Timeline for Reunite the Stars
Current License: CC BY-SA 4.0
46 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Aug 28, 2021 at 2:59 | vote | accept | Eric | ||
Aug 26, 2021 at 13:06 | comment | added | loopy walt | @Eric I think otherwise there would have to be a limiting cycle with positive sum of dot products which I can't picture, | |
Aug 26, 2021 at 5:31 | comment | added | Eric | If g=3, will the phase I square size grow to infinity? | |
Aug 25, 2021 at 23:29 | comment | added | loopy walt | @Eric Oscillations are also refleclted in the sum of dot product which goes negatice during contractins whiiich is only possible if _g_>3. | |
Aug 25, 2021 at 15:36 | comment | added | Eric | Is the oscillation always there or just for g slightly larger than 3? So it seems if you have N stars whose unit velocity vectors sum to zero, then $g^*\leq N-1$. And if they don't sum to zero, then $g^*\lt N-1$. If true, I feel there must be some intuitive explanation for this. A conservation law or something. | |
Aug 25, 2021 at 14:14 | comment | added | loopy walt | @JaapScherphuis Yep, it's the same principle that lets you, for example, calculate sample covariances centring only one factor.. As soon as one factor is zero mean it kills any bias the other may still have. | |
Aug 25, 2021 at 14:01 | comment | added | Jaap Scherphuis | @loopywalt I got it now. I was being a bit slow, but I am now fully convinced that the proof works. I didn't realise that the fact that the four star vectors balance out means that it does not matter how the positional vectors are defined. My mental picture of what the sum of those dot products represented was off when the origin or meeting point was elsewhere. A sign error basically. | |
Aug 25, 2021 at 13:49 | comment | added | loopy walt | @Retudin did it now.Doesn't look too promising. Mind you I didn't try too hard; it may still be possible. | |
Aug 25, 2021 at 13:34 | comment | added | Retudin | @loopywalt: Just wondering: Did you also simulate the other always-carrying option NSEW? I would not be surprised if that would converge faster (for any g>3) - but I may be terribly wrong. | |
Aug 25, 2021 at 12:24 | comment | added | loopy walt | @JaapScherphuis Hm, I'd normally take your doubts very seriously, but in this case I'm 99.99% sure you are just having a really bad day. The point is that the four vectors come in pairs E/W and S/N that are opposite therefore E+W+S+N = (E+W) + (S+N) = 0 + 0 = 0. Now, if the stars unite at point P then the sum <E,P> + <W,P> + <S,P> + <N,P> the scalar product being linear in either argument evaluates to <0,P> = 0 | |
Aug 25, 2021 at 12:00 | comment | added | Jaap Scherphuis | Clearly? Well, not to me yet. If it must sum to zero, then you are using position vectors that are centred on the final meeting point. In my example that meeting point is off to the East, so that the initial sum is already negative, giving time for the guardian to set things up. It may seem obvious but I don't think your argument rigorously proves that this kind of thing won't work for four stars. | |
Aug 25, 2021 at 10:21 | comment | added | loopy walt | @JaapScherphuis ah, but in your example the sum needn't go to zero for the stars to reunite; in the original situation it clearly must (pair E/W and S/N to see). | |
Aug 25, 2021 at 10:17 | comment | added | Jaap Scherphuis | I am not convinced by the proof that g>3. Imagine a similar problem but with just 3 stars - 2 stars going East, one going West. Any speed g>1 suffices to bring those stars together, while your argument would suggest g>2 is needed. | |
Aug 25, 2021 at 7:57 | comment | added | justhalf | Ah, I see, so it's more meta description on what we are doing, ok. Nice solution! So the answer is g* = 3 then, based on the last clarification in the question. | |
Aug 25, 2021 at 7:57 | comment | added | loopy walt | ... but stabilise the shrinking trend. Purely empirical, I have no proof whatsoever for that. | |
Aug 25, 2021 at 7:55 | comment | added | loopy walt | @justhalf Begin by picturing the "square equilibrium" the guardian visits the corners ccw for eample moving the S star from the SE corner back to the NE corner where he drops it and picks up the E star etc. this is stable at g=3. if g is slightly faster the guardian could have afforded to go slightly more W instead of straight N, meeting the E star slightly earlier and reducing the S stars lateral (i.e.perpendicular to its own movement preference N->S) displacement. Now, by not going quite as far W as we could (thereby reducing the reduction) we give away a bit of the guardian's speed ... | |
Aug 25, 2021 at 7:47 | comment | added | justhalf | "we reduce the reduction by 50%" what do you mean by this? Surely it's not about reducing the speed of the guardian. But what? | |
Aug 25, 2021 at 7:24 | history | edited | loopy walt | CC BY-SA 4.0 |
more simulations
|
Aug 25, 2021 at 7:00 | history | edited | loopy walt | CC BY-SA 4.0 |
rewrite
|
Aug 25, 2021 at 6:44 | comment | added | loopy walt | @Eric have you been reading my mind again? I'm in the middle of updating my answer with exactly that. | |
Aug 25, 2021 at 6:42 | comment | added | Eric | What about going in circles carrying one star after another, the guardian always being loaded. Will this scheme be better? | |
Aug 23, 2021 at 12:56 | history | edited | loopy walt | CC BY-SA 4.0 |
added 303 characters in body
|
Aug 22, 2021 at 4:12 | comment | added | Eric | Another interesting twist: if you allow the four stars to start with different initial positions, or even different preferred directions (other than due East, due West etc.), will $g^*$ ever increase with respect to the $g^*$ in the original problem? | |
Aug 21, 2021 at 12:09 | comment | added | loopy walt | @Eric yes, that is actually easy to see in both scenarios. We have the lower bound g > 2m-1 (take sum of the dot products of the stars' positions and velocity vectors with no guardian this will increase at a rate of 2m with guardian at a rate >= 2m-1 - g) for an upper bound 4m + epsilon will allow you to go for one star after the other and bring it back to the origin. over time the individual distances will equilibrate and shrink. | |
Aug 21, 2021 at 10:22 | comment | added | Eric | We can also generalize to higher dimensions m, by having 2m stars evenly distributed on the unit circle centered at the origin. Do you think $g^*$ will grow linearly as dimensions increase? | |
Aug 21, 2021 at 10:03 | comment | added | Eric | If we generalize a bit and let the Prime Star break into N Sub Stars, evenly distributed on a circle and with directions of velocities fixed in the spirit of the 4 Stars case. Do you think $g^*(N)$ grows linearly? | |
Aug 19, 2021 at 13:27 | comment | added | loopy walt | @Steve cool! In case it helps: I have observed that once the simulatin is in the shrinking phase g can be switched to 3+epsilon and it will still converge. At this point S and W will stay nonpositive all the time. Do your calculations give any clue why 10/3 should be the magic number? | |
Aug 19, 2021 at 12:56 | comment | added | Steve | I'd got some calculations in Excel towards trying out a different idea, and have modified them to confirm this cycle, and also make some further tweaks - it's possible to drop off the stars a little "early" or "late" if desired, and the pattern still converges if not too much "slack" is taken away. I strongly suspect that using this "wiggle room" would allow for finding a solution for a specific value of $g > g^*$ that is not only completed within a finite time, but also a finite number of steps, but I've not yet found a specific procedure... | |
Aug 19, 2021 at 9:46 | comment | added | justhalf | @loopywalt, do you have analytical solution for the 10/3? | |
Aug 19, 2021 at 9:33 | comment | added | justhalf | @Eric, what do you mean by original problem? | |
Aug 19, 2021 at 7:52 | comment | added | Eric | If you can do it for for some initial distances, I guess $g^*=3$ for the original problem, too. The question is how to get to the desired distances. But of course it is entirely possible that for some initial distances we'll not be able to do that. | |
Aug 19, 2021 at 6:11 | comment | added | justhalf | Yes, I mean the behavior of the system has two phases. | |
Aug 19, 2021 at 6:07 | comment | added | loopy walt | @justhalf the strategy itself doesn't change; only how the system responds. It will converge only once the distances are close enough to a certain ratio, During the expanding phase the system moves towards this ratio. Why this behaviour emerges at g=10/3 I don't really know, that's why I call this answer "experimental". | |
Aug 19, 2021 at 6:03 | comment | added | justhalf | I see. Also, so this strategy has two phases? The first hundred cycles (as you mention) will increase the total distance, but afterward it keeps decreasing? Is this accurate? Or can you describe the behavior of the system as a whole? | |
Aug 19, 2021 at 6:00 | comment | added | loopy walt | @justhalf If we are allowed to distribue the inital distances at will we can. The trick is to make E arrive at zero the moment we deliver S and similarly for N and W. | |
Aug 19, 2021 at 5:58 | comment | added | loopy walt | @Eric see justhalf's comment or my comment to the deleted answer. | |
Aug 19, 2021 at 5:57 | comment | added | justhalf | For 2 points, the limiting cycle is at g=1. Can we find a limiting cycle at g=3 for 4 points? | |
Aug 19, 2021 at 5:53 | comment | added | justhalf | @loopywalt, and while you're at it, can you say something as well about how even though there are three stars moving, that the hard limit is 3 even though we can make such that one of the stars we are not carrying is actually going towards the point. | |
Aug 19, 2021 at 5:53 | history | edited | loopy walt | CC BY-SA 4.0 |
added 115 characters in body
|
Aug 19, 2021 at 5:51 | comment | added | justhalf | Hm, I get the math, but sounds counterintuitive lol. I guess the resolution is similar to Achilles and the tortoise. | |
Aug 19, 2021 at 5:51 | comment | added | Eric | Why is 3 the hard limit? | |
Aug 19, 2021 at 5:46 | comment | added | loopy walt | @justhalf I don't because it is not necessary. The cycle time decays exponentially; therefore the total time, i.e. the sum of cycles converges meaning after a finite time we will have performed infinitely many cycles and all stars be at zero. | |
Aug 19, 2021 at 5:43 | comment | added | justhalf | Also I get it now about your strategy. Basically we consider the value $V$ to be the sum of speed of the stars moving away from our designated rendezvous point. Since we can carry 1, maximum $V$ is at least 3, so g should at least be 3 as well to "counter" this effect. | |
Aug 19, 2021 at 5:40 | comment | added | justhalf | Can you explain how to perform the last cycle to make them converge at the same location? | |
Aug 19, 2021 at 5:25 | history | edited | loopy walt | CC BY-SA 4.0 |
added missing factor of 2
|
Aug 19, 2021 at 5:18 | history | answered | loopy walt | CC BY-SA 4.0 |