approximation to that optimal result

YOUSEFY

Hi, I want to know how to calculate the margin for some ratio. Suppose that we have have achieved a ratio of 2.5 and the best is 2. So we have 20% margin. I want to know how to calculate this margin percent? Is it as following: 2.5-2/(2)*100 = 25%. It is not exactly 20% so this is why i want to check here

Kevin Driscoll

The constant force, changing direction problem I dont think would be toooooo bad using calculus-of-variations/least-action. But its not trivial. Again, I bet someones worked it out

@YOUSEFY Thats kind of ambiguous. It depends on waht you mean by margin

For example, if 2 represents your costs, and 2.5 the revenue, then you'd say you have a profit margin of (2.5-2)/2.5 = 20%

YOUSEFY

@KevinDriscoll Agree! this is why I don't understand how he calculate it. I read this paper and says "and the best known integrality gap of the LP is 2 [8], leaving a 20% margin for improvement." arxiv.org/abs/1412.0681

Kevin Driscoll

But if instead 2 represents last quarters revenue, and 2.5 this quarters revenue, then to find the 'percent increase' youd want (2.5-2)/2 = 25%

Aruka J

(i gtg but thanks for the input @KevinDriscoll and @Secret )

Kevin Driscoll

07:04

@ArukaJ Good luck. Sorry it turned out to be more complicated than I realized at first

YOUSEFY

@KevinDriscoll "2.5" represents my best achievement while 2 represents that I can reduce 2.5 to 2 by some "unknown technique".

Kevin Driscoll

@YOUSEFY I mean hes saying the lower bound is 2 and the current best known is 2.5. So you can decrease teh current best known factor by 20%. Or one could say, the current best known factor is 25% larger than the optimal one

That is 20% is the maximum possible percent decrease from the current best

So by using the term 'margin' its clear enough that theyre referring to the margin with reference ot the current best, rather than with respect to the optimal

YOUSEFY

@KevinDriscoll you are exactly right. 2 is a lower bound and 2.5 is so far an upper bound. Yea, you are right about that margin is for current best, not to the optimal.

Kevin Driscoll

Honestly, you could say it either way

And which one you choose is just a conventional choice

As long as its clear in context what you mean

YOUSEFY

@KevinDriscoll I understand that to calculate the maximum possible percent decrease from current best is (2.5-2)/2.5=20%. which is (gap between lower bound and upper bound)/best upper bound = margin for improving!

Really thank you @KevinDriscoll for answering this question. Best!

Kevin Driscoll

07:23

No problem, good luck

Time for me to go to sleep

YOUSEFY

Have a good sleep! here in Russia we are in afternoon! we are already in good time to study!

user486600

07:40

Hello, Does anyone here know why in the end of the following answer, the uniqueness of limits implies that $\{f(x_j)\}\to f(x)$ math.stackexchange.com/questions/2535356/…

user104729

$\mathsf{CRing} \mathcal{CRing} \mathscr{CRing} \mathbf{CRing}$

Jake Shellman

Hey, I could really use some help understanding this post: math.stackexchange.com/questions/417365/…

In particular, could someone help me understand lemma 2?

(symplectic stuff)

Secret

09:36

[Random]

$S=\omega$

$S \subsetneq f(S)$

$T=f(S)-S$

$g : f(S) \mapsto f(S)$

$g(x) = \begin{cases}x,x\in S \\ ?,x \in f(S)-S \end{cases}$

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$Mathematics$ Mathematics